Important Questions

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Very short answer type questions

Question 1.
The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region at which 5 the semiconductor has a negative resistance.
Answer:
Resistance of a material can be found out by the slope of the curve V versus I. Part BC of the curve u shows the negative resistance as with the increase in current and decrease in voltage.

Question 2.
Sn, C, SI and Ge are all group 14 elements, Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why? (NCERT Exemplar)
Answer:
If the valence and conduction bands overlap (no energy gap), the substance is referred as a conductor. For insulators the energy gap is large and for semiconductors, the energy gap is moderate. The energy gap of Sn is O eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV. Accordingly, their electrical conductivity varies.

Question 3.
Why are elemental dopants for Silicon or Germanium usually chosen from group 13 or group 15? (NCERTExemplar)
Answer:
The size of dopant atoms should be such as not to distort the pure semiconductor lattice structure and yet easily contribute a charge camer on forming covalent bonds with Si or Ge.

Question 4.
What happens to the width of depletion Layer of a p.n junction when it is
(i) forward biased,
(ii) reverse biased
Answer:
(i) When forward biased, the width of depletion layer decreases.
(ii) When reverse biased, the width of depletion layer increases.

Question 5.
In a transistor, doping level in base is increased slightly. How will it affect
(i) collector current and
(ii) base current?
Answer:
When doping level in base is increased slightly;
(i) Collector current decreases slightly and
(ii) Base current increases slightly.

Question 6.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction (NCERT Exemplar)
Answer:
No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.

Question 7.
Explain why elemental semiconductors cannot be used to make visible LEDs.(NCERT Exemplar)
Answer:
Elemental semiconductor’s band-gap is such that’ electromagnetic emissions are in infrared region.

Question 8.
Draw the logic circuit of a NAND gate and write its truth table.
Answer:
Logic circuit of a NAND gate

Truth table

A	B	Y = A.B
0	0	1
0	1	1
1	0	1
1	1	0

Short answer type questions

Question 1.
Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
or
Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators. How does the change in temperature affect the behaviour of these materials? Explain briefly.
Answer:

Distinguishing Features
(a) In conductors: Valence band and conduction band overlap each other.
In semiconductors: Valence band and conduction band are separated by a small energy gap.
In insulators: Valence band and conduction band are separated by a large energy gap.
(b) In conductors: Large number of free electrons are available in conduction band.
In semiconductors: A very small number of electrons are available for electrical conduction.
In insulators: Conduction band is almost empty i.e., no electron is available for conduction.

Effect of temperature

• In conductors: At high temperature, the collision of electrons become more frequent with the atoms/molecules at lattice site in the metals as a result the conductivity decreases (or resistivity increases).
• In semiconductors: As the temperature of the semiconducting material increases, more electron-hole pairs becomes available in the conduction band and valence band, and hence the conductivity increases or the resistivity decreases.
• In insulators: The energy band between conduction band and valence band is very large, so it is unsurpassable for small temperature rise. So, there is no change in their behaviour.

Question 2.
Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors?
Answer:

Intrinsic semiconductor	Extrinsic semiconductor
1. It is a pure semiconductor material with no impurity atoms in it.	It is prepared by doping a small quantity of impurity atoms to the pure semiconductor.
2. The number of free electrons in the conduction band and the number of holes in valence band is exactly equal.	The number of free electrons and holes is never equal. There is an excess of electrons in n-type semicoñductors and excess of holes in p-tpe semiconductors.

Question 3.
The circuit shown in the figure has two oppositely connected ‘ ideal diodes connected in parallel. Find the current flowing through each diode in the circuit.

Answer:
(i) Diode D₁ is reverse biased, so it offers an infinite resistance. So, no current flows in the branch of diode D₁.
(ii) Diode D₂ is forward biased and offers no resistance in the circuit. So, current in the branch,
I = \(\frac{V}{R_{e q}}=\frac{12 \mathrm{~V}}{2 \Omega+4 \Omega}=\frac{12 \mathrm{~V}}{6 \Omega}\) = 2A.

Question 4.
A Zener of power rating 1W is to be used as a voltage regulator. If Zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7 V, what should be the value of Rs for safe operation (see figure)? (NCERT Exemplar)

Answer:
Here, P =1W, V_z = 5 V
V_s = 3V to 7 V
I_Zmax = \(\frac{P}{V_{Z}}=\frac{1}{5}=0.2 \mathrm{~A}\) = 200 mA
R_s = \(\frac{V_{s}-V_{z}}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}\) = 10Ω

Question 5.
Draw V-I characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.
Answer:

(i) In the reverse biasing, the current of order of μA is due to movement/drifting of minority charge carriers from one region to another through the junction. A small applied voltage is sufficient to sweep the minority charge carriers through the junction. So, reverse current is almost independent of critical voltage.

(ii) At critical voltage (or breakdown voltage), a large number of covalent bonds break, resulting in the increase of large number of charge carriers. Hence, current increases at critical voltage. Semiconductor device that is used in reverse biasing, is Zener diode.

Question 6.
How is a light-emitting diode fabricated? Briefly state its working. Write any two important advantages of LEDs over the conventional incandescent low power lamps.
Answer:
LED is fabricated by:
(i) heavy doping of both the p and n regions.
(ii) providing a transparent cover so that light can come out.

Working: When the diode is forward biased, electrons are sent from n →b p and holes from p → n. At the junction boundary, the excess minority carriers on either side of junction recombine with majority carriers. This releases energy in the form of photon hv = E_g
Advantages:

• Low operational voltage and less power consumption.
• Fast action and no warm-up time required.
• Long life and ruggedness.
• Fast on-off switching capability.

Question 7.
Explain, with the help of a circuit diagram, the working of a photodiode. Write briefly how it is used to detect the optical signals.
Or
(a) How is photodiode fabricated?
(b) Briefly explain its working. Draw its VI characteristics for two different intensities of illumination.
Or
With what considerations in view, a photodiode is fabricated?
State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be
more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason?
Answer:
A photodiode is fabricated using photosensitive semiconducting material with a transparent window to allow light to fall on the junction of the diode.

Working: In diode (any type of diode), an electric field ‘E’ exists across the junction from n-side to p-side, when light with energy hv greater than energy gap E₂ (hv> Eg ) illuminate the junction, then electron-hole pairs are generated due to absorption of photons, in or near the depletion region of the diode. Due to existing electric fIeld, electrons are collected on n-side and holes are collected on p-side, giving rise to an emf. Due to the generated emf, an electric current of μA order flows through the external resistance.

Detection of Optical Signals: It is easier to observe the change in the current with change in the light intensity if a reverse bias is applied. Thus, photodiode can be used as a photodetector to detect optical signals. The characteristic curves of a photodiode for two different illuminations I₁ and I₂ (I₂ > I₁)are shown below

Question 8.
Describe briefly using the necessary circuit diagram, the three basic processes which take place to generate the emf in a solar cell when light falls on it. Draw the V-I characteristics of a solar cell. Write two important criteria required for the selection of a material for solar cell fabrication.
Answer:
The three basic processes which take place to generate the emf in a solar cell are generation, separation and collection.
(i) Generation of electron-hole pairs due to the light incident (with hv>Eg) close to the junction.
(ii) Separation of electrons and holes due to the electric field of the depletion region.
(iii) Collection of electrons and holes by n-side and p-side, respectively.

Important criteria for the selection of a material for solar cell fabrication are:
(i) bandgap (~ 1.0 to 1.8 eV).
(ii) high optical absorption (-10 4 cm^-1),
(iii) electrical conductivity.
(iv) availability of the raw material, and
(v) cost

Long answer type questions

Question 1.
(a) State briefly the process involved in the formation of p-n junction explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in:
(i) Forward biasing
(ii) Reverse biasing
How are these characteristics made use of in rectification?
Or
Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode
(i) in forward bias and
(ii) in reverse bias. Draw the typical V-I characteristics of a silicon diode.
Describe briefly the following terms:
(i) “minority carrier injection” in forward bias
(ii) “breakdown voltage” in reverse bias.
Answer:
(a) Two processes occur during the formation of a p-n junction are diffusion and drift. Due to the concentration gradient across p and n-sides of the junction, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This movement of charge carriers leaves behind ionised acceptors (negative charge Φ -immobile) on the p-side and donors (positive charge immobile) on the n-side of the junction. This space charge region on either side of the junction together is known as depletion region.

(b) The circuit arrangement for studying the V-I characteristics of a diode are shown in figure (a) and (b). For different values of voltages, the value of current is noted. A graph between V and I is obtained as in figure (c). From the V-I characteristic of a junction diode, it is clear that it allows current to pass only when it is forward biased. So, if an alternative voltage is applied across a diode, the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages.


(i) Minority Carrier Injection: Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carriers). Similarly, holes from p-side cross this junction and reach the n-side (where they are minority carriers). This process under forward bias is known as minority carrier injection.

(ii) Breakdown Voltage: It is critical reverse bias voltage at which current is independent of applied voltage.

Question 2.
State the principle of working of p-n diode as a rectifier. Explain with the help of a circuit diagram, the use of p-n diode as a full-wave rectifier. Draw a sketch of the input and output waveforms.
Or
Draw a circuit diagram of a full-wave rectifier. Explain the working principle. Draw the input/output waveforms indicating clearly, the functions of the two diodes used.
Or
With the help of a circuit diagram, explain the working of a junction diode as a full-wave rectifier. Draw its Input and output waveforms. Which characteristic property makes the junction diode suitable for rectification?
Answer:
Rectification: Rectification means conversion of AC into DC. A p-n diode acts as a rectifier because an AC changes polarity periodically and a p-n diode allows the current to pass only when it is forward biased. This makes the diode suitable for rectification.

Working: The AC input voltage across secondary s₁ and s₂ changes polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal S₁ is positive relative to centre tap O and s₂ is negative relative to O. Then diode D₁ is forward biased and diode D₂ is reverse biased. Therefore, diode D₁ conducts while diode D₂ does not. The direction of current (i₁) due to diode D₁ in load resistance R_L is direction from A to B. In next half cycle, the terminal s₁ is negative and s₂ is positive relative to centre tap O.

The diode D₁ is reverse biased and diode D₂ is forward biased. Therefore, diode D₂ conducts while D₁ does not.
The direction of current (i₂) due to diode D₂ in load resistance R_L is still from A to B. Thus, the current in load resistance R_L is in the same direction for both half cycles of input AC voltage. Thus for input AC signal the output current is a continuous series of unidirectional pulses. In a full-wave rectifier, if input frequency is f hertz, then output frequency, will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

Question 3.
Why is a Zener diode considered as a special purpose semiconductor diode? Draw the I-V characteristics of Zener diode and explain briefly, how reverse current suddenly Increase at the breakdown voltage? Describe briefly with the help of a circuit diagram, how a Zener diode works to obtain a constant DC voltage from the unregulated DC output of a rectifier.
Answer:
Zener diode works only in reverse breakdown region that is why it is, considered as a special purpose semiconductor. I – V characteristics of Zener diode is given below.

Reverse current is due to the flow of electrons from n → p and holes from p → n. As, the reverse-biased voltage increase the electric field across the junction, increases significantly and when reverse bias voltage V = V_Z, then the electric field strength is high enough to pull the electrons from p-side and accelerated it to n-side. These electrons are responsible for the high current at the breakdown.

Voltage regulator converts an unregulated DC output of rectifier into a constant regulated DC voltage, using Zener diode. The unregulated voltage is connected to the Zener diode through a series resistance Rs such that the Zener diode is reverse biased. If the input voltage increase, then-current through R_s and Zener diode increases.

Thus, the voltage drop across R_s increase without any change in the voltage drop across zener diode. This is because of the breakdown region, Zener voltage remains constant even though the current through Zener diode changes.
Similarly, if the input voltage decreases, the current through R_s and Zener diode decreases. The voltage drop across Rs decreases without any change in the voltage across the Zener diode. Now, any change in input voltage results the change in voltage drop across Rs, without any changes; in voltage across the Zener diode. Thus, Zener diode acts as a voltage regulator.

Question 4.
(a) Differentiate between three segments of a transistor on the basis of their size and level of doping.
(b) How is a transistor biased to be in active state?
(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for AC current gain.
Or
Explain with the help of a circuit diagram the working of n-p-n transistor as a common emitter amplifier.
Or
Draw the circuit diagram of a common-emitter amplifier using an n-p-n transistor. What is the phase difference between the input signal and output voltage? Draw the input and output waveforms of the signal. Write the expression for its voltage gain. State two reasons why a common emitter amplifier is preferred to a common base amplifier.
Answer:
(a) Emitter: It is of moderate size and heavily doped.
Base: It is very thin and lightly doped.
Collector: The collector side is moderately doped and larger in size as compared to the emitter.

(b) Transistor is said to be inactive state when its emitter-base junction is suitably forward biased and base-collector junction is suitably reverse biased.
(c) The circuit of common emitter amplifier using n-p-n transistor is shown below :

Working: If a small sinusoidal voltage, with amplitude V_S is superposed on DC basic bias (by connecting the sinusoidal voltage in series with base supply V_BB), the base current will have sinusoidal variations superimposed on the base current I_B.

As a consequence the collector current is also sinusoidal variations superimposed on the value of collector current I_c, this will produce corresponding amplified changes in the value of output voltage V₀.
The AC variations across input and output terminals may be measured by blocking the DC voltage by large capacitors. The phase difference between input signal and output voltage is 180°. The input and output waveforms are shown in figure.

Voltage gain,
A_v = \(\beta \frac{R_{L}}{R_{i}}\)
AC current gain, β_AC = \(\left(\frac{\Delta I_{C}}{\Delta I_{B}}\right)_{V_{C E}}\)
Reasons for Using a Common Emitter Amplifier
(i) Voltage gain is quite high.
(ii) Voltage gain is uniform over a wider frequency range or power gain is high.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 1 Electric Charges and Fields Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 1 Electric Charges and Fields

Very short answer type questions

Question 1.
Define electrostatics.
Answer:
Electrostatics deals with the study of forces, fields and potentials arising from static charges.

Question 2.
Define charge.
Answer:
It is defined as the basic and characteristic property of elementary particles of matter in the form of which certain force of interaction energies may be explained.

Question 3.
What are the behaviour of charges?
Answer:
Like charges repel and unlike charges attract each other.

Question 4.
Define the polarity of charge.
Answer:
The property which differentiates the two kinds of charges is called the polarity of charge.

Question 5.
Who first assigned the positive and negative signs to charge?
Answer:
Benjamin Franklin

Question 6.
What are conductors?
Answer:
The substances which allow electricity to pass through them easily are called conductors e.g., metals, human, earth etc.

Question 7.
What are insulators?
Answer:
Most of the non-metals like glass, porcelain, plastic, nylon, wood offer high resistance to the passage of electricity through them, they are called insulators.

Question 8.
Define the semiconductors.
Answer:
The substances which offer resistance to the movement of charges are called semiconductors. They are intermediate between conductors and insulators,

Question 9.
What are point charges?
Answer:
If the sizes of charged bodies are very small as compared to the distances between them, they are called point charges.

Question 10.
Write the law of conservation of charges.
Answer:
“The total charge of the isolated system is always conserved.” It is not possible to create or destroy net charge carried by isolated system although the charge carrying particles may be created or destroyed in a process.

Question 11.
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why then is the electrostatic field inside a conductor zero? (NCERT Exemplar)
Answer:
The electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inner surface of an isolated conductor. So, the electrostatic field inside a conductor is zero.

Question 12.
An arbitrary surface encloses a dipole. What is the electric flux through this surface? (NCERT Exemplar)
Answer:
Net charge on a dipole = -q + q = 0. According to Gauss’s theorem, electric flux through the surface,

Question 13.
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure. (NCERT Exemplar)

Answer:

Short answer type questions

Question 1.
Describe the gold-leaf electroscope.
Question
A simple apparatus to detect charge on a body is the gold-leaf electroscope. It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows on to the leaves and they diverge. The degree of divergence is an indicator of the amount of charge.

Question 2.
Define the grounding or earthing.
Answer:
When we bring a charged body in contact with the earth, all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor (such as our body). This process of sharing the charges with the earth is called grounding or earthing.

Question 3.
What is the importance of earthing in buildings?
Answer:
Earthing provides a safety measure for electrical circuits and appliances. A thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply. The electric wiring in our houses has three wires; live, neutral and earth. The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate. Metallic bodies of the electric appliances such as electric iron, refrigerator, TV are connected to the earth wire. When any fault occurs or live wire touches the metallic body the charge flows to the earth without damaging the appliance and without causing any injury to the humans; this would have otherwise been unavoidable since the human body is a conductor of electricity.

Question 4.
Describe the additivity of charges in brief.
Answer:
If a system contains two point charges q₁ and q₂, the total charge of the system is obtained simply by adding algebraically qi and q2, i. e., charges add up like real numbers or they are scalars like the mass of a body. If a system contains n charges q₁, q₂, q₃,……, q_n, then the total charge of the system is q₁ + q₂ + q₃ + … + q_n. Charge has magnitude but no direction, similar to the mass. However, there is one difference between mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative. Proper signs have to be used while adding the charges in a system. For example, the total charge of a system containing five charges +1, +2, -3, +4 and -5, in some arbitrary unit, is (+1) + (+2) + (-3) + (+4) + (-5) = -1 in the same unit.

Question 5.
(a) Define electric flux. Write its SI unit.
(b) A small metal sphere carrying charge +Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P₁ and P₂.

(c) Draw the pattern of electric field lines in this arrangement.
Answer:
(a) Electric flux over an area in an electric field represents the total number of electric field lines crossing this area and is given by the product of surface area and the component of electric field intensity normal to the area.
The SI unit of flux is Nm²C^-1.

(b) Let point P₁ is at distance R from the centre 0.
S₁ is the Gaussian surface, then according to Gauss’s theorem

Inside the shell the charge is zero, so the field is also zero

(c) The direction of electric field is shown in figure.

Question 6.
A metallic spherical shell has an inner radius R₁ and outer radius R₂– A charge Q is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)
Answer:
When a charge +Q is placed at the centre of spherical cavity,

The charge induced on the inner surface = -Q
The charge induced on the outer surface = +Q
∴ Surface charge density on the inner surface = \(\frac{-Q}{4 \pi R_{1}^{2}}\)
Surface charge density on the outer surface = \(\frac{+Q}{4 \pi R_{2}^{2}}\)

Question 7.
Two charges q and -3q are placed fixed on x-axis separated by distance id\ Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:

Let the charge 2q be placed at point P as shown. The force due to q is to the left and that due to -3q is to the right.

Long answer type questions

Question 1.
How can you charge a metal sphere positively without touching it?
Answer:
Figure (a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positiye charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig.(d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge.

Similar steps are involved in charging a metal sphere negatively by induction, by bringing a positively charged rod near it. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire.

Question 2.
Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Answer:
Electric Dipole Moment : The strength of an electric dipole is measured by the quantity electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.
Electric dipole moment,
p = q × d
It is a vector quantity.
In vector form it is written as \(\), where the direction of \(\) is from negative charge to positive charge.
Electric field of dipole at points on the equatorial plane:

The magnitudes of the electric field due to the two charges + q and – q are given by,

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 15 Communication Systems Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 15 Communication Systems

Very short answer type questions

Question 1.
Name the essential components of a communication system.
Answer:
Transmitter, medium or channel, and receiver.

Question 2.
What is the function of a transducer used in a communication system?
Answer:
Transducer used as a sensor or detector in communication system. It converts the physical signal into electrical signal.

Question 3.
What is the function of a repeater in a communication system?
Answer:
Repeater pick up the signals from the transmitter, amplifies it and transmits it to the receiver. Thus, repeater comprises up of receiver, transmitted and amplifier. Its function is to extend the range of communication.

Question 4.
Define bandwidth and describe briefly its importance in communicating signals.
Answer:
It is defied as the frequency range over which given equipment operates.
Importance: To design the equipment used in communication system for distinguishing different message signals.

Question 5.
Which basic mode of communication is used for telephonic communication?
Answer:
Point to point is a basic mode of communication, which is used for telephonic conversation. In this mode of communication, communication takes place over a link between a single transmitter and a receiver.

Question 6.
How are microwaves produced?
Answer:
A type of electromagnetic wave is microwave whose wavelength ranging from as long as metre to as short as millimeter and having the frequency range 3000 MHz to 300 GHz. This also includes UHF, EHF and various sources with different boundaries.

Question 7.
What is sky wave propagation?
Answer:
Skywave propagation is a mode of propagation in which communication of radiowaves in frequency range 2 MHz-20 MHz takes place due to reflection from the ionosphere.

Question 8.
Would sky waves be suitable for transmission of TV signals of 60 MHz frequency? (NCERT Exemplar)
Answer:
A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz. But, here the frequency of TV signals are 60 MHz which is beyond the required range. So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

Question 9.
How are sidebands produced?
Answer:
Sidebands are produced due to the superposition of carrier wave of frequency ω_c over modulating or audio signal of frequency ω_m. The frequency of lower sideband is ω_c -ω_m and the upper side band is ω_c+ω_m

Question 10.
Why are broadcast frequencies (carrier waves) sufficiently spaced in amplitude modulated wave?
Answer:
To avoid mixing up of signals from different transmitters. This can be done by modulating the signals on high-frequency carrier waves, e.g., frequency band for satellite communication is 5.925 – 6.425 GHz.

Question 11.
Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?
Answer:
The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid distortion.

Question 12.
What is the function of a bandpass filter used in a modulator for obtaining AM signal?
Answer:
Bandpass filter rejects DC and sinusoid of frequency ω_m, 2ω_m, and 2ω_c and retains frequencies ω_c + ω_m.
Thus, it allows only the desired frequencies to pass through it.

Question 13.
On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by ω_c, ω_c -ω_m and ω_c +ω_m. Suggest ways to minimise cost of radiation without compromising on information. (NCERT Exemplar)
Answer:
In amplitude modulated signals, only sideband frequencies contain information.
Thus only (ω_c +ω_m) and (ω_c – ω_m) contain information.

Now, according to question, the total radiated power is due to energy carried by
ω_c, (ω_c – ω_m) and (ω_c +ω_m)
Thus to minimize the cost of radiation without compromising on information, ω_c can be left and transmitting
(ω_c +ω_m), (ω_c – ω_m) or both (ω_c +ω_m) and (ω_c – ω_m).

Question 14.
Two waves A and B of frequencies %MHz and 3 MHz, respectively are beamed in the sartie direction for communication via skywave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection? (NCERT Exemplar)
Answer:
As the frequency of wave B is more than wave A, it means the refractive index of wave B is more than refractive index of wave A (as refractive index increases with frequency increases). For higher frequency wave (i.e., higher refractive index) the angle of refraction is less i.e., bending is less. So, wave B travels longer distance in the ionosphere before suffering total internal reflection.

Short answer type questions

Question 1.
Draw a block diagram of a generalized communication system. Write the functions of each of the following:
(a) Transmitter
(b) Channel
(c) Receiver
Answer:
The block diagram of a generalized communication system is shown in figure

Functions are as follows :
(a) Transmitter: It comprises of message signal source, modulator and transmitting antenna. Transmitter makes signals compatible for communication channel via modulator and antenna.
(b) Channel: It is a link for propagating the signal from transmitter to receiver.
(c) Receiver: It recovers the desired original message signals from the received signals at the end of channel.

Question 2.
Explain the terms
(i) Attenuation and
(ii) Demodulation used in communication system.
Answer:
Attenuation: The loss in strength of a signal while propagating through a medium is known as attenuation.
Demodulation: The process of retrieval of information, from the carrier wave at the receiver end. This is the reverse process of modulation.

Question 3.
(a) Distinguish between ‘Analog and Digital signals’.
(b) Explain briefly two commonly used applications of the Internet.
Answer:
(a) A signal that varies continuously with time (e. g., sine waveform) is called an analog signal.
A signal that is discrete is called a digital signal. The presence of signal is denoted by digit 1 and absence is denoted by digit 0.

(b) Uses of Internet: email, e-banking, e-shopping, e-ticketing, charting, surfing, file transfer, etc.

Question 4.
What is ground wave communication? Explain why this mode cannot be used for long-distance communication using high frequencies.
Answer:
The mode of wave propagation in which wave guided along the surface of the earth is called ground wave communication.
The maximum range of propagation in this mode depends on
(i) transmitted power and
(ii) frequency (less than a few MHz)
At high frequencies, the rate of energy dissipation of the signal increases and the signal gets attenuated over a short distance.

Question 5.
Define the term modulation. Draw a block diagram of a simple modulator for obtaining AM signals.
Or
Draw a block diagram of a simple modulator to explain how the AM wave is produced. Can the modulated signal be transmitted as such? Explain.
Answer:
Modulation is the process in which low-frequency message signal is superimposed on high-frequency carrier wave so that they can be transmitted over long distances. The block diagram for a simple modulator for obtaining AM signal is shown as below :

Question 6.
Define modulation index. Why is it kept low? What is the role of a bandpass filter? Give its physical significance.
Answer:
Modulation index is the ratio of the amplitude of modulating signal to that of carrier wave. Mathematically,
µ = \(\frac{A_{m}}{A_{c}}\)
Reason: It is kept low to avoid distortion.
Role: A bandpass filter rejects low and high frequencies and allows a band of desired frequencies to pass through it.
Physical Significance: It signifies the level of distortion or noise. A lower value of modulation index indicates a lower distortion in the transmitted signal.

Question 7.
State the concept of mobile telephony and explain its working.
Answer:
Concept of mobile telephony is to divide the service area into a suitable number of cells centered on an office MTSO (Mobile Telephone Switching Office). Mobile telephony means that you can talk to any person from anywhere.
Explanation:

• Entire service area is divided into smaller parts called cells or cell zones.
• Each cell has a base station to receive and send signals to all the mobile phones present inside that cell.
• Each base station is linked to MTSO. MTSO coordinates between base station and TCO (Telephone Control Office).

Question 8.
What is space wave propagation? State the factors which limit its range of propagation. Derive an expression for the maximum line of sight distance between two antennas for space wave propagation.
Answer:
Space Wave Propagation
The mode of propagation in which radio waves travel, along a straight line, from the transmitting to the receiving antenna.
Limiting Factors
(i) Curvature of the earth
(ii) Insufficient height of the receiving antenna
(iii) LOS distance (> 40 MHz) travel in straight line

Derivation : In right angled triangle BOD, ∠BDO =90°
∴BO² = (OD)² +(BD)²
le., (R_e+h)² =R_e² +(BD)² ………………. (1)
As height h of the tower is very small as compared to radius (R_e) of earth the point B will be very close to A, so that
BD ≈AD = d(say)
∴ Equation (1) given (R_e + h)² = R_e² + d²

or height of transmitting antenna, h =\(\frac{d^{2}}{2 R_{e}}\)
or covering range of TV transmitting tower, d = \(\sqrt{2 R_{e} h}\)
Thus, covering range of TV signal can be increased by increasing the height of transmission antenna.
For a transmitting antenna of height h_T, and a receiving antenna of height h_R, the maximum line of sight distance
becomes
d_M= \(\sqrt{2 R h_{T}}+\sqrt{2 R h_{R}}\)

Question 9.
(a) Explain any two factors which justify the need of modulating a low-frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation
Answer:
(a) (i) If λ is the wavelength of the signal then the antenna should have a length at least \(\frac{\lambda}{4}\)
For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Such a long antenna is not possible to construct and operate. So, there is need to modulate the wave in order to reduce the height of antenna to a reasonable height.

(ii) The power radiated by a linear antenna (length l) is proportional to \(\left(\frac{l}{\lambda}\right)^{2}\).
This shows that power radiated increases with decreasing λ. So, for effective power radiation by antenna, there is need to modulate the wave.
(b)

• High frequency
• Less noise
• Maximum use of transmitted power

Question 10.
Given reasons for the following :
(i) For ground wave transmission, size of antenna (l) should the comparable to wavelength (λ) of signal, i.e. I = \(\frac{\lambda}{4}\)
(ii) Audio signals converted into an electromagnetic wave are not directly transmitted.
(iii) The amplitude of a modulating signal is kept less than the amplitude of carrier wave.
Answer:
(i) To radiate the signals with high efficiency.
(ii) Because they are of large wavelength and power radiated by antenna is very small as
P ∝ \(1 / \lambda^{4}\)
(iii) It is so to avoid making over modulated carrier wave. In that situation, the negative half cycle of the modulating signal is dipped and distortion occurs in reception.

Question 11.
Which of the following would produce analog signals and which would produce digital signals?(NCERT Exemplar)
(a) A vibrating tuning fork
(b) Musical sound due to a vibrating sitar string
(c) Light pulse
(d) Output of NAND gate
Answer:
Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is transformed into electric signals. The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct 4 amplitudes. Thus, (a) and (b) would produce analog signals and (c) and (d) would produce digital signals.

Question 12.
Why is AM signal likely to be more noisy than a FM signal upon transmission through a channel? (NCERT Exemplar)
Answer:
In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal.
In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal.

Long answer type questions

Question 1.
What does the term LOS communication mean? Name the types of waves that are used for this communication. What is the range of their frequencies? Give typical, examples, with the help of suitable figure of communication systems that use space wave mode propagation.
Answer:
LOS Communication: It means “Line of sight communication”. Space wave are used for LOS communication.
In this communication, the space waves (radio or microwaves) travel directly from transmitting antenna to receiving antenna. Frequency for LOS communication must be more than 40 MHz.

If transmitting antenna and receiving antenna have heights h_T and h_R respectively, then radio horizon of transmitting antenna.
d_T = \( \sqrt{2 R_{e} h_{T}}\)
where R_e is radius of earth and radius horizon of receiving antenna,
d_R = \(\sqrt{2 R_{e} h_{R}}\)
∴ Maximum line of sight distance,
d_M = d_T + d_R = \(\sqrt{2 R_{e} h_{T}}+\sqrt{2 R_{e} h_{R}}\)

Television, broadcast, microwave links and satellite communication.
The satellite communication is shown in figure. The space wave used is microwave.

Question 2.
(a) Distinguish between sinusoidal and pulse-shaped signals,
(b) Explain, showing graphically, how a sinusoidal carrier wave is superimposed on a modulating signal to obtain the resultant amplitude modulated (AM) wave.
Answer:
(a) In the process of modulation, some specific characteristics of the carrier wave is varied in accordance with the information or message signal. The carrier wave maybe
(i) Continuous (sinusoidal) wave, or
(ii) Pulse, which is discontinuous.
A continuous sinusoidal carrier wave can be expressed as,
E = E_o sin (ωt +Φ)
Three distinct characteristics of such a wave are amplitude (E₀), angular frequency (ω) and phase angle fcj)).
Any one of these three characteristics can be varied in accordance with the modulating baseband (AF) signal, giving rise to the respective Amplitude Modulation;
Frequency Modulation and Phase Modulation.

Again, the significant characteristics of a pulse are Pulse Amplitude, Pulse Duration or Pulse Width and Pulse Position (representing the time of rise or fall of the pulse amplitude). Any one of these characteristics can be varied in accordance with the modulating baseband (AF) signal, giving rise to the respective. Pulse Amplitude
Modulation (PAM), Pulse Duration Modulation (PDM), Pulse Width Modulation (PWM) and Pulse Position Modulation (PPM).

(b) Amplitude Modulation: When a modulating AF wave is superimposed on a high-frequency carrier wave in a manner that the frequency of modulated wave is same as that of the carrier wave, but its amplitude is made proportional to the instantaneous amplitude of the audio, frequency modulating voltage, the process is called amplitude modulation (AM).

Let the instantaneous carrier voltage (e_c) and modulating voltage (e_m) be represented by,
e_c = E_c sinω_c t …………………… (1)
e_m = E_m sinω_mt …………………….. (2)
Thus, in amplitude modulation, amplitude A of modulated wave is made proportional to the instantaneous modulating voltage e_m i.e.,
A = E_c+ke_m ……………………….. (3)
where k is a constant of proportionality.
In amplitude modulation, the proportionality constant k is made equal to unity. Therefore, maximum positive amplitude of AM wave is given
by.
A = E_c +e_m =E_c +E_m sinω_m t …………………….. (4)
It is called top envelope.
The maximum negative amplitude of AM wave is given by,
-A = – E_c – e_m
= – (E_c +E_m sinω_m t) …………………………. (5)

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 14 Ecosystem Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 14 Ecosystem

Very Short Answer Type Questions

Question 1.
How is ‘stratification’ represented in a forest ecosystem?
Answer:
Stratification is the vertical distribution of species at different levels. Trees occupy top vertical strata or layer, shrubs the second layer and herbs/grasses occupy the bottom layers.

Question 2.
Write a difference between net primary productivity and gross primary productivity.
Answer:
Gross primary productivity (GPP) is the rate of production of organic matter during photosynthesis. Net primary productivity (NPP) is the available biomass for the consumption by heterotrophs.
GPP – R = NPP.

Question 3.
Why is the rate of assimilation of energy at the herbivore level called secondary productivity? [NCERT Exemplar]
Answer:
It is because the biomass available to the consumer for consumption is a resultant of the primary productivity from plants.

Question 4.
Why is an earthworm called a detritivore?
Answer:
This is because earthworm breaks down detritus into smaller particles.

Question 5.
Justify the pitcher plant as a producer. [NCERT Exemplar]
Answer:
Pitcher plant is chlorophyllous and is thus capable of photosynthesis and act as producer.

Question 6.
Name any two organisms which occupy more than one trophic level in an ecosystem? [NCERT Exemplar]
Answer:
Man and sparrow.

Question 7.
What is common to earthworm, mushroom, soil mites, and dung beetle in an ecosystem? [NCERT Exemplar]
Answer:
They are all detritivores, i.e., decomposing organisms which feed on dead remains of plants and animals.

Question 8.
“Man can be a primary as well as a secondary consumer.” Justify this statement.
Answer:
Man has a varied diet. When on a vegetarian diet, they are primary consumers, and when on a non-vegetarian diet, they are secondary consumers.

Question 9.
Name an omnivore which occurs in both grazing food chain and the decomposer food chain. [NCERT Exemplar]
Answer:
Sparrow/crow.

Question 10.
Differentiate between standing state and standing crop in an ecosystem.
Answer:
In an ecosystem, standing crop is the mass of living material in each trophic level at a particular time. Whereas standing state refers to the amount- of nutrients in the soil at any given time.

Question 11.
Under what conditions would a particular stage in the process of succession revert back to an earlier stage? [NCERT Exemplar]
Answer:
Natural or human-induced disturbances like fire, deforestation, etc.

Question 12.
Climax stage is achieved quickly in secondary succession as compared to primary succession. Why? [NCERT Exemplar]
Answer:
The rate of succession is much faster in secondary succession as the substratum (soil) is already present as compared to primary succession where the process starts from a bare area (rock).

Short answer type questions

Question 1.
What is an incomplete ecosystem? Explain with the help of a suitable example. [NCERT Exemplar]
Answer:
An ecosystem is a functional unit with biotic and abiotic factors interacting with one another resulting in a physical structure. Absence of any component will make an ecosystem incomplete as it will hinder the functioning of the ecosystem. Examples of such an ecosystem can be a fish tank or deep aphotic zone of the oceans where producers are absent.

Question 2.
Justify the importance of decomposers in an ecosystem.
Answer:
Decomposers which are heterotrophic organisms, mainly fungi and bacteria break down complex organic matter into inorganic substances like carbon dioxide, water, and nutrients. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs. Decomposers secrete digestive enzymes that break down dead and waste into simple, inorganic materials, which are subsequently absorbed by them.

Question 3.
“In a food chain, a trophic level represents a functional level, not a species.” Explain.
Answer:
Trophic level in a food chain is a level at which organisms obtain their food. Each trophic level has a specific mode of obtaining food. Thus, trophic level represents the mode of obtaining food and not the species. The species has no significance in the food chain. In the food chain, there are generally four trophic levels-producers or autotrophs, the first trophic level; primary consumer/herbivore secondary trophic level; secondary consumer/carnivore third trophic level and finally tertiary consumers (top carnivores) representing fourth trophic level. Thus, in a food chain, trophic levels represent a functional level and which species represent the trophic level, does not matter.

Question 4.
How does primary succession start in water to the climax community? Explain.
Answer:
In primary succession in water, the pioneers are the small phytoplanktons. They are replaced with time by free-floating angiosperms, then by rooted hydrophytes; sedges, grasses, and finally the trees. The climax again would be a forest. With time the water body is converted into land.

Question 5.
Why are nutrient cycles in nature called biogeochemical cycles? [NCERT Exemplar]
Answer:
Nutrient cycles are called biogeochemical cycles because ions/molecules of a nutrient are transferred from the environment (rocks, air and water) to organisms (life) and then brought back to the environment in a cyclic pathway.
The literal meaning of biogeochemical is bio-living organisms and geo-rocks, air, and water.

Question 6.
(a) State any two differences between phosphorus and carbon cycles in nature.
(b) Write the importance of phosphorus in living organisms.
Answer:
(a)

Phosphorus cycle	Carbon cycle
1. It is a sedimentary cycle.	It is a gaseous cycle.
2. Atmospheric inputs through rainfall are much smaller.	Atmospheric inputs through rainfall are more.
3. Gaseous exchange of phosphorus between organism and environment is nil.	Gaseous exchange of carbon between organism and environment is much more.

(b) Phosphorus is a major constituent of biological membranes, nucleic acids, and cellular energy transfer systems.

Long answer type questions

Question 1.
Explain succession of plants in xerophytic habitat until it reaches climax community.
Answer:
Xerarch Succession:

• It starts in primary bare, dry area such as rocks or sand dunes, etc.
• The pioneer species on the rock are usually lichens and blue-green algae under humid conditions.
• They secrete acids to dissolve rocks, help in weathering and soil formation.
• Little soil formation makes the way to the appearance of small plants like bryophytes (mosses). They accumulate more soil and organic matter.
• With time, they are succeeded by bigger plants; perennial grasses and shrubs.
• After several more stages, ultimately a stable climax forest community is formed.
• Climax community remains stable as long as environment remains unchanged.
• With time, the xerophytic habitat gets converted into mesic habitat.

Question 2.
Describe the advantages for keeping the ecosystems healthy.
Answer:
An healthy ecosystem is stable and have a functional balance amongst different populations found in ecosystem. Ecosystem advantages are benefits provided by ecosystem processes to environment in its cleaning and maintenance, enhancing aesthetic beauty, maintenance of biodiversity, protection of soil, water and land sources besides providing a habitat to wildlife, tribals and grazing areas. The important advantages are given ahead :

(i) Oxygen Release : (Purify air) Suspended particulate matter is intercepted by vegetation and made to settle down. Air is thus removed of its pollutants. It is further purified by removal of carbon dioxide and release of oxygen during photosynthetic activity of vegetation.

(ii) Water: Most of rainwater is held over the soil by plant litter like a sponge. It slowly percolates down and becomes the source of all springs, rivulets and rivers. The water is clean and fresh.

(iii) Prevention of Floods: As there is little runoff, flood water is not formed.

(iv) Protection of Soil: Maintenance of soil fertility depends upon a good soil cover and optimum nutrient cycling. Soil cover also protects the soil from air and water erosion. Soil particles remain bound together by plant roots. Landslides are rare.

(v) Biodiversity: Natural ecosystems are a source of biodiversity with a variety of genes, gene pools, species and habitats.

(vi) Climate: Ecosystems, especially forests, maintain good climatic conditions by increasing humidity, reducing extremes of temperature and increasing periodicity of rainfall.

(vii) Nutrient Cycling: It is one of the most important ecosystem services which maintains the continuity of life on earth. Through cycling, biogenetic nutrients are made available all the time for absorption.

(viii) Pollination: Insects, especially bees, and birds visit areas around the forests for pollination of crop plants, bushes, and trees.

(ix) Pest Control: In natural ecosystem, pests are kept under control by their natural predators. Maintenance of natural ecosystems will allow the predators to free the nearby areas of pests.

(x) Wildlife: Ecosystems provide habitats to wildlife.

(xi) Aesthetic Value: Ecosystems also provide aesthetic, cultural, and spiritual value.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 12 Biotechnology and its Applications Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 12 Biotechnology and its Applications

Very short answer type questions

Question 1.
State the role of transposons in silencing of mRNA in eukaryotic cells.
Answer:
Transposons or mobile genetic elements in viruses are the sources of the complementary dsRNA, which in turn bind to specific mRNA and cause RNA interference of the parasite.

Question 2.
How do interferons protect us?
Answer:
Interferons protect non-infected cells from further viral infections, by creating cytokine barriers.

Question 3.
State the role of C peptide in human insulin.
Answer:
C-peptide (extra stretch of polypeptide) makes the insulin inactive.

Question 4.
How are two short polypeptide chains of insulin linked [ together?
Answer:
Two short polypeptide chains of insulin are limced together by disulphide bridges.

Question 5.
How can bacterial DNA be released from the bacterial cell for biotechnology experiments?
Answer:
The bacterial cell wall is digested by the enzyme lysozyme to release 1 DNA from the cell.

Question 6.
Suggest any two possible treatments that can be given to a patient exhibiting adenosine deaminase deficiency.
Answer:

• Enzyme replacement therapy (in which functional ADA is injected)
• Bone marrow transplantation
• Gene therapy/Culturing the lymphocytes followed by introduction of functional ADA cDNA into it and returning it into the patient’s body. (Any two)

Question 7.
Name a molecular diagnostic technique to detect the presence
of a pathogen in its early stage of infection.
Answer:
ELISA (Enzyme Linked Immunosorbent Assay)

Question 8.
What are transgenic animals. Uive an example.
Answer:
The transgenic animals are those that have their DNA manipulated to possess and express/foreign genes e.g.. transgenic cow-Rosie, rats, pigs fish, rabbits and mice.

Question 9.
What was the speciality of the milk produced by the transgenic cow, Rosie?
Answer:
The first transgenic cow, Rosie, produced milk with human alpha-lactalbumin which was nutritionally, more balanced product for human babies than natural cow milk.

Question 10.
What is biopiracy?
Answer:
Biopiracy refers to the use of bioresources by multinational companies and other organisations without proper authorisation from the country and people without compensatory payment.

Question 11.
Name the following:
(a) The semi-dwarf variety of wheat which is high-yielding and disease-resistant.
(b) Any one inter-specific hybrid mammal.
Answer:
(a) Kalyan Sona/Sonalika
(b) Mule/Hinny/Liger/Tigon

Question 12.
For which variety of Indian rice, patent was filed by a USA Company? [NCERT Exemplar]
Answer:
Indian Basmati was crossed with semi-dwarf variety and was claimed as a new variety for which the patent was filed by a USA company.

Question 13.
Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for recombinant DNA technology.
Answer:
Bacteria: lysozyme; fungi : chitinase.

Question 14.
Bt cotton is resistant to pest, such PS lepidopteran, dipterans and coleopter&ns. Is Bt cotton resistant to other pests as well? [NCERT Exemplar]
Answer:
Bt cotton is made resistant to only certain specific taxa of pests. It is quite likely that in future, some other pests may infest the Bt cotton plants. It is similar to immunisation against small-pox which does not provide immunity against other pathogens like those that cause cholera, typhoid, etc.

Short answer type questions

Question 1.
Explain the application of biotechnology in producing Bt-cotton.
Or One of the major contributions of biotechnology is to develop pest-resistant varieties of cotton plants. Explain how it has been made possible.
Or One of the main objectives of biotechnology is to minimise the use of insecticides on cultivated crops. Explain with the help of a suitable example how insect resistant crops have been developed using techniques of biotechnology.
Answer:
Production of Bt-cotton, a Pest Resistant Crop : Soil bacterium Bacillus thuringiensis possess gene called Cry-gene which synthesises an endotoxin protein called Cry-protein. Now, by biotechnology technique, the Cry gene from B. thuringiensis have been isolated, cloned, introduced and incorporated into cotton-plant using recombinant DNA technology. In the genetically modified cotton crop plants, the Cry or Bt-toxin gene expresses to produce a toxic insecticidal protein in an inactive form called prototoxin in crystalline state.

As an insect feeds over the plant, the inactive prototoxin crystals pass into the gut where alkaline pH and digestive enzymes solubilise the crystals and convert the prototoxin into an active toxin. The activated toxin creates pores in the midgut epithelial cells by lysing that cause death of the insect. Thus, the GM cotton plants do not require protection of expensive insecticides as they themselves act as bioinsecticides.

Since there are a number of Cry genes and Cry IAc and Cry II Ab controls the cotton bollworms while Cry IAb controls corn borer.

Question 2.
Write the functions of
(a) cry lAc gene
(b) RNA interference (RNAi)
Or Explain the process of RNA interference.
Answer:
(a) Cry IAc gene is present in Bacillus thuringiensisThe gene encodes for a toxic insecticidal protein during a particular phase of their growth. Cry genes are isolated and incorporated in the crop plant. The insect feeding on transgenic crop die because of the presence of toxin protein. Cry IAc produces Bt-toxins specific for cotton bollworm insect group.

(b) RNAi involves the silencing of a specific m-RNA due to the complementary ds-RNA molecule that binds to and prevents translation of m-RNA. As a result, parasite a favourable protein are not produced
and it could not infest, multiply and survive in a transgenic host expressing specific RNA interference. The transgenic plant, therefore, gets itself protected from the parasite such as nematode.

Question 3.
How did Eli Lilly synthesise the human insulin? Mention one difference between this insulin and the one produced by the human pancreas.
Or How did Eli Lilly Company go about preparing the human insulin? How is the insulin thus produced different from that produced by the functional human insulin gene?
Or How did an American Company, Eli Lilly use the knowledge of r-DNA technology to produce human insulin?
Or Explain how Eli Lilly, an American Company, Produced insulin by Recombinant DNA technology?
Answer:
Eli Lilly prepared two DNA sequences corresponding to A and B chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulphide bonds to form human insulin.

Insulin in human pancreas is synthesised as a pro-hormone containing the C peptide, which is removed to form mature hormone. The synthesised insulin did not contain C peptide and was directly prepared
in mature form.

Question 4.
How is a mature functional insulin hormone different from its pro-hormone form? [NCERT Exemplar]
Answer:
Mature functional insulin is obtained by processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during maturation of pro-insulin to insulin.

Question 5.
How does a transgenic organism differ from the rest of its population? Give any two examples of such organism for human advantage.
Answer:
A transgenic organism contains foreign gene, hence it differs from the rest of the population in having one or more extra genes apart from the gene pool of that population showing an additional phenotype. Example, (i) Transgenic E. coli, with gene for human insulin, (ii) Transgenic mouse with gene for human growth hormone.

Question 6.
What is GEAC and what are its objectives? [NCERT Exemplar]
Or Mention two objectives of setting up GEAC by our Government.
Or State the purpose for which the Indian Government has set up GEAC.
Or Describe the responsibility of GEAC, set up by the Indian Government.
Answer:
GEAG (Genetic Engineering Approval Committee) is an Indian government organisation. Its objective are to:
(a) examine the validity of GM (Genetic modification of organism) research.
(b) inspect the safety of introducing GM for public services.

Long answer type questions

Question 1.
List the disadvantages of insulin obtained from the pancreas of slaughtered cow and pigs. [NCERT Exemplar]
Answer:

• Insulin being a hormone is produced in very little amounts in the body. Hence, a large number of animals need to be sacrificed for obtaining small quantities of insulin. This makes the cost of insulin very high, demand being manyfold higher than supply.
• Slaughtering of animal is also not ethical.
• There is potential of immune response in humans against the administered insulin which is derived from animals.
• There is possibility of slaughtered animals being infested with some infectious micro-organism which may contaminate insulin.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 11 Biotechnology: Principles and Processes Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 11 Biotechnology: Principles and Processes

Very short answer type questions

Question 1.
Write the two components of the first artificial recombinant DNA molecule constructed by Cohen and Boyer.
Answer:

• Antibiotic resistance gene and
• Plasmid vector of Salmonella typhimurium.

Question 2.
Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer:
Gel Electrophoresis.

Question 3.
Mention the uses of cloning vector in biotechnology.
Answer:
Cloning vectors are used for transferring fragments of foreign DNA into a suitable host. They are also used to select recombinants from non-recombinants.

Question 4.
Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
Answer:
Alien DNA must be linked to ori or origin of replication site to start replication.

Question 5.
Why is it essential to have a ‘selectable marker’ in a cloning vector?
Answer:
Selectable markers are essential to identify and eliminate non-transformants, by selectively permitting the growth of the transformant.

Question 6.
Biotechnologists refer to Agrobacterium tumifaciens as a natural genetic engineer of plants. Give reasons to support the statement.
Answer:
This is because A. tumifaciens can transfer genes naturally by delivering a piece of T-DNA to plant cells. It has a tumour inducing plasmid.

Question 7.
Name the host cells in which micro-injection technique is used to introduce an alien DNA.
Answer:
Animal cells.

Question 8.
How does an alien DNA gain entry into a plant cell by ‘biolistics’ method?
Answer:
In biolistics method cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

Question 9.
What is the host called that produces a foreign gene product? What is this product called?
Answer:
The host that produces a foreign gene product is called competent host. The product is called recombinant protein.

Question 10.
Give any two microbes that are useful in biotechnology. [NCERT Exemplar]
Answer:
E. coli and Saccharomyces cerevisiae.

Question 11.
What is EcoRI? How does EcoRI differ from an exonuclease?
Answer:
EcoRI is restriction endonuclease enzyme. Exonuclease removes nucleotides from the ends of DNA while EcoRI makes cut at specific position within the DNA.

Question 12.
What is the significance of adding proteases at the time of isolation of genetic material (DNA)? [NCERT Exemplar]
Answer:
Role of proteases is to degrade the proteins present inside a cell (from which DNA is being isolated). If the proteins are not removed from DNA preparation then they could interfere with any downstream treatment of DNA.

Question 13.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process? [NCERT Exemplar]
Answer:
If denaturation of double-stranded DNA does not take place, then primers will not be able to anneal to the template, no extension will take place, hence no amplification will occur.

Question 14.
What would happen when you grow a recombinant in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing? [NCERT Exemplar]
Answer:
In the absence of antibiotic, there will be no pressure on recombinants to retain the plasmid (containing the gene of your interest). Since, maintaining a high copy number of plasmids is a metabolic burden to the microbial cells, it will thus tend to lose the plasmid.

Short answer type questions

Question 1.
What is meant by gene cloning? [NCERT Exemplar]
Answer:
Gene cloning refers to a process in which a gene of interest is ligated to a vector. The recombinant DNA thus produced is introduced in a host cell by transformation. Each cell gets one DNA molecule and when the transformed cell grows to a bacterial colony, each cell in the colony has a copy of the gene.

Question 2.
List the key tools used in recombinant DNA technology.
Answer:
The key tools used in recombinant DNA technology are as follows:

• Restriction enzymes
• Polymerase enzyme
• Ligase enzyme ‘
• Vectors
• Host organism/cell.

Question 3.
(a) What are “molecular scissors”? Give one example.
(b) Explain their role in recombinant DNA technology.
Or Why are molecular scissors so called? Write their use in biotechnology.
Answer:
(a) The restriction endonucleases are called molecular scissors, as they cut the DNA segments at particular locations, e.g., EcoRI. They are so called because they cut DNA at specific points.

(b) The restriction enzymes cut the DNA strands a little away from the centre of the palindromic sites, but between the same two bases on the opposite strands. This leaves single stranded portions with overhanging stretches called sticky ends on each strand as they form hydrogen bonds with their complementary cut counterparts. This stickiness at the ends facilitates the action of the enzyme DNA ligase.

Question 4.
Name the natural source of agarose. Mention one role of agarose in biotechnology.
Answer:
The natural source of agarose is sea weed. Agarose is a natural polymer. It is used to develop the matrix for gel electrophoresis. It helps in the separation of DNA fragments based on their size.

Question 5.
For producing a recombinant protein (for therapeutic purpose) in large scale, which vector would you choose – a low copy number or high-copy number? [NCERT Exemplar]
Answer:
High-copy number, because higher the copy number of vector plasmid, higher the copy number of gene and consequently, protein coded by the gene is produced in high amount.

Question 6.
What modification is done in the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector? [NCERT Exemplar]
Answer:
T-DNA is the only essential part required to make Ti plasmid a cloning vector. The plasmid is disarmed by deleting the tumour inducing genes in the plasmid so that it becomes an effective cloning vector and remove it harmful effect.

Question 7.
Describe the role of CaCl₂ in preparation of competent cells.
[NCERT Exemplar]
Answer:
CaCl₂ is known to increase the efficiency of DNA uptake to produce transformed bacterial cells. The divalent Ca⁺² ions create transient pores in the bacterial cell wall, by which the entry of foreign DNA is facilitated into the bacterial cells.

Question 8.
(a) Why must a cell be made ‘competent’ in biotechnology experiments? How does calcium ion help in doing so?
(b) State the role of ‘biolistic gun’ in biotechnology experiments?
Answer:
(a) A recombinant DNA transfer into the host cell, needs that the recipient cell must be made competent in order to receive and absorb the DNA, present in the surrounding. The calcium ions in the medium increase the efficiency with which DNA enters the bacterium through the pores in the cell wall.

(b) Biolistic or Gene Gun : It is a vectorless method in which DNA is directly introduced in the nucleus of plant cell. Plant cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA.

Long answer type questions

Question 1.
(a) Name the selectable markers in the cloning vector pBR322? Mention the role they play. ‘
(b) Why is the coding sequence of an enzyme β-galactosidase a preferred selectable marker in comparison to the ones nam£d above?
Answer:
(a) In pBR322, amp^R and Tet^R, the two antibiotic resistant genes act as selectable markers. If an alien DNA ligates at the Bam HI site of tetracycline resistant gene in the vector pBR322, the recombinant loses the tetracycline resistance. Non-recombinant will grow on both the media containing tetracycline/ampicillin whereas recombinant will grow on ampicillin medium but not on medium containing tetracycline. In this case, one antibiotic resistance gene helps in selecting the transformants and the other antibiotic gene gets inactivated due to insertion of alien DNA.

Thus, selectable markers (amp^R) help in indentifying and eliminating non-transformants and help in selecting those host cells which contain the recombinant vector i.e., transformants.

(b) The selection of recombinant by the inactivation of one of the antibiotic resistance gene is a cumbersome, complicated, time consuming technique involving plating both the recombinant and non-recombinant on the ampicillin medium and then on tetracycline containing medium. In insertional inactivation, the DNA inserted in the coding sequence of an enzyme p-galactosidase results in inactivation of the enzyme and the bacterial colony, with insert shows no colouration while those without inserted plasmid, form blue colour colonies. This is a simple, less cumbersome technique.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 10 Microbes in Human Welfare Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 10 Microbes in Human Welfare

Very short answer type questions

Question 1.
Give an example of a rod-shaped virus,
Answer:
Tobacco mosaic virus.

Question 2.
Which one of the following is the baker’s yeast used in fermentation?
Saccharum barberi, Saccharomyces cerevisiae, Sonalika.
Answer:
Saccharomyces cerevisiae.

Question 3.
Why is distillation required for producing certain alcoholic drinks? [NCERT Exemplar]
Answer:
Distillation increases the alcohol content in alcoholic drinks.

Question 4.
What would have happened if antibiotics were not discovered? s [NCERT Exemplar]
Answer:
If antibiotics were not discovered, bacterial and fungal diseases would , not have been controllable.

Question 5.
Give the scientific name of the source organism from which the first antibiotic was produced.
Answer:
Penicillium notatum.

Question 6.
Write the scientific name of the microbe used for fermenting
malted cereals and fruit juices.
Answer:
Saccharomyces cerevisiae.

Question 7.
Name a microbe used for statin production. How do statins lower blood cholesterol level? [NCERT Exemplar]
Answer:
Monascus purpureus is used for statin production. Statins lower blood cholesterol level by competitively inhibiting the enzyme responsible for . synthesis of cholesterol.

Question 8.
Name the group of organisms and the substrate they act on to produce biogas.
Answer:
Name of the group of organisms – Methanogens.
Substrate – Cellulosic material/cow dung/agriculture waste.

Question 9.
Name any genetically modified crop. [NCERT Exemplar]
Answer:
Bt. cotton.

Question 10.
What are Nucleopolyhedroviruses being used for nowadays? [NCERT Exemplar]
Answer:
Nucleopolyhedroviruses are used for the biological control of insect pests.

Question 11.
Which of the following is a free-living bacteria that can fix nitrogen in the soil?
Answer:
Spirulina, Azospirillum, Sonalika Ans. Azospirillum.

Question 12.
Mention the role of cyanobacteria as a biofertiliser.
Answer:
It is a biological organism that fixes atmospheric nitrogen.

Short answer type questions

Question 1.
Name the source of streptokinase. How does this bioactive molecule function in our body?
Answer:
Source: Streptococcus.
Streptokinase is a clot buster, i.e., it removes clot from the blood vessels of patients who had a heart attack.

Question 2.
Why are some molecules called bioactive molecules? Give two examples of such molecules.
Answer:
This is because microbes like bacteria or fungi are used in their production, e.g.,
Citric acid – Acetic acid
Butyric acid – Lactic acid
Ethanol – Lipases
Streptokinase – Cyclosporin A (Any two)

Question 3.
Given below is a list of six microorganisms. State their usefulness to humans.
(a) Nucleopolyhedrovirus
(b) Saccharomyces eerevisiae
(c) Monascus purpureus
(d) Trichoderma polysporum
(e) Penicillium notatum
(f) Propionibacterium sharmanii
Answer:
(a) Nucleopolyhedrovirus: Biocontrol agents for pest.
(b) Saccharomyces eerevisiae: Bread making and alcohol/wine production.
(c) Monascus purpureus: Statins-blood cholesterol lowering agents.
(d) Trichoderma polysporum: Cyclosporin-A, immune suppressing medicine in organ transplant patients.
(e) Penicillium notation: Penicillin, a anti-bacterial antibiotic.
(f) Propionibacterium sharmanii: Preparation of Swiss cheese with large holes.

Question 4.
What are methanogens? How do they help to generate biogas?
Or What are methanogens? Nameanimals they are present in and the role they play there?
Answer:
Methanogens are the bacteria which are grown anaerobically on cellulose material and produce large amount of methane along with CO₂ and H₂S.

Methanobacterium is the common methanogen found in the anaerobic sludge during sewage treatment. These bacteria are also present in the rumen (a part of the stomach) of cattle. In rumen, these bacteria help in the breakdown of cellulose present in the food of cattle. Hence, they play an important role in the nutrition of cattle.

The excreta (dung) of cattle commonly called gobar is rich in cellulosic material and these bacteria. Dung can be used for generation of biogas commonly called Gobar gas.
Note: Human beings are unable to digest cellulose in the vegetable food because the enzyme cellulose is not secreted. Cellulose forms the roughage in the body.

Question 5.
(a) How do organic fanners control pests? Give two examples, (b) State the difference in their approach from that of conventional pest control methods.
Answer:
(a) Organic farmers control pests by use of biological methods (biocontrol). They use natural enemy of the pest such as predator or parasite, by introducing these biocontrol agents in the field.
Example:
(i) Ladybird – a beetle and dragonflies are used to get rid of aphids and mosquitoes.
(ii) Bacillus thuringiensis, a bacterium is introduced into crop plants to control butterflies and caterpillars.

(b) In conventional pest control, toxic chemical pesticides are used which are expensive, persist, pollute the environment and kill both the target and non-target pests (beneficiary and parasitic insects which depend upon pests).

Long answer type questions

Question 1.
Explain the process of sewage water treatment before it can be discharged into natural water bodies. Why is this treatment essential?
Or Explain the different steps involved in sewage treatment before it can be released into natural water bodies.
Or Secondary treatment of the sewage can also called biological treatment. Justify this statement and explain the process.
Or Explain the different steps involved during primary treatment phase of sewage.
Or Explain the process of secondary treatment given to the primary effluent up to the point it shows significant change in the level of biological oxygen demand (BOD) in it.
Answer:
Sewage water treatment involves two steps:
(i) Primary treatment and
(ii) Secondary treatment.

(i) Primary Treatment: It is a physical process of removing small and large particles through filtration and sedimentation. Firstly, the sewage is passed through the wire mesh of screens of sequentially smaller pore sizes to remove floating objects (like polythene bags etc). Then the grit is sedimented by passing the sewage into a grit chamber. The sewage is then kept in settling tanks, where the suspended materials settle down to form the primary sludge. The effluent is then taken for secondary treatment.

(ii) Secondary Treatment: It is a biological process by the heterotrophic bacteria naturally present in the sewage. The primary effluent is passed into large aeration tanks where it is constantly agitated and air is pumped out. This causes the rapid growth of aerobic microbes into ‘floes’ which consume the organic matter of sewage leading to the reduction in biochemical oxygen demand (BOD). After the significant reduction in BOD of sewage, the effluent is passed into settling tanks where floes are sedimented leading to the formation of activated sludge. A part of this activated sludge is used as inoculum which is pumped back into the aeration tanks. The major part of this sludge is pumped into anaerobic sludge digesters, where its digestion occurs by the anaerobic bacteria producing methane, hydrogen sulphide and carbon dioxide. These gases form biogas. After secondary treatment the effluent is released into natural water bodies like streams and rivers.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 9 Strategies for Enhancement in Food Production Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 9 Strategies for Enhancement in Food Production

Very short answer type questions

Question 1.
Suggest the breeding method most suitable for animals1 that are below average in milk productivity.
Answer:
Outcrossing.

Question 2.
How is a mule produced?
Answer:
The mule is produced by breeding between male donkey and female horse (mare).

Question 3.
Write the name of the following:
(a) The most common species of bees suitable for apiculture.
(b) An improved breed of chicken.
Answer:
(a) Apis indica/Apis mellifera/Apis dorsata
(b) Leghorn/Rhode island red/Minorcha.

Question 4.
List any two economically important products for humans obtained from Apis indica.
Answer:
Honey and beeswax.

Question 5.
Which of the following is the semi-dwarf wheat that is high yielding and disease resistant?
Pusa Shubhra, Kalyan Sona, Ratna
Answer:
Kalyan Sona.

Question 6.
Write the names of two semi-dwarf and high yielding rice varieties developed in India after 1966.
Answer:
Jaya, Ratna.

Question 7.
Why is the South Indian sugarcane preferred by agriculturalists?
Answer:
South Indian sugarcane has thicker stem and higher sugar content.

Question 8.
Name any two diseases the ‘Himgiri’ variety of wheat is resistant to.
Answer:
Leaf and stripe rust; Hill bunt.

Question 9.
What is meant by ‘hidden hunger’? [NCERT Exemplar]
Answer:
Consumption of food deficient in nutrients particularly, micronutrients, proteins and vitamins is called hidden hunger.

Question 10.
What is protoplast fusion? [NCERT Exemplar]
Answer:
The merging of protoplasts obtained from two different cells to form a hybrid protoplast is called protoplast fusion.

Question 11.
What is the economic value of Spirulina?
Answer:
Spirulina can .serve as food rich in proteins, minerals, vitamins, fats and’ carbohydrates.

Question 12.
Identify two correct statements from the following:
(i) Apiculture means apical meristem culture.
(ii) Spinach is iron-enriched.
(iii) Green revolution has resulted in improved pulse-yield.
(iv) Aphids cannot infect rapeseed mustard.
Answer:
(ii) and
(iv) are correct.

Short answer type questions

Question 1.
(a) Name any two fowls other than chicken reared in a poultry farm.
(b) Enlist four important components of poultry farm management.
Answer:
(a) Pigeon, Turkey, Duck, Geese (any two)
(b) Poultry farm management includes :

• Proper feed and water
• Hygiene and health care of birds
• Proper and safe farm conditions
• Selection of disease free and suitable breeds.

Question 2.
In animal husbandry, if two closely related animals are mated for a few generations, it results in loss of fertility and vigour. Why is this so? [NCERT Exemplar]
Answer:
The phenomenon being referred to is called ‘inbreeding depression’ and results in loss of fertility and vigour. This happens because the recessive alleles tend to get together and express harmful effects in the progeny.

Question 3.
Give the scientific name of the most common species of honey bee reared in India. Why is it advantageous to keep beehives in crop-fields during flowering periods?
Or Honey collection improves when beehives are kept in crop- fields during the flowering season. Explain.
Answer:
The most common species of honey bee reared in India is Apis indica. Honeybees are good pollinators of almost all the plants. The flowers in turn offer floral rewards like nectar and pollen grains. So, when beehives are kept in crop-fields during the flowering season, honey collection increases and in turn the yield also increases.

Question 4.
How has mutation breeding helped in improving the production of mung bean crop?
Answer:
Mutation breeding produced disease resistant varieties against yellow mosaic virus and powdery mildew.
(a) Write the desirable characters a farmer looks for in his sugarcane crop.
How did plant breeding techniques help north Indian farmers to develop cane with desired characters?
(a) The desirable characters that should be present in sugarcane crop are as follows :

• High yield
• Thick stem
• High sugar content
• Ability to grow in North India.

(b) With the help of plant breeding, the two varieties of sugarcane t.e. Saccharum barberi [sugarcane of North India] and Saccharum officinarum [sugarcane of South India] were crossed to obtain sugarcane varieties having desirable, qualities. So that a good quality sugarcane variety could be grown in North India.

Question 6.
Differentiate between somaclones and somatic hybrids. Give one example of each.
Answer:
Somaclones are the genetically similar plants, similar to the original parent plant from which explant was taken, to start the tissue culture. Somatic hybrids are the structures produced by the process of fusing protoplasts of somatic cells derived from two different varieties/species of plants on suitable culture/nutrients medium under aseptic conditions.

Question 7.
Why is it necessary to emasculate a bisexual flower in a plant breeding programme? Mention the condition under which emasculation is not necessary.
Answer:
Emasculation is necessary to ensure that only the desired pollen grains are used for pollination and the stigma is protected from contamination (from unwanted self pollen). The anthers are removed followed by bagging so the plant now behaves as a female plant. The pollen grains from the anthers of the desired male plant can be dusted on the stigma of flower of the female plant to obtain desired results.
Emasculation is not required if the plant produces unisexual flowers.

Question 8.
How does culturing Spirulina solve the food problems of the growing human population?
Or Large scale cultivation of spirullina is highly advantageous for human population. Explain giving two reasons.
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants, straw, molasses, animal manure and even sewage, to produce large quantities of biomass. It serves as a food rich in protein, minerals, fats, carbohydrate and vitamins, being environment friendly.

Long answer type questions

Question 1.
Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Answer:
Inbreeding: involves the mating between closely related animals of same breed for 4-6 generations.
The breeding strategy in inbreeding includes following steps:

• Identification of the superior males and superior females of the same breed followed by mating.
• Progeny obtained is evaluated and superior males and females among them are identified for further mating, e.g., in case of cattle, a superior female is a cow or buffalo that produces more milk per lactation and the superior male is the bull which gives rise to superior progeny as compared to other males.
• Inbreeding increases homozygosity, which is necessary to evolve a pure line in any animal.
• Inbreeding exposes harmful recessive genes that are eliminated by selection.
• It helps in accumulation of superior genes and elimination of less desirable genes.

Disadvantages: Continued inbreeding causes inbreeding depression, reduced fertility and low productivity.

Question 2.
A sugarcane has been affected by virus. How can a virus-free cane be developed from it? Explain the procedure.
Answer:
Tissue culture method has application in production of healthy plants from the virus infected sugarcane plants by using meristems (apical and axillary buds) as explants which are free from virus.

Meristem Culture: It is in vitro culture of meristem containing regions being present over the shoot apex and very young buds over the nodes. With the help of fine scalpel shoot tip of 0.1-1.0 mm length is removed. Alternatively, sections of 3rd and 4th nodes from stem apex are removed. They are 1-2 cm long. Their leaves are detached from the tips of petioles. The sections are surface sterilised in 0.5% sodium hypochlorite for ten minutes, rinsed in sterile distilled water and thinly paired to develop fresh surface. They are placed over solid culture medium having cytokinin (generally BAP). Cytokinin is known to overcome apical dominance and induce growth of axillary buds.

Each explant will develop either a number of shoots or a single shoot. It is known as multiple shoot or single shoot cultures respectively. The single shoot is again cut into nodal segments for further culturing. Ultimately shoots of 2-3 cm are exercised and transferred to medium promoting root formation (with extra auxin). The plantlets are then hardened and transferred to field.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 8 Human Health and Disease Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 8 Human Health and Disease

Very short answer type questions

Question 1.
Malaria, typhoid, pneumonia and amoebiasis are some of the human infectious diseases. Which ones of these are transmitted through mechanical carriers?
Answer:
Malaria and amoebiasis are transmitted through mechanical carriers.

Question 2.
How does haemozoin affect the human body when released in blood during malarial infection?
Answer:
Haemozoin is responsible for the chill and high fever recurring every three to four days during malarial infection.

Question 3.
What causes swelling of the lower limbs in patients suffering from filariasis?
Answer:
Wuchereria (W. bancrofti and W. malayi).

Question 4.
Why is Gambusia introduced into drains and ponds?
Answer:
To feed on mosquito larvae so as to eliminate the vectors responsible for causing malaria.

Question 5.
Recently chikungunya cases were reported from various parts of the country. Name the vector responsible.
Answer:
Aedes mosquito is responsible for chikungunya cases.

Question 6.
What role do macrophages play in providing immunity to humans?
Answer:
Macrophages destroy the microbes (by phagocytosis) and provide protection against diseases.

Question 7.
In what way are monocytes a cellular barrier in immunity?
Answer:
Monocytes can phagocytose (by the process called phagocytosis) and thereby destroy the pathogens.

Question 8.
How does colostrum provides initial protection against diseases to new bom infants? Give one reason.
Answer:
Colostrum contains several antibodies which are absolutely essential for developing resistance in the new-born babies.

Question 9.
State the functions of mast cells in allergy response.
Answer:
Mast cells release chemicals like histamine and serotonin in allergic response.

Question 10.
What is an autoimmune disease? Give an example.
Answer:
It is an abnormal immune response in which the immune system of the body starts rejecting its own body cells or ‘self cells and molecules. For example, rheumatoid arthritis.

Question 11.
State two different roles of spleen in the human body.
Answer:
Spleen is the secondary lymphoid organ that stores lymphocytes, it filters microbes and acts as a reservoir to store erythrocytes.

Question 12.
Why sharing of injection needles between two individuals is not recommended?
Answer:
Sharing of needles can transmit diseases like HIV, AIDS, Hepatitis B or C from infected to non-infected individuals.

Question 13.
Retroviruses have no DNA. However, the DNA of the infected host cell does possess viral DNA. How is it possible?
Answer:
On infecting the host cell, the viral RNA transforms into viral DNA by reverse transcription. This viral DNA then incorporates into the host DNA.

Question 14.
Suggest any two techniques which can help in early detection of bacterial and viral infections much before the symptoms appear in the body.
Answer:
Enzyme Linked Immunosorbent Assay (ELISA), Polymerase Chain Reaction (PCR).

Question 15.
Mention the useful as well as the harmful drug obtained from the latex of Poppy plant.
Answer:
Useful drug – morphine.
Harmful drug – heroin.

Short answer type questions

Question 1.
Define the term health. Mention any two ways of maintaining it.
Answer:
Health is a state of complete physical, mental and social well-being. Good health can be maintained through balanced diet and regular exercise.

Question 2.
List the specific symptoms of typhoid. Name its causative agent.
Answer:
Specific symptoms of typhoid are as follows:

• Constant high fever (39° to 40°C)
• Weakness
• Stomach pain
• Loss of appetite

Its causative agent is Salmonella typhi.

Question 3.
Identify a, b, c and d in the following table:

Name of the human disease	Name of the causal bacteria/virus	Specific organ or its part affected
(i) Typhoid	Salmonella typhi	a
(ii) Common cold	b	c
(iii) Pneumonia	Streptococcus pneumoniae	d

Answer:
(a) Small intestine
(b) Rhino virus
(c) Nose and respiratoiy passage
(d) Alveoli of lungs

Question 4.
At what stage does Plasmodium gain entry into the human body? Write the different stages of its life-cycle in the human body.
Or Trace the life-cycle of malarial parasite in the human body when bitten by an infected female Anopheles.
Answer:
Plasmodium falciparum is the malarial parasite.
Plasmodium life-cycle:
The gametocyte develops in the red blood cells of human.

Question 5.
Explain the role of the following in providing defence against infection in human body :
(i) Histamines
(ii) Interferons
(iii) B-cells
Answer:
(i) Histamines: These are chemicals which cause inflammatory responses.
(ii) Interferons: These are glycoproteins which protect non-infected cells from further viral infection.
(iii) B-cells: These produce proteins called antibodies in response to pathogens into the blood to fight with them.

Question 6.
(a) What is the functional difference between B cells and T cells?
(b) Name the source used to produce hepatitis-B vaccine using rDNA technology.
Answer:

(a)

B-Lymphocytes	T-Lymphocytes
(i) They arise from bone marrow.	They arise from bone marrow and thymus.
(ii) B-cells form humoral or antibody-mediated immune system (AMIS).	T-cells form cell-mediated immune system (CMIS).
(iii) They defend against viruses and bacteria that enter the blood and lymph.	They defend against pathogens including protists and fungi that enter the cells.
(iv) They form plasma cells and memory cells by the division.	They form killer, helper and suppressor cells by the division of lymphoblasts.

(b) Hepatitis-B vaccine is produced from surface antigens of transgenic yeast by r-DNA technology. The antigens represent whole protein vaccine.

Question 7.
In the metropolitan cities of India, many children are suffering from allergy/asthma. What are the main causes of this problem. Give some symptoms of allergic reactions. [NCERT Exemplar]
Answer:
Allergy is the exaggerated response of the immune system to certain antigens present in the environment. In metropolitan cities life style is responsible in lowering of immunity and sensitivity to allergens. More polluted environment increases the chances of allergy in children. Some symptoms of allergic reactions are sneezing, watery eyes, running nose and difficulty in breathing.

Question 8.
Differentiate between benign and malignant tumours.
Answer:

Benign tumour	Malignant tumour
(i) It is a non-cancerous tumour.	It is a cancerous tumour.
(ii) Benign tumour does not show metastasis and is non-invasive.	It shows metastasis and thus invades other body parts.
(iii) It stops growth after reaching a certain size.	Malignant tumour shows indefinite growth.
(iv)Limited	There is no adherence amongst cells. They tend to slip past one another.
(v) It is less fatal to the body.	It is more fatal to the body.

Question 9.
Write the source and the effect on the human body of the following drugs:
(i) Morphine
(ii) Cocaine
(iii) Marijuana
Answer:
(i) Morphine: It is obtained from poppy plant Papaver somniferum. It binds to specific opioid receptors present in central nervous system and ‘ gastrointestinal tract.
(ii) Cocaine: It is obtained from coca plant Erythroxylum coca. It interferes with the transport of the neurotransmitter dopamine.
(iii)Marijuana : It is obtained from Cannabis sativa. It affects the cardiovascular system of the body.

Question 10.
(a) Why is there a fear amongst the guardians that their adolescent wards may get trapped in drug/alcohol abuse?
(b) Explain ‘addiction’ and dependence’ in respect of drug/alcohol abuse in youth.
Answer:
(a) Reasons for alcohol abuse in adolescents:

• Social pressure
• Curiosity and need for adventure, excitement and experiment.
• To escape from stress, depression and frustration.
• To overcome hardships of life.
• Unstable or unsupportive family structure

(b) Addiction: The psychological attachment to certain effects such as euphoria and a temporary feeling of well-being, associated with drugs and alcohol is called addiction.
Dependence: The tendency of the body to manifest a characteristic and unpleasant withdrawal syndrome on abrupt discontinuation of regular dose of drug/alcohol is called dependence.

Long answer type questions

Question 1.
(a) Cancer is one of the most dreaded diseases of humans. Explain ‘Contact inhibition’ and ‘Metastasis’ with respect to the disease.
(b) Name the group of genes which have been identified in normal cells that could lead to cancer and how they do so?
(c) Name any two techniques which are useful to detect cancers of internal organs.
(d) Why are cancer patients often given a-interferon as part of the treatment?
Answer:
(a) Contact inhibition is the property of normal cells in which contact with other cells inhibits their uncontrolled growth.
Metastasis is the property in which tumour cells reach distant sites in the body, through blood.
(b) Proto oncogenes or Cellular oncogenes.
These genes when activated under certain condition could lead to oncogenic transformation of the cells.
(c) Biopsy/radiography/CT/MRI
(d) a-interferon activates immune system and destroys the tumour.

Question 2.
Why do some adolescents start taking drugs? How can the situation be avoided? [NCERT Exemplar]
Answer:
Many factors are responsible for motivating youngsters towards alcohol or drugs. Curiosity, need for adventure and excitement, experimentation are the initial causes of motivation. Some youngsters start consuming drugs and alcohol in order to overcome negative emotions (such as stress, pressure, depression, frustration) and to excel in various fields. Several mediums like television, internet, newspaper, movies etc. are also responsible for promoting the idea of alcohol to the younger generation. Amongst these factors, reasons such as unstable and unsupportive family structures and peer pressure can also lead an individual to be dependant on drugs and alcohol.

Preventive measures against addiction of alcohol and drugs are as follows:
(a) Parents should motivate and try to increase the willpower of their child.
(b) Parents should educate their children about the ill-effects of alcohol. They should provide them with proper knowledge and counselling regarding the consequences of addiction to alcohol.
(c) It is the responsibility of the parent to discourage a child from experimenting with alcohol. Youngsters should be kept away from the company of friends who consume drugs.
(d) Children should be encouraged to devote their energy in other extra¬curricular and recreational activities.
(e) Proper professional and medical help should be provided to a child if sudden symptoms of depression and frustration are observed.

Question 3.
A person shows strong unusual hypersensitive reactions when exposed to certain substances present in the air, identify the condition. Name the cells responsible for such reactions. What precaution should be taken to avoid such reactions? [NCERT Exemplar]
Answer:
The condition is called allergy. Mast cells are responsible for such reactions.
To avoid such reactions following precautions must be taken:

1. Use of drugs like antihistamine, adrenalin and steroids quickly reduces the symptoms.
2. Avoid contact with substances to which a person is hypersensitive.

Question 4.
What would happen to immune system, if thymus gland is removed from the body of a person? [NCERT Exemplar]
Answer:
Thymus is the primary lymphoid organ. In thymus gland, immature lymphocytes differentiate into antigen-sensitive lymphocytes. If thymus gland is removed from the body of a person, his immune system becomes weak. As a result the person’s body becomes prone to infectious diseases.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 7 Evolution Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 7 Evolution

Very short answer type questions

Question 1.
Name the scientist who disproved spontaneous generation theory.
Answer:
Louis Pasteur disproved the theory of spontaneous generation.

Question 2.
What did Louis Pasteur’s experiment on ‘killed yeast’ demonstrate? Name the theory that got disproved on the basis of his experiment.
Answer:
Louis Pasteur demonstrated that life comes only from pre-existing life. The theory of spontaneous generation was disproved on the basis of his experiment.

Question 3.
Write the hypothetical proposals put forth by Oparin and Haldane.
Or List two main propositions of Oparin and Haldane.
Or State two postulates of Oparin and Haldane with reference to origin of life.
Answer:
Oparin and Haldane proposed that life originated from pre-existing non-organic molecules and the diverse organic molecules were formed from these inorganic constituents by chemical evolution.

Question 4.
When we say “survival of the fittest”, does it mean that
(a) those which are fit only survive, or
(b) those that survive are called fit. Comment. [NCERT Exemplar]
Answer:
Those individuals which survive and reproduce in their respective environment are called fit.

Question 5.
Why are analogous structures a result of convergent evolution?
Answer:
They are not anatomically similar structures though they perform similar functions.

Question 6.
Identify the examples of convergent evolution from the following:
(i) Flippers of penguins and dolphins
(ii) Eyes of octopus and mammals
(iii) Vertebrate brains
Answer:
(i) Flippers of penguins and dolphins
(ii) Eyes of octopus and mammals

Question 7.
What does Hardy-Weinberg equation p² + 2pq + q² = 1 convey?
Answer:
Hardy-Weinberg equation convey genetic equilibrium, i.e., sum total of all allelic frequencies is 1.

Question 8.
What is founder effect? [NCERT Exemplar]
Answer:
Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. The original drifted population becomes founder and the effect is called founder effect.

Question 9.
State the significance of Coelacanth in evolution.
Answer:
It is an ancestor of amphibians.

Question 10.
Name the first human like hominid. Mention his food habit and brain capacity.
Answer:
Homo habilis was the first human-like hominid. Homo (man) habili (skilful) was carnivorous and hunted large animals. He had a brain capacity of 650-800 cc.

Question 11.
Name the common ancestor of the great apes and man.
Answer:
Dryopithecus/Ramapithecus.

Question 12.
By what Latin name the first hominid was known?
[NCERT Exemplar]
Answer:
Homo habilis.

Short answer type questions

Question 1.
Describe the experiment that helped Louis Pasteur to dismiss the theory of spontaneous generation of life.
Answer:

Louis Pasteur (1864) boiled broth in flasks having bent swan or S-shaped necks. No microorganisms were observed in broth after keeping for several days though broth was connected to air through the bent neck. It is because the dirt carrying microorganisms got settled in the bent part of neck. When the neck was broken, colonies of microorganisms soon developed over the broth showing the microorganisms have come from air.

Question 2.
Mention the contribution of S.L. Miller’s experiments on Origin of Life.
Answer:
S.L. Miller created an environment in laboratory similar to the one that existed before life originated. In a closed flask containing CH₄, H₂, NH₃ and water vapour at 800°C, electric discharge was created. The conditions were similar to those in primitive atmosphere. After a week, they observed presence of amino acids and complex molecules like sugars, nitrogen bases, pigments and fats in the flask. This provided experimental evidence for the theory of chemical origin.

Question 3.
How does the study of fossils support evolution? Explain.
Answer:
Different aged rock sediments contain fossil of different types. Early rocks contain fossils of simple organisms while recent rocks contain fossils of complex organisms, e.g., dinosaur.

By studying fossils occurring in different strata of rocks, geologists are able to reconstruct the geological period in which they existed and the cause of evolutionary change. Hence, new forms of life originated at different times in the history’ of earth.

Question 4.
Explain “fitness of a species” as mentioned by Darwin.
Answer:
“Fitness of a species” according to Darwin means reproductive fitness. All organisms after reaching reproductive age have varying degree of reproductive potential. Some organisms produce more offspring and some organism produce only few offspring. This phenomenon is also called as differential reproduction.
Hence the species which produces more offsprings are selected by nature.

Question 5.
While creation and presence of variation is directionless, natural selection is directional as it is in the context of adaptation. Comment. [NCERT Exemplar]
Answer:
Creation and variation occur in a sexually reproducing population as a result of crossing-over during meiosis and random fusion of gametes. It is however the organisms that are selected over a period of time which are determined by the environmental conditions. In other words, the environment provides the direction with respect to adaptations so that the organisms are more and more fit in terms of survival.

Question 6.
Branching descent and natural selection are the two key concepts of Darwinian theory of evolution. Explain each concept with the help of a suitable example.
Answer:
Branching Descent : Different species descending from the common ancestor get adapted in different habitats, e.g., Darwin’s finches-varieties of finches arose from grain eaters; Australian marsupials evolved from common marsupial.

Natural Selection: It is a process in which heritable variations enable better survival of the species to reproduce in large number, e.g., white moth surviving before the industrial revolution and black moth surviving after industrial revolution; long-necked giraffe survived the evolution process; DDT-resistant mosquitoes survive.

Question 7.
Explain the salient features of Hugo de Vries theory of mutation.
Answer:
Salient features of Hugo de Vries theory of mutation are as follows:

• Mutations cause evolution.
• New species originate due to large mutations.
• Evolution is a discontinuous process and not gradual.
• Mutations are directionless.
• Mutations appear suddenly.
• Mutations exhibit their effect immediately.

Question 8.
What does Hardy-Weinberg Principle of equilibrium indicate? List any two factors that could alter the equilibrium. What would such an alteration lead to?
Answer:
Hardy-Weinberg principle states that in a given population, the frequency of occurrence of alleles of a gene is supposed to remain fixed and even remains same through generations. This is also called genetic equilibrium. Sum total of all the alleles is 1. Hence, p2 + 2pq + q2 = 1 (p, q represent the frequency of gene A and allele a).
Factors affecting Hardy-Weinberg equilibrium are gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selections.

(i) Gene Flow/Migration: The movement of a section of population from one place to another, results in the addition of new alleles to the local gene pool of the host population. This is called gene migration. Migration causes variations at the genetic level.

(ii) Genetic Drift: The random changes in gene frequency in a population occurring by chance alone rather than by natural selection is called genetic drift. The effects of genetic drift are more prominent in small populations.

Long answer type questions

Question 1.
(a) Differentiate between analogy and homology giving one example each of plant and animal respectively.
(b) How are they considered as an evidence in support of evolution?
Or
Differentiate between homology and analogy. Give one example of each.
Answer:
(a) Analogy is the phenomenon where different structures evolving for the same function and hence having similarity are the result of convergent evolution. These structures are called analogical structures. Example are
(i) Tendril of pea (leaf-let modified) and grapevine/ cucurbita (stem modified),
(ii) Flippers of penguins (wing modified) and dolphin (fore arm modified) both help in swimming.

Homology is the phenomenon where same structure developed along different lines due to adaptation to different needs/habitats as a result of divergent needs/habitats are a result of divergent evolution. These structures are called homologous structure.

Examples: (i) Thorns of Bougainvillea and tendrils of cucurbita represent homology as both are modified stems, (ii) Forelimbs of man, cheetah, whale and bat.

(b) Analogy and homology of the structures represent anatomical and morphological evidence of evolution. Analogy shows that similar habitat result in the selection of similar adaptive features in different groups of organisms but toward same function. This is a result of convergent evolution.

Homology include the same structures developed to have different forms to perform different functions in different animals. It is a result of divergent evolution. It indicate towards common ancestory. The comparative anatomy of forelimbs in all the mammals show similarities in the pattern of bones and pentadactyl organisation.

Question 2.
The evolutionary story of moths in England during industrialisation reveals, that ‘evolution is apparently reversible’. Clarify this statement. [NCERT Exemplar]
Answer:
The peppered moth occurs in two forms, i.e., light coloured (Biston betularia typoica) and dark coloured (Biston betrularia carbonaria). Before Industrial Revolution : Only light coloured moths were prevalent. Light coloured moths camouflaged well with the lichens that covers tree trunks, on the contrary dark moths were easy prey on the tree trunks and were very rare.

During the Industrial Revolution: The population of dark coloured moth increased. While, that of light coloured moth decreased. This change was due to the burning of coal in factories.
The smoke from the factories killed the lichens and the tree trunks turned black due to the deposition of soot. The black moths had an advantage against soot, therefore, escaped predation of birds while on the other hand, white moths were identified in sharp contrast and become easy prey.

With the Progression of Industrial Revolution: The coal was replaced by oil and electricity.
This resulted in reduction of soot deposits on the tree trunk. Gradually, the population of black moth decreased and that of light moth began to increase.