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Vector Algebra Class 12 Important Questions with Solutions Previous Year Questions

Algebra of Vectors

Question 1.
Find the position vector of a point which divides the join of points with position vectors \(\vec{a}-2 \vec{b}\) and 2\(2 \vec{a}+\vec{b}\) externally in the ratio 2:1. (Delhi 2016)
Answer:
Let given position vectors are \(\overrightarrow{O A}=\vec{a}-2 \vec{b}\) and \(\overrightarrow{O B}=2 \vec{a}+\vec{b}\).

Let \(\overrightarrow{O A}\) be the position vector of a point C which divides the join of points, with position vectors
\(\overrightarrow{O A}\) and \(\overrightarrow{O B}\), externally in the ratio 2:1.
∴ \(\overrightarrow{O A}\) = \(\frac{2 \overrightarrow{O B}-1 \overrightarrow{O A}}{2-1}=\frac{2(2 \vec{a}+\vec{b})-1(\vec{a}-2 \vec{b})}{1}\) [by external section formula]
= 4\(\vec{a}\) + 2\(\vec{b}\) – \(\vec{a}\) + 2\(\vec{b}\) = 3\(\vec{a}\) + 4\(\vec{b}\)

Question 2.
If \(\vec{a}\) = 4î – ĵ + k̂ and \(\vec{b}\) = 2î – 2ĵ + k̂, then find a unit vector parallel to the vector \(\vec{a}+\vec{b}\). (All India 2016)
Answer:
Given vectors are
\(\vec{a}\) = 4î – ĵ + k̂, \(\vec{b}\) = 2î – 2ĵ + k̂.
Now, \(\vec{a}+\vec{b}\) == (4î – ĵ + k̂) + (2î – 2ĵ + k̂)
= 6î – 3ĵ + 2k̂

and \(|\vec{a}+\vec{b}|=\sqrt{(6)^{2}+(-3)^{2}+(2)^{2}}\)
= \(\sqrt{36+9+4}=\sqrt{49}\) = 7units

∴ The unit vector parallelto the vector \(\vec{a}+\vec{b}\) is
\(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{6 \hat{i}-3 \hat{j}+2 \hat{k}}{7}\)

Question 3.
The two vectors ĵ + k̂ and 3î – ĵ + 4k̂ represent the two sides \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) respectively of triangle ABC. Find the length of the median through A. (Delhi 2016; Foreign 2015)
Answer:
Given, \(\overrightarrow{A B}\) = ĵ + k̂ and \(\overrightarrow{A C}\) = 3î – ĵ + 4k̂

Alternate Method:
Given \(\overrightarrow{A B}\) = ĵ + k̂ and \(\overrightarrow{A C}\) = 3î – ĵ + 4k̂

Question 4.
Write the direction ratios of the vector 3\(\vec{a}\) + 2\(\vec{b}\), where \(\vec{a}\) = î + ĵ – 2k̂ and \(\vec{b}\) = 2î – 4ĵ + 5k̂ (All India 2015C)
Answer:
Clearly, 3\(\vec{a}\) + 2\(\vec{b}\) = 3 (î + ĵ – 2k̂) + 2 (2î – 4ĵ + 5k̂)
= (3î + 3ĵ – 6k̂) + (4î – 8ĵ + 10k̂)
= 7î – 5ĵ + 4k̂
Hence, direction ratios of vectors 3\(\vec{a}\) + 2\(\vec{b}\) are 7, – 5 and 4.

Question 5.
Find the unit vector in the direction of the sum of the vectors 2î + 3ĵ – k̂ and 4î – 3ĵ + 2k̂. (Foreign 2015)
Answer:
Let \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = 4î – 3ĵ + 2k̂
Now, sum of two vectors,
\(\vec{a}+\vec{b}\) = (2î + 3ĵ – k̂) + (4î – 3ĵ + 2k̂) = 6î + k̂

Question 6.
Find a vector in the direction of vector 2î – 3ĵ + 6k̂ which has magnitude 21 units. (Foreign 2014)
Answer:
To find a vector in the direction of given vector, first of all we find unit vector in the direction of given vector and then multiply it with given magnitude.

Let \(\vec{a}\) = 2î – 3ĵ + 6k̂
Then, |\(\vec{a}\)| = \(\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}\)
= \(\sqrt{4+9+36}=\sqrt{49}\) = 7 units

The unit vector in the direction of the given vector \(\vec{a}\) is

Now, the vector of magnitude equal to 21 units
and in the direction of a is given by
21â = 21\(\left(\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\right)\) = 6î – 9ĵ + 18k̂

Question 7.
Find a vector a of magnitude 5√2, making an angle of \(\frac{\pi}{4}\) with X-axis, \(\frac{\pi}{2}\) with Y-axis and an acute angle 0 with Z-axis. (All India 2014)
Answer:
Here, we have l = cos \(\frac{\pi}{4}\), m = cos \(\frac{\pi}{2}\) and n = cosθ
⇒ l = \(\frac{1}{\sqrt{2}}\), m = 0 and n = cosθ

Question 8.
Write a unit vector in the direction of the sum of the vectors \(\vec{a}\) = 2î + 2ĵ – 5k̂ and \(\vec{b}\) = -2î + ĵ – 7k̂. (Delhi2014C)
Answer:
\(\frac{1}{13}\)(4î + 3ĵ – 12k̂)

Question 9.
Find the value of p for which the vectors 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are parallel. (All India 2014)
Answer:
Given, 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are two parallel vectors, so their direction ratios will be proportional.

Question 10.
Write the value of cosine of the angle which the vector \(\vec{a}\) = î + ĵ + k̂ makes with Y-axis. (Delhi 2014C)
Answer:
Given, \(\vec{a}\) = î + ĵ + k̂
Now, unit vector in the direction of \(\vec{a}\) is

∴ Cosine of the angle which given vector makes with Z-axis is \(\frac{1}{\sqrt{3}}\)

Question 11.
Find the angle between X-axis and the vector î + ĵ + k̂. (All India 2014C)
Answer:
Let \(\vec{a}\) = î + ĵ + k̂
Now, unit vector in the direction of \(\vec{a}\) is

So, angle between X-axis and the vector
î + ĵ + k̂ is cos α = \(\frac{1}{\sqrt{3}}\) ⇒ α = cos^-1 (\(\frac{1}{\sqrt{3}}\))
[∵ â = lî + mĵ + nk̂ and cos α = l ⇒ α = cos^-1l]

Question 12.
Write a vector in the direction of the vector î – 2ĵ + 2k̂ that has magnitude 9 units. (Delhi 2014C)
Answer:
3î – 6ĵ + 6k̂

Question 13.
Write a unit vector in the direction of vector \(\vec{P Q}\), where P and Q are the points (1, 3, 0) and (4, 5, 6), respectively. (Foreign 2014)
Answer:
First, find the vector \(\overrightarrow{P Q}\) by using the formula (x₂ – x₁)î + (y₂ – y₁)ĵ + (z₂ – z₁)k̂, then required unit vector is given by \(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}\)

Given points are P (1, 3, 0) and Q (4, 5, 6).
Here, x₁ = 1, y₁ = 3, z₁ = 0
and x₂ = 4, y₂ = 5, z₂ = 6
So, vector PQ = (x₂ – x₁)k̂ + (y₂ – y₁)ĵ + (z₂ – z₁)k̂
= (4 – 1)î + (5 – 3)ĵ + (6 – 0)k̂
= 3î + 2ĵ + 6k̂
∴ Magnitude of given vector

Hence, the unit vector in the direction of \(\vec{P Q}\) is

Question 14.
If a unit vector \(\vec{a}\) makes angle \(\frac{\pi}{3}\) with î, \(\frac{\pi}{4}\) with ĵ and an acute angle θ with k̂, then find the value of θ. (Delhi 2013)
Answer:
Here, we have
l = cos\(\frac{\pi}{3}\), m = cos \(\frac{\pi}{4}\) and n = cos θ

Question 15.
Write a unit vector in the direction of the sum of vectors \(\vec{a}\) = 2î – ĵ + 2k̂ and \(\vec{b}\) = – î + ĵ + 3k̂. (Delhi 2013)
Answer:
\(\frac{1}{\sqrt{26}} \hat{i}+\frac{5}{\sqrt{26}} \hat{k}\)

Question 16.
If \(\vec{a}\) = xî +2ĵ – zk̂ and \(\vec{b}\) = 3î – yĵ + k̂ are two equal vectors, then write the value of x + y + z. (Delhi 2013)
Answer:
Two vectors are equal, if coefficients of their components are equal.
Given, \(\vec{a}=\vec{b}\) ⇒ xî + 2ĵ – zk̂ = î – yĵ + k̂

On comparing the coefficient of components, we get
x = 3, y = -2, z = -1
Now, x + y + z = 3 – 2 – 1 = 0

Question 17.
P and Q are two points with position vectors 3\(\vec{a}\) – 2\(\vec{b}\) and \(\vec{a}\) + \(\vec{b}\), respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2 : 1 externally. (All India 2013)
Answer:
\(-\vec{a}+4 \vec{b}\)

Question 18.
L and M are two points with position vectors 2\(\vec{a}\) – \(\vec{b}\) and \(\vec{a}\) + 2\(\vec{b}\), respectively. Write the position vector of a point N which divides the line segment LM in the ratio 2 : 1 externally. (All India 2013)
Answer:
5\(\vec{b}\)

Question 19.
A and B are two points with position vectors 2\(\vec{a}\) – 3\(\vec{b}\) and 6\(\vec{b}\) – \(\vec{a}\), respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1:2. (All India 2013)
Answer:
Given, A and B are two points with position vectors 2\(\vec{a}\) – 3\(\vec{b}\) and 6\(\vec{b}\) – \(\vec{a}\), respectively. Also, point P divides the line segment AB in the ratio 1 : 2 internally.

Question 20.
Find the sum of the vectors \(\vec{a}\) = î – 2ĵ + k̂ \(\vec{b}\) = – 2î + 4ĵ + 5k̂ and \(\vec{c}\) = î – 6ĵ – 7k̂. (Delhi 2012)
Answer:
Given vectors are \(\vec{a}\) = î – 2ĵ + k̂ \(\vec{b}\) = – 2î + 4ĵ + 5k̂ and \(\vec{c}\) = î – 6ĵ – 7k̂.
Sum of the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) is
\(\vec{a}+\vec{b}+\vec{c}\) = (î – 2ĵ + k̂) + (- 2î + 4ĵ + 5k̂) + (î – 6ĵ – 7 k̂)
= – 4ĵ – k̂

Question 21.
Find the sum of the following vectors. \(\vec{a}\) = î – 3k̂, \(\vec{b}\) = 2ĵ – k̂, \(\vec{c}\) = 2î – 3ĵ + 2k̂. (Delhi 2012)
Answer:
3î – ĵ – 2k̂

Question 22.
Find the sum of the following vectors. \(\vec{a}\) = î – 2ĵ, \(\vec{b}\) = 2î – 3 ĵ, \(\vec{c}\) = 2î + 3k̂. (Deihi 2012)
Answer:
5î – 5ĵ + 3k̂

Question 23.
Find the scalar components of \(\vec{AB}\) with initial point A (2,1) and terminal point B(- 5, 7). (All India 2012)
Answer:
Given initial point is A (2,1) and terminal point is B (- 5, 7), then scalar component of \(\overrightarrow{A B}\) are
x₂ – x₁ = – 5 – 2 = – 7and y₂ – y₁ = 7 – 1 = 6.

Question 24.
For what values of \(\vec{a}\), the vectors 2î – 3ĵ + 4k̂ and aî + 6ĵ – 8k̂ are collinear? (Delhi 2011)
Answer:
If \(\vec{a}\) and \(\vec{b}\) are collinear, then use the condition \(\vec{a}\) = λ\(\vec{b}\), where λ is some scalar.

Let given vectors are \(\vec{a}\) = 2î – 3ĵ + 4k̂ and \(\vec{a}\) = aî + 6ĵ – 8k̂
We know that, vectors \(\vec{a}\) and \(\vec{b}\) are said to be collinear, if
\(\vec{a}\) = k. \(\vec{b}\), where k is a scalar.
∴ 2î – 3ĵ + 4k̂ = k(aî + 6ĵ – 8k̂)

On comparing the coefficients of î and ĵ, we get
2 = ka and -3 = 6k ⇒ k = –\(\frac{1}{2}\)
∴ 2 = –\(\frac{1}{2}\)a ⇒ a = -4

Question 25.
Write the direction cosines of vector -2î + ĵ – 5k̂. (Delhi 2011)
Answer:
Direction cosines of the vector aî + bĵ + ck̂ are

Question 26.
Write the position vector of mid-point of the vector joining points P(2, 3, 4) and Q (4, 1, – 2). (Foreign 2011)
Answer:
Mid-point of the position vectors
\(\vec{a}\) = a₁î + a₂ĵ + a₃k̂ and
\(\vec{b}\) = b₁î + b₂ĵ + b₃k̂ is \(\frac{\vec{a}+\vec{b}}{2}\) or \(\frac{\left(a_{1}+b_{1}\right) \hat{i}+\left(a_{2}+b_{2}\right) \hat{j}+\left(a_{3}+b_{3}\right) \hat{k}}{2}\)

Given points are P(2, 3, 4) and Q(4,1,-2) whose position vectors are \(\overrightarrow{O P}\) = 2 î + 5ĵ + 4k̂ and \(\overrightarrow{O Q}\) =4î + ĵ – 2k̂.
Now, position vector of mid-point of vector joining points P(2, 3, 4) and Q(4, 1, – 2) is

Question 27.
Write a unit vector in the direction of vector \(\vec{a}\) = 2î + ĵ + 2k̂. (All India 2011; Delhi 2009)
Answer:
We know that, unit vector in the direction of â is â = \(\frac{\vec{a}}{|\vec{a}|}\)

Required unit vector in the direction of vector
\(\vec{a}\) = 2î + ĵ + 2k̂

Question 28.
Find the magnitude of the vector \(\vec{a}\) = 3î – 2ĵ + 6k̂. (All India 2011C: Delhi 2008)
Answer:
Magnitude of a vector r = xî + yĵ + zk̂ is |\(\vec{r}\)| = \(\sqrt{x^{2}+y^{2}+z^{2}}\)

Given vector is a = 3i – 2/ + 6fc.
∴ Magnitude of \(\vec{a}\) = \(|\vec{a}|=\sqrt{(3)^{2}+(-2)^{2}+(6)^{2}}\)
= \(\sqrt{9+4+36}=\sqrt{49}\) = 7 units

Question 29.
Find a unit vector in the direction of vector \(\vec{a}\) = 2î + 3ĵ + 6k̂. (All India 2011C)
Answer:
\(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 30.
If A, B and C are the vertices of a ΔABC, then what is the value of \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}\) ? (Delhi 2011C)
Answer:
Let ΔABC be the given triangle.

Now, by triangle law of vector addition,

Question 31.
Find a unit vector in the direction of \(\vec{a}\) = 2î – 3ĵ + 6k̂. (Delhi 2011c)
Answer:
\(\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 32.
Find a vector in the direction of \(\vec{a}\) = 2î – ĵ + 2k̂, which has magnitude 6 units. (Delhi 2010C)
Answer:
4î – 2ĵ + 4k̂

Question 33.
Find the position vector of mid-point of the line segment AB, where A is point (3, 4, -2) and Bis point (1, 2, 4). (Delhi 2010)
Answer:
2î + 3ĵ + k̂

Question 34.
Write a vector of magnitude 9 units in the direction of vector -2î + ĵ + 2k̂. (All India 2010)
Answer:
-6î + 3ĵ + 6k̂

Question 35.
Write a vector of magnitude 15 units in the direction of vector î – 2ĵ + 2k̂. (Delhi 2010)
Answer:
5î – 10ĵ + 10k̂

Question 36.
What is the cosine of angle which the vector √2î + ĵ + k̂ makes with Y-axis? (Delhi 2010)
Answer:
\(\frac{1}{2}\)

Question 37.
Find a vector of magnitude 5 units and parallel to the resultant of \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = î – 2ĵ + k̂. (Delhi 2011)
Answer:
First, find resultant of the vectors a and o, which is \(\vec{a}\) + \(\vec{b}\). Then, find a unit vector in the direction of \(\vec{a}\) + \(\vec{b}\). After this, the unit vector is multiplying by 5.

Given, \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = î – 2ĵ + k̂.
Now, resultant of above vectors = \(\vec{a}\) + \(\vec{b}\)
= (2î + 3ĵ – k̂) + (î – 2ĵ + k̂) = 3î + ĵ

Question 38.
Let \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = 4î – 2ĵ + 8k̂ and \(\vec{c}\) = î – 2ĵ + k̂. Find a vector of magnitude 6 units, which is parallel to the vector 2\(\vec{a}\) – \(\vec{b}\) + 8 \(\vec{c}\). (All India 2010)
Answer:
First, find the vector 2 a – b + 3c, then find a unit vector in the direction of 2a-b + 3c.
After this, the unit vector is multiplying by 6.
Given, \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = 4î – 2ĵ + 8k̂ and \(\vec{c}\) = î – 2ĵ + k̂
∴ \(2 \vec{a}-\vec{b}+3 \vec{c}\)
= 2 (î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3 (î – 2ĵ + k̂)
= 2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂ + 3î – 6ĵ + 3k̂
⇒ \(2 \vec{a}-\vec{b}+3 \vec{c}\) = î – 2ĵ + 2k̂

Now, a unit vector in the direction of vector

Hence, vector of magnitude 6 units parallel to the Vector \(2 \vec{a}-\vec{b}+3 \vec{c}\) = 6\(\left(\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)\)
= 2î – 4ĵ + 4k̂

Question 39.
Find the position vector of a point R, which divides the line joining two points P and Q whose position vectors are 2\(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – 8\(\vec{b}\) respectively, externally in the ratio 1 : 2. Also, show that P is the mid-point of line segment RO. (Delhi 2010)
Answer:
Given, \(\overrightarrow{O P}\) = Position vector of P = 2\(\vec{a}\) + \(\vec{b}\)
and \(\overrightarrow{O Q}\) = Position vector of Q = \(\vec{a}\) – 3\(\vec{b}\)

Let OR be the position vector of point R, which divides PQ in the ratio 1 : 2 externally

Now, we have to show that P is the mid-point of RQ,

Hence, P is the mid-point of line segment RQ.

Product of Two Vectors and Scalar Triple Product

Question 1.
Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\). (CBSE 2018)
Answer:

Question 2.
Find the value of [î, k̂, ĵ], (CBSE 2018C)
Answer:
[î, k̂, ĵ] = î ∙ (k̂× ĵ)
= -[[î, k̂, ĵ] = –\(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = – 1

Question 3.
Find λ and μ, if (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂) = 0. (All India 2016)
Answer:
Given, (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂)

= î(3μ + 9λ) ĵ k̂
On comparing the coefficients of î, ĵ and k̂ , we get
3μ + 9λ = 0, – μ + 27 = 0 and – λ – 9 = 0
⇒ μ = 27 and – λ = 9
⇒ μ = 27 and λ = – 9
Also, the values of μ and λ satisfy the equation
3μ + 9λ = 0.
Hence, μ = 27 and λ = – 9.

Question 4.
Write the number of vectors of unit length perpendicular to both the vectors \(\vec{a}\) = 2î + ĵ + 2k̂ and \(\vec{b}\) = ĵ + k̂. (All India 2016)
Answer:
We know that, unit vectors perpendicular to \(\vec{a}\)
and \(\vec{b}\) are ±\(\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)\)
So, there arc two unit vectors perpendicular to the given vectors.

Question 5.
If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) = 0, then write the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (Foreign 2016)
Answer:

Question 6.
If \(|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}\) = 400 and \(|\vec{a}|\) = 5, then write the value of \(|\vec{b}|\). (Foreign 2016)
Answer:

Question 7.
Find λ, if the vectors \(\vec{a}\) = î + 3ĵ + k̂, \(\vec{b}\) = 2 î – ĵ – k̂ and \(\vec{c}\) = λĵ + 3 k̂ are coplanar. (Delhi 2015)
Answer:

⇒ 1(- 3 + λ) – 3(6) + 1(2λ) = 0
[expanding along R₁]
⇒ – 3 + λ – 18 + 2λ = 0
⇒ 3λ = 21
∴ λ = 7

Question 8.
If \(\vec{a}\) = 7î + ĵ – 4k̂ and \(\vec{b}\) = 2î + 6ĵ + 3k̂, then find the projection of \(\vec{a}\) on \(\vec{b}\). (Delhi 2015)
Answer:

Question 9.
If â, b̂ and ĉ are mutually perpendicular unit vectors, then find the value of |2â + b̂ + ĉ |. (All India 2015)
Answer:
Given â, b̂ and ĉ are mutually perpendicular unit vectors, i.e.

Question 10.
Write a unit vector perpendicular to both the vectors \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + ĵ. (All India 2015)
Answer:
First, determine perpendicular vectors of \(\vec{a}\) and \(\vec{b}\), i.e., \(\vec{a} \times \vec{b}\). Further , determine perpendicular unit vector by using formula \(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\).

Given vector are \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + ĵ
As we know the, vectors \(\vec{a} \times \vec{b}\) is perpendicular to both the vectors, so let us first evaluate \(\vec{a} \times \vec{b}\).
Then, \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= î(0 -1) – ĵ(0 – 1) + k̂(1 – 1)
= – î + ĵ
Then , the unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is given by

Question 11.
Find the area of a parallelogram whose adjacent sides are represented by the vectors 2 î – 3 k̂ and 4 ĵ + 2 k̂. (Foreign 2015)
Answer:
Let adjacent sides of a parallelogram bc
\(\vec{a}\) = 2 î – 3 k̂ and \(\vec{b}\) = 4 ĵ + 2 k̂.

Question 12.
If \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors, |\(\vec{a}\) + \(\vec{b}\)| = 13 and |\(\vec{a}\)| = 5, then find the value of |\(\vec{b}\)|. (All India 2014)
Answer:

Question 13.
If \(\vec{a}\) and \(\vec{b}\) are two unit vectors such that \(\vec{a}\) + \(\vec{b}\) is also a unit vector, then find the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014)
Answer:

Question 14.
Find the projection of the vector î + 3ĵ + 7k̂ on the vector 2î – 3 ĵ + 6k̂. (Delhi 2014)
Answer:
let \(\vec{a}\) = î + 3ĵ + 7k̂ and \(\vec{a}\) = 2î – 3 ĵ + 6k̂

Question 15.
Write the projection of vector î + ĵ + k̂ along the vector ĵ. (Foreign 2014)
Answer:
1

Question 16.
Write the value of the following. î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ). (Foreign 2014)
Answer:
we have, î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ)
= î × ĵ + î × k̂ × ĵ × k̂ + ĵ × î + k̂ × î + k̂ × ĵ
[∵ cross product is distributive over addition]
= k̂ – ĵ + î – k̂ + ĵ – î = \(\vec{0}\)
[∵ î × ĵ = k̂, î × k̂ = – ĵ, ĵ × k̂ = î, ĵ × î = – k̂, k̂ × î = ĵ, k̂ × ĵ = – î ]

Question 17.
If vectors \(\vec{a}\) and \(\vec{b}\) are such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 2/3 and \(\vec{a}\) × \(\vec{b}\) is a unit vector, then write the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014: All India 2010)
Answer:

Question 18.
Find \(\vec{a} \cdot(\vec{b} \times \vec{c})\), if \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂, and \(\vec{c}\) = 3î + ĵ + 2k̂. (All India 2014)
Answer:
Given, \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂, and \(\vec{c}\) = 3î + ĵ + 2k̂.

= 2(4 – 1) – 1 (- 2 – 3) + 3( – 1 – 6)
= 2 × 3 – 1 × (-5) + 3 × (- 7)
= 6 + 5 – 21 = 11 – 21 = – 10

Question 19.
If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then find the angle between \(\vec{a}\) and \(\vec{b}\), given that (√3\(\vec{a}\) – \(\vec{b}\)) is a unit vector. (Delhi 2014C)
Answer:

Question 20.
If |\(\vec{a}\)| = 8, |\(\vec{b}\)| = 3 and|\(|\vec{a} \times \vec{b}|\)| = 12, find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2014C)
Answer:
let θ be the angle between latex]\vec{a}[/latex] and \(\vec{b}\).

Question 21.
Write the projection of the vector \(\vec{a}\) = 2 î – ĵ + k̂ on the vector \(\vec{b}\) = î + 2ĵ + 2k̂. (Delhi 2014 C)
Answer:
\(\frac{2}{3}\)

Question 22.
Write the value of λ, so that the vectors a = 2î + λĵ + k̂ and b = î – 2ĵ + 3k̂ are perpendicular to each other. (Delhi 2013C, 2008)
Answer:
Given vectors are \(\vec{a}\) = 2î + λĵ + k̂
and \(\vec{b}\) = î – 2ĵ + 3k̂
Since, vectors are perpendicular.
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ (2î + λĵ + k̂) ∙ (î – 2ĵ + 3k̂)
⇒ 2 – 2λ + 3 = 0
∴ λ = 5/2

Question 23.
Write the projection of (\(\vec{b}\) + \(\vec{c}\)) on \(\vec{a}\), where \(\vec{a}\) = 2î – 2ĵ + k̂, \(\vec{b}\) = î + 2ĵ – 2k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. (All India 2013 C)
Answer:

Question 24.
Write the projection of the vector 7î + ĵ – 4k̂ on the vector 2î + 6 ĵ + 3k̂. (Delhi 2013C)
Answer:
\(\frac{8}{7}\)

Question 25.
If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\)|, then prove that vector 2\(\vec{a}\) + \(\vec{b}\) is perpendicular to vector b. (Delhi 2013)
Answer:

Question 26.
Find |\(\vec{x}\)|, if for â unit vector a, \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15. (All India 2013)
Answer:
Given, \(\vec{a}\) is a unit vector. Then, |\(\vec{a}\)| = 1.

Question 27.
Find λ, when projection of \(\vec{a}\) = λî + ĵ + 4k̂ on \(\vec{b}\) = 2 î + 6 ĵ + 3k̂ is 4 units. (Delhi 2012)
Answer:
Given, \(\vec{a}\) = λî + ĵ + 4k̂ on \(\vec{b}\) = 2 î + 6 ĵ + 3k̂ and projection of \(\vec{a}\) and \(\vec{b}\) = 4.

⇒ 2λ + 18 = 28
⇒ 2λ = 10
∴ λ = 5

Question 28.
Write the value of (k̂ × ĵ) . î + ĵ . k̂. (All India 2012)
Answer:
Use the results k̂ × ĵ = – î
ĵ ∙ k̂ and î ∙ î = 1 and simplify it.

Given, (k̂ × î) ∙ î + ĵ ∙ k̂ = (- î) ∙ î + ĵ ∙ k̂
= – (î ∙ î) + 0 = – 1 [∵ (î ∙ î) = 1]

Question 29.
If \(\vec{a} \cdot \vec{a}\) = 0 and \(\vec{a} \cdot \vec{b}\) = 0, then what can be concluded about the vector \(\vec{b}\)? (Foreign 2011)
Answer:

From Eqs. (i) and (ii). it may be concluded that \(\vec{b}\) is either zero or non-zero perpendicular vector.

Question 30.
Write the projection of vector î – ĵ on the vector î + ĵ. (All India 2011)
Answer:
0

Question 31.
Write the angle between vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes √3 and 2 respectively, having \(\vec{a}\). \(\vec{b}\) = √6. (All India 2011)
Answer:
let θ be the angle between \(\vec{a}\) and \(\vec{b}\), then use the following formula
cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\).

Question 32.
For what value of λ are the vectors î + 2λĵ + k̂ and 2î + ĵ – 3k̂ perpendicular? (All India 2011C)
Answer:
\(\frac{1}{2}\)

Question 33.
If |\(\vec{a}\)| = √3, |\(\vec{b}\)| = 2 and angle between \(\vec{a}\) and \(\vec{b}\) is 60°, then find \(\vec{a}\).\(\vec{b}\). (Delhi 2011C)
Answer:

Question 34.
Find the value of λ, if the vectors 2î + λĵ + 3k and 3î + 2ĵ – 4k̂ are perpendicular to each other. (All India 2010c)
Answer:
3

Question 35.
If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a}\).\(\vec{b}\) = 3, then find the projection of \(\vec{b}\) on \(\vec{a}\). (All India 2010C)
Answer:

Question 36.
If \(\vec{a}\) and \(\vec{b}\) are two vectors, such that \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\), then find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2010)
Answer:
Use the following formulae:
\(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
and \(|\vec{a} \times \vec{b}|\) = \(|\vec{a}||\vec{b}|\) sin θ
where, θ is the angle between \(\vec{a}\) and \(\vec{b}\).

Question 37.
Find λ, if (2î + 6ĵ + 14k̂) × (î – λĵ + Ik̂) = 0. (All India 2010)
Answer:
– 3

Question 38.
If the sum of two unit vectors a and b is a unit vector, show that the magnitude of their difference is √3. (Delhi 2019, 2012c)
Answer:
let \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\). Then, according to given condition \(\vec{c}\) is a unit vector, i.e. |\(\vec{c}\)| = 1.

[taking positive square root, as magnitude cannot be negative]

Question 39.
If \(\vec{a}\) = 2î + 5ĵ + k̂, \(\vec{b}\) = î – 2 ĵ + k̂ and \(\vec{c}\) = – 3î + ĵ + 2k̂, find \([\vec{a} \vec{b} \vec{c}]\). (Delhi 2019)
Answer:

= 2(- 4 – 1) – 3(2 + 3) + 1(1 – 6)
= – 10 – 15 – 5 = – 30

Question 40.
If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 7 and \(\vec{a} \times \vec{b}\) = 3î + 2ĵ + 6k̂, find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2019)
Answer:
let θ be the angle between \(\vec{a}\) and \(\vec{b}\).

Question 41.
Find the volume of cuboid whose edges are given by -5î + 7ĵ + 5k̂, -5î + 7ĵ – 5k̂ and 7î – 5 ĵ – 5k̂. (All India 2019)
Answer:

= |- 3 (- 21 – 15) – 7 (15 + 21) + 5(25 – 49)|
= |1108 – 252 – 120|
= 264 cubic units

Question 42.
Show that the points A(-2î + 5ĵ + 5k̂), B(î + 2 ĵ + 5k̂) and C(7î – k̂) are collinear. (All India 2019)
Answer:

Question 43.
Find \(|\vec{a} \times \vec{b}|\), if \(\vec{a}\) = 2î + ĵ + 5k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂. (All India 2019)
Answer:
We have, \(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂
∴ \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|\)
= î ( – 2 – 15) – ĵ (- 4 – 9) + k̂(10 – 3)
= – 17î + 13ĵ + 7k̂

Question 44.
If θ is the angle between two vectors î – 2 ĵ + 3k̂ and 3î – 2 ĵ + k̂, find sin θ. (CBSE 2018)
Answer:
let \(\vec{a}\) = î – 2 ĵ + 3k̂ and \(\vec{b}\) 3î – 2 ĵ + k̂

Question 45.
If \(\vec{a}+\vec{b}+\vec{c}\) = 0 and |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 6 and |\(\vec{c}\)| = 9, then find the angle between \(\vec{a}\) and \(\vec{b}\). (CBSE 2018C)
Answer:

(5)² + 2 × 5 × 6 × cos θ + (6)² = (9)²
⇒ 25 + 60 cos θ + 36 = 81
⇒ cos θ = \(\frac{20}{60}=\frac{1}{3}\)
⇒ θ = cos^-1\(\left(\frac{1}{3}\right)\)

Question 46.
If î + ĵ + k̂, 2î + 5ĵ, 5î + 2ĵ – 5k̂ and î – 6ĵ – k̂ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear or not. (Delhi 2019)
Answer:

Question 47.
The scalar product of the vector \(\vec{a}\) = î + ĵ + k̂ with a unit vector along the sum of the vectors \(\vec{b}\) = 2î + 4ĵ – 5k̂ and \(\vec{c}\) = λî + 2ĵ + 5k̂ is equal to 1. Find the value of λ and hence find the unit vector along \(\vec{b}\) + \(\vec{c}\). (All India 2019)
Answer:

Question 48.
Let \(\vec{a}\) = 4 î + 5ĵ – k̂, \(\vec{b}\) = î – 4ĵ + 5k̂ and \(\vec{c}\) = 3î + ĵ – k̂. Find a vector which is perpendicular to both \(\vec{c}\) and \(\vec{b}\) and \(\vec{d} \cdot \vec{a}\) = 21. (CBSE 2018)
Answer:
We have, \(\vec{a}\) = 4 î + 5ĵ – k̂, \(\vec{b}\) = î – 4ĵ + 5k̂ and \(\vec{c}\) = 3î + ĵ – k̂.
Since, \(\vec{d}\) is perpendicular to both \(\vec{c}\) and \(\vec{b}\).

= λ[î(5 – 4) – ĵ(15 + 1) + k̂(- 12 – 1)]
= λ(î – 16ĵ – 13k̂)
Also, it is given that \(\vec{a} \cdot \vec{a}\) = 21
∴ λ(î – 16ĵ – 13k̂) ∙ (4î + 5ĵ – k̂) = 21
⇒ λ(4 – 80 + 13) = 21
⇒ λ(- 63) = 21
⇒ λ = \(\frac{-1}{3}\)
Now from Eq. (j), we get
\(\vec{d}\) = –\(\frac{-1}{3}\)(î – 16ĵ – 13k̂)

Question 49.
Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D (7, 7, 2) are coplanar. (CBSE 2018C)
Answer:
Given points are A(4, 4, 4), B (5, x, 8), C(5, 4, 1) and D(7, 7, 2), then position vectors of A, B, C and D respectively, are

⇒ 1(0 + 9) – (x – 4) (- 2 + 9) + 4(3 – 0) = 0
⇒ 9 – (x – 4) (7) + 12 = 0
⇒ 9 – 7x + 28 + 12 = 0
⇒ 49 – 7x = 0
⇒ 7x = 49
⇒ x = 7

Question 50.
Find the value of x such that the points A(3, 2, 1), B(4, x, 5), C(4, 2,- 2) and D (6, 5, -1) are coplanar. (All India 2017)
Answer:
5

Question 51.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three mutually perpendicular vectors of the same magnitude, then prove that \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined with the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). (Delhi 2017, 2013C, 2011)
Answer:
If three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular to each other, then \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}\) = \(\vec{c} \cdot \vec{a}\) = 0 and if all three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are equally inclined with the vector \((\vec{a}+\vec{b}+\vec{c})\) that means each vector \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) makes equal angle with \((\vec{a}+\vec{b}+\vec{c})\) by using formula
cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\).

Question 52.
Using vectors, find the area of the ΔABC, whose vertices are A(1, 2, 5), 5(2, -1, 4) and C(4, 5, -1). (Delhi 2017; All India 2013)
Answer:
Let the position vectors of the verices A, B and C of ΔABC be

Question 53.
Let \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 0 ∙ ĵ + 0 ∙ k̂ and \(\vec{c}\) = c₁î + c₂ĵ + c₃k̂, then
(a) Let c₁ = 1 and c₂ = 2, find c₃ which makes \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
(b) If c₂ = – 1 and c₃ = 1, show that no value of c₁ can make \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar. (Delhi 2017)
Answer:
Given, \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 0 ∙ ĵ + 0 ∙ k̂ and \(\vec{c}\) = c₁î + c₂ĵ + c₃k̂
The given vectors are coplanar iff \([\vec{a} \vec{b} \vec{c}]\) = 0

(a) If c₁ = 1 and c₂ = 2,
Then, from Eq.(i), we get
\(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
1 & 2 & c_{3}
\end{array}\right|\) = 0

⇒ – 1(c₃ – 0) + 1(2 – 0) = 0
⇒ – c₃ + 2 = 0
⇒ – c₃ = – 2
⇒ c₃ = 2

(b) If c₂ = – 1 and c₃ = 1, then from Eq. (i), we get
\(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
c_{1} & -1 & 1
\end{array}\right|\) = 0
⇒ 1(0) – 1(1 – 0) + 1(- 1 – 0) = 0
⇒ 0 – 1 – 1 = 0
⇒ – 2 ≠ 0
∴ No value of c1 can make 1’ and coplanar.
Hence proved

Question 54.
Show that the points A, B, C with position vectors 2î – ĵ + k̂, î – 5ĵ – 5k̂ and 5î – 4ĵ – 4k̂ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle. (All India 2017)
Answer:

Question 55.
Show that the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar, if a + b, 6+ c and c+ a are coplanar. (Delhi 2016, Foreign 2014)
Or
Prove that, for any three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c},[\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]=2[\vec{a} \vec{b} \vec{c}]\). (Delhi 2014)
Answer:

Question 56.
Show that the four points A (4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D (-4, 4, 4) are coplanar. (All India 2016)
Or
Show that the four points A, B, C and D with position vectors 4î + 5ĵ + k̂, – ĵ – k̂, 3î + 9ĵ + 4k̂ and 4(- î + ĵ + k̂), respectively are coplanar. (All India 2014)
Answer:
Let the position vector of points A, B, C and D are

= – 4(12 + 3) + 6 (- 3 + 24) – 2(1 + 32)
= – 60 + 126 – 66 = 0
Hence, the four points A, B, C and D are coplanar.

Question 57.
The two adjacent sides of a parallelogram are 2î – 4ĵ – 5k̂ and 2î + 2 ĵ + 3k̂. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. (All India 2016)
Answer:
Let ABCD be the given parallelogram with

Question 58.
If \(\vec{a} \times \vec{b}=\vec{c} \times \vec{d}\) and \(\vec{a} \times \vec{c}=\vec{b} \times \vec{d}\), then show that \(\vec{a}-\vec{d}\) is parallel to \(\vec{b}-\vec{c}\), where \(\vec{a} \neq \vec{d}\) and \(\vec{b} \neq \vec{c}\). (Foreign 2016; Delhi 2009)
Answer:
Use the result, if two vectors are parallel, then their cross-product will be a zero vector.

Question 59.
If \(\vec{r}\) = xî + yĵ + zk̂, find \((\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})\) + xy. (Delhi 2015)
Answer:

Question 60.
If \(\vec{a}\) = î + 2ĵ + k̂, \(\vec{b}\) = 2î + ĵ and \(\vec{c}\) = 3î – 4 ĵ – 5k̂, then find a unit vector perpendicular to both of the vectors \((\vec{a}-\vec{b})\) and \((\vec{c}-\vec{b})\). (All India 2015)
Answer:

Question 61.
Find the value of λ so that the four points A, B,C and D with position vectors 4 î + 5ĵ + k̂, -ĵ – k̂,3i + Xj+4k and – 4 î + 4ĵ + 4 k̂, respectively are coplanar. (Delhi 2015C)
Answer:
Use the condition that four points with position vectors \(\vec{A}, \vec{B}, \vec{C}\) and \(\vec{D}\) are coplanar, if
\([\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=\overrightarrow{0}\) = 0.

On expanding along R₁, we get
⇒ – 4(3λ – 15 + 3) + 6(- 3 + 24) – 2(1 + 8λ – 40) = 0
⇒ – 4(3λ – 12) + 6(21) – 2(8λ – 39) = 0
⇒ – 12λ + 48 + 126 – 16λ + 78 = 0
⇒ – 28λ + 252 = 0
λ = 9

Question 62.
Prove that \(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{d}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{d}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{d}
\end{array}\right]\). (All India 2015C)
Answer:

Question 63.
If \(\vec{a}\) = 2î – 3ĵ + k̂, \(\vec{b}\) = – î + k̂, \(\vec{c}\) = 2 ĵ – k̂ are three vectors, find the area of the parallelogram having diagonals \(\) and \(\). (Delhi 2014C)
Answer:

Question 64.
Vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\) are such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5 and |\(\vec{c}\)| = 7. Find the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014,2008; All India 2008)
Answer:
\(\frac{\pi}{3}\)

Question 65.
The scalar product of the vector \(\vec{a}\) = î + ĵ + k̂ with a unit vector along the sum of vectors \(\vec{b}\) = 2î + 4ĵ – 5k̂ and \(\vec{c}\) = λî + 2ĵ + 3k̂ is equal to one. Find the value of λ and hence, find the unit vector along \(\vec{b}\) + \(\vec{c}\). (All India 2014)
Or
The scalar product of vector i + j + k with the unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ. (All India 2009,2008C)
Answer:
First, determine the unit vector of \(\vec{b}+\vec{c}\), i.e. \(\frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}\). Further put \(\vec{a} \cdot \frac{(\vec{b}+\vec{c})}{|\vec{b}+\vec{c}|}\) = 1 and then determine the value of λ.

⇒ (λ + 6)² = λ² + 4λ + 44 [squaring both sides]
⇒ λ² + 36 + 12λ + 4λ + 44
⇒ 8λ = 8
⇒ λ = 1
Hence, the value of λ is 1.
On substituting the value of λ in Eq. (1), we get Unit vector along \(\vec{b}+\vec{c}\)

Question 66.
Find the vector \(\vec{p}\) which is perpendicular to both \(\vec{α}\) = 4î + 5ĵ – k̂ and \(\vec{β}\) = î – 4ĵ + 5k̂ and \(\vec{p}\). \(\vec{q}\) = 21, where \(\vec{q}\) = 3i + j – k. (All India 2014C)
Answer:

= î (25 – 4) – ĵ (20 + 1) + k̂(- 16 – 5)
= î(21) – ĵ(21) + k̂(- 21)
= 21î – 21ĵ – 21k̂
So, \(\vec{p}\) = 21λk̂—21λ?—21λk [fromEq.(i)] ….. (ii)
Also, given that \(\vec{p} \cdot \vec{q}\) = 21
∴ (21λî – 21λĵ – 21λk̂) . (3î + ĵ – k̂) = 21
⇒ 63λ – 21λ + 21λ = 21
⇒ 63λ = 21
⇒ λ = \(\frac{1}{3}\)
On putting λ = \(\frac{1}{3}\) in Eq. (ii), we get

which is the required vector.

Question 67.
Find the unit vector perpendicular to both of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) where, \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + 2ĵ + 3k̂. (Foreign 2014)
Answer:

⇒ 2x + 3y + 4z = 0 …… (ii)
and (xî + yĵ + zk̂) . (- ĵ – 2k̂) = 0
⇒ – y – 2z = 0
⇒ y = – 2z
On putting the value of yin Eq. (ii), we get
2x + 3 (- 2z) + 4z = 0
⇒ x = z
On substituting the value of x and y in Eq. (1),
we get
⇒ z² + 4z² + z² = 1
⇒ 6z² = 1
⇒ z = ± \(\frac{1}{\sqrt{6}}\)
then, x = ± \(\frac{1}{\sqrt{6}}\)
and y = ± \(\frac{2}{\sqrt{6}}\)
Hence, the required vectors are

Question 68.
Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, î + ĵ + 2k̂ and 2î + 3k̂, respectively. (All India 2014C)
Answer:
A unit vector perpendicular to plane ABC is
\(\frac{\overrightarrow{A B} \times \overrightarrow{A C}}{|\overrightarrow{A B} \times \overrightarrow{A C}|}\)

Let O be the origin of reference.

Question 69.
Dot product of a vector with vectors î – ĵ + k̂, 2î + ĵ – 3k̂ and î + ĵ + k̂ are respectively 4, 0 and 2. Find the vector. (Delhi 2013C)
Answer:

⇒ a₁ + a₂ + a₃ = 2
On subtracting Eq. (iii) from Eq. (i), we get
– 2a₂ = 2
⇒ a₂ = – 1
On substituting a₂ = – 1 in Eq. (ii) and (iii),
we get
2a₂ – 3a₃ = 1 …… (iv)
⇒ a₁ + a₃ = 3
On multiplying Eq. (v) by 3 and then adding with Eq. (iv), we get
5a₁ = 1 + 9 = 10
⇒ a₁ = 2
On substituting a₁ = 2 in Eq. (v), we get
a₃ = 1
Hence, the vector is \(\vec{a}\) = 2î – ĵ + k̂

Question 70.
Find the values of λ for which the angle between the vectors \(\vec{a}\) = 2λ²î + 4λĵ + k̂ and \(\vec{b}\) = 7î – 2ĵ + λk̂ is obtuse. (All India 2013C)
Answer:
let θ be the obtuse angle between the vectors

14λ² – 7λ < 0
= 2λ² – λ < 0
Either λ < 0, 2λ – 1 > 0 or λ > 0, 2λ – 1 < 0
= Either λ < 0, λ > \(\frac{1}{2}\) or λ > 0, λ < \(\frac{1}{2}\) Clearly, first option is impossible. ∴ λ > 0, λ < \(\frac{1}{2}\)
0 < λ < \(\frac{1}{2}\)
λ ∈ \(\left(0, \frac{1}{2}\right)\)

Question 71.
If a, b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5, then prove that |\(\vec{a}+\vec{b}+\vec{c}\)| = 5√2. (All India 2013C, 2010C)
Or
If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three vectors, such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5 and each one of these is perpendicular to the sum of other two, then find |\(\vec{a}+\vec{b}+\vec{c}\)|. (All India 2011C, 2010C)
Answer:

Question 72.
If \(\vec{a}\) = 3î – ĵ and \(\vec{b}\) = 2î + ĵ – 3k̂, then express \(\vec{b}\) in the form \(\vec{b}=\vec{b}_{1}+\vec{b}_{2}\), where \(\vec{b}_{1} \| \vec{a}\) and \(\vec{b}_{2} \perp \vec{a}\). (All India 2013C)
Answer:
Given \(\vec{a}\) = 3î – ĵ and \(\vec{b}\) = 2î + ĵ – 3k̂
Let \(\overrightarrow{b_{1}}\) = x₁î + y₁ĵ + z₁k̂ are two vectors such that \(\overrightarrow{b_{1}}+\overrightarrow{b_{2}}=\vec{b}, \overrightarrow{b_{1}} \| \vec{a}\) and \(\overrightarrow{b_{2}} \perp \vec{a}\)

Consider, \(\vec{b}_{1}+\vec{b}_{2}=\vec{b}\)
⇒ (x₁ + x₂)î + (y₁ + y₂)ĵ + (z₁ + z₂)k̂ = 2î + ĵ – 3k̂

On comparing the coefficient of î ĵ and k̂ both sides, we get
x₁ + x₂ = 2
y₁ + y₂ = 1
z₁ + z₂ = -3

Now, consider \(\overrightarrow{b_{1}} \| \vec{a}\)
⇒ \(\frac{x_{1}}{3}=\frac{y_{1}}{-1}=\frac{z_{1}}{0}\)
⇒ x₁ = 3λ, y₁ = -λ,and z₁ = 0 …(iv)

On substituting the values of x, y and z, from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get
x₂ = 2- 3λ, y₂ = -1 + λ and z₂ = -3 …(v)
Since, b2 ± a , therefore b2 a = 0
⇒ 3x₂ – y₂ = 0
⇒ 3 (2 – 3λ) – (1 + λ) = 0
⇒ 6 – 9λ – 1 – λ = 0
⇒ 5 – 10λ = 0
⇒ λ = \(\frac{1}{2}\)

On substituting λ = \(\frac{1}{2}\) in Eqs. (iv) and (v), we get

Question 73.
If \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = ĵ – k̂, then find a vector \(\vec{c}\), such that \(\vec{a} \times \vec{c}=\vec{b}\) and \(\vec{a} \cdot \vec{c}\) = 3. (Delhi 2013, 2008)
Answer:
Given \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = ĵ – k̂
Let \(\vec{c}\) = xî + yĵ + zk̂

= î (z – y) – ĵ(z – x) + k̂(y – x)
Now, \(\vec{a} \times \vec{c}=\vec{b}\) [given]
= î(z – y) + ĵ(x – z) + k̂(y – x)
= 0î + 1ĵ + (-1)k̂ [∵ \(\vec{b}\) = ĵ – k̂]

On comparing the coefficients from both sides, we get
z – y = 0,x – z = 1, y – x = -1
⇒ y = z and x – y = 1…(i)

Also given, \(\vec{a} \cdot \vec{c}\) = 3
⇒ (î + ĵ + k̂) . (xî + yĵ + zk̂) = 3
⇒ x + y + z = 3 (1)
⇒ x + 2y = 3 [∵ y = z] …(ii)

On subtracting Eq. (i) from Eq. (ii), we get
3y = 2
⇒ y = \(\frac{2}{3}\) = z [∵ y = z]

From Eq. (i),
x = 1 + y + 1 = 1 + \(\frac{2}{3}=\frac{5}{3}\)
Hence, \(\vec{c}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\)

Question 74.
If \(\vec{a}\) = î – ĵ + 7k̂ and \(\vec{b}\) = 5î – ĵ + λk̂, then find the value of λ, so that \(\) and \(\) are perpendicular vectors. (All India 2013)
Answer:
Use the result that if \(\vec{a}\) and \(\vec{b}\) are perpendicular, then their dot product should be zero and simplify it.
Given, \(\vec{a}\) = î – ĵ + 7k̂ and \(\vec{b}\) = 5î – ĵ + λk̂
Then, \(\vec{a}+\vec{b}\) = (î – ĵ + 7k̂) + (5î – ĵ + λk̂)
= 6î – 2ĵ + (7 + λ) k̂
and a – = (î – ĵ + 7k̂) – (5î – j ̂+ λk̂)
= -4î + (7 – λ)k̂

Since, \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular
vectors, then \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 0
⇒ [6î – 2ĵ + (7 + λ)k̂]- [-4î + (7 – λ)k̂] = 0 (1)
⇒ -24 + (7+ X)(7 – X) =0
⇒ 49 – λ² = 24
⇒ λ² = 25
∴ λ = ± 5

Question 75.
If p = 5î + λĵ – 3k̂ and q = î + 3ĵ – 5k̂, then find the value of λ, so that \(\vec{p}+\vec{q}\) and \(\vec{p}-\vec{q}\) are perpendicular vectors. (All India 2013)
Answer:
λ = ± 1

Question 76.
If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three vectors, such that |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 12, |\(\vec{c}\)| = 13 and \(\vec{a}+\vec{b}+\vec{c}\) = 0, then find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (Delhi 2012)
Answer:
-169

Question 77.
Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î – 2ĵ + 7k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. Find a vector \(\vec{p}\), which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{p}.\vec{c}\) = 18. (All India 2012,2010)
Answer:
Given vectors are \(\vec{a}\) = î + 4ĵ + 2k̂,
\(\vec{b}\) = 3î – 2ĵ + 7k̂
and \(\vec{c}\) = 2î – ĵ + 4k̂

Let \(\vec{p}\) = xî + yĵ + zk̂
We have, \(\vec{p}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
\(\vec{p} \cdot \vec{a}\) = 0
⇒ (xî + yĵ + zk̂) – (î + 4ĵ + 7k̂) = 0
⇒ x + 4y + 2z = 0 ………….(i)

and \(\vec{p} \cdot \vec{b}\) = 0
⇒ (xî + yĵ + zk̂) . (3î – 2ĵ + 7k̂) = 0
⇒ 3x – 2y + 7z = 0 …(ii)

Also, given ~p-~c =18 (1)
⇒ (xî + yĵ + zk̂) . (2î – ĵ + 4k̂) = 0
⇒ 2x – y + 4z = 18 …(iii)

On multiplying Eq. (i) by 3 and subtracting it from Eq. (ii), we get
– 14y + z = 0 ..(iv)

Now, multiplying Eq. (i) by 2 and subtracting it from Eq. (iii), we get
– 9y = 18
⇒ y = -2

On putting y = -2 in Eq. (iv), we get
-14 (-2) + z = 0
⇒ 28 + z = 0
⇒ z = -28

On putting y = -2 and z = -28 in Eq. (i), we get
x + 4 (-2) + 2 (-28) = 0
⇒ x – 8 – 56 = 0
⇒ x = 64

Hence, the required vector is
\(\vec{p}\) = xî + yĵ + zk̂
i.e. \(\vec{p}\) = 64î – 2ĵ – 28k̂

Question 78.
Find a unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where a = 3î + 2 ĵ + 2k̂ and b = î + 2ĵ – 2k̂. (Delhi 2011)
Answer:
\(\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\)

Question 79.
If a and 6 are two vectors, such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 1 and \(\vec{a}\).\(\vec{b}\) = 1, then find \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\). (Delhi 2011)
Answer:

Question 80.
If vectors \(\vec{a}\) = 2î + 2ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂ and \(\vec{c}\) = 3î + ĵ are such that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\), then find the value λ. (Foreign 2011; All India 2009C)
Answer:
Given, \(\vec{a}\) = 2î + 2ĵ + 3k̂,
\(\vec{b}\) = -î + 2ĵ + k̂
and \(\vec{c}\) = 3î + ĵ

Also, \(\vec{a}\) + λ\(\vec{b}\) is perpendicular to \(\vec{c}\).
∴ (\(\vec{a}\) + λ\(\vec{b}\)).\(\vec{c}\) = 0 …(i) [∵ when \([latex]\)[/latex], then \(\vec{a} \cdot \vec{b}\) = 0]
Now, \(\vec{a}\) + λ\(\vec{b}\) = (2î + 2ĵ + 3k̂) + λ (-î + 2ĵ + k̂)
⇒ \(\vec{a}\) + λ\(\vec{b}\) = î(2 – λ) + ĵ(2 + 2λ) + k̂(3 + λ)
Then, from Bq. (i), we get
[î (2 – λ) + ĵ (2 + 2λ) + k̂(3 + λ)].[3î + ĵ] = 0
⇒ 3(2 – λ) + 1(2+ 2k) = 0
⇒ 8 – λ = 0
∴ λ = 8

Question 81.
Using vectors, find the area of triangle with vertices A (1, 1, 2), 5(2, 3, 5) and C (1, 5, 5). (All India 2011)
Answer:
\(\frac{1}{2}\)\(\sqrt{61}\) sq.units

Question 82.
Using vectors, find the area of triangle with vertices A (2, 3, 5), B (3, 5, 8) and C(2, 7, 8). (Delhi 2010C)
Answer:
\(\frac{1}{2}\)\(\sqrt{61}\) sq.units

The post Vector Algebra Class 12 Maths Important Questions Chapter 10 appeared first on Learn CBSE.

We have given these Important Questions for Class 10 Science Chapter 12 Electricity to solve different types of questions in the exam. Previous Year Questions & Important Questions of Electricity Class 10 Science Chapter 12 will help the students to score good marks in the board examination.

Important Questions of Electricity Class 10 Science Chapter 12

Question 1.
A current of 10 A flows through a conductor for two minutes.
(i) Calculate the amount of charge passed through any area of cross section of the conductor.
(ii) If the charge of an electron is 1.6 × 10^-19 C, then calculate the total number of electrons flowing. (Board Term I, 2013)
Answer:
Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s
(i) Amount of charge Q passed through any area of cross-section is given by I = \(\frac { Q }{ t }\)
or Q = I × t ∴ Q = (10 × 120) A s = 1200 C

(ii) Since, Q = ne
where n is the total number of electrons flowing and e is the charge on one electron
∴ 1200 = n × 1.6 × 10^-19
or n = \(\frac { 1200 }{ 1.6×10^{-19} }\) = 7.5 × 10²¹

Question 2.
Define electric current. (1/5, Board Term 1,2017)
Answer:
Electric current is the amount of charge flowing through a particular area in unit time.

Question 3.
Define one ampere. (1/5, Board Term 1,2015)
Answer:
One ampere is constituted by the flow of one coulomb of charge per second.
1 A = 1 C s^-1

Question 4.
Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so. (Board Term I, 2013)
Answer:
A cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor.

Question 5.
Draw the symbols of commonly used components in electric circuit diagrams for
(i) An electric cell
(ii) Open plug key
(iii) Wires crossing without connection
(iv) Variable resistor
(v) Battery
(vi) Electric bulb
(vii) Resistance (4/5, Board Term 1,2017)
Answer:

Question 6.
A student plots V-I graphs for three samples of nichrome wire with resistances R₁, R₂ and R₃. Choose from the following the statements that holds true for this graph. (2020)

(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₁ > R₃
Answer:
(d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of -Rj is highest. Thus, R₂ > R₁ > R₃

Question 7.
State Ohms law. (AI 2019)
Answer:
It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.

Question 8.
A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph. (2020)

Answer:
As graph is a straight line, so it is clear from the graph that V ∝ I.

The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line.
According to ohms law,

V = IR or R = \(\frac { V }{ I }\)
So, the slope of V’-/ graph at any point represents the resistance of the given conductor.

Question 9.
Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph? (Board Term I, 2016)

Answer:
Since, the graph is straight line so we can either extrapolate the data or simply mark the value from graph as shown in figure.

Hence, the value of current for V = 10 volts is 5 amperes (or 5 A).
From Ohm’s law, V = IR,
We can write, R = \(\frac { V }{ I }\)
At any point on the graph, resistance is the ratio of values of V and I. Since, the given graph is straight line (ohmic conductor) so, the slope of graph will also give the resistance of the resistor
R = \(\frac { 10 V }{ 5A }\) = 2Ω
Alternately, R = \(\frac { (8-2)V }{ (4-1)A }\) = \(\frac { 6 V }{ 3A }\) = 2 Ω

Question 10.
V-I graph for a conductor is as shown in the figure

(i) What do you infer from this graph?
(ii) State the law expressed here. (Board Term I, 2014)
Answer:
(i) Refer to answer 8.
(ii) Refer to answer 7.

Question 11.
State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor. (Board Term I, 2016)
Answer:
Refer to answer 7 and 8.

Question 12.
State and explain Ohm’s law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features. (Board Term I, 2014)
Answer:
Ohm’s law: Refer to answer 7.
Resistance : It is ihe properly of a conductor lo resist the How of charges through it.
Its SI unit is ohm (Ω). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 Ω). lvolt
1 ohm = \(\frac { 1 volt }{ 1 ampere }\)
V-I graph for an ohmic conductor can be drawn as given in figure.

Important feature of V-I graph are:
(i) It is a straight line passing through origin.
(ii) Slope of V-I graph gives the value of resistance of conductor slope = R = \(\frac { V }{ I }\)

Question 13.
Assertion (A) : The metals and alloys are good conductors of electricity.
Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. (2020)
Answer:
(c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity.

Question 14.
A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l’ is (2020)
(a) \(\frac { A }{ 2 }\)
(b) \(\frac { 3A }{ 2 }\)
(c) 2A
(d) 3A
Answer:
(c) : The resistance of a conductor of length!, and area of cross section, A is
R = ρ\(\frac { l }{ A }\)
where ρ is the resistivity of the material.
Now for the conductor of length 21, area of cross-section A’ and resistivity ρ.
R’ = ρ\(\frac { l’ }{ A’ }\) = ρ\(\frac { 2l }{ A’ }\)
But given, R = R’ ⇒ ρ\(\frac { l }{ A }\) = ρ\(\frac { 2l }{ A }\) or A’ = 2A

Question 15.
Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. (2020)
Answer:
(a)

Question 16.
How is the resistivity of alloys compared with those of pure metals from which they may have been formed? (Board Term I, 2017)
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals.

Question 17.
(i) List three factors on which the resistance of a conductor depends.
(ii) Write the SI unit of resistivity. (Board Term 1, 2015)
Answer:
(i) Resistance of a conductor depends upon the following factors:
(1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R).
R ∝ I

(2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance.
R ∝ \(\frac { 1 }{ A }\)

(3) Nature of conductor.
(ii) SI unit of resistivity is Ω m.

Question 18.
Calculate the resistance of a metal wire of length 2m and area of cross section 1.55 × 10⁶ m², if the resistivity of the metal be 2.8 × 10^-8 Ωm. (Board Term I, 2013)
Answer:
For the given metal wire,
length, l = 2 m
area of cross-section, A = 1.55 × 10^-6 m²
resistivity of the metal, p = 2.8 × 10^-8 Ω m
Since, resistance, R = ρ\(\frac { l }{ A }\)
So R = (\(\frac { 2.8×10^{-8}×2 }{ 1.55×10^{-6} }\))Ω
= \(\frac { 5.6 }{ 1.55 }\) × 10^-2 Ω = 3.6 × 10^-2Ω or R = 0.036Ω

Question 19.
(a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.
(c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018)
Answer:
(a) Refer to answer 17 (i).
(b) Metal have very low resistivity and hence they are good conductors of electricity.
Whereas glass has very high resistivity so glass is a bad conductor of electricity.
(c) Alloys are commonly used in electrical heating devices due to the following reasons
(i) Alloys have higher resistivity than metals
(ii) Alloys do not get oxidised or burn readily.

Question 20.
Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms. (Board Term I, 2017)
Answer:
We are given, the length of wire, l = 1 m, radius of wire, r = 0.01 cm = 1 × 10^-4 m and resistance, R = 20Ω As we know,
R = ρ\(\frac { l }{ A }\), where ρ is resistivity of the material of the wire.
∴ 20Ω.= ρ\(\frac{l}{\pi r^{2}}\) = ρ\(\frac{1 \mathrm{~m}}{3.14 \times\left(10^{-4}\right)^{2} \mathrm{~m}^{2}}\)
∴ ρ = 6.28 × 10^-7 Ω m

Question 21.
A copper wire has diameter 0.5 mm and resistivity 1.6 × 10^-8 Ω m. Calculate the length of this wire to make it resistance 100 Ω. How much does the resistance change if the diameter is doubled without changing its length? (Board Term I, 2015)
Answer:
Given; resistivity of copper = 1.6 × 10^-8 Ω m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 Ω
Radius of wire, r = \(\frac {d }{ 2}\) = \(\frac {0.5 }{ 2 }\) mm
= 0.25 mm = 2.5 × 10^-4 m
Area of cross-section of wire, A = nr²
∴ A = 3.14 × (2.5 × 10^-4)²
= 1.9625 × 10^-7 m²
= 1.9 × 10^-7 m²
As, R = ρ\(\frac { l }{ A }\)
∴ 100 Ω = \(\frac{1.6 \times 10^{-8} \Omega \mathrm{m} \times l}{1.9 \times 10^{-7} \mathrm{~m}^{2}}\)
l = 1200 m
If diameter is doubled (d’ = 2d), then the area of cross-section of wire will become
A’ = πr² = π(\(\frac { d’ }{ 2 }\))² = π(\(\frac { 2d }{ 2 }\))² = 4A
Now R ∝ \(\frac { 1 }{ A }\), so the resistance will decrease by four times or new resistance will be
R’ = \(\frac { R }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25Ω

Question 22.
The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10^-8 ohm meter, find the length of the wire. (Board Term I, 2014)
Answer:
Here, r = 0.01 cm = 10^-4 m, ρ = 50 × 10^-8 Ω m and R = 10 Ω
As, R = ρ\(\frac { l }{ A }\)
or l = \(\frac{R A}{\rho}=\frac{R}{\rho}\left(\pi r^{2}\right)\)
so l = \(\frac{10}{50 \times 10^{-8}} 3.14 \times\left(10^{-4}\right)^{2}\)
= 0.628 m = 62.8 cm

Question 23.
A wire has a resistance of 16 Ω. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance? (Board Term I, 2013)
Answer:
When wire is melted, its volume remains same, so,
V’ = V or A’l’ = Al
Here, l’ = \(\frac { l }{ 2 }\)
Therefore, A’ = 2 A
Resistance, R = ρ\(\frac { l }{ A }\) = 16 Ω
Now, R’ = \(\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{(l / 2)}{2 A}=\frac{1}{4} \rho \frac{l}{A}\)
So, R’ = \(\frac { R }{ 4 }\) = \(\frac { 16 }{ 4 }\) = 4 Ω
Percentage change in resistance,
= \(\left(\frac{R-R^{\prime}}{R}\right) \times 100=\left(\frac{16-4}{16}\right)\) × 100 = 75%

Question 24.
If the radius of a current carrying conductor is halved, how does current through it change? (2/5 Board Term I, 2014)
Answer:
If the radius of conductor is halved, the area of cross-section reduced to (\(\frac { 1 }{ 4 }\)) of its previous value.
Since, R ∝ \(\frac { 1 }{ A }\), resistance will become four times
From Ohm’s law, V = IR
For given V, I ∝ \(\frac { 1 }{ R }\)
So, current will reduce to one-fourth of its previous value.

Question 25.
Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10^-8 Ω m. (Board Term I, 2014)
Answer:
Resistance is the property of a conductor to resist the flow of charges through it.
Factors affecting resistance of a conductor:
Refer to answer 17(i)
Rheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit.
We are given, length of wire, l = 50 cm = 50 × 10^-2 m cross-sectional area, A = 0.01 mm²
= 0.01 × 10^-6 m²
and resistivity, ρ = 5 x 10^-8 Ω m.
As, resistance, R = ρ\(\frac { l }{ A }\)
∴ R = \(\left(\frac{5 \times 10^{-8} \times 50 \times 10^{-2}}{0.01 \times 10^{-6}}\right)\) Ω
= 2.5 Ω

Question 26.
If a person has five resistors each of value \(\frac { 1 }{ 5 }\) Ω, then the maximum resistance he can obtain by connecting them is
(a) 1 Ω
(b) 5 Ω
(c) 10 Ω
(d) 25 Ω (2020)
Answer:
(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,
R_s = \(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\) 1 Ω

Question 27.
The maximum resistance which can be made using four resistors each of 2 Ω is
(a) 2 Ω
(b) 4 Ω
(c) 8 Ω
(d) 16 Ω (2020)
Answer:
(c) : A group of resistors can produce maximum resistance when they all are connected in series.
∴ R_s = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω

Question 28.
The maximum resistance which can be made using four resistors each of resistance \(\frac { 1 }{ 2 }\) Ω is
(a) 2 Ω
(b) 1 Ω
(c) 2.5 Ω
(d) 8 Ω (2020)
Answer:
(a) The maximum resistance can be produced from a group of resistors by connecting them in series.
Thus, R_s = \(\frac { 1 }{ 2 }\) Ω + H \(\frac { 1 }{ 2 }\) Ω + \(\frac { 1 }{ 2 }\) Ω + \(\frac { 1 }{ 2 }\) Ω = 2 Ω

Question 29.
Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance. (Board Term I, 2014)
Answer:
Here, R₁ = 10 Ω, R₂ =15 Ω, R₃ = 5 Ω.
In parallel combination, equivalent resistance, (R_eq) is given by

Question 30.
List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013)
Answer:
(a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.

(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.

Question 31.
Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω (2018)
Answer:
(i) The resistance of the series combination is higher than each of the resistances. A parallel combination of two 9 Ω resistors is equivalent to 4.5 Ω. We can obtain 13.5 Ω by coupling 4.5 Ω and 9 Ω in series. So, to obtain 13.5 Ω, the combination is as shown in figure (a).

(ii) To obtain a equivalent resistance of 6 Ω, we have to connect two 9 Ω resistors in series and then connect the third 9 Ω resistor in parallel to the series combination as shown in the figure (b).

Question 32.
Three resistors of 3 Ω each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery. (Board Term I, 2017)

Answer:
As given in circuit diagram, two 3 Ω resistors are connected in series to form R₁; so R₁ = 3 Ω + 3 Ω = 6 Ω
And, R₁ and R₂ are in parallel combination, Hence, equivalent resistance of circuit (R_eq) given by

R_eq = 2 Ω
Using Ohm’s law, V = IR
We get,
3 V = I × 2 Ω
or I = \(\frac { 3 }{ 2 }\) A = 1.5 A
Current drawn from the battery is 1.5 A.

Question 33.
Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. (Board Term I, 2013)
Answer:
Let resistance of each resistor be R.
For series combination,
R_s = R₁ + R₂
So, R_s = R + R = 2R
For parallel combination,

Question 34.
(a) A 6 Ω resistance wire is doubled on itself. Calculate the new resistance of the wire.
(b) Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer. (2020)
Answer:
(a) Given resistance of wire, R = 6 Ω
Let l be the length of the wire and A be its area of
cross-section. Then
R = \(\frac { ρl }{ A }\) = 6 Ω
Now when the length is doubled, l’ = 2l and A’ = \(\frac { A }{ 2 }\)
∴ R’ = \(\frac{\rho(2 l)}{A / 2}=\frac{4 \rho l}{A}\) = 4 × 6 Ω = 24 Ω

(b) Given the total resistance of the combination = 3 Ω
In order to get a total resistance of 3 Ω, the three resistors has to be connected as shown.

Such that, \(\frac{1}{R_{P}}=\frac{1}{2}+\frac{1}{2}\) = 1
⇒ R_p = 1 Ω
and R_s = 2 Ω + 1 Ω = 3 Ω

Question 35.
Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2 V each, a combination of three resistors of 10 Ω, 20 Ω and 30 Ω connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following :
(a) Current through each resistor
(b) Total current in the circuit
(c) Total effective resistance of the circuit. (2020)
Answer:
The circuit diagram is as shown below.

(a) Given, voltage of the battery = 2V + 2V + 2V = 6 V
Current through 10 Ω resistance,
I₁₀ = \(\frac{V}{R}=\frac{6}{10}\) = 0.6 A
Current through 20 Ω resistance,
I₂₀ = \(\frac{V}{R}=\frac{6}{20}\) = 0.3 A
Current through 30 Ω resistance,
I₃₀ = \(\frac{V}{R}=\frac{6}{30}\) = 0.2 A
(b) Total current in the circuit, 1= I₁₀ + I₂₀ + I₃₀
= 0.6 + 0.3 + 0.2 = 1.1 A
(c) Total resistance of the circuit,
\(\frac{1}{R_{P}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30}=\frac{11}{60}\)

Question 36.
(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b) In an electric circuit two resistors of 12 Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery. (Delhi 2019)
Answer:
(a) Resistors in parallel : When resistors are connected in parallel.

(i) The potential difference across their ends is the same.
(ii) The sum of current through them is the current drawn from the source of energy or cell.
I = I₁ + I₁ + I₃ or \(\frac{V}{R_{P}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}\)
(iii) The equivalent resistance is given by,
\(\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
Hence equivalent resistance in parallel combination is equal to the sum of reciprocals of the individual resistances.

Question 37.
For the series combination of three resistors current in each resistor, establish the relation R = R₁ + R₂ + R₃ where the symbols have their usual meanings. Calculate the equivalent resistance of the combination of three resistors of 6 Ω, 9 Ω and 18 Ω joined in parallel. (Board Term I, 2016)
Answer:

Given figure shows the series combination of three resistors R₁, R₂ and R₃ connected across a voltage source of potential difference V.
Let current I is flowing through the circuit.
V₁, V₂ and V₃ are the potential differences across resistors R₁, R₂ and R₃ respectively.

Since, the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.
∴ v = v₁ + v₂ + v₃ …(i)
In series current through each resistor is same. Applying the Ohms law,
V₁ = IR₁, V₂ = IR₂ and V₃ = IR₁ ……..(ii)
If R_s is the equivalent resistance of the circuit, then
V = IR_s …(iii)
From eqns. (i), (ii) and (iii),
we can write IR_s = IR₁ + IR₂ + IR₃
or R_s = R₁ + R₂ + R₃
We can conclude that when several resistors are joined in series, the resistance of the combination R_s equals the sum of their individual resistances,
R₁, R₂ and R₃
Given : R₁ = 6 Ω, R₂ = 9 Ω,
R₃ = 18 Ω are connected in parallel.
Equivalent resistance, R_eq, is given by

or R_eq = 3 Ω

Question 38.
State ohms law. Represent it graphically. In the given circuit diagram calculate
(i) the total effective resistance of the circuit.
(ii) the current through each resistor.

Answer:
Ohm’s law: Refer to answer 7.
Graphical representation of Ohm’s law

For the given circuit
R₁ = 3 Ω, R₂ = 4 Ω., R₃ = 6 Ω and V = 6V.
(i) Total effective resistance of the circuit, R_eq is given by
\(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{9}{12}\)
or R_eq = \(\frac { 12 }{ 9 }\) Ω = \(\frac { 4 }{ 3 }\) Ω = 1.33 Ω

(ii) Since, potential difference across each resistor connected in parallel is same.
So, V₁ = V₂ = V₃ = 6 V
Applying Ohm’s law,
V₁ = I₁R₁ or I₁ = \(\frac { V_1 }{ R_1 }\) or I₃ = \(\frac { 6 }{ 3 }\) A = 2A
Similarly, I₂ = \(\frac { 6A }{ 4 }\) = 1.5 A and I₃ = \(\frac { 6 }{ 6 }\) A = 1 A

Question 39.
(a) Prove that the equivalent resistance of three resistors R₁, R₂ and R₃ in series is R₁ + R₂ + R₃
(b) You have four resistors of 8 Ω each. Show how would you connect these resistors to have effective resistance of 8 Ω? (4/5, Board Term I, 2015)
Answer:
(a) Refer to answer 37.
(b) If you have four 8 Ω resistors and the effective resistance is also 8 Ω then the two 8 Ω resistors are connected in series. Now you have pair of two 16 Ω resistors (8 Ω + 8 Ω). If you connect these resistors in parallel, you will have net resistance 8 Ω.

Question 40.
Draw a labelled circuit diagram showing three resistors R₁, R₂ and R₃ connected in series with a battery (E), a rheostat (Rh), a plug key (K) and an ammeter (A) using standard circuit symbols. Use this circuit to show that the same current flows through every part of the circuit. List two precautions you would observe while performing the experiment. (Board Term I, 2014)
Answer:

Change the positions of ammeter and note the reading of ammeter each time. You will find that all the reading obtained are same.
So, the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in this circuit (series combination) the current is the same in every part of the circuit.

Precautions:
(i) All the connections are neat and tight.
(ii) Ammeter is connected with the proper polarity, i.e., positive terminal of the ammeter should go to positive terminal and negative terminal of ammeter to the negative terminal of the battery or cell used.

Question 41.
Two wires A and B are of equal length and have equal resistances. If the resistivity of A is more than that of B, which wire is thicker and why ? For the electric circuit given below calculate:

(i) current in each resistor
(ii) total current drawn from the battery, and
(iii) equivalent resistance of the circuit. (Board Term I, 2014)
Answer:
Let l_A, a_A and R_A be the length, area of cross-section and resistance of wire A and l_B, a_B and R_B are that of wire B.
Here, l_A = l_B and R_A = R_B
If ρ_A and ρ_B are the resistivities of wire A and B respectively then
R_A = ρ_A \(\frac{l_{A}}{a_{A}}\) and R_B = ρ_B \(\frac{l_{B}}{a_{B}}\), As R_A = R_B
∴ ρ_A \(\frac{l_{A}}{a_{A}}\), ρ_B \(\frac{l_{B}}{a_{B}}\)
or \(\frac{ρ_{A}}{ρ_{B}}\) = \(\frac{a_{A}}{a_{B}}\)
Since ρ_A > ρ_B therefore a_A > a_B Hence, wire A is thicker than wire B.
For parallel combination,
V₁ = V₂ = V₃ = 6V
(i) Using Ohm’s law
I₁ = V₁/R₁ = 6/30 = 0.2 A
I₂ = V₂/R₂ = 6/10 = 0.6 A
I₃ = V₃/R₃ = 6/5 = 1.2 A

(ii) Total current drawn from battery,
I = I₁ + I₂ + I₃ = 0.2 + 0.6 + 1.2 = 2 A
(iii) Equivalent resistance of the circuit, R_eq can be obtained by Ohm’s law
V= I R_eq
So, 6 V = 2 A × R_eq or, R_eq = \(\frac { 6 }{ 2 }\) = 3 Ω
Aliter, \(\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
\(\frac{1}{30}+\frac{1}{10}+\frac{1}{5}=\frac{1+3+6}{30}=\frac{10}{30}=\frac{1}{3}\)
or R_eq = 3 Ω

Question 42.
(a) Derive an expression to find the equivalent resistance of three resistors connected in series. Also draw the schematic diagram of the circuit.
(b) Find the equivalent resistance of the following circuit.

Answer:
(a) Refer to answer 37.
(b) For the given circuit,
R₁ = 6 Ω, R₂ = 10 Ω, R₃ = 15 Ω.
As \(\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
\(\frac{1}{R_{e q}}=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\)
= \(\frac{5+3+2}{30}=\frac{10}{30}=\frac{1}{3}\)
R_eq = 3 Ω

Question 43.
Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, an ammeter and a plug key, all connected in series. Also connect a voltmeter to record the potential difference across the 15 Ω resistor and calculate
(i) the electric current passing through the above circuit and
(ii) potential difference across 5 Ω resistor when the key is closed. (Board Term 1, 2013)
Answer:

Potential of the battery, V = (2 × 5) V = 10 V
Equivalent resistance,
R_eq = R₁ + R₂ + R₃
= (5 + 10 + 15)Ω = 30 Ω
(i) Current through circuit, I = \(\frac{V}{R}=\frac{10}{30} \mathrm{~A}=\frac{1}{3} \mathrm{~A}\)
(ii) Potential across 5 Ω resistor, V₁ = IR₁
= \(\frac { 1 }{ 3 }\) × 5 = \(\frac { 5 }{ 3 }\) V = 1.67 V

Question 44.
The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become
(a) two times
(b) half
(c) one-fourth
(d) four times (2020)
Answer:
(a) : We know, H = I²Rt = \(\frac { V^2 }{ 4 }\). t
Now when, R’ = \(\frac { R }{ 24 }\), V’ = V and t’ = t
H’ = \(\frac{V^{\prime 2} t^{\prime}}{R^{\prime}}=\frac{V^{2} t}{R / 2}=\frac{2 V^{2} t}{R}\) = 2H

Question 45.
(a) Write the mathematical expression for Joules law of heating.
(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V. (2020)
Answer:
(a) The Joule’s law of healing implies that heat produced in a resistor is
(i) directly proportional to the square of current lor a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor.
i.e., H = I² Rt
(b) Given, charge q = 96000 C, time t = 2 h = 7200 s and potential difference V = 40 V
We know, H = I²Rt = \(\frac{Q^{2}}{t^{2}} \times \frac{V}{Q}\) × t × t = VQ
= 40 × 96000 = 3.84 × 10⁶ J = 3.84 MJ

Question 46.
Write Joules law of heating. (1/3, 2018)
Answer:
Refer to answer 45(a).

Question 47.
Explain the use of an electric fuse. What type of material is used for fuse wire and why? (Board Term I, 2016)
Answer:
Electric fuse protects circuits and appliances by stopping the flow of any unduly high electric current. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

Question 48.
(a) Why is tungsten used for making bulb filaments of incandescent lamps?
(b) Name any two electric devices based on heating effect of electric current. (2/5, Board Term I, 2015)
Answer:
(a) (i) Tungsten is a strong metal and has high melting point (3380°C).
(ii) It emits light at high temperatures (about 2500°C).
(b) Electric laundry iron and electric heater are based on heating effect of electric current.

Question 49.
A fuse wire melts at 5 A. If it is desired that the fuse wire of same material melt at 10 A, then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reasons for your answer. (3/5, Board Term I, 2014)
Answer:
Let the resistance of the wire be R, heat produced in the fuse at 5 A in Is is
H=(5)²R ( H – I²Rt)
50. fuse melts at (5)²R joules of heat.
Let, the resistance of new wire is R’
So, heat produced in 1 second = (10)²R’
To prevent it from melting
(5)²R = (10)²R’ or R’ = \(\frac { R }{ 4 }\)
As R ∝ \(\frac { 1 }{ A }\)
∴ cross-sectional area of new fuse wire is four times the first fuse.
Now, A = πr², so new radius is twice the previous one. So, at 10 A, the new fuse wire of same material and length has larger radius than the earlier one.

Question 50.
What is heating effect of current? List two electrical appliances which work on this effect. (2/5, Board Term I, 2013)
Answer:
If only resislors are connected to the battery, the source energy continually gets dissipated entirely in the form of heal. This is known as healing effect of current, ’file amount of heat (77) produced in time t is given by Joule’s law of heating.
H = I²Rt
Where, 7 is current flowing through resistor R.
The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some common devices based on heating effect of current.

Question 51.
Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be
(a) 0.4 A
(b) 0.6 A
(c) 0.8 A
(d) 1A (2020)
Answer:
(d) : Given power of first bulb, P₁ = 100 W and second bulb P₂ = 40 W
Current through 100 W bulb, I₁ = 1 A
Current through 40 W bulb, I₂ = ?
Since both the bulbs are connected in series, the electric current passing through both the bulbs are same i.e., I₂ = 1 A.

Question 52.
Write the relation between resistance (R) of filament of a bulb, its power (P) and a constant voltage V applied across it. (Board Term I, 2017)
Answer:
P = \(\frac { V^2}{ R }\)

Question 53.
Power of a lamp is 60 W. Find the energy in joules consumed by it in Is. (Board Term I, 2016)
Answer:
Here, power of lamp, P = 60 W time,
t = 1 s
So, energy consumed = Power × time = (60 × 1)J = 60 J

Question 54.
Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V. (2/3, 2018, Board Term I, 2014)
Answer:
Since both the bulbs are connected in parallel and to a 220 V supply, the voltage across each bulb is 220 V. Then
Current drawn by 100 W bulb,
I₁ = \(\frac { power rating}{ voltage applied }\) = \(\frac { 100 W}{ 220 V }\) = 0.454 A
Current drawn by 60 W bulb,
I₂ = \(\frac { 60 W}{ 220 V }\) = 0.273 A
Total current drawn from the supply line,
I = I₁ + I₂ = 0.454 A + 0.273 A = 0.727 A = 0.73 A

Question 55.
How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts. (Board Term I, 2016)
Answer:
Here, V = 220 V, R = 55 Ω
By Ohm’s law V = IR
∴ 220 = 7 × 55 or I = 4A
Wattage of electric iron = Power
= \(\frac{V^{2}}{R}=\frac{(220)^{2}}{55}\) = 880 W

Question 56.
An electric iron has a rating of 750 W; 200 V. Calculate:
(i) the current required.
(ii) the resistance of its heating element.
(iii) energy consumed by the iron in 2 hours. [Board Term 1, 2015]
Answer:
Here, P = 750 W, V = 200 V
(i) As P = V7
I = P/V= (750/200) A = 3.75A
(ii) By Ohm’s law V = IR or R = V/I
∴ R = \(\frac { 200}{ 3.75 }\) Ω = 53.3 Ω
(iii) Energy consumed by the iron in 2 hours
= P × t = 750 W × 2h = 1.5 kWh
or E = (750 × 2 × 3600) J = 5.4 × 10⁶ J

Question 57.
An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb. (1/3, Board Term I, 2015)
Answer:
Here, V= 220 V,/= 2.5 A
Power of the bulb P = VI = 220 × 2.5 W = 550 W

Question 58.
(a) Define power and state its SI unit.
(b) A torch bulb is rated 5 V and 500 mA. Calculate its
(i) power
(ii) resistance
(iii) energy consumed when it is lighted for 2 \(\frac { 1 }{ 2 }\) hours.
Answer:
(a) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.
P = VI = I²R = V²/R
The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of IV.
1 W = 1 volt × 1 ampere = 1 V A

(b) Given, V = 5 V and I = 500 mA = 0.5 A
(i) Power, P = V × 7 = 5 × 0.5 = 2.5 W
(ii) As, P = \(\frac{V^{2}}{R} \Rightarrow R=\frac{V^{2}}{P}=\frac{25}{2.5}\) = 10 Ω
(iii) Given, time t = 2.5 hrs = 9000 s
∴ The energy consumed, E = P × t
= 2.5 × 9000 = 2.25 × 10⁴ J
= 6.25 Watt hour

Question 59.
Two identical resistors, each of resistance 15 Ω, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case. (2020)
Answer:
Given, R₁ = R₁ = 15 Ω, V = 6 V
(i) When connected in series,
R_s = R₁ + R₂ = 15 Ω + 15 Ω = 30 Ω
Power, P_S = \(\frac{V^{2}}{R_{S}}=\frac{36}{30}\) W
(ii) When connected in parallel,

Question 60.
An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω. are connected to a 6 V battery as shown in the circuit. Calculate.

(a) the total resistance of the circuit
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp. (Delhi 2019)
Answer:
Resistance of the lamp = 20 Ω
External resistance = 4 Ω
(a) As both the lamp and external resistance are connected in series, therefore the total resistance,
R = 20 + 4 = 24 Ω
(b) Current, I = \(\frac { V}{ R }\) = \(\frac { 6 }{ 24 }\) = 0.25 A
(c) (i) Potential difference across the electric lamp
\(\frac { Total voltage }{ Total resistance }\) × resistance of lamp
= \(\frac { 6}{ 24 }\) × 20 = 5 V

(ii) Potential difference across conductor
\(\frac { Total voltage }{ Total resistance }\) × resistance of conductor
= \(\frac { 6}{ 24 }\) × 4 = 1 V

(d) Power of the lamp
= (current)² × resistance of lamp
= (0.25)² × 20 = 1.25 W

Question 61.
Compare the power used in 2 Ω. resistor in each of the following circuits. (AI 2019)

Answer:
In circuit A,
Total resistance, R = l + 2 = 3Ω
Voltage across 2 Ω = \(\frac { V_{Total}}{ R_{Total} }\) × 2 Ω = \(\frac { 6}{ 3 }\) × 2 = 4 V
∴ Power used in 2 Ω resistor,
p = \(\frac { V^2}{ R }\) = \(\frac { (4)^2}{2 }\) = 8 w
In circuit B, Voltage across both the resistance is same i.e. 4 V and both are connected in parallel combination.
∴ Power used in 2 Ω resistor = \(\frac { V^2}{ R }\) = \(\frac { (4)^2}{2 }\) = 8 w
∴ Power used in 2 Ω resistor in each case is same i.e. 8 W.

Question 62.
A bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. (AI 2019)
Answer:
In first case, P = 40 W, V = 220 V
Current drawn l = \(\frac { P}{ V }\) = \(\frac { 40}{ 220 }\) = 0.18 A
Also, resistance of bulb,
R = \(\frac{V^{2}}{P}=\frac{(220)^{2}}{40}\) = 1210 Ω
In second case, P = 25 W, V = 220 V
Current drawn, I = \(\frac { P}{ V }\) = \(\frac { 25}{ 220 }\) = 0.11 A
Also, resistance of the bulb,
R = \(\frac { V^2}{ P }\) = \(\frac { (220)^2}{ 25 }\) = 1936 Ω
Hence, by replacing 40 W bulb to 25 W bulb, having same source of voltage the amount of current flows decreases while resistance increases.

Question 63.
(a) How two resistors, with resistances R₁ Ω and R₁ Ω respectively are to be connected to a battery of emf V volts so that the electrical power consumed is minimum?
(b) In a house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 3.60 per kWh. (Board Term I, 2017)
Answer:
(a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series.
(b) Power of each bulb P₁ = 100 watt
Total power of 3 bulbs, P₁ = 3 × 100 = 300 watt
Energy consumed by bulbs in 1 day
E₁ = P₁ × t = 300 watt × 5 hours.
= 1500 Wh = 1.5 kWh
Power of each fan = 50 watt
Total power of 2 fans = 2 × 50 watt
P₂ = 100 watt
Energy consumed by fans in 1 day
E₂ = P₂ × t = 100 watt × 10 hours
= 1000 watt hour = 1 kWh
Energy consumed by heater,
E₃ = 1 kW × 1/2 h = 0.5 kWh
Total energy consumed in one day
E = E₁ + E₂ + E₃ = (1.5 + 1 + 0.5) kWh = 3 kWh
Total energy consumed in a month of 31 days
= E × 31 = (3 × 31) kWh = 93 kWh
Cost of energy consumed = Rs (93 × 3.60) = Rs 334.80

Question 64.
(a) An electric bulb is connected to a 220 V generator. If the current drawn by the bulb is 0.50 A, find its power.
(b) An electric refrigerator rated 400 W operates 8 hours a day. Calculate the energy per day in kWh.
(c) State the difference between kilowatt and kilowatt hour. (3/5, Board Term I, 2013)
Answer:
(a) Here, V = 220 V, I = 0.50 A
Power of the bulb, P = VI = (220 × 0.5)W = 110 W
(b) Energy consumed by electric refrigerator in a day = Power x time
= 400 W × 8 h = 3200 Wh = 3.2 kWh
(c) Kilowatt is unit of power and kilowatt hour is a unit of energy.

Question 65.
(i) State one difference between kilowatt and kilowatt hour. Express 1 kWh in joules.
(ii) A bulb is rated 5V; 500 mA. Calculate the rated power and resistance of the bulb when it glows. (Board Term I, 2013)
Answer:
(i) Refer to answer 64(c).
1 kWh = 1000 W × 1 h
= 1000 W × 3600 s = 3600000 J = 3.6 × 10⁶ J

(ii) Here, V = 5 V, I = 500 mA = 0.5 A
Power rating of bulb is
P = VI = ( 5 × 0.5)W = 2.5W
Resistance of the bulb is R = V/I = (5/0.5) Ω = 10 Ω

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Class 12 Biology Chapter 1 Important Questions Reproduction in Organisms

Question 1.
Identify the picture and mention the vegetative part that helps it to propagate. (All India 2015C)

Answer:
The given figure shows the rhizome of a ginger. Rhizome is an underground stem that takes part in vegetative propagation.

Question 2.
Name the vegetative propagules in the following: (All India 2014C)
(i) Agave
(ii) Bryophyllum
Answer:
The vegetative propagules in given plants are as follows
(i) In Agave – Bulbil
(ii) In Bryophyllum – Leaf buds/Adventitious buds

Question 3.
Write the name of the organism that is referred to as ‘Terror of Bengal’. (Delhi 2014)
Answer:
Water hyacinth is referred to as the ‘Terror of Bengal’.

Question 4.
Give one example each of a fungus which reproduces by (Delhi 2014C)
(i) budding
(ii) conidia
Answer:
Fungus that reproduces by
(i) budding – Yeast
(ii) conidia – Penicillium

Question 5.
Give one example of a plant that reproduces by (Delhi 2014C)
(i) runner
(ii) offset
Answer:
Below are is the example of a plant that reproduces by
(i) runner – Oxalis
(ii) offset -Pistia

Question 6.
Which of the following statements is true of Hydra? (All India 2013C)
(i) It produces asexual gemmules.
(ii) It produces unicellular buds.
(iii) It produces multicellular buds.
Answer:
(iii) Hydra produces multicellular buds.

Question 7.
Which one of the following statements is true of ginger? (All India 2013C)
(i) Germinating bud appears from the eye of the stem tuber.
(ii) Germinating bud appears from the node of rhizome.
(iii) Germinating bud appears from the notch of the leaf margin.
Answer:
(ii) The germinating bud appears from the node of rhizome in ginger.

Question 8.
Name an organism, where cell division is itself a mode of reproduction. (All India 2013; Foreign 2010)
Answer:
In unicellular organisms like Amoeba, bacteria, etc., cell division itself is a mode of reproduction.

Question 9.
How does Penicillium reproduce asexually? (Delhi 2013)
Answer:
Penicillium reproduces asexually by conidia formation, which are non-motile spores produced singly or in chains by the constriction of tip of hyphal branches. These spores germinate by giving out germ tubes.

Question 10.
Name an alga that reproduces asexually through zoospores. Why are these reproductive units so called? (All India 2013)
Answer:
Chlamydomonas is an alga that reproduces asexually through zoospores. Due to mobility (motile) like animals, these reproductive units are referred to as zoospores.

Question 11.
Which one of the following statements is true for yeast? (Delhi 2013C)
(i) The cell divides by binary fission.
One of them develops into a bud.
(ii) The cell divides unequally. The smaller cell develops into a bud.
(iii) The cell produces conidia, which develop into a bud.
Answer:
Statement (ii) is true for yeast. The cell divides unequally. The smaller cell develops into a bud.

Question 12.
Which one of the following organisms exhibit binary fission?
Bacillus, Penicillium, Yeast, Amoeba (Delhi 2012C)
Answer:
Bacillus and Amoeba reproduce by binary fission. In this mode, the body of an individual divides into two equal halves.

Question 13.
Offsprings produced by asexual reproduction are called clones. Justify giving two reasons. (All India 2010)
Answer:
Offsprings produced by asexual reproduction are called clones, because

• they are morphologically similar to their parent.
• they have same genetic composition as their parent.

Question 14.
Identify this reproductive structure and name the organism they are being released from. (Delhi 2010)

Answer:
The reproductive structure in the figure are zoospores. These are motile, microscopic and thin-walled structures released by Chlamydomonas for asexual reproduction.

Question 15.
Name the organism and the mode of reproduction represented in the diagram given below (All India 2010)

Answer:
Organism Yeast
Mode of reproduction Asexual by budding.

Question 16.
Banana crop is produced by farmers without sowing of seeds. Explain how the plant is propagated? (All India 2014C)
Answer:
Banana crop is cultivated by farmer’s through vegetative propagation.
For this purpose, a rhizome of banana is taken instead of a seed. Each piece from a rhizome is able to give rise to a new plant.

Question 17.
Unicellular organisms are immortal, whereas multicellular organisms are not. Justify. (Delhi 2011C)
Answer:
Unicellular organisms are considered immortal mainly because, in them the parent body as a whole constitutes the reproductive unit and after reproduction it continues to live as daughter cells. The multicellular organisms produce their reproductive structures in specialised organs and their whole body dies due to ageing and senescence.

Question 18.
(i) Name the organism that reproduce
through the following reproductive structures
(a) Conidia
(b) Zoospores
(ii) Mention similarity and one difference between these two reproductive units, (Delhi 2011C)
Answer:
(i) (a) Conidia – Penicillium
(b) Zoospore – Chlamydomonas

(ii) Similarity between conidia and zoospore is that the both are asexual reproductive structures. The difference between conidia and zoospore is that, the conidia are non-motile while zoospores are motile structures.

Question 19.
At what stage does the meiosis occur in an organism exhibiting haplontic life cycle and mention the fate of the products thus produced. (Delhi 2019)
Answer:
In organisms with haplontic life cycle, meiosis occur in the diploid cell, i.e., zygote which divides to form haploid spores. These spores grow into haploid individuals.

Question 20.
Write the number of chromosomes body cells of honeybee workers and drone have. (Delhi 2019)
Or
A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason. (Delhi 2016)
Answer:
In case of honeybees, the male (drones) have 16 chromosomes and females (queen and workers) have 32 chromosomes, because of incomplete parthenogenesis. Fertilised diploid eggs give rise to females by sexual reproduction and unfertilised haploid eggs develop into drones (males) by parthenogenesis.

Question 21.
Explain the significance of meiocytes in a diploid organism. (Delhi 2016)
Answer:
Most of the animals including human beings have diploid parental body and their gametes are formed by meiosis. In such organisms, specialised cells called meiocytes are present, which take part in the production of gametes.

At the time of gamete formation, meiocytes undergo meiotic division. As a result, the number of chromosomes in the daughter cells (i.e. in the gametes) get reduced to half and thus from diploid meiocytes haploid gametes are formed.

Question 22.
Explain the importance of syngamy and meiosis in a sexual life cycle of an organism. (Delhi 2016)
Answer:
Syngamy It helps to restore the diploid condition in sexually reproducing organism by the formation of diploid zygote and thus ensuring the continuity of species.
Meiosis It helps in producing haploid gametes in sexually reproducing organisms.

Question 23.
A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain. (All India 2015)
Answer:
A pea plant is monoecious or bisexual, hence even a single plant can produce seeds while papaya is dioecious, hence an individual plant alone cannot produce viable seeds.

Question 24.
How many chromosomes do drones of honeybees possess? Name the type of cell division involved in the production of sperms by them. (All India 2015)
Answer:
Drones of honeybees possess 16 chromosomes (haploid). They produce sperms by a mitosis¬like division, rather than meiosis.

Question 25.
Meiosis is an essential event in the sexual life cycle of any organism. Give two reasons. (Foreign 2015)
Answer:
Meiosis is an essential event in the sexual life cycle of any organism because

• It is a source of genetic variability.
• It helps to conserve the chromosome number of species from generation to generation.

Question 26.
Give one example of an animal which exhibits oestrus cycle. (Delhi 2014C)
Answer:
Example of an animal that exhibits oestrus cycle is cow.

Question 27.
Write the two pre-fertilisation events from the list given below Syngamy, Gametogenesis, Embryogenesis, Pollination (All India 2014C)
Answer:
Gametogenesis and pollination are pre-fertilisation events.

Question 28.
In which two of the following organisms is the fertilisation external?
Bony fishes, Ferns, Frogs, Birds (Delhi 2014C)
Answer:
Bony fishes and frogs exhibits external fertilisation.

Question 29.
Differentiate between oviparous and viviparous animals. (Delhi 2014C)
Answer:
In oviparous animals, embryo develops outside the body of female and they lay eggs, e.g. reptiles. On the contrary in viviparous animals, embryo develops inside the body of female. Thus, they give birth to babies, e.g. humans.

Question 30.
Give the name of the common phenomenon with reference to reproduction in rotifers, honeybees and turkeys. (Delhi 2013C)
Answer:
Parthenogenesis is the common phenomenon with reference to reproduction in rotifers, honeybees and turkeys.

Question 31.
Which of the following statements is true of date palm?
(i) It is monoecious producing both staminate flowers and pistillate flowers on the same plant.
(ii) It is monoecious producing staminate flowers on one tree and pistillate flowers on another tree.
(iii) It is dioecious producing staminate flowers on one tree and pistillate on another tree. (All India 2013C)
Answer:
(iii) Date palm is a dioecious plant which produces staminate flowers on one tree and pistillate flowers on another tree.

Question 32.
Cucurbits and papaya plants bear staminate and pistillate flowers.
Mention the categories they are put under separately on the basis of the type of flowers they bear. (Delhi 2012)
Or
How are Cucurbita plants different from papaya plant with reference to flowers they bear? (All India 2011C)
Answer:
Papaya is dioecious because the staminate and pistillate flowers are borne on two different plants, while Cucurbita is monoecious because it bears both staminate and pistillate flowers on the same plant.

Question 33.
Mention the unique flowering phenomenon exhibited by Strobilanthes kunthiana (Neelakurinji). (Delhi 2012)
Answer:
In the shrub plant, Strobiknthes kunthiana flowers bloom only once in twelve years.

Question 34.
Name the mode of reproduction that ensures the creation of new variants. (Delhi 2012C)
Answer:
The process of sexual reproduction ensures the creation of new variants. It involves the fusion of gametes from different parents (male and female) resulting into genetic recombination that leads to variations.

Question 35.
Name the type of cell division that takes place in the zygote of an organism exhibiting haplontic life cycle. (Delhi 2011)
Answer:
Meiosis is the type of cell division that takes place in the zygote of an organism exhibiting haplontic life cycle.

Question 36.
Name the phenomenon and the cell responsible for the development of a new individual without fertilisation as seen in honeybees. (Foreign 2011)
Answer:
The phenomenon of the development of a new individual without fertilisation is called parthenogenesis. In honeybees, egg cell undergoes parthenogenesis to form haploid drones (n).

Question 37.
Mention the site where syngamy occurs in amphibians and reptiles, respectively. (Delhi 2010)
Answer:
In amphibians, syngamy takes place outside the body of organism in water while in reptiles, it takes place inside the body of female organism.

Question 38.
A moss plant produces a large number of antherozoids but relatively only a few egg cells. Why? (Delhi 2010)
Answer:
The antherozoids or male gametes in moss plants are motile and depend on water for their transport towards female gametes (non-motile). During gamete transfer, large number of male gametes are lost. Thus, to ensure fertilisation, large number of male gametes are released in water so as to reach the non-motile female gametes.

Question 39.
Why are papaya and date palm plants said to be dioecious, whereas cucurbits and coconut palms monoecious, inspite of all of them bearing unisexual flowers? (Foreign 2010)
Answer:
Papaya and date palm plants are said to be dioecious because male and female flowers are borne on separate plants, whereas cucurbits and coconut palms are monoecious because male and female flowers are borne on the same plant.

Question 40.
(i) Why do organisms like algae and
fungi shift from asexual mode of reproduction to sexual mode ?
(ii) What is a juvenile phase in organisms ? (2018C)
Or
Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions? (Delhi 2014)
Answer:
(i) Organisms such as fungi and algae switch to sexual mode of reproduction during adverse conditions because asexual reproduction produces a large population that may not survive due to lack of resources. Sexual reproduction brings variations among the individuals, that might help the individuals to better adapt to the changed conditions and to survive. This ensures the continuity of species.

(ii) The period during which an organism grows to attain sexual maturity is called juvenile phase in animals. In plants, it is known as vegetative phase.

Question 41.
Why do moss plants produce very large number of male gametes? Provide one reason. What are these gametes called? (All India 2015)
Answer:
Refer to Answer No. 20.

Question 42.
Coconut palm is monoecious, while date palm is dioecious. Why are they so called? (Delhi 2014)
Answer:
Refer to Answer No. 21.

Question 43.
The cell division involved in gamete formation is not of the same type in different organisms. Justify. (Delhi 2011)
Answer:
The gametes involved in sexual reproduction are usually haploid. Thus, in diploid organism cell division involved in gamete formation is meiosis, as it leads to the production of haploid gametes. However, in case of haploid organisms, gametes are formed due to mitotic cell division.

Question 44.
A list of three flowering plants is given below Date palm, cucurbits and pea.
Which ones out of them are
(i) monoecious
(ii) bearing pistillate flowers? (Foreign 2011)
Answer:
(i) Cucurbits and pea plants are monoecious bearing both staminate and pistillate flowers on the same plant.
(ii) Date palm, cucurbits and pea, all three bear pistillate flowers.

Question 45.
Name any two organisms and the phenomenon involved where the female gamete undergoes development to form new organisms without fertilisation? (Foreign 2011)
Answer:
Parthenogenesis helps in the development of new organisms without fertilisation, e.g. lizards, honeybees, etc. Refer to Answer. No. 2.

Question 46.
Differentiate between an annual and biennial plants. Provide one example of each. (Delhi 2017)
Answer:
Differences between annual and biennial plants are as follows

Question 47.
(i) List the three states the annual and biennial angiosperms have to pass through during their life cycle.
(ii) List and describe any two vegetative propagules in flowering plants. (Delhi 2017)
Answer:
(i) The three states through which the annual and biennial angiosperms pass during their life cycle are

• Vegetative or Juvenile phase
• Reproductive phase
• Senescent phase

(ii) Vegetative propagules are the parts/units of a plant which can be used for vegetative propagation, e.g. roots, stems, leaves, etc.
Root propagules are used for the production of a new plant via roots, e.g. fleshy roots in case of sweet potato, tapioca and Dahlia.
Vegetative propagation through leaves Many plant leaves have adventitious buds which help in the development of a new plant, e.g. Begonia, Bryophyllum, etc.

Question 48.
Vijay’s mother was preparing batter for dosa and added a small packet of dried yeast to the mixture. She left it for sometime and Vijay noticed that the level of batter has risen in the bowl. He asked his mother about his observation.
(i) Why was yeast added to the dosa batter?
(ii) Name and explain the process by which yeast multiplies.
Answer:
(i) Yeast was added to dosa batter to make it light and fluffy. Yeast divides very quickly and releases carbon dioxide which raises the batter in quick time.
(ii) Yeast reproduces by budding which is a type of asexual reproduction during budding unequal and small projections called buds are produced. These buds initially remain attached to parent body but later they detach to form new individuals.

Question 49.
Arjun saw an article on ‘Reproductive behaviours of various mammals in a magazine. For example, how some animals can reproduce throughout their life, while others cannot. He took the article to his teacher for further clarification.
(i) What is meant by term continuous breeders and seasonal breeders?
(ii) In non-primates, what is the phase of morphological and physiological changes known as?
(iii) List the values observed in Arjun.
Answer:
(i) Animals that remain reproductively active throughout their reproductive cycle are called continuous breeders, e.g. humans. Animals which attain reproductive activity only during favourable seasons are called seasonal breeders, e.g. dogs, birds, etc.
(ii) Oestrus cycle.
(iii) Arjun is intelligent, observant and inquisitive.

The post Reproduction in Organisms Class 12 Important Questions and Answers Biology Chapter 1 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 2 Sexual Reproduction in Flowering Plants to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 2 Important Questions, Sexual Reproduction in Flowering Plants Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 2 Important Questions Sexual Reproduction in Flowering Plants

Question 1.
Angiosperms bearing unisexual flowers are said to be either monoecious or dioecious. Explain with the help of one example each. (All India 2016)
Answer:
Plant bearing flowers of both sexes, i.e. staminate and pistillate flowers called monoecious, e.g. lea mays (maize).
When both sexes, i.e. staminate and pistillate flowers, are present on different plants; these plants are called dioecious, e.g. Carica papaya (Papaya).

Question 2.
These pictures show the gynoecium of (A) Papaver and (B) Michelia flowers. Write the difference in the structure of their ovaries. (Delhi 2015C)

Answer:
The gynoecium of Papaver is multicarpellary and syncarpous (pistils are fused together), the ovary can be unilocular to multilocular. The gynoecium of Michelia is multicarpellary and apocarpus (pistils are free) and the ovary is always unilocular.

Question 3.
Name the parts of the flower which the tassels of corn cob represent. (All India 2014)
Answer:
The parts of the flower that represent the tassels of corn cob are stigma and style which wave in the wind to trap pollen grains flowing with the wind.

Question 4.
Draw a diagram of a mature microspore of an angiosperm. Label its cellular components only. (Foreign 2014)
Or
Draw a labelled diagram of at mature pollen grain. (Delhi 2013C)
Answer:
The labelled diagram of a mature microspore of an angiosperm with its cellular components is given below

Question 5.
State the function of filiform apparatus found in mature embryo sac of an angiosperm. (Foreign 2014)
Answer:
The special cellular thickenings present in synergids at the micropylar tip called filiform apparatus, found in mature embryo sac of an angiosperm help in guiding the entry of pollen tubes up to the synergids.

Question 6.
Give an example of a plant which came into India as a contaminant and is a cause of pollen allergy. (All India 2014)
Answer:
Parthenium or Carrot grass is a major contaminant which came to India and caused pollen allergy.

Question 7.
A bilobed, dithecous anther has 100 microspore mother cells per microsporangium. How many male gametophytes this anther can produce? (Delhi 2010)
Answer:
An anther is a four-sided (tetragonal) structure consisting of four microsporangia.
Each microsporangium has 100 microspore
mother cells, so total number of microspore
mother cells in anther = 4 × 100 = 400
microspore mother cells. Meiosis in each microspore mother cell produces 4 male gametes, so 400 cells will produce = 4 × 400 = 1600 male gametes.

Question 8.
An anther with malfunctioning tapetum often fails to produce viable male gametophytes. Give one reason. (Delhi 2010)
Or
Write the function of tapetum in anthers. (Delhi 2012)
Answer:
The anther with malfunctioning tapetum cannot provide complete nutrition to the developing microspores or male gametophytes. So, it fails to produce viable male gametophyte.

Question 9.
In the TS of a mature anther given below, identify ‘a’ and ‘b’ and mention their functions. (All India 2019)

Answer:
In the given figure, a is sporogenous tissue and b is tapetum.

• Sporogenous tissue has cell which are potential Pollen Mother Cell (PMC) or microspore mother cell and give rise to microspore tetrad after meiotic cell division.
• Tapetum nourishes the developing microspores or pollen grains.

Question 10.
A pollen grain in angiosperm at the time of dehiscence from an anther could be 2-celled or 3-celled. Explain, how are the cells placed within the pollen grain when shed at a 2-celled stage? (All India 2017)
Answer:
A pollen grain is partly germinated microspore representing the male gametophyte. It divides by unequal mitotic division and forms two cells. Thus, each mature pollen grain in angiosperms have a generative cell and a vegetative cell.

In about 60% of angiosperms, pollen grains are shed at this 2-celled stage. However, in about 40% flowering plants, the generative cell may further divide mitotically to give rise to two male gametes and pollen grains are shed at this 3-celled stage.
The placement of cells within the pollen grain when shed at 2-celled stage can be visualised as shown below

Question 11.
A mature embryo sac in a flowering plant may possess 7-cells, but 8-nuclei. Explain with the help of a diagram only. (Delhi 2017)
Answer:
A typical angiospermic embryo sac is 8-nucleated and 7-celled.

Question 12.
In a flowering plant, a microspore mother cell produces four male gametophytes while a megaspore mother cell forms only one female gametophyte. Explain. (Delhi 2017)
Answer:
In flowering plants, microspore mother cells are found embedded in the spOrophytic tissue of anther. These cells undergo meiosis and give rise to four microspores that remain together in a microspore tetrad. After attaining maturity, these microspores separate from each other and each microspore develops into a male gametophyte or pollen grain.

On the other hand, megaspore mother cell develops in the ovary of a flower and divides by meiotic division to produce four megaspores. From these, three degenerate while, the one undergoes further development and mitotic divisions to produce female gametophyte. Thus, in a flowering plant, a microspore mother cell produces four male gametophytes while, megaspore mother cell produces one female gametophyte.

Question 13.
Gynoecium of a flower may be apocarpous or syncarpous. Explain with the help of an example each. (Delhi 2016)
Answer:
Gynoecium of a flower is called as apocarpous when the carpels are free, e.g. apocarpus in Ranunculus. Whereas it is called syncarpous when the carpels are fused, e.g. syncarpous in Petunia.

Question 14.
Draw a diagram of a section of a megasporangium of an angiosperm and label funiculus, micropyle, embryo sac and nucellus. (All India 2016)
Answer:
Diagrammatic view of a megasporangium and pollen grains are shed at this 3-celled stage. (anatropous ovule)

Question 15.
Differentiate between the two cells enclosed in a mature male gametophyte of an angiosperm. (All India 2013)
Answer:
Haploid pollen grains represent the male gametophyte. It contains two cells, i.e. vegetative cell and generative cell.
The vegetative or tube cell is larger in size as compared to generative cell and have a vacuolated cytoplasm. The generative cell on the other hand have thin dense cytoplasm with prominent nuclei that give rise to two male gametes, while vegetative cell does not.

Question 16.
Name all the haploid cells present in an unfertilised mature embryo sac of a flowering plant. Write the total number of cells in it. (All India 2013)
Or
How many haploid cells are present in mature female gametophyte of a flowering plant? Name them. (Delhi 2013C)
Answer:
An unfertilised embryo sac of angiosperm is composed of 7 cells, i.e. 7-celled and 8-nucleated. Among 8-nuclei, 6 are enclosed by cell walls and organised into cells, which are haploid in number (3 antipodals, 2 synergids and 1 egg cell) and a large central cell with 2 polar nuclei.

Question 17.
Draw a labelled schematic diagram of the transverse section of a mature anther of an angiosperm plant. (Delhi 2013)
Answer:

Question 18.
Explain the function of germ pores. (All India 2012 )
Answer:
Germ pores are prominent apertures of pollen grain exine where sporopollenin is absent. These are the regions where intine comes out forming a pollen tube to release male gamete in the embryo sac.

Question 19.
Identify and label the parts in the given anatropous ovule. (All India 2010C)

Answer:
A – Micropyle B – Outer integument
C – Inner integument D – Embryo sac

Question 20.
Pollen banks are playing a very important role in promoting plant breeding programme the world over.
How are pollens preserved in the pollen banks? Explain. How are such banks benefitting our farmers? Write any two ways. (Delhi 2019)
Answer:
Pollen grains are stored for years in liquid nitrogen (-196°C) in pollen banks for later use in plant breeding programmes. Plant breeding is a technique of manipulation of plant species in order to create desired plant types that are better suited for cultivation, give better yield and are disease resistant.
The objectives of such pollen banks include incorporation of certain traits or characters into crop plants in order to enhance the food production such as
(i) Increased tolerance to environmental stresses such as salinity, extreme temperature, drought, etc.
(ii) Resistance to pathogens like viruses, fungi and bacteria.

Question 21.
(i) Name the organic material exine of the pollen grain is made up of. How is this material advantageous to pollen grain?
(ii) Still it is observed that it does not form a continuous layer around the pollen grain. Give reason.
(iii) How are ‘pollen banks’ useful? (All India 2016C)
Answer:
(i) The organic material of exine of pollen grain is sporopollenin. This is most resistant biological material known so far. It protects pollen grains from damages.
(ii) Exine on pbllen grain is not a continuous layer. It is thin at places and pollen tube germinates by growth of intine through these thin parts of exine called germ pores.
(iii) Pollen banks are used to store pollen grains for short as well as long period of time in viable conditions.

Question 22.
Draw a labelled diagram of a section of an enlarged view of microsporangium of an angiosperm. (All India 2016C)
Or
Why are angiosperm anthers called dithecous? Describe the structure of its microsporangium. (Delhi 2014)
Or
Describe the structure of a mature microsporangium of an angiosperm. (Delhi 2014)
Answer:
The structure of a mature microsporangium can be described with the help of given diagram.

Since, the angiosperm’s anthers are bilobed, each lobe bearing two thecae, they are referred to as dithecous. Microsporangium appears circular in outline and is usually surrounded by four wall layers. The outer three layers epidermis, endothecium and middle layers are protective in function. They also help in dispersal of pollens by dehiscing themselves. While, the innermost layer tapetum is nutritive in function and nourishes the developing pollen grdins. The centre of the microsporangium comprises of compact
sporogenous tissue. The cells of this sporogenous tissue undergo meiotic divisions to form microspore tetrads, that further develop to form pollen grains.

Question 23.
Draw a labelled diagram of a typical anatropous ovule. (Delhi 2014)
Answer:
For structure or labelled diagram of anatropous ovule,
Refer to Answer No. 14.

Question 24.
Explain the process of microsporogenesis in angiosperms. (Delhi 2013C)
Answer:
The formation of microspores from a pollen mother cell through meiosis is called microsporogenesis.

Microspores are arranged as tetrad. As the anthers mature and dehydrate they dissociate from each other and develop into mature pollen grains. Pollen grains or the male gametophytes are released by dehiscence of anther.

Question 25.
Draw a diagram of a mature embryo sac of an angiosperm and label the following parts in it. (Delhi 2013)
(i) Filiform apparatus
(ii) Synergids
(iii) Central cells
(iv) Egg cell
(v) Polar nuclei
(vi) Antipodals
Answer:
For mature embryo sac of angiosperm, Refer to figure in Answer No. 11.

Question 26.
Explain the process of megasporogenesis in angiosperms. (Delhi 2013C)
Or
Trace the development of megaspore mother cell up to the formation of a mature embryo sac in a flowering plant. (Delhi 2012)
Answer:
The process of formation of megaspores from the megaspore mother cell (2n) by the meiosis division in the ovule is called megasporogenesis. Refer to Answer No. 28.

Question 27.
Describe the structure of a 3-celled pollen grain of an angiosperm. (Delhi 2012C)
Answer:
The 3-celled pollen grain structure in an angiosperm consists of two male gametes and one vegetative cell. The vegetative cell is bigger, has abundant food reserve and a large irregularly-shaped nucleus. The generative cell is small and floats in the cytoplasm of the vegetative cell. It is spindle-shaped with dense cytoplasm and a nucleus. In over 60% of angiosperms, pollen grains are shed at this 2-celled stage. In most of the angiospermic species, the generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage).

Question 28.
Describe the process of megasporogenesis up to fully developed embryo sac formation in an angiosperm. (All India 2019)
Or
Where does the process of megasporogenesis start in an angiosperm? Describe the process up to the formation of embryo sac. (Delhi 2019)
Answer:
In angiosperms, the process of megasporogenesis starts inside the nucellus of the ovule. During megasporogenesis, the Megaspore Mother Cell (MMC) undergoes meiosis resulting in the production of four megaspores. Out of the four megaspores, only one is functional while the other three degenerate.

The functional megaspore undergoes mitosis to form two nuclei, which migrate to opposite poles, forming a 2-nucleate embryo sac.

Further, mitotic divisions lead to the formation of 4-nucleate and 8-nucleate stages of the embryo sac. In these mitotic divisions, nuclear division is not followed by cell division. After the 8-nucleate stage, cell walls are laid down and a typical female gametophyte or embryo sac is formed.
Among the 8 nuclei, 6 are enclosed by cell walls and organised into cells, while the remaining 2 nuclei (polar nuclei) are situated above the egg apparatus in a large central cell.

Out of the six cells, three are grouped at the micropylar end and constitute the egg apparatus. It is made up of two synergids and one egg cell. The other three cells are located at the chalazal end and are called antipodals. Thus, a typical angiosperm embryo sac after maturity is 8-nucleate and 7-celled.

Question 29.
(i) Draw a diagrammatic sketch of a transverse section of an anther of an angiosperm. Label its different walls and the tissue forming microspore mother cells.
(ii) Describe the process of microsporogenesis up to the formation of a microspore.
(iii) Write the function of ‘germ pore’ in a pollen grain of an angiosperm. (2018C)
Answer:
Answer:

(ii) For sporogenesis, Refer to Answer No. 24.
(iii) For function of germ pore, Refer to Answer No. 18.

Question 30.
(i) Describe the sequence of the process of microsporogenesis in angiosperms.
(ii) Draw a labelled diagram of a 2-celled final structure formed. (Delhi 2015C)
Or
Trace pollen grain development from sporogenous tissue in the anther. (Delhi 2012)
Answer:
(i) Development of pollen grain from Pollen Mother Cell (PMC)

• Pollen mother cell or microspore mother cell undergoes meiosis to form microspore tetrad or haploid microspores.
• As the anther matures, the microspores dissociate from the tetrad and develop into pollen grains.
• Nucleus of the microspores undergoes mitosis to form a large vegetative cell and small spindle-shaped generative cell.
• They develop a two-layered wall, the outer exine made of sporopollenin and the inner intine made of cellulose and pectin.
• Usually the pollen grains are liberated at this 2-celled stage.

In certain species, the generative cell divides mitotically to form two male gametes and the pollen grains are 3-celled during liberation.

Question 31.
(i) Draw a labelled diagram of the sectional view of microsporangium of an angiosperm. (Delhi 2015)
(ii) Explain the development of male gametophyte in the microsporangium. (Delhi 2015)
Answer:
(i) Refer to Answer No. 22 for figure of microsporangium.
(ii) Refer to Answer No. 30 for development of male gametophyte.

Question 32.
(i) Describe the formation of mature female gametophyte within an ovule in angiosperms.
(ii) Describe the structure of cell that guides the pollen tube to enter the embryo sac. (All India 2014)
Answer:
(i) The functional megaspore undergoes mitosis to form 2 nuclei, which migrate to opposite poles, forming a 2-nucleate embryo sac. Further, mitotic divisions lead to the formation of 4-nucleate and 8-nucleate stages of the embryo sac. In these mitotic divisions, nuclear division is not followed by cell wall formation. After the 8-nucleate stage, cell walls are laid down and a typical female gametophyte or embryo sac is formed. Among the eight nuclei, six are enclosed by cell wall and organised into cells, while the remaining two nuclei (polar nuclei) are situated above the egg apparatus in a large central cell.

Out of the six cells, three are grouped at the micropylar end and constitute the egg apparatus made up of two synergids and one egg cell. The other three cells are located at the chalazal end and are called antipodals. Thus, a typical angiosperm embryo sac after maturity is 8-nucleate and 7-celled.

(ii) The egg apparatus present towards the micropylar end, comprises of two synergids and an egg cell.
These synergids possess special cellular thickenings at their micropylar tip and called filiform apparatus. This filiform apparatus guides the pollen tube to enter into embryo sac.
For figure. Refer to Answer No. 28 fig (e).

Question 33.
Draw a labelled diagram of sectional view of a mature embryo sac of an angiosperm. (Delhi 2014)
Answer:
Refer to figure of embryo sac in Answer No. 11.

Question 34.
How does the megaspore mother cell develop into 7-celled and 8-nucleate embryo sac in an angiosperm? Draw a labelled diagram of a mature embryo sac. (Delhi 2012)
Or
Explain with the help of diagram the development of mature embryo sac from a megaspore mother cell in angiosperm. (Foreign 2012, All India 2010C)
Or
Describe the process of megasporogenesis in angiosperm until 8-nucleate stage. (All India 2013C)
Answer:
For development of megaspore mother cell. Refer to Answer No. 32 (i).

Question 35.
(i) Draw a diagram of an enlarged view of TS of one microsporangium of an angiosperm and label the following parts
(a) Tapetum
(b) Middle layers
(c) Endothecium
(d) Microspore mother cell
(ii) Mention the characteristic features and function of tapetum.
(iii) Explain the following giving reasons
(a) Pollen grains are well-preserved as fossils.
(b) Pollen tablets are in use of people these days. (Foreign 2011)
Answer:
(i) For diagram. Refer to Answer No. 22.
(ii) Tapetum is the inner nourishing layer of microsporangial wall. The cells of tapetum have dense cytoplasm and more than one nucleus. These cells nourish the developing pollen grains.
(iii) (a) The outer exine layer of pollen grain is highly resistant because of sporopollenin. It is an organic material which can withstand harsh conditions, action of alkalis and acids. No enzyme can degrade sporopollenin. Thus, pollen grains are well-preserved as fossils.

(b) Pollen grains are rich in nutrients. So, used by people as health tablets or food supplements

Question 36.
Write one advantage and one disadvantage of cleistogamy to flowering plants. (2018C)
Answer:
The advantage of cleistogamy is that it ensures pollination in the absence of pollinators. Disadvantage of cleistogamy is that there is no chance of variation to occur.

Question 37.
What is pollen-pistil interaction and how is it mediated? (Foreign 2014)
Answer:
Pollen-pistil interaction is a chain or group of • events that take place from the falling of pollen over the stigma to the formation of pollen tube and its entry into the ovule. It is mediated by chemical components of pollen grain, interacting with that of pistil.

Question 38.
Differentiate between xenogamy and geitonogamy. (Delhi 2014C)
Answer:
Xenogamy is the transfer of pollen grains from anther of one flower to the stigma of another flower of a different plant, while geitonogamy is the transfer of pollen grains from anther of one flower to the stigma of another flower on same plant.

Question 39.
How do the pollen grains of Vallisneria protect themselves? (All India 2012)
Or
Why do the pollen grains of Vallisneria have a mucilaginous covering? (Delhi 2010)
Answer:
As the pollination of Vallisneria takes place by means of water, the pollen grains are covered by mucilaginous coating that protects them from damage and desiccation.

Question 40.
What is cleistogamy? Write one advantage and one disadvantage of it, to the plant. (All India 2019)
Answer:
Cleistogamy is a type of self-pollination that occurs in a permanently closed flower. Advantage and disadvantage of cleistogamy are as follows

• Advantage Cleistogamous flowers produce assured seed-set even in the absence of pollinators.
• Disadvantage Cleistogamous flowers are invariably autogamous. So, there is no chance of cross-pollination. Hence, less variations are generated in the progeny.

Question 41.
You are conducting artificial hybridisation on papaya and potato. Which one of them would require the step of emasculation and why ? However for both you will use the process of bagging. Justify giving one reason. (Delhi 2019)
Answer:
Papaya produces unisexual flowers and potato produces bisexual flowers. Therefore, the step of emasculation will be done on potato because emasculation is done on bisexual flower to avoid self-pollination. But, bagging is done on unisexual flowers, so to dust suitable pollen grains op the stigma when the stigma turns receptive and the flowers are rebagged.

Question 42.
Express the process of pollination in Vallisneria. (Delhi 2019)
Answer:
Vallisneria is a water pollinated plant. In this plant, the process of pollination involves reaching of female flower at the surface of water by the long stalk and release of pollen grains onto the surface of water. These pollen grains are carried water currents to reach the stigma eventually.

Question 43.
A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain. (All India 2016)
Or
Out of mafty papaya plants growing in your garden, only a few bear fruits. Give reason.
Answer:
A single pea plant produces pods with viable seeds because the pea plant is autogamous, i.e they have the ability of self-pollination. Whereas the individual papaya plant is prevented from both autogamy and geitonogamy. In this plant, male and female flowers are present on different plants, i.e. each plant is either male or female.

Question 44.
Explain the process of pollination in Vallisneria. (Delhi 2016)
Answer:
Pollination in Vallisneria Refer to Answer No. 7.

Question 45.
List the different types of pollination depending upon the source of pollen grain. (Delhi 2016)
Or
Differentiate between autogamy, geitonogamy and xenogamy. (All India 2012)
Answer:
Depending on the source of pollen grain, pollination can be classified into

• Autogamy It is the transfer of pollen grain from anther to the stigma of the same flower.
• Geitonogamy It is the transfer of pollen grains from anther of one flower to the stigma of another flower on the same plant. Geitonogamy is functionally cross-pollination involving pollinating agent, but genetically it is equivalent to autogamy since the pollen grains come from the same plant.
• Xenogamy It is the transfer of pollen grains from anther to the stigma of different plants of same species. It brings genetically different types of pollen grains to the stigma.

Question 46.
In angiosperms, zygote is diploid, while primary endosperm cell is triploid. Explain. (All India 2013)
Or
Mention the reasons for difference in ploidy of zygote and primary endosperm nucleus in an angiosperm. (Delhi 2010)
Answer:
In angiosperms or flowering plants, zygote is diploid and primary endosperm nucleus is triploid. It is because in these plants, one of the male gametes fuses with egg cell, which results in the formation of zygote. So, zygote is diploid. While primary endosperm cell is triploid because the nucleus of the second male gamete (n) fuses with the two haploid polar nuclei or diploid secondary nucleus (2n) of the central cell to form a triploid primary endosperm nucleus (3n). This process is referred to as triple fusion. The central cell is now called primary endosperm cell.

Question 47.
Explain triple fusion in angiosperm. (Delhi 2013)
Answer:
For triple fusion. Refer to Answer No. 11.

Question 48.
State one advantage and one disadvantage of cleistogamy. (All India 2012)
Answer:
For cleistogamy. Refer to Answer No. 5.

Question 49.
Why should a bisexual flower be emasculated and bagged prior to artificial pollination? (Foreign 2010)
Or
Why does a breeder need to emasculate a bisexual flower? Mention a condition in a flower where emasculation is not necessary. (Delhi 2011)
Answer:
Emasculation in a bisexual flower is required to prevent contamination of the stigma with self-pollen grains. Bagging is done to prevent contamination of the stigma of the emasculated flower with any other unwanted pollen grains. That is why a bisexual flower should be emasculated and bagged prior to artificial pollination. Emasculation is not required in unisexual flowers.

Question 50.
(i) Draw a LS of pistil showing pollen tube entering into the embryo sac. Label the following. (All India 2019)
(a) Nucellus
(b) Antipodals
(c) Synergids
(d) Micropyle

(ii) Write the functions of the following
(a) Synergids
(b) Micropyle
Answer:
(i) For pistil-pollen diagram, Refer to Answer No. 19 (ii).
(ii) (a) Synergids These possess special cellular thickenings at their micropylar tip called filiform apparatus. This filiform apparatus guides the pollen tube to enter embryo sac.
(b) Micropyle It facilitates the entry of pollen tube and thus fertilisation.

Question 51.
Emasculation and bagging are the two important steps carried during artificial hybridisation to obtain superior varieties of desired plants. Explain giving reasons, in which types of flowers and at what stages are the two processes carried out. (All India 2019)
Answer:
For emasculation and bagging. Refer to Answer No. 14. If the plant bears bisexual flowers, emasculation and bagging are carried out before the anther dehisces.
If the plant bears unisexual flowers, emasculation is not required. The female flower buds are bagged before the opening of flowers are

Geitonogamy result sun progenies which are animals and even humans. It is believed that pollination by wind (anemophily) is more coihmon because transport by wind-current can take pollens to distant places.

Question 52.
(i) Differentiate between geitonogamy and xenogamy.
(ii) Write the difference in the characteristics of the progeny produced as a result of the two processes. (Delhi 2019)
Answer:
(i) Differences between geitonogamy and xenogamy purelines, e.g. homozygous. They are genetically similar. Xenogamy results in hybrids, e.g. heterozygous. They show variations in characters.

Question 53.
How does a bisexual flowering plant ensure cross-pollination ? Explain. (Delhi 2019)
Answer:
Bisexual flowers ensuring cross-pollination Refer to Answer No. 22.

Question 54.
(i) Can a plant flowering in Mumbai be pollinated by pollen grains of the same species growing in New Delhi? Provide explanation to your answer,
(ii) Draw the diagram of a pistil where pollination has successfully occurred. Label the parts involved in reaching the male gametes to its desired destination. (All India 2017)
Answer:
(i) Yes, a plant flowering in Mumbai can be pollinated by pollen grains of the same species growing in New Delhi. It is mainly because there are certain agents of pollination that can carry pollen grains to long distance. Plants can use either abiotic or biotic agents for pollination. Abiotic pollinators include wind and water while biotic pollinators are insects, birds,

(ii) The parts involved in transferring the male gametes to its desired destination are stigma, style, micropyle, filiform apparatus and synergids. Longitudinal Section (LS) of a post-pollinated pistil is given below

Question 55.
What does an interaction between pollen grains and its compatible stigma result in after pollination? List two steps in sequence that follow after the process. (Delhi 2016)
Answer:
When the pollen grains fall on the stigma, the pollen tube enters one of the synergids and releases two male gametes.

• One of the male gametes moves towards the egg cell and fuses with it to complete syngamy to form the zygote.
• The other male gamete fuses with the two polar nuclei and forms triploid Primary Endosperm Nucleus (PEN). This is termed as triple fusion.
• Since, two kinds of fusion syngamy and triple fusion takes place, the process is known as double fertilisation and is characteristics of flowering plants.

Question 56.
As a senior biology student you have been asked to demonstrate to the students of secondary level in your school, the procedure(s) that shall ensure cross-pollination in a hermaphrodite flower. List the different steps that you would suggest and provide reasons for each one of them. (All India 2016)
Answer:
Cross-pollination is done to mix two desired characters of two different species of a plant. For example, purple and white flower of a pea.

• Select two pea plants one with white and other with purple flower.
• Label them as male (white flowered) and female (purple flowered) plant.
• Cut anthers from purple flower with the help of scissors before their dehiscence to avoid self-pollination and cover it with white paper bag.
• Now collect pollens from the white flower (male plant) with the help of brush.
• Dust the pollens on the stigma of female (purple fewer) flower.
• Cover it again with paper bag till seed formation.

Question 57.
Flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination. Explain three such devices. Delhi 2016C
Or
Why do hermaphrodite angiosperms develop outbreeding devices? Explain any two such devices with the help of examples. (All India 2015)
Or
Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination. (All India 2014)
Answer:
Hermaphrodites or bisexual flowers develop outbreeding devices to ensure cross-pollination and avoid self-pollination. The three outbreeding devices that flowering plants have developed to discourage self-pollination are
(i) Unisexuality (Dicliny) Flowers are unisexual, so that self-pollination is not possible. The plants may be monoecious (bearing both male and female flowers, e.g. maize) or dioecious (bearing male and female flowers on different plants, e.g. mulberry, papaya).

(ii) Dichogamy Anthers and stigmas mature at different times in a bisexual flower for preventing self-pollination.
(a) Protandry Anthers mature earlier than stigma of the same flower. The pollens thus btfcome available to stigmas of the older flowers, e.g. sunflower, Salvia.
(b) Protogyny Stigmas mature earlier, so that they get pollinated before the anthers of the same flower develop pollen grains, e.g. Mirabilis jalapa,
Gloriosa, Plantago.

(iii) The third device to prevent self-pollination is self-incompatibility. It is a genetic mechanism that prevents self-pollen from fertilising the ovules by preventing pollen germination or pollen tube growth in the pistil. All these methods encourage cross-pollination thus causing genetic variations among them.

Question 58.
Explain the phenomenon of double fertilisation. (Delhi 2014)
Answer:
The phenomenon of double fertilisation occurs in following steps

• In an angiospermic plant, two male gametes are discharged by a pollen tube into the cytoplasm of a synergid of the embryo sac.
• One of the male gametes fuses with the egg to form a zygote. This process is called syngamy.
• Other male gamete fuses with the secondary nucleus to form the primary endosperm nucleus, this process is called triple fusion.
• Since, there are two fusions (syngamy and triple fusion) inside an ovule during fertilisation, it is known as double fertilisation.

Question 59.
Write the differences between wind pollinated and insect pollinated flowers. Give an example of each type. (Foreign 2014)
Answer:
The differences between wind pollinated and insect pollinated flowers are

Question 60.
Differentiate between geitonogamy and xenogamy in plants. Which one between the two will led to inbreeding depression and why? (Delhi 2011)
Answer:
For difference. Refer to Answer No. 17 (i).
Geitonogamy will lead to inbreeding depression because the pollen grains are genetically similar resulting into inbreeding. Continuous inbreeding reduces fertility.

Question 61.
(i) Write the characteristic features of anther, pollen and stigma of wind pollinated flowers.
(ii) How do flowers reward their insect pollinator? Explain. (All India 2010)
Answer:
(i) In wind pollinated flowers,

• Anthers are well-exposed for easy dispersal of pollen grains.
• Pollen grains are light and non-sticky, so that they can be transported by wind currents.
• Stigma is large and feathery to trap pollens.

(ii) Flower rewards their insect pollinators easily by offering

• Nectar and edible pollen grains.
• Safe place for insects to lay eggs by some flowers, e.g. Amorphophallus and Yucca.

Question 62.
(i) Describe any two devices in a flowering plant which prevent both autogamy and geitonogamy.
(ii) Explain the events up to double fertilisation after the pollen tube enters one of the synergids in an ovule of an angiosperm. (Delhi 2011)
Answer:
(i) The two devices that prevent both autogamy and geitonogamy in flowering plants are as follows
(a) Self-incompatibility In some plants when pollen from same flower or other flower of the same plant comes on the stigma, it is incapable of bringing about fertilisation.
It is due to the presence of similar self-sterile gene, e.g. tobacco, potato, etc. It prevents autogamy and geitonogamy.

(b) Dioecy In several species such as papaya, male and female flowers are present on different plants. Thus, each plant is either male or female. This condition also prevents both autogamy and geitonogamy.

(ii) In the ovule, the pollen tube is attracted by secretions of synergids. Usually the pollen tube enters the embryo sac by passing into one of the two synergids which starts degenerating. The pollen tube bursts up by absorbing hydrolytic substances secreted by degenerating synergids. It is followed by double fertilisation in flowering plants.

Double fertilisation It is the fusion of two male gametes to two different cells of the same female gametophyte in order to produce two different structures.

Nawaschin (1898) was the first to show that both the male gametes are involved in fertilisation in Fritillaria and Lilium. Double fertilisation consists of two events that are as follows

• Syngamy or Amphimixis Fusion of the egg nucleus with one male gamete is called syngamy. This fusion results in the formation of diploid cell, the zygote.
• Triple fusion Along with syngamy, the other’male gamete moves towards the two polar nuclei located in the central cell and fuses with them to produce a triploid Primary Endosperm Mother (PEM) cell. In this way, fertilisation occurs in flowering plants.

Question 63.
Draw the longitudinal section of a flower showing growth of pollen tube up to the embryo sac. Label the following parts
(i) Stigma
(ii) Pollen tube
(iii) Integument
(iv) Chalazal end
(v) Nucellus
(vi) Synergids
Answer:
For LS of flower. Refer to Answer No. 19 (ii).

Question 64.
(i) Plan an experiment and prepare a flow chart of the steps that you would follow to ensure that the seeds are formed only from the desired sets of pollen grains. Name the type of experiment that you carried out.
(ii) Write the importance of such experiments. (All India 2015)
Answer:
(i) Artificial hybridisation is carried out to ensure that seeds are formed from the desired set of pollen grains. This is done by emasculation and bagging.
The flow chart below shows the steps to be followed

(ii) Importance of such experiments are
(a) Creation of new genetic recombination with better qualities.
(b) Incorporation of a large number of desirable characters into a single variety.

Question 65.
Angiospermic flowers may be monoecious, cleistogamous or show self-incompatibility. Describe the characteristic features of each one of them and state which one of these flowers promotes inbreeding and outbreeding, respectively. (All India 2014)
Answer:
The characteristic features of angiospermic flowers
(i) Monoecious flowers are unisexual, i.e. they have either the male reproductive or female reproductive part in separate flowers, both produced on same plant. The flowers (male and female) are separate. It prevents self¬pollination and promotes cross-pollination.

(ii) Cleistogamous flowers are those flowers in which anthers and stigmas lie close to each other and do not open at all, even at maturity. These flowers are invariably autogamous and promote inbreeding depression as there is no chance for cross-pollination at all.

(iii) Self-incompatible In angiospermic flowers, there is a genetic mechanism, wherein the flowers prevent the self-pollens from fertilising the ovules or inhibit their germination on stigma. This device or mechanism promotes outbreeding.

Question 66.
(i) Draw a longitudinal section of a pistil of an angiosperm showing the growth of pollen tube up to the micropyle of ovule. Label
(a) stigma,
(b) embryo sac
(c) pollen tube
(d) micropyle.

(ii) Explain the events that occur, up to fertilisation, when the compatible pollen grain lands on the stigma. (Delhi 2014)
Answer:
(i) For LS of flower. Refer to Answer No. 19 (ii).
(ii) The events that occur when compatible pollen grains fall on stigma in the sequence are as follows

(a) Pollen-pistil interaction Once the compatible pollen grains fall on stigma which is receptive, it recognises and accepts the pollen with the aid of chemical components interacting with pollen.

(b) Germination of pollen grain Once the pollen is recognised, it germinates on the stigma of flower. The tube cell of pollen grain protrudes out through germ pores to form a pollen tube. The generative cell divides to form two male gametes that are released into the tube.

(c) Growth of pollen tube The pollen tube grows down through the tissues of stigma and style and enters ovule, usually through micropyle. Inside ovule, the filiform apparatus guides the pollen tube, carrying gametes to the egg cell.

(d) Double fertilisation After releasing the two male gametes into the synergids, one of them fuses with egg to form a diploid zygote (syngamy) and other male gamete fuses with 2 polar nuclei to form triploid primary endosperm cell (triple fusion). Because of the occurrence of these two types of fusions, it is called double fertilisation.

Question 67.
Why is fertilisation in an angiosperm referred to as double fertilisation? Mention the ploidy of the cells involved. (All India 2012)
Answer:
In fertilisation (in angiosperm), two types of fusion occur, i.e. syngamy and triple fusion, in the embryo sac. That is why it is called double fertilisation.
Ploidy of cells involved in double fertilisation: Zygote is diploid (2n). It is formed as a result of syngamy, i.e. fusion of two haploid gametes (male gamete + egg). Primary endosperm nucleus (3M) is formed as a result of triple fusion, i.,e. fusion of two haploid polar nuclei with male gamete.

Question 68.
Draw a diagram of a fertilised embryo sac of a dicot flower. Label all its cellular components. (Delhi 2015)
Or
(i) Why is the process of fertilisation in angiosperms termed as double fertilisation? Explain.
(ii) Draw a diagram of an angiospermic embryo sac, where fertilisation is just completed.

Label the following parts

• Micropylar end of embryo sac
• The part that develops into an embryo
• The part that develops into an endosperm
• The degenerating cells at the chalazal end

(iii) Draw a labelled diagram of globular embryonic stage of an angiosperm. (Foreign 2011)
Answer:
(i) Refer to Answer No. 31 (d) for double fertilisation.
(ii) Fertilised angiospermic embryo sac with its cellular component is given below

Question 69.
(i) Geitonogamy is functionally a cross-pollination, but genetically similar to autogamy. Explain.
(ii) Why do flowering plants need to develop outbreeding devices? Explain any three such devices developed by flowering plants. (All India 2010)
Answer:
(i) Transfer of pollen grains from the anther to stigma of another flower of the same plant is called geitonogamy. It is functionally cross-pollination as it involves a pollinating agent, but genetically similar to autogamy since, the pollen grains come from the same plant (genetically same parent).

(ii) Continued self-pollination results in inbreeding depression because majority of flowering plants produce hermaphrodite flowers and pollen grains generally come in contact with the stigma of same flower.

(iii) To discourage this, flowering plants developed many devices. For some of these devices, Refer to Answer No. 22.

Question 70.
Given below is a section of maize grain. Identify A and state its function.

Answer:
A is endosperm. It provides nutrition to the developing embryo.

Question 71.
Identify A in the figure showing a stage of embryo development in a dicot plant and mention its function. (All India 2016)

Answer:
A is cotyledon. It is the storehouse of food.

Question 72.
Mention the function of coleorhiza. (Delhi 2015C, All India 2012)
Answer:
Coleorhiza is a protective sheath covering the young root of the embryo in plants of the grass family.

Question 73.
The meiocyte of rice has 24 chromosomes. Write the number of chromosomes in its endosperm. (All India 2013)
Answer:
The meiocyte is a diploid cell and have 24 chromosomes. Thus, its haploid chromosome number is 12.
Number of chromosomes in endosperm is
12 × 3 = 36

Question 74.
Write the function of scutellum. (Delhi 2012)
Answer:
The cotyledon of embryo of grass family is called scutellum. It is an embryonic leaf.

Question 75.
Banana is a true fruit but is also a parthenocarpic fruit. Give reason. (Foreign 2010C)
Or
Why is banana referred to as a parthenocarpic fruit? (All India 2013)
Or
Why is banana considered a good example of parthenocarpy? (All India 2012)
Answer:
The fruit of banana is formed from the ovary, so it is a true fruit. It is a parthenocarpic fruit because the ovary develops into fruit without fertilisation and is thus seedless.

Question 76.
Why is apple referred to as a false fruit? (All India 2010C)
Answer:
In apple, the thalamus also contributes to fruit formation. So, apples are called false fruits.

Question 77.
Name the mechanism responsible for the formation of seed without fertilisation in angiosperms. Give an example of a species of flowering plants with such seed formation. (Delhi 2010)
Answer:
Apomixis is the mechanism responsible for the formation of seeds without fertilisation in angiosperms, e.g. grasses.

Question 78.
It is said apomixis is a type of asexual reproduction. Justify. (Delhi 2019)
Answer:
Through apomixis, viable seeds can be produced without fertilisation and zygote formation through gametic fusion. This is not the case during sexual reproduction. In sexual reproduction, seeds are produced through gametic fusion following fertilisation. So, apomixis is called a type of asexual reproduction.

Question 79.
Name the type of fruit apple is categorised under and why? Mention two other examples which belong to the same category as apple. (Delhi 2015C)
Answer:
Apple is categorised as false fruit because it does not develop from the ovary, but thalamus. Cashew and strawberry belong to the same category as apple.

Question 80.
Write the difference between the tender coconut water and the thick, white kernel of a mature coconut and their ploidy. (Delhi 2015C)
Answer:
The tender coconut water represents the free nuclear endosperm while, the white kernel is the cellular endosperm. The water and kernel of the endosperm are both triploid.

Question 81.
Suggest two advantages to a farmer using apomictic seeds of hybrid varieties.(Foreign 2015)
Answer:
Two advantages of apomictic seeds to a farmer are as follows

• It lowers the cost of production.
• Apomictic seeds do not have to be produced every year.

Question 82.
List the post-fertilisation events in angiosperms. (Delhi 2014)
Answer:
The post-fertilisation events in angiosperms include

• Endosperm and embryo development.
• Maturation of ovule into seed.
• Maturation of ovary into fruit.

Question 83.
Draw a labelled diagram of a matured embryo of a dicotyledonous plant. (All India 2014C)
Answer:
The labelled diagram of a mature embryo of a dicotyledonous plant is as given below

Question 84.
Write the fate of egg cell and polar nuclei after fertilisation. (Delhi 2013)
Or
Write the fate of triple fusion in mature fruit of coconut. (Delhi 2012)
Answer:
The egg cell after fertilisation with one of the two male nuclei forms zygote which further develops into an embryo. Polar nuclei fuse with other male nuclei to form primary endosperm nucleus which further develops into endosperm. In case of coconut, the endosperm is watery.

Question 85.
Some angiospermic seeds are said to be ‘albuminous’, whereas few others are said to have a perisperm. Explain each with the help of an example. (Foreign 2012)
Answer:
Some angiospermic seeds are albuminous as they retain endosperm even after embryo development, i.e. not completely consumed by embryo, e.g. wheat, maize, castor. nucellus are persistent which is referred to as perisperm, e.g. black pepper and beet.

Question 86.
Differentiate between albuminous and non-albuminous seeds, giving one example of each. (Delhi 2011)
Answer:
Difference between albuminous and non-albuminous seeds is Albuminous seed

Question 87.
(i) Given below is a TS of an apple.
(ii) Why is an apple categorised as a false fruit?

Answer:
(i) In the given figure of TS of an apple,
A – Thalamus, B – Seed, C – Endocarp
(ii) For false fruit, Refer to Answer No. 7.

Question 88.
State what is apomixis? Write its significance. How can it be commercially used? (All India 2019)
Answer:
For apomixis, Refer to Answer No. 24.

Question 89.
Differentiate between parthenocarpy and parthenogenesis. Give one example of each. (2018C)
Answer:
The differences between parthenocarpy and parthenogenesis are

Question 90.
If the meiocyte of a maize plant contains 20 chromosomes. Write the number of chromosomes in the endosperm and embyro of the maize grain and give reasons in support of your answer. (2018C)
Answer:
If the meiocyte of a maize plant contains 20 chromosomes, then the number of chromosomes in the endosperm and the embryo of maize grain will be 30 and 20, respectively.
The meiocyte is a diploid cell and have 20 chromosomes. Thus, its haploid chromosome is 10.
Number of chromosome in endosperm is 10 × 3 = 30.
Number of chromosomes in embryo is 2n,
2 × 10 = 20.

Question 91.
Do you think apomixis can be compared with asexual reproduction? Support your answer, giving one reason. How is apomixis beneficial to farmers? Explain. (2018 C)
Answer:
Yes, apomixis can be compared with asexual reproduction. It is also called a form of asexual reproduction that mimics sexual reproduction.
In apomixis, seeds are produced without fertilisation and zygote formation through gametic fusion. So, it can be called as a form of asexual reproduction.
Two advantages of apomictic seeds (apomixis) to farmers are

• It lowers the cost of production.
• Apomictic seeds do not have to be produced every year.

Question 92.
How are parthenocarpic fruits produced by some plants and apomictic seeds by some other? Explain. (Delhi 2016)
Answer:
Parthenocarpic Seeds:
Some fruits develop without undergoing fertilisation, these are called parthenocarpic fruits, e.g. banana and this process of formation of fruit without fertilisation is called parthenocarpy. It can be induced through the application of growth hormones and such fruits are seedless.

Apomictic Seeds:
It is a form of asexual reproduction that mimics sexual reproduction, but produces viable seeds without fertilisation. It does not involve formation of zygote through the gametic fusion. It occurs in some species of Asteraceae and grasses.

These are produced from segments of fruit (mango stem), male gametic content of pollen (Cyperus) and other vegetative parts. In some species, the diploid egg cell is formed without reduction division and develops into embryo without fertilisation. It is an asexual reproduction in the absence of pollinators and takes place in extreme environments.

In some species like Citrus, some of the nucellar cells surrounding the embryo sac start dividing and develop into embryo. It occurs in the megaspore mother cell that does not undergo meiosis, thus produces diploid embryo sac through mitotic divisions.

Farmers pick apomictic seeds in plants which do not reproduce sexually. So that farmers use this method for the production of cloned seed.

Question 93.
State what is apomixis. Comment on its significance. How can it be commercially used? (All India 2015)
Answer:
Apomixis is the special mechanism to produce seeds without fertilisation. It is a form of asexual reproduction that mimics sexual reproduction.
Significance:

• It helps in fixing heterosis or hybrid vigour in plants permanently.
  Rapid multiplication of genetically uniform individuals can be achieved without risk of segregation.

Commercial use Apomixis is used in plant breeding. It increases the chance of developing superior gene combinations and facilitates the rapid incorporation of desirable traits.

Question 94.
Double fertilisation is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post-fertilisation events that are responsible for it. (Delhi 2015)
Answer:
Double fertilisation is reported in both castor and
groundnut, but their mature seeds are different in terms of endosperm. The primary endosperm nucleus formed after fertilisation divides mitotically without cytokinesis to initiate the formation of endosperm. At this stage, the endosperm is called free nuclear endosperm.

Then, cell wall formation occurs and the endosperm becomes cellular type. The number of free nuclei formed before cellularisation varies greatly. Endosperm may be completely utilised by the developing embryo before the maturation of seeds as in groundnut. Such seeds are called non-albuminous or non-endospermic seeds.

When a portion of endosperm remains in seeds and is used up during seed germination, such seeds are called albuminous or endospermic seeds, e.g. castor.

Question 95.
Describe endosperm development in angiosperm. Foreign (2014)
Answer:
Endosperm development takes place by three methods
(i) In nuclear type, which is a common method, the Primary Endosperm Nucleus (PEN) undergoes repeated mitotic division without cytokinesis. At this stage, the endosperm is called free nuclear endosperm.

(ii) In cellular type, cell wall formation occurs and the endosperm becomes cellular. The number of free nuclei formed before cellularisation varies greatly, e.g. in coconut, the water is free nuclear endosperm and surrounding white kernel is cellular endosperm.

(iii) In helobial type endosperm formation, one half of endosperm is nuclear type and other half is cellular type.

Question 96.
(i) How is apomixis different from parthenocarpy?
(ii) Describe any two modes by which apomictic seeds can be produced. (Delhi 2014C)
Answer:
(i) Parthenocarpy is the development and production of seedless fruits in the absence of fertilisation, whereas apomixis refers to development of seeds and fruits, without fertilisation. So, the main difference between apomixis and parthenocarpy is that seeds are formed in former, while absent in later.

(ii) The two modes by which apomictic seeds can be produced are
(a) Agamospermy in which the seed or embryo is derived from diploid egg cell, formed without meiosis and syngamy. This diploid egg cell develops into embryo without undergoing fertilisation, e.g. apple, Rubus.
(b) Adventive embryony The method in which diploid cells surrounding the embryo sac, e.g nucellus and integument protrude into the sac and develop into embryo. This may also lead to the formation of more than one embryos in an embryo sac or ovule, leading to condition called polyembryony, e.g. Citrus, Opuntia.

Question 97.
Differentiate between parthenogenesis and parthenocarpy. (All India 2014C)
Answer:
For Parthenocarpy and Parthenogenesis, Refer to Answer No. 20.

Question 98.
(i) Describe the endosperm development in coconut.
(ii) Why is tender coconut considered as healthy source of nutrition?
(iii) How are pea seeds different from castor seeds with respect to endosperm? (All India 2013)
Answer:
(i) Coconut endosperm formation is nuclear type.
The primary endosperm nucleus undergoes free nuclear division without cell wall formation.
(ii) Soft coconut is an endosperm. It is rich in nutrients like fats, proteins, carbohydrates, minerals, vitamins, etc. Hence, it is considered as a healthy source of nutrition. [1)
(iii) The seeds of pea are non-endospermic, while castor seeds are endospermic. The endosperm in pea seeds is consumed completely during embryo development, but endosperm is not utilised in castor seeds.

Question 99.
Differentiate between perisperm and endosperm giving one example of each. (All India 2012)
Answer:
Differences between perisperm and endosperm are

Question 100.
LS of a maize grain is given below. Label the parts A, B, C and D in it. (All India 2012)

Answer:
A-Pericarp
B – Scutellum (cotyledon)
C – Coleoptile
D – Coleorhiza

Question 101.
With the help of an example of each explain the following Apomixis, Parthenocarpy, Polyembryony. (India 2012C)
Answer:
Apomixis The phenomenon in which seeds are produced without fertilisation is called apomixis or agamqspermy, e.g. grass.
Parthenocarpy It is a commercially important process in which seedless fruit is formed without fertilisation, e.g. banana.
Polyembryony The occurrence of more than one embryo in a seed is known as polyembryony, e.g. orange.

Question 102.
(i) Explain any two ways by which apomictic seed can develop.
(ii) List one advantage and one disadvantage of an apomictic crop.
(iii) Why do farmers find production of hybrid seeds costly? (Delhi 2019)
Answer:
(i) Apomictic seed can develop in two ways as follows
(a) These are produced from segments of fruit (mango stem), male gametic content of pollen (Cyperus) and other vegetative parts.
(b) In some species, the diploid egg cell is formed without reduction division and develops into embryo without fertilisation.

(ii) Advantage of apomictic crop These are the products of asexual reproduction in the absence of pollinators and takes place in extreme environment.
Disadvantage of apomictic crop During the production of apomictic crop, there is no segregation of characters in the hybrid progeny.

(iii) Apomixis produces hybrid seeds and it is a costly process. Since, hybrid seeds do not maintain hybrid characters, the apomictic seeds are to be produced year after year. This is the reason why farmers find production of hybrid seeds costly.

Question 103.
(i) When a seed of an orange is squeezed, many embryos, instead of one are observed. Explain, how it is possible.
(ii) Are these embryos genetically similar or different? Comment. (India 2017)
Answer:
(i) It is true that when we squeeze a seed of an orange, many embryos, instead of one are observed. It occurs mainly due to a process named polyembryony. It is a phenomenon of occurrence of more than one embryo in a seed. It was first described by Antony van Leeuwenhoek (1719) in Citrus. Polyembryony can be spontaneous or induced experimentally. The polyembryony may arise by the following reasons Formation of additional embryos from synergids or from antipodals and polar nuclei (very rare). Some embryos can also be derived by the activation of some sporophytic cells of ovule such as nucellus or integument.
Embryos can also be developed from an additional embryo sac in the same ovule.

(ii) The embryos formed as a result of polyembryony are genetically similar to one another. However, the embryos arising from gametophytic tissues are similar to each other, but not to their parents.

The embryos that are derived from sporophytic tissues are usually similar to each certain amount of endosperm is formed. It is an other as well as their parents.

Question 104.
(i) Explain the different ways apomictic seeds can develop. Give an example of each.
(ii) Mention one advantage of apomictic seeds to farmers.
(iii) Draw a labelled mature stage of a dicotyledonous embryo. (All India 2014)
Answer:
(i) For apomictic seeds, Refer to Answer No. 27 (ii)
(ii) For advantage of apomictic seeds, Refer to Answer No. 12.
(iii) For diagram. Refer to Answer No. 14.

Question 105.
(i) Why does endosperm development precede embryo development in angiosperm seeds? State the role of endosperm in mature albuminous seeds.
(ii) Describe with the help of three labelled diagrams of the different embryonic stages that include mature embryo of dicot plants. (Delhi 2014)
Answer:
(i) The embryo development starts only after a adaptation for assured nutrition of the developing embryo therefore, endosperm development precedes embryo development. The role of endosperm in mature albuminous seeds is storage of reserve food for growing embryo.

(ii) The embryonic stages during the development of mature embryo sac.
For figure. Refer to Answer No. 5 in Miscellaneous section on page no. 44.

Question 106.
(i) Mature seeds of legumes are non-albuminous. Then, can it be assumed that double fertilisation does not occur in legumes? Explain your answer.
(ii) List the differences between the embryos of dicot (pea) and monocot (grass) families. (Delhi 2014C)
Answer:
(i) Seeds of legumes are non-albuminous that implies that endosperm in such seeds is completely used up in providing nutrition to developing embryo. The endosperm is formed as a result of triploid fusion, i.e. between a male gamete and two polar nuclei. This makes it obvious that it cannot be formed in the absence of double fertilisation.
Therefore, though the seeds of legumes are non-albuminous, it clearly states the occurrence of double fertilisation in them.

(ii) The differences between the embryos of pea and grass can be summarised as follows

Question 107.
(i) Why are seeds of some grasses called apomictic? Explain.
(ii) State two reasons to convince a farmer to use an apomictic crop. (Delhi 2014C)
Answer:
(i) The seeds of some grasses develop without fertilisation. It may be because a diploid egg cell develops into an embryo directly (without undergoing meiosis and syngamy) or some diploid cells of nucellus or integument surrounding the embryo sac, protrude inside and develop into embryos. This phenomenon of developing embryo and seeds without fertilisation is called apomixis and such seeds produced are referred to as apomictic.

(ii) For advantage of apomictic seeds. Refer to Answer No. 12.

Question 108.
Give reasons why?
(i) Most zygotes in angiosperms divide only after certain amount of endosperm is formed.
(ii) Groundnut seeds are exalbuminous and castor seeds are albuminous.
(iii) Micropyle remains as a small pore in the seed coat of a seed.
(iv) Integuments of an ovule hardens and the water content is highly reduced as the seed matures.
(v) Apple and cashewnuts are not called true fruits. (All India 2011)
Answer:
(i) Zygotes in angiosperms mostly divide only after a certain amount of endosperm is formed as an adaptation strategy to assure nutrition for the developing embryo.

(ii) Groundnut seeds are exalbuminous because the developing embryo utilises the endosperm completely. So, there is no endosperm left in the seed.
Castor seeds are albuminous because endosperm is not completely used up by the developing embryo. There is some amount of endosperm left in the seeds always.

(iii) Micropyle allows the entry of water and oxygen during seed germination.

(iv) During unfavourable conditions, seeds become dormant. The loss of water reduces the metabolic activity of seeds and hardens the integuments.

(v) In these fruits, thalamus contributes in fruit formation. So, they are not called true fruits.

Question 109.
(i) Draw a labelled longitudinal view of an albuminous seed.
(ii) How are seeds advantageous to flowering plants? (All India 2010)
Answer:
(i) LS of an albuminous seed (caster) is

(ii) Advantages of seeds to flowering plants are

• Provide protection to embryo in most delicate stage.
• Help in dispersal and spread in new habitats.
• Contain sufficient food reserves.
• Produce genetic variations.
• Seeds are related to pollination and fertilisation.

Question 110.
Explain the development of the zygote into an embryo and of the primary endospermic nucleus into an endosperm in a fertilised embryo sac of a dicot plant.
Answer:
Development of Endosperm
Refer to Answer No. 26.
Embryo Development in Dicot Seed

• Embryo formation starts after a certain amount of endosperm is formed.
• Zygote divides by mitosis to form a proembryo.
• Formation of globular and heart-shaped embryo occurs, which finally becomes horseshoe-shaped mature embryo.
• In dicot plant, embryo consists of two cotyledons and an embryonal axis between them.
• The portion of embryonal axis above the level of attachment of cotyledons is epicotyl and terminates in the plumule.
• The portion of embryonal axis below the level of attachment of cotyledon is the hypocotyl, it becomes radicle (root tip).

Question 111.
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shape and size are seen. Mention how it has happened. (Delhi 2011)
Answer:
In orange seed, embryos originate by adventive embryony from diploid cells of nucellus or integuments and thus if orange seed is squeezed, many embryos of different sizes can be observed.

Question 112.
Name two end products of double fertilisation in angiosperms. How are they formed? Write their fate during the development of seed. (Delhi 2014C)
Answer:
The two end products of double fertilisation in angiosperms are diploid zygote and a triploid Primary Endosperm Nucleus (PEN).
Diploid zygote is formed by the fusion of haploid gametes, i.e. male gamete and egg, while another male gamete and two polar nuclei of central cell fuse to form triploid primary endosperm nucleus.

During the development of seed, the zygote undergoes mitotic divisions to form a mature embryo while, the primary endosperm cell gives rise to nutritive tissue called endosperm, which provides nourishment to growing embryo.

Question 113.
(i) Explain the process of double fertilisation in angiosperms.
(ii) Why does the development of endosperm precedes that of embryo?
(iii) List the parts of a typical dicot embryo.
Answer:
(i) Double fertilisation After releasing the two male gametes into the synergids from a pollen tube, one of them fuses with egg to form a diploid zygote (syngamy) and other male gamete fuses with 2 polar nuclei to form triploid primary endosperm cell (triple fusion). Because of the occurrence of these two types of fusions this process is called double fertilisation.
Refer to figure in Answer No 19. (ii) in Topic 2.

(ii) The embryo development starts only after a certain amount of endosperm is formed. It is an adaptation to assure availability of nutrition for the developing embryo.
Therefore, endosperm development precedes embryo development.
The role of endosperm in mature albuminous seeds is storage of reserve food for growing embryo.

(iii) Parts of a typical dicot embryo are
(a) Plumule (b) Radicle (c) Cotyledons (d) Hypocotyl

Question 114.
Read the following statement and answer the questions that follows.
‘A guaVa fruit has 200 viable seeds’.
(i) What are viable seeds?
(ii) Write the total number of
(a) Pollen grains
(b) Gametes are producing 200 viable guava seeds.
(iii) Prepare a flow chart to depict the post-pollination events leading to viable seed production in a flowering plant. (Delhi 2017)
Answer:
(i) Viable seeds are seeds having the ability to germinate when favourable conditions are present.
(ii) (a) Number of pollen grains = 200
(b) Gametes are producing 200 viable guava seeds = 400
(iii) Post-pollination events include
(a) Double fertilisation (b) Formation of embryo (c) Formation of endosperm
Flow chart depicting post-pollination events is as follows

Question 115.
Read the statement and answer the questions that follows
A flower of brinjal has 520 ovules in its ovary. However, it produces a fruit with only 480 viable seeds.
(i) What could have prevented the rest of the 40 ovules from maturing into viable seeds? Explain giving a reason.
(ii) Describe the development of a dicot embryo in a viable seed.
(iii) Why certain angiospermic seeds are albuminous, while others are exalbuminous? Explain. (Delhi 2017)
Answer:
(i) Ovules can develop into viable seeds having the ability to germinate under suitable environmental conditions, however, often ovules do not develop in viable seeds because of the following reasons

• Excessive dry weather or high temperature.
• Damage to embryo.
• Starvation due to exhaustion of food in ovule.

(ii) For development of a dicot embryo. Refer to Answer No. 41 in Topic 3.
The embryonic stages during the development of a mature embryo sac are

(iii) In flowering plants, during the development of seeds, along with embryo, a nutritive tissue called endosperm is also formed. During embryo development, endosperm degenerates to release its content that helps in the growth of embryo. In some plants, the endosperm is completely used up in embryo development.
In such seed, endosperm is absent, e.g. pea, gram, beans, etc. These seeds are called exalbuminous seeds. On the other hand, in some plants, the endosperm tissue persists in the viable seeds. These are called as albuminous seeds, e.g. castor.

Question 116.
(i) Explain the post-pollination events leading to seed production in angiosperms.
(ii) List the different types of pollination depending upon the source of pollen grain. (Delhi 2016)
Answer:
(i) Post-pollination events include pollen-pistil interaction, double fertilisation, formation of endosperm, development of embryo, seed and fruit.
For pollen-pistil interaction. Refer to Answer No. 2 of Topic 2.
For double fertilisation, Refer to Answer No. 23 of Topic 2.

Development of an endosperm Endosperm development precedes embryo development. The primary endosperm cell divides repeatedly to form a triploid endosperm tissue.

In most of the cases, the PEN undergoes successive nuclear divisions without cytokinesis, to give rise to free nuclear endosperm. Subsequently, cell wall formation starts from the periphery and the endosperm becomes completely cellular, e.g. coconut, rice, maize, sunflower, etc.

For figure. Refer to Answer No. 4 in Miscellaneous. Development of an embryo Embryo develops at the micropylar end of the embryo sac where the zygote is situated. Most zygotes divide only after certain amount of endosperm is formed.

The nutrition for the development of embryo is provided by the endosperm. The zygote gives rise to proembryo and subsequently to the globular, heart-shaped and mature embryo. Development of a seed Double fertilisation in angiosperms triggers the transformation of ovule into a seed. Seeds are formed inside the fruits. Formation of a fruit A fruit is formed as a result of cell division and differentiation in the ovary, which is transformed into fruit as a result of stimuli received from pollination as well as from developing seed.

(ii) Depending upon the source of pollen grains, pollination is of following three types
(a) Autogamy (Self-pollination) It is the kind of pollination achieved within the same flower. The pollen from the anthers of a flower is transferred to the stigma of the same flower, e.g. wheat, rice, pea, etc.

Autogamy is further classified as

• Cleistogamy In some plants, flowers never open up and the anthers dehisce inside the closed flowers to ensure pollination.
  Thus, cleistogamous flowers are invariably autogamous as there is no chance of cross-pollination. These flowers produce assured seed sets even in the absence of pollinators, e.g. Oxalis, Viola, etc.
• Homogamy In this method, both the anthers and the stigma of bisexual flowers mature at the same time, e.g. Mirabilis.

(b) Geitonogamy The pollination where the pollen grains from the anther of a flower are transferred to the stigma of another flower borne on the same plant but at different branches. It usually occurs in plants, which show monoecious condition, e.g. Cucurbita. It is functionally cross-pollination (involves a pollinating agent) but genetically it is similar to autogamy (since pollen grains come from same plant).

(c) Xenogamy (Cross-pollination) It involves the transfer of pollen grains from the anther of one plant to the stigma of another plant. This is the only type of pollination, which brings genetically different types of pollen grains to the stigma during pollination, e.g. papaya, maize, etc.

Question 117.
(i) Explain the events after pollination leading to the formation of a seed in angiosperms.
(ii) Mention the ploidy levels of the cells of different parts of an albuminous seed. (Foreign 2015)
Answer:
(i) The following events take place between pollination and formation of a seed in angiosperms
(a) Double fertilisation
Refer to Answer No. 23 of Topic 2.

(b) Endosperm formation
Refer.to Answer No. 6 (i) of this section only.

(c) Embryo development
Refer to Answer No. 6 (i) of this section only.

(d) Seed formation Seeds are fertilised ovule that are developed inside a fruit. Each seed consists of a seed coat, an embryonal axis and cotyledon.

(ii) The ploidy of the endosperm of an albuminous seed is 3rt and ploidy of the embryo is 2n.

Question 118.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds.
Answer the following questions giving reasons
(i) What would have been the minimum number of ovules present in per pollinated pistil?
(ii) How many microspore mother cells would minimally be required to produce requisite number of pollen grains?
(iii) How many pollen grains must have minimally pollinated the carpel?
(iv) How many male gametes would have used to produce these 200 viable seeds?
(v) How many megaspore mother cells were required in this process? (Delhi 2015)
Answer:
(i) The minimum number of ovules would also have been 240, as 240 viable seeds are formed. After fertilisation, the ovule turns into seeds.
(ii) The minimum number of microspore mother cells in the above case would be 60. This is because each microspore mother cell gives rise to 4 microspores.
Thus, to obtain 240, 60 cells are required.
(iii) The minimum number of pollen grains that must have been involved are 240, as 240 viable seeds are formed. This is because each pollen grain contains 2 male gametes, out of which one fuses with the egg forming zygote that gives rise to seeds.
(iv) The number of male gametes involved would be 240. Each male gamete fuses with one egg nucleus to form zygote that gives rise to seed.
(v) 240 megaspore mother cells were involved. They undergo meiotic division to form 4 haploid megaspores. Out of them, only 1 becomes functional megaspore, rest 3 degenerate.

Question 119.
A flower of brinjal plant following the process of sexual reproduction produces 300 viable seeds.
Answer the following questions giving reasons
(i) How many ovules are minimally involved?
(ii) How many megaspore mother cells are involved?
(iii) What is the minimum number of pollen grains that must land on stigma for pollination?
(iv) How many male gametes are involved in the above case?
(v) How many microspore mother cells must have undergone reduction division prior to dehiscence of anther in the above case? (Delhi 2015)
Answer:
(i) The minimum number of ovules would be 300 as 300 viable seeds are formed.
(ii) 300 megaspore mother cells were involved.
(iii) The minimum number of pollen grains that must have been involved are 300.
(iv) The number of male gametes involved in seed formation are 300.
(v) Minimum 75 microspore mother cells.must have been involved to form 300 pollen grains.

Question 110.
(i) Draw a LS of a pistil showing pollen tube entering the embryo sac in an angiosperm and label the six parts other than stigma, style and ovary.
(ii) Write the changes a fertilised ovule undergoes within the ovary in an angiosperm plant. (All India 2013)
Answer:
(i) Diagram Refer to Answer No. 19 (ii) of Topic 2.
(ii) Changes taking place in a fertilised ovule within the ovary in an angiospermic plant are
Fertilised ovule – Seed
Funiculus – Stalk of ovule
Integument – Seed coat
(a) outer – Testa
(b) inner – Tegman
Polar nuclei – Endosperm
Nucellus – Utilised or remaining perisperm
Antipodal – Degenerate
Synergid – Degenerate
Egg – Embryo

Question 111.
(i) Name the structures which the parts A and B shown in the diagram given below, respectively develop into.

(ii) Explain the process of development which B undergoes in albuminous and exalbuminous seeds. Give one example of each of these seeds. (Foreign 2011)
Answer:
(i) The part A is zygote which develops into the embryo. The part B is primary endosperm nucleus which develops into the endosperm.
(ii) Endosperm Formation

• Primary endosperm cell divides repeatedly and forms triploid endosperm nucleus.
• Primary endosperm nucleus undergoes successive free nuclear divisions to give rise to a number of free nuclei. At this stage, it is called free nuclear endosperm.
• Wall formation takes place from the periphery and proceeds towards the centre and the endosperm becomes cellular.
• In albuminous seeds, some amount of endosperm persists in the mature seed as the developing embryo does not consume it completely, e.g. wheat, maize.
• In exalbuminous seeds, the endosperm is completely consumed by the developing embryo before seed maturation, e.g. in pea/groundnut.

Question 112.
(i) Draw a labelled diagram of LS of an embryo of grass (any six labels).
(ii) Give reason for each of the following
(a) Anthers of angiosperm flowers are described as dithecous.
(b) Hybrid seeds have to be produced year after year. (All India 2011)
Answer:
(i) LS of grass embryo is given below

(ii) (a) A typical angiosperm anther is bilobed with each lobe having two thecae.
So, anther is called dithecous.
(b) Hybrid seeds show segregation of traits and do not maintain the hybrid character in plants. So, they need to be produced every year and cannot be stored.

Question 113.
Explain double fertilisation and trace the post-fertilisation events in sequential order leading to seed formation in a typical dicotyledonous plant. (Foreign 2010)
Answer:
For double fertilisation. Refer to Answer No. 23 of Topic 2.

Post-fertilisation events can be traced as

• Development of endosperm, enlargement of seeds and fruit formation.
• Zygote develops into an embryo.
• Central cell becomes primary endosperm cell and the primary endosperm nucleus develops into the endosperm.
• Antipodals and synergids degenerate.
• Integuments develop into seed coat.
• Ovules ripen into seeds.
• Ovary ripens to form the fruit.

For development of embryo, Refer to Answer No. 6 (i) of this section only.

Question 114.
In a group discussion during laboratory class, the process of seed formation in flowering plants was being discussed. One student pointed out that pollination is major step involved in seed formation. Another student asked how does pollination occur in plants present in water. The teacher explained the process of pollination to solve their queries.
(i) What is pollination? Give the different types of pollination.
(ii) How do flowers present submerged in water bodies adapt themselves for pollination?
Answer:
(i) The transfer of pollen grains from the anther of the stamen to the receptive stigma of a flower is called pollination. It is of two types, self-pollination and cross-pollination.
(ii) In submerged plants, the pollen grains are released in water. In such plants, pollen grains are long, ribbon-like and are carried passively inside the water to stigma for pollination.

Question 115.
A biology student Ashu, read an article on apple being a false fruit. He asked his teacher about how a fruit can be called false and was explained about the development of fruits.
(i) What are false fruits? Give an example.
(ii) What is meant by parthenocarpy?
Name a fruit develops by this method.
(iii) What are the values shown by Ashu?
Answer:
(i) The fruit derived from ovary along with other accessory floral parts like thalamus is called false fruit, e.g. apple, strawberry, etc.
(ii) In parthenocarpy, fruits develop without undergoing fertilisation. Such fruits are seedless and called as parthenocarpic fruits, e.g. banana.
(iii) Ashu is intelligent, observant, inquisitive bo9857y who is curious about new developments around him.

The post Sexual Reproduction in Flowering Plants Class 12 Important Questions and Answers Biology Chapter 2 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 3 Human Reproduction to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 3 Important Questions, Human Reproduction Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 3 Important Questions Human Reproduction

Question 1.
Why are human testes located outside the abdominal cavity? Name the pouch in which they are present. (all India 2014)
Answer:
Human testes are located outside the abdominal cavity as it helps in maintaining low temperature (2-2.5%) lower than body temperature) required for spermatogenesis.
Testes are enclosed in a pouch called scrotum.

Question 2.
Write the location and functions of following in human testes
(i) Sertoli cells
(ii) Leydig cells (All India 2014)
Or
Name the cells that nourish the germ cells in the testes. Where are these cells located in the testes? (All India 2013C)
Answer:
(i) Location of Sertoli cells Within the lining of seminiferous tubule of testis.
Function of Sertoli cells They provide nutrition to the developing sperms or germ cells.

(ii) Location of Leydig cells In the interstitial spaces between the seminiferous tubules. Function of Leydig cells They synthesise and secrete male hormones, i.e. androgens, testosterone.

Question 3.
Write the function of the seminal vesicle. (Delhi 2012)
Answer:
Seminal vesicle produces an alkaline secretion containing prostaglandins, proteins and fructose. The high fructose content provides energy to the spermatozoa. These secretions form 60-70% of the fluid found in the semen.

Question 4.
List the different parts of human oviduct through which the ovum travels till it meets the sperm for fertilisation. (Delhi 2014C)
Answer:
The different parts of human oviduct through which the ovum travels, till it gets fertilised are given below in the sequence

• Fimbriae, finger-like projections Collect the ovum from ovary after ovulation.
• Infundibulum Ovum from fimbriae is guided into funnel-shaped infundibulum, part of Fallopian tube.
• Ampulla A wider part of oviduct that leads ovum into isthmus.
• Isthmus It has a narrow lumen which opens into uterus. In the junction of ampulla- isthmus, the ovum gets fertilised.

Question 5.
Draw a labelled diagram to show inter-relationship of four accessory ducts in a human male reproductive system. (Delhi 2019)
Answer:
Refer to figure 3.1 (b) on page no. 48.

Question 6.
The below diagram shows human male reproductive system (one side only).

(i) Identify ‘X’ and write its location in the body.
(ii) Name the accessory gland ‘Y’ and its secretion.
(iii) Name and state the function of ‘Z’. (Delhi 2015C)
Answer:
(i) .X-Testis. It is located outside the abdominal cavity within a pouch called scortum.
(ii) K-Seminal vesicle. It produces an alkaline secretion rich in fructose and constitutes 60% of the volume of the semen.
(iii) Z-Epididymis. It stores the sperms and secretes a fluid which helps in the maturation of sperms.

Question 7.
(i) Draw a labelled diagrammatic view of human male reproductive system,
(ii) Differentiate between vas deferens and vasa efferentia. (Delhi 2014)
Answer:
(i) Refer to figure 3.1 on page no. 48.
(ii) Distinction between vas deferens and vasa efferentia

Question 8.
Name and explain the role of inner and middle walls of human uterus. (Delhi 2014)
Answer:
The innermost wall of uterus is called endometrium.
Role of Endometrium
(i) It lines the uterine cavity and is glandular in nature.
(ii) It undergoes cyclic changes during menstrual cycle.
The middle wall or layer of uterus is called myometrium.

Role of Myometrium
(i) It is made up of thick layer of smooth muscles.
(ii) It shows strong contractions during the delivery of baby.

Question 9.
Draw a labelled diagram of reproductive system in a human female. (All India 2011)
Answer:
Refer to figure 3.2 on page no. 49.

Question 10.
When do the oogenesis and the spermatogenesis initiate in human females and males, respectively? (Delhi 2012)
Answer:
Oogenesis initiates during foetal or embryonic stage in females, whereas spermatogenesis in males starts at puberty.

Question 11.
Mention the differences between spermiogenesis and spermiation. (Delhi 2012)
Answer:
Difference between spermiogenesis and spermiation is

Question 12.
Where is acrosome present in humans? Write its function. (All India 2012)
Answer:
In humans, the acrosome is present in the anterior portion of the head of human sperm. Function Hydrolytic enzymes or sperm lysins present in acrosome help in the penetration of sperm into egg, during fertilisation.

Question 13.
List the changes the primary oocyte undergoes in the tertiary follicular stage in human embryo. (Foreign 2011)
Answer:
The primary oocyte within the tertiary follicle grows in size. The fully grown primary oocyte completes its first meiotic division to produce two daughter nuclei in which larger haploid cell is called secondary oocyte and the tiny one is called first polar body. The secondary oocyte retains bulk of nutrient ricfi cytoplasm of primary oocyte. (1)

Question 14.
Draw a sectional view of human ovary showing the different stages of developing follicles, corpus luteum and ovulation. (Delhi 2019)
Or
(i) Draw a sectional view of human ovary and label the following parts
(a) Primary follicle
(b) Secondary oocyte
(c) Graafian follicle
(d) Corpus luteum
(ii) Name the hormones influencing follicular development of corpus luteum. (All India 2019)
Answer:
Refer to figure 3.6 on page no. 53.
The hormones influencing follicular development of corpus uteum are FSH, oestrogen and progesterone.

Question 15.
Explain the role of pituitary and sex hormones in the process of spermatogenesis. (All India 2015C)
Or
Spermatogenesis in human males is a hormone regulated process. Justify. (Delhi 2014)
Or
Explain the hormonal regulation of the process of spermatogenesis in humans. (All India 2013)
Answer:
Hormonal control of spermatogenesis in human males is as follows

• Gonadotropin Releasing Hormone (GnRH) is released significantly from the hypothalamus during puberty.
• GnRH stimulates anterior pituitary to secrete gonadotropins, i.e. LH and FSH or Interstitial Cell Stimulating Hormone (ICSH).
• Luteinising Hormone (LH) acts on Leydig cells to stimulate the synthesis and secretion of androgens which then stimulate the process of spermatogenesis.
• Follicle Stimulating Hormone (FSH) acts on Sertoli cells and stimulates them to secrete inhibin which then stimulates the process of spermiogenesis.

Question 16.
What happens to corpus luteum in human female if the ovum is (i) fertilised, (ii) not fertilised? (Delhi 2015)
Or
Mention the fate of corpus luteum and its effect on the uterus in the absence of fertilisation of the ovum in human female. (Foreign 2010)
Answer:

• In case of fertilisation, the corpus luteum continues secreting progesterone which is required for the maintenance of endometrium during pregnancy.
• In the absence of fertilisation, the corpus luteum degenerates and gets converted into corpus albicans. Deficiency of progesterone causes disintegration of the endometrium leading to menstruation and thus, a new cycle starts.

Question 17.
Draw a diagram of a mature human sperm. Label any three parts and write their functions.
Or
Draw a labelled diagram of a human sperm. (Foreign 2015, 2011)
Or
Draw a diagram of human sperm. Label only those parts along with their functions that assist the sperm to reach and gain entry into the female gamete. Foreign 2014
Answer:
For structure of a human sperm Refer to figure 3.5 on page no. 52.
The parts that assist sperm to reach and gain entry into female gamete are,
Tail Its wiggling movement helps the sperm to swim in female reproductive tract.
Middle piece It contains mitochondria that provide energy for sperm movement.
Head It contains acrosome loaded with sperm lysins. These help to dissolve the layers of female gamete or egg.

Question 18.
Write the effect of high concentrations of LH on a mature Graafian follicle. (All India 2014)
Answer:
The high concentration of LH (also known as LH surge) induces the rupture of Graafian follicle, which results in the release of secondary oocyte hence, causing ovulation in females.

Question 19.
Differentiate between spermatogenesis and spermiogenesis. (Delhi 2014)
Answer:
Differences between spermatogenesis and spermiogenesis are

Question 20.
Draw and label the parts of the head region only of a human sperm. (Delhi 2014C)
Answer:
Diagram showing head of human sperm is given below

Question 21.
Identify A, B, C and D with reference to gametogenesis in humans, in the flow chart given below. (All India 2012C)

Answer:
A – Leydig’s cell
B – Sertoli cell
C – Spermatogenesis
D – Spermiogenesis

Question 22.
Name the labels A, B, C, D, E and F in the diagram of seminiferous tubule. (Delhi 2011)

Answer:
In the given diagram,
A – Spermatogonium
B – Primary spermatocyte
C – Secondary spermatocyte
D – Spermatid
E – Spermatozoa
F – Sertoli cell

Question 23.
Differentiate between menarche and menopause. (All India 2010)
Answer:
Differences between menarche and menopause are

Question 24.
Study the sectional view of human testis showing seminiferous tubules given below.

(i) Identify A, B and C.
(ii) Write the function of A and D. (Foreign 2010)
Answer:
(i) A-Spermatogonia
B-Interstitial cells
C-Spermatozoa.

(ii) A-Spermatogonia undergo meiosis to produce spermatozoa (sperms).
D-Sertoli cells provide nutrition to the germ cells.

Question 25.
How and at what stage of menstrual cycle is corpus luteum formed in human females? When does it regress? (Delhi 2010C)
Answer:
After ovulatory phase (ovulation), the luteal phase starts. The remaining parts of ruptured Graafian follicle transform into corpus luteum in this phase. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of endometrium.

In the absence of fertilisation, the corpus luteum degenerates in the ovary and gets converted into corpus albicans.

Question 26.
Draw a labelled diagrammatic sectional view of a human seminiferous tubule. (Delhi 2017)
Answer:
Refer to figure 3.3 on page no. 51.

Question 27.
(i) How many primary follicles are left in each ovary in a human female at puberty?
(ii) Draw a sectional view of the ovary showing the different follicular stages of a human female in her preovulatory phase of menstrual cycle. (Outside Delhi 2016C)
Answer:
(i) A large number of primary follicles degenerate in females during the period from birth to puberty by the process called follicular atresia. As a result, about 60000-80000 primary follicles are left in each ovary at puberty.

(ii) Refer to figure 3.6 on page no. 53.

Question 28.
Explain the events in a normal woman during her menstrual cycle on the following days (Delhi 2015C)
(i) Ovarian event from 13-15 days.
(ii) Ovarian hormones level from 16-23 days.
(iii) Uterine events from 24-29 days.
Answer:
(i) In the ovarian event from 13-15 days, a immature ovum (egg cell) is released from the Graafian follicle. Both LH and FSH attain maximum peak. FSH helps Graafian follicle to attain maturity and LH helps in its rupture. Ovum covered by a number of layer and a yellow fat layer forms corpus luteum. It releases (secretes) progesterone.

(ii) During menstrual cycle, the period level from 16-23 days is called luteal phase (secretory phase). The corpus luteum secretes large amount of progesterone which is essential for the maintenance of endometrium.

(iii) Uterine events from 24-29 days are under the influence of progesterone hormone. It influences the maintenance of the endometrium for any pregnancy to occur. In the absence of pregnancy, the corpus luteum degenerates and endometrium sheds off. It causes the menstrual flow or bleeding.

Question 29.
Explain the events in a normal woman during her menstrual cycle on the following days (All India 2015C, Outside Delhi 2015C)
(i) Pituitary hormone levels from 12 days.
(ii) Uterine events from 13-15 days.
(iii) Ovarian events from 16-23 days.
Answer:
(i) The period of 8-12 days after the onset of menstruation is the follicular phase. During this phase, GnRH from hypothalamus stimulates anterior pituitary to release FSH and LH. FSH stimulates the ovarian follicles to secrete oestrogen, which in turn stimulates the proliferation of the endometrium of the uterine wall. This causes the endometrial lining to thicken.

(ii) The uterine events between day 13 and 15 are governed by the high LH and FSH levels. The endometrium is intact due to the effect of these gonadotropin hormones and also prepares itself for pregnancy, if fertilisation occurs.

(iii) During 16-23 days, ruptured Graafian follicle gets converted into corpus luteum in the pvary.
It starts secreting progesterone which maintains the endometrium, necessary for the implantation of fertilised ovum followed by other events of pregnancy.

Question 30.
Explain the hormonal control of spermatogenesis in humans. (Foreign 2014)
Answer:
Refer to Answer No. 6. (3)

Question 31.
Explain the steps in the formation of an ovum from oogonium in humans. (All India 2013)
Answer:
In human females, primary oocytes are formed from the oogonia during the embryonic developmental stages in the foetal ovaries.

• Oogonial cells start dividing and enter prophase-I of meiosis. They remain suspended at this stage as primary oocytes.
• Each primary oocyte is surrounded by a layer of granulosa cells and becomes the primary follicle.
• The primary follicle when surrounded by more layers of granulosa ceils, is called a secondary follicle.
• Secondary follicle transforms into a tertiary follicle, with the development of a fluid-filled cavity (antrum) around the primary oocyte
• Granulosa cells become organised into an outer theca externa and an inner theca interna.
• Now, primary oocyte completes meiosis-I and forms a larger haploid secondary oocyte and a tiny first polar body.
• Tertiary follicle grows and becomes a mature follicle called Graafian follicle.
• Secondary oocyte secretes a new membrane called zona pellucida around it.
• At this stage, follicle ruptures to release the secondary oocyte, which moves into the cytoplasm.
• Secondary oocyte completes meiosis-II only when a sperm enters its cytoplasm. It forms a larger cell, the ootid and a small pell, the second polar body. This event occurs in the ampulla of Fallopian tube.

Question 32.
(i) Explain menstrual cycle in human females.
(ii) How can the scientific understanding of the menstrual cycle of human females help as a contraceptive measure ? (2018)
Answer:
(i) Menstrual Cycle The inner lining of uterus called endometrium, grows and thickens each month and prepares itself for the implantation of an embryo. If tht) pregnancy does not occur, the endometriumisheds off. The monthly development and shedding of the functional layer of the uterus is called the menstrual phase and the monthly maturation of an egg and its release is called the ovarian cycle. A typical menstrual cycle completes in an average of about. 28 days. It starts at the age of 13 or 15 and continues till about 50 years of age. Menstrual cycle occurs in three major phases namely menstrual phase, follicular phase and secretory phase.
(a) Menstrual Phase It lasts for 3-4 days. It occurs due to the breakdown of endometrium lining cf uterus and blood vessels.
(b) Follicular Phase or Proliferative Phase. It is regulated by the hormones secreted by anterior pituitary gland whose secretions stimulate the ovarian follicle to secrete oestrogens.

During the second week, most of the developing follicles die and one follicle continues to grow and gets mature to form Graafian follicle.

(c) Secretory Phase or Luteal phase The phase of menstrual cycle with possibility of fertilisation is the initial luteal phase. It is marked by the presence of corpus luteum (yellow body). During pregnancy, this yellow body secretes progesterone. In the absence of pregnancy, it regresses to form corpus albicans and menstruation starts (menstrual phase). Due to the hormones secreted by corpus luteum, i.e. oestrogen and progesterone, the release of FSH and LH is inhibited. This prevents the development of new follicles.
This phase lasts for 14 days.

Maintenance of endometrium by progesterone is necessary for the implantation of embryo and pregnancy. During pregnancy, menstrual cycle stops due to high level of progesterone.

(ii) The scientific understanding of the menstrual cycle of human females helps as a contraceptive method. This method is known as periodic abstinence. In this method, the couples avoid or abstain the coitus from day 10 to 17 of the menstrual cycle when ovulation could be expected. The chances of fertilisation are very high during this period.

Question 33.
(i) Arrange the following hormones in the sequence of their secretion in a pregnant woman. hCG, LH, FSH, Relaxin
(ii) Mention their source and the function they perform. (Delhi 2017)
Answer:
(i) The sequence of secretion of the given hormones in a pregnant woman is
FSH → LH → hCG → Relaxin

Question 34.
(i) Explain the menstrual phase in a human female. State the level of ovarian and pituitary hormones during this phase.
(ii) Why is follicular phase in the menstrual cycle also referred as proliferative phase? Explain.
(iii) Explain the events that occur in a Graafian follicle at the time of ovulation and thereafter.
(iv) Draw a Graafian follicle and label antrum and secondary oocyte. (Delhi 2016)
Answer:
(i) The reproductive cycle in female primates, e.g. monkeys, apes and human beings is called menstrual cycle.
Menstrual phase starts from 3rd day and ends on 5th day of the menstrual cycle. It is initiated due to the reduced secretion of progesterone and oestrogen from the regressing corpus luteum in the ovary. The endometrium breaks down and blood along with degenerated ovum constitutes the menstrual flow. The secretion of pituitary hormones, i.e. FSH and LH is also reduced during this phase.

(ii) In follicular phase, primary follicles in the ovary grow under the influence of Follicle Stimulating Hormone (FSH). It starts from 6th day and ends on 13 th or 14th day of 28 day cycle. FSH also stimulates the ovarian follicles to secrete oestrogen which in turn stimulates endometrium to proliferate, so that it becomes thicker and highly vascularised. Thus, it is also called proliferative stage of menstrual cycle.

(iii) Graafian Follicle at Ovulation:
At the time of ovulation following events occur

• LH and FSH reach at their peak levels (about 14th-16th day of cycle).
• High level of LH induces Graafian follicle to rupture and the release of secondary oocyte from it.
• After ovulation, the remaining cells of Graafian follicle are stimulated by LH to develop corpus luteum (an endocrine gland which secrete progesterone hormone).

(iv) Diagrammatic sectional view of Graafian follicle

Question 35.
(i) Explain the process of spermatogenesis in humans.
(ii) Draw a human sperm and label acrosome and middle piece. Mention their functions. Outside (Delhi 2016C)
Answer:
(i) Spermatogenesis is the production of sperms in males.

• In testis, the immature male germ cells (spermatogonia) produce sperms by spermatogenesis. It begins at puberty due to significant increase in the secretion of gonadotropins, i.e. luteinising hormone and follicle stimulating hormone under the influence of Gonadotropin Releasing Hormone (GnRH) released from hypothalamus.
• Spermatogonia (sing, spermatogonium) present on the inside wall of seminiferous tubules multiply by mitotic division and increases in numbers.
• Each spermatogonium is diploid and contains 46 chromosomes. Some of the spermatogonia transform to primary spermatocytes.
• The primary spermatocyte undergoes meiosis -1 and forms two haploid secondary spermatocytes containing 23 chromosomes each.
• The secondary spermatocytes undergo meiosis – II and form four equal sized haploid spermatids.
• Spermatids transform into the spermatozoa by spermiogenesis.
• After spermiogenesis, the sperm heads get embedded in the Sertoli cells and released from the seminiferous tubules via spermiation process.

(a) The acrosome is filled with hydrolytic enzymes that help the sperm to penetrate the ovum.
(b) Middle piece possesses many mitochondria to produce energy for the movement of tail to facilitate sperm motility.

Question 36.
Describe the roles of pituitary and ovarian hormones during the menstrual cycle in a human female. (Delhi 2015)
Or
Explain the ovarian and uterine events that occur during a menstrual cycle in a human female under the influence of pituitary and ovarian hormones, respectively. (Delhi 2014)
Or
Explain the different phases of menstrual cycle and correlate the phases with the different levels of pituitary hormones in a human female. (Delhi 2014C)
Or
With the help of graphical illustration only, show the changes in the levels of the pituitary hormone during menstrual cycle in humans. (Delhi 2011C)
Answer:
The cycle of events starting from one menstruation till next in female primates is called menstrual cycle. It comprises of four phases which are regulated by both pituitary (LH and FSH) and ovarian (oestrogen and progesterone) hormones that affect ovaries and uterus, respectively. The events occurring in a menstrual cycle are as follows

For graphical illustration of menstrual cycle;
Refer to figure 3.J0 on page no. 55.

Question 37.
Explain the development of a secondary oocyte (ovum) in a human female from the embryonic stage up to its ovulation. Name the hormones involved in this process. (Delhi 2015)
Or
When and where are primary oocytes formed in a human female? Trace the development of these oocytes till ovulation (in menstrual cycle). How do gonadotropins influence this developmental process? (Delhi 2010)
Answer:
Embryonic Stages Refer to Answer No. 22.
Influence of Gonadotropins on Oogenesis:

• Gonadotropins, i.e., LH and FSH stimulate follicular development and secretion of oestrogen by the growing follicles.
• Both LH and FSH attain a peak level in the middle of the cycle (14th day).
• Rapid release of LH during mid-cycle causes ovulation.
• LH also stimulates the formation of corpus luteum from the ruptured follicle and secretion of progesterone from corpus luteum.

Question 38.
(i) How is ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the growth till puberty in the humans?
(ii) Draw a sectional view of human ovary and label the different follicular stages, ovum and corpus luteum. (Delhi 2014)
Answer:
(i) Oogenesis is markedly different from spermatogenesis in the following aspects

(ii) Refer to figure 3.6 on page no. 53.

Question 39.
Schematically represent and explain the events of spermatogenesis in humans. (Delhi 2014C)
Answer:
Refer to Answer No. 26 (i)

Question 40.
(i) Draw a transverse section of a human ovary showing the sequential development of different follicles up to the corpus luteum.
(ii) Comment on the corresponding ovarian and pituitary hormone levels during these events. (Delhi 2014C)
Answer:
(i) Refer to figure 3.6 on page no. 53.
(ii) Refer to Answer No. 28.

Question 41.
The following is the illustration of the sequence of ovarian events (A-I) in a human female.

(i) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(ii) Name the ovarian hormone and the pituitary hormone that have caused the above mentioned event.
(iii) Explain the changes that occur in the uterus simultaneously in anticipation.
(iv) Write the differences between C and H.
(v) Draw a labelled sketch of the structure of a human ovum prior to fertilisation. (Delhi 2012)
Answer:
(i) Figure F illustrates ovulation. It represents the ovulatory stage of oogenesis.
(ii) Ovarian and pituitary hormones involved in causing ovulation are
Ovarian hormone Oestrogen.
Pituitary hormone LH and FSH.
(iii) In anticipation of receiving the fertilised egg, the endometrium of the uterus gets thickened and also the blood supply to the endometrium increases.
(iv) In the figure, C stage represents the secondary follicle and the H stage represents the degenerating corpus luteum.

(v) Structure of a human ovum

Question 42.
The following is the illustration of the sequence of ovarian events A-I in a human female.

Identify the figure that illustrates corpus luteum and name the pituitary hormone that influences its formation.
(ii) Specify the endocrine function of corpus luteum. How does it influence the uterus? Why is it essential?
(iii) What is the difference between D and E?
(iv) Draw a neat and labelled sketch of Graafian follicle. (Delhi 2012)
Answer:
(i) Gis corpus luteum. LH secreted by anterior pituitary’ influences its formation.
(ii) The endocrine function of corpus luteum is to secrete progesterone which is essential for the maintenance of endometrium layer of uterus. Thickened endometrium is necessary for the implantation of fertilised ovum and other events of pregnancy.
(iii) D is a developing secondary follicle containing primary oocyte and undifferentiated thecal cells. E is a mature tertiary follicle that contains a secondary oocyte and differentiated theca cells.
(iv) Structure of a Graafian follicle

Question 43.
(i) Describe the stages of oogenesis in human females.
(ii) Draw a labelled diagram of a human ovum released after ovulation. (Delhi 2012)
Answer:
(i) For oogenesis, Refer to Answer No. 22.
(ii) Human ovum Refer to Answer No. 32 (v).

Question 44.
(i) Draw a diagrammatic labelled sectional view of a seminiferous tubule of a human.
(ii) Describe in sequence the process of spermatogenesis in humans.
Answer:
(i) Sectional view of a seminiferous tubule
Refer to figure 3.3 on page no. 51.
(ii) Refer to Answer No. 26 (i).

Question 45.
State from where do the signals for parturition originate in human females. (All India 2019)
Answer:
Foetal-ejection reflex generated by fully developed foetus and placenta stimulates pituitary to release the hormone oxytocin responsible for parturition.

Question 46.
Mention the function of zona pellucida. (Delhi 2015C, 2013)
Answer:
During (ertilisation, a sperm comes in contact with the zona pellucida layer of ovum and induces changes in the membrane and forms fertilisation envelope. It helps block the entry of additional sperms. So, zona pellucida ensures that only one sperm fertilises an ovum.

Question 47.
How does the sperm penetrate through the zona pellucida in human ovum? (Delhi 2013C)
Answer:
The sperm penetrates through the zona pellucida by releasing sperm lysins (hyaluronidase and protease) present in the acrosome.

Question 48.
How is the entry of only one sperm ensured into an ovum during fertilisation in humans? (All India 2012)
Answer:
Presence of zona pellucida layer ensures that only one sperm can fertilise an ovum.
Refer to Answer No. 2.

Question 49.
Write the function of oxytocin. (Delhi 2012)
Answer:
Oxytocin secreted by posterior pituitary helps in parturition by inducing and enhancing uterine muscle contractions.

Question 50.
Mention the function of trophoblast in human embryo. (Delhi 2011)
Answer:
Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst to the endometrium of the uterus.

Question 51.
Name the embryonic stage that gets implanted in the uterine wall of a human female. (All India 2011)
Answer:
Blastocyst gets implanted in the uterine wall of human female.

Question 52.
What stimulates pituitary to release the hormone responsible for parturition? Name the hormone. (All India 2011)
Answer:
Foetal-ejection reflex stimulates pituitary to release the hormone responsible for parturition. This hormone is oxytocin.

Question 53.
Why is breastfeeding recommended during the initial period of an infant’s growth? Give reasons. (Delhi 2016)
Or
(i) Why is mother’s milk considered very essential for the healthy growth of infants?
(ii) What is the milk called that is produced in the initial days of lactation? (All India 2016)
Answer:
The first milk which comes out from the mother’s mammary glands just after childbirth is called colostrum. This is a thin, yellowish, opalescent fluid.

It is rich in proteins and energy along with several antibodies (IgA) that provide passive immunity to the newborn. This immunity helps the newborn to attain resistance to many infections, but it is low in fat and iron content. Hence, breastfeeding during the initial period of infant growth is highly recommended by the doctors for bringing up a disease-free, healthy baby.

Question 54.
Where does fertilisation occur in humans? Explain the events that occur during this process. (All India 2014)
Or
Explain the events that occur during fertilisation of an ovum in humans. How is it that only one sperm enters the ovum? (All India 2014C)
Answer:
In humans, the fertilisation of ovum takes place in ampullary-isthmic junction of the Fallopian tube.
The events that occur during the process of fertilisation are

• The sperm reaches the junction of ampulla and isthmus and comes in contact with zona pellucida layer of ovum.
• Acrosome of sperm head releases sperm lysin enzyme that dissolves corona radiata and digests zona pellucida layer to enter the cytoplasm of ovum.
• After the entry of sperm into the ovum, zona pellucida induces changes in membrane and blocks the entry of additional sperms
• Entry of sperm stimulates the secondary oocyte to complete its suspended second meiotic division thus, producing haploid egg or ovum and second polar body.
• Nucleus of sperm and of ovum fuse to form a diploid zygote.

Question 55.
When and where do chorionic villi appear in humans? State their function. (Delhi 2013)
Answer:
After implantation in the uterus, finger-like projections called chorionic villi appear on the trophoblast of blastocyst. It interdigitates with uterine tissue to form placenta that transports oxygen and nutrients to foetus and remove waste products and carbon dioxide produced by the foetus.

Question 56.
(i) Where do the signals for parturition originate in humans?
(ii) Why is it important to feed the newborn babies on colostrum? (All India 2012)
Answer:
(i) The signals for parturition originate from the fully developed foetus and the placenta, which induces mild uterine contraction called foetal-ejection reflex.
(ii) Colostrum contains necessary antibodies (IgA) that provide protection against disease to infants.

Question 57.
Explain the function of umbilical cord. (All India 2012)
Answer:
The placenta is connected to developing embryo through an umbilical cord, which helps in the transport of substances like oxygen, nutrients, etc. to and from the developing embryo. Blood in umbilical cord can be collected during delivery and stored to treat various diseases in future because it is rich in stem cells.

Question 58.
Mention the number of cells in the following stages.

Answer:
(i) 1 (ii) 16 (iii) 64 (2)

Question 59.
Name the hormones produced only during pregnancy in human female. Mention their source organs. (Foreign 2011)
Answer:
The hormones produced only during pregnancy in human female are human Chorionic and placenta Gonadotropin (hCG), human Placental Lactogen (hPL) and relaxin.
The source of hCG and hPL is placenta and that of relaxin is ovary and placenta.

Question 60.
Why is parturition called a neuroendocrine mechanism? (All India 2011C)
Answer:
Process of parturition is induced by both neural system and endocrine system therefore, it is called a neuroendocrine mechanism.
Vigorous contraction of the uterus at the end of pregnancy causes expulsion/delivery of the foetus. The process of delivery of foetus (childbirth) is called parturition.
Relaxin hormone is secreted by the ovary and placenta to facilitate parturition by softening the connective tissue of symphysis pubica. Foetus and the placenta induce mild uterine contractions called foetal-ejection reflex.

This initiates the release of oxytocin hormone from the posterior pituitary. Oxytocin acts on uterine muscles to cause stronger contractions, which in turn stimulate further secretion of oxytocin.
This causes more stronger contractions leading to the expulsion of the baby out of the uterus, through the birth canal. After the baby is delivered, the placenta is also expelled out of the uterus.

Question 61.
(i) Explain the process of fertilisation in human.
(ii) Name the embryonic stage that gets implanted in human females. Explain the process of implantation. (Delhi 2019)
Answer:
(i) Refer to Answer No. 23 (i). (3)
(ii) Blastocyst gets implanted in the uterine wall.
Process of implantation, Refer to Answer No. 23 (i).

Question 62.
The given diagram shows a part of the human female reproductive system.

(i) Name the gamete cells that would be present in X’ if taken from a newborn baby.
(ii) Name ‘Y’ and write its function.
(iii) Name ‘Z’ and write the events that take place here. (All India 2015C)
Answer:
(i) Primary oocytes are present in A, i.e. ovary of a newborn baby.
(ii) Y is fimbriae. It helps in the collection of ovum from ovary after ovulation.
(iii) Z is the ampullary-isthmic junction. It is the site of fertilisation in humans.

Question 63.
Describe the process of parturition in human. (Delhi 2015)
Answer:
Refer to Answer No. 16

Question 64.
(i) How is placenta formed in human female?
(ii) Name any two hormones which are secreted by it and are also present in a non-pregnant woman. (Foreign 2014)
Answer:
(i) After implantation of blastocyst, the finger-like projections called chorionic villi, appear on the trophoblast. They get surrounded by uterine tissue and maternal blood which become interdigitated with each other to form placenta.
(ii) The two hormones secreted by placenta that are also present in a non-pregnant woman are oestrogen and progesterone.

Question 65.
Given below is the diagram of a human ovum surrounded by a few sperms. Study the diagram and answer the following questions.

(i) Which one of the sperms would reach the ovum earlier?
(ii) Identify ‘D and ‘E. Mention the role of ‘E.
(iii) Mention what helps the entry of sperm into the ovum and write the changes occurring in the ovum during the process.
(iv) Name the specific region in the female reproductive system where the event represented in the diagram takes place. (All India 2019)
Answer:
(i) The sperm ‘A’ would reach the ovum earlier.
(ii) D-Cororia radiata, E-Zona pellucida-During fertilisation, a sperm comes in contact with zona pellucida layer of the ovum and induces changes in the membrane that block the entry of additional sperms. Hence, E helps to prevent polyspermy.
(iii) The hydrolytic secretions of the acrosome help the sperm to enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane. This induces the completion of the meiotic division of the secondary oocyte.
(iv) These events take place in the ampullary-isthmic junction of the Fallopian tube of uterus.

Question 66.
(i) Draw a diagram of the adult human female reproductive system and label the different
(a) parts of Fallopian tube
(b) layers of uterus wall
(ii) Explain the events during fertilisation of an ovum in humans.
Answer:

(ii) Refer to Answer No. 23 (i).

Question 67.
Briefly explain the events of fertilisation and implantation in an adult human female.
(ii) Comment on the role of placenta as an endocrine gland. (Delhi 2016)
Or
Explain the process of fertilisation and implantation in human. (Foreign 2015)
Answer:
(i) Fertilisation is the process of fusion of a sperm with an ovum.

• The motile sperms move through the cervix’, enter the uterus and reach the junction of the isthmus and ampulla (ampullary-isthmic junction) of the Fallopian tube.
• The ovum released from the ovary also reaches the ampullary-isthmic junction where fertilisation takes place.
• Fertilisation can only occur if the ovum and sperms are transported simultaneously to this junction. This explains why all copulations do not lead to fertilisation and pregnancy.
• The sperm comes in contact with the zona pellucida of the ovum and releases sperm lysins which induces changes in the egg membrane. It blocks the entry of the other sperms into the egg. Thus, it ensures that only one sperm can fertilise an ovum.
• The secretions of the acrosome help the sperm to enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane.
• This induces the completion of meiotic division of the secondary oocyte. The secondary meiotic division results in the formation of a secondary polar body and a haploid ovum (ootid).
• The haploid nucleus of the sperm and that of ovum fuse together to form a diploid zygote.

Implantation: The mitotic division starts as the zygote moves through the isthmus of the oviduct towards the uterus called cleavage thus, forming 2, 4, 8, 16 daughter cells called blastomeres.

• The embryo with 8-16 blastomeres is called morula. But, it is not larger than a zygote.
• The morula continues to divide and transforms into blastocyst as it moves further into the uterus.
• The blastomeres in the blastocyst gets arranged into an outer layer called trophoblast and the inner group of cells attached to trophoblast called the inner cell mass.
• The trophoblast layer then gets attached to the endometrium and the inner cell mass differentiates into the embryo. After attachment, the uterine cells divide rapidly and cover the blastocyst.

As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy.

(ii) Placenta also acts as an endocrine tissue and produces several hormones like human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL) oestrogen, progesterone, etc.
The hCG stimulates and maintains corpus luteum to secrete progesterone. hPL stimulates the growth of mammary glands during pregnancy. Relaxin helps in parturition by softening the connective tissue of pubic symphysis.

Question 68.
During the reproductive cycle of a human female when, where and how does placenta develop? What is the function of placenta during pregnancy and embryo development? (Delhi 2015)
Answer:
Placenta is an organ that connects the developing foetus to the uterine wall of mother for supporting pregnancy.

After implantation, finger-like projections appear on the trophoblast called chorionic villi, which are surrounded by the uterine tissue and maternal blood. The chorionic villi and the uterine tissue become interdigitated with each other and jointly form a structural and functional unit between foetus and maternal body, i.e. placenta.

Functions of Placenta:

• It facilitates the supply of oxygen and nutrients to the developing embryo.
• It also facilitates the removal of carbon dioxide and waste materials produced by the foetus.
• The placenta is connected to the developing embryo through the umbilical cord, which helps in the transport of substances to and from the developing embryo.
• Placenta also acts as an endocrine tissue and produces several hormones like human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL), oestrogen, progesterone, etc.

The production of these hormones is necessary to support foetal growth, metabolic changes in the mother and maintenance of pregnancy.

Question 69.
(i) Explain the events taking place at the time of fertilisation of an ovum in a human female.
(ii) Trace the development of the zygote up to its implantation in the uterus.
(iii) Name-and draw a labelled sectional view of the embryonic stage that gets implanted. (Delhi 2010)
Answer:
(i) Refer to Answer No. 23 (i).
(ii) Refer to Answer No. 23 (i).
(iii) Blastocyst is the stage implanted in the uterus.

Question 70.
Describe the post-zygotic events leading to implantation and placenta formation in humans. Mention any two functions of placenta. (All India 2010)
Or
ExplainExplain the process of fertilisation in human female and trace the post-fertilisation events in a sequential order up to implantation of the embryo. (Foreign 2010)
Answer:
Post-zygotic events leading to implantation and placenta formation.
Refer to Answer No. 23 (i).
Events leading to formation of placenta and functions of placenta-Refer to Answer No. 24.

Question 71.
(i) Mention the event that induces the completion of the meiotic division of the secondary oocyte.
(ii) Trace the journey of the ovum from . the ovary, its fertilisation and further development until the implantation of the embryo. (All India 2010C)
Answer:
(i) Secondary oocyte completes meiosis-II only
when a sperm enters into its cytoplasm. It forms a larger cell, the ootid and a small cell, the second polar body.

(ii) (a) The secondary oocyte is released by the rupture of the Graafian follicle in the process called ovulation.
(b) It is moved into the Fallopian tube with the help of fimbriae.
(c) It reaches the ampullary-isthmic junction of the Fallopian tube where fertilisation takes place.
(d) After fertilisation, cleavage starts in the zygote.
(e) Cleavage results in the formation of 2, 4, 8 and 16 daughter cells called blastomeres. The embryo with 8-16 blastomeres is a solid spherical structure called morula.
(f) Morula continues to divide and blastomeres rearrange themselves as it moves further into the uterus.
(g) As a result, blastocyst is formed, which contains trophoblast (outer layer) and inner cell mass.
(h) The trophoblast attaches to endometrium and blastocyst gets embedded in it (implantation).

Question 72.
Draw a diagrammatic sectional view of the female reproductive system of human and label the parts.
(i) Where does the secondary oocyte develop?
(ii) Which help in the collection of ovum after ovulation?
(iii) Where does fertilisation occur?
(iv) Where does implantation of embryo occur? (Delhi 2013)
Answer:
For diagram, refer to the figure 3.2 on page no. 48.
(i) The secondary oocyte develops in ovary.
(ii) Fimbriae help in the collection of ovum after ovulation.
(iii) In the junction of ampulla-isthmus, the ovum gets fertilised.
(iv) Implantation of embryo occurs in the cavity of uterus.

Question 73.
Write the function of each one of the following
(i) Fimbriae (oviducal)
(ii) Coleoptile
(iii) Oxytocin (Delhi 2012)
Answer:
(i) Fimbriae are the feathery finger-like projections present at the end of Fallopian tubes. They help to collect the ovum after its release from the ovary during ovulation.
(ii) Coleoptile is a conical sheath present in the monocot seeds and its function is to protect the developing plumule.
(iii) Oxytocin is a hormone secreted by the posterior pituitary that stimulates the contraction of uterine muscles during child birth (parturition).

Question 74.
Write the function of each of the following
(i) Middle piece in human sperm
(ii) Tapetum in anthers
(iii) Luteinising hormone in human males (Delhi 2012)
Answer:
(i) Middle piece in human sperm contains several mitochondria which produce energy for the motility of the sperm.
(ii) Tapetum in anthers It is the innermost layer of the anther. The main function of tapetum is to provide nourishment to the developing pollen grains.
(iii) Luteinising hormone in human males stimulates the Leydig cells to produce testosterone which is necessary to complete the process of spermatogenesis.

Question 75.
Write the function of each of the following
(i) Seminal vesicle
(ii) Seutellum
(iii) Acrosome of human sperm (Delhi 2012)
Answer:
(i) Seminal vesicle During spermatogenesis, it secretes an alkaline fluid rich in fructose and prostaglandins. High level of fructose provides energy to sperms.
(ii) Scutellum It is a tissue present in seed to absorb food from the adjacent endosperm and develops into growing embryo.
(iii) Acrosome of human sperm It is the cap-like structure that is present at the tip of the sperm (male gamete) in head region. The acrosome contains hydrolytic enzyme, which helps the sperm to enter into the cytoplasm of the ovum and thus, helps in fertilisation.

Question 76.
Give reasons for the following
(i) The human testes are located outside the abdominal cavity.
(ii) Some organisms like honeybees are called parthenogenetic animals. (All India 2012)
Answer:
(i) Testes are located outside the abdominal cavity within a pouch called scrotum. It maintains low temperature of the testes (2-2.5°C lower than normal body temperature) required for spermatogenesis.
(ii) In honeybees, the drone bees are developed parthenogenetically from an unfertilised egg. Since, they develop from unfertilised diploid eggs (and do not undergo fertilisation), they are called parthenogenetic animals.

Question 77.
(i) Explain the following phases in the menstrual cycle of a human female.
(a) Menstrual phase
(b) Follicular phase
(c) Luteal phase
(ii) A proper understanding of menstrual cycle can help immensely in family planning. Do you agree with the statement? Provide reasons for your answer. (All India 2017)
Answer:
(i) (a) Menstrual phase The menstrual cycle starts with menstrual phase. It lasts for about 3-5 days. The menstrual flow results due to the breakdown of endometrial lining of the uterus and its blood vessels that gome out through vagina.
(b) Follicular phase It lasts till about 13th day of menstrual cycle. In this phase, the primary follicles in the ovary grow to become a fully mature Graafian follicle. The secretion of gonadotropins (LH and FSH) from anterior pituitary increases gradually during the follicular phase. They stimulate follicular development as well as secretion of oestrogen by the growing follicles.
(c) Luteal phase This phase lasts for about 10-14 days. In this phase, the ruptured Graafian follicle transforms into corpus luteum. It secretes large amount of progesterone which is essential to maintain endometrium.

(ii) Yes, a proper understanding of menstrual cycle can help in family planning as this knowledge can be used to avoid the meeting of sperms and ovum. This is known as periodic abstinence or rhythm method of birth control, i.e. temporary avoidance of sex. In this method, a couple can avoid or abstain the coitus from day 10 to 17 of the menstrual cycle because ovulation occurs during this period. The chances of fertilisation are very high during this period.

Question 78.

(i) One of the sperms is observed to penetrate ‘A’ of the ovum, as shown in the above diagram. Name ‘A’.
(ii) How is the sperm able to do so?
(iii) Where exactly in the Fallopian tube does this occur?
(iv) Explain the events thereafter up to morula stage. (All India 2013C)
Answer:
(i) A is zona pellucida.
(ii) Sperm is able to penetrate the egg with the help of sperm lysins contained in acrosome.
(iii) Ampullary-isthmic junction.
(iv) As soon as the sperm enters the egg, meiotic division of secondary oocyte gets completed. As a result a second polar body and a haploi ovum are formed. Then, haploid sperm and ovum fuse together to form a diploid zygote which undergoes cleavage via mitotic division. At 8-16 cell stage, the embryo is called morula.

Question 79.
(i) Draw a labelled diagram of the human female reproductive system.
(ii) Enumerate the events in the ovary of a human female during
(a) follicular phase
(b) luteal phase of menstrual cycle (Delhi 2012)
Answer:
(i) Refer to figure 3.2 on page no. 49.
(ii) (a) Follicular phase The primary follicle in the ovary grows to become fully mature Graafian follicle. Endometrium gets thickened and the follicular cells secrete oestrogen.
(b) Luteal phase The remaining part of ruptured Graafian follicle transforms into corpus luteum that secretes progesterone.

Question 80.
(i) Write the specific location and the functions of the following cells in human males
(a) Leydig cells
(b) Sertoli cells
(c) Primary spermatocytes
(ii) Explain the role of any two accessory glands in human male reproductive system. (Delhi 2011)
Answer:
(i) Refer to Answer No. 2 of Topic 1.
Primary spermatocytes are located inside the seminiferous tubule and are involved in the formation of spermatozoa.
(ii) Accessory glands of human male reproductive system are
(a) Prostate and seminal vesicles Their secretions provide a fluid medium for the sperms to swim towards the ovum. They provide nutrition to sperms.
(b) Bulbourethral glands Their secretion helps in the lubrication and neutralising the acidity of urogenital tract.

Question 81.
(i) When does oogenesis begin?
(ii) Differentiate between the location and function of Sertoli cells and Leydig cells. (All India 2010)
Answer:
(i) Oogenesis begins during the embryonic development stage when a million gamete mother cells (oogonia) are formed within each foetal ovary.
(ii) (a) Sertoli cells are located on the inside lining of seminiferous tubule. These cells provide nutrition to the germ cells.
(b) Leydig cells or interstitial cells are located in the regions outside the seminiferous tubule called interstitial spaces. These cells synthesise and secrete testicular hormone called androgens.

Question 82.
Smita was studying in class 6th. Her mother wanted to tell her daughter about the menstrual cycle. Was she correct in doing so?
Answer:
Smita’s mother was correct as it is necessary to inform girls about the onset of menstrual cycle, so that they do not misunderstand it. It shows the care and responsibility cf mother towards her daughter.

Question 83.
Charu got pregnant with her first baby. She asked her husband (who is a doctor) to know about the changes which would take place in her body during pregnancy.
(i) What is implantation?
(ii) What values does the Charu’s husband show?
Answer:
(i) The implantation is the embedding of the blastocyst in the endometrium of the uterus.
(ii) Being a doctor, Charu’s husband shows positive attitude towards her wife’s and he should provide step by step explantation for her queries.

Question 84.
Rita was 9 months pregnant, she wanted to know that how the foetus gets delivered after the human pregnancy period. So, she asked her gynaecologist to explain it to her, so that she can mentally prepared for her delivery. Her gynaecologist explained to her about the process in a very precise manner.
(i) What is the average duration of human pregnancy known as?
(ii) What is the term given to the process of delivery of foetus?
(iii) What are the values exhibited by Rita’s gynaecologist?
Answer:
(i) The average duration of human pregnancy is called as gestation period. It is approximately 270 days in humans.
(ii) The term given to the process of delivery of foetus is called parturition.
(iii) Rita’s gynaecologist has a scientific temper and patience to deal with her patients.

The post Human Reproduction Class 12 Important Questions and Answers Biology Chapter 3 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 4 Reproductive Health to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 4 Important Questions, Reproductive Health Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 4 Important Questions Reproductive Health

Question 1.
Our government has intentionally imposed strict conditions for MTP in our country. Justify giving a reason. (Delhi 2017)
Answer:
MTP or induced abortion is the termination of pregnancy due to certain medical reasons. Government of India legalised MTP in 1971 with strict conditions to avoid its misuse, i.e. to check indiscriminate and illegal female foeticides.

Question 2.
Name an IUD that you would recommend to promote the cervix hostility to sperms. (Delhi 2014C)
Answer:
The hormone releasing IUDs, e.g. progestasert and LNG-20 are recommended to promote the cervix hostility to sperms.

Question 3.
State one reason, why breastfeeding the baby acts as a natural contraceptive for the mother. (Delhi 2014C)
Answer:
Lactation or Breastfeeding the baby delays the onset or return of menstruation and ovulation cycle due to interference of hormone prolactin. Therefore, the chances of conception are nil during this period, i.e. up to six months. Hence, breastfeeding the baby may act as a natural contraceptive (lactational amenorrhea) for mother. (1)

Question 4.
Mention one positive and one negative application of amniocentesis. (Delhi 2010)
Answer:
Applications of amniocentesis are

• Positive,application It can be used to diagnose any chromosomal abnormality or genetic disorder in foetus.
• Negative application It can be used to determine the sex of foetus and lead to female foeticide.

Question 5.
Why is tubectomy considered a contraceptive method? (Foreign 2010)
Answer:
In tubectomy, a small part of Fallopian tube or oviduct is cut and tied up to block the passage of ova from ovary to the site of fertilisation in Fallopian tube. It prevents fertilisation. So, it is considered as a contraceptive method.

Question 6.
Mention the problems that are taken care of by Reproduction and Child Healthcare Programme. (All India 2016)
Answer:
Reproduction and Child Healthcare (RCH) programmes cover wide range of reproduction related areas. They include

• Creating awareness among people about various reproduction related aspects.
• Support for building up a reproductively healthy society by providing increased medical facilities, better postnatal care, better detection and cure of diseases like STDs, etc.

Question 7.
What is amniocentesis? How is it misused? (Delhi 2014C)
Or
Why there is a statutory ban on amniocentesis? Why is this technique so named? (Delhi 2012C)
Or
What is amniocentesis? Why has the government imposed a statutory ban inspite of its importance in medical field? (Foreign 2010)
Answer:
Amniocentesis is a prenatal diagnostic test. It is named so, because it is based on the chromosomal pattern of the cells in the amniotic fluid that surrounds the developing foetus in the womb.
It is misused to detect the sex of pre-born child that leads to female foeticide. Hence, there is statutory ban on amniocentesis.

Question 8.
What do oral pills contain and how do they act as effective contraceptives? (Delhi 2014C)
Answer:
Oral contraceptives or pills contain either progestogens or progesterone-oestrogen combinations. They function as contraceptives by

• inhibiting ovulation.
• inhibiting implantation.
• altering the quality of cervical mucus to prevent the motility of sperms in female reproductive tract.

Question 9.
Why is Cu-T considered a good contraceptive device to space children? (Delhi 2011)
Answer:
Copper-T (Cu-T) is an Intra Uterine Device (IUD) that is inserted by experts and it serves as an effective contraceptive in the following ways

• Increases phagocytosis of sperms within the uterus.
• Copper ions released by Cu-T suppress the motility of sperms and their fertilising ability.

Question 10.
Name an oral pill used as a contraceptive by human females. Explain, how does it prevent pregnancy? (Delhi 2011)
Or
Why is Saheli a well-accepted contraceptive pill? (Foreign 2010)
Answer:
‘Saheli’ is a non-steroidal contraceptive pill used by females to space children.
Saheli inhibits ovulation and implantation. It alters the quality of cervical mucus to prevent the entry of sperms into cervix.

Question 11.
Describe the lactational amenorrhea method of birth control. (All India 2011)
Answer:
Lactation amenorrhea refers to the absence of menstruation during the period of intense lactation following parturition. It is a birth control method because

• ovulation and other events of menstrual cycle are stopped at this time.
• as long as the mother breastfeeds her child, chances of conception are nil because of the suppressed gonadotropin activity. However, this method is generally reliable upto only six months after delivery.

Question 12.
At the time of Independence, the population of India was 350 million, which exploded to over 1 billion by May 2000.
List any two reasons for this rise in population and any two steps taken by the government to check this population explosion. (Foreign 2011)
Answer:
Reasons for rise in population include

• All-round development in various fields and increased health facilities along with better living conditions.
• Reduced maternal and infant mortality rate.

Two major steps taken by the government to check this population growth are

• People are educated and support the idea of small family by using various contraceptive methods.
• There is statutory raising of marriageable age of females to 18 years and of males to 21 years.

Question 13.
How do copper and hormone releasing IUDs act as contraceptives? Explain. (All India 2010)
Answer:
Copper and hormone releasing IUDs act as contraceptives because:

• Copper IUDs (Cu-T, Cu-7) release Cu ions, which suppress sperm motility and the fertilising capacity of sperms.
• Hormone releasing IUDs (progestasert, LNG-20)

(a) inhibit ovulation.
(b) make the cervix unreceptive to sperms.
(c) make the uterus unsuitable for implantation.

(iii) They both increase the phagocytosis of sperms within the uterus.

Question 14.
(i) List any four characteristics of an ideal contraceptive.
(ii) Name two intrauterine contraceptive devices that affect the motility of sperms. (All India 2016)
Answer:
(i) An ideal contraceptive must have the following four characteristics

• It must be safe and user friendly.
• It must be easily available.
• It must be reversible with little or no side effects.
• It must not interfere with the sexual drive, desire or sexual act of the user.

(ii) Cu-T and Cu-7 are two examples of IUDs that affect the motility of sperms.

Question 15.
Name two hormones that are constituents of contraceptive pills. Why do they have high and effective contraceptive value? Name a commonly prescribed non-steroidal oral pill. (All India 2014)
Answer:
The two hormones that are the constituents of oral pills are

• progesterone
• oestrogen

They inhibit ovulation and fertilisation and also modify the quality of cervical mucus to prevent or retard the entry of sperms. Hence, they have high and effective contraceptive value. Saheli is the most commonly prescribed new oral contraceptive pill for females. It contains a non-steroidal preparation called centchroman. It is once a week pill with few side effects and high contraceptive value. It was developed by CDRI (Centra! Drug Research Institute), Lucknow.

Question 16.
If implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, then why is there a statutory ban on amniocentesis? Write the use of this technique and give reason to justify the ban. (All India 2014)
Answer:
Though implementation of better techniques and new strategies is required to provide more efficient care and assistance to people still, there is a statutory ban on amniocentesis. Amniocentesis helps to determine any chromosomal abnormalities or genetic disorders, sex of foetus and foetal infections, etc., by using minute amount of amniotic fluid surrounding the foetus.

This prenatal diagnostic test is particularly useful for those women who are at increased risk or have genetic disorders or chromosomal problems. However, this is also misused to determine the sex of foetus which had ultimately lead to increased female foeticides. Therefore, government has imposed a statutory ban to prevent its further misuse and to balance the unequal sex-ratio prevailing in human population. (3)

Question 17.
A woman has certain queries as listed below, before starting with contraceptive pills. Answer them.
(i) What do contraceptive pills contain and how do they act as contraceptives?
(ii) What schedule should be followed for taking these pills? (Delhi 2014)
Answer:
(i) Refer to Answer No. 8. (2)
(ii) The oral contraceptive pills are to be taken daily for 21 days, preferably within the first five days of menstrual cycle. After the onset of menstruation cycle, i.e. 5-7 days, the process is to be repeated in the same pattern (again for 21 days). This schedule is to be followed till the woman wants to avoid conception. (1)

Question 18.
(i) Name any two copper releasing IUDs.
(ii) Explain, how do they act as effective contraceptives in human females. (All India 2014)
Answer:
(i) The copper releasing IUDs are Cu-T, Cu-7 and multiload.-375.
(ii) Refer to Answer No. 9.

Question 19.
Name and explain the surgical method advised to human males and females as a mean of birth control. Mention its one advantage and one disadvantage. (Foreign 2014)
Answer:
The surgical or sterilisation methods advised to human males and females as effective means of birth control are

• Vasectomy (In males) A sterilisation method in which a small portion of vas deferens is removed or tied up through a cut or incision on scrotum, thus blocking the transport of sperms from the testes to the copulatory organ.
• Tubectomy (In females) A sterilisation method in which small part of Fallopian tube is removed or tied up through incision in abdomen or through vagina. It blocks the passage of ova from ovary to the site of fertilisation.

The advantage of these two sterilisation methods in both human males and females is that it is a very effective method for preventing conception as it blocks the transport of gametes. The disadvantage of this method is that this surgical procedure cannot be reversed, so it is helpful for only those who already have children and do not want to extend their family further.

Question 20.
How do ‘implants’ act as an effective method of contraception in human females? Mention its one advantage over contraceptive pills. (Delhi 2012)
Answer:
Subcutaneous ‘implants’ contain synthetic progesterone and are placed under the skin. They are an effective contraceptive method as they check ovulation and thicken cervical mucus to prevent sperm transport.

‘Subcutaneous implants’ are more advantageous than contraceptive pills as they are long lasting and once implanted, they are effective for up to 5 years.

Question 21.
Your school has been selected by the Department of Education to organise and host an interschool seminar on ‘Reproductive Health Problems and Practices’. However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is ‘too embarrassing’.
Put forth four arguments with apporpriate reasons and explanation to justify the topic to be very essential and timely. (All India 2015)
Answer:
Parents should encourage their children to attend such seminar as they will get right information regarding myths and misconceptions about sex related aspects. Following are the four points to justify this topic to be essential

• Awareness of problems due to uncontrolled population growth, social evils like sex abuse and sex related crimes, etc., need to be created so that children should think and take up necessary steps to prevent them and thereby build up a reproductively healthy society.
• Large group of school students comprises of adolescents who have attained puberty. Therefore, these seminars are necessary to provide medical help and care for reproduction related problems like menstrual problems, infertility, pregnancy, delivery, contraception, abortions, etc.
• Knowledge about Sexually Transmitted Diseases (STDs) is essential as children should be aware that unprotected sex with multiple partners results in the transmission of sex related problems.
• Increasing population is a major problem of India which is directly related with reproductive health. Children should be aware of family planning programmes such as Reproductive and Child Healthcare (RCH) programmes.

Question 22.
Name any two assisted reproductive technologies that help infertile couples to have children. (Delhi 2012C)
Answer:
The assisted reproductive technology methods that can help infertile couple to have children are

• Zygote Intra Fallopian Transfer (ZIFT) and
• Artificial Insemination (AI) Technique.

Question 23.
Expand GIFT and ICSI. (All India 2012C)
Answer:
GIFT – Gamete Intra Fallopian Transfer.
ICSI – Intra Cytoplasmic Sperm Injection.

Question 24.
After a brief medical examination a healthy couple came to know that both of them are unable to produce functional gametes and should look for an ‘ART’ (Assisted Reproductive Technique). Name the ‘ART’ and the procedure involved that you can suggest to them to help them bear a child. (Delhi 2015)
Answer:
The ART that would help the couple to bear a child is IVF (In Vitro Fertilisation) or Test tube baby programme. In this process, ova from wife/donor female and sperms from the husband/donor male are collected and fused to form zygote in the laboratory under same conditions as in the body. This is in vitro fertilisation (fertilisation outside the body).

Zygote or early embryo is transferred into Fallopian tube or uterus for further development. This is called Embryo Transfer (ET). It can be Zygote Intra Fallopian Transfer (ZIFT) or Intra Uterine Transfer (IUT).

Question 25.
An infertile couple is advised to adopt test tube baby programme. Describe two principal procedures adopted for such technologies. (Delhi 2015)
Or
Explain the Zygote Intra Fallopian
Transfer Technique (ZIFT). How is Intra Uterine Transfer (IUT) Technique different from it? (All India 2010)
Answer:
ZIFT (Zygote Intra Fallopian Transfer) is the technique in which zygote or early embryo with up to 8 blastomeres is transferred into the Fallopian tube of female.
On the other hand in IUT, embryo with more than 8 blastomeres is transferred into the uterus.
These are the two principal procedures adopted for test tube baby programme.

Question 26.
A childless couple has agreed for a test tube baby programme. List only the basic steps the procedure would involve to conceive the baby. (Delhi 2015C)
Or
(i) Give any two reasons for infertility among young couple.
(ii) Test tube baby programme is a boon to such couples. Explain the steps followed in the procedure. (All India 2010C)
Answer:
(i) The reasons of infertility in young people can be physical, congenital diseases, use of drugs, immunological or even psychological factors.
(ii) In test tube programme,
(a) Ova from the wife or a donor female and the sperms from the husband or a donor male are allowed to fuse under simulated conditions in the laboratory. It is called in vitro fertilisation.
(b) Embryo is then transferred into the uterus or Fallopian tube for further development.

The process of embryo transfer is done in following ways Zygote or embryo up to 8 blastomeres is 1 transferred into Fallopian tube (ZIFT). Embryo with more than 8 blastomeres is transferred into uterus (IUT).

Question 27.
‘Intra Cytoplasmic Sperm Injection (ICSI)’ and ‘Gamete Intra Fallopian Transfer (GIFT)’ are two assisted reproductive technologies. How is one different from the other? All (India 2014C)
Answer:
In Gamete Intra Fallopian Transfer (GIFT), the ovum from a healthy donor female is transferred to a female, who cannot produce ova. However, she can provide suitable environment for fertilisation and embryo development (in vivo fertilisation).

In Intra Cytoplasmic Sperm Injection (ICSI), the fertilisation is done in vitro by injecting sperms directly into the ovum from a donor female, under simulated conditions. The embryo is thus, formed in laboratory and which is later transferred to the uterus or Fallopian tube for further development.

Question 28.
Why is ZIFT a boon to childless couples? Explain the procedure. (Delhi 2013C)
Answer:
ZIFT (Zygote Intra Fallopian Transfer) is a boon to childless couples as it helps them to become parents. In this technique, ova from wife/donor female and sperms from husband/donor male are fused to form zygote in laboratory. Zygote is allowed to divide up to 8 blastomeres stage and it is at this stage, a zygote or early embryo is transferred into the Fallopian tube. Implantation takes place in the uterus where further development takes place.

Question 29.
State any four methods to overcome infertility in human couples. (Delhi 2011C)
Answer:
Following are the four methods to overcome infertility problems in human couples

• Test tube baby programme In this method, the fusion of ovum and sperm is done outside the body of a woman (in vitro fertilisation) to form zygote which divides to form embryo. The embryo is then implanted in the uterus where it develops into a foetus and then into the child.
• Intra Cytoplasmic Sperm Injection (ICSI) In this technique, embryo is formed in the laboratory by directly injecting the sperm into the ovum followed by embryo transfer.
• Artificial Insemination Technique (AIT) Semen (containing sperms) from husband or donor is artificially introduced into the vagina hr uterus (IUI).
• Gamete Intra Fallopian Transfer (GIFT) Sperm and unfertilised ova are transferred into the Fallopian tube of the female and they are allowed to fuse naturally.

Question 30.
A couple where both husband and wife are producing functional gametes, but the wife is still unable to conceive, is seeking medical aid. Describe any one method that you can suggest to this couple to become happy parents. (All India 2014)
Answer:
In case, if both husband and wife are producing functional gametes, but wife is not able to conceive, the IVF technique can be employed to bless them with child.
Method of in vitro fertilisation is given below

• Gametes from both husband and wife are collected, i.e. sperm and ova. These are fused to form zygote under laboratory conditions. As the fertilisation takes place outside the female body, it is referred to as in vitro fertilisation.
• The zygote or embryo is then either transferred to Fallopian tubes (if up to 8 blastomeres), i.e. ZIFT or to the uterus (more than 8 blastomeres), i.e. IUT.

Question 31.
Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile couple. (All India 2013)
Answer:
Refer to Answer No. 3 and 4.

Question 32.
(i) Explain one application of each one of the following (Delhi 2019)
(a) Amniocentesis
(b) Lactational amenorrhea
(c) ZIFT
(ii) Prepare a poster for the school programme depicting the objectives of ‘Reproductive and Child Healthcare Programme’.
Answer:
(i) (a) Amniocentesis The benefits of amniocentesis include the diagnosis of chromosomal abnormalities and developmental disorders of foetus.
(b) Lactational amenorrhea It is the absence of menstruation during the period of intense lactation following parturition. Because ovulation does not occur in this period, the chances of conception are nill.
(c) ZIFT (Zygote Intra Fallopian Transfer) is related to embryo transfer in the test tube programme. In this technique, the zygote or embryo up to 8 blastomeres is transferred into the Fallopian tube.

Question 33.
Reproductive and Child Healthcare (RCH) programmes are currently in operation. One of the major tasks of these programmes is to create awareness amongst people about the wide range of reproduction related aspects. As this is important and essential for building a reproductively healthy society.
(i) ‘Providing sex education in schools is one of the way to meet this goal’. Give four points in support of your opinion regarding this statement.
(ii) List any two ‘indicators’ that indicate a reproductively healthy society. (Delhi 2016)
Answer:
(i) (a) Introducing sex education in schools is a good step for providing useful information to the adolescents, so as to discourage them from believing in myths and misconceptions about sex related issues.
(b) Better awareness about sex related matters.
(c) Better detection and cure of STDs.
(d) Awareness of problems due to uncontrolled population growth.

(ii) (a) Increased number of medically assisted deliveries and better postnatal care leading to decreased maternal and infant mortality rates.
(b) Awareness of problems due to uncontrolled population growth, social evils like sex abuse and sex related crimes, etc., should be created to enable people to think and take necessary steps to prevent them.

Question 34.
A large number of married couples in the world are childless. It is shocking to know that in India, the female partner is often blamed for the couple being childless.
(i) Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as a biology student can promote to check this social evil.
(ii) State any two reasons responsible for the cause of infertility.
(iii) Suggest a technique that can help the couple to have a child where the problem is with the male partner. (All India 2016)
Answer:
(i) There is a common myth prevailing in our society that inability of a couple to have child is due to the infertility of female partner only, it is because female carries the child in her womb. Being a biology student we should create awareness among people that both male and female partners equally contribute for the birthy of a child. This is because baby is formed from zygote that is formed by the fusion of both male and female gametes. Hence, infertility in either male or female can be responsible for the failure of conception. Infertility in both males and females can be easily cured as there are so many specialised infertility clinics which provide treatments to childless couples. In case treatment is not possible, the couples can be assisted to have children through certain special techniques called the assisted reproductive technologies.

(ii) Causes of infertility could be

• Sexually transmitted diseases both in males and females.
• Some physiological problems in females/males, so gametes (sperm/ova) are not produced.

(iii) IVF (In Vitro Fertilisation) and Artificial Insemination (AI) can be done if the sperm count of male is low.

The post Reproductive Health Class 12 Important Questions and Answers Biology Chapter 4 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 15 Biodiversity and Conservation to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 15 Important Questions, Biodiversity and Conservation Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 15 Important Questions Biodiversity and Conservation

Question 1.
Identify A and B in the figure given below representing proportionate number of major vertebrate taxa. (Delhi 2014)

Answer:
In the mentioned diagram, among the major vertebrate taxa
A-represents mammals.
B-represents ambhibians.

Question 2.
Write the level of biodiversity represented by a mangroves. Give another examples falling in the same level. (Delhi 2014C)
Answer:
The mangroves represent biodiversity at ecological level. Other examples of ecological diversity are deserts, rainforests, coral reefs, etc.

Question 3.
Name the type of biodiversity represented by the following
(i) 1000 varieties of mangoes in India.
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing in different regions of “Himalayas. (All India 2013)
Answer:
(i) Genetic diversity
(ii) Genetic diversity.

Question 4.
Name the type of biodiversity represented by the following
(i) 50000 different strains of rice in India,
(ii) Estuaries and alpine meadows in India. (Delhi 2013)
Answer:
(i) Genetic diversity
(ii) Ecological diversity.

Question 5.
Why is tropical environment able to support greater species diversity? (Delhi 2011C)
Answer:
Tropical latitudes have less seasonal variations and constant environment which promote niche and greater species diversity.

Question 6.
Eichhornia crassipes is an alien hydrophyte introduced in India. Mention the problem posed by this plant. (All India 2010C)
Answer:
Water hyacinth (Eichhomia) introduced in India is threatening the existing aquatic life in ponds and lakes, etc., as it clogs the stagnant waterbodies very fast, thus, the native species are threatened.

Question 7.
The Amazon rainforest is referred to as the lungs of planet. Mention any one human activity which causes loss of biodiversity in this region. (All India 2010C)
Answer:
Human activities causing loss of biodiversity in Amazon rainforest are as follows

• Many plants are cut in Amazon rainforest for cultivation of soybeans.
• Forests are converted to grasslands for raising beef cattle.

Question 8.
India has more than 50000 strains of rice. Mention the level of biodiversity it represents. (Delhi 2010; Outside Delhi 2010)
Answer:
India has more than 50000 strains of rice which shows the genetic level of biodivesity of rice.

Question 9.
How is an alien species invasion considered as one of the cause of biodiversity loss? Support you answer with the help of an example. (All India 2019)
Or
Alien species invasions is one of the causes of biodiversity loss. Explain with the help of an example. (All India 2012C)
Or
With the help of an example, explain how alien species invasion causes biodiversity loss (Delhi 2011)
Or
Alien species are a threat to native species. Justify taking examples of an animal and a plant alien species. (All India 2010)
Or
Sometimes alien species affect the indigenous organisms leading to their extinction. Substantiate this statement with the help of any two examples. (Delhi 2010C)
Answer:
Alien species become invasive, compete with the native species and cause extinction of indigenous species.

• Introduction of Nile perch into Lake Victoria leads to extinction of more than 200 species of cichlid fish in that lake.
• Carrot grass (Parthenium) and Lantana introduced in our country have become invasive and cause environmental damage. They pose a threat to the native species of plants in our forests.

Question 10.
Substantiate with the help of one example that in an ecosystem mutualists.
(i) tend to coevolve and
(ii) are also one of the major causes of biodiversity loss. (Delhi 2019)
Answer:
(i) In nature mutualists often co-evolve such as mediterranean orchid Ophrys. Ophrys employs sexual deceit to get pollinated by a species of bee. One petal of flower resembles to female bee. If female bee changes its colour pattern even slightly the success of pollination will be reduced unless orchid flower coevolves to maintain resemblance with female bee.

(ii) Coextinction is the one of the Evil Quartet in which organism with obligatory relationship like plant pollinator mutualism will result in extinction of one partner, if other is eliminated in nature.

Question 11.
Identify the areas labelled i, ii, iii and iv in the pie chart given below representing the biodiversity of plants showing their proportionate number of species of major taxa. (2018C)

Answer:

• Lichens
• Algae
• Fungi
• Mosses

Question 12.
Mention the kind of biodiversity of more than a 1000 varieties of mangoes in India represent. How is it possible? (Delhi 2016)
Answer:
The different varieties of mangoes in India represent genetic diversity.
It occurs because India lies within tropical latitudes where, the environment is constant and predictable. Also, more solar energy is available which leads to higher productivity.

Question 13.
List four causes of biodiversity loss. (Delhi 2014C)
Answer:
The four causes of biodiversity loss are as follows

• Habitat loss and fragmentation
• Overexploitation
• Alien species invasions
• Coextinctions.

Question 14.
What is meant by alien species invasion? Name one plant and one animal alien species that are a threat to our Indian native species. (All India 2013)
Answer:
Intentional or chance introduction of exotic species into new Islands or countries by man is called alien species invasion.

For example, Nile perch introduced into Lake Victoria in East Africa caused loss of more than 200 species of cichlid fish. Plant alien species-Lantana camara and animal alien species-Clarias gariepinus have posed threat to our Indian native species.

Question 15.
“Stability of a community depends on its species richness.’ Write how did David Tilman show this experimentally. (Outside Delhi 2013)
Answer:
The stability of a community depends on species richness. This was confirmed by David Tilman. Through his ecology experiments Tilman showed

• The plots with more species showed less year-to-year variations in total biomass.
• Plots with increased diversity showed higher productivity.

Question 16.
Where would you expect more species biodiversity in tropics or in polar regions? Give reasons in support of your answer. (Outside Delhi 2013)
Or
Giving two reasons explain why there is more species biodiversity in tropical latitudes than in temperate ones. (All India 2010)
Answer:
Biodiversity is more in tropical latitudes than in temperate or in polar regions. The reasons are (any two)

• Speciation is a function of time. The temperate regions were subjected to frequent glaciation in the past, while the tropics have remained undisturbed and so had longer time to evolve more species diversity.
• More solar radiation is available in tropical region. This leads directly to more productivity and indirectly to greater species diversity.
• The environment of tropics is less seasonal and relatively more constant and predictable, which encourages niche specialisation and species diversity.

Question 17.
State the use of biodiversity in modern agriculture. (Outside Delhi 2011)
Answer:
The use of biodiversity in agriculture is immense. It is a source of hybrids, GM plants, biopesticides, organic farming, biofertiliser, improved varieties of plants, disease resistant plants. It also promotes sustainable management of agricultural resources, conservation and farming of all wild and native varieties of plants, etc.

Question 18.
Justify with the help of an example where a deliberate attempt by humans has led to the extinction of a particular species. (Delhi 2011)
Answer:
Overexploitation of natural resources or over hunting of animals has led to extinction of particular species, e.g. Steller’s sea cow and passenger pigeon.
Other examples of human driven extinction include Dodo bird due to habitat fragmentation.

Question 19.
The given graph shows species-area relationship. Write the equation of the curve A and B explain. (All India 2011)

Answer:
The equation for the curve A is S = C4^Z where, S- Species richness,
A – Area,
C – Y-intercept,
Z – Slope of line (regression coefficient)
(i) Alexander von Humboldt observed that within a region, species richness increased with increasing explored area, but only up to a limit.
(ii) The relation between species richness and area for a wide variety of taxa like angiosperms, birds, fishes, etc., turns out to be a rectangular hyperbola.

Question 20.
Explain, taking one example, the effect of coextinction on biodiversity. (Delhi 2011c)
Answer:
Coextinction means that when a species is going to become extinct, the plant and animal species associated with it in an obligatory relationship also gets extinct. For example, coevolved plant-pollinator mutualism as in case of Pronuba yuccasella and Yucca. If one partner is on the verge of extinction, the other partner will also get extinct.

Question 21.
How does overexploitation of beneficial species affect biodiversity? Explain with the help of one example. (Delhi 2011C)
Answer:
Human beings when overexploite any plant/animal species, it affects their population size. The marine fish population around the world are declining due to overharvesting. It results in endangering the continued existence of some commercially important species. In the last 500 years, extinction of Stellar’s sea cow and passenger pigeon were due to the overexploitation of these animals by humans.

Question 22.
In the biosphere, immense biological diversity exists at all levels of biological organisation. Explain any two levels of biodiversity. (All India 2010)
Answer:
Levels of Biodiversity in Biosphere

• Genetic diversity It refers to the diversity of genes within a species. For example, there are more than 50000 genetically different strains of rice in India.
• Species diversity It refers to the number of different species within a given region. For example. Western Ghats have a greater amphibian species diversity than Eastern Ghats.

Question 23.
List the features that make a stable biological community. (All India 2010)
Answer:
Features of Stable Biological Community

• It should not show much variation in productivity from year-to-year.
• It should be resistant or resilient to occasional disturbances both natural and man-made.
• It must be resistant to invasions of alien species. (Any two)

Question 24.
Write any two hypothesis put forth by ecologists explaining the existence of greater biodiversity in tropical regions than in temperate regions. (Foreign 2010)
Answer:
For two hypothesis put forth by ecologists explaining the existence of greater biodiversity in tropical regions, than in temperates. Refer to Answer No. 16.

Question 25.
Explain any two most important levels of biological organisation showing biodiversity with the help of an example each. (Delhi 2019, 2018)
Or
Explain the levels of biodiversity at genetic, specific and ecological levels with the help of one example each. (Delhi 2016)
Answer:
Genetic diversity A single species shows diversity at the genetic level over its distributional range. For example, the genetic variation shown by the plant Rauwolfia vomitoria growing in the different Himalayan ranges might be due to its potency and concentration of reserpine.

Species diversity It is the diversity at species level and it is affected by species richness and species evenness. For example, the Western Ghats have a greater amphibian species diversity than the Eastern Ghats.
Ecological diversity It is the diversity at ecological level and is dependent on genetic and species diversity.
At the ecosystem level, India shows diversity with its deserts, rainforests, mangroves, coral reefs and alpine meadows.

Question 26.
(i) India has greater ecosystem diversity than Norway. Do you agree with the statement? Give reasons in support of your answer.
(ii) Write the difference between genetic biodiversity and species biodiversity that exists at all the levels of biological organisation. (2018)
Answer:
(i) India is one of the twelve megadiversity regions of the world with 8.1% of genetic resources of the world. Since, India is placed in the tropical part of the world it exhibits more species diversity than the regions of temperate zones.
India, with much of its land area in the . tropical latitudes, has more than 1200 species of.birds. Greenland, Norway is away from tropics, so it exhibits less biodiversity.

(ii) Differences between genetic and species biodiversity are as follows
Table
Genetic Diversity
It is related to the number of genes and their alleles found in organisms.
It is a trait of a species.
It influences biotic adaptability and distribution of species in diverse habitats.

Species Diversity
It is related to the number and distribution of species found in an area.
It is a trait of community.
It influences interactions and stability of the community.

Question 27.
Name and describe any three causes of biodiversity losses. (Delhi 2017)
Or
Name and explain any two ways that are responsible for the loss of biodiversity. (Foreign 2014)
Causes of biodiversity losses are as follows

• Habitat Loss and Fragmentation : The main cause of extinction of species is the v destruction of their habitat. For example, tropical rainforests once covering more than 14% of the land surface, now cover only 6% of land area.
• Overexploitation of natural resources byhumans results in degradation and extinction of the resources. For example, Steller’s sea cow, passenger pigeon and many marine fishes have become extinct in last 500 years.

Question 28.
Write the importance of species diversity to the ecosystem. Support your answer with the finding of Tilman. (Delhi 2016)
Answer:

• Ecologists for many years, believe that communities with more species are more stable than those with less species. Greater species diversity ensures natural sustainability for all life forms.
• David Tilman’s long term ecosystem experiments using outdoor plots provided answers to stability of a biological community and species richness in that community. Tilman found that plots with more species showed less year-to-year variation in total biomass. He also showed in his experiments that increased diversity contributed to higher productivity.

Question 29.
Since the origin of life on earth, there were five episodes of mass extinction of species.
(i) How is the ‘Sixth extinction’, presently in progress, different from the previous episodes?
(ii) Who is mainly responsible for ‘Sixth Extinction’?
(iii) List any four points that can help to overcome this disaster, outside (Delhi 2014)
Answer:
(i) The current species extinction rate is estimated to be 100-1000 time faster than in the pre-human era.
(ii) All activities performed by human beings for survival and maintenance of their lifestyle.
(iii) Point that can help to overcome this disaster are as follows

• Preventing habitat loss and fragmentation.
• Checking overexploiting.
• Preventing alien species invasion.
• Preventing coextinction.
• Conservation/protection of species.

Question 30.
The following graph shows the species-area relationship. Answer the following questions as directed.
(i) Name the naturalist who studied the kind of relationship shown in the graph. Write the observations made by him.
(ii) Write the situations as discovered by the ecologists when the value of Z (slope of the line) lies between

(a) 0.1 and 0.2
(b) 0.6 and 1.2
What does Z stand for?
(iii) When would the slope of the line B become steeper? (All India 2014)
Answer:
(i) Alexander von Humboldt studied the relationship shown in above graph. He observed that the species richness in an area increased with an increase in exploring area, up to a certain limit only.
(ii) (a) Ecologists have observed that when the value of Z lies between 0.1-0.2 then the Species are considered for a small or average area.
(b) When the value of Z lies between 0.6-1.2, the area considered is very large. Z represents the slope of the line, i.e. regression coefficient.
(iii) The slope of the line B will become steeper when very large areas such as continents are considered for species area relationship.

Question 31.
Explain giving three reasons, why tropics show greatest levels of species diversity? (All India 2014)
Or
List the reasons that account for the greater biological diversity in tropics. (Foreign 2012)
Answer:
For reasons why tropics have greatest species diversity. Refer to Answer No. 16.

Question 32.
Alien species are highly invasive and are a threat to indigenous species. Substantiate this statement with any three examples. (All India 2012)
Answer:
For Allien species invasion. Refer to Answer No. 27 (iii).

Question 33.
Taking one example each of habitat loss and fragmentation, explain how are the two responsible for biodiversity loss. (All India 2012)
Answer:
For fragmentation and habitat loss, Refer to Answer No. 27 (i).

Question 34.
(i) Explain the species-area relationship using the graphical representation given below.

(ii) Explain giving reasons why there is greater biodiversity in tropical regions of the earth. (All India 2012c)
Answer:
(i) Alexender von Humboldt studied species-area relationship. He observed that within a region, the species richness increased with increasing area upto a certain limit.
(ii) Ecologists have discovered that the value of Z lies in the range of 0.1-0.2 regardless of taxonomic group or region. When species-area relationship is considered for a large area like a whole continent, regression coefficient Z or slope of the line becomes steeper with Z-values in the range of 0.6-1.2.

For example, for fruit eating birds and mammals in tropical forests of different continent, the slope is found to be 1.15.

Question 35.
Explain by giving example, how coextinction is one of the causes of loss of biodiversity. List the other three causes also (without description). (foreign 2011)
Answer:
Coextinction is one of the causes of loss of biodiversity as when a species becomes extinct, the plant and,animal species associated with it in an obligatory manner, also become extinct.
For example,

• In plant pollinator mutualism, extinction of one results in the extinction of other.
• If a host fish becomes extinct, the unique parasites depending on it would also become extinct.

The other causes of loss of biodiversity are

• Habitat loss fragmentation
• Overexploitation
• Invasion of alien species.

Question 36.
Explain rivet popper hypothesis. Name the ecologist who proposed it. (Foreign 2011)
Answer:
Rivet Popper Hypothesis

• The hypothesis was proposed by Paul Ehrlich.
• In an airplane (ecosystem), all parts are joined together using thousands of rivets (species).
• If every passenger travelling in it, starts popping a rivet to take home (causing a species to become extinct), it may not affect the flight safety (proper functioning of ecosystem) initially, but as more and more rivets are removed, the plane becomes dangerously weak after some time.
• Further, loss of rivets on the wings. (Key species that drive major ecosystem function) is obviously a more serious threat to flight safety than loss of a few rivets on the seats or windows inside the plane.

Question 37.
Why Western Ghats in India have been declared as biological hotspot? (Delhi 2015)
Answer:
Western Ghats have been declared as biological hotspot due to it is rich biodiversity and ecological condition favoring many species of plants and animals. Many endemic species of amphibians, reptiles and fishes are found in these regions.

Question 38.
Write the importance of cryopreservation in conservation of biodiversity. (Delhi 2011)
Answer:
The importance of cryopreservation in conservation of biodiversity is that gametes of threatened species can be preserved in viable and fertile conditions for long periods by cryopreservation.

Question 39.
Suggest two practices giving one example of each, that help to protect rare or threatened species. (All India 2017)
Answer:
Practices that help to protect rare threatened species are as follows

• In situ (on-site) conservation involves protection of species in their natural habitat. In involves biosphere reserves, national parks, wildlife sanctuaries, sacred groves, etc,
• Ex situ (off-site) conservation involves placing threatened animals and plants in special care units for their protection. Zoological parks, botanical gardens and wildlife safari parks serve this purpose.

Question 40.
Why are sacred groves highly protected? (All India 2016)
Answer:
Sacred groves are small patches of forests with special religious importance in a particular culture.

• These are also mythologically important.
• These are undisturbed forests without any human interventions.
• These are highly protected hence, they include a number of rare, endangered and endemic species. (2)

Question 41.
list any four techniques where the principle of ex situ conservation of biodiversity has been employed. (All India 2015)
Answer:
Four techniques where principle of ex situ conservation of biodiversity has been employed are

• Tissue culture
• Cryopreservation
• Botanical gardens
• Zoological parks.

Question 42.
Why is there a need to conserve biodiversity? (Foreign 2014)
Answer:
The main reasons to conserve biodiversity are as follows

• Narrowly utilitarian reasons These are obvious reasons as human derives multiple economic benefits like food, fibre, medicinal and industrial products, etc.
• Broadly utilitarian reasons Biodiversity plays a major role in providing ecosystem services which cannot be given a price-tag, e.g. oxygen, pollination, aesthetic pleasure, etc.

Question 43.
State the uses of biodiversity in modern agriculture. (All India 2011)
Answer:
Uses of biodiversity in modern agriculture are –

• Humans obtain food, fibres, medicines and many industrial products from plants.
• Wild varieties of plants are used for breeding to obtain disease and pest resistant crops with many desirable traits.
• By exploring molecular, genetic and species level diversity for economically important products, rich biodiversity can be obtained.

Question 44.
Differentiate between in situ and ex situ approaches of conservation of biodiversity. (All India 2011)
Answer:
Differences between in situ and ex situ approaches of conservation of biodiversity are as follows
Table
In situ conservation
This method involves protection of endangered species in their natural habitat.
It helps in recovering populations in the surroundings where they have developed their distinct features.
e.g. national parks, biosphere reserves, wildlife sanctuaries, etc.

Ex situ conservation
It involves placing of threatened animals and plants in special care unit out of their natural habitat for their protection.
It helps in recovering populations or preventing their extinction under stimulated conditions that closely resemble their natural habitats.
e.g. botanical gardens, zoological parks.

Question 45.
Biodiversity must be conserved as it plays an important role in many ecosystem services that nature provides. Explain any two services of the ecosystem. (Delhi 2010)
Answer:
The two ecosystem services are

• Forest ecosystem mitigates droughts and floods and provides oxygen.
• The wildlife helps in pollination of crops, without which fruits/seeds are not produced.

Question 46.
Why certain regions have been declared as biodiversity hotspots by environmentalists of the world? Name any two hotspot regions of India. (Delhi 2010)
Answer:
Certain regions are declared hotspots by the environmentalists, because these regions have very high levels of species richness and high degree of endemism.
Hotspots of India are Western Ghats and Sri Lanka, Himalayas and Indo-Burma.

Question 47.
White Bengal tigers are protected in special settings in zoological parks. Tiger reserves are maintained in Western Ghat.
(i) How do these two approaches differ from each other? Mention the advantages of each one.
(ii) What is the significance of cryopreservation technique? (All India 2010C)
Answer:
(i) Zoological parks are man-made places where endangered species are kept and taken care of by specilised skilled persons. It helps to . prevent species extinction.
Tiger reserves are natural habitats of tigers in which they are protected in natural habitat. It helps in recovering population of the species where they have developed their distinctive features.
(ii) Cryopreservation is ex situ conservation technique in which tissues, organs, embryos, seeds gametes, pollens, etc., are stored at very low temperature of -196°C for future use.

Question 48.
List six advantages of ‘ex situ’ approach to conservation of biodiversity. (Delhi 2019)
Answer:
Ex situ conservation strategies help

• To conserve those animals that have become extinct in wild, but can be maintained in zoological parks.
• To preserve gametes of threatened species in viable condition through cryopreservation.
• To propagate threatened plants via tissue culture.
• To grow plants with recalcitrant seeds in orchards where all possible varieties are maintained.
• To conserve seed of commercially important plants in seed banks.
• To save endangered or threatened plant that needs urgent measure to save it from extinction in botanical gardens.

Question 49.
Explain the ‘ex situ conservation’ of biodiversity. How is the in situ conservation different from it? (2018C)
Answer:
For ex situ conservation, Refer to Answer No. 3 (ii) and for difference Answer No. 8.

Question 50.
In situ conservation can help endangered threatened species. Justify the statement. (Delhi 2017)
Answer:
In situ conservation involves the protection of species in their natural habitats. It helps in the conservation of threatened/endangered species via following means

• Biodiversity hotspots are regions with high levels of species richness and high degree of endemism (i.e. species confined to that region are not found anywhere else). Hotspots cover less than 2% of the earth’s land area.
• Protected areas are ecologically unique and biodiversity rich regions. These are legally protected as biosphere reserves, national parks and sanctuaries.
• Sacred groves are forest areas set aside, all the trees and wildlife within it are venerated . and given total protection. These are religious and cultural places, which are protected.

Question 51.
Why should biodiversity be conserved? Explain giving three reasons. (Outside Delhi 2016C)
Answer:
Biodiversity should be conserved for the following reasons

• The broadly utilitarian argument says that biodiversity plays a major role in many ecosystem services that nature provides.
• The narrow utilitarian argument says that humans derive countless direct economic benefits from nature and products of medicinal importance.
• The ethical argument for conserving biodiversity relates to what we owe to millions of plant, animal and microbe species with whom we share this planet.

Question 52.
Many plant and animal species are on the verge of their extinction because of loss of forest land by indiscriminate use by the humans. As a biology student what method would you suggest along with its advantages that can protect such threatened species from getting extinct. (Delhi 2015)
Answer:
As a biology student, I would suggest ex situ conservation approach for such animals. It involves placing the threatened animals and plants in special care units for their protection. It includes off-site collections (botanical gardens, zoological parks, etc.) and (gene banks, seed banks, tissue culture, etc).

Advantages of ex situ conservation are

• Off-site collection can be used to restock depleted population, reintroduce species in the wild and restore degraded habitats.
• It is useful in maintaining a large number of genotypes in small area, rapid multiplication of endangered species, through tissue culture, etc.

Question 53.
Compare narrowly utilitarian and broadly utilitarian approaches to conserve biodiversity, with the help of suitable examples. (Foreign 2015)
Answer:
Comparison between narrowly utilitarian and broadly utilitarian approaches are as follows
Table
Narrowly utilitarian approach
It includes most of the resources required for our day-to-day life.
e.g. food, oil, clothes, wood and drugs.

Broadly utilitarian approach
It includes most of the ecosystem services provided to us by nature.
e.g. release of O₂ and fixation of CO₂.

Question 54.
There are many animals that have become extinct in the wild but continue to be maintained in zoological parks.
(i) What type of biodiversity conservation is observed in this case?
(ii) Explain any other two ways which help this type of conservation. (Delhi 2014)
Answer:
(i) Zoological parks are an example of ex situ conservation, in which threatened animals are kept in special conditions, away from natural habitat, in order to protect them.
(ii) The other ways of maintaining the endangered animals or species by ex situ conservation are botanical gardens and cryopreservation.

In botanical garderns, plants are grown in special conditions. Threatened and critically endangered plant species are grown here.
In cryopreservation, gametes and seeds of threatened species are preserved in viable condition at the temperature of around -196° C.

Question 55.
The sacred groves of Aravalli Hills and Ooty botanical garden both aim at biodiversity conservation. How do they differ in their approaches? Explain. (All India 2013C)
Answer:
Sacred groves are undisturbed forest patches, surrounded by highly degraded landscapes where not even a “single human activity is allowed, e.g. sacred groves of Aravalli hill is protecting many endemic species.
In botanical gardens, the plants are raised outside their natural habitat under the supervision of skilled persons.

Question 56.
What are the two types of desirable approaches to conserve biodiversity? Explain with examples bringing out the difference between the two types. (Delhi 2012, All India 2012)
Answer:
Conservation of biodiversity meant the preservation of existing variable forms of life in their natural habitat, while deriving sustainable benefits from nature. It is done in situ (on-site) and ex situ (off-site). In situ preservation and conservation always aim to protect organisms in their natural habitats (e.g. sacred groves, biosphere reserves).

Ex situ conservation helps to protect threatened plants and animals. In this method, these organisms are placed outside the natural habitat and are promoted to reproduce.

Question 57.
(i) Why should we conserve biodiversity? How can we do it?
(ii) Explain the importance of biodiversity hotspots and sacred groves. (Delhi 2016)
Answer:
(i) For reasons, Refer to Answer No. 15 and for methods, Refer to Answer No. 18.
(ii) Importance of biodiversity hotspots and sacred groves Biodiversity hotspots are the regions of accelerated habitat loss. These can reduce the ongoing mass extinctions by almost 30%. Sacred groves are important in biodiversity conservation, as these are, naturally protected by the native population without much effort. In Meghalaya, the sacred groves are the last refuges for large number of rare and threatened plants.

Question 58.
(i) Explain the narrow utilitarian, broadly utilitarian and ethical arguments in favour of conservation of biodiversity.
(ii) How is designation of certain areas as hotspots a step towards biodiversity conservation? Name any two hotspots in India. (Delhi 2014C)
Answer:
(i) Refer to Answer No. 15.
(ii) Hotspots are regions exhibiting high degree of endemism and great species richness, therefore designating these areas as ‘biodiversity hotspots’ allows their maximum protection and reduce the ongoing extinction by about 30%. Such hotspot regions in India are Western Ghats and Himalayas.

Question 59.
(i) Why is there a need to conserve biodiversity?
(ii) Name and explain any two ways that are responsible for the loss of biodiversity. (All India 2014)
Answer:
(i) The biodiversity needs to be conserved because of three reasons

• Narrow utilitarian includes most of the resources required for our day-to-day life, e.g. food, oil, clothes, firewood, drugs and medicines, industrial products all are derived from nature, thus needs to be conserved to reap more benefits.
• Broadly utilitarian includes most of the ecosystem services provided to us by nature. Such as release of oxygen and fixation of C02 by photosynthesis in plants, pollination and dispersal of seeds, etc. Therefore, for the continuation of these services biodiversity needs to be conserved.
• Ethical reasons as it becomes our moral duty to take care of all living species in our surroundings irrespective of their economic importance and pass this biological legacy to our future generations.

(ii) The two ways that are responsible for the loss of biodiversity are

• Habitat loss and fragmentation of natural habitats due to the natural reasons or human activities and pollution results in degradation of habitats, thereby threatening the survival of many species concerned.
• Coextinction leads to loss of biodiversity as when a species becomes extinct, the plant and animal species associated with it in obligatory way also become extinct, e.g. when a host organism (fish) becomes extinct, the parasites exclusive to it also become extinct.

Question 60.
(i) Taking one example each of habitat loss and fragmentation, explain how are the two responsible for biodiversity loss.
(ii) Explain two different ways of biodiversity conservation. (Outside Delhi 2012)
Answer:
(i) (a) When large-sized habitats are broken or fragmented due to human settlements, building of roads, digging of canals, etc., animals requiring large territories and some animals with migratory habitats are badly affected.
(b) The Amazon rainforest (called the ‘lungs of the planet’) is being cut and cleared for cultivation of soybean or for conversion into grasslands for raising beef cattle.

(ii) There are two basic approaches for conservation of biodiversity
(a) In situ conservation (on-site conservation)
(b) Ex situ conservation (off-site conservation)

In situ conservation: It is the conservation and protection of biodiversity in its natural habitat.
It helps in recovering population in the surroundings where they have developed their distinctive features.
Example National parks, biosphere reserves, wildlife sanctuaries.

Ex situ conservation: It is the conservation of selected rare plants and animals in places outside their natural habitat.
It helps in recovering population or preventing their extinction under conditions that closely resemble their natural habitats.
Example Botanical gardens, zoological parks, wildlife safari parks, gene banks, etc.

Question 61.
A construction worker was not allowed to cut trees in a village due to protests made by villagers that the trees were sacred to them. Would you support them? What values do you gather from this instance?
Answer:
Yes, I would support them. The sacred groves are protected areas which are of religious and cultural value to native people.
Values shown by villagers are care and concern towards environment and courage.

Question 62.
During a class tour to a botanical garden, Mamta saw that many plant species were present there, which are no more found in the wild. She asked her teacher about how these rare species cultivated and protected in such areas?
(i) Botanical gardens and zoos, etc., represent which mode of biodiversity conservation?
(ii) Mention the factors responsible for the extinction of biodiversity.
(iii) What values do you observe in Mamta?
Answer:
(i) Ex situ mode of conservation, where threatened animals and plants are placed in special care units.
(ii) Overexploitation of resources, fragmentation of habitats and introduction of alien species with native flora and fauna.
(iii) Environment friendly, intelligent, observant and curious.

The post Biodiversity and Conservation Class 12 Important Questions and Answers Biology Chapter 15 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 16 Environmental Issues to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 16 Important Questions, Environmental Issues Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 16 Important Questions Environmental Issues

Question 1.
An electrostatic precipitator in a thermal power plant is not able to generate high voltage of several thousands. Write the ecological implication because of it. (All India 2017)
Answer:
If an electrostatic precipitator is not generating high voltage, it would not be able to release electrons. When air will be passed through this thermal power plant; dust particles would not be able to get collected separately. Hence, it will not be able to control pollution.

Question 2.
Mention two advantages for preferring CNG over diesel as an automobile fuel. (All India 2016, 2015)
Answer:
The two advantages for preferring CNG over diesel as an automobile fuel are as follows

• It bums more effectively leaving no unbumt remains behind.
• It tends to be cheaper than diesel and petrol.

Question 3.
List two advantages of the use of unleaded petrol in automobiles as fuel. (All India 2015)
Answer:
The two advantages of using unleaded petrol in automobiles as fuel are as follows

• There will be no lead pollution in the environment.
• It allows the use of catalytic converter as the exhaust will not harm the catalyst like leaded petrol.

Question 4.
State the causes of accelerated eutrophication. (Delhi 2014)
Answer:
Effluents from industries and domestic sewage and other human activities are the causes of accelerated eutrophication.

Question 5.
Excessive nutrients in a freshwater body cause fish mortality. Give two reasons. (Delhi 2014C)
Answer:
The reasons for fish mortality in a freshwater body due to excessive nutrients are given below.

• Increase in Biological Oxygen Demand (BOD).
• Eutrophication.

Question 6.
Inspite of being non-polluting, why are there great apprehensions in using nuclear energy for generating electricity? (Foreign 2014)
Answer:
Despite being non-polluting, there are great apprehensions in using nuclear energy for generating electricity because of

• danger of accidental leakage.
• lack of safe disposal methods of radioactive waste.

Question 7.
Write the name of the organism that is referred to as the ‘Terror of Bengal.’ (Delhi 2014)
Or
Why is Eichhornia crassipes nick named as Terror of Bengal? (Delhi 2012)
Or
Name the world’s most problematic aquatic weed. What is the nature of the water body in which the weed grow abundantly? (Delhi 2012C)
Or
Eichhornia crassipes is an alien hydrophyte introduced in India. Mention the problem pased by this plant? (All India 2010C)
Answer:
The name of organism that is referred to as ‘Terror of Bengal’ is Eichhornia crassipes. It is because it grows at an alarming rate and spreads on the surface of the eutrophic water body. It is world’s most problematic weed. This cuts out light and also causes an increase in the oxygen demand. Thus, causing the death of fishes and other aquatic organisms.

Question 8.
Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic convertors? (All India 2013, 2012 Foreign 2010)
Answer:
Lead found in petrol corrodes the catalytic rods of platinum, rhodium, in catalytic convertor and decreases its efficiency. Thus, unleaded petrol is recommended for motor vehicles equipped with catalytic convertors.

Question 9.
How do algal blooms affect the life in water bodies? (All India 2011)
Answer:
Algal blooms cause deterioration of the water quality. They reduce the dissolved oxygen content of water and cause mortality (death) of aquatic animals like fish. (1)

Question 10.
Mention the information that the health derive by measuring BOD of a water body. (All India 2010)
Answer:
BOD (Biochemical Oxygen Demand) is the amount of the oxygen required by microorganisms to decompose the organic matter present in one litre of water during a particular time period. The higher BOD value indicates high level of pollution. It directly gives an idea of the health of water body. ID

Question 11.
Write the effective remedy found by Ahmed khan of Bengaluru for the efficient use of the plastic waste generated by big cities. (All India 2019)
Or
How did Ahmad Khan, plastic sac manufacturer from Bengaluru, solve the ever-increasing problem of accumulating plastic waste? (All India 2012)
Or
What is polyblend? Why did the plastic manufactures think of producing it? Write its usefulness? (All India 2011)
Answer:
Ahmad Khan developed a fine powder called polyblend of recycled modified plastic. This mixture is mixed with the bitumen and used to lay roads. It enhanced the bitumen’s water repellant properties and helped to increase road life by a factor of three. By this way, the problems created by plastic waste was solved and plastic manufacturers thought of producing polyblend.

Question 12.
How did a citizen group called Friends of Areata Marsh, Areata, California, USA, help to improve water quality of the marsh land using integrated waste water treatment? Explain in four steps. (2018)
Answer:
A citizen group called ‘Friends of Areata Marsh’ used integrated waste water treatment to improve the water quality of the Areata marsh land. This integrated waste water treatment process involved following steps

• The first step involved conventional sedimentation and filtration of water.
• t was followed by addition of chlorine in the water.
• Then, a series of six connected marshes was developed.
• Appropriate plants, algae, fungi and bacteria were seeded in that area to absorb, neutralise and assimilate the pollutants.

Question 13.
Plenty of algal bloom is observed in a pond in your locality.
(i) Write, what has caused this bloom and how does it affect the quality of water.
(ii) Suggest a preventive measure. (Delhi 2017)
Or
What is an ‘algal bloom’? State, its cause and any two harmful effects. (Outside Delhi 2014C)
Answer:
(i) The presence of large amounts of nutrients in water leads to algal bloom.
Algal bloom affects the quality of water in the following ways

• It causes deterioration of water quality making it unfit to use.
• Eutrophication (natural ageing of water bodies) is accelerated, resulting in hypoxic conditions (i.e. low oxygen), killing off the aquatic diversity.

(ii) To prevent the algal bloom, the farmers instead of using chemical nutrients must start using natural products like manures produced using vermicomposting. Awareness should be created among people about the effects of dumping waste in water bodies.

Question 14.
Public transport in Delhi uses CNG since 2002. List the advantages of this fuel policy. (All India 2015)
Answer:
The advantages of fuel policy involving use of CNG in public transport in Delhi are
(i) CNG, i.e. compressed natural gas is a better fuel than petrol or diesel, because it is cheaper, burns more effeciently, does not pollute the environment.
(ii) It cannot be siphoned off by thieves. In addition to this, it cannot be adulterated like petrol and diesel.

Question 15.
Name two metals used in a catalytic convertor. How do they help in keeping the environment clean? (Delhi 2014C)
Or
State the function of a catalytic convertor in an automobile. (All India 2011C)
Answer:
The metals used in a catalytic convertor are platinum, palladium and rhodium (any two).
The catalytic convertors fitted in automobiles reduce the emission of poisonous gases by converting the unbumt hydrocarbons into CO₂ and water, carbon monoxide to CO₂ and nitric oxide (NO) to nitrogen, while the exhaust is passed through them.

Question 16.
‘Fish mortality increases with influx of nutrients in a freshwater body’. Write two reasons. How will the influx of nutrients affect the BOD level of this water body? (All India 2014C)
Or
Write the harmful effects of the presence of large amount of nutrients in a water body. (Delhi 2012C)
Answer:
Fish mortality increases with influx of nutrients in a freshwater body, because

• Abundant nutrients encourage the growth of free-floating algae and plants, thereby increasing organic matter.
• The influx of nutrients increases the BOD level of water body, as the microbes flourishing in it consume up all the oxygen in degrading the organic matter. As the BOD increases, the Dissolved Oxygen (DO) decreases, making the water body unfit for aquatic life.

Question 17.
Explain, how does the inflow of large amount of nutrients like phosphates and nitrates into the water body drastically affects the aquatic life there? Name the phenomenon responsible. (Delhi 2014C)
Answer:
The inflow of large amount of nutrients like nitrates and phosphates initially encourages growth of aquatic organisms, plants and animals. With time as the organic matter increases, debris along with silt piles up at bottom making the water body shallower and warmer.

This leads to gradual appearance of floating and marshy plants and is eventually converted into land. Hence, the aquatic life gets severely affected and disrupted. This phenomenon is called eutrophication. It may be natural or artificial due to result of human activities.

Question 18.
BOD was measured in two different places A and B of a river in the direction of its flow. BOD value was higher at A than at B. What do you infer from this observation and why? (All India 2013C)
Answer:
Higher BOD in river A indicates the discharge of domestic waste is done near A. The organic matter increases the BOD as decomposers need oxygen for the microbial activity. As sewage is decomposed towards B, then there is gradual rise in the dissolved oxygen. Reappearance of fishes and other clean water organisms is indicating recovery of river from sewage pollution.

Question 19.
Name any two sources of e-waste and write two different ways for the disposal. (Delhi 2013)
Or
Mention how e-waste is produced and disposed off ? Write the solution for its treatment? (All India 2010)
Answer:
Electronic wastes (e-wastes) are generated from electronic appliances. It includes broken mobile phones, televisions, irreparable computers, etc. The sources of e-waste come from a vast area of our daily life as each and every field is dependant in electronic appliances. Incineration and land filling are two ways of e-waste disposal. Recycling is the solution for the treatment.

Question 20.
Explain, biomagnification of DDT in an aquatic food chain. How does it affect the bird population? (Outside Delhi 2012)
Answer:
There can be up to 1000 times increase in concentration of DDT in phytoplankton as compared to water, in zooplankton as compared to phytoplankton, in different fish as compared to zooplankton and more DDT in fish eating birds as compared to fish. This continuous increase of DDT concentration is called biomagnification. Higher amounts of pesticide disturb calcium metabolism of birds resulting in thinning of egg shells and their premature breaking that kills the embryos. Thus, causing decline in the bird population.

Question 21.
Explain, how does the algal bloom eventually choke the water body in an industrial area? (Delhi 2012)
Answer:
The nutrient enrichment of water bodies near industrial area is due to the passage of industrial effluents sewage, etc. This causes increased growth of planktonic algae that causes colouration of water called algal bloom. Excessive growth of algae cut off light for submerged plants which kills the latter and causes organic loading. This leads to decreased oxygen level and pilling of aquatic animals and plants which eventually chokes the water body.

Question 22.
Study the graph given below. Explain, how is oxygen concentration affected in the river when sewage is discharged into it.

Answer:
When sewage containing organic waste is discharged into the river, the microorganisms consume a large amount of O₂ for degrading organic matter. This causes decline in O₂ content in water downstream from the point of sewage discharge. After a short distance, there is a gradual rise in dissolved oxygen downstream because of reduced BOD.

Question 23.
Why is it difficult to get rid of ‘water hyacinth’ from a water body? Name one abiotic component and one biotic component of the ecosystem that get affected by its spread in the water body. (Delhi 2011C)
Answer:
The eutrophic bodies support excessive growth of water hyacinth (Terror of Bengal), therefore, it is very difficult to get rid of it and hence, known as an aquatic weed. Abiotic and biotic components affected by the growth of water hyacinth are light and the growth of the submerged plants respectively.

Question 24.
Study the graph given below. Explain, how oxygen concentration is affected in the river, when sewage is discharged into it? (Delhi 2011)

Answer:
The graph shows the decrease in dissolved oxygen due to sewage decomposition. When sewage is discharged into river, microorganisms present in water help in biodegradation of organic matter. They consume a lot of oxygen. Therefore, there is a sharp decline in dissolved oxygen. When the sewage is completely degraded, oxygen concentration again increases.

Question 25.
How does an Electrostatic Precipitator (ESP) work to remove particulate pollutants released from the thermal power plants? (Delhi 2010)
Answer:
Electrostatic Precipitator (ESP) has electrode wires and a stage of collecting plates. Wires are provided with an electric current of several thousands volts, which produces a corona that releases electrons.

These electrons attach to dust particles, giving them a negative charge within a very small fraction of a second. Collecting plates are earthed, so that they attract charged dust particles. The velocity of air passing through plates is slow enough to allow the dust particles released from thermal power plants to fall on them.

Question 26.
Mention the major causes of air pollution in metro cities. Write any three ways, which it can be reduced. (All India 2010)
Answer:
Major causes of air pollution are

• Smoke from thermal power plants, forest fires, volcanic eruptions.
• Use of unleaded petrol.
• Excessive use of fossil fuels by automobiles and industries releasing particulate and air pollutants.

Three control methods for air pollution include

• Using electrostatic precipitators to remove particulate matter from exhaust of industries and thermal power plant.
• Using scrubber to remove gases like S02.
• Use of catalytic convertors in automobiles for reducing emission of poisonous gases.

Question 27.
Study the given aquatic food chain. Answer the questions that follow

(i) Give reason, why there is a continuous increase in the DDT content in different trophic levels of the chain?
(ii) Name the phenomenon responsible for the increase in DDT content. (Delhi 2010)
Answer:
(i) Continuous increase in the DDT concentration is because of following reasons

• It cannot be metabolised.
• It is not excreted, but it is passed on to next higher tropic levels.

(ii) Biomagnification is the phenomenon responsible for increase in the DDT content.

Question 28.
Explain giving reasons the causes of appearance of peaks A and B in the graph shown below. (All India 2010)

Answer:
A – High BOD due to sewage discharge.
B – Increase in dissolved oxygen due to sewage decomposition.
Microorganisms involved in biodegradation of organic matter consume a lot of oxygen.
Therefore, there is a sharp decline in dissolved oxygen. When the sewage is completely degraded, oxygen concentration again increases.

Question 29.
While on a visit to a pond in the city-neighbourhood, the visitors were delighted to find large expanse of water covered with colourful algal mass.
(i) As a student of biology do you agree with their delight? Give reasons in support of your answer.
(ii) Explain the cause of such algal growth. (Delhi 2019)
Answer:
(i) As a student of biology, I do not agree with the delight of visitors. Large expanse of water covered with colourful algal mass represents excessive growth of algae in pond. This excessive growth is called algal bloom. Because algal bloom deteriorates water quality and causes fish mortality in the water body, it is not possible to agree with the delight of visitors.

(ii) The presence of large amounts of nutrients in water leads to algal bloom. These nutrients (nitrogen and phosphorus) are added to waterbodies from fertilisers that get dissolved in soil and are washed off by waterbodies. Excessive growth of phyptoplankton occurs due to the presence of large amounts of nutrients in water.

Question 30.
Explain the effect on the characteristics of a river when urban sewage is discharged into it. (2018)
Or
Explain the changes that can be observed in the characteristics of river water when sewage is discharged into it and a few weeks after the discharge with respect to
(i) Level of dissolved oxygen.
(ii) Population of freshwater organisms. (Delhi 2015C)
Answer:
(i) Discharge of domestic sewage into a river water results in the rise of BOD, because decomposers consume a lot of oxygen. If sewage quantity is large, then the dissolved oxygen will be completely consumed thus leaving nothing for aquatic organisms present in water body. However, as sewage is decomposed, there will be gradual rise in dissolved oxygen downstream.

(ii) With depletion in level of dissolved oxygen, population of freshwater organisms (plants and animals) would decline. But, as levels of dissolved ‘oxygen would increase, fish and other aquatic organisms reappear indicating the recovery of river from sewage pollution.
The effect on the characteristics of a river when urban sewage is discharged into it can be plotted on graph as shown below.

Question 31.
(i) Name any two places where it is essential to install electrostatic precipitators. Why is it required to do so?
(ii) Mention one limitation of the electrostatic precipitator. (Delhi 2016)
Answer:
(i) Electrostatic precipitators are needed to be installed in thermal power plant and industries. It is the most widely used method for controlling the air pollution or removing particulate matter. In this process, small particles present in dirty air are first charged electrically and then allowed to settle down over differently charged collecting plates. This method allows smaller particles to be removed from the emissions.

(ii) Limitation of using electrostatic precipitator is that it only works on gravitational settling method. Velocity of air between the plates is maintained low to allow the dust to fall in.

Question 32.
Our farmers still use DDT. How is this affecting the local bird population? (Delhi 2015C)
Answer:
Our farmers still use DDT and which is well-known cause for biological magnification. Due to the excessive use of DDT, calcium metabolism in birds is disturbed, which results in thinning of egg shell. Due to this, premature breaking of eggs occurs and leads to decline in birds population. If DDT leaches from the agricultural field, it gets into the water body (the concentration is 0.003 ppb) and enters the food chain : Zooplanktons (0.04 ppm)- Small fish (0.5 ppm)-Large fish (2 ppm)-Any fish eating birds (5 ppm). Concentration of DDT increases along the food chain, reaching a high level in the top carnivore bird.

Question 33.
With the help of a flow-chart exhibit the events of eutrophication. (Outside Delhi 2015)
Answer:
Eutrophication is the natural ageing of the water. The flow chart exhibiting the events of eutrophication in lake is given below

Question 34.
With the help of a flowchart, show the phenomenon of biomagnification of DDT in an aquatic food chain. (All India 2015)
Or
Explain biomagnification. How does the biomagnification of DDT affect the population of fish eating birds? (Foreign 2010)
Or
Why is there a decline in the population of the fish eating birds, when the water body is amidst agricultural fields? (Delhi 2010c)
Answer:
Biological magnification (Biomagnification) can be defined as the increase in concentration of toxicants at successive trophic levels, e.g. DDT. This is because these toxicants cannot be metabolised or excreted. Therefore, it gets accumulated in an organism and pass on to higher trophic levels.

DDT accumulates in the fish eating birds and disturbs calcium metabolism, which results in thinning of egg shell. This results in decline in the population of the fish eating birds.
Biological magnification of DDT is shown in the flowchart below

Question 35.
Drinking water problem in our urban areas is caused mainly because, we fail to protect our water bodies. Explain, how accelerated eutrophication chokes our water bodies to death? (outside Delhi 2015C)
Answer:
The accelerated eutrophication is the nutrient enrichment of water bodies due to human activities like passage of sewage, industrial effluents, domestic wastes and agricultural run offs rich in nitrates and phosphates.
The lake’s fertility increases due to these nutrients leading to dense growth of plants and the planktonic algae, i.e. algal bloom.
Algal blooms and floating plants cut off light for submerged plants due to which the latter die. There is drastic decrease in oxygen replenishment inside water. It causes organic loading of water. Decreased oxygen level also kills aquatic animals, further adding to organic loading. Decomposition is replaced by putrefaction which is anaerobic.
It produces secondary pollutants that kill the bloom promoting plants as well and thus choking water bodies to death. (3)

Question 36.
Two types of aquatic organisms in a lake show specific growth patterns as shown below, in a brief period of time. The lake is adjacent to an agricultural land extensively supplied with fertilisers.

Answer the questions based on the facts given above
(i) Name the organisms depicting the patterns A and B.
(ii) State the reason for the growth pattern seen in A.
(iii) Write the effects of the growth patterns seen above. (All India 2014)
Answer:
(i) The organisms depicting pattern A are
microorganisms, while B depicts- zooplanktons or fishes.
(ii) With the increase in organic matter of lake due to influx of nutrients from agricultural land, the number of microorganisms increases which degrade the organic matter.
(iii) From the above growth patterns, it can be concluded that

• with the increase in microbes in water b’ody the BOD increases, making it unfit for aquatic life.
• dissolved oxygen reduces drastically leading to mortality of aquatic organisms, i.e. fishes.

Question 37.
A few residents in your locality, for business gains, have established small scale indsutrial/commercial activities such as pathological labs and fabric dyeing centres without obtaining ‘No objection certificates’ from municipal authorities. Would you support these activities? Give any three reasons in support of your answer. (All India 2014C)
Answer:
No, these activities would not be supported, because

• The industrial/commercial activities will release its toxic gases into the atmosphere and other chemicals into the nearby water bodies without treating it and thus, cause pollution of air, water and soil.
• The land acquired for operating these activities should be at a location where people or neighbourhood are leasi affected, so prior permission is mandatory.
• Disposable waste of pathological labs can be infectious and toxic, including the same from other industries, so it is necessary to keep a check on how the wastes are disposed off, segregated and recycled by these industries.

Question 38.
Why should the spraying of DDT as an insecticide on vegetable crops be banned? Explain. (Delhi 2013C)
Answer:
Spraying of DDT as an insecticide on vegetable crops should be banned, because our food grains such as wheat, rice, fruits and vegetables accumulate the varying amount of insecticide residues and that enter the human body through food chain. As these chemicals are toxic and non-biodegradable, they get accumulated in organism’s body and are difficult to excrete and their concentration keeps on increasing at successive trophic levels. The organisms, thus present at highest level are severely affected.

Question 39.
How does algal bloom destroy the quality of a freshwater body? Explain. (Delhi 2013)
Or
How does an algal bloom cause eutrophication of a water body? Name the weed that can grow in such a eutrophic lake. (Delhi 2012)
Answer:
Algal bloom destroy the quality of freshwater body and causes eutrophication because run off of nutrients such as animal wastes, fertilisers (N and P) and sewage from land leads to an increase in the fertility of the lake.
It causes a tremendous increase in the primary productivity of the water body leading to increased growth of algae, resulting into algal bloom. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life. This leads to eutrophication of fresh water body and destroys its quality.
Eichhomia crassipes (water hyacinth) also called ‘Terror of Bengal’, as it grows abundantly in such eutrophic lake.

Question 40.
By the end of 2002, the public transport of Delhi switched over to a new fuel. Name the fuel. Why is this fuel considered better? Explain. (Delhi 2012)
Answer:
Delhi had been categorised as the fourth most polluted city of the world in a list of 41 cities in the 1990s. Vehicular pollution has added to the pollution of air in Delhi.

So, in the year 2002, it switched over to a new fuel CNG (Compressed Natural Gas).
Reasons of CNG being considered better fuel, Refer to Answer No. 14.

Question 41.
(i) Why are colourful polystyrene and plastic packaging used for protecting the food, considered an environmental menace?
(ii) Write about the remedy found for the efficient use of plastic wastes by Ahmad Khan of Bengaluru. (Delhi 2012)
Answer:
(i) Because polysterene and plastic are non-biodegradable, they accumulate and add to environmental pollution hence, are considered a menace.
(ii) For remedy for the efficient use of plastic wastes by Ahmad Khan, Refer to Answer No. 11.

Question 42.
(i) State the consequence if the electrostatic precipitator of a thermal plant fails to function.
(ii) Mention any four methods by which the vehicular air pollution can be controlled. (All India 2011)
Answer:
(i) In the absence of Electrostatic Precipitator (ESP) the particulate pollutants and gaseous pollutants of the exhaust will enter into atmosphere. In case of ESPs, about 99% of particulate matter is absorbed.
(ii) General steps to reduce vehicular air pollution are

• Phasing out of old vehicles.
• Use of unleaded petrol and low sulphur petroband diesel.
• Use of catalytic convertors in vehicles.
• Application of pollution control norms for vehicles.

Question 43.
Eutrophication is the natural ageing of a lake. Explain. (Foreign 2011)
Or
How does eutrophication in a lake take place? Explain. (Delhi 2011C)
Answer:
The natural ageing of a lake is called eutrophication. In a young lake, there is very little life. Over the time, streams drain into it and water becomes enriched with nutrients like phosphate and nitrates.

As a result, phytoplanktons and some other plants flourish well in this water. Due to this, the organic matter increases in lake, water becomes warmer and shallower and decomposers also start growing.

The high number of decomposers make use of large quantity of oxygen for decomposition. This leads to depletion of dissolved oxygen of the water and mortality of fish and other aquatic organisms.

The eutrophied water bodies can lead to algal blooms, which further add organic matter in lake. These consume more oxygen leading to its depletion. Finally the water bodies gets converted to land.

Question 44.
Biomagnification and accelerated eutrophication are both caused due to indiscriminate use of chemicals and irresponsible human activities. Do you agree? Support your answer with explanation and an example of each. (All India 2019)
Answer:
I agree that biomagnification and accelerated eutrophication are both caused due to the indiscriminate uses of chemicals and irresponsible human activities. For explanation and example of each, Refer to Answer No. 34 and 35, respectively.

Question 45.
(i) Why are catalytic convertors recommended for vehicles?
(ii) Why should such vehicles use only unleaded petrol?
(iii) Why is CNG preferred to diesel as a fuel in vehicles? (Delhi 2010C)
Answer:
(i) For catalytic converters in vehicles, Refer to Answer No. 15.
(ii) Such vehicles (with catalytic convertor) should use unleaded petrol as leaded petrol inactivates the catalyst.
(iii) For preference of CNG as a fuel in vehicles. Refer to Answer No. 14.

Question 46.
Name the layer of the atmosphere that is associated with good ozone. (Delhi 2019)
Answer:
Good ozone is found in stratosphere, which is formed by UV radiations and oxygen molecules. This type of ozone is beneficial for living organisms.

Question 47.
Name the greenhouse gases that contribute to global warming. (Delhi 2014)
Answer:
The greenhouse gases that contribute to global warming are CO₂, CH₄, N₂O and CFCs.

Question 48.
State the purpose of signing the Montreal Protocol. (Foreign 2014)
Answer:
The purpose of signing the Montreal Protocol is to control the emission of ozone depleting substances.

Question 49.
Where is good ozone present? Why is it called SO? (All India 2014C)
Answer:
Good ozone is present in stratosphere (i.e. the upper part of atmosphere).
It is called ‘good ozone’, because it acts as a shield for absorbing ultraviolet radiations from the sun.

Question 50.
Name the two gases contributing maximum to the greenhouse effect. (Delhi 2014C)
Answer:
The two gases contributing maximum to greenhouse effect are CO₂ and CH₄.

Question 51.
Mention two harmful effects of UV-B exposure on human eye. (Delhi 2012)
Answer:
UV-B exposure on human eye causes inflammation of cornea and cataract.

Question 52.
Write the unit used for measuring ozone thickness. (Delhi 2011)
Answer:
Dobson Unit (DU) is used for measuring ozone thickness.

Question 53.
How does Jhum cultivation promote deforestation? (All India 20110)
Answer:
In Jhum cultivation, the farmers cut down the trees of forest and land is used for farming. After cultivation, the area is left free for several years for recovery. The recovery phase is often ignored leading to deforestation.

Question 54.
How is snow blindness caused in humans? (All India 2010)
Answer:
Snow blindness is caused by the absorption of UV-B radiation.

Question 55.
Name two greenhouse gases produced by anaerobic microbes. (Foreign 2010)
Answer:
Carbon dioxide and methane are the greenhouse gases produced by anaerobic microbes.

Question 56.
Mention the causes of thinning of ozone layer. (Delhi 2010c)
Answer:
CFCs release chlorine atoms, when UV-rays act on them. Chlorine degrades the ozone into molecular oxygen. Thus, causing thinning of ozone layer.

Question 57.
Justify the need for signing of ‘Montreal Protocol’ by the participating nations in 1987. (All India 2019)
Answer:
Montreal Protocol was signed at Montreal (Canada) in 1987 (effective in 1989) to control the emission of ozone depleting substances like CFCs and aerosols. The balance of ozone production and degradation has been disrupted due to enhancement of ozone degradation by Chlorofluorocarbons (CFCs). Once CFCs are added to the stratosphere, they have permanent and continuous effects on ozone levels.

Question 58.
List four benefits to human life by eliminating the use of CFCs. (All India 2017)
Answer:
Benefits of eliminating CFCs from human life are as follows

• Prevention of global warming.
• Reduction in ozone degradation.
• Reduction in greenhouse effect,
• Beneficial for human health.

Question 59.
Explain the relationship between CFC and ozone in the atmosphere. (All India 2016)
Answer:
The chlorofluorocarbon releases active chlorine (Cl, CIO). This active chlorine gets perched over atmospheric ice crystals and remains functional for a long time. A single chlorine atom converts 1 lakh molecules of ozone into O₂.

Question 60.
What is joint forest management? How can it help in conservation of forest? (Foreign 2015)
Answer:
The concept of Joint Forest Management (JFM) was introduced by the Government of India. In this programme, support of local communities was taken for conservation of forests and in return, the local people were made to use the products obtained from the forest free of cost. In this programme, local people protect forest, which helps in conservation of forest and its biodiversity.

Question 61.

(i) What are the after effects of the degradation of ozone?
(ii) How does it affect human health? (Delhi 2015C)
Answer:
(i) The degradation of ozone will lead to higher mortality of young ones of animals, high number of mutations, inhibition of photosynthesis, global warming and disturbance in aquatic and terrestrial food chains.
(ii) Ozone degradation allows the harmful UV-B exposure on human eyes that causes imflammation of cornea and cataract. It also causes various types of skin cancer.

Question 62.

(i) Expand CFC.
(ii) How does it reduce ozone to oxygen? (All India 2015C)
Answer:
(i) CFC-Chlorofluorocarbon
(ii) For reduction of ozone to oxygen. Refer to Answer No. 14.

Question 63.
Write, what was the percentage of forest cover of India at the beginning and at the end of 20th century? How different is it from the one recommended by National Forest Policy? (Foreign 2014)
Answer:
In the beginning of the 20th century, the forest cover was about 30%, while towards its end it is cover for hilly regions including Himalayas, while 33% for plains. However, the situation is contrastingly different as the production of forests in India is very slow.

Question 64.
(i) State the cause of depletion of ozone layer.
(ii) Specify any two ill effects that it can cause in the human body. (Foreign 2014)
Answer:
(i) The main cause of depletion of ozone layer is imbalance between production and degradation of ozone in the stratosphere. This balance is disrupted mainly due to increase in ozone degradation by CFCs.

(ii) The ill effects that ozone depletion can cause in human body are
(a) Allows UV-B radiation to enter atmosphere that increases mutation and damages DNA.
(b) Causes ageing of skin and damage to skin cells and cancer.
(c) Inflammation of cornea (snow blindness), cataract, etc.

Question 65.
Refrigerants are considered to be a necessity in modern living, but are said to he responsible for ozone holes detected in Antarctica. Justify. (Foreign 2012)
Or
Chlorofluorocarbons (CFCs) are widely used as refrigerants. Then why is it suggested to reduce its emission as far as possible? Explain. (Delhi 2010)
Answer:
Refrigerants contain CFCs, which are released in the lower part of the atmosphere. Their emissions into the atmosphere is suggested to be reduced because:

• They move upwards and reach the stratosphere.
• In stratosphere, UV-rays act on them releasing Cl atoms. Cl degrades ozone and releases molecular oxygen.
• Cl atoms act only as catalysts, the CFCs have a permanent damaging effect on ozone.

Question 66.
The figure given below shows the relative contribution of four greenhouse gases to global warming.

(i) Identify the gases A and C.
(ii) Why are these four gases called greenhouse gases? (Foreign 2011)
Answer:
(i) A – Carbon dioxide (CO₂)
C – Chlorofluorocarbons (CFCs)

(ii) These four gases (CO₂, CH₄, N₂O) and CFCs) are called greenhouse gases, as they absorb infrared radiations emitted by the earth’s surface and cause greenhouse effect, i.e. heating of earth’s surface and atmosphere.

Question 67.
Enumerate the consequences of deforestation. (All India 2012C)
Answer:
The consequences of deforestation are as follows

• Deforestation leads to reduced rainfall, increased drought, hotter summer and colder winters.
• It also leads to soil erosion.
• Deforestation increases atmospheric CO₂ content by releasing carbon stored in organic matter and reduced primary activity.

Question 68.
(i) Why and how must the ozone layer in the stratosphere be protected? Explain.
(ii) How do deforestation and greenhouse gases negatively affect our environment? Explain. (All India 2019)
Answer:
(i) The ozone layer in the stratosphere must be protected, because the ozone in the stratosphere is good ozone that acts as a shield and absorbs ultraviolet radiations from the sun. UV rays are extremely harmful to living organisms.

Ozone gas is continuously formed by the action of UV rays on molecular oxygen and also degraded into molecular oxygen in the stratosphere. In order to protect the ozone layer, there should be a balance between production and degradation of ozone in the stratosphere.

In the recent years, the balance has been disrupted due to enhancement of ozone degradation by chlorofluor carbons (CFCs). So the emission of ozone degrading substances like CFCs need to be controlled to protect the ozone layer in stractosphere.

(ii) For effects of deforestation. Refer to text on page no. 405-406.

Effects of Greenhouse Gases
(i) Greenhouse gases include CO₂, CFCs, etc. Their high level in the atmosphere allow the heat waves to reach earth, but prevent their escape and earth becomes warm. This increased temperature in last three ecades had caused changes in precipitation patterns.
(ii) The increased temperature also cause EL-nino effect, melting of the polar ice caps and Himalayan snow.

Question 69.
(i) Why is the ozone layer required in the stratosphere? How does it get degraded? Explain.
(ii) Why is the ozone depletion a threat to mankind? (Delhi 2013C)
Answer:
(i) Ozone found in stratosphere is required because it acts as a shield absorbing harmful ultraviolet radiations coming from the sun.
Ozone layer gets depleted by the ozone depleting compounds mainly Chlorofluorocarbons (CFCs), which travel to the stratosphere after being released from the refrigerants and industrial emissions. UV-rays from the sun acts on CFCs releasing atoms, which degrade ozone releasing molecular oxygen.

(ii) The ozone depletion in a threat to mankind due to ill effects that ozone deplation can cause in human body. For ill effects of ozone depletion in human body, Refer to Answer No. 19 (ii).

Question 70.
Mohit while buying a car insisted on CNG with better mileage whereas Anil leaned towards diesel version of a high end car with relatively low mileage.
What qualities are being displayed by Mohit’s behaviour? Mention any two reasons behind Mohit’s decision.
Answer:
Mohit displays concern towards his environment and responsibility.

• CNG burns efficiently wihout leaving residues.
• CNG is cheaper.

Question 71.
Looking at the deteriorating air quality because of air pollution in many cities of the country, the citizens are very much worried and concerned about their health. The doctors have declared health emergency in the cities where the air quality is very severely poor.
(i) Mention any two major causes of air pollution.
(ii) Write any two harmful effects of air pollution to plants and humans.
(iii) As a captain of your school Eco-club, suggest any two programmes you would plan to organise in the school, so, as to bring awareness among the students on how to check air pollution in and around the school. (2018)
Answer:
(i) The two major causes of air pollution are:

• Particulate matter released from thermal power plants and other industries.
• Poisonous gases (e.g. CO₂,CO, NO₂) released from automobiles.

(ii) Effect on plants Reduces growth and yield of crops and causes premature death of plants. Effects on humans Irritation and inflammation of the lungs and premature deaths.

(iii) As a captain of my school Eco-club, I will suggest following two programmes to bring awareness among students on how to check air pollution.

• Organisation of ‘Van-Mahotsava’ programme in my school campus. In this progamme, there would be lectures on cures of air pollution and how planting trees can help against it. Plants will also be planted in school campus.
• By organising a debate on ‘air pollution and its causes.’In this programme, students who have given their any contribution in controlling air pollution will be rewarded.

Question 72.
A national newspaper reported that a 50 metre high ‘Sanitary landfill’ the dumping site of city’s garbage in one of the metro-cities crashed and caused heavy damage and disaster in and around the area. A couple of cars, two-wheelers and cattle were swept away in the nearby overflowing canal. Three persons including a young girl were crushed under the garbage and died.
(i) Write any two points that in your opinion could have caused this landfill crash.
(ii) Mention any four preventive measures to be adhered to as a policy which could have avoided this accident.
(iii) Write any two suggestions that you would like to give to the citizens, so as to help in preventing such a disaster in future. (2018C)
Answer:
(i) Two points that in my opinion could have caused landfills crash are

• Increase in weight of garbage due to overloading of landfill.
• Poor monitaring and management of landfill area.

(ii) Four preventive measures to be adhered as policy are

• Prevent overloading of landfill area.
• Garbage dumped should be compacted regularly.
• Engineers to design landfills.
• Regular monitoring of landfill area.

(iii) TWo suggestions to citizens so as to help them in preventing such disaster in future are

• Citizens should recycle the waste as much as possible.
• Citizens should avoid creating trash.

Question 73.
Public all over India is very much concerned about the deteriorating air quality in large parts of North India. Alarmed by this situation the Resident’s Welfare Association of your locality organised an awareness programme entitled ‘Bury Not Bum’. They invited you, being a biology student to participate.
(i) How would you justify your arguments that promote burying and discourage burning? (Give two reasons)
(ii) With the help of flow charts, one for each practice, depict the chain of events that follow. Delhi 2017
Answer:
(i) Following arguments can be put forward to promote burying and discourage burning

• When we bum wastes, pollutants such as carbon dioxide, carbon monoxide, etc., are released in the atmosphere. These pollutants damage the environment. Due to burning, smoke is released into atmosphere, which causes problems; like asthma, emphysema, etc.
• Burying the waste does not have any harmful effect. In fact, when we bury organic wastes, the soil becomes enriched with nutrients.

(ii) The solid wastes are of two types

• Biodegradable wastes These can be degraded by the microbes, e.g. organic wastes, paper, etc.
• Non-biodegradable wastes These cannot be degraded by the microbes and can remain as such for long periods, e.g. plastic.

Flow charts depicting effects of burying and burning are as follows

Question 74.
Since 2nd October, 2014 ‘Swachh Bharat Abhiyan’ has been launched in our country.
(i) Write your views on this initiative giving justification.
(ii) As a biologist name two problems that you may face while implementing the programme in your locality.
(iii) Suggest two remedial methods to overcome these problems. Delhi 2015
Answer:
(i) Mahatma Gandhi said, ‘Sanitation is more important than independence’ and it is so true, when the present situtation of our country is seen. In our country, more than 72% of mral population lacks adequate sanitation facilities while 55% of population has no access to basic amenities like toilets in various cities of India.

I am completely in support of Swachh Bharat Abhiyan. It is our primary duty to keep our nation clean, because wastes are the biggest evil that hinder the progress and the development of a country. An unclean environment leads to a lot of problems, starting from affecting the health of citizens to the shortage of land.

On a large scale, it is responsible for the pollution, be it air and water, which creates problems both on development and economic fronts of the country.

(ii) (a) It is very difficult to change the mind set of people, not to spit on roads, staircase, etc.
(b) Many people living even in good societies do have habit of keeping their garbage in open areas, which get spilled by animals (dogs, cats), thus polluting the colony.

(iii) These problems can be overcome by implementing following strict measures

• A committee of few members can be established and whenever anyone spitting, throwing garbage on roads or urinating in open is spotted should be punished.
• We should keep sweepers on monthly wages for regular cleaning of roads.
• Separate dustbins (like blue, red, green) for specific waste disposal, i.e. biodegradable and non-biodegradable.

Question 75.
Presently, air quality of Delhi has significantly improved in comparison to what existed before 1997.
This is the result of a lot of conscious human efforts. You are being asked to conduct an awareness programme in your locality wherein you will comment on the steps taken by the Delhi Government to improve the air quality.
(i) Write any two of your comments.
(ii) List any two ways that you would include in your programme so as to ensure the maintenance of good quality of air.
(iii) State any two values your programme will inculcate in the people of your locality. Foreign 2014
Answer:
(i) The comments on the steps taken by Delhi Government to improve air quality are

• Switching over of the entire public transport from diesel or petrol to CNG.
• Use of unleaded and low-sulphur petrol by vehicles.

(ii) The other two ways that can be included in the programme, so as to ensure maintenance of good quality of air can be

• Create awareness among people to switch over to CNG as it is less polluting as well as cost effective and phasing out of old vehicles.
• Follow the routine pollution check according to pollution-level norms for vehicles and use catalytic convertors.

(iii) The programme can inculcate following values

• The people will become more responsible and guard their contributions towards air pollution.
• After knowing the hazards and various consequences of pollutants, they will not indulge in activities especially unauthorised and without having permission and will prevent others from doing so.

The post Environmental Issues Class 12 Important Questions and Answers Biology Chapter 16 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 12 Biotechnology and its Applications to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 12 Important Questions, Biotechnology and its Applications Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 12 Important Questions Biotechnology and its Applications

Question 1.
Write the two specific ‘cry’ genes that encode the proteins which control cotton boll worms. (All India 2019)
Answer:
cry IAc and cry IIAb control cotton bollworm and cry I Ab controls corn borer.

Question 2.
Mention the chemical change that proinsulin undergoes, to be able to act as mature insulin. (2018)
Answer:
The C-peptide present in proinsulin is removed during its maturation.

Question 3.
What are cry genes? In which organisms are they present? (All India 2017)
Answer:
‘cry genes’ are genes found in Bacillus thuringiensis, a bacterium. These genes encode for protein crystals that contain a toxic insecticidal protein called Bt toxin.

Question 4.
Suggest any two possible treatments that can be given to patient exhibiting adenosine deaminase deficiency. (Delhi 2015)
Or
A boy has been diagnosed with ADA (Adenosine Deaminase) deficiency.
I Suggest any one possible treatment. (Delhi 2014C)
Answer:

• Bone marrow transplantation.
• Gene therapy (not permanent cure).

Question 5.
Why do children cured by enzyme replacement therapy for adenosine deaminase deficiency need periodic treatment? (All India 2015)
Answer:
Children cured by enzyme replacement therapy for adenosine deaminase deficiency need periodic treatment because in such type of treatment genetically engineered lymphocytes are used and these cells are mortal.

Question 6.
State the role of C-peptide in human insulin. (All India 2014)
Answer:
The C-peptide is an extra stretch of the peptides that connect the A and B-polypeptide chains of insulin in prohormone. During processing to release mature and functional insulin, this C-peptide is removed.

Question 7.
Write the possible source of RNA interference (RNAi) gene. (Delhi 2013C)
Or
State the role of transposons in silencing of mRNA in eukaryotic cells. (All India 2012)
Answer:
Mobile genetic elements, i.e. transposons are the possible source of RNA interference (RNAi) gene which is further involved in the silencing of the specific mftNA and prevents translation.

Question 8.
Name any two techniques that serve the purpose of early diagnosis of some bacterial/viral human diseases. (Foreign 2011)
Or
Name a molecular diagnostic technique to detect the presence of a pathogen in its early stage of infection. (Delhi 2010)
Answer:
Techniques that serve the purpose of early diagnosis of some bacterial/viral human diseases are as follows

• Polymerase Chain Reaction (PCR).
• DNA recombinant technology
• ELISA

Question 9.
How does dsRNA gain entry into eukaryotic cell to cause RNA interference? (Delhi 2011C)
Answer:
dsRNA gains entry into eukaryotic cell either through

• infection by virus having RNA genome.
• mobile genetic elements (transposons) that replicate via an RNA intermediate.

Question 10.
Name the source organism of the gene cry IAc and its target pest. (Foreign 2011)
Answer:
Source of gene cry IAc is Bacillus thuringiensis and its target pest is cotton bollworm.

Question 11.
Name the source used to produce hapatitis-B vaccine using rDNA technology. (Delhi 2015C)
Answer:
The source of hepatitis-B vaccine is yeast. Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines produced using this approach allow large scale production and hence greater availability for immunisation.

Question 12.
Write the functions of
(i) cry IAc gene
(ii) RNA interference (RNAi) (Outside Delhi 2015C)
Answer:
(i) cry IAc codes for toxic insecticidal protein as inactive protoxins in Bacillus thuringiensis. This toxin kills the cotton bollworm.
(ii) RNA interference is associated with silencing of specific ntRNA and is a method of cellular defence in eukaryotes.

Question 13.
What is gene therapy? Name the first clinical case in which it was used. (Delhi 2014)
Or
Name the disease that was first to get the gene therapy treatment. Write the cause of the disease and the effect it has on the patient. (Delhi 2014C, 2013)
Answer:
Gene therapy is a corrective therapy or technique of genetic engineering that is used to replace a faulty or non-functional gene with a normal healthy functional gene.

The first clinical gene therapy was given to a 4 year old girl with ADA (Adenosine Deaminase) deficiency in 1990. It is caused due to the deletion of the gene coding for ADA, which adversely . affects the functioning of imnune system.

Question 14.
Why does Bt toxin not kill the bacterium that produces it, but kills the insect that ingests it? (Delhi 2014)
Or
Why do the toxic insecticidal proteins secreted by Bacillus thuringiensis kill the insect and not the bacteria itself? (Foreign 2010)
Answer:
Bt toxin does not kill bacteria because in bacteria it exists in inactive state.
When Bt toxin is ingested by an insect, it gets converted into its active form due to the alkaline pH of the gut.
The activated toxin binds to the surface of the epithelial cells of the midgut and creates pores. Water enters through these pores and causes swelling and lysis of cells in insect body.

Question 15.
Explain how Eli Lilly, an American company, produced insulin by recombinant DNA technology. (Foreign 2014)
Answer:
Steps involved in insulin production by Eli Lilly company are as follows

• DNA sequences corresponding to the two polypeptide, A and B-chains of insulin were synthesised in vitro.
• They were introduced into plasmid DNA of E. coli.
• This bacterium was cloned under suitable conditions.
• The transgene was expressed in the form of polypeptides A and B, secreted into the medium.
• They were extracted and combined by creating disulphide bridge to form human insulin.

Question 16.
What do ‘cry genes’ in Bacillus thuringiensis code for? State its importance for cotton crop. (All India 2014C)
Or
Name the soil bacterium that produces a protein/chemical that is toxic to insect pests. Show with example that these are encoded by different forms of the genes. (All India 2012)
Answer:
‘cry genes’ in Bacillus thuringiensis code for toxic insecticidal proteins called Cry proteins which are encoded by different forms of a gene called cry gene, e.g. cry IAc and cry II Ab control the cotton bollworm whereas cry IAb controls corn borer.

Cry proteins when expressed in cotton crops through genetic engineering confer pest resistance against cotton bollworms and prevent damage. As the larva of these insects when feed upon cotton plant parts, the toxin gets activated in their gut, lysing their cells and leads to death thus, making them pest resistant.

Question 17.
Human insulin when synthesised in the body needs to be processed before it can act. Explain giving reasons. (Delhi 2014C)
Or
Why is proinsulin so called? How is proinsulin different from functional insulin in humans? (All India 2013, 2012C)
Answer:
Human insulin when initially synthesised in human body consists of three peptide chains-A, B and C. The C-peptide is an extra stretch of amino acids joining the A and B-chains. This is called proinsulin or prohormone. It undergoes processing or splicing to release the functional mature insulin that can carry out its normal functions.
During processing, the C-peptide is removed. Only A and B-chains contribute to form the functional insulin.

Question 18.
Write any two ways how genetically modified plants are found to be useful? (All India 2014C)
Answer:
The genetically modified plants are found to be useful as they

• reduce or minimise the use of agrochemicals, i.e. fertilisers, insecticides, herbicides, etc.
• reduce post-harvest losses and enhance nutritional value of crop.

Question 19.
(i) State the role of DNA ligase in biotechnology.
(ii) What happens when Meloidogyne incognita consumes cells with RNAi gene? (Delhi 2012)
Answer:
(i) DNA ligase enzyme is used to join two DNA fragments from their ends.
(ii) When Meloidogyne incognita (parasite) consumes cells with RNA/ gene, parasite cannot survive and the infection is prevented. It is mainly because introduced DNA forms both sense and anti-sense RNA. These two strands being complementary to each other form dsRNA. This dsRNA binds and prevents translation of nematode OTRNA. Thus, the wRNA of nematode is silenced and the parasite cannot survive there. This produces Meloidogyne incognita resistant tobacco plants.

Question 20.
(i) Mention the cause and the body system affected by ADA deficiency in humans.
(ii) Name the vector used for transferring ADA-DNA into the recipient cells in humans. Name the recipient cells. (All India 2012)
Answer:
(i) For ADA deficiency, Refer to Answer No. 13.
(ii) Retroviral vector is used to transfer ADA-DNA into the recipient cells of human, i.e. lymphocytes.

Question 21.
How does ‘RNA interference’ take place in eukaryotes? Mention its importance. (Delhi 2012C)
Or
Explain the process of RNA interference. (Delhi 2011)
Answer:
The strategy to prevent nematode infestation in tobacco plant is RNA interference.
It involves silencing of a specific mRNA due to a complementary dsRNA which binds and prevents translation.

Question 22.
Explain how a hereditary disease can be corrected. Give an example of the first successful attempt made towards correction of such disease. (Delhi 2011)
Answer:
Hereditary disease can be corrected by gene therapy. It is a collection of methods that allow correction or replacement of defective genes. The first gene therapy was given in 1990 to a 4 year old girl with Adenosine Deaminase (ADA) deficiency. It is caused due to the deletion of gene for adenosine deaminase.
The treatment involves following steps

• Lymphocytes from the blood of patient are grown on culture outside the body.
• A functional ADA, cDNA (using a retroviral vector) is then introduced into these lymphocytes.
• Such genetically engineered lymphocytes are returned to the blood of patient.
• Periodic infusion of such genetically engineered lymphocytes is required by the patient.

Question 23.
How does recombinant DNA technology help in detecting the presence of mutant gene in cancer patients? (All India 2o11c)
Answer:
A single-stranded DNA or RNA, tagged with a radioactive molecule (probe) is allowed to hybridise with its complementary DNA in a clone of cells followed by detection using
autoradiography. The clone having the mutated gene will not appear on photographic film, because probe will not be complementary with mutated gene thus, helpful in defecting the presence of mutated gene in cancer patients.

Question 24.
Why is the introduction of genetically engineered lymphocytes into an ADA deficiency patient not a permanent cure? Suggest a possible permanent cure. (Delhi 2010)
Answer:
The genetically engineered lymphocytes have a definite lifespan. Hence, the patient requires periodic infusion of genetically engineered lymphocytes, so the cure is not permanent. The cure can be permanent, if the gene isolated from marrow cells producing ADA is introduced into the cells at early embryonic stages.

Question 25.
How did Eli Lilly synthesise the human insulin? Mention one difference between this insulin and the one produced by the human pancreas. (All india 2010)
Answer:
(i) For insulin synthesis, Refer to Answer No. 15.
(ii) Differences between insulin produced by rDNA and insulin produced by human pancreas are as follows

Question 26.
How is Bt cotton made to attain resistance against bollworm? (Delhi 2010C)
Answer:
Bt toxin genes cry IAc and cry IIAb control cotton bollworms. These genes are isolated from the bacterium and are incorporated into cotton plants. For action of Bt toxin, Refer to Answer No. 14 for methodology.

Question 27.
How has the use of Agrobacterium as vectors helped in controlling Meloidogyne incognita infestation in tobacco plants? Explain in correct sequence. (2018)
Answer:
Several nematodes infect a wide variety of plants and animals including human beings.
A nematode Meloidogyne incognita infects the roots of tobacco plants which reduces the production of tobacco.
The strategy adopted to prevent this infection is based on the process of RNA interference (RNAi). RNA; mechanism takes place in all eukaryotic organisms as a method of cellular defence.

This method involves the following steps

• Silencing of a specific mRNA due to the complementary dsRNA molecule that binds to and prevents translation of the mRNA.
• Agrobacterium vectors are used to introduce nematode specific genes into the host plant. It produces both sense and antisense RNA in the host cells.
• These two RNAs are complementary to each other and form a double-stranded RNA (dsRNA) that initiates RNAi and hence, silence the specific mRNA of the nematode.
• The parasite cannot survive in a transgenic host, therefore the transgenic plant gets itself protected from the parasite.

Question 28.
Explain the various steps involved in the production of artificial insulin. (All India 2017)
Or
Recombinant DNA technology is of great importance in the field of medicine. With the help of a flow chart, show how this technology has been used in preparing genetically engineered human insulins. (Delhi 2015)
Answer:
An American company Eli Lilly produced insulin via recombinant DNA technology in 1983.
Insulin production by using recombinant DNA technology is shown in flow chart below

Question 29.
Why do lepidopterans die when they feed on Bt cotton plant? Explain how does it happen. (Delhi 2017)
Or
Explain the application of rDNA technology to produce insulin. (Delhi 2015C)
Answer:
Bt cotton plants are the transgenic plants that express a Bacillus thuringiensis gene called cry gene. This gene, encodes for protein crystals having insecticidal properties against insects of group Lepidoptera, Diptera and Coleoptera. Inside the bacterium, these proteins remain inactive and do not harm the bacteria. However, these inactive crystals can get activated in the alkaline pH of the gut of insects upon ingestion.

After activation, these crystals can bind to the receptors present on the membranes of gut epithelial cells. Due to this binding, the membrane swells and pores are created on them. These pores lead to bursting of cell and soon the lepidopteran dies due to starvation.

Question 30.
Explain how Eli Lilly, an American company produced insulin by recombinant DNA technology. (2018)
Answer:
For DNA technology for insulin production, Refer to Answer No. 28.

Question 31.
What is GMO? List any five possible advantages of a GMO to a farmer. (All India 2018)
Answer:
The plants, bacteria, fungi and animals whose
genes have been altered are called Genetically Modified Organisms (GMOs). GM plants are useful in many ways.
Genetic modification has done the following changes to the phenotypic expression of the plants,

• Crops become more tolerant to abiotic stresses like cold, drought, salt and heat.
• Dependence on chemical pesticides has reduced, i.e. pest-resistant crops.
• Helped to reduce post-harvest losses.
• Efficiency of mineral usage increased in plants, preventing early exhaustion of fertility of soil.
• Enhanced nutritional value of food, e.g. vitamin-A enriched rice.

Question 32.
How has the study of biotechnology helped in developing pest-resistant cotton crop? Explain. (Delhi 2016C)
Or
One of the major contributions of biotechnology is to develop pest-resistant varieties of cotton plants. Explain how it has been made possible. (Foreign 2015)
Or
Name the pest that destroys the cotton bolls. Explain the role of Bacillus thuringiensis in protecting the cotton crop against the pest to increase the yield. (All India 2013)
Or
How is the Bt cotton plant created as a GM plant? How is it protected against bollworm infestation? (Delhi 2013C)
Or
Name the genes responsible for making Bt cotton plants resistant to bollworm attack. How do such plants attain resistance against bollworm attacks? Explain. (Delhi 2012)
Answer:
Biotechnology had hepled in generating plant varieties against the pests that destroy the cotton balls, i.e. cotton bollworms and cotton borer.
Bt cotton is created by using some strains of a bacterium. Bacillus thuringiensis (Bt is short form).

• This bacterium produces protein that kills certain insects such as lepidopterans (tobacco budworm and armyworm), coleopterans (beetles) and dipterans (flies and mosquitoes).
• Bacillus thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein.
• Bt toxin protein exists as inactive protoxin in bacteria, but once an insect ingests this inactive toxin, it is converted into an active form due to the alkaline pH of the gut, which solubilises the crystals.
• The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis leading to death of insect.
• Specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated into several crop plants.
• Most Bt toxins are insect-group specific. The toxin is coded by a gene named cry. For example, the proteins encoded by the genes cry IAc and cry IIAb control the cotton bollworms and cry IAb controls corn borer.

Question 33.
Why is molecular diagnosis preferred over conventional methods? Name any two techniques giving one use of each. (Delhi 2016C)
Answer:
Using conventional methods of diagnosis (serum and urine analysis), early detection of diseases is not possible. To overcome this problem, some molecular diagnosis techniques were developed that provide early detection of diseases. These are as follows
(i) Polymerase Chain Reaction (PCR) helps in early detection of diseases or pathogens by the amplification of their nucleic acid.
PCR can amplify nucleic acids of pathogens even when their concentration is very low.
This technique can be used for detecting HIV in suspected AIDS patients, genetic mutations in suspected cancer patients and in identifying genetic disorders.

(ii) Enzyme Linked Immunosorbent Assay (ELISA) is based on the principle of antigen-antibody interaction. Infection by pathogen can be detected by the presence of antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesised against the pathogen.

Question 34.
How has RNAi technique helped to prevent the infestation of roots in tobacco plants by a nematode Meloidogync incognita? (Delhi 2016)
Or
How did the process of RNA interference help to control the nematode from infecting the roots of tobacco plants. (Foreign 2014)
Answer:
A nematode Meloidogyne incognita infects the roots of tobacco plants, which reduces the production of tobacco. The strategy adopted to prevent this infection is based on the process of RNA interference (RNAf). RNAi takes place in all eukaryotic organisms as a method of cellular defence. This method involves silencing of a specific mRNA due to complementary rfsRNA molecule that binds to and prevents translation of mRNA.
Thus, resulting into the death of the nematode.
Refer to Answer No. 19 (ii) for methodology.

Question 35.
Explain enzyme replacement therapy to treat adenosine deaminase deficiency. Mention two disadvantages of this procedure. (All India 2016)
Answer:
In enzyme replacement therapy, the patient is given functional ADA (Adenosine Deaminase) by injection.
Refer to Answer No. 22 for methodology.
Disadvantages

• The patient does not completely recover from the disease.
• It needs periodic injections of the enzyme to the patients.

Question 36.
People are quite apprehensive to use GM crops. Give three arguments in support of GM crops so as to convince the people in favour of such crops. (Outside Delhi 2016C)
Answer:
The GMOs are the plants, animals, bacteria, etc. whose genes have been altered by genetic manipulation. They were created because

• The agrochemicals used in agriculture are too costly.
• Excess use of chemicals in field adversely affects our environment by causing pollution. GMOs are beneficial in many ways.

Advantages of GMOs are as follows

• Tolerance against abiotic stresses, such as cold, drought, salt, heat.
• Reduce dependence on chemical pesticides.
• Reduce post-harvest losses.
• Increased efficiency of mineral usage by plants.

Question 37.
Mention the cause of ADA deficiency in humans. How has genetic engineering helped patients suffering from it? (Outside Delhi 2015C)
Or
(i) Explain the effect of deletion of the gene for ADA in an individual.
(ii) How does the gene therapy help in this case? (All India 2010C)
Answer:
(i) Deletion of the gene for ADA in an individual leads to ADA deficiency disorder. Adenosine Deaminase (ADA) enzyme is crucial for immune system to function.
(ii) Gene therapy is helpful in the treatment of ADA deficiency.
Refer to Answer No. 22 for methodology.

Question 38.
Describe any three potential applications of genetically modified plants. (All India 2015)
Answer:
Potential applications of genetically modified plants are

• Nutritional enhancement, e.g. vitamin-A enriched rice.
• Stress tolerant crops became more tolerant to abiotic stresses such as cold, drought, etc.
• Creation of tailor made plants by using GM plants to supply alternative resources of industries in the form of starches, biofuels, etc.

Question 39.
Name the host plant and its part that Meloidogyne incognita infects. Explain the role of Agrobacterium in the production of dsRNA in the host plant. (Delhi 2014C)
Answer:
The nematode Meloidogyne incognita infects the roots of tobacco plants.
The Agrobacterium is used as vectors carrying nematode specific genes to be introduced in host plant. These genes when expressed inside host plant produce sense and anti-sense RNA strands, complementary to nematode’s functional mRNA. This binding results in the formation of double-stranded RNA and inhibiting or silencing the translation of RNA specified. This process is called RNA interference.

Question 40.
(i) Tobacco plants are damaged severely when infested with Meloidogyne incognita. Name and explain the strategy that is adopted to stop this infestation.
(ii) Naipe the vector used for introducing the nematode specific gene in tobacco plant. (All India 2012)
Or
How does RNA interference help in developing resistance in tobacco plant against nematode infection? (Delhi 2010)
Answer:

• Infestation of tobacco plant with the nematode can be stopped by using RNA interference (RNAi) process.
  For the process of RNAi, Refer to Answer No. 21.
• Vector used for introducing the nematode specific gene in tobacco plant is Agrobacterium.

Question 41.
(i) How has biotechnology helped in producing Meloidogyne incognita resistant tobacco plant?
(ii) Why does this nematode die on eating such a GM plant? (Delhi 2010C)
Answer:
(i) For developing Meloidogyne incognita resistant tobacco plant, Refer to Answer No. 34.
(ii) Due to the RNAi process, specific mRNA of nematode is silenced. The result is that the parasite could not survive after eating such GM or transgenic plant (host), expressing, v specific interfering RNA.

Question 42.
Explain the application of biotechnology in producing Bt cotton. (Delhi 2014)
Answer:
For the use of Biotechnology to develop St cotton, Refer to Answer No. 32.

Question 43.
(i) Name the source from which insulin was extracted earlier. Why is this insulin no more in use by diabetic people?
(ii) Explain the process of synthesis of insulin by Eli Lilly company. Name the technique used by the company.
(iii) How is the insulin produced by human body different from the insulin produced by the above mentioned company? (All India 2011)
Answer:
(i) Insulin was extracted earlier from pancreas of slaughtered pigs and cattle animals. Insulin obtained from these sources caused some allergy or some other reactions to the foreign protein.
(ii) For production of human insulin by Eli Lilly company. Refer to Answer No. 28.
(iii) For different insulin produced by pancreas and by using rDNA technology, Refer to Answer No. 25 (ii).

Question 44.
What is biopiracy? (Delhi 2017)
Answer:
Biopiracy refers to the use of bioresources by multinational companies and other organisations without proper authorisation from people and countries concerned.

Question 45.
What are transgenic animals? Give an example. (All India 2016)
Answer:
Animals with manipulated DNA to possess and express the characters of an extra (foreign) gene are known as transgenic animals, e.g. transgenic
rats, rabbits, pigs, etc.

Question 46.
Mention two objectives of setting up GEAC by our government. (All India 2015)
Answer:
GEAC is an organisation setup by the Indian government to make decisions pertaining with genetic research and experiments. The two objectives of setting up Genetic Engineering Approval Committee (GEAC) by the Indian Government are

• To make decisions regarding the validity of GM research.
• To make decisions regarding the safety of introducing GM organisms for public services.

Question 47.
What is biopiracy? State the initiative taken by the Indian Parliament against it. (Delhi 2014)
Answer:
For biopiracy, Refer to Answer No. 1.
The Indian Parliament has cleared a second amendment of Indian Patents Bill as an initiative step against biopiracy. This bill considers issues including patent terms, emergency provisions as well as research and development initiative.

Question 48.
How have transgenic animals proved to be beneficial in
(i) production of biological products?
(ii) chemical safety testing? (Delhi 2014, 2013)
Answer:
(i) The transgenic animals have been proved beneficial in the production of biological products like human protein a-1 antitrypsin (by coding genes for that protein only), for the treatment of emphysema and production of human protein (a-lactalbumin) enriched milk by transgenic cow, i.e. Rosie. This milk was more nutritionally balanced for. human babies than natural cow’s milk.
(ii) Transgenic animals are studied for testing toxicity of drugs and other chemicals, as they carry genes that make them more sensitive to toxic substances.

Question 49.
Mention any four benefits derived from transgenic animals. (Delhi 2012C)
Answer:
Benefits derived from transgenic animals are as follows

• They are specially made to serve as models for studying human diseases.
• They produce useful biological products that can be created by the introduction of a portion of gene, which codes for a particular product such as human protein, a-1-antitrypsin is used to treat emphysema.
• Transgenic mice are being developed for testing the safety of vaccine before they are used in humans.
• They can be used to study the effects of certain toxic substances.

Question 50.
How is ‘Rosie’ considered different from a normal cow? Explain. (All India 2011; Foreign 2010)
Answer:
The transgenic cow Rosie, produced human protein-enriched milk (2.4 gm/L). It contained the human a-lactalbumin and was nutritionally more balanced for human babies than natural cow’s milk.

Question 51.
(i) What are transgenic animals ?
(ii) Name the transgenic animal having the largest number amongst all the existing transgenic animals.
(iii) Mention any three purposes for which these animals are produced. (2018 C)
Answer:
(i) Transgenic animals Refer to text on page no. 310.
(ii) Transgenic mice have the largest number (over 95%) amongst all the existing transgenic animals.
(iii) For purpose of producing transgenic animals.
Refer to text on page no. 310.

Question 52.
Biopiracy should be prevented. State why and how. (All India 2011)
Answer:
Biopiracy should be prevented because

• The countries and people concerned are not given adequate compensatory payment.
• The countries/people also lose their right to grow and use breeding experiments to improve the other varieties of the same species.
  It may be prevented by implementing specific laws that take into consideration all the biopatents and biopiracy related issues.

Question 53.
Aditya’s father requested to the farmer’s in his village to stop using chemical fertilisers in crop fields. Was he correct? Give reason.
Answer:
Yes, agrochemicals like fertilisers and pesticides have harmful effects on the health of living beings and the environment.

Question 54.
A group of activists was protesting without violence against the release of a GMO in the market, stating that production of GMO is an unethical practice.
(i) What is meant by GMO? Can the production of GMOs be considered unethical? Give reason.
(ii) What values do you observe in the activists involved in the protest?
Answer:
(i) GMO refers to Genetically Modified Organism. Yes, manipulation of living organism without showing concern about them is considered unethical.
(ii) Value observed is concern towards animal safety.

Question 55.
Anil’s father is suffering from high blood sugar. Doctors had advised regular insulin injections, which were declined by the family due to objections on the process used in insulin production. The doctor explained the process now used in insulin production and its advantages.
(i) Why did the family declined the use of insulin injection despite its necessity ?
(ii) Name the process by which insulin is produced now-a-days and the organism involved.
(iii) What values are shown by the doctor?
Answer:
(i) Earlier the insulin was produced by slaughtering of animals like pigs. They also cause allergic reactions in patients. The family declined because they were against such process and products.
(ii) The process is recombinant DNA technology and organism used is E.coli.
(iii) The doctor is responsible, aware of recent developments in science and honest towards his duty.

The post Biotechnology and its Applications Class 12 Important Questions and Answers Biology Chapter 12 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 14 Ecosystem to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 14 Important Questions, Ecosystem Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 14 Important Questions Ecosystem

Question 1.
How does a detritivore differ from a decomposer? Explain with an example of each. (Delhi 2015C)
Answer:
Difference between detritivores and decomposers is as follows

Question 2.
How is ‘stratification’ represented in a forest ecosystem? (Delhi 2014)
Answer:
The stratification, i.e. the vertical distribution of species at different levels in a forest ecosystem. It can be represented-as

Question 3.
What does ‘R’ represent in the given equation for productivity in an ecosystem?
GPP – R = NPP (All India 2014C)
Answer:
In the given equation for productivity in an ecosystem, GPP – R = NPP. ‘R’ represents the energy utilised by plants or producers in respiration. It is also referred to as respiratory losses.

Question 4.
Mention any two reasons, why the primary productivity varies in different types of ecosystems. (All India 2014C)
Answer:
The primary productivity varies in different types of ecosystems because

• It depends upon plant species (producers) of a given ecosystem and their photosynthetic capacity.
• It is dependent on various environmental factors, availability of nutrients.

Question 5.
Write the equation that helps in deriving the net primary productivity of an ecosystem. (Delhi 2013)
Or
How is the net primary productivity of an ecosystem derived ? (All India 2012C)
Answer:
Net Primary Productivity (NPP) is the weight of the organic matter stored by the producers which is available to heterotrophs for consumption. It is equal to the rate of organic matter produced during photosynthesis. Gross Primary Productivity (GPP) minus respiratory losses (R) yields NPP, i.e. NPP = GPP – R.

Question 6.
Why are green algae not likely to be found in the deepest strata of the ocean? (All India 2013)
Answer:
Green algae survive by utilisation of food synthesised by themselves through photosynthesis. At deepest layer in ocean, light is absent. So, photosynthesis does not take place and hence, green algae are not likely to be found in deep strata of ocean.

Question 7.
Write a difference between net and gross primary productivity. (All India 2011)
Answer:
Difference between NPP and GPP is as follows

Question 8.
Write the relationship between productivity, gross primary productivity, net primary productivity and secondary productivity. (All India 2019)
Answer:
The rate of biomass production is called productivity in term of gm-2 yr-1 for an ecosystem. It can be divided in Gross Primary Productivity (GPP) and Net Primary Productivity (NPP). GPP of an ecosystem is the rate of production of organic matter during photosynthesis. NPP is the available biomass for the consumption of heterotrophs (herbivores and decomposers). Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Question 9.
How does the dead organic matter get decomposed in nature? Explain. (All India 2012C)
Answer:
In nature, the decomposer organisms like bacteria and fungi act on the dead organic matter and decompose it by relearing various digestive enzymes.
The process of breaking down of complex organic matter into inorganic substances like CO₂, water and nutrients by bacteria and fungi is called decomposition.

Question 10.
Describe how do oxygen and chemical composition of detritus control decomposition. (Delhi 2011)
Answer:
Effect of oxygen The decomposition of detritus is largely an oxygen requiring process. Most of the decomposers (bacteria and fungi) are aerobic organisms. They require oxygen for their cellular activities while acting on dead organic matter.
Effect of chemical composition Decomposition rate is slower, if detritus is rich in lignin and chitin. It is quicker if detritus is rich in nitrogen and water soluble substances like sugars.

Question 11.
Describe the inter-relationship between productivity, gross primary productivity and net productivity. (All India 2017)
Answer:
The rate of biomass production is called productivity. It is expressed in terms of g^-2 yr^-1 or (kcal m^-2) yr^-1.
Productivity of an ecosystem can be categorised as primary and secondary productivity.
Primary Productivity (PP) is the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis. It can be divided into Gross Primary Productivity (GPP) It is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration.

Net Primary Productivity (NPP) It is the amount of energy left in the producers after utilisation of some energy for respiration.

Inter-relationship between GPP and NPP : Gross primary productivity minus the respiration losses is net primary productivity. It is actually the available mass for consumption by heterotrophs.
GPP – R = NPP where, R = Respiration losses.

Question 12.
Why earthworm is considered a farmer’s friend? Explain humification and mineralisation occurring in a decomposition cycle. (Foreign 2015)
Answer:
Earthworms are farmer’s friend because these help in fragmentation of detritus and loosening of the soil. Both these processes are helpful for the decomposition of dead organic matter at a faster rate into inorganic substances like carbon dioxide, water and nutrients. These can be used by crop plants for their growth.

Humification and mineralisation occurring during decomposition cycle are as follows Humification It is the process of accumulation of dark coloured amorphous substance called humus. Humus is highly resistant to microbial action and undergoes decomposition at very slow rate.
Mineralisation It is the process by which humus is degraded by some microbes to release inorganic substances.

Question 13.
How does a detritivore differ from a decomposer? Explain with an example. (Delhi 2015C)
Answer:
The differences between detritivores and decomposers are as follows

Question 14.
Justify the importance of decomposer in an ecosystem. (Foreign 2015)
Answer:
When a plant or animal dies, it leaves behind nutrients and energy in the organic material present in it’s body. Decomposers complete the decomposition process by breaking down this organic matter into NH₄, CO₂ and nutrients. This releases raw nutrients in a form, which is usable to plants again. This process resupplies nutrients to the ecosystem. When the decomposers feed on dead organisms, they leave behind nutrients. These nutrients become part of the soil.

Question 15.
(i) What is primary productivity? Why does it vary in different types of ecosystems?
(ii) State the relation between gross and net primary productivity. (Delhi 2014)
Answer:
(i) For primary productivity. Refer to answer no. 11.
For causes of its variation in different types of ecosystems, Refer to Answer No. 4.
(ii) For relation between gross and net primary productivity. Refer to Answer No. 11.

Question 16.
How is detritus decomposed step-by-step by different agents and made available as nutrients to the plants? Explain. (Delhi 2013C)
Or
Explain the different steps involved in the process of decomposition of detritus. (Delhi 2 one)
Or
Describe the process of decomposition of detritus under the following heads fragmentation, leaching, catabolism, humification and mineralisation. Delhi 2010
Answer:
Detritus It is the raw material for decomposition. It includes dead remains of plants (leaves, bark and flowers) and of animals including faecal matter.
It is largely an aerobic process, i.e. requires oxygen for its processing. Different steps involved in the process of decomposition are

• Fragmentation is the process of breaking down of detritus into smaller particles.
• Leaching is the process by which water soluble inorganic nutrients go down into the soil horizons and get precipitated as unavailable salts.
• Catabolism is the process of degradation of detritus into simple organic material by the action of bacterial and fungal enzymes and their further conversion into inorganic compounds.
• Humification is a process that leads to an accumulation of a dark coloured, amorphous and colloidal substance called humus, which is highly resistant to microbial action and decomposes at a very slow rate. It serves as a reservoir of nutrients.
• Mineralisation is the process by which humus is further degraded by microbial action and releases inorganic nutrients.

Question 17.
(i) Explain primary productivity and the factors that influence it.
(ii) Describe, how do oxygen and chemical composition of detritus control decomposition? (Delhi 2011)
Answer:
(i) For primary productivity, Refer to Answer No. 11.
Factors affecting primary productivity are

• Availability of nutrients.
• Quality and duration of sunlight.
• Water availability.
• Temperature of given place.
• Type of plant species inhabiting a particular area.
• Photosynthetic capacity of plants.

(ii) For control of decomposition by oxygen and chemical composition of detritus. Refer to Answer No. 10.

Question 18.
‘Man can be a primary as well as secondary consumer’. Justify this statement. (Foreign 2015)
Answer:
Man can be a primary as well as secondary consumer because man eats plants as well as meat of other animals. So, man is herbivore as well as carnivore.

Question 19.
State what does standing crop of a trophic level represent. (Delhi 2013)
Answer:
Standing crop of a trophic level represents the total mass of living material or energy content of all the organisms of a trophic level at a particular time and location.

Question 20.
What is a detritus food chain made up of ? How do they meet their energy and nutritional requirements? (All India 2013)
Answer:
Detritus food chain is made up of decomposers, i.e. some bacteria and fungi.
They meet their energy and nutrient requirements by degrading the dead organic matter of detritus.

Question 21.
Mention the role of pioneer species in primary succession on rocks. (Foreign 2012)
Answer:
Lichens are the pioneer species in the primary succession on rocks. They secrete acids to dissolve the rock and help in weathering and soil formation and pave way for small plants like bryophytes.

Question 22.
List any two ways of measuring the standing crop of a trophic level. (Foreign 2010)
Answer:
Standing crop can be measured as

• Biomass of living organisms in a unit area.
• Number of living organisms in a unit area.

Question 23.
It is possible that a species may occupy more than one trophic level in the same ecosystem at the same time. Explain with the help of one example. (All India 2013)
Answer:
A single species may occupy more than one trophic level at the same time in ecosystem because, the trophic level of species represents the functional level of species in the energy flow. The preference of food habit depends on the available food.
For example, sparrow is primary consumer when it eats seeds and secondary consumer when feeds on insects and worms.

Question 24.
Why the pyramid of energy is always upright? Explain. (Delhi 2013)

Answer:
Pyramid of energy is always upright because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat at each step.
Each bar in the energy pyramid indicates the amount of energy present at each trophic level in a given time.

Question 25.
Discuss the relationship between detritus food chain and grazing food chain in a terrestrial ecosystem. (All India 2012C)
Answer:
The producers occupy the first trophic level in terrestrial ecosystem and thus, make base of food chain. These producers (plants) are grazed upon by the grazers (herbivores). Primary carnivores prey upon herbivores and are eaten up by secondary carnivores.

The decomposers act on all dead animals/plants and return nutrients stored in them to the soil. These nutrients are again used up by the plants and cycle goes on.
This way the grazing food chain and detritus food chain are linked up.

Question 26.
Identify the type of given ecological pyramid and give one example each of pyramid of number and pyramid of biomass in such cases. (All India 2011)

Or
Explain standing crop in an ecosystem. Draw a pyramid of biomass when a small standing crop of phytoplanktons supports a large standing crop of zooplankton in the sea. (Delhi 2010)
Answer:
Given ecological pyramid represents the inverted pyramid of biomass wherein small standing crop of phytoplanktons supports large standing crop of zooplankton.

Pyramid of number is inverted in tree ecosystem and that of biomass in pond ecosystem.
Or
For standing crop in an ecosystem. Refer to Answer No. 2.

Question 27.
Name the type of food chains responsible for the flow of larger fraction of energy in an aquatic and a terrestrial ecosystem, respectively. Mention one difference between the two food chains. (Delhi 2010)
Answer:
Food chain responsible for the flow of large fraction of energy in an aquatic ecosystem is grazing food chain. In terrestrial ecosystem, it is detritus food chain.
Difference between grazing and detritus food chain is as follows
Grazing food chain:
Transfer of energy starts from producers.

Detritus food chain:
Transfer of energy starts from detritus/decomposing organic matter.

Question 28.
‘In a food chain, a trophic level represents a functional level, not a species’. Explain. (Delhi 2016)
Answer:
In a food chain, a trophic level represents a functional level, not a species. It is a specific position of an organism in the food chain. All organisms occupy a particular place in their natural surrounding or in a community according to their feeding relationship with other organisms.

This specific position is based on the organism’s feeding relationship with other organisms and the source of their nutrition or food.

Producers occupy the first trophic level, primary consumers (herbivores) occupy the second and the secondary consumers (carnivores) occupy the third level on the basis of their function in food chain.

Question 29.
Differentiate between primary and secondary succession. Provide one example of each. (All India 2016)
Answer:
Differences between primary succession and secondary succession are as follows

Question 30.
In a botanical garden of a city there is a huge banyan tree growing on which hundreds of birds and thousands of insects live. Draw the pyramids of numbers and also biomass represented by this community. Comment giving reasons on the two different pyramids drawn. (Delhi 2018C)
Answer:
Pyramid of number and pyramid of biomass showing interaction between trees, insects, birds and big birds is as follows

Pyramid of number in this ecosystem is spindle-shaped as the number of insects is maximum. The number of trees and birds are less than the insects. The number is gradually decreasing at each trophic level.

The pyramid of biomass in this ecosystem is erect because the biomass decreases at each trophic level.

Question 31.
What is ecological succession? Where and why would the rate of succession be faster in newly created pond or a forest destroyed by a forest fire? (Delhi 2015C)
Answer:
Ecological succession The sequential, gradual and predictable changes in the species composition in an area are called succession or ecological succession. The entire sequence of communities that successively changes in a given area is called sere (s). The species that invades a bare area and starts the ecological succession are called pioneer species.

Types of Ecological Succession

• Hydrarch succession The plant succession which takes place in wet area or water, leading to a successional series progress from hydric to the mesic conditions.
• Xerarch succession The plant succession which takes place in dry area, leading to a successional series from xeric to mesic conditions.

The rate of succession will be faster in a forest destroyed by fire because here some soil or sediment is present leading to secondary succession while in a newly created pond the succession is primary, therefore, it is slow.

Question 32.
Explain succession of plants in xerophytic habitat until it reaches climax community. (Delhi 2015C)
Or
Explain, how does a primary succession start on a bare rock and reach a climax community. (Delhi 2012)
Answer:
Primary Succession on Rocks (Xerarch succession)

• Lichens are the pioneer species on a bare rock.
• Lichens secrete acids to dissolve rock, help in weathering and soil formation.
• Later, small plants like bryophytes appear, which hold in small amount of soil.
• Bryophytes are succeeded by bigger plants.
• After several more stages of succession, ultimately a stable climax forest community is formed.
• In this way, xerophytic habitat gets converted into a mesophytic climax community. All successions whether taking place in water or on land, proceed to form a similar mesic climax community.

Question 33.
Differentiate between two different types of pyramids of biomass with the help of an example. (Delhi 2013)
Answer:
Pyramid of biomass refers to the relationship between producers and consumers in terms of biomass. It can be upright, e.g. in grasslands ecosystem or inverted, e.g. in pond ecosystem.

Question 34.
Explain the differences and the similarities between hydrarch and xerarch succession in plants. (Delhi 2011)
Answer:
Differences between hydrarch and xerarch succession are as follows

Similarity between hydrarch and xerarch succession Both these successions result in medium water conditions, i.e. mesic conditions.

Question 35.
Trace the succession of plants in dry bare rock. (All India 2010)
Answer:
Xerarch succession starts on bare rocks.
For the succession of plants in dry bare rock.
Refer to Answer No. 15.

Question 36.
(i) What is hydrarch succession ?
(ii) Compare the pioneer species and climax communities of hydrarch and xerarch succession respectively.
(iii) List the factors upon which the type of invading pioneer species depend in secondary hydrarch succession. Why is the rate of this succession faster than that of primary succession? (Delhi 2019)
Answer:
(i) Hydrarch succession occurs in wet areas or water, leading to successional series that progresses from hydric to the mesic condition.

(iii) The type of invading pioneer species in secondary hydrarch succession depends on the factor like condition of the soil, availability of water, the environment and also the seeds or other propagules present. The secondary succession is faster than primary succession because it begins in areas where natural biotic communities have been destroyed such as abandoned farm lands, burned or cut forests. Since, some soil is present, succession is faster than primary succession.

Question 37.
(i) The pyramid of energy is always upright. Explain.
(ii) Explain with the help of labelled diagrams the difference between an upright pyramid of biomass and an inverted pyramid of biomass. (All India 2019)
Or
Draw the pyramids of biomass in sea and in a forest. Explain giving reasons why are the two pyramids different ?
Or
It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass can be both upright and inverted. Explain with the help of examples and sketches. (All India 2015)
Or
(i) Draw the pyramids of hiomass in a sea and in a forest. Explain giving reason, why are the two pyramids different.
(ii) Pyramid of energy is always upright. Explain. (Foreign 2010)
Answer:
(i) Pyramid of Energy It represent the total energy of the organism in each trophic level. Pyramid of energy is always upright, i.e., it can never be inverted because when enegy is transferred from a particular trophic level to the next trophic level, some energy is always lost as heat at each step.

(ii) (a) The pyramid of biomass in a sea ecosystem is inverted. Because, the sum total of the weight of phytoplankton (producer) is far less than a few fishes feeding at higher trophic levels.

(b) Pyramid of biomass in a forest ecosystem is upright because producers are more in biomass than primary consumers.
Primary consumers are more than secondary consumers and secondary consumers are more than tertiary consumers (top).

Question 38.
(i) What is a trophic level in an ecosystem ? What is ‘standing crop’ with reference to it ?
(ii) Explain the role of the ‘first trophic level’ in an ecosystem.
(iii) How is the detritus food chain connected with the grazing food chain in a natural ecosystem? (2018)
Answer:
(i) Trophic level Organisms occupy a place in the natural surroundings or in a community according to their feeding relationship with the other organisms. Organisms also occupy a specific place in the food chain on the basis of their nutrition or food. It is known as trophic level.
Standing crop refers to the mass of a living material at a particular time in a trophic level.

(ii) First trophic level consists of producers. These are autotrophs. They trap the light energy and convert it into chemical energy.
The organisms of other trophic levels, i.e. consumers, depend on them for their energy and food requirement. Thus, first trophic level plays a very important role in the ecosystem.

(iii) Connection between grazing food chain and detritus food chain. Grazing food chain and detritus food chain are connected to each other.
Nutrients released by decomposers of detritus food chain are utilised by plants or autotrophs of grazing food chain. These plants in turn transfer these nutrients to other organisms. The dead remains of these organisms are source of food for the decomposers.

Question 39.
(i) What is an ecological pyramid? Compare the pyramids of energy, biomass and number.
(ii) Write any two limitations of ecological pyramids. (All India 2017)
Answer:
(i) Ecological pyramid It is the diagrammatic illustration of connection between different trophic levels in terms of energy, biomass and number of organisms. The base of each pyramid represents the producers or the first trophic level. Apex represents tertiary or top level consumers. In general, all pyramids are upright, but there are few exceptions.

(ii) Limitations of ecological pyramids:

• It never takes into account the same species belonging to two or more trophic levels.
• Saprophytes are not given any place in ecological pyramid even though, they play an important role in ecosystem.

Question 40.
(i) Taking example of a small pond, explain how the four components of an ecosystem function as a unit.
(ii) Name the type of food chain that exists in a pond. (All India 2016)
Answer:
(i) Following are the four components of pond ecosystem which function as unit

• Abiotic (light, temperature, water).
• Autotrophs (producers mainly phytoplanktons).
• Heterotrophs (zooplankton, small fishes, big fishes).
• Decomposers (bacteria, fungi).

Pond ecosystem is self-sustainable unit.
Abiotic components are water, inorganic nutrients dissolved in water, light, temperature. Biotic components are producers like algae and aquatic plants. Consumers are zooplanktons, larvae, tadpole and some fishes. Primary carnivores are water beetles, scorpions, dragonfly larvae, Hydra and some fishes. Decomposers include fungi, bacteria and some flagellates.

This pond system performs all the functions of an ecosystem, which are given below

• Autotrophs They convert inorganic materials with the help of solar energy.
• Heterotrophs They consume autotrophs.
• Decomposers They decompose and mineralise dead organic materials to release them back for reuse by the autotrophs. The above events are repeated again and again.

(ii) Grazing food chain exists in a pond.

Question 41.
(i) Differentiate between primary and secondary ecological successions.
(ii) Explain the different steps of xerarch succession occurring in nature. (Foreign 2014)
Answer:
(i) For differences between primary and secondary ecological successions, Refer to Answer No. 12.
(ii) For different steps of xerarch succession, Refer to Answer No. 15.

Question 42.
(i) With suitable examples, explain the energy flow through different trophic levels. What does each bar in this pyramid represent?
(ii) Write any two limitations of ecological pyramids. (Delhi 2014C)
Answer:
(i) The energy flows unidirectionally from the first trophic level (producers) to last trophic level (consumers) and as the energy flows from one trophic level to another, some energy is always lost as heat into the surrounding environment. So, the amount of energy flowing decreases at each successive trophic level. This can be explained with the help of a diagram of a grazing food chain. Refer to text on page no. 352.

The pyramid of energy is always upright and each bar in the pyramid indicates the amount of energy present at each trophic level in a given time or per unit area.

(ii) The limitations of ecological pyramids are

• It does not consider the same single species operating at two or more trophic levels.
• It assumes simple food chains that do not exist in nature and do not accommodate food web.
• Saprophytes, detritivores and decomposers are not given any place in pyramids, despite their vital role in ecosystem (any two).

Question 43.
(i) Draw a pyramid of numbers of a situation, where a large population of insects feed upon a very big tree. The insects in turn, are eaten by small birds which in turn are fed upon by big birds.
(ii) Differentiate giving reasons, between the pyramid of biomass of the above . situation and the pyramid of numbers that you have drawn. (Delhi 2012)
Answer:
(i) For pyramid of numbers, Refer to Answer No. 13.
(ii) For differentiation between the pyramid of biomass and pyramid of numbers of given situation, Refer to Answer No. 22(i).

Question 44.
(i) Explain the significance of ecological pyramids with the help of an example.
(ii) Why are the pyramids referred to as upright or inverted? Explain. (All India 2012)
Answer:
(i) Significance of ecological pyramids: They graphically represent the relation between producers and consumers in order to calculate energy content, biomass and number of organisms of that trophic level. A trophic level represents only a functional level not a species as such. A given species may occupy more than one trophic level in the
same ecosystem at the same time. The ecological pyramids provide an overall idea of the trophic levels occupied by an organism in an ecosystem.
Example A sparrow is a primary consumer when it eats seeds, fruits, peas and a secondary consumer when it eats insects and worms.

(ii) Upright pyramids: When producers are more in number and biomass than the herbivores and herbivores are more in number and biomass than the carnivores, pyramids are referred to as upright.
Energy at a lower trophic level is always more than at a higher trophic level. Pyramid of energy is referred to as always upright.
Inverted pyramids When the numbers and biomass of producers are less and consumers increase and become largest in top consumer level, Pyramid of number and biomass is referred to as inverted.

Question 45.
(i) Explain, how a hydrarch succession progresses from hydric to mesic condition and forms from a stable community?
(ii) Why is the rate of secondary succession faster than that of primary succession? (Delhi 2010C)
Answer:
(i) Hydrarch succession occurs in wet areas or water, leading to successional series that progresses from hydric to the mesic condition.
In hydrarch succession, pioneers are phytoplanktons. These phytoplanktons with time are replaced by free-floating angiosperms, followed by rooted hydrophytes, sedges, grasses and finally the trees, leading to mesic condition.

(ii) The secondary succession is faster than primary succession because it begins in areas where natural biotic communities have been destroyed such as abandoned farm lands, burned or cut forests. Since, some soil is present succession is faster than primary succession. At any time during primary or secondary succession, natural and human induced disturbance can convert a particular serai stage of succession to an earlier stage.

Question 46.
Explain how xerarch succession progresses from xeric to mesic condition and forms a stable climax community.
You may use a flow chart. (All India 2010c)
Answer:
Xerarch succession occurs in dry areas and the series progresses from xeric to mesic condition. The climax community remains stable as long as environment remains unchanged. With time, the xerophytic habitat gets converted into a mesophytic one.

For further detail about xerarch succession, Refer to Answer No. 15.

Question 47.
Mention four significant services that a healthy forest ecosystem provide. (Delhi 2019)
Answer:
Four significant services that a healthy forest ecosystem provide are as follows

• Provisioning services Fruits, vegetables, trees, fishes and livestock are directly available to us by ecosystem. The other benefits in this category are timber, oils, medicines, etc.
• Regulating services These are benefits that moderate natural phenomenon. These include pollination, decomposition, flood control, etc.
• Cultural services It contributes towards the development and advancement of people, e.g. recreation facilities, spiritual healing, etc.
• Supporting services Allow the earth to sustain basic life forms, e.g. photosynthesis, water cycle, etc.

Question 48.
Explain the function of ‘reservoir’ in nutrient cycle. List the two types of nutrient cycle in nature. (Foreign 2o11)
Answer:
Reservoir in an ecosystem meets the deficit that arises due to the imbalance in the rate of influx and efflux of nutrients.
The two types of nutrient cycles are

• Gaseous cycle
• Sedimentary cycle

Question 49.
Name the two types of nutrient cycles exist in nature. Where are their reservoirs present? State the function of reservoirs. (All India 2010C)
Answer:
Two types of nutrient cycles in nature
(i) (a) Gaseous cycles (carbon and nitrogen cycle).
(b) Sedimentary cycles (phosphorus and sulphur cycle).

(ii) (a) Reservoir for gaseous cycle is atmosphere.
(b) Reservoir for sedimentary cycle is earth’s crust.

Function of Reservoir It meets the deficit which occurs due to the imbalance in the rate of influx and efflux of nutrients.

Question 50.
How does phosphorus cycle differ from carbon cycle? (All India 2010)
Answer:
Differences between phosphorus cycle and carbon cycle are as follows

Question 51.
Global carbon is fixed in the biosphere through photosynthesis. Explain any two ways by which carbon is returned to the atmosphere. (All India 2010C)
Answer:
The carbon is returned to atmosphere by two ways

• All living organisms breath and release CO₂ into the atmosphere. So, carbon is replenished in the atmosphere in the form of carbon dioxide.
• Through combustion of the organic matter, the CO₂ is released into the atmosphere. The organic fossil fuels when burn also release CO₂ back to the atmosphere.

Question 52.
(i) State any two differences between phosphorus and carbon cycle in nature.
(ii) Write the importance of phosphorus in living organisms. (Foreign 2015)
Answer:
(i) For differences between phosphorus and carbon cycle, Refer to Answer No. 4.
(ii) Phosphorus is an important component of nucleic acids, biomembranes as phospholipids. It is important element of energy molecule ATP and energy producing ATP systems. In an animal’s body also, its salts are present in shells, bones and teeth. It takes part in metabolic reactions involved in release of energy from food and utilisation of this energy in various functions of the body.

Question 53.
State the function of a reservoir in a nutrient cycle. Explain the simplified model of carbon cycle in nature. (All India 2014)
Answer:
For function of a reservoir in a nutrient cycle.
Refer to Answer No. 2.
For simplified model of carbon cycle in nature. Refer to the diagram on page no. 363 and 364.

Question 54.
Draw and complete the following model of carbon cycle filling A, B, C, D, E and F. (Foreign 2009)

Answer:

Question 55.
(i) Name the specific cellular components where phosphorus is in abundance in living organisms.
(ii) Name the natural reservoir of phosphorus.
(iii) Explain the phosphorus cycle.
Answer:
(i) Phosphorus is a major constituent of biological membranes, nucleic acids and cellular energy transfer systems.
(ii) The natural reservoir of phosphorus is rock, in the form of phosphates.
(iii) For phosphorus cycle. Refer to text on page no. 363.

Question 56.
Discuss the role of healthy ecosystem services as a prerequisite for a wide range of economic, environmental and aesthetic goods and services. (Delhi 2017)
Answer:
Ecosystem services are the products of ecosystem processes. Forests are the major source of ecosystem services and are prerequisite for environmental, aesthetic goods and indirect economic values in the following ways
1. Environmental Values:

• Carbon-fixation Huge amount of CO₂ in the atmosphere is removed naturally and fixed by plants into organic molecules and energy through photosynthesis. All the other trophic levels, i.e. consumers depend upon this energy produced by them.
• Release of oxygen by the producers as a byproduct in the process of photosynthesis, improves the air quality and supports life on earth.
• Soil formation and soil protection are the major ecosystem services accounting for nearly 50% of their total worth. Plant cover protects the soil from drastic changes in the temperature. There is little wind or water erosion as soil particles are not exposed to them. The soil remains spongy and fertile. There occurs no landslides and floods.
• Nutrient cycling There is no depletion of nutrients, but the same are repeatedly circulated.

2. Economic Values:

• Ecosystem services provide certain economical important products also. These include timber, paper, rubber, wax, medicines, cosmetics, resins, etc.
• Ecosystem also provides other benefits such as aid in pollination. Bees and other insects of natural ecosystem visit nearby farmlands to pollinate crops.

3. Other Aesthetic Values Natural ecosystems are a source of spiritual, cultural and aesthetic values.

Question 57.
Describe the advantages for keeping the ecosystems healthy. (Delhi 2015)
Answer:
The various benefits that humans obtain from the ecosystem are collectively called ecosystem services.
The advantages of keeping an ecosystem healthy can be grouped into the following types

• Provisioning services Fruits, vegetables, trees, fishes and livestock are directly available to us by ecosystem. The other benefits in this category are timber, oils, medicines, etc.
• Regulating services These are benefits that moderate natural phenomenon. These include pollination, decomposition, flood control, etc.
• Cultural services It contributes towards the development and advancement of people, e.g. recreation facilities, spiritual healing, etc.
• Supporting services allow the earth to sustain basic life forms, e.g. photosynthesis, water cycle, etc.

Question 58.
(i) Draw a simplified model of phosphorus cycle in a terrestrial ecosystem.
(ii) Write the importance of such cycles in ecosystems. (All India 2014C)
Answer:
(i) Refer to figure 14.9 on page no. 363.
(ii) The importance of nutrient cycle in ecosystem can be briefed as

• It allows exchange, storage and transfer of biogenetic nutrients through various biotic components of ecosystem.
• It also allows exchange of nutrients between abiotic and biotic components of ecosystem.
• Recycling of nutrients allows them to be used repeatedly and indefinitely.

Question 59.
Explain the carbon cycle with the help of a simplified model. Delhi 2012
Answer:
For Carbon cycle, Refer to text on page no. 363 and 364.

Question 60.
(i) (a) Name the given biogeochemical (nutrient) cycle.

(b) Name an activity of the living organisms not depicted in the cycle by which this nutrient is returned to the atmosphere.
(ii) How would the flow of nutrient in the cycle be affected due to the large scale deforestation? Explain giving reasons. (Delhi 2011)
Answer:
(i) (a) It is a carbon cycle.
(b) Decomposition of organic wastes by decomposers cycles the carbon back into the atmosphere.

(ii) Large scale deforestation will negatively affect the flow of nutrients in the cycle by increasing the level of CO₂ in the air. The CO₂ level will increase because the CO₂ present in the atmosphere will not be fixed by photosynthesis in the absence of plants. Plants annually fix 4 × 10³ kg of carbon through photosynthesis.

If such a huge amount of carbon will not be utilised in photosynthesis, the organism will die at large scale due to lack of food. The decomposition of these dead organisms will further add to amount of CO₂ available in the atmosphere.

Question 61.
Differentiate between standing state and standing crop in an ecosystem. (Foreign 2010)
Answer:
Differences between standing state and standing crop are as follows

Question 62.
(i) Healthy ecosystems are the base of wide range of (ecosystem) services. Justify.
(ii) Explain the differences and the similarities between hydrarch and xerarch successions of plants. (Delhi 2011)
Answer:
(i) Healthy ecosystem provides following ecological services and acts as a base of such services in the following ways

• Purification of air and maintenance of gaseous composition.
• Mitigation of droughts and floods.
• Cycling of nutrients.
• Storehouse of carbon.
• Maintenance of biodiversity.
• Habitat for a number of wildlife.
• Influence on hydrological cycle.

(ii) Differences between hydrarch and xerarch succession are as follows

Similarities Both hydrarch and xerarch succession lead to mesic conditions.

Question 63.
(i) Trace the succession of plants on a dry bare rock.
(ii) How does phosphorus cycle differ from carbon cycle ? (All India 2010)
Answer:
(i) Primary succession on rocks The organisms that first invade a bare area are called pioneer species. The pioneer species on a bare rock are usually lichens. Lichens secrete acids which dissolve rocks, thereby leading to weathering and soil formation. This paves the way for small plants or bryophytes which hold the soil. They are succeeded by bigger plants and ultimately an entire forest gets established. Forests represent the climax community in this succession.

(ii) Differences between phosphorus cycle and carbon cycle are as follows

Question 64.
In a food chain, grass is consumed by goat which in turn is consumed by humans. Anant states that man is the secondary consumer in this food chain. Is Anant correct? It yes, give reason.
Answer:
Yes, Anant is correct. In this food chain grass is the producer which feeds energy into the food chain and goat is the primary consumer. Since, goat is eaten by humans, they are called secondary consumers.

Question 65.
Aryan worked on the farm with his father.
He saw that his father added earthworms to the soil. He came to his father and told him not to do so. His father explained why he was doing so and Aryan was satisfied with the explanation.
(i) Why were earthworms added to the soil?
(ii) What values do you observe in Aryan’s father?
Answer:
(i) Earthworms are detritivores which cause the breakdown of detritus into small, simpler forms. This helps in improving soil quality.
(ii) Aryan’s father is intelligent, hard worker and environment friendly.

Question 66.
Rajni was watching a programme on TV about the earth. The host was explaining how the increased cutting of trees and forests is affecting the ecosystem, its components and services, etc. She got curious and asked her teacher about ‘what services the ecosystem provides to human beings?
(i) How does ecosystem services benefit us?
(ii) Name the value put forth by Robert Constanza and his colleagues on ecosystem services.
(iii) What values are showed by Rajni?
Answer:
(i) Healthy ecosystems provide wide range of economic, environmental, aesthetic goods and services. These services include purification of air, mitigation of droughts and floods, cycling of nutrients, providing habitats for wildlife, maintenance of biodiversity, etc.
(ii) Robert Constanza and his colleagues have estimated the average value of fundamental ecosystem service at US $ 33 trillion a year.
(iii) Values shown by Rajni are awareness and curiosity towards environmental affairs.

The post Ecosystem Class 12 Important Questions and Answers Biology Chapter 14 appeared first on Learn CBSE.

We have given these Class 12 Biology Important Questions Chapter 13 Organisms and Populations to solve different types of questions in the exam. Go through these Class 12 Biology Chapter 13 Important Questions, Organisms and Populations Important Questions & Previous Year Questions to score good marks in the board examination.

Class 12 Biology Chapter 13 Important Questions Organisms and Populations

Question 1.
Why are mango trees unable to grow in temperate climate? (Outside Delhi 2016)
Answer:
Mango trees are unable to grow in temperate climate because low temperature affects the kinetics of enzyme’s functioning and through this basal metabolism, activity and other physiological functions of the organism get affected.

Question 2.
Give an example of an organism that enters ‘diapause’ and why? (Delhi 2014)
Answer:
Many zooplanktons in lakes and ponds enter diapause so, as to escape unfavourable environmental conditions and to delay their overall development.

Question 3.
Mention how do bears escape from stressful time in winter? (Delhi 2013C)
Answer:
Bears escape from stressful time in winter by going into hibernation.

Question 4.
Write the basis on which an organism occupies a space in its community/natural surroundings. (All India 2013)
Answer:
An organism occupies individual or species level in its community. This level is occupied on the basis of ecological level of organisation or ecological hierarchy.
Individual → Population → Biotic community → Biome

Question 5.
How do seed bearing plants tide over dry’ and hot weather conditions? (All India 2013C)
Answer:
During unfavourable conditions, seed-bearing plants reduce their metabolic activities and undergo a state of dormancy. The seeds germinate under favourable moisture and temperate conditions.

Question 6.
Why are some organisms called as eurythermal and some others as stenohaline? (Foreign 2011)
Answer:
Organisms, which can tolerate and thrive in a wide range of temperature are called as eurythermal, while organisms, which can tolerate and thrive in a narrow range of salinity are stenohaline.

Question 7.
Why are green plants not found beyond a . certain depth in the ocean? (Delhi 2011)
Answer:
Beyond a certain depth, green plants are not found, because light is unavailable in that zone.

Question 8.
Mention any two activities of animals, which get cues from diurnal and seasonal variations in light intensity. (Delhi 2011c)
Answer:
The two activities of animals, which get cues from diurnal and seasonal variations in light intensity are

• Timings of their foraging
• Migratory activities
• Reproduction

Question 9.
How do animals like fishes and snails avoid summer related unfavourable conditions? (Delhi 2010)
Answer:
Fishes migrate and snails undergo into aestivation or summer sleep to avoid summer related problems.

Question 10.
How do prickles help cactus to survive in desert? Give two methods. (All India 2010C)
Answer:
The two methods by which prickles help cactus to survive in desert are

• By reducing and altering outer surface, they reduce evaporation and transpiration of water.
• By providing defence against grazing animals.

Question 11.
Plants that inhabit a rainforest are not found in a wetland. Explain. (Delhi 2016)
Answer:
Rainforest plants are not found in wetland as they have adapted to growing in very wet soil. The rainforest plants have a thin, smooth bark because they do not need thick bark to prevent moisture. The smoothness of the bark makes it difficult for other plants to grow on their surface. While in wetland, plants have roots in the soil under the water, but they grow above the water for respiration (pneumatophores).

Question 12.
Why the plants that inhabit a desert are not found in a mangrove? Give reasons. (Delhi 2016C)
Answer:
Desert plants are not found in a mangrove because in xerophytic (desert) condition, plant roots grow deeply to explore any possibility of underground water, lehves are highly reduced (spines) to minimise transpiration. While mangroves grow in marshy areas. Their roots grow vertically upwards from the soil for the absorption of oxygen from the atmosphere as the soil is poorly aerated. These roots are called pneumatophores.

Question 13.
Heat loss or heat gain depends upon the surface area of the organism’s body. Explain with the help of a suitable example. (All India 2016C)
Or
Why are small animals rarely found in the polar regions? Explain. (Foreign 2010)
Answer:
Small animals have a large surface area relative to their volume. So, they tend to lose body heat very fast during cold conditions. They need to spend more energy to generate body heat. Due to this, smaller animals are rarely found in polar regions.

Question 14.
How do mammals living in colder regions and seals living in polar regions able to reduce the loss of their body heat? (Delhi 2015C)
Answer:
Mammals living in colder climates generally have shorter ears and limbs to minimise heat loss. This is called Allen’s rule. In polar seas, aquatic mammals like seals have a thick layer of fat (blubber) below their skin that acts as an insulator and reduces loss of body heat.

Question 15.
Shark is eurythermal, while polar bear is stenothermal. What is the advantage the former has and what is the constraint the later has? (Delhi 2015C)
Answer:
The organisms that can tolerate and thrive in wide range of temperature are called eurythermal and those restricted to narrow range of temperatures are called stenothermal.
The levels of thermal tolerance of different species determine to a large extent their geographical distribution. Hence, eurythermal organisms (e.g. sharks) have vast geographical distribution as compared to stenothermal organisms (e.g. polar bears) that are limited to certain areas only.

Question 16.
Many freshwater animals cannot survive in marine environment. Explain. (Delhi 2015)
Answer:
Sea water contains high quantity of salt that is not favourable for freshwater animals. They face osmotic problems, hence they cannot survive in sea water for long.

Question 17.
When you go for a trek/trip to any high altitude places, you are advised to take it easy and rest for first two days. Comment, giving reason. (Foreign 2015)
Or
Why do people suffer from altitude sickness after reaching the high altitude regions? How does their body acclimatised after a couple of days? (Delhi 2015C)
Or
How does our body adapt to low oxygen availability at high altitudes? Foreign 2011
Answer:
When we go for a trip to any high altitude places, we are advised to take it easy and rest for first two days because we may suffer from high altitude sickness. A! low atmospheric pressure of high altitudes, body does not get enough oxygen, hence symptoms like nausea, fatigue start appearing. So, the body requires time to acclimatise to these conditions.
Some body adaptations at high altitudes are

• To compensate low oxygen, the production of red blood cells is increased.
• Faster breathing rate.

Question 18.
Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples. (Delhi 2012)
Answer:
Examples of organisms that suspend their metabolic activities in unfavourable conditions.

• Bacteria, fungi and lower plants They form thick-walled spores, which help them to survive in unfavourable conditions. Spores germinate on the return of favourable conditions.
• Higher plants Seeds and some other vegetative reproductive structures serve as the means to tide over periods of stress. They reduce their metabolic activity and undergo dormancy.
• Animals They undergo hibernation or aestivation, if unable to migrate, e.g. some snails and fishes.
• Zooplanktons They enter diapause (suspended development) under unfavourable conditions.

Question 19.
Explain the responses of all communities to environment over time. (All India 2011)
Answer:
Responses of communities to environment are

• Some organisms maintain homeostasis by physiological or behavioural means (regulate).
• The internal environment in majority of animals and nearly all plants change with the change of external environment (conform).
• Some organisms leave their habitats temporarily during unfavourable conditions and return back, when conditions become favourable (migrate).
• Some organisms suspend their metabolic activities to avoid stress by timely escaping, e.g. hibernation and aestivation. (1/2 x 4)

Question 20.
Bear hibernates, whereas some species of zooplanktons enter diapause to avoid stressful external conditions. How are these two ways different from each other? (Foreign 2011)
Answer:
Differences between diapause and hibernation is

Question 21.
How do seals adapt to their natural habitat? Explain. (Foreign 2010)
Answer:
Seals adapt to the natural habitat (cold climate) by developing a thick layer of fat (blubber) below their skin that acts as an insulator and reduces excess loss of body heat,

Question 22.
Hummingbirds live among the bushes in tropics, while penguins live on icebergs. They cannot survive if their habitats are reversed. Justify. (Delhi 2010C)
Answer:
Hummingbirds live in their natural habitats (tropics). They have large surface area relative to their volume. So, they tend to lose heat very fast, even when it is warm outside.

Penguins live on icebergs (natural habitat). They have less surface area to volume ratio. The lesser the ratio, more effective will be the thermoregulation. Also, they hide in group to escape from cold conditions. Therefore, both of them will not survive, if their habitat are reversed.

Question 23.
(i) ‘Organisms may be conformers or regulators.’ Explain this statement and give one example of each.
(ii) Why are there more conformer than regulators in the animal world? (All India 2017)
Answer:
(i) Organisms may be regulators or conformers depending upon the response they show against abiotic stresses.
Regulators are organisms that have the ability to maintain a relatively constant internal environment by physiological and behavioural means.
These organisms ensure a constant body temperature (thermoregulation), constant osmotic concentration (osmoregulation), etc.
For example, in all birds, mammals, etc., sweating occurs profusely and the evaporation brings down the temperature of the body during summers.

Conformers Some organisms cannot.maintain a constant internal environment, i.e. their osmotic concentration and body temperature changes with the ambient surrounding. Such organisms are called as conformers. In aquatic animals such as Asterias, the osmotic concentration of body fluids changes according to the osmotic concentration of the surrounding water.

(ii) In the animal world, conformers are more than the regulators because thermoregulation is energetically expensive for many organisms. This is specially true for small animals like shrews and hummingbirds. Heat loss and heat gain is a function of surface area. Since, small animals have a larger surface area relative to their volume, they tend to loose body heat very fast.

Question 24.
During a school trip to ‘Rohtang Pass’, one of your classmate suddenly developed ‘altitude sickness’. But, he/she recovered after some time.
(i) Mention one symptom to diagnose the sickness.
(ii) What caused the sickness?
(iii) How could she/he recover by her/himself after some time? (Delhi 2016)
Answer:
(i) The primary symptoms of ‘altitude sickness’ are headache, nausea, fatigue, etc.
(ii) Sickness was caused due to the low atmospheric pressure prevailing at high altitude. Due to this, the body gets deprived of sufficient oxygen.
(iii) Ascending slowly is the best way to avoid altitude sickness. This way the body compensates low oxygen availability by increasing red blood cells production, decreasing the binding capacity of haemoglobin and by increasing breathing rate. Thus, she/he recovers by herself after some time.

Question 25.
Why do tribes who live in high altitude of Himalayas experience discomfort in respiration? How do they get adapted to survive in such a situation? (All India 2015C)
Answer:
The tribes who live in the high altitudes of Himalayas, experience discomfort in respiration because of the low atmospheric pressure prevailing there. Due to this, the body gets deprived of sufficient oxygen. The body copes up with the low oxygen stress by the following adaptations

• Increasing red blood cell production.
• Decreasing the binding affinity of haemoglobin.
• increasing the breathing rate.

Question 26.
In certain seasons we sweat profusely, while in some other season we shiver. Explain. (Delhi 2016C)
Answer:
In certain seasons, we sweat profusely while in some other season we shiver to maintain homeostasis. Homeostasis ensures constant body temperature and constant osmotic concentration of the body. It is maintained by following means. Human body maintains constant body temperature (37°C) by following way.
In summers, the outside temperature is very high than our body temperature. Due to this, profuse sweating occurs. This causes evaporation and cooling effect on the body.
In winters, the outside temperature is much lower than our body temperature. This causes shivering, a kind of exercise that produces heat and raises the body temperature.

Question 27.
Explain with the help of suitable examples the three different ways by which organisms overcome their stressful conditions lasting for short duration. (Delhi 2016)
Or
Explain by taking three different examples how do certain organisms pull through the adverse conditions when unable to migrate under stressful period. (Delhi 2016C)
Answer:
Organisms either migrate or suspend their metabolic activities, when conditions are stressful for short duration.
However, in the absence of migration, following mechanisms Help an organism to overcome adverse conditions

• Metabolic activities are reduced to undergo a state of dormancy, e.g. higher plants.
• Hibernates (frogs) or aestivates (snails) or undergo diapause (zooplanktons).
• Thick-walled spores are formed in stressful conditions and germinate under suitable conditions, e.g. bacteria, fungi and lower groups of plants.

Question 28.
Why are certain organisms called regulators or conformers? Explain with the help of one example of each. (All India 2016C)
Answer:
Some organisms are called conformers because they cannot maintain constant internal environment, e.g. lower organisms. On the other hand, certain organisms are called regulators as they can maintain their internal environment in response to external conditions, e.g. humans.

Question 29.
How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to conditions unfavourable for their survival? (All India 2015)
Or
How do organisms like fungi, zooplanktons and bears overcome the temporary short-lived climatic stressful conditions? Explain. (All India 2010)
Answer:
Ways to adapt to unfavourable conditions are

• Snails undergo aestivation (summer sleep) to avoid summer related problems.
• Bears undergo hibernation during winter.
• Zooplanktons enter a stage of suspended development called diapause.
• Bacteria and fungi slow down their metabolic rate and form a thick-walled spores to overcome stressful conditions. These spores germinate after the onset of suitable environment.
• Seeds undergo a state of dormancy during unfavourable conditions and germinate under favourable conditions.

Question 30.
(i) State how the constant internal environment is beneficial to organisms.
(ii) Explain any two alternatives by which organisms can overcome stressful external conditions. (All India 2014)
Answer:
(i) Constant internal environment (homeostasis) is beneficial to organisms as it permits all biochemical reactions and physiological functions to proceed with its optimum efficiency, thereby enhancing the overall efficiency of an organism under wide geographical range.
(ii) The two alternatives by which organisms can overcome stressful external conditions are

• Migration Organisms move temporarily to a favourable area under stressful conditions and return back when the period is over.
• Hibernation and aestivation are ways to escape the stress during winters and summers, respectively.

Question 31.
Water is very essential for life. Write any three features both for plants and animals, which enable them to survive in water scarce environment. (Delhi 2012)
Answer:
Three features in plants and animals that enable them to survive in water scarce environment are as follows
Adaptations in Plants

• Thick cuticle on their leaf surface.
• Stomata are arranged in deep pits (sunken) to minimise water loss through transpiration.
• Leaves are reduced to spines. The photosynthetic function is carry out by thick, fleshy, flattened stems.

Adaptations in Animals

• Kangaroo rat meets the water requirement through internal oxidation of fats. They concentrate their urine, so that minimum volume of water is excreted.
• Snails undergo aestivation during summers.

Question 32.
The following graph represents the organismic response to certain environmental condition (e.g. temperature)
(i) Which one of these A or B depicts conformers?
(ii) What does the other line graph depict?
(iii) How do these organisms differ from each other with reference to homeostasis?
(iv) Mention the category to which humans belong. (All India 2009)

Answer:
(i) A depicts conformers.
(ii) The other line B depicts regulators.
(iii) Differences between conformers and regulators are

(iv) Humans are regulators.

Question 33.
(i) Following are the responses of different animals to various abiotic factors. Describe each one with the help of an example.
(a) Regulate
(b) Conform
(c) Migrate
(d) Suspend
(ii) If 8 individuals in a population of 80 butterflies die in a week, calculate the death of population of butterflies during that period. (2018)
Answer:
(a) Regulate Some organisms are able to maintain a constant body temperature and osmotic concentration despite change in the external environment. All birds and mammals and a very few lower vertebrate and invertebrate species are capable of such regulation by physiological adjustment through thermoregulation and osmoregulation.

For example, most of the mammals regulate their body temperature to about 37°C. The body may respond to it by sweating (at high temperatures) and shivering (at low temperatures). Most reptiles regulate their body temperature by staying in burrows during summer and by basking in sun during winters. However, plants do not have such mechanisms to maintain an internal temperature.

(b) Conform A majority of organisms cannot maintain a constant internal environment, i.e. their osmotic concentration and their body temperature change with their environment (ectotherms). Such organisms are conformers. The efficiency of conformers is reduced under stressful conditions. Conformers had not evolved mechanisms like regulators because thermoregulation and osmoregulation mechanisms are energetically expensive for many small-sized organisms.

(c) Migrate The organisms can move away temporarily from the stressful habitat to a more favourable one. They return back when the unfavourable (stressful) period is over.

(d) Suspend To survive under unfavourable conditions, the organisms like bacteria, fungi and lower plants produce various kinds of thick-walled spores, which germinate on availability of suitable environment. Similarly higher plants, produce dormant structures like seeds to cope up with unfavourable conditions. In case of animals, if the organism is unable to migrate, it avoids unfavourable conditions by escaping in time. During unfavourable conditions, zooplanktons may enter a stage of suspended development, termed as diapause.

(ii) Refer to Answer No. 18 (Topic 2).

Question 34.
(i) Explain giving reasons why the tourists visiting Rohtang Pass or Mansarovar are advised to resume normal active life only after a few days of reaching there.
(ii) It is impossible to find small animals in the polar regions. Give reasons. (All India 2012)
Answer:
(i) Tourists visiting to Rohtang Pass near Manali (> 3500 m) may suffer from altitude sickness. They resume normal active life only after, a week because in low atmospheric pressure at high altitudes, the body does not get enough oxygen.
Gradually, the body compensates low oxygen availability by

• Increasing red blood cell production.
• Decreasing the binding affinity of haemoglobin.
• Increasing the breathing rate.

(ii) For reasons why small animals are not found in polar regions, Refer to Answer No. 13.

Question 35.
Mention the term used to describe a population interaction between an orchid growing on a forest tree. (Delhi 2019)
Or
What is an interaction called when an orchid grows on a mango plant? (Delhi 2012)
Answer:
An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them and use them only for support. Hence, the relationship between a mango tree and an orchid is an example of commensalism.

Question 36.
Provide an instance where the population size of species can be estimated indirectly, without actually counting them or seeing them. (Delhi 2016C)
Answer:
Tiger population in National Parks is calculated on the basis of pugmarks and faecal matter.

Question 37.
Name the interaction that exists between sucker fish and shark. (Delhi 2016C)
Or
Name the type of interaction that exists between barnacles and whale. (Delhi 2015C)
Answer:
Sucker fish and shark show commensalism. Other example includes whale and barnacles growing on its back.

Question 38.
State the type of interaction that exists between ticks and dogs. (All India 2015C)
Or
Name the interaction that exists between Cuscuta and shoe-flower plant. (Delhi 2015C)
Answer:
The type of interaction that exists between ticks and dog is parasitism. Ticks are ectoparasite.
Or
The interaction between Cuscuta and shoe-flower bush is called parasitism. Here, Cuscuta is the parasite, which infests the shoe-flower bush and derives nutrition from it.

Question 39.
Name the type of interaction seen between fig and wasps. (All India 2015C)
Answer:
Mutualism is the interaction that exists between fig and wasps.

Question 40.
Name the two intermediate hosts on which the human liver fluke depends to complete its life cycle, so as to facilitate parasitisation of its primary host. (Delhi 2014)
Answer:
The human liver fluke requires two intermediate hosts, i.e. freshwater snail and fish to complete its life cycle and facilitate parasitisation of its primary host.

Question 41.
State Gause’s competitive exclusion principle. (All India 2014)
Answer:
Gause’s competitive exclusion principle states that two competiting species for same resource cannot coexist, if all other ecological factors are constant.

Question 42.
Name the interspecific interactions in which one is detrimental while other is neutral. (All India 2013C)
Answer:
Amensalism, e.g. Penicillium produces toxin and kills bacteria, but the former does not get affected.

Question 43.
Write, what do phytophagous insects feed on? (Delhi 2012)
Answer:
Phytophagous insects feed on sap mainly, but can consume the other parts of plants as well.

Question 44.
Mention the unique feature with respect to flowering and fruiting in bamboo species. (Delhi 2012)
Answer:
Bamboo plants flower only once in their lifetime, generally after 50-100 years. They produce large number of fruits and then die.

Question 45.
In a pond, there were 20 Hydrilla plants. Through reproduction, 10 new Hydrilla plants were added in a year. Calculate the birth rate of the population. (Delhi 2012)
Answer:
The birth rate of Hydrilla
=\(\frac{\text { Number of individuals born }}{\text { Total number of individuals }}\)
= \(\frac { 10 }{ 20 }\) = 0.5 per Hydrilla plant per year.
Birth rate is 0.5 per Hydrilla plant or 500 per thousand/year.

Question 46.
Pollinating species of wasps show mutualism with specific fig plants.
Mention the benefits the female wasps derive from the fig trees from such an interaction. (All India 2011)
Answer:
The wasp uses the fig plant ovary for oviposition. It also uses the developing seeds of the fruit to nourish its larvae.

Question 47.
Why are cattle and goats not seen browsing on Calotropis growing in the fields? (Foreign 2011)
Answer:
Cattle or goat do not graze on Calotropis plant because it produces poisonous cardiac glycosides.

Question 48.
In a pond, there were 200 frogs. 40 more were born in a year. Calculate the birth rate of the population. (Delhi 2010)
Answer:
The birth rate of frog population
= \(\frac { 40 }{ 200 }\) = 0.2 per frog/year
or 200 per thousand/year.

Question 49.
Why do predators avoid eating monarch butterfly? How does the butterfly develop this protective feature? (Foreign 2010)
Answer:
Predators avoid the monarch butterfly as it is highly distasteful to its predators (birds) because of a special chemical present in its body. It acquires this chemical during the caterpillar stage by feeding on a poisonous plant (milkweed).

Question 50.
Comment on the interaction between a clown fish living among the tentacles of a sea anemone. (Delhi 2010)
Answer:
The interaction between a clown fish living among the tentacles of sea anemone is called commensalism.

Question 51.
Comment on the interaction between certain species of fig trees and wasps. (Delhi 2010C)
Answer:
The relation between fig trees and wasps is mutualism.

Question 52.
If 8 individuals in a laboratory population of 80 fruitflies died in a week, then what would be the death rate for population for the said period? (Delhi 2010)
Answer:
Death rate = \(\frac{\text { Number of individual died }}{\text { Total number of individuals }}\)
= \(\frac { 8 }{ 80 }\) = 0.1 individuals/week 80
Death rate will be 0.1 individuals /week.

Question 53.
State Gause’s ‘competitive exclusion’ principle. How have the recent studies modified this principle? (All India 2019)
Answer:
Gause’s competitive exclusion principle states that two closely related species competing for the same resources cannot coexist indefinitely and the competitively inferior one will be eliminated eventually (true if resources are limiting). This principle has been modified to ‘complete competitors cannot coexist’ one of the primary ways niche-sharing species can coexist is the competition-colonisation trade-off. In simple terms, species that are better competitors will be specialists, whereas species that are better colonisers are more likely to be generalists.

Question 54.
Name and explain the interaction that is seen between clown fish and sea anemones. (All India 2019)
Or
Why dd clown fish and sea anemone pair up? What is this relationship called? (Delhi 2012; All India 2008)
Answer:
Clown fish maintains commensalism with the sea anemone. In this interaction, one species is benefitted and the other is neither harmed nor benefitted. Sea anemone has stinging tentacles that provide protection to clown fish from predators. The anemone does not appear to derive any benefit from the clown fish.

Question 55.
What is mutualism? Mention any two examples where the organisms involved are commercially exploited in agriculture. (All India 2015)
Or
Explain mutualism with the help of an example. (All India 2014)
Answer:
Mutualism is an interaction that confers benefits to both the interacting species. Two examples where organisms involved are commercially exploited in agriculture are

• Mycorrhizae represent a close mutual association between fungi and the roots of higher plants. Fungi help the plants in absorption of nutrients, specially phosphorus, while the plants provides food for the fungus.
• Lichens represent an intimate mutualistic relationship between a fungus and photosynthesising algae or cyanobacteria.
  Here, the fungus helps in the absorption of nutrients and provides protection, while algae prepare the food.

Question 56.
Koel is clever enough to lay eggs in a crow’s nest. Write the reason for this peculiar behaviour. Name the type of interaction. (All India 2015C)
Or
Mention the changes the koel must have undergone in order to achieve brood parasitism, during the course of evolution. (All India 2010C)
Answer:
Brood parasitism in birds is a fascinating example of parasitism in which the parasitic bird lay its eggs in the nest of its host and let the host incubate them.
During the course of evolution, the egg of the koel have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird, detecting the foreign eggs and ejecting them from the nest.

Question 57.
Draw the labelled diagrams of stable and declining age pyramids of human population. (All India 2015C)
Answer:
For stable and declining age pyramids of human population diagrams, Refer to figure 13.3 on page no. 325.

Question 58.
Describe the mutual relationship between . fig tree and wasp and comment on the phenomenon that operates in their relationship. (All India 2014)
Answer:
The relationship between fig tree and wasp shows mutualism. The wasp while searching for sites to lay its eggs, pollinates the fig’s inflorescence. On the other hand, the fig not only provides shelter (fruit) for oviposition to wasp, but also allows its larvae to feed on seeds.

Question 59.
Construct an age pyramid which reflects an expanding growth status of human population. (All India 2014)
Answer:
The age pyramid geometrically represents the proportions of different age groups (pre-reproductive, reproductive and post-reproductive) in population. The triangular shape of age pyramid represents the expanding growth status of human population, because it has maximum number of individuals in pre-reproductive age group.

Question 60.
Construct an age pyramid which reflects a stable growth status of human population. (All India 2014)
Answer:
The age pyramid that reflects a stable growth status of human population can be represented as follows

Question 61.
Differentiate between commensalism and mutualism by taking one example each from plants only. (All India 2014)
Answer:
Commensalism is the kind of interaction between species in which one is benefitted and other is neither benefitted nor harmed. Example of such association is orchid growing as an epiphyte on a mango tree, which remains unaffected by its growth.

Whereas mutualism is the type of interaction in which both the species involved are benefitted. e.g. lichen representing mutual association between algae and fungi, in which algae is protected by fungi, whi,ch also provides nutrients for synthesis of food, while algae provides food to fungi, as they are incapable of synthesising their own food.

Question 62.
Explain Verhulst-Pearl logistic growth of a population. (All India 2014)
Answer:
Verhulst-Pearl logistic growth is a type of population growth in which the population growing in a habitat with limited resources initially shows a lag phase, followed by exponential phase and finally a declining or stationary phase, when the growth or density of population reaches carrying capacity.
It can be explained by following equation
dN/dt = rN\(\left(\frac{K-N}{K}\right)\)
Where, N – Population density at time t
r – Intrinsic rate of natural increase
K – Carrying capacity

Question 63.
Apart from being a part of food chain, predators play other important roles. Mention any two such roles supported by examples. (Delhi 2014)
Answer:
Apart from being a part of food chain, predators play important role in

• Maintaining the prey population and regulate interspecies population.
• Indicating ecological disturbances as predators are small in number and highly sensitive to ecological changes owing to their excessive adaptation.

Question 64.
How does monarch butterfly defend itself from predators? Explain. (Delhi 2013C)
Answer:
For defence mechanism evolved by the monarch butterfly, Refer to Answer No. 15.

Question 65.
Explain brood parasitism with the help of an example. (All India 2012)
Answer:
The phenomenon in which one organism (parasite) lays its eggs in the nest of another organism is called brood parasitism.

For example, cuckoo (parasite) lays eggs which resemble the host’s (crow) egg in size and colour in crow’s nest and let it be incubated by them till they hatch and grow.

Question 66.
Why are herbivores considered similar to predators in an ecological context? Explain. (All India 2010)
Answer:
In ecological context, herbivores are considered similar to predators because they feed on plants and their products to meet their food requirements, just like predators, i.e. they feed on prey for their food.

Question 67.
How does the floral pattern of mediterranean orchid Ophrys guarantee cross-pollination? (Delhi 2010)
Answer:
In the flowers of Ophrys, one petal resembles the female of a bee species in size, colour, markings, etc. Male bee perceives it as a female and pseudocopulates with it.

During the process, the pollen grains from the anthers get dusted on the body of the bee. When the bee is attracted to another flower of this orchid species, the. process is repeated and the pollen grains from the body of the bee get dusted on the stigma, thus cross-pollination is achieved.

Question 68.
How do plants benefit from having mycorrhizal symbiotic association? (Foreign 2010)
Answer:
Benefits to plants having mycorrhizal symbiotic association are

• The fungus absorbs nutrients from the soil and passes it on to the plant.
• Mycorrhiza provides resistance to root-borne pathogens.
• They show increased tolerance to salinity and drought.
• An overall increase occurs in plant growth and development.

Question 69.
Explain the two defence mechanisms evolved in preys to avoid overpopulation of their predator. (All India 2010C)
Answer:
Defence mechanisms evolved in preys are

• To avoid being detected easily by the predators, some species of insects and frogs are cryptically coloured (camouflaged).
• Some plants have thorns or spines for defence mechanism, e.g. Acacia, cactus.

Question 70.
Study the graph given below and answer the questions that follows

(i) The curve W is described by the following equation :
\(\frac { dN }{ dt }\) = rN \(\frac { K-N }{ K }\)
What does ‘K’ stand for in this equation? Mention its significance.
(ii) Which one of the two curves is considered a more realistic one for most of the animal populations?
(iii) Which curve would depict the population of a species of deer if there are no predators in the habitat? Why is it so? (All India 2019)
Answer:
(i) ‘K’ stands for ‘carrying factor’. The carrying capacity signifies the limit of habitat, i.e. limited resources in a given habitat to support growth upto a certain level beyond which no further growth can take palce.
(ii) The curve ‘V’ is considered a more realistic one for most of the animal populations. It is so, becasue in the curve ‘b’, the sources of food and space are limited and it supports the growth curve of animal populations.
(iii) The curve ‘b’ would depict the population of a species of deer, if there are no predators in the habitat. In the absence of predators, prey population will increase. Thus, the competition will increase due to the limited food and shelter resources within the prey population.

Question 71.
Mention the special adaptations evolved in parasites and why? (Delhi 2019)
Or
Explain parasitism and coevolution with the help of one example of each. (All India 2016)
Or
Explain coevolution with reference to parasites and their hosts. Mention any four special adaptive features evolved in parasites for their parasitic mode of life. (All India 2015C)
Answer:
Coevolution is a phenomenon where many parasites have evolved to be host-specific in such a way that both the host and parasite tend to co-evolve. If the host develops special mechanism to resist the parasite, the parasite also has to evolve the mechanisms to counteract the host’s resistance.

Many living organisms live as a parasite on host (plant or animals). Parasite ensures free-lodging and meals from the host and for this it evolves mechanisms to encounter and neutralise the resistance coming from host. This can be understood by the following examples
(i) Parasites evolve special adaptations like

• loss of’unnecessary sense organs.
• presence of adhesive organs like suckers.
• loss of digestive system.
• high reproductive capacity.

(ii) The host in turn evolves special mechanism for rejecting or resisting the parasite. So, when the parasite is not able to complete its life cycle on single host it finds alternative host or secondary host to complete their life cycle in order to perpetuate its species and ensures its own reproduction. Two examples are

• Liver fluke-hosts are man and snail.
• Malaria parasite-hosts are man and mosquito.

Question 72.
Explain brood parasitism with the help of an example. (Delhi 2016C; 2013C)
Answer:
For brood parasitism, Refer to Answer No. 31.

Question 73.
Explain Gause’s competitive exclusion principle with the help of a suitable example. (All India 2016C)
Answer:
Cause’s competitive exclusion principle Refer to Answer No. 7.
For example, The abingdon tortoise in Galapagos Islands became extinct within a decade after goats were introduced on the Island, apparently due to their greater browsing efficiency.