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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 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.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 6 Molecular Basis of Inheritance Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 6 Molecular Basis of Inheritance

Question 1.
Name the specific components and the linkage between them that form deoxyadenosine.
Answer:
Nitrogenous base (Adenine) and pentose sugar and N-glycosidic linkage.

Question 2.
Why is RNA more reactive in comparison to DNA?
Answer:
RNA is more reactive because:

• It is single stranded.
• Every nucleotide has an additional OH^– group present at position 2 in the ribose.
• Mutates faster as compared to DNA.

Question 3.
In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix, which has 2 × 10⁹ bp in saturating the presence of this compound. [NCERT Exemplar]
Answer:
The new length of DNA helix would be
= 2 × 10^-9 × 0.44 × 10^-9bp.

Question 4.
In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates,
still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why?
[NCERT Exemplar]
Answer:
DNA polymerase is highly specific to recognise only deoxyribonucleoside r triphosphates. Therefore, it cannot hold RNA nucleotides.

Question 5.
Name the enzyme involved in the continuous replication of DNA strand. Mention the polarity of the template strand.
Answer:
DNA polymerase is involved in continuous replication of DNA strand. The polarity of template strand is 3′ → 5′.

Question 6.
What is a cistron?
Answer:
Cistron is a segment of DNA coding for a polypeptide chain.

Question 7.
Name the transcriptionally active region of chromatin in a nucleus.
Answer:
Euchromatin or Exon.

Question 8.
Write the function of RNA polymerase n.
Answer:
RNA polymerase II transcribes precursor of mRNA or hnRNA.

Question 9.
Mention the two additional processings which /mRNA needs to
undergo after splicing so as to become functional.
Answer:
Capping and tailing.

Question 10.
Give an example of a codon having dual function.
Answer:
AUG has dual function. It acts as initiation codon and also codes for methionine.

Question 11.
Sometimes cattle or even human beings give birth to their young ones that have extremely different sets of organs like limbs/position of eye(s), etc. Why? [NCERT Exemplar]
Answer:
This is due to a disturbance in coordinated regulation of expression of sets of genes associated with organ development or due to mutations.

Question 12.
How does a degenerate code differ from an unambiguous one?
Answer:
Degenerate code means that one amino acid can be coded by more than one codon. Unambiguous code means that one codon codes for only one amino acid.

Question 13.
What is aminoacylation? State its significance.
Answer:
Aminoacylation of t-RNA involves activation of amino acids by ATP which gets linked to OH’ present at the 3′ end of specific t-RNA. The process is also called charging of t-RNA.

Question 14.
Mention how c|oes DNA polymorphism arise in a population?
Answer:
DNA polymorphism in a population arise due to presence of inheritable mutations at high frequency.

Question 15.
How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments?
Answer:
By using density gradient centrifugation, where satellite DNA forms small peaks.

Short answer type questions

Question 1.
Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the 12-strain into virulent strain? Explain.
[NCERT Exemplar]
Answer:
RNA is more labile and prone to degradation, owing to the presence of 2’OH group in its ribose. Hence, heat-killed S-strain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

Question 2.
Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them. [NCERT Exemplar]
Answer:
The enzymes involved in DNA replication other than DNA polymerase and ligase are listed below with their functions:\

• Helicase – Opens the helix
• Topoisomerases – Removes the tension caused due to unwinding
• DNA ligase – Joins the cut DNA strands

Question 3.
Following are the features of genetic codes. What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon.
Answer:

• Stop codon is a codon, which does not code for any amino acid and here the polypeptide chain is released e.g., 3 codons – UAA, UAG, UGA.
• Unambiguous codon : One codon codes for only one specific amino acid.
• Degenerate codon : Here, one amino acid is coded by more than one codon e.g., amino acid glycine is coded by four codons (GGU, GGC, GGA, GGG).
• Universal codon means a codon specifies the same amino acid in all
  the organisms even in a virus. ,

Question 4.
A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your answer. [NCERT Exemplar]
Answer:
The statement is correct because of degeneracy of codons, mutations at third base of codon, usually does not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.

Question 5.
(a) Name the scientist who called tRNA an adapter molecule.
(b) Draw a clover leaf structure of fRNA showing the following:
(i) Tyrosine attached to its amino acid site.
(ii) Anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(c) What does the actual structure of JRNA look like?
Answer:
(a) Francis Crick
(b)
(c) The actual structure of t-RNA looks like inverted L.

Question 6.
A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena. [NCERT Exemplar]
Answer:
In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducer. Hence, cannot relieve the lac operon from its repressed state. Therefore, lac operon is always expressed.

Question 7.
Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage? [NCERT Exemplar]
Answer:
Bacteriophage does not have repetitive sequences such as VNTRs in its genome, as its genome is very small (i.e., 5386bp) and have all the coding sequence. Therefore, DNA fingerprinting is not done for phages.

Long answer type questions

Question 1.
Describe Meselson and Stahl’s experiment that was carried in 1958 on E.coil. Write the conclusion they arrived at after the experiment.
Matthew Meselson and Franklin Stahl (1958) grew E. coli in a medium containing ¹⁵NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as
the only nitrogen source for many generations. The result was that ¹⁵N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

They transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently as CsCl gradients to measure the densities of DNA.

Thus, the DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium (that is after 20 minutes; E.coli divides in 20 minutes) had a hybrid or intermediate density. DNA extracted from the culture after another generation (that is after 40 minutes, II generation) was composed of equal amounts of hybrid DNA and of light DNA.

Matthew Meselson and Franklin Stahl’s experiment demonstrated that DNA replication is semi-conservative.

Generation I Generation II

Question 2.
(a) Describe the series of experiments of F. Griffith. Comment on the significance of the results obtained.
(b) State the contribution of MacLeod, McCarty and Avery.
Answer:
(a) Frederick Griffith (1928), a British doctor, was studying the pathogenicity of different strains of Streptococcus pneumoniae. It has two strains- (i) virulent cause pneumonia, has S-type of bacteria, covered by sheath of mucilage, (ii) non-virulent do not produce the disease, has R-type of bacteria, devoid of sheath of mucilage.

Griffith found that on injecting live R-type bacteria did not produce the disease while live S-type caused pneumonia and the death in mice. However, when heat-killed S-type injected, they did not produce the disease. Finally, Griffith injected a combination of live-R-type and heat-killed S-type bacteria into mice. Some mice survived while others developed the disease of pneumonia and died. Autopsy of dead mice showed that they possessed both the type of bacteria (virulent-S-type and non-virulent-R-type) in living form through the mice that had been injected with dead virulent (S-type) and living non-virulent (R-type) bacteria.

From the above experiment, Frederick Griffith concluded that the occurrence of living S-type virulent bacteria is possible only by their formation from R-type non-virulent bacteria which pick-up the trait of virulence from dead bacteria. This phenomenon is called Griffith effect or transformation. Griffith proposed that the transforming principle is a chemical substance released by heat-killed S-type, which changed the S-type into S-bacteria. It was a permanent change as the new S-type formed only S-type progeny.

(b) Avery, MacLeod, McCarty discovered that DNA from the heat-killed S-strain caused the living R-strain bacteria to become transformed into living S-type. They found proteases and RNAases did not affect transformation while DNAases inhibit transformation. They concluded that DNA is the hereditary material.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 5 Principles of Inheritance and Variation Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 5 Principles of Inheritance and Variation

Question 1.
Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer:
Round/Wrinkled, Yellow/Green.

Question 2.
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.
Answer:
Test Cross.

Question 3.
State a difference between a gene and an allele.
Answer:
Gene is a unit of heredity passed from’one generation to next generation and determine the expression of any morphological or physiological inheritable character of an organism.

Alleles are alternative form of a gene, occupying the same position on homologous chromosomes and affecting the alternative forms (contrasting traits) of the same character.

Question 4.
A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits.
Answer:
Inflated green pod is the dominant trait.

Question 5.
Mention the type of allele that expresses itself only in homozygous state in an organism.
Answer:
Recessive allele.

Question 6.
A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer:
Axial, violet flower.

Question 7.
What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants?
Answer:
True breeding lines are plants which have undergone continuous self-pollination for several generations. These are homozygous for traits.

Question 8.
Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer:
Anaphase-I of Meiosis-I.

Question 9.
A male honeybee has 16 chromosomes whereas its female has 32 chromosomest Give one reason.
Answer:
A male honeybee with 16 chromosomes develops parthenogenetically from an unfertilised egg and is haploid (n) whereas the female honeybee with 32 chromosomes develops from a fertilised egg, the zygote and is diploid (2n).

Question 10.
Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
Answer:
Haemophilia is a sex-linked recessive disease and the defective gene is present on X chromosome only and not on Y chromosome. Father never passes X chromosome to the son as father only contributes Y chromosome to the son.

Question 11.
Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer:
Klinefelter syndrome is caused due to the presence of an additional X-chromosome in the genotype of an individual i.e., 44 + XXY.

Question 12.
State the chromosomal defect in individuals with Turner’s syndrome.
Answer:
In Turner’s syndrome, the karyotype of the individual is 44 + XO. The X-chromosome is missing. It is due to the non-disjunction of X-chromosomes during oogenesis/spermatogenesis.

Short answer type questions

Question 1.
Mendel published his work on inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work.
Answer:

• The communication was not easy in those days and his work could not be widely publicised.
• His concept of genes as stable and discrete units that controlled the expression of traits and of the pair of alleles which did not ‘blend’ each other was not accepted by contemporaries as an explanation for the apparently continuous variation seen in nature.
• Mendel’s approach of using mathematics to explain biological phenomena was totally new and unacceptable to many of the biologists of his time.
• Though Mendel’s work suggested that factors (genes) were discrete units, he could not provide any physical proof for the existence of factors what they were made of.

Question 2.
The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes.
Draw your conclusion on the basis of the pedigree. [NCERT Exemplar]

Answer:
The pedigree chart shows that the trait is autosomal linked and recessive in nature. But, the parents are carriers (i.e. heterozygous) hence, among the offsprings only few show the trait irrespective of sex. The other offsprings are either normal or carrier.

Question 3.
What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
[NCERT Exemplar]
Answer:
Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome 21. Such individuals are anploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness, etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum.

The chance of having a child with Down’s syndrome increase with the age of the mother (40+) because ova are present in females since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physicochemical exposures during the mother’s life-time.

Question 4.
Differentiate between male and female heterogamety.
Answer:
In male heterogamety, the male is heteromorphic and have XY or XO type of sex chromosomes and produce two types of sperms, 50% with X- chromosome and 50% with Y-chromosome or none. The sex of the offspring depends upon the type of sperm, which fuses with egg e.g., mammals, Drosophila, grasshopper.

In female heterogamety, the female is heteromorphic and heterogametic and have ZW or ZO type of sex chromosomes and produce two types of eggs. The sex of the offspring depends upon the type of egg, which is fertilised, e.g., bird and some reptiles, butterflies and moths.

Question 5.
Explain mechanism of sex-determination in birds.
Answer:
Sex-determination in birds is opposite to human beings. Here the females contain heteromorphic sex chromosomes (AA + ZW) while the males have homomorphic (AA + ZZ) chromosomes. Thus, there is female heterogamety. The females are heterogametic and produce two types of eggs (A + Z) and (A + W). The male gametes are of one type (A + Z).

Question 6.
Why are human females rarely haemophilic? Explain. How do haemophilic patients suffer?
Answer:
Haemophilia is a sex-linked recessive disorder. The females have XX chromosomes and the males have XY chromosomes. If one of the two X chromosomes is normal, she remains a carrier and not diseased. Females will haemophilic only when both the X chromosomes carry the haemophilia gene, and this is possible only when the mother is a carrier and father is haemophilic. Haemophilic patients suffer from non-stop bleeding and nd clotting.

Question 7.
How do genes and chromosomes share similarity from the point of view of genetical studies? [NCERT Exemplar]
Answer:
By 1902, the chromosome movement during meiosis had been worked out.
Walter Sutton and Theodore Boveri (1902) noted that the behaviour of chromosomes was parallel to the behaviour of genes and used chromosome movement to explain Mendel’s laws.
They studied the behaviour of chromosomes during mitosis (equational division) and during meiosis (reduction division). The chromosomes as well as genes occur in pairs and the two alleles of gene pair are located on homologous sites of homologous chromosomes.

Chromosomes segregate when germ cells are formed.

Question 8.
Write short notes on –
(i) Phenylketonuria
(ii) Aneuploidy
Answer:
(i) Phenylketonuria : It is an inborn error of metabolism. The affected individual lack an enzyme called phenylalanine hydroxylase that converts the amino acid phenylalanine into tyrosine. As a result, phenylalanine gets accumulated and converted into phenylpyruvic acid and other derivatives in brain, causing mental retardation. These are also excreted through urine due to their absorption by kidney.

(ii) Aneuploidy: It is a phenomenon which occurs due to non¬disjunction resulting into gain or loss of one or more chromosomes, during meiosis.

Long answer type questions

Question 1.
(a) A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing axial, violet flowers, produced axial, violet flowers and axial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny.
(b) Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer:

(b) Law of Independent Assortment : When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of character.

Question 12.
(a) Write the blood group of people with genotype IAIB. Give reasons in support of your answer.
(b) In one family, the four children each have a different blood group. Their mother has blood group A and their father has blood group B. Work out a cross to explain how it is possible.
Answer:
(a) Blood group AB. Both the alleles I^A and I^B are co-dominant and express themselves completely.
(b) A cross is carried out between heterozygous father (of blood group B) and heterozygous mother (of blood group A) to get four children with different blood groups.

All the four blood groups are controlled by three allelic genes I^A, I^B, i
and thus it shows phenomena of multiple allelism. Both I^A and I^B are dominant over i. However, when together, both are dominant and show the phenomena of co-dominance forming the blood group AB. Six genotypes are possible with combination of these three alleles.

Question 3.
(A) Why are colourblindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people suffering from them.
(B) About 8% of human male population suffers from colourblindness whereas only about 0.4% of human female population suffers from this disease. Write an explanation to show how it is possible.
Answer:
(A) Colourblindness and thalassemia are categorised as Mendelian disorders because they are (i) due to alteration or mutation in a single gene (ii) transmission to the offspring follow principle of inheritance (iii) can be studied by pedigree analysis.
Symptoms of Colourblindness
(a) Difficulty in distinguishing between colours

• Protanopia-red colourblindness
• Deuternopia-green colourblindness
• Tritanopia-blue colourblindness

(b) Rapid eye movement (in rare cases)
(c) Inability to see shades or tones of the same colour.
(d) In rare cases, some people see only black, white and grey.

Symptoms of Thalassemia
(a) Formation of abnormal haemoglobin molecules resulting into haemolytic anaemia.
(b) Slow growth, delayed puberty.
(c) Bones become wider than normal, brittle and break easily
(d) Poor appetite
(e) A pale and listless appearance
(f) Dark urine
(g) Enlarged spleen, liver or heart.

(B) Colourblindness is a recessive sex-linked disorder in which the patient cannot distinguish red-green colour. The gene for colourblindness is present on X-chromosome. Presence of colourblindness in 8% human male population, is due to the presence of a single X-chromosome in male. There is no chance of dominant and recessive condition, as there is a single gene. Hemizygous condition is enough for the occurrence of defect in males. In females, due to presence of two X-chromosomes, the single gene of colourblindness cannot express and such females are carriers (XX^c) for a female to be colourblind, both of her X-chromosome, carry gene for colourblindness(X^cX^c).

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)

(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.