Mahesh

Students get through AP Inter 2nd Year Physics Important Questions 1st Lesson Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 1st Lesson Waves

Very Short Answer Questions

Question 1.
Write the formula for the speed of sound in solids and gases.
Answer:
Speed of sound in solids,
V_s = [latex]\sqrt{Y / \rho}[/latex] [y = Young’s modulus of solid, ρ = density of solid]
Speed of sound in gases,
V_s = [latex]\sqrt{\gamma P / \rho}[/latex] [γP = Adiabatic Bulk modulus of gas, ρ = density of gas] .

Question 2.
What does a wave represent?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 3.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare factions are formed alternatively.

Question 4.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = [latex]\frac{2 \pi}{T}[/latex]; k = [latex]\frac{2 \pi}{\lambda}[/latex]
Parameters:

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 5.
What is the principle of superposition of waves ? .
Answer:
When two or more waves are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.
If y₁, y₂, ……………… y_n be the individual displacements of the particles,then resultant displacement
y = y₁ + y₂ + ……………… + y_n.

Question 6.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 7.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary ?
Answer:
Phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary is radian or 180°.

Question 8.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transeverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 9.
What do you understand by the terms node’ and ‘antinode’ ?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes: The points at which the amplitude is maximum, are called antinodes.

Question 10.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is [latex]\frac{\lambda}{4}[/latex].

Question 11.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 12.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics..
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 13.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + [latex]\frac{1}{2}[/latex]) [latex]\frac{v}{2l}[/latex] where n = 0, 1, 2, 3, ……….

Question 14.
If the air column in a long tube, closed at one end, is set in vibration, what harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n +1][latex]\frac{v}{4 l}[/latex] where n = 0, 1, 2, 3,

Question 15.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible ?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by no
v_n = [latex]\frac{\mathrm{nv}}{21}[/latex]
where n = 1, 2, 3, ……………….

Question 16.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.
Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.
It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard, ∆υ = υ₁ ~ υ₂
Importance:
1. It can be used to tune musical instruments.
2. Beats are used to detect dangerous gases.
Explanation for tuning musical instruments with beats:
Musicians use. the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 4.
What is ‘Doppler effect’ ? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between .the observer and the source of sound is called doppler effect.
Examples:

1. The frequency of whistling, engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Sample Problem on Doppler effect:
Two trucks heading in opposite direction with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330m/s). After the two trucks have passed each other, what frequency does the driver of the second truckhear?
Answer:
Speed of first truck = 60 kmph
= 60 × [latex]\frac{5}{18}[/latex] = 16.66 m/s;
Speed of second truck = 70 kmph 5
= 70 × [latex]\frac{5}{18}[/latex] = 19.44 m/s
Frequency of horn of first truck = 400 Hz;
Velocity of sound, (V) =330 m/s
Frequency of sound heard by the driver of the second truck when approaching each other,
v¹ = [latex]\left(\frac{V+V_0}{V-V_s}\right) v=\left(\frac{330+19.44}{330-16.66}\right)[/latex] × 400 = 446 Hz
Frequency of sound heard by the driver of the second truck when approaching each other,
V¹¹ = [latex]\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{v}=\left(\frac{330-19.44}{330+16.66}\right)[/latex] × 400 = 358.5 Hz

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. [IPE]
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length λ and frequency v, travelling in opposite direction be given by

y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω = 2πv and k = [latex]\frac{2 \pi}{\lambda}[/latex]
The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, [latex]\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}[/latex] ……………… etc, the amplitude = zero
These positions are known as “Nodes”.
If x = [latex]\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}[/latex] …………… etc., the amplitude = maximum (2a)
These positions are called “Antinodes”.
If the string vibrates in ‘P’ segments and T is its length, then length of each segment = [latex]\frac{l}{\mathrm{P}}[/latex]
Which is equal to [latex]\frac{\lambda}{2}[/latex]
∴ [latex]\frac{l}{\mathrm{P}}=\frac{\lambda}{2} \Rightarrow \lambda=\frac{2 l}{\mathrm{P}}[/latex]
Harmonic frequency v = [latex]\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}[/latex]
v = [latex]\frac{v \mathrm{P}}{2 l}[/latex] ………………. (1)
If’ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is
v = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}[/latex] …………… (2)
From the Eqs (1) and (2) :
Harmonic frequency v = [latex]\frac{\mathrm{P}}{2 l} \sqrt{\frac{\Gamma}{\mu}}[/latex]
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = [latex]\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex] …………….. (3)

Laws of Transverse Waves Along Stretched String:
Fundamental frequency of the vibrating string v = [latex]\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex]
First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ [latex]\frac{1}{l}[/latex] ⇒ vl = constant, when ‘T’ and ‘μ’ are constant. .

Second Law: When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ [latex]\sqrt{T}[/latex] ⇒ [latex]\frac{v}{\sqrt{T}}[/latex] = constant, when ‘l’ and ‘m’ are constant.
JT .
Third Law: When the length (l) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ [latex]\frac{1}{\sqrt{\mu}}[/latex] ⇒ [latex]v \sqrt{\mu}[/latex] = constant, when ‘l’ and T are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. [A.P. 17; IPE 2015, 2016 (TS)]
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super-imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two anti nodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength [latex]\left(\frac{\lambda_1}{2}\right)[/latex]
l = [latex]\frac{\lambda_1}{2}[/latex] ⇒ λ₁ = 2l
fundamental frequency v₁ = [latex]\frac{\mathrm{v}}{\lambda_1}[/latex] where v is velocity,of sound in air, v₁ = [latex]\frac{v}{21}[/latex] = v
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = [latex]\frac{2 \lambda_2}{2}[/latex] ⇒ λ₂ = [latex]\frac{21}{2}[/latex]
If ‘v₂’ is frequency of second harmonic then v₂ = [latex]\frac{v}{\lambda_2}=\frac{v \times 2}{2 l}[/latex] = 2v
v₂ = 2v ……………… (2)
Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = [latex]\frac{3 \lambda_3}{2}[/latex]
λ₃ = [latex]\frac{2l}{3}[/latex]
If ‘v₂‘ is frequency of third harmonic then
v₃ = [latex]\frac{v}{\lambda_3}=\frac{v \times 3}{2 l}[/latex] = 3V
v₃ = 3v …………… (3)
Similarly we can find the remaining or higher harmonic frequencies i.e., v3, v4 etc., can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : V₁ : v₂ = 1 : 2 : 3

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. [IPE 2015, 2016(A.P.), (T.S) A.P. & T.S. Mar. 15]
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = [latex]\frac{\lambda_1}{4}[/latex] ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = [latex]\frac{v}{\lambda_1}[/latex] where ‘υ’ is velocity of sound in air
v₁ = [latex]\frac{v}{4 l}[/latex] = v …………….. (1)
To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to [latex]\frac{3}{4}[/latex] of the wavelength.
∴ l = [latex]\frac{3 \lambda_3}{4}[/latex] where ‘λ₃‘ is wave length of third harmonic
λ₃ = [latex]\frac{4l}{3}[/latex]
where ‘X3’ is wave length of third harmonic.
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = [latex]\frac{v}{\lambda_3}=\frac{3 v}{41}[/latex]
v₃ = 3v …………….. (2)
Similarly the next overtone in the close pipe is only fifth harmonic, it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to [latex]\frac{5}{4}[/latex] of wave length (λ₅)
∴ l = [latex]\frac{5 \lambda_5}{4}[/latex] where ‘λ₅‘ is wave length of fifth harmonic.
λ₅ = [latex]\frac{4l}{5}[/latex]
If ‘v₅‘ is frequency of fifth harmonic (second overtone)
v₅ = [latex]\frac{v}{\lambda_5}=\frac{5 v}{4 l}[/latex]
v₅ = 5v …………….. (3)
∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [Mar. 17, BMP, 2016 (AP) Mar. 14, (TS)]
Answer:
Doppier effect: The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppier effect. When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = Listener
Let ‘S’ be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T = υ_s. T Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength λ’ = λ – υ_sT.
λ’ = λ – [latex]\frac{v_s}{v}[/latex] [∵ υ = [latex]\frac{1}{T}[/latex]]
= [latex]\frac{\lambda v-v_s}{v}=\frac{v-v_s}{v}[/latex] [∵ υ = vλ]
If “v'” is apparent frequency heard by the listener then v’ = [latex]\frac{v}{\lambda^{\prime}}[/latex] where ‘υ’ is Velocity of sound in air
v’ = [latex]\frac{v . v}{v-v_S}[/latex]
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency v’ = [latex]\frac{v . v}{v+v_s}[/latex], which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity.

Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg / m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [IPE 2016 (T.S)
Answer:
v = 30 Hz; I = 0.6 m; μ = 0.05 kg m^-1
υ = ?; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N .

Question 2.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;
μ = [latex]\frac{\mathrm{M}}{1}=\frac{0.16 \times 10^{-3}}{0.4}[/latex] = 0.4 × 10^-3 kg/m;
T = 70 N; v_n = [latex]\frac{\mathrm{P}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex]
v₁ = [latex]\frac{1}{21} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 \times 0.4} \sqrt{\frac{70}{0.4 \times 10^{-3}}}[/latex] = 523 Hz
v₂ = 2v₁ = 2 × 523 = 1046 Hz
v₃ = 3v₁ = 3 × 523 = 1569 Hz

Question 3.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
l = 70 cm = 70 × 10^-2m; v = 331 m/s ;
v = ?
v = [latex]\frac{v}{4 l}=\frac{331}{4 \times 70 \times 10^{-2}}[/latex] = 118.2 Hz.

Question 4.
A steel cable of diameter 3 cm is kept under a tension of lOkN. The density of steel is 7.8 g/ cm3. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = [latex]\frac{D}{2}=\frac{3}{2}[/latex] cm
= [latex]\frac{3}{2}[/latex] × 10^-2m;

Question 5.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle. [T.S. Mar. 17]
Solution:
When a whistling train away from rest observer,
v’ = [latex]\left[\frac{v}{v-v_s}\right] v[/latex] ……………… (1)
When a whistling train away from rest observer,
v” = [latex]\left[\frac{v}{v+v_s}\right] v[/latex] ……………… (2)
Here v’ = 219 Hz; v” = 184 Hz;
v = 340 m/s

Question 6.
A rocket is moving at a speed of 200 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ [latex]\left[\frac{1+v_{\mathrm{S}}}{v}\right]^{-1}[/latex] and not the approximate
v = v₀ [1 – [latex]\frac{v_s}{v}[/latex]]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [1 – [latex]\frac{v_s}{v}[/latex]]^-1
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v₀, Therefore, the frequency as registered by the rocket is
v’ = [latex]v\left(\frac{v+v_0}{v}\right)[/latex]
= 2540 Hz × [latex]\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)[/latex]
≃ 4080Hz

Textual Examples

Question 1.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both.
a) Motion of kink in a longitudinal spring produced by displacing one end of the spring sideways.
b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
c) Waves produced by a motorboat sailing in water.
d) Ultrasonic waves in air produced by a vibrating quartz crystal.
Solution:
a) Transverse and longitudinal
b) Longitudinal
c) Transverse and longitudinal
d) Longitudinal.

Question 2.
A wave travelling along a string is des-cribed by, y(x, t) = 0.005 sin (80.0 x – 3.01), in which the numerical constants are in SI untis (0.005 m, 80.0 rad m^-1, and 3.0 rad s^-1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ?
Solution:
On comparing this displacement equation with Eq. y (x, t)n = a sin(kx – ωt + Φ)
y(x, t) = a sin (kx – ωt).
We find
a) the amplitude of the wave is
0. 005 m = 5 mm.

b) the angular wave number k and angular frequency ω are k = 80.0 m^-1 and ω = 3.0 s^-1
We then relate the wavelength λ to k through Eq.
λ = [latex]\frac{2 \pi}{K}[/latex]
= [latex]\frac{2 \pi}{80.0 \mathrm{~m}^{-1}}[/latex] = 7.85 cm

c) Now we relate T to ω by the relation
T = [latex]\frac{2 \pi}{\omega}[/latex]
= [latex]\frac{2 \pi}{3.0 \mathrm{~s}^{-1}}[/latex]
= 2.09 s
and frequency, v = [latex]\frac{1}{T}[/latex] = 0.48 Hz
The displacement y at x = 30.0 cm and time t= 20s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (-36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) ≃ 5 mm

Question 3.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the .wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? [A.P. Mar. 19]
Solution:
Mass per unit length of the wire,
μ = [latex]\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}[/latex] = 6.9 × 10^-3 kg m^-1
Tension, T = 60 N
The speed of wave on the wire is given by
υ = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}[/latex] = 93 m s^-1

Question 4.
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.
Solution:
We know that 1 mole of any gas occupies 22.4 litres at STP Therefore, density of air at STP is:
ρ₀ = (mass of one mole of air) / (Volume of one mole of air at STP)
= [latex]\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}[/latex] = 1.29 kgm^-3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
υ = [latex]\left[\frac{1.01 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1.29 \mathrm{~kg} \mathrm{~m}^{-3}}\right]^{1 / 2}[/latex] = 280 m s^-1

Question 5.
A pipe, 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source ? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s^-1.
Solution:
The first harmonic frequency is given by
v¹ = [latex]\frac{v}{\lambda_1}=\frac{v}{2 L}[/latex] (open pipe)
Where L is the length of the pipe. The frequency of its n^th; harmonic is
v_n = [latex]\frac{n v}{2 L}[/latex] for n = 1, 2, 3, ……………… (open pipe)
First few modes of an open pipe are shown in Fig.
For L = 30.0 cm. υ = 330 m s^-1
v_n = [latex]\frac{\mathrm{n} \times 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}[/latex] = 550 s^-1

Clearly, for a source of frequency 1.1 kHz the air column will resonate at υ₂, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig.), the fundamental frequency is

and only to odd numbered harmonics are present :
v₃ = [latex]\frac{3 v}{4 L}/latex], v₅ = [latex]\frac{5 v}{4 L}[/latex] and s0 0n.
For L = 30 cm and υ = 330 m s^-1;, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Question 6.
Two sitar strings A and B playing the note ‘Dha’ are slightly put of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Solution:
Increase in the tension of a string increases its frequency. It the original frequency of B (v_B) were greater than that of A(v_A) .further increase in v_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that v_B < v_A. Since v_A – v_B = 5 Hz, and v_a = 427 Hz, we get v_B = 422 Hz.

Question 7.
A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 ms^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ [latex]\left[\frac{1+\mathrm{v}_{\mathrm{S}}}{\mathrm{v}}\right]^{-1}[/latex] and not the approximate
v = v₀ [latex]\left[1-\frac{v_s}{v}\right][/latex]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [latex]\left(1-\frac{v_{\mathrm{S}}}{v}\right)^{-1}[/latex]
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v₀, the frequency intercepted by the target and not v0, Therefore, the frequency as registered by the rocket is
v’ = [latex]v\left(\frac{v+v_0}{v}\right)[/latex]
= 2540 Hz × [latex]\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)[/latex]
≃ 4080Hz

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 14th Lesson Microbes in Human Welfare Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 14th Lesson Microbes in Human Welfare

Very Short Answer Questions

Question 1.
Why does ‘Swiss cheese’ have big holes? Name the Bacteria responsible for it.
Answer:
The large holes in ‘Swiss cheese’ are due to the production of large amounts of CO₂ by Propionibacterium shamanic.

Question 2.
What are fermenters?
Answer:
These are large vessels in which microbes are grown in large numbers on an industrial scale.

Question 3.
Name a microbe used for statin production. How do statins lower blood cholesterol levels?
Answer:
Monascus purpureus statins act by competitively inhibiting the enzyme responsible for the synthesis of cholesterol.

Question 4.
Why do we prefer to call secondary waste water treatment as Biological treatment?
Answer:
The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful microbes into floes while growing these microbes consume the major part of the organic matter in the effluent, which reduces the BOD (Biochemical oxygen demand).

Question 5.
What is Nucleo polyhedro virus is being used for now-a-days?
Answer:
These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications. They have been shown to have no negative impacts on plants, mammals, birds, fish or even on non-target insects.

Question 6.
Write the most important characteristic that Aspergillus niger, Clostridium butylicum and Lactobacillus share.
Answer:
The microbes are acid producers. E.g. : Aspergillus niger (a fungus) produce citric acid, Clostridium butylicum (a bacterium) produce butyric acid and Lactobacillus (a bacterium) produce lactic acid.

Question 7.
Name any two genetically modified crops.
Answer:
Bt-cotton, Bt-brinjal.

Question 8.
Name any two industrially important Enzymes.
Answer:
Lipases, Streptokinase.

Question 9.
Name an Immuno suppressive Agent. From where it is obtained?
Answer:
Cyclosporin-A, Trichoderma polysporum.

Question 10.
What is the group of Bacteria found in both the rumen of cattle and sludge of sewage treatment?
Answer:
Methanogens – Methanobacterium.

Question 11.
Name the scientists who were credited for showing the role of Pencillin as an antibiotic. Ernst chain and Howard florey after Alexander Fleming.

Short Answer Questions

Question 1.
How do mycorrhizal fungi help the plants harbouring them?
Answer:
The symbiotic association between fungal members and roots of vascular plants is called mycorrihiza. Many members of the genus Glomus forms Mycorrhiza. The fungal symbiont in these associations facilitates absorption of phosphorous by the plant from the soil. Plants having such associations show resistance to root-borne pathogens, tolerance to salinity and drought and an overall increase in plant growth and development.

Question 2.
What is the chemical Nature of Biogas? Explain the process of Biogas production.
Answer:
Biogas is a mixture of methane (CH₄), Co₂, traces of H₂S and moisture. It is generated by the decomposition of excreta or dung of cattle, domestic waste material and industrial and agricultural sewage due to the activity of anaerobic bacteria (Methanococcus and Methanobacillus) present in them. It is used as fuel. The Bacteria which grow anaerobically on cellulosic material product large amount of Biogas, and the Bacteria are called Methariogens.

E.g : Methano bacterium commonly found in the anaerobic sludge during sewage treatment and also present in the food of cattle and in the rumen of cattle. In rumen, this bacteria help in the breakdown of cellulose and play an important role in the nutrition of cattle. The excreta or dung can be used for generation of Biogas (Gobar gas).

Question 3.
What ire Biofertilisers? Write a brief note on them.
Answer:
Organisms that enrich the nutrient quality of the soil are called Bio fertilisers.
E.g.: Bacteria, Fungi and Cyanobacteria.
1) Bacteria :
In symbiotic Bacteria, Rhizobium is present in the root nodules and legumes. These Bacteria fix atmospheric Nitrogen into organic forms which is used by the plant as a nutrient. Other Bacteria can fix atmospheric Nitrogen, while free living in the soil are Azospirillum and Azotobacter thus enriching the Nitrogen content of’the soil.

2) Fungi :
The symbiotic association between fungal members and roots of vascular plants is called mycorrhiza. Many members of the genus Glomus forms Mycorrhiza. The fungal symbiont in these associations facilitates absorption of phosphorous by the plant from the soil. Plants having such associations show resistance to root-borne pathogens, tolerance to salinity and drought and an overall increase in plant growth and development.

Long Answer Questions

Question 1.
Write brief essay on microbes in sewage treatments?
Answer:
Municipal waste-water is also called sewage. It contains large amount of organic matter and microbes, many of which are pathogenic. This sewage cannot be discharged into natural water bodies before disposal, sewage is treated in sewage treatment plants to make it less polluting treatment of waste water is done by heterotrophic microbes naturally present in the sewage. This treatment is carried out in two stages.

1) Primary treatment:
It involves physical removal of particles from the sewage through filteration and sedimentation. Floating debris is removed by seanential Alteration. Then the grit (soil & small pebbles) is removed by sedimentation.

All solids that settle form the primary sludge and the supernatant forms the effluent. The effluent is taken for the secondary treatment.

2) Sedbndary treatment or Biological treatment:
The primary effluent is taken into large tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of aerobic microbes into floes. While growing, these microbes consume the major of the organic matter in the effluent. This significantly reduces the BOD of the effluent. The sewage water is treated till the BOD is reduced. Now the effluent is passed into a settling tank where the bacterial floes are allowed to sediment. This sediment is called activated sludge.

This activated sludge is pumped into aeration tanks (small amount) and anaerobic sludge digesters (large amount). Here other kind of bacteria which grow anaerobically, digest the bacteria & fungi in the sludge. During this digestion bacteria produces a mixture of gases such as methane, H₂S & CO₂. The effluent from the secondary treatment plant is released into natural water bodies like rivers and streams.

Intext Questions

Question 1.
Bacteria cannot be seen with the naked eye, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small cup of curd and contains millions of bacteria which can be easily observed under a microscope.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
Bacteria and fungi carry out the process of Fermentation and during this process, they release CO₂. Fermentation is the process of converting a complex organic substances into a simples substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of CO₂ and little energy. The dough used for making idli and dosa gives a pubbed appearance due to the release of CO₂.

Question 3.
Name the states involved in Ganga action plan.
Answer:
Uttarakhand, Uttar Pradesh.

Question 4.
Name some Indian traditional Indian foods made of wheat, rice and bengals gram (or their products). Which of these food involve the use of microbes?
Answer:
a) Wheat products – Bread, Cake
b) Rice products – Idli, Dosa
c) Bengal gram products – Dhokla, Khandvi

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Answer:
Several microorganisms are used for preparing medicines. Antibiotics are medicines produced by certain microorganisms to kill other disease causing microorganisms. These medicines are commonly obtained from bacteria and fungi. They kill or stop the growth of microorganisms.
E.g.: Pencillium notatum produces, pencillin which checks the growth of staphylococci bacteria.

Question 6.
Do you think microbes can also be used as a source of energy? If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as methane bacterium used for the generation of gobar gas. The dung is mixed with water to form the slurry and thrown into the tank. The tank is filled with numerous anaerobic methane bacteriun which produce bio gas from slurry. Bio gas can be removed through the pipe and is then used as a source of energy, while the spent slurry is removed from the outlet and js used as a fertilizer.

Question 7.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished?
Answer:
Microbes play an improtant role in organic farming, which is done without the use of chemical fertilizers and pesticides. Biofertilisers are living organisms which help increasing the fertility of soil. Biofertilisers are introduced in seeds, roots or soil to mobilize the availability of nutrients. Thus they are extremely beneficial in enriching the soil with organic nutrients. Thus cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of legumes.

Biofertilizers are ecofriendly microbes can also act as biopesticides to control insect pests in plants. E.g.: Bacillus thuringiensis which produce a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural field, which larvae of insects feed on crops, these bacterial spore enter the gut of the larvae and release toxins, thereby it dies. Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens. Baculoviruses is another biopesticide that is used as a biological control agent against insets and arthropods.

Question 8.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which’. The BOD values of the three samples A, B and C were recorded as 20 mg/L, 8 mg/L and 400 mg/L respectively. Which sample of the water is most polluted ? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
BOD (Biological Oxygen Demand) is the method of determining the amount of oxygen required by microorganisms to decompose the waste present in water supply. If the quantity of organic waste in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value also will increase.

Therefore it can be concluded that if the water supply is more polluted, then it will have a higher BOD value out of the above three samples, sample C is most polluted since it has the maximum BOD value of 400 mg/L. After this, secondary effluent discharge from a sewage treatment plant is most polluted. Thus sample A has the BOD value of 20 mg/L while sample B is river water and has the BOD value of 8 mg/L. Hence correct label for each sample is.

A – 20 mg/L – Secondary effluent discharge from a sewage treatment plant
B – 8 mg/L – River water
C – 400 mg/L – Untreated sewage water.

Question 9.
Name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agent) are obtained.
Answer:
Cyclosporin A – Trichoderma polysporum
Statin – Monascus purpureus

Question 10.
Find out the role of microbes in the following and discuss it with your teacher.
a) single cell protein (SCP) b) Soil.
Answer:
a) Single cell protein :
It is a protein obtained from contain microbes which forms an altenate source of proteins in animal feeds. The microbes involved are algae, fungi, yeast or bacteria. These microbes are grown On an industrial scale to obtain the desired protein. E.g.: Spirulina can be grown oh waste materials obtained from molasses, sewage and animal manures. It seves a rich supplement of dietary nutrients such as proteins, carbohydrates, fats, minerarls and vitamins. Methylophilous methylotrophus have a large rate of Biomass production, produce large amounts of proteins.

b) Soil:
Soil microbes play an important role in maintaining soil fertility. They help in the formation of nutrient rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rlizobium is a symbiotic bacteria found in the root nodules of legumes. Azospirillum and Azetobacter are free living nitrogen fixing bacteria.

Question 11.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Bio gas, Citric acid, Penicillin and Curd.
Answer:
Penicillin → Biogas → Citric acid → Curd
Pencillin:
It is the most important product (Antibiotic) for the welfare of Human society. It is used for controlling various bacterial diseases.

Biogas :
It is an eco-friendly source of energy.

Citric acid :
It is used as a food preservative.

Curd:
A food item obtained by the action of lactobacillus bacteria of milk.

Question 12.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste matter that is carried away in sewars and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water borne diseases. Sewage water is a major cause of polluting drinking water. Hence it is essential that sewage water is properly collected, treated and disposed.

Question 13.
What is the key difference between primary and secondary sewage treatment?
Answer:

Primary sewage tratement	Second sewage tratement
1) It is a mechanical process involving the removal of coarse solid particles.	1) It is a biological process involving the action of microbes.
2) It is inexpensive and relatively less complicated.	2) It is a very expensive and compli- catedprocess.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 13th Lesson Strategies for Enhancement in Food Production Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 13th Lesson Strategies for Enhancement in Food Production

Very Short Answer Questions

Question 1.
What is meant by “hidden hunger”?
Answer:
It is a chronic lack of vitamins and minerals that often has no visible warning signs. It leads to mental impairment, poor health, and productivity or even death.

Question 2.
Name two Semi-Dwarf varieties of rice developed in India.
Answer:
Jaya and Ratna.

Question 3.
Give two examples of wheat varieties introduced in India, which are high-yielding and disease resistant.
Answer:
Sonalika and Kalyan Sona.

Question 4.
Give two examples of fungi used in SCP production.
Answer:
Candida utilis (Torula yeast), Saccharomyces cerevisiae (Baker’s yeast), Chaetomium cellulolyticum.

Question 5.
Which two species bf sugarcane were crossed for better yield?
Answer:
Saccharum barberi’and Saccharum officinarum were crossed for better yield, high sugar etc.

Question 6.
Define totipotency and explant.
Answer:
Totipotency:
The ability of a cell (or) an explant to regenerate into a complete plant is called totipotency.

Explant:
It is a part of a plant taken out and grown in a test tube under sterile conditions in special nutrient media.

Question 7.
Define micropropagation and somaclones.
Answer:
Micropropagation:
The production of large number of plants in a very short time and limited space is known as micropropagation.

Somaclones :
Plants grown through tissue culture which are genetically identical with the original plant are called somaclones.

Question 8.
What is meant by germplasm collection?
Answer:
The entire collection of plants/seeds, having all the diverse alleles for all genes in a given crop is called germplasm collection.

Question 9.
What is meant by Biofortification?
Answer:
Development of crops with higher levels of vitamins and minerals or higher protein and healthier fats to improve public health.

Question 10.
Which part of the plant is best suited for making virus-free plants and why?
Answer:
Meristem. Here the cells are in dividing state.

Short Answer Questions

Question 1.
Give few examples of biofortified crops. What benefits do they offer to the society?
Answer:
1) Wheat variety, Atlas 66, having a high protein content.
2) Rice variety, Golden Rice – β carotene containing variety.

6) Broad, lablab, French and garden peas – Protein enriched beans.

Benefits :

1. It aims at breeding crops with higher levels of vitamins and minerals.
2. Higher protein and healthier fats to improve public health.

Question 2.
Write a short note on SCR
Answer:
Dried biomass of a single cell species of microbes that can be used as proteins are called single cell protein.

1. SCP is an alternative sources of proteins for animal and human nutrition.
2. Microbes like algae, fungi and bacteria are used in SCP production.
3. Spirulina maxima, chlorella pyrenoidosa and scenedesmos acutus are the examples of Algal SCP.
4. Spirulina can be grown easily on materials like waste water from potato processing plants, straw, molasses and animal manure and can serve as food which is rich in protein, minerals, fats, carbohydrates & vitamins.
5. SCP utilization can reduces environmental pollution.
6. It is possible to produce large amount of protein in short period of time. For example 250 kg cow produces 200 g of protein per day. In the same period 250 g of a protein produces by methylophilus methylotrophus bacterial SCP.
7. Candida utilis, Bakers yeast, chaetomium cellulolyticum are good examples of Fungal SCP.

Long Answer Questions

Question 1.
You are a Botanist working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety.
Answer:
The main steps in breeding a new genetic variety of a crop are :

1) Collection of Variability:
Genetic variability is the root of any breeding programme. Collection and preservation of all the different wild varieties, species and relatives of cultivated species is a prerequisite for effective exploitation of natural genes available in the populations.

The entire collection having all the diverse alleles for all genes in a given crop is called Germplasm collection.

2) Evaluation and selection of parents :
The germplasm is evaluated so as to identify plants with desirable characters. The selected plants are multiplied and are used. Purelines are created wherever desirable.

3) Cross Hybridisation among the selected parents :
After emasculation (Removal of Anthers from bisexual flower of a fpmale parent) the female flowers are enclosed in a polythene bag to prevent undesired cross pollination.

pollen grains are collected from the male parent with the help of a brush and are transferred to the surface of the stigma and thus cross pollination is affected artificially.

4) Selection and Testing of superior recombinants :
It involves selecting among the progeny of hybrids, those plants that have the desired character combination. The selection process requires careful scientific evaluation of the progeny. Due to this, plants that are superior to both the parents are obtained. These are self pollinated for several generations till they reach a homozygosity.

5) Testing, release and commercialisation of new characters :
The newly selected lines are evaluated for their yield and other traits of quality, disease resistance etc. It is done by growing these in research fields and recording their performance under ideal fertilizer application, irrigation and other crop management practices. It is followed by testing the materials in farmers fields for at least 3 growing seasons at several places in the country, in all agroclimatic zones. Finally they are distributed to farmers as a new variety.

Question 2.
Describe the tissue culture technique and what are the advantages of tissue culture over conventional method of plant breeding in crop improvement programmes?
Answer:
Tissue culture Technique : It involves
a) Preparation of Nutrient medium :
The nutrient medium is a mixture of various essential nutrients, aminoacids, vitamins and carbohydrates. These are mixed in distilled water and PH is adjusted to 5.6 to 6.0. Growth regulators like auxins, cytokinins are added to the medium. The nutrient medium is poured in glass vessels and closed tightly with cotton plugs before sterilizing them in an autoclave.

b) Sterilisation :
The nutrient medium is rich in nutrients and therefore attracts the growth of microorganisms. The culture medium is auto claved for 15 mins, at 121°c or 15 pounds of pressure to make aseptic.

c) Preparation of explant:
Any living part of plant can be used as explant. The explants must be cleaned with liquid detergent and in running water and surface sterilised with sodium hypo chlorite and rinsed with distilled water.

d) Inoculation of explants :
The transfer of explants on to the sterilized nutrient medium is called inoculation. It is carried out under sterilized conditions.

e) Incubation:
The culture vessels with inoculated explants are incubated in a culture room under controlled temperature, optimum light and humidity. The cultures are incubated for 3 – 4 weeks, the cells of the explant divide arid redivide, producing a mass of tissue called callus. The callus is transferred to another medium containing growth regulators to initiate the formation of roots and leafy shoot (organogenesis). Sometimes embryo like structures develop directly from the callus which are referred as somatic embryos. These can be encapsulated with sodium alginate to form synthetic or artificial seeds.

f) Acclamatization and transfer to pots :
The plants produced through tissue culture are washed gently and are planted in pots kept in glass house for 1-2 weeks. Finally they are transferred to field.

Advantages:

1. The production of exact copies of plants that produce particularly good flowers, fruits or have other desirable traits.
2. To quickly produce mature plants.
3. The production of multiples of plants in the absence of seeds or necessary pollinators to produce seeds.
4. The regeneration of whole plants froih plant cells that have been genetically modified.
5. The production of plants from seeds that otherwise have very low chances of germinating and growing i.e., orchids and nepenthes.
6. To clean particular plants of viral and other infections and to quickly multiply these plants as cleaned stock for Horticulture and Agriculture.

Intext Questions

Question 1.
Describe in brief, various steps involved in plant breeding.
Answer:
The main steps in breeding a’new genetic variety of a crop are :
1) Collection of Variability: Genetic variability is the root of any breeding programme . Collection and preservation of all the different wild varieties, species and relatives of cultivated species is a prerequisite for effective exploitation of natural genes available in the populations.

The entire collection haring all the diverse alleles for all genes in a given crop is called Germplasm collection.

2) Evaluation and selection of parents :
The germplasm is evaluated so as to identify plants with desirable characters. The selected plants are multiplied and are used. Purelines are created wherever desirable.

3) Cross Hybridisation among the selected parents :
After emasculation (Removal of Anthers from bisexual flower of a female parent) the female flowers are enclosed in a polythene bag to prevent uridesired cross pollination.

pollen grains are collected from the male parent with the help of a brush and are transferred to the surface of the stigma and thus cross pollination is affected artificially.

4) Selection and Testing of superior recombinants :
It involves selecting among the progeny of hybrids, those plants that have the desired character combination. The selection process requires careful scientific evaluation of the progeny. Due to this, plants that are superior to both the parents are obtained. These are self pollinated for several generations till they reach a homozygosity.

5) Testing, release and commercialisation of new characters :
The newly selected lines are evaluated for their yield and other traits of quality, .disease resistance etc. It is done by growing these in research fields and recording their performance under ideal fertilizer application, irrigation and other crop management practices. It is followed by testing the materials in farmers fields for at least 3 growing seasons at several places in the country, in all agroclimatic zones. Finally they are distributed to farmers as a new variety.

Question 2.
What is the major advantage of producing plants by micropropagation?
Answer:
Plants produced are genetically identical to the original source of plant.

Question 3.
Find out what are the various components of the medium used for propagation of an explant in vitro?
Answer:
Sucrose, Inorganic salts, vitamins, amino acids, growth regultaors like Auxins, Cytokinins, water, agar-agar.

Question 4.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:
Rice – Semidwarf varities – Jaya, Ratna.

Question 5.
The term ‘desirable trait’ can mean different things for different plants. Justify the statement with suitable examples.
Answer:
Different plants have different qualities in expressing desirable traits/characters. Eg.: Bacillus thermogenesis is used to kill other insect larvae acting as pesticide is a desirable trait. Another examples are Bt Cotton, Bt Com, rice, tomato, potato all have good qualities in increasing food production.

Question 6.
Is there any relationship between dedifferentiation and the higher degree of success achieved in plant tissue culture experiments?
Answer:
Yes, cultured tissue must certain competent cells or cells capable of reasoning competance (dedifferentiation).
Ex : an Explant → dedifferentiation → allay redifferentiation (whole plant) = cellular Totipotency.

Question 7.
“Give me a living cell of any plant and I will give you a thousand plants of the same type” is this only a slogan or is it scientifically possible? Write your comments and Justify them.
Answer:
It is possible to produce large number of plants from a living cell with in a short period is called micropropagation. Because each living cell of any plant has the capacity to develop into a new plant called totipotency. Based on this principle. It is possible to produce several plants which are identical to parents.

Question 8.
What are the physical barriers of a cell in the protoplast fusion experiment? How are the barriers overcome?
Answer:
Cell wall is the physical barrier of cell. To Isolate protoplast from the cells, the cells are to be treated with cellulages and pectinages to break the cell walls. Then protoplasts are taken and are fused with the help of polyethylene Glycol to obtain a hybrid plants.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 12th Lesson Biotechnology and its Applications Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 12th Lesson Biotechnology and its Applications

Very Short Answer Questions

Question 1.
Give different types of cry genes and pests which are controlled by the proteins encoded by these genes.
Answer:
The proteins encoded by the genes cry IAC and cry II Ab control the cotton bollworms, cry IAb controls com borer. [Cry protein is a protein toxin produced by Bacillus thuringiensis that kills insects]

Question 2.
Can a disease be detected before its symptoms appear? Explain the principle involved
Answer:
Yes. Very low concentrations of bacteria or viruses (when the symptoms of the disease are not visible) can be detected by amplification of their nucleic acid through PCR. rDNA technology, Enzyme Linked Immunosorbent Assay are some of the techniques that serve the purpose of early diagnosis.

Question 3.
What is GEAC and what are its objectives?
Answer:
GEAC stands for Genetic Engineering Approval Committee. It make decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.

Question 4.
Name the nematode that infects the roots of tobacco plants. Name the strategy adopted to prevent this infestation.
Answer:
Meloidegyne incognitia. A novel strategy adopted to prevent this infestation is process of RNA interference [RNj/fii]

Question 5.
For which variety of Indian rice, has a patent been filed by a USA company?
Answer:
Basmati Rice.

Question 6.
Give one example for each of transgenic plants which are suitable for food processing and those with improved nutritional quality.
Answer:
‘Flavr Savr’ – tomato, Golden Rice from Taipei is rich in vitamin A and prevent blindness.

Question 7.
What is green revolution? Who is regarded as father of green revolution?
Answer:
Green revolution: Substantial and dramatic increase in agricultural production which was termed as green revolution by William Gaud, the director of United States Agency for International development (USAID). Norman Borlaug is regarded as father of Green Revolution. Dr. M.S. Swaminathan and his team is the success of green revolution in our country.

Short Answer Questions

Question 1.
List out the beneficial aspects of transgenic plants.
Answer:
Plants with desirable characters created through gene transfer methods are called Transgenic plants. Beneficial aspects are :
a) Transgenic crop plants are efficient because they have many beneficial traits like virus resistance, insect resistance and herbicide resistance.

1. Papaya in resistant to papaya ring spot virus
2. Bt. cotton is resistant to insects.
3. Transgenic tomato plants are resistant to bacterial pathogen pseudomonas.
4. Transgenic potato plants are resistant to fungus phytophthora.

b) Transgenic plants which are suitable for food processing are produced with improved nutritional quality.
Eg: i) Transgenic tomato “Flavr Savr” are bruise resistant i.e., suitable for storage and transport due to delayed ripening and offers longer shelf life,
ii) Transgenic Golden Rice Taipei’is rich in vitamin A and prevents blindness.

c) Transgenic plants are used for hybrid seed production.
Eg., : Male sterile plants of Brassica napus are produced. This will eliminate the problem of manual emasculation and reduce the cost of hybrid seed production.
d) Transgenic plants have been shown to express the genes of insulin, interferon, human growth hormones, antibiotics, antibodies etc.
e) Transgenic plants are used as bio-reactors for obtaining, commercially useful products, specialised medicines and antibodies on large scale is called molecular farming.
f) Transgenic plants tolerant to abiotic stresses caused by chemicals, cold, drought, salt, heat etc.,
i) Basmati variety of rice was made resistant against biotic and abiotic stresses.
ii) Round up ready soyabean is herbicide tolerant.

Question 2.
What are some bio-safety issues concerned with genetically modified crops?
Answer:

1. There is fear of transferring allergins or toxins to humans and animals as side effects.
2. There is a risk of changing the fundamental nature of vegetables.
3. They may pose a harmful effect on biodiversity and have an adverse impact on environment.
4. There is a risk of gene pollution due to transfer of the new genes into related wild species through natural out crossing. This may result in the development of super weeds which may be fast growing than the crops and may be resistant to weedicides.
5. They may bring about changes in natural evolutionary pattern.

Question 3.
Give a brief account of Bt. Cotton
Answer:
Bt.Cotton : Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tobacco budworm armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes). Bacillus thuringiensis forms protein crystals which contain a toxic insecticidal protein. The gene responsible for the production of this toxic protein is introduced genetically into the cotton seeds, protects the plants from Bollworm, a major pest of cotton. The worm feeding on the leaves of Bt Cotton plant becomes lethargic and sleepy thereby causing less damage to the plant! Use of Bt Cotton has led to 3 – 27% increase in cotton yield in countries where it is grown.

The toxin is coded by a gene named ‘cry’. The proteins encoded by the genes cry IAc and cry II Ab control the cotton bollworms and Cry IAb controls com borer.

Question 4.
Give a brief account of pest resistant in plants.
Answer:
Pest resistant plants: Several nematodes parasitize a wide variety of plants and animals. A nematode. Meloidegyne incognitia infects the roots of tobacco plants and causes a great reduction in yield. To prevent this infestation, process of RNA interference was adopted. Using Agrobacterium vectors, nematode specific genes were introduced into the host plant. The introduction of DNA was such that it produced both sense and anti-sense RNAs in the host cells. These two RNAs formed a double stranded RNA that initiated RNAi and thus silenced the specific mRNA of the nematode. The parasite could not survive in a transgenic host expressing specific interfering RNA. The transgenic plant therefore got itself protected from the parasite.

Intext Questions

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
a) Bacteria are resistant to the toxin
b) Toxin is immature
c) Toxin is inactive
d) Bacteria encloses toxin in a special sac
Answer:
c) Toxin is inactive

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Answer:
Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products. Eg.: E.coli. In the plasmid of E.coli, the two DNA sequences corresponding to A and B chain of human Insulin are inserted so as to produce human Insulin chains. Later on these chains are extracted from E.coli and combined to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
Advantages :

1. Most of the GM crops have been developed for pest resistance which increases the crop productivity and therefore reduces the reliance on chemical pesticides.
2. Many varieties of GM food crops have been developed which Have enhanced nutritional quality. Eg.: Golden rice, rich in vitamin A.
3. These plants present the loss of fertility of soil by increasing the efficiency of mineral usage.
4. They are highly tolerant to unfavourable abiotic conditions.
5. The use of GM crops decreases the pest harvesting loss of crops.

Disadvantages :

1. They can affect the native biodiversity in an area.
2. They can cause genetic pollution in the wild relative of crop plants hence it is affecting our natural environment.
3. They are affecting human health.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Answer:
It is a protein toxin produced by Bacillus thuningiensis that kills certain insects. Man by using there proteins, controls the insects and are minimising the usage of insecticides.

Question 5.
List out the advantages of recombinant Insulin.
Answer:

1. It is supplied continuously.
2. Stabilizes the market price.
3. Humulin is absorbed more rapidly.
4. Humulin is less expensive.

Question 6.
What is meant by the term Biopesticide? Name and explain the mode of action of a popular biopesticide?
Answer:
The organism which kills the other pathogenic organisms called Biopesticides. Eg. : Bacillus thuringiensis. It produces inactive crystals but once on insect in guts, it is converted into an active form due to alkaline pH of the gut and cause cell swelling and lysis finally the death of the insect.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes

Very Short Answer Questions

Question 1.
Define Biotechnology.
Answer:
It is a science that utilizes the properties and uses of microorganisms or exploits cells and the cell constituents at the industrial level for generating useful products essential to life and human welfare.

Question 2.
What are molecular scissors? Where are they obtained from?
Answer:
Molecular scissors are the restriction enzymes that cut the DNA at specific locations. Usually, they are obtained from Bacteria.

Question 3.
Name any two artificially restructured plasmids.
Answer:
PBR₃₂₂, PUC_{19 101}.

Question 4.
What is E coRI? How does it function?
Answer:
E.coRI is a restriction enzyme obtained from Escherichia coli. It specifically recognises GAA sites on the DNA and cuts it between G and A.

Question 5.
What are cloning vectors? Give an example.
Answer:
Vectors used for multiplying the foreign DNA sequences are called cloning vectors.
Ex: Plasmids, Bacteriophages, Cosmids.

Question 6.
What is recombinant DNA?
Answer:
The hybrid DNA formed by the fusion of DNA with desirable genes with the vector DNA by DNA ligase is called r-DNA.

Question 7.
What is palindromic sequence?
Answer:
The specific nucleotide sequence of DNA which is recognised by restriction enzyme is called palindrome, which is 4-6 base pairs in length.

Question 8.
What is the full form of PCR? How it is useful in Biotechnology?
Answer:
PCR stands for Polymerase chain reaction. It can be used for the diagnosis of diseases like AIDS, middle ear infection, and tuberculosis.

Question 9.
What is down stream processing?
Answer:
Separation and purification of products before they are ready for marketing is called
downstream processing.

Question 10.
How does one visualize DNA on an agar – gel?
Answer:
To make the DNA visible in the gel, Ethidium bromide is added to the gel solution and the buffer. This positively charged polycyclic compound binds to .DNA inserting itself between the base pairs. Southern blotting may also be used as visualization technique for agarose gels.

Question 11.
How can you differentiate between exonucleases and endonucleases?
Answer:

Exonuclease	Endonuclease
1. They remove nucleotides from the ends of the DNA	1. They make cuts at specific locations with in the DNA

Short Answer Questions

Question 1.
Write short notes on restriction enzymes.
Answer:
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut DNA molecules at a particular point by recognising a specific Sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the Genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.
a) Exonucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cut’s at specific location with in the DNA-

Question 2.
Give an account of amplification of gene of interest using PCR.
Answer:
PCR stands for Polymerase chain reaction. In this reaction, multiple copies of the gene of interest are synthesized in vitro using two sets of primers(oligonucleotides) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the. genomic DNA as template. As the process of replication of DNA is repeated many times, the segments of DNA can be amplified to approximately billion times.

Such repeated amplification is achieved by the use of a thermostable DNA polymerase such as Taq polymerase (isolated from thermus aquaticus) which remain active even during high temperature induced denaturation of ds DNA. The amplified fragment, if desired can: now be used to ligate with a vector for further cloning.

Question 3.
What is a bio-reactor? Describe briefly the stirring type of bio-reactor.
Answer:
Bio-reactor is a large vessel which is used for biological conversion of raw material into specific product.

A stirred-tank bio-reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents.

The stirrer facilitates even mixing and oxygen availability throughout the bio-reactor. Alternatively air can be bubbled through the reactor. It has an agitator system, an oxygen delivery system, a foam control system, a temperature .control system, pH control system and sampling ports, so that small volumes of the culture can be with drawn periodically.

Question 4.
What are the different methods of insertion of recombinant DNA into the host cell?
Answer:
There are several methods of introducing the ligated DNA into recipient cells. Recipient cells after making them competent to receive, take up the DNA present in their surrounding. R-DNA can be forced into such cells by incubating the cells with r-DNA on ice followed by placing them briefly at 42eC -(heat shock) and then putting them back on ice. This enables the bacteria to fake up the r-DNA.

In Micro-injection method, r-DNA is directly injected into the nucleus of an animal cell.

In Bio listic of gene gun method – cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA.

In another method, ‘Disarmed pathogen’ vectors are used which when allowed to infect the cell, transfer the r-DNA into the host.

Long Answer Questions

Question 1.
Explain briefly the various processes of recombinant DNA technology.
Answer:
The important methods in recombinant DNA technology are performed through genetic engineering. They are :
i) Isolation of a desired gene,
ii) Insertion of isolated gene into a suitable vector,
iii) Introduction of recombinant vector into a host and
iv) Selection of the transformed host cells.

I) Isolation of a desired gene :

1. The desired gene is isolated from the donor cell. Normally bacteria are the source of desired genes.
2. The cell walls of bacteria are degraded with the help of enzymes.
3. The cell membranes are lysed with the help of detergents.
4. By treating the cellular constituents with phenols and suitable nucleases and by subjecting to gradient centrifugation, pure DNA is isolated.
5. The purified DNA is cut into a number of fragments by restriction endonucleases.
6. The restriction enzymes cleave DNA molecules in two ways.

i) In one way they cut both strands of DNA at exactly opposite points to each other. This results in DNA fragments with blunt ends or flush ends, where two strands end at the same point. Such cut is generally termed as even cut.

ii) But commonly, most enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is generally termed as staggered cut. This generates protruding ends i.e., one strand of DNA double helix extends some bases beyond the other. Since the target site is palindromic in nature, the protruding ends generated by such a cleavage have complimentary base sequence.

As a result, they readily pair with each other and such ends are called cohesive or sticky ends. This stickyness of the ends facilitates the action of the enzyme DNA ligase. When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of ’sticky ends’ and these can be joined together readily by using DNA ligases. E.g.: The restriction enzyme E coRI.
E – The first letter, represents the name of genus Escherichia.
Co – The next two letters, represent the species Escherichia Coli.

The letter R is derived from the name of strain.

Roman numbers following the names indicate the order in which the enzymes were isolated from the strain of bacteria.

This enzyme specifically recognises GAA sites on the DNA and cuts it between G and A (G ↓ A)

7) The resultant fragments are separated from each other by gel electrophoresis.
8) The desired fragments are selected by Southern blotting technique.

II) Insertion of isolated gene into a suitable vector :

1. The selected fragments of DNA are inserted into a suitable vector to produce a large number of copies of genes. This is called gene cloning.
2. There are two major types of vectors, namely plasmids and bacteriophages.
3. Among the two types, plasmids are the ideal cloning vectors.
4. To isolate a plasmid, the Bacterial cell is treated with EDTA (Ethylene diamine tetra acetic acid) along with lysozyme enzyme to digest the cell wall.
5. Then the bacterial cell is subjected to centrifugation in sodium lauryl sulphate to separate the plasmid.
6. The plasmid DNA is cut with the help of restriction endonuclease.
7. The circular plasmid is converted into a linear molecule having sticky ends.
8. The two sticky ends of linear plasmid are joined to the ends of desired gene by DNA ligase.
9. The plasmid containing foreign DNA segments is called recombinant DNA (r DNA) or Chimeric DNA.

III) Introduction of recombinant vector into a suitable host :

1. The r DNA molecule is introduced into suitable bacterial host cell by transfor-mation.
2. The cell containing r DNA is called transformed cell.
3. Bacterial cell walls are not permeable to recombinant vectors, but keeping in dil. Calcium chloride renders the bacterial cell wall permeable to recombinant vectors.
4. The r DNA replicates with in the host cell.
5. The transformed cell grows on the culture medium. Each daughter cell contains r DNA.

IV) Selection of transformed host cells :
1) Selection of transformed cells depends on the nature of gene which is cloned. 2) It can be done in two ways. The are :
a) Without using probes,
b) By using probes.

a) Without using probes:

1. If the gene is cloned for antibiotic resistance, the cells are first incubated on a medium without antibiotic for one hour, to allow the antibiotic resistance gene to be expressed.
2. Then the cells are placed on a medium with an antibiotic for selection of colonies containing rDNA.
3. The cells which have expressed the gene will survive and the others die.

b) By using probes:
When transformed cells are cultured on the nutrient medium, several cells are produced. To select the cells containing the desired gene colony hybridization method is used. In this gene specific probes are used. A probe is a small fragment of single stranded RNA or DNA which is tagged with radioactive, molecule. It can search out complimentary DNA sequences from an organism.

Question 2.
Give a brief account of the tools of recombinant DNA technology.
Answer:
Key tools are :
1) Restriction enzymes :
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut.

DNA molecules at a particular point by recognising a specific sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.

a) ExOnucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cuts at specific location with in the DNA.

Most restriction enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is know as staggered cut. E CoRI recognises 5′ GAATT 3′ sites on the DNA and cuts it between G and A results in the formation of sticky ends or cohesive end pieces. This stickyness of the ends facilitates the action of enzyme DNA ligase.

Cloning vectors :
The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host is called vector. Vectors used for multiplying the foreign DNA sequences are called cloning vectors. Commonly used cloning vectors are plasmids, bacteriophages, cosmids. Plasmids are extrachromosomal circular DNA molecules present in almost all bacterial species. They are inheritable and carry few genes are easy to isolate and reintroduce into the bacterium (Host).

Features required to facilitate cloning into a vector:
a) Origin of replication:
(ori) This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within host cells. It is also responsible for controlling the copy number of the linked DNA.

b) Selectable marker :
In addition to ‘ori’, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the any transformants normally, the genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin etc., are useful selectable markers for E.Coli.

c) Cloning sites :
In order to link the alien DNA, the vector needs to have very few, preferably single recognition sites for the restriction enzymes.

d) Molecular weight:
The cloning vector should have low molecular weight.

e) Vectors for cloning genes in plants and animals:
The tumour inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector such that it is no more pathogenic to plants. Similarly retroviruses have also been disarmed and are now used to deliver desirable genes into animal cells.

Intext Questions

Question 1.
Do Eukaryotic cells have restriction Endonucleases? Justify your answer.
Answer:
No. only prokaryotic cells have restriction Endonucleases. Today we know 900 or more restriction enzymes have been Isolated from over 230 strains of Bacteria.

Question 2.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:

1. Small volumes of the culture can be withdrawn periodically.
2. Air can be bubbled through the reactor.
3. It has a control system that regulates the temperature and pH.

Question 3.
Can you recall Meiosis and indicate at what stage a recombinant DNA is made?
Answer:
Pachytene of Meiosis – I.

4. Describe briefly the following :
a) Origin of replication b) Bioreactors c) Down stream processing
Answer:
a) Origin of replication :
It is the specific DNA sequence which is responsible for initiating replication.

b) Bioreactors:
They are large vessels which are used for biological conversion of raw material into specific products.

c) Down stream processing :
Separation and purification of products is called down stream processing.

Question 5.
Explain briefly a) PCR b) Restriction enzymes and DNA c) Chitinase
Answer:
a) PCR:
Polymerase chain reaction: Is a biochemical technology in molecular biology to amplify a single or few copies of a piece of DNA generating thousands of copies of a particular DNA sequence.

b) Restriction enzymes and DNA :
Restriction enzymes are DNA cutting enzymes found in Bacteria because they cut with in the molecule.

c) Chitinase :
It is an enzyme obtained from the fungus, used in the digesting the bacterial cell walls.

Question 6.
Discuss with your teacher and find out how to distinguish between
a) Plasmid DNA and Chromosomal DNA
b) RNA and DNA
c) ExonuCleases and Endonucleases

a)

Plasmid DNA	Chromosomal DNA
1. Plasmid DNA is a small circular molecule found in Bacteria.	1. It is main genetic material which is a circular, double stranded DNA
2. It carry a few genes.	2. It carry several genes.

b)

RNA	DNA
1. It is made up of only polynucleotide chain.	1. It is made up of 2 polynucleotide chains.
2. Purines or pyTamidines do not exist in 1 : 1 ratio.	2. Purines and pyramidines exist in 1 : 1 ratio.
3. Nitrogen bases are Adenine, Guanine, uracil and cytosine.	3. Nitrogen bases are Adenine, Guanine, Thymine and cytosine.

c)

Exonuclease	Endonuclease
1. They remove nucleotides grom the ends of the DNA.	1. They make cuts at specific positions with in the DNA.

Question 7.
What does ‘H’ in’d’ and HI refer to in the enzyme Hind III?
Answer:
‘H’ – The genes from which it was isolated – Haemophilus
in – Influenza (the species name) .
d – Refers the bacterial strain from which enzyme was isolated.
III – Third isolated enzyme from the strain.

Question 8.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
The presence of more than one recognition site for the same enzyme with in the vector will generate several fragments which will complicate the gene cloning. So vector needs to have very few, preferably single, recognition site for the commonly used restriction enzymes.

Question 9.
What does ‘competent’ refer to ‘incompetent cells’ used in transformation experiments?
Answer:
DNA is a hydrophilic molecule. It cannot pass through cell membranes. In order to force Bacteria to take up the plasmid, the bacterial cells must be made competent to take up DNA by treating them with a bivalent cation such as calcium which increaes the efficiency with which DNA enters the bacterium.

Question 10.
What is the significance of adding proteases at the time of Isolation of Genetic material (DNA)?
Answer:
Proteins can be removed by treatment with proteases.

Question 11.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?
Answer:
The segment of DNA cannot be amplified.

Question 12.
What modification is done on the Ti plasmid of Agrobacterium tumefascians to convert it into a cloning vector?
Answer:
It is no more pathogenic to plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Question 13.
What is meant by gene cloning?
Answer:
It is the replication of DNA fragments by the use of a self-replicating genetic material. It duplicates only individual genes of organism’s DNA.

Question 14.
Decide the ratio between ester bonds and hydrogen bonds that are broken in each palindromic sequence of DNA when treated with E coRI during the formation of sticky ends.
Answer:
1 : 4.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance

Very Short Answer Questions

Question 1.
Distinguish between Heterochromatin and Euchromatin. Which of the two is transcriptionally active?
Answer:
The chromatin that is more densely packed and stains dark is called heterochromatin. The chromatin that is loosely packed and stained light is called Euchromatin. Euchromatin is transcriptionally active chromatin.

Question 2.
Who proved that DNA is Genetic Material? What is the organism they worked on?
Answer:
Alfred Hershey and Martha chase (1952). They worked with viruses that infect Bacteria, and bacteriophages.

Question 3.
What is the function of DNA polymerase?
Answer:
DNA polymerise is a highly efficient enzyme which catalyse polymerisation of a large number of nucleotides in a very short time. It also catalyse the reaction with a high degree of accuracy.

Question 4.
What are the components of a nucleotide?
Answer:
A nucleotide has three components – a nitrogenous base, a pentose sugar and a phosphate group’.

Question 5.
Given below is the sequence of coding strand of DNA in a transcription unit.
5’AATGCAGCTATTAGG – 3 . Write the sequence of
a) Its complementary strand
b) The mRNA
Answer:
a) 5’TTACGTCGATAATCC-3′
b) 5’ AAUGCAGCUAUUAGG – 3’

Question 6.
Name any three viruses which have RNA as the Genetic Material.
Answer:
Tobacco mosaic virus, QB bacteriophage, HIV.

Question 7.
What are the components of a transcription unit?
Answer:
a) A promotor b) The structural Gene c) A terminator.

Question 8.
What is the difference between exons and Introns?
Answer:

Exons	Introns
1) Coding or expressed sequences.	1) Intervening sequences.
2) They appear in nature or processed RNA.	2) They do not appear in mature or processed RNA.

Question 9.
What is meant by capping and tailing?
Answer:
Adding of an unusual nucleotide (methyl guanosine triphosphate) to the 5′ -end of heterogenous nuclear RNA [hnRNA) is called capping.

Adding of adenylate residues to the 3’ – end in a template independent manner is called tailing.

Question 10.
What is meant by point mutation? Give an example.
Answer:
Change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

Question 11.
What is meant by charging of tRNA?
Answer:
Activation of aminoacids in the presence of ATP and linked to their cognate tRNA is known as charging of tRNA or amino acylation of tRNA.

Question 12.
What is the function of the codon AUG?
Answer:
It has dual functions. It codes for methionine and also act as the initiator codon.

Question 13.
Define stop codon. Write the codons.
Answer:
Codons which do not code for any aminoacids are called stop codons. They are UAA, UAG, UGA.

Question 14.
What is the difference between the template strand and a coding strand in a DNA molecule?
Answer:
The two strands have opposite polarity and the DNA-dependant RNA polymerise also catalyses the polymerisation in only one direction that is, 5′ → 3′, the strand that has the polarity 3′ → 5′ acts as a template and is also called template strand. The other strand which has polarity 5′ → 3′ and does not code for anything is called coding strand.

Question 15.
Write any two differences between DNA and RNA.
Answer:

DNA	RNA
1) Nitrogen bases are Adenine, Guanine Thymine and cytosine.	1) Nitrogen bases ate Adenine, Guanine, Uracil and Cytosine.
2) Deoxyribose sugar is present.	2) Ribose sugar is present.

Question 16.
In a typical DNA molecule the proportion of thymine is 30% of the N bases. Find out the percentages of other N bases.
Answer:
Adenine = 30%
Guanine = 20%
Cytosine = 20%

Question 17.
The proportion of nucleotides in a given nucleic acid are Adenine 18%, Guanine 30%, Cytosine 42% and uracil 10%. Name the nucleic acid and mention the number of strands in it.
Answer:
RNA. Only one strand is present.

Short Answer Questions

Question 1.
Define transformation in Griffith’s Experiment. Discuss how it helps in the identifi¬cation of DNA as genetic material.
Answer:
Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with ‘s’ shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia.

He injected heat killed ‘s’ strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed ‘s’ strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

Question 2.
Discuss the significance of heavy isotope of Nitrogen in Meselson and Stahl’s experiment.
Answer:
Matthew Meselson and Franklin Stahl, grow E.Coli in a medium containing ¹⁵NH₄Cl and observed 15_N was incorporated in newly synthesized DNA. This heavy DNA molecule could be distinguished from themormal DNA by centrifugation in a cesium chloride density gradient.

They transformed the cells into a medium with ¹⁵NH₄Cl (normal) and took samples at various time intervals and extracted the DNA that remained as double stranded helices. The various samples were separated independently on cesium chloride (cscl) gradients to measure the densities of DNA. Thus the DNA that was extracted from the culture, one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid density. DNA extracted from the culture after another generation (40 mts) was composed of equal amounts of this hybrid DNA and of ‘Light’ DNA.

Question 3.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defined your answer.
Answer:
A single base mutation in a gene may result in loss or gain of a gene and so a function.
E.g.: A change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

E.g.: 2) Consider a statement that is made like a genetic code is – RAM HAS RED CAP.
If we remove a letter ‘S’ in HAS, it will be RAM HAR EDC AP.

Question 4.
How many types of RNA polymerases exist in cells ? Write their names and functions.
Answer:
Three RNA polymerases exist in Cells. They are :

1. RNA polymerase I – It transcribes r RNAs (28S, 18S, and 5.8S)
2. RNA polymerase II – It transcribes the precursor of RNA, the heterogeneous nuclear RNA (hn RNA)
3. RNA polymerase III – It is responsible for transcription of tRNA 5 Sr RNA and Sn RNAs (small nuclear RNAs)

Question 5.
What are the contributions of George Gamow, H.G. Khorana, Marshall Nirenberg in deciphering the genetic code?
Answer:
George Ganow, suggested that, in order to code for all the 20 Amino’ acids, the code should be made up of three nucleotides. This was a very bold proposition, because a permutation and combination of 4³ would generate 64 codons, generating more codons than required.

H.G. Khorana developed chemical method in synthesising RNA molecules with defined combinations of bases.

Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. The enzyme polynucleotide phosphorylase was also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA)

Question 6.
On the diagram of the secondary structure of tRNA shown below label the location
of the following parts.
Answer:

a) Anticodon
b) Acceptor stem
c) Anticodon stem
d) 5′ end
e) 3′ end

Question 7.
Draw the schematic / diagrammatic presentation of the lac operon.
Answer:

Question 8.
What are the differences between DNA and RNA.
Answer:

DNA	RNA
1. It contsists of two strands of nucleotides.	1. It consists on only strand of nucleotides.
2. It is present more in nucleus and very little in chloroplasts and mitochondria.	2. It is present more incytoplasm and little in nucleus.
3. Deoxyribose sugar (C5H10O4) is present.	3. Ribose sugar (C5H10O5) is present.
4. Thymine and cytosine are pyrimidines.	4. Uracil and cytosine are pyrimidines.
5. DNA is made up of 4 millions nucleotides.	5. RNA is made up of 75 – 2000 nucleotides.
6. It undergoes self-replication.	6. It does not undergo self replication except in RNA viruses.
7. DNA is the genetic material.	7. It is non-genetic material (except in RNA) viruses.
8. It does not participate directly in protein synthesis.	8. RNA participate directly in protein synthesis.
9. Metabolically DNA is of one type.	9. Metabolically RNA is of three types.
10. The base puring is A = T and G ≡ C.	10. The base puring is A = U and G = C.

Question 9.
Write the important features of Genetic code?
Answer:

1. The codon is triplet. 61 codons code for aminoacids and 3 codons donot code for any aminoacids called stop codons.
2. One codon codes for only one aminoacid, hence it is unambiguous and specific.
3. Some aminoacids are coded by more than one codon, hence the code is degenerated.
4. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal: For Ex: UUU code for phenylalanine (Phe) in Bacteria or Humans.
6. AUG has dual functions. It codes for methionine and also acts as initiator codon.

Question 10.
Write briefly on nucleosomes.
Answer:
Nucleosome is a bead like structure of chromosomes. It consists of eight histone molecules and a DNA segment of about 150 base pairs. Each Nucleosome is separated from one another by a linker DNA sequence of about 50 base pairs. Nucleosome helps to fold DNA into a compact form in the interphase nucleus. Otherwise the length of a chromosome, when linear is many orders of magnitude greater than the diameter of the nucleus.

Intext Questions

Question 1.
Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases : Adenine, Thymine, Uracil, Cytosine,
Nucleosides : Cytidine, Guanosine.

Question 2.
If a double stranded DNA has 20% of cytosine, calculate the percent of adenine in the DNA.
Answer:
Cytosine = 20%
Guamine = 20%
Adenine = 30%
Thymine = 30%

Question 3.
If the sequence of one strand of DNA is written as follows : Write down the sequence of complementary strand in 3′ → 5′ direction.
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3‘
Answer:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows.
5 – ATGCATGCATGCATGCATGCATGCATGC – 3 write down the sequence of m RNA
Answer:
3′ AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 5

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi conservative mode of DNA replication? Explain.
Answer:
The two strands would separate and acts as a template for the synthesis of new complementary strands. After completion of replication, each DNA would have one parental and one newly synthesised strand.

Question 6.
Depending upon the chemical nature of the template [DNA or RNA] and the nature of nucleic acids synthesized from it [DNA or RNA], List the types of nucleic acid polymerases.
Answer:
DNA polymerases
RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and chase purified biochemicals (proteins, DNA, RNA etc) from the heat killed ‘S’ cells. They discovered the DNA alone from S bacteria caused R bacteria to become transformed. They also discovered thal protein digesting enzymes and RNA digesting enzymes did not affect transformation. So the transforming principle was not a protein or RNA. Digestron with DNAse did inhibit transformation suggesting that the DNA caused the transformation.

Question 8.
Differentiate between the followings :
a) mRNA and tRNA
b) Template strand and Coding strand
Answer:

a)

RNA	tRNA
1) It contains more nucleotides.	1) It contains lesser nucleotides.
2) It moves important information from the DNA to ribosome.	2) It transports aminoacids into a growing protein chain.

b)

Template strand	Coding strand
1) It is complementary strand it serves as the template for making the coding strand.	1) It contains coding genes. It is to be transcribed that is the side make ‘sense’.
2) It runs from 3’to 5′.	2) It runs from 5′ to 3′.

Question 9.
List two essential roles of ribosomes during translation.
Answer:

1. Ribosome acts as the site where protein synthesis takes place from individual aminoacids.
2. Ribosome acts a catalyst for forming peptide bond.
   E.g. : 23 S r-RNA in bacteria acts as ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added. Which induced the Lac operon. Then, why does Lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes namely, an operator gene, a promoter gene and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose. In Lac operon lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promotor region.

Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose or metabolise it into glucose and galactose. After sometime, when the level of the inducer decreases, it causes the synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence the transcription is stopped.

Question 11.
Explain (in one or two lines) the functions of the followings :
a) Promoter b) tRNA c) Exons.
Answer:
a) Promoter:
It is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase. :

b) tRNA :
It is a small RNA that reads the genetic code present on mRNA it carries specific aminoacids to mRNA on ribosome during translation of proteins.

c) Exons:
Exons are coding sequences of DNA in Eukaryotes that transcribe for proteins.

Question 12.
Briefly describe the following :
a) Transcription b) Translation.
Answer:
a) Transcription:
It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promotor region of the template DNA and terminates at the terminator region The segment of DNA between these two regions is known as transciption unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of nucleotides, and certain cofactors such as Mg²⁺

During transcription, three events occur. They are : a)Initiations b)Elongation, c) Termination. The DNA dependant RNA polymerase and certain initiation factors bind at the double stranded DNA at the promotor region of the template strand and initiate the process of transcription, RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex in two separate strands.

Then one of the strands, called sense strand acts as template for mRNA synthesis. The epzyme RNA polymerase utilises nucleoside triphosphates as raw material and polymerizes them to form m RNA according to the complimentary bases present on the template DNA. This process of opening of helix and elongation of polynucleotide chain continuous until the enzyme reaches terminator region. As RNA polymerase reaches the terminator region, the newly synthesized mRNA transcripted alpng with enzyme is released. Another factor called terminator factor (p) required for the termination of the transcription.

b) Translation :
It is the process of polymerizing amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain. This process invoivs 3 steps, a) Initiation b) Elongation c) Termination. During the initiation, tRNA gets charged when the aminoacid binds by using ATR The start codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in.a large subunit for the attachment of subsequent aminoacid.

The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon dowonstream along with mRNA so as to leave the space for binding of another charged tRNA, The aminoacid brought by tRNA get linked with the previous aminoacid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (UAA, UAG and UGA), the process of translation gets terminated. The polypeptide chain is released and ribosomes get detached from mRNA.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule?
Answer:
Due to Lack of Helicase enzyme, unwounding does not occurs. The single stranded Binding proteins cover the DNA strands preventing them from annealing into a double strand.

Question 14.
In an experiment, DNA is treated with a compound which trends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases. From 0.34 nm to 0.44 nm calculate the length of DNA double helix (which has 2 × 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
2 × 10⁹ × 0.44 × 10^-9 bp.

Question 15.
Recall the experiments done by Frederick Griffith. Where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of pneumococcus have transformed the R – strain into virulent strain? Explain?
Answer:
RNA is more labile and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence heat killed ‘s’ strain may not have retained its ability to transform the ‘R’ strain into virulent form if RNA was its genetic material.

Question 16.
You are repeating the Hershey – Chase experiment and are provided with two Isotopes : ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:
Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive Isotope while ¹⁵N is not radioactive but is the heavier Isotope of Nitrogen). Even if ¹⁵N was radioactive, them its presence would have been detected both inside the cell as well as in the supernatant because ¹⁵N would also get incorporated in amino group of aminoacids in proteins. Hence the use of ¹⁵N would not give any conclusive results.

Question 17.
Do you think that the alternate splicing of exons may enable a structural gene to code for several Isoproteins from one and the same gene? If yes, how? If not, why so?
Answer:
Functional m RNA of structural genes need not always include all of its exons. This alternate splicing of exons in sex-specific, tissue – specific, and even developmental stage specific. By such alternate splicing of exons a single gene may encode for several Isoproteins and/or proteins of similar class’. In absence of such a kind of splicing, there should have been new genes for every protein/Isoprotein. Such an extravagency has been avoided in natural phenomena by way of alternate splicing.

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass/ Density would sediment faster?
Answer:
Proteins have lower density when compared to others. So a molecule with higher mass would sediment faster than molecules with light weight (density).

Question 19.
Do Retroviruses follow central Dogma? Give one example.
Answer:
Genetic material of Retroviruses in RNA. At the time of synthesis of protein RNA is reverse transcribed to its complementary DNA first/which is opposite to the central Dogma. Hence Retroviruses are not known to follow central Dogma.

Question 20.
If there are 2.9 × 10⁹ complete turns in a DNA molecule. Estimate the length of the molecule (1 angstrom = 10^-8 cm).
Answer:
1 turn of DNA = 3.4 nm.
Number of turns are = 2.9 × 10⁹
The length of the DNA molecule = 2.9 × 10⁹ × 3.4 nm.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation

Very Short Answer Questions

Question 1.
What is the cross between the F₁ Progeny and the Homozygous recessive parent called? How is it useful?
Answer:
Test cross. It is used to test whether an Individual is Homozygous (pure) or Heterozygous (Hybrid).

Question 2.
Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
Answer:
Yes, it depends on the linkage of genes on the same chromosome.

Question 3.
Who proposed the Chromosome theory of inheritance?
Answer:
Walter Sutton and Theodore Boveri.

Question 4.
Define true breeding. Mention its significance.
Answer:
All offsprings has the same phenotype as its parents is called the breeding. They express characters for several generations.

Question 5.
Explain the terms Phenotype and Genotype.
Answer:
The Physical or external appearance of a character is called Phenotype. The genetic makeup of an individual is called Genotype.

Question 6.
What is point mutation? Give an example.
Answer:
Mutation which occurs in a single base pair of DNA is called point mutation.
E.g.: Sickle cell anemia.

Question 7.
What is the genotype of wrinkled phenotype of pea seeds?
Answer:
rr is the genetic nature of the phenotypic wrinkled pea seeds.

Question 8.
What will be the phenotypic ratio in the offsprings obtained from the following crosses.
a) Aa x aa b) AA x aa c) Aa x Aa d) Aa x AA
Note : Gene A is dominant over gene a.
Answer:

Question 9.
In garden pea, the gene T for tall is dominant over its allele for dwarf. Give the genotypes of the parents in the following crosses.
a) Tall x Dwarf producing all tall plants.
b) tall x tall producing 3 tall and 1 dwarf plants.
Answer:

Short Answer Questions

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:

1. It is an annual plant that has well defined characteristics.
2. It can be grown and crossed easily.
3. It has bisexual flowers containing both male and female parts.
4. It can be self fertilized conveniently.
5. It has a short life cycle and produces large number of offsprings.

Question 2.
Differentiate between the following.
a) Dominant and Recessive b) Homozygous and Heterozygous
Answer:

Dominant	Recessive
a) The characters which are expressed phenotypically in both homozygotes and heterozygotes or”The allele or gene that shows up”.	The character which is not expressed phenotypically in heterozygous condition or the allele that is masked.

 

Homozygous	Heterozygous
b) An individual having two similar or identical alleles for a single character. Hence it will produce only one kind of gametes.	An individual having two different alleles for a single character. Hence it will produce two different types of gametes.

Question 3.
Explain the law of Dominance using a Monohybrid cross.
Answer:
Characters are controlled by discrete units called factors. Factors occur in pairs. In a dissimilar pair of factors pertaining to a character one number of the pair dominates the other. The law of Dominance is used to explain the expression of only one of the parental characters in a Monohybrid cross in the F₁ generation and the expression of both in F₂ generation in the ratio of 3 : 1.

Question 4.
Define and design a test cross.
Answer:
Crossing between Fx indivi – duals with the recessive parent is called test cross. It is used to test whether an individual is homozygous or heterozygous. A mono hybrid test cross gives a phenotype ratio of 1 : 1 and a dihybrid test cross gives a ratio of 1 : 1 : 1 : 1.

Question 5.
Explain the co-dominance with example.
Answer:

a) Co-dominance:
The phenomenon where heterozygotes have features of both the homozygotes, that is, an allele is neither dominant nor recessive to the other. 1) Different types of red blood cells that determine ABO blood grouping in humans 2) Seed coat pattern size in lentil plants. A cross between pure breeding spotted lentils with pure breeding dotted lentils produce heterozygotes. These Fx hybrids show the phenotypic features of both parents which means that neither the spotted nor the dotted, allele is dominant or recessive to the other. Because both traits show up equally in the heterozygote’s phenotype, the alleles are termed co-dominant. F₁ plants are self pollinated produce F₂ progeny in the ratio of (lspotted : 2 spotted and dotted : 1 dotted).

Question 6.
Explain the Incomplete Dominance with example.
Answer:

Incomplete Dominance: It is the condition where one allele of a gene is not completely dominant over the other allele and results in the heterozygous having phenotype different from the dominant and recessive homozygotes.

In a cross between true beeding red flowered plants (RR) and true white flowered plants, (rr) the F₁ was pink (Rr). When the F₁ was self pollinated, the F₂ resulted in the ratio of 1(RR) : 2(Rr) : 1(rr)

Red Pink White

Here the genotype ratios were as in monohybrid cross of Mendel but the phenotypic ratio had changed from the 3 : 1 because, R was not completely dominant over r and is possible to distinguish Rr as pink from RR and rr. Thus the phenotypic and genotypic ratios in F₂ progeny are the same, that is 1 : 2 : 1.

Question 7.
Write a brief note on chromosomal mutations and gene mutations.
Answer:
a) Chromosomal mutations :
Any change in the number or structure of chromosomes are called chromosomal mutations. They may be

a) Structural changes :
During prophase I of Meiosis, homologous regions of chromosomes show many changes in structure inorder to form pairs or bivalents. They are :
1) Deletion : A part of chromosome is lost or broken.
2) Duplication : A particular part of chromosome is repeated.

b) Gene Mutations :
Mutations which occur due to change in a single base pair of DNA are called gene mutations or point mutations.

Question 8.
Define Law of Segregation and Law of Independent Assortment.
Answer:
Law of segregation :
The two alleles of a gene when present together in a heterozygous state, they do not fuse or blend in any way but remain distinct and segregate during meiosis or in the formation of gametes, so that each meiotic product of gamete will carry only one of them.

Law of Independent Assortment:
When two pairs of traits are combined in a hybrid, seggregation of one pair of characters is independent of the other pair of characters.

Long Answer Question

Question 1.
Explain the dihybrid cross with the help of punnel square board by taking contrasting traits, seed colour and seed shape.
Answer:
The cross made between individuals differing in two characters is known as Dihybrid cross. Mendel studied the inheritance of two characters at a time. Mendel crossed a homozygous pea plant having yellow and round seeds with another pea plant having green and wrinkled seeds. The F₁ hybrids were found to have yellow and round seeds. This shows that are dominant.

When F₁ plants allowed to cross among themselves, four distinct types of seeds appeared in F₂ generation. Two of these were similar to parental combinations, while the other two were new combinations. The phenotypic ratio of 9 : 3 : 3 : 1 can be derived as a combination series of 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.

Intext Questions

Question 1.
Mendel crossed pea plants producing round seeds with those producing wrinkled seeds. From a total of 7324 F₂ seeds, 5474 were round and 1850 were wrinkled. Using the symbols R and r for genes, predict the a) the parental (p) genetypes, b) the gametes, c) F₁ progeny, d) the cross between F₁ hybrids, e) genotypes, phenotypes, genotypic frequency, phenotypic ratio of F₂ progeny.
Answer:
a)

d) Cross between F₁ hybrids

e) 3 : 1 – phenotypic ratio of F₂ progeny

Question 2.
The following data was obtained from an experiment on peas. The grey coloured seed is dominant over white coloured seed. Use the Letter G for grey and g for white traits. Predict genotypes of the parents in each of the following crosses.

Parent	Progeny
	Grey	White
a) Grey × White	164	156
b) Grey × Grey	59	19
c) White × White	0	100
d) Grey × Grey	180	0

Answer:

Question 3.
In tomatoes red fruit colour (R) is dominant yellow (r). Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearence of following.
a) The F₁, b) The F₂, c) The offspring of a cross of the F₁ back to the red parent, d) The offspring of a cross of the F₁ back to the yellow parent.
Answer:

Question 4.
In pea, axillary position of flowers T is dominant over its terminal position (t). Coloured flowers (C) are dominant to white flowers (c). A true breeding plant with coloured flowers in axils is crossed to one with white terminal flowers. Give the phenotypes, genotypes and expected ratios of F₁, F₂ back cross and test cross progenies. What genotype ratio is expected in the F₂ progeny?
Answer:

In Back cross 1 : 1 : 1 : 1
Test cross 1 : 1 : 1 : 1
Genotypic ratio of this cross = 1 : 2 : 2 : 4 : 1 : 2 : 1: 2 : 1
Phenotypic ratio of this cross = 9 : 3 : 3 : 1

Question 5.
In summer squash, a plant with white flowers and disc shaped fruits is crossed to a plant with yellow flowers and sphere shaped fruits. The F₁ hybrids had white flowers and disc shaped fruits. Which phenotypes are obminant? Give the genotypes of the parents and the hybrids. If these hybrids were selfed and 256 progeny were obtained, what would be the frequency of the various phenotypes?
Answer:

In dihybrid cross 9 : 3 : 3 : 1 phenotypic ratio is obtained.
The hybrids were selfed gives 256 progeny. Among them
White disc shaped fruits are 144 [[latex]\frac{256}{19}[/latex] × 9
White sphere shaped fruits = 48 [latex]\frac{256}{10}[/latex] × 3
Yellow; disc shaped fruits = 48 [latex]\frac{256}{16}[/latex] × 3
Yellow sphere shaped fruits = 16 [latex]\frac{256}{16}[/latex] × 1]

Question 6.
Give the ratios of the following
a) Monohybrid test cross
b) Dihybrid test cross
c) F₂ Phenotypic ratio of mono-hybrid cross
d) F₂ Phenotypic ratio of dihybrid cross
e) F₂ Genotypic ratio of monohybrid cross
f) F₂ Genotypic ratio of dihybrid cross.
Answer:
a) 1 : 1
b) 1 : 1 : 1 : 1
c) 3 : 1
d) 9 : 3 : 3 : 1
e) 1 : 2 : 1
f) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Question 7.
A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
Answer:
Locus is a fixed position on a chromosome which is occupied by a single or more genes. A diploid organisms which is heterozygous at four loci, will have over different contratling characters at four different loci. For example the organisms have Aa, Bb, Cc, Dd then during meiosis, it will seggregate to form 8 separate gametes.

If the genes are not linked, then the diploid organism will produce 16 different gametes. However if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited togethers during the process of meiosis.

Question 8.
What is crossing over? In which stage of cell division crossing over occurs? What is its significance?
Answer:
Exchange of chromatid segments between two homologous chromosomes is called crossing over. It ccurs in pachytene sub stage of meiosis. It leads to the evolution of new genetic recombinations.

Question 9.
“Genes contain the information that is required to express a particular trait”. Explain.
Answer:
Genes are the functional segments of DNA, which is a part of chromosome, controls traits. If proteins work efficiently then trait get expressed in a better way, height depends on the release of growth hormones by proteins. Thus through proteins present in genes, genes controls the traits.

Question 10.
For the expression of traits, genes provide only the potentiality and the environment provides the opportunity. Comment on the varaeity of the statement.
Answer:
Phenotype = Genotype + Environment
Trait = (Potentiality) (Opportunity)

Question 11.
Two heterozygolis parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer:

The above condition explained by using dihybrid crosses in drosophila were carried out by Morgan. He crossed yellow bodied (y) and white eyed (w) female with brown bodied (y⁺) and red eyed (w⁺) male, which produced F₁ with brown bodied red eyed female and yellow bodied white eyed male.

Question 12.
In peas, tallness is dominant over dwarfness and violet colour of flowers is dominant over the white colour. When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, violet = 138
Tall, white = 132
Dwarf, violet = 136
Dwarf, white = 128
Mention the genotypes of the two parents and of the four offsprings.
Answer:
Tall plant violet flowers
Dwarf plant white flowers
Test cross:

Question 13.
How do genes and chromosomes share similarities from the view point of genetical studies?
Answer:
The genes are located on chromosomes so therefore they both contain genetic information.

Gene :
A gene is an ordered sequence of nucleotides located in a particular position on a particular chromosome that encodes a specific functional product.

Chromosome components in a cell that contain genetic information. Each chromosome contains genes. Chromosomes occur in pairs, one from mother and one from father.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 8th Lesson Viruses Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 8th Lesson Viruses

Very Short Answer Questions

Question 1.
What is the shape of the T₄ phage? What is its genetic material?
Answer:
Tad pole-shape. Its Genetic material is double-stranded DNA.

Question 2.
What are virulent phages? Give an example.
Answer:
T-even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages Eg: Bacteriophage.

Question 3.
What is lysozyme and what is its function?
Answer:
Lysozyme is an viral enzyme, which breaks or lysed the bacterial cell wall to release their newly produced phage particles.

Question 4.
Define ‘lysis’ and ‘burst size’ with reference to viruses and their effects on host cells.
Answer:
The cycle in which the host cell wall ruptures to release the virions is called lysis. The number of newly synthesized bacteriophage particles released from a single affected host cell is called burst size.

Question 5.
What is a prophage?
Answer:
The phage DNA which is incorporated into bacterial DNA is called prophage.

Question 6.
What are temperate phages? Give an example.
Answer:
The phage DNA gets integrated into the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. Eg : A [Lambda] phage.

Question 7.
Mention the difference between virulent phages and temperate phages.
Answer:

Virulent Phage	Temperate Phage
1) Phages that attack the Bacterium E.coli cause lysis of the cells and are called virulent phages.	1) The phage DNA gets integrated into the bacterial DNA, becomes part of it and remains latent are called temperate phages.
2) They show lytic cycle.	2) They show lysogenic cycle.

Question 8.
What is the shape of Tmv? What is its genetic material?
Answer:
The shape of Tmv is rod and genetic material is single stranded RNA consisting of 6500 nucleotides.

Short Answer Questions

Question 1.
What is ICTV? How are viruses named?
Answer:
International Committee on Taxonomy of Viruses [ICTV] regulates the norms of classification and nomenclature of viruses. The ICTV has only three hierarchial levels, The family, Genus and Species. The family names end with the suffix viridae’ while the Genus name ends with virus and the Species names are common English expressions . Viruses are named after the disease they cause.
Eg : Polio virus.

Question 2.
Explain the chemical structure of viruses.
Ans. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to virus and also protects the genome. Capsid is made up of protein sub-units called capsomeres.

The nucleic acid may be either a double stranded DNA (ds DNA) or single stranded DNA [ss DNA]. In general, viruses that infect plants have ss RNA and viruses that infect animals have ds DNA. Most viruses have a single nucleic acid molecule but a few have more than one. Eg: HIV has two identical RNA molecules.

Question 3.
Explain the structure of TMV.
Answer:

1. It is a rod shaped virus. It is about’ 300 nm long and 18 or 19 nm in diameter with a molecular weight of 39 × 10⁶ daltons.
2. The capsid is made up of 2,130 sub-units called capsomeres.
3. Capsomeres are arranged in a helical manner around a central core of 4 nm.
4. Each protein sub-unit is made up of a single polypeptide chain with 158 amino acids.
5. Single stranded RNA is present inside the capsid and is spirally coiled.
6. RNA of TMV consists of 6,500 nucleotides.

Question 4.
Explain the structure of T-even Bacteriophages.
Answer:

1. The viruses which attack bacteria are called Bacteriophages. They were discovered by Twort (1915).
2. Felix’d’ Herelle (1917) coined the term Bacteriophage.
3. Bacteriophages are tadpole-shaped with a large head and a tail.
4. The head is hexagonal and is capped by hexagonal pyramid, measures about 65 × 95 nm.
5. The head is formed with several cap-someres, each of which is a single protein.
6. The head protein forms a semipermeable membrane enclosing the folded double stranded DNA which is 1000 times longer than the phage.
7. The tail is composed of several parts present around central core.
8. The tail core is 95 nm long and 8 nm in diameter. It is surrounded by tail sheath composed of 144 sub-units which are arranged in 24 rings of 6 each.
9. The head and tail are joined by collar whose function is not known.
10. At the tip of the tail, hexagonal tail plate is present with six tail pins and tail fibres.
11. The tail fibres help in attachment of the phage on to the host cell.

Question 5.
Explain the lytic cycle with reference to certain viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail Core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:
The plasma membrane of the host cell gets dissolved or lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or virions.

Long Answer Questions

Question 1.
Write about the discovery and structural organization of viruses.
Answer:
In 1892, the Russian Pathologist Dmitri Iwanowski, while studying Tobacco mosaic disease, he filtered the sap of diseased tobacco leaf through filters and is injected into a healthy plant. He found the symptoms of Mosaic disease in it. Finally he reported that, a filterable agent was responsible for the disease. Later Martinus Beijerinck repeated Iwanowski’s experiments and concluded that, the disease causing agent was a contagious living fluid’ (contagium vivium fluidum).

W.M Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized and was named TMV. Fralnke conrael (1956) confirmed that the genetid material of TMV is RNA.

Structure:-
Viruses range in size from 300 nm as in TMV to 20 nm as in Parvoviruses. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to the virus and provides a protective covering for the genome. It is made up of protein sub-units called, capsomeres. A virus contains its genetic information in either a double stranded (ds) DNA or single stranded (ss) DNA. In general, phytophages have ss RNA and zoophages have ds DNA. Bacteriophages are usually ds DNA viruses.

Shape:-
a) Helical viruses resemble long rods.
Eg: Rabies Virus, Tobacco Mosaic Virus.

b) Polyhedral viruses. They resemble polyhedral shape (many sided)
Eg : Herpes simplex and polio viruses.

c) Enveloped viruses: The capsid is covered by an envelope which are roughly spherical.
Eg : Influenza virus.

d) Complex viruses: Viruses which infect bacteria have complicated structures.
Eg : Bacteriophages have polyhedral symmetry in the head and helical symmetry in the tail sheath.

Question 2.
Describe the process of multiplication of viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:-
The plasma membrane of the host cell gets dissolved hr lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or varions.

Some Bacteriophages such as λ(Lambda) phages do not cause lysis and death of host cell when they multiply. Instead, the phage DNA, upon penetration into an E.coli cell gets integrated in to the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. The inserted phage DNA is now called prophages.

The prophages replicate along with the bacterial genetic material. The prophage remains latent with in the progny cells. In some rare spontaneous events, or when the host cell is exposed to UV light or some chemicals the phage DNA separates from the bacterial genetic material leading to the initiation of the lytic cycle. This lysogenic cycle facilitates the phenomenon of transduction.

Intext Questions

Question 1.
When discussing the multiplication of viruses, virologists prefer to call the process as replication, rather than reproduction. Why?
Answer:
Viruses must invade a host cell and take over the hosts metabolic machinery. So it is called replication.

Question 2.
In dealing with Public Health, the approach to deal with bacterial diseases is treatment. Canyou guess the nature of the general Public Health approach to viral diseases? What example do you cite to support your answer?
Answer:
“Prevention is better than cure” – in AIDS.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria

Very Short Answer Questions

Question 1.
Write briefly on the occurrence of microorganisms.
Answer:
Microorganisms are omnipresent in vast numbers and are described as ubiquitous. Among them, Bacteria are found in soil, water, air and on or inside living organisms. They also occur in a variety of foods, can withstand extreme cold, heat and drought conditions.

Question 2.
Define Microbiology.
Answer:
Microbiology is a branch of biology that deals with the scientific study of microorganisms. That are too small to be seen with the naked eye.

Question 3.
Name the Bacteria which is a common inhabitant of human intestine. How is it used in biotechnology?
Answer:
Escherichia coli (E.coli). It is used in biotechnology for the production of Insulin hormone.

Question 4.
What are Pleomorphic Bacteria? Give an example.
Answer:
The Bacteria which change their shape depending up on the type of environment and nutrients available are called Pleomorphic Bacteria.
Eg : Acetobacter.

Question 5.
What is sex pilus? What is its function?
Answer:
Sex pilus is a conjugation apparatus that pulls two cells together prior to DNA transfer. It help in binding two conjugants and also help in passing F plasmid from F⁺ cell to F^– cell.

Question 6.
What is a genophore?
Answer:
Genophore is the main genetic material of Bacteria which is circular, double stranded DNA without histone proteins. It is also called Bacterial chromosome.

Question 7.
What is a plasmid? What is its significance?
Answer:
“Plasmid is a small self-replicating circular, double stranded DNA molecule present in a Bacteria in addition to main genetic material”. They are used as agents (vectors) in modem Genetic Engineering techniques.

Question 8.
What is conjugation? Who discovered it and in which organism?
Answer:
Transfer of DNA from one Bacterium (Donor) to the recipient Bacterium through direct contact is called conjugation. It was first observed in Escherichia coli by Lederberg and Tatum (1946).

Question 9.
What is transformation? Who discovered it and in which organism?
Answer:
Uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient cell is called transformation. It was discovered by Frederick Griffith in Streptococcus pneumoniae (1928).

Question 10.
What is transduction? Who discovered it and in which organism?
Answer:
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction. It was observed by Lederberg and Zinder in Salmonella typhimurium (1951).

Short Answer Questions

Question 1.
How are bacteria classified on the basis of Morphology?
Answer:
Bacteria are classified into four types based on their shape. They are
I) Cocci : Spherical Bacteria.
Based on the number of cells and their arrangement, the Cocci are classified into
(a) Monococcus : A single cell
(b) Diplococcus : A pair of cells
(c) Tetracocci : A group of four cells
(d) Streptococcus : A linear chain of cells
(e) Staphylococci : A bunch of cells
(f) Sarcina : A group of 8 cells, arranged in cubes

II) Bacilli :
Rod-shaped Bacteria : The Bacillus forms exist as
a) Monobacillus : A single rod-shaped ceil.
b) Diplobacilli : A pair of rod-shaped cells.
c) Streptobacilli : A chain of rod-shaped cells.

III) Spirillum Helical rod-shaped cell.
It may be a distinct helical shape (spirillum) or Slender long and cork-screw shaped (Spirochete).

IV) Vibrio : Comma-shaped Bacteria.

Question 2.
How are Bacteria classified on the basis of number and distribution of flagella?
Answer:
Bacteria are classified into four types based on the number and arrangement of flagella.
They are
1) Monotrichous : A single polar flagellum is present at one end of Bacterial cell.
2) Lophotrichous : A tuft of flagella at one pole of the cell.
3) Amphitrichous : A single flagellum at each end of the cell.
4) Peritrichous : Flagella distributed over the entire cell.

Question 3.
What are the nutritional groups of Bacteria based on their source of .energy and carbon?
Answer:
1) Photoautotrophs :
They are photosynthetic i.e., they capture light energy and transform it into chemical energy and obtain carbon from atmospheric CO₂.
Eg: Chlorobium, Chromatium,

2) Chemoautotrophs :
They derive energy from the oxidation of inorganic substances and carbon from atmospheric CO₂.
Eg: Nitrosomonas, Nitrobacter, Beggiaotoaand Methanogens.

3) Photoheterotrophs :
They capture light energy and carbon from other organic sources.
Eg : Rhodospirillum and Rhodopseudomonas.

4) Chemoheterotrophs:
They derive both carbon and energy from organic compounds.
Eg: Xanthomonas, Salmonella.

Question 4.
Explain the conjugation in Bacteria.
Answer:
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F^– cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

Long Answer Questions

Question 1.
Explain different methods of sexual reproduction in Bacteria.
Answer:
True sexual reproduction is absent in Bacteria. However, the exchange of genetic material takes place in three ways. They are 1) Conjugation 2) Transformation 3) Transduction

1) Conjugation:-
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

2) Transformation:-
It is uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient chromosome in a heritable form is known as transformation. It was discovered by Frederick Griffith (1928) in Streptococcus pneumoniae.

3) Transduction:-
The transfer of genetic material from one Bacterium to another through Bacteriophage is known as transduction. It was discovered by Lederberg and Zinder in Salmonella typhimurium.

Question 2.
“Bacteria are friends and foes of man” – Discuss.
Answer:
Many Bacteria are directly or indirectly beneficial and harmful to humans. Hence they are considered both as friends and foes of man.
1) Beneficial activities:-
Some Bacteria are important to man in the field of agriculture, industry, medicine and biotechnology.

I) Agriculture :
Bacteria play a significant role in maintaining the fertility of the soil.

They are :
a) Ammonifying Bacteria:
These Bacteria convert the proteins, and amino acids of the dead bodies into Ammonia. This process is called Ammonification.
Eg: Bacillus
b) Ammonia is oxidised to nitrates, by nitrifying bacteria, This process is called Nitrification Eg: Nitrosomonas, Nitrobacter.
c) Symbiotic Bacteria like Rhizobium, non-symbiotic Bacteria like Clostridium and photosynthetic Bacteria like Rhodospirillum, Rhodomicrobium and Chlorobacterium fix the atmospheric gaseous nitrogen and enrich the soil.
d) Bacillus thuringiensis is used as an agent against insect pests. Hence popularly used as a bioinsecticide.

II) Industry:
Industrially, Bacteria are employed in a number of processes. They are :
a) Clostridium butyricum and C. felcinium are used in retting of sunhemp and flax respectively to extract the fibre.
b) Some Bacteria are used in tanning industry.
c) Bacteria such as Bacillus megaterium and Micrococcus are used in curing of tobacco and tea respectively.
d) Some Bacterm are used in fermentation process, Eg: Lactobacillus.
e) Bacteria like Methanococcus and Methanobacillus ferment the dung anaerobically and produce methane that is commonly referred to as gobar gas.
f) In various chemical industries, Bacteria are extensively used for the production of chemicals. Few, examples are given below.

Chemical	Bacterium
Acetone, Butanol	Clostridium acetobutylicum
Vinegar	Acetobacter aceti
	A. pasteurianum
Lactic Acid	Lactobacillus delbruckii
Propionic Acid	Propionibacterium propionum
Ethanol	Zymomonas mobilis
	Thermoanaerobacter ethanolicus

III) Medicine:
a) Corynebacterium glutamicum produces lysine, an essential amino acid.
b) A large number of antibiotics are produced by Bacteria. Species of Streptomyces and Bacillus produce important antibiotics as mentioned below.

Antibiotic	Bacterium
Streptomycin, Cyclohexamide	Streptomyces griseus
Chloramphenicol	S. venezuelae
Neomycin	S. fradiae
Kanamycin	S. kanamyceticus
Amphotericin	S. nodosus
Oxytetracycline	S. rimosus
Polymyxin B	Bacillus polymyxa
Bacitracin	B. licheniformis

IV) Biotechnology:
a) With the help of recombinant DNA technology, it was made possible to produce insulin hormone from Escherichia coli.
b) Some Bacteria store large quantities of proteins inside their cells. These are used as a source of single cell protein (SCP).
Eg : Brevibatiterium.
c) Agrobacterium tumefaciens is used as vector in genetic engineering.

2) Harmful activities :
A few saprophytic and all parasitic Bacteria carry on some processes which are harmful to man.

i) Spoilage of food materials :
Bacteria grow on different types of food materials and render them unsuitable for human consumption. Some of these Bacteria produce powerful toxins while growing on the food materials. For instance, Clostridium botulinum releases a very potential toxin, ‘botulin’ which causes ’botulism’, a type of food poisoning.

ii) Plant diseases :
A number of species of Bacteria are reported to cause different plant diseases as mentioned below.

Disease	Bacterium
Blight of rice	Xanthomonas oryzae
Citrus canker	X. axonopodis PV citri
Crown gall of apples and pear	Agrobacterium tumefaciens

Some important crop diseases caused by Bacteria

Some Bacteria cause Human diseases. They are :

Disease	Bacterium
Tetanus	Clostridium tetani
Botulism	Clostridium botulinum
Cholera	Vibrio cholera
Typhoid	Salmonella typhi
Diptheria	Corynebacterium diphtheriae
Tuberculosis	Mycobacterium tuberculosis
Leprosy	Mycobacterium leprae
Pneumonia	Diplococcus pneumoniae
Gonorrhoea	Neisseria gonorrhoea
Syphilis	Treponema pallidum

Intext Questions

Question 1.
Many people believe that bacteria do little more than cause human illness and infectious diseases. How does the information in the chapter help you correct that misconception?
Answer:
Bacteria are directly or indirectly beneficial to Humans. Hence they are considered both as “friends and foes of man”.

Question 2.
An organism is described as a peritrichous bacillus. How might you translate this bacteriological language into a description of the organism.
Answer:
It is Rod shaped Bacteria.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development

Very Short Answer Questions

Question 1.
Define plasticity. Give an example.
Answer:
Ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures in called plasticity.
Ex : Heterophily.

Question 2.
What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.
Answer:
Bakane (foolish seedling) disease in rice seedlings. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it.
Answer:
Growing Apical bud inhibits, the growth of Axillary buds is called Apical dominance. It is caused by Auxins.

Question 4.
What is meant by bolting ? Which hormone causes bolting?
Answer:
Sudden elongation of internodes prior to flowering is called bolting. It is caused by Gibberellins.

Question 5.
Define respiratory climactic. Name the PGR associated with it.
Answer:
The rise in the rate of respiration during the ripening of fruits is known as respiratory climatic. It is associated by ethylene.

Question 6.
What is Ethephon? Write its role in agricultural practices.
Answer:
It is an Ethylene releasing chemical formulation. It hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits. It promotes female flowers in cucumbers, there by increasing the yield.

Question 7.
Which of the PGRs is called stress hormone and why?
Answer:
ABC [Abscisic acid] is called stress hormone. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 8.
What do you understand by vernalisation?
Answer:
The method of inducing flowering quantitatively or qualitatively on exposure to low temperature is called vernalisation. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity. It specially refers to the promotion of flowering. It also stimulates a subsequent photoperiodic flowering response in biennials, [cabbage, carrot]

Question 9.
Define the terms quiescence and dormancy.
Answer:
The condition of a seed when it is unable to germinate only because favourable external condition normally required for growth are not present is called quiescence.

The condition of a seed when it fails to germinate because of internal conditions even though external conditions are suitable is called dormancy.

Short Answer Questions

Question 1.
Write a note on agricultural / horticultural applications of auxins.
Answer:

1. IBA, NAA and IAA help to initiate rooting in stem cuttings, widely used for plant propagation in horticulture.
2. Auxins like 2,4-D, 2,4,§-T acts as herbicides and kills broad leaved dicot weeds to prepare weed – free lawns.
3. Auxins stimulates fruit growth. Ex : Tomatoes.
4. Auxins induces flowering in pineapple.
5. Prevents pre harvest fruit drop.

Question 2.
Write the physiological responses of Gibberellins in plants.
Answer:

1. Gibberellins delay senescence. Thus fruits can be left on the tree longer so as to extend the market period. .
2. Spraying of Gibberellins on sugarcane crop, increases the length of the stem, thus increasing the yield as much as 20 tonns per acre.
3. GA hastens the maturity period of conifers thus leading to early seed production.
4. GA also promotes bolting in cabbages, beet etc.,
5. They also produce parthenocarpic fruits in grapes and tomato.
6. Gibberellins favour the formation of male flowers in cucurbita.

Question 3.
Write any four physiological effects of cytokinins in plants.
Answer:

1. Cytokinins induces cell division.
2. They help to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
3. Cytokinins help to overcome apical dominance.
4. They promote nutrient mobilisation which helps in the delay of senescence.
5. Cytokinins help in the opening of stomata by increasing the concentration of k+ ions in guard cells.

Question 4.
What are the physiological prodesses that are regulated by ethylene in plants?
Answer:

1. Ethylene promotes the ripening of fruits.
2. Ethylene promotes the senescence and abscission of leaves and flowers.
3. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds and sprouting of potato tubers.
4. Ethylene promotes rapid intemode/petiole elongation in deep water rice plants.
5. It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
6. Ethylene is used to initiate flowering (mango) and for synchronising fruit set in pineapples.
7. It promotes female flowers in cucumbers, thereby increasing the yield.

Question 5.
Write short notes on seed dormancy.
Answer:
The inability of seed to germinate or grow is called dormancy. It may be due to either external factors or internal factors.

Internal factors:
1) Immature embryo :
The embryo has not reached morphological maturity to germinate.
Ex : Ranunculus.

2) Hard seed coat:
In fabacea members, seeds have hard seed coats which prevent uptake of oxygen or water. It can be broken by scarification in which the hard seed coat is ruptured or weakened.

3) Chemicals :
Seeds of some plants (tomato) contain chemical compounds which inhibit their germination.

External factors:
1) Low temperature treatment :
Many seeds (polygonum) with not germinate until they have been exposed to low temperatures in moist conditions in the presence of oxygen for weeks to months. The practice of layering the seeds during winter in moist sand and peat is called stratification or prechilling.

2) Seeds of many domestic plants may be limited only by lack of moisture or warm temperature.

Question 6.
Which one of the plant growth regulators would you use if you are asked to.
a) Induce rooting in a twig
Answer:
Auxins

b) Quickly ripen a fruit
Answer:
Ethylene

c) Delay leaf senescence
Answer:
Cytokinins

d) Induce growth in axillary buds
Answer:
Cytokinins

e) ‘Bolt’ a rosette plant
Answer:
Gibberellins

f) Induce immediate stomatal closure in leaves
Answer:
Abscisic acid

g) Overcome apical dominance
Answer:
Cytokinins

h) Kill dicotyledonous weeds
Answer:
Auxins – 2, 4 – D

Question 7.
Describe briefly, a) Sigmoid growth curve, b) Absolute and relative growth rates.
Answer:

a) Sigmoid Growth Curve : It consists of 3 phases namely
1) lag phase
2) log pase
3) stationary phase.

1) Lag phase :
Growth is slow

2) Log phase :
Growth increases rapidly at an exponential rate

3) Stationary phase :
Limited nutrient supply, slows down the growth leading to a stationary phase. If we plot the parameter of growth against time, we get a typical sigmoid or S-curve.

b) Absolute and relative growth rates :
Measurement and comparision of the total growth per unit time is called absolute growth rate. The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called relative growth rate.

Long Answer Question

Question 1.
List five natural plant growth regulators. Write a note on discovery, physiological functions and agricultural / horticultural applications of any one of them.
Answer:
Auxins, Gibberellins, Cytokinins, Abscisic acid, Ethylene

Auxins:
Discovery :
Observation of Charles Darwin and his son Francis Darwin, that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light. It was concluded that the tip of the coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from the tips of coleoptiles of oat seedlings.

Physiological functions :

1. Auxins help of initiate rooting in stem cuttings.
2. Auxins promote flowering in pineapples.
3. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Auxins promote apical dominance.
5. Auxins also induce parthenocarpy in tomatoes.
6. They are widely used as herbicides 2,4-D widely used to kill dicotyledonous weeds.
7. Auxins also controls zylem differentiation and help in cell division.

Agricultural / Horticultural applications ;

1. Auxins help to initiate rooting in stem cuttings, widely used for plant propagation in Horticulture.
2. Auxins, 2, 4-D widely used as herbicide ie, kills dicotyledonous weeds to prepare a weed free-lawns.

Intext Questions

Question 1.
Fill in the blanks with appropriate word/words.
a) The phase in which growth is most rapid is ………… .
b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more ………… in the apical bud than in the lateral ones.
c) In addition to ouxin, a …………. must be supplied to the culture medium to obtain a good callus in plant tissue culture.
d) ………….. of vegetative plants are the sites of photoperiodic perception.
Answer:
a) LOg phase or exponential phase
b) Auxins
c) Cytokinin
d)Shootapices of plants

Question 2.
Gibberellins promote the formation of Male flowers on genetically Modified (dwarf) plants in cannabis whereas Ethylene promotes formation of Female flowers on genetically Modified (dwarf) plants.

Question 3.
Classify the following plants into long day plants(LDP), short day plants(SDP) and Day Neutral plants(DNP). Xanthium, Spinach, Henbane (Hyoscyamus), Rice, Strawberry, Bryophyllum, niger, Sunflower, Tomato, Maize.
Answer:
Long day plants – Henbane, Spinach,
Short day plants – Xanthium, Sunflower, Rice, Tomato
Day nutral plants : Maize, Bryophyllum

Question 4.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers. Which plant growth regulator can be applied to achieve this?
Answer:
Ethyline.

Question 5.
Where are the following hormones synthesized in plants?
a) IAA b) Gibberellins c) Cytokinins.
Answer:
a) IAA – growing apices of the stems and roots.
b) Gibberellins : Fungal pathogen, Gibberella fujikuroi.
c) Cytokinins – roots apices, shoot…, young fruits.

Question 6.
Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are influenced by light.
Answer:
Growth, differentiation and development.

Question 7.
Growth is one of the characteristics of all living organisms. Do unicellular organisms also grow? If so, what all the parameters?
Answer:
Yes, unicellar organisms also grow, some parameters are increase in fresh weight, length, area, volume and cell number.

Question 8.
Rice seedlings infected with fungus Gibberella fujikuroi are called Foolish seedlings. What is the reason?
Answer:
The fungus cause foolish seeding disease in rice which show, plant grow. Very tall, become pale, produce less tillers and less yield.

Question 9.
Why isn’t any one parameter good enough to demonstrate growth throughout the life of a flowing plant?
Answer:
One parameter is not good enough to demonstrate growth because various parts show growth… is estimated in different parameters Eg: Growth of the pollentube is measured interms of the length, increase in surface are dexoles growth in a dorsiventral leaf.

Question 10.
‘Both growth and differentiation in higher plants are open’ Comment.
Answer:
Both growth and diffentitation in higher plants are open because, cells/tissues arising out of the same meristem have different structures of maturity and their location. Eg. : Cells away from root apical meristem differentiate as root cap cells, while those pushed to the periphery nature as epidermis.

Question 11.
Both a short day plant and a long day plant can produce flowers simutaeneously in a given place’. Explain.
Answer:
5 me plants require the exposure to light for a period exceeding critical duration and some require a light for a period less than critical duration for flowing.

Question 12.
Would a defoliated plant respond to photoperiodic cycle? Why?
Answer:
No, because some hormonal substance migraterwom leaver to shoot apices for induce flowing.

Question 13.
What would be expected to happen if
a) GAj is applied to rice seedlings.
b) Dividing cells stop differentiating.
c) A rotten fruit gets mixed with unripe fruits.
d) You forget to add cytokinnin to culture medium.
Answer:
a) Appearance of disease symptoms (Bakane disease) in uninfected rice seedings.
b) Growth of the plant body in stopped.
c) Unripe fruits became ripened due to ethylene.
d) Initiation of heafy shoots is inhibited.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants

Very Short Answer Questions

Question 1.
Different substrates get oxidised during respiration. How does respiratory quotient (RQ) indicate which type of substrate (i.e) carbohydrate, fat or protein is getting oxidised?
RQ = A/B. What do A & B stand for?
What type of substrates have RQ of 1, <1; > 1?
Answer:

carbohydrates have RQ = 1
Fats have RQ value = < 1
Organic acids have RQ = > 1

Question 2.
What is the specific role of F₀ – F₁ particles in Respiration?
Answer:
The F₁ headpiece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate. F₀ is an intergral membrane protein complex that forms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP

Question 3.
When does anaerobic respiration occur in man and yeast?
Answer:
When oxygen is inadequate for cellular respiration, anaerobic respiration occurs.

Question 4.
What is the common pathway for aerobic and anaerobic respirations? Where does it take place?
Answer:
Glycolysis. It occurs in cytoplasm of the cell.

Question 5.
What cellular organic substances are never used as respiratory substrates?
Answer:
Pure proteins or fats are never used as respiratory substrates.

Question 6.
Why is the respiratory quotient (RQ) of fats less than that of carbohydrates?
Answer:
During the oxidation of fats, the amount of CO₂ evolved is less than the amount of O₂ consumed so respiratory quotient value is less than one. Where as in oxidation of carbohydrates, the amount of CO₂ evolved is equal to the amount of O₂ consumed. So the respiratory quotient value is one.

Question 7.
What is meant by Amphibolic pathway?
Answer:
Respiratory pathway involves both anabolism and catabolism so it is considered as amphibolic pathway. In this, the breakdown and the synthesis of fatty acids and proteins took place.

Question 8.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:
Ubiquinone and cytochrome – C.

Question 9.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:
Oxygen. It receives electrons from complex IV

Question 10.
Do you know of any step in Kreb’s cycle where there is a substrate level phosphorylation? Explain.
Answer:
In Kreb’s cycle during the conversion of succinyl – CoA to succinic acid a molecule of GTP is synthesised. This step is a substrate level phosphorylation.

Short Answer Questions

Question 1.
Why is the respiratory pathway referred to as amphibolic pathway? Explain.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, j they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 2.
Write about two ATP yielding reactions of glycolysis.
Answer:
1) 1, 3 BPGA (bisphosphoglyceric acid) looses phosphate group in the presence of phosphoglycerokinase to form 3-phosphoglyceric acid. ADP accepts phosphate group and gets converted to ATP.

2) Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of pyruvic kinase results in the formation of pyruvic acid. ADP accepts phosphate group and gets converted to ATP.

Question 3.
The net gain of ATP for the complete aerobic oxidation of glucose is 36. Explain.
Answer:
Balance sheet of ATP production in aerobic oxidation of Glucose.
1) Glycolysis :
1. ATP produced by substrate level phosphorylation
Bisphosphoglyceric acid to phosphoglyceric acid : 2 × 1 = 2ATP
phosphoenol pyruvic acid to pyruvic acid : 2 × 1 = 2ATP
ATP consumed : for the phosphorylation of glucose
and fructose-6-phosphate : -2 ATP
Net gain of ATP : +2 ATP

2. ATP from NADH generated in glycolysis :
G-3-P to BPGA (2NADH, each worth 2ATP) 2 × 2 = 4ATP
Total ATP gain from glycolysis in the presence of O₂ : ^(a)6 ATP

2) Oxidative decarboxylation of pyruvic acid
Pyruvic acid to acetyl COA
(2 NADH, each worth 3 ATP) : ^(b)2 × 1 = 6 ATP

3) Krebs cycle
1. ATP produced in substrate level phosphorylation:
Succinyl CoA to succinic acid : 2 × 1 = 2 ATP
2. ATP from NADH: Isocitric acid to Oxalosuccinic acid : 2 × 3 = 6 ATP
a-Ketoglutaric acid to succinyl CoA : 2 × 3 = 6 ATP
Malic acid to Oxalocaetic acid : 2 × 3 = 6 ATP
3. ATP from FADH2 : Succinic acid to fumaric acid : 2 × 2 = 4 ATP
Total ATP value of krebs cycle : ^(a)24 ATP
Net gain of ATP in aerobic respiration per mole glucose
(a + b + c) : 36 ATP

Question 4.
Define RQ. Write a short note on RQ.
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ cosumed in respiration is called the Respiratory Quotient (RQ).

The respiratory quotient depends upon the type of respiratory substrate used during respiration.
1) When carbohydrates are used as substrate, the RQ is 1, because equal amounts of CO₂ and O₂ are evolved and consumed.
Ex : C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + Energy
RQ = [latex]\frac{6CO_2}{6O_2}[/latex] = 0.1

2) When fats are used in respiration the RQ is less than 1.
Ex : 2(C₅₁ H₉₈ O₆) + 145 O₂ → 1 O₂ CO₂ + 98 H₂O + Energy Tripalmitin
RQ = [latex]\frac{102CO_2}{145O_2}[/latex] = 0.7

3) When protein are respiratory substrates the ratio would be about 0.9

Question 5.
Describe briefly the process of fermentation.
Answer:
In fermentation, the incomplete oxidation of glucose is achieved under anaerobic conditions by set of reactions where pyruvic acid is converted to CO₂ & Ethanol. The ezymes, pyruvic acid decarboxylase, and alcohol dehydrogenase catalyse these reactions. Two types of fermetations are :

1. Alcoholic fermentation
2. Lactic acid fermentation

1) Alcoholic fermentation :
This is most common type of fermentation process. In alcoholic fermentation, pyruvic acid is broken to ethyl alcohol & CO₂ in two steps.
i) The pyruvic acid is decarboxylated to acetaldehyde in the presence of enzyme pyruvic decarboxylase.

ii) Acetaldehyde is then reduced to ethyl alcohol by NADH⁺ + H⁺ produced in glycolysis acetaldehyde is final hydrogen acceptor. The reaction is catalysed by alcohol dehydrogenase.

2) Lactic acid fermentation:
Pyruvic acid formed at the end of glycolysis is converted to lactic acid by lactobacillus lactibacteria.

Question 6.
Explain various complexes involved in electron transport system of respiration.
Answer:
Electron flow through the mitochondrial electron transport (Aerobic respiration) is carried out by five enzyme complexes. These complexes are the integral proteins of the inner mitochondrial membrane. They are complex I, II, III, IV & V out of these first four complexes involved in Electron transport chain.
i) Complex I :
NADH – dehydrogenase (or) NADH-Q-reductase. This complex I transfers electrons from NADH to ubiquinone.

ii) Complex II :
Succinate dehydrogenase (or) ubiquinone.This complex trasfers electrons from succinate to ubiquinone via Fe – S – centres.

iii) Complex III :
Cytochrome ‘c’ Reductase or cyt-b-c, complex. This complex contains cytochrome ‘b’ and cytochrome ’c’ cyt-c(mobile carrier). The ultimate electron acceptor of complex-III. This complex works through Q-cycle mechanism.

iv) Complex IV :
Cytochrome -c- oxidase. It contains two heme proteins called cyta and cyta₃ and two copper proteins. This complex transfers electrons to oxygen.

v) Complex V :
ATP synthase or F₀ – F₁ complex. This complex involved for the production of ATP from ADP & inorganic phosphate.

Question 7.
Describe the structure of Complex-V and explain the process of oxidative phospho-rylation as explained by chemosmotic hypothesis.
Answer:

Complex – V is an ATP synthase. This complex-V consists of two major components, F₁ and F₀. The F₁ head piece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. F₀ is an integral membrane protein complex that froms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP For each ATP produced, 3H⁺ passes through F₀ from the intermembrane space to the matrix down the electrochemical proton gradient.

In photophosphorylation light energy is utilised for the production of proton gradient required for phosphorylation, but in respiration it is the energy of oxidation reduction utilized for the proton gradient. It is for this reason that the process is called oxidative phosphorylation. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that one molecule of FADH₂ produces 2 molecules of ATP.

Long Answer Questions

Question 1.
Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Answer:
Glucose is broken down into 2 molecules of pyruvic acid is called glycolysis. It was given by gustav embden, Otto Mayerhof and J.Parnas so called EMP pathway. It occurs in the cytoplasm of the cell and takes place in all living orgnisms. In this, 4 ATP are formed of which two are utilised and 2 NADPH⁺ K⁺ are formed. A⁺ the end of glycolysis, 2PA, 2ATP and 2NADPH⁺ H⁺ are formed as end products. The ATP and NADPH⁺ H⁺ are utilised for fixation of CO₂.

Glycolysis occur in cytoplasm. Pyruvic acid, 2ATP, 2NADPH⁺H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH + H⁺ are completely oxidised through TCA cycle, ETS pathway and produce 36 ATP molecules. In Anaerobic respiration, pyruvic acid is partially oxidised results in the formation of Ethyol alcohol and CO₂.

Question 2.
Explain the reactions of Krebs cycle.
Answer:

The acetyl CoA enters into the [mito chondrial matrix] a cyclic pathway tricarboxylic acid cycle, more commonly called krebs cycle after the scientist Hans Krebs who first elucidated it.

1) Condensation :
In this acetyl CoA condenses with oxaloacetic acid and water to yield citric acid in the presence of citrate synthetase and CoA is released.

2) Dehydration :
Citric acid looses water molecule to yield cisaconitic acid in the presence of aconitase.

3) Hydration :
A water molecule is added to cis aconic acid to yield isocitric acid in the presence of a conitase.

4) Oxidation I:
Isocitric acid undergoes oxidation in the presence of dehydrogenase to yield succinic acid

5) Decarboxylation :
Oxalosuccinic acid undergoes decarboxylation in the presence ofdecarboxylase to form a-keto glutaric acid.

6) Oxidation II, decarboxylation :
α – keto glutaric acid undergoes oxidation and decarboxylation in the presence of dehydrogenase and condenses with co.A to form succinyl co. A.

7) Cleavage :
Succinyl co.A splits into succinic acid and co.A in the presence of thiokinase to form succinic acid. The energy released is utilised to from ATP from ADP and PI.

8) Oxidation – III:
Succinic acid undergoes oxidation and forms Fumaric acid in the presence of succinic dehydrogenase.

9) Hydration :
A water molecule is alcohol to Fumaric acid in the presence of Fumarase to form Malic acid.

10) Oxidation IV :
Malic acid undergoes oxidation in the presence of malic dehydrogenase to form oxaloacetic acid.

In TCA cycle, for every 2 molecules of Acetyl co.A undergoing oxidation, 2 ATP, 8 NADPH+ H⁺, 2FADH₂ molecules are formed.

Intext Questions

Question 1.
Differentiate between
a) Respiration and Combustion
b) Glycolysis and Krebs cycle
c) Aerobic respiration and Fermentation
Answer:
a)

Respiration	Combustion
1) Oxygen is utilised, CO2, water and energy are released.	1) It requires oxygen.

b)

Glycolysis	Krebs cycle
1) It occurs in the cytoplasm.	1) It occurs in the mitochondrial matrix.
2) It occurs in all living organisms.	2) It occurs in higher plants onlv.

c)

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Substances which oxidises in respiration are called respiratory substrates. Ex: Glucose, Fats, Proteins and Organic acids. Among them glucose is the common respiratory substrate.

Question 3.
Give the schematic representation of glycolysis.
Answer:

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
Glycolysis = Cytoplasm
PAOD = Matiix (Mitochondria)
Krebs cycle = Matrix (Mitochondria)
Electron transport system = Matrix (Mitochondria)

Question 5.
Give the schematic representation of an overall view of Krebs cycle.
Answer:

Question 6.
Explain ETS
Answer:
NADPH + H⁺ or FADH₂ formed in glycolysis and Mitochondrial matrix are oxidised through the electron transport system and the electrons are passed on to O₂ resulting in the formation of H₂O. The metabolic pathway through which an electron passes from one carrier to another is called electron transport system.

NADPH + H⁺ is oxidised and releases electrons in the presence of NADH dehydrogenase (complex I), which are transferred to ubiquinone located with in the inner membrane. Ubiquinone also receives reducing equivalents via FADH₂ (Complex II). The reduced ubiquinone (Ubiquinol) is oxidised and releases electrons which are accepted by cytochrome c₁ Cytochrome_c (Complex III) cytochrome C is a small protein, acts as a mobile carrier for the transfer of electrons between complex III and TV to cytochromes a and a₃, finally reaches [latex]\frac{1}{2}[/latex]O₂ along with 2H⁺, produce 1 H₂O molecule.

During this electron flow, 10H⁺ moves from matrix to inner mitochondrial membrane [4 H⁺ at complex I, 4H⁺ at complex III, 2 at complex IV]. As a result H⁺ concentration increases towards the inner membrane of mitochondria. So the H⁺ comes back to the matrix side through ATPage [F₀, F₁], involves in the synthesis ATP.

Question 7.
Distinguish between the following
a) Aerobic respiration and Anaerobic respiration
b) Glycolysis and Fermentation
c) Glycolysis and Citric acid cycle
Answer:
a)

Aerobic respiration	Anaerobic respiration
1. It occurs in the presence of O2.	1. It occurs in the absence of O2.
2. It occurs in four steps. a) Glycolysis b) PAOD c) Krebs cycle d) ETS pathway	2. It occurs in two steps. a) Glycolysis b) Fermentation
3. End products are CO2 and H2O.	3. End products are CO2 and Ethyl alcohol.
4. 686 K.Cal of energy is released.	4. 56 K.Cal of energy is released.
5. 36 ATP are formed.	5. 2 ATP are formed.

b)

Glycolysis	Fermentation
1. It is the 1st step in both aerobic anaerobic respiration.	1. It is the 2nd step of anaerobic repsiration.
2. End products are PA, ATP, NADPH+ H+	2. End products are CO2 and C2H5OH

c)

Glycolysis	Citric acid cycle
1. It occurs in the cytoplasm.	1. It occurs in the mitochondrial matrix.
2. It occurs in all living organisms.	2. It occurs in higher plants only.

Question 8.
What are the assumptions made during the calculations of net gain of ATP?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose is oxidised.
These calculations can be made only on certain assumptions that:

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle, ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesize any other compound.
4. Only glucose is being transferred – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But these kind of assumption are not really valid in a living system all pathways work simultaneously and do not take place one after another.

Differences

Fermentation	Aerobic respiration
1) Partial breakdown of glucose.	1) Complete oxidation of glucose.
2) End products are CO2 and Ethyl alcohol.	2) End products are CO2 and H2O.
3) 2 ATP molecules are formed.	3) 36 ATP are formed.

Question 9.
Discuss “The respiratory pathway is an amplibolic pathway”.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the 1 pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 10.
Define respiratory quotient (RQ). What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient.
Respiratory quotient value of fats is less than one
2C₅₁ H₉₈ O₆ + 145 O₂ → 1 02 CO₂ + 98 H₂O + energy
(Tripalmitin) Respiratory quotient = [latex]\frac{102}{145}[/latex] = 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
Synthesis of ATP from ADP and Pi coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough such that the energy is released can be coupled to ATP synthesis. All the energy entained in a substrate is not released free into the cell or in a single step. It is release in a series of slow step wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

13. Find the correct ascending sequence of the following, on the basis of energy released in respiratory oxidation.
a) 1 gm pf fat
b) 1 gm of protein
c) 1 gm of glucose
d) 0.5 gm of protein + 0.5 gm of glucose
Answer:
c → b → d → a

Question 14.
Name the products, respectively in aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast.
Answer:
In aerobic glycolysis :
Pyruvic acid, ATP, NADPH + H⁺

In skeletal muscle :
Lactic acid

In anaerobic fermentation :
CO₂ and Ethanol

Question 15.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Why?
Answer:
Glucose or fruit juice. They contain more sugars which oxidise in the presence of O₂ and releases more energy.

Question 16.
In a way green plants and cyanobacteria have synthesised all the food on earth. Comment.
Answer:
Only green plants and cyanobacteria can prepare their own food but not all food on earth.

Question 17.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Answer:
Red muscle fibres utilise fats or carbohydrates as their food. They can work for long time but with little force.