CBSE Class 12 Biology Question Paper (Outside Delhi 2016) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Outside Delhi 2016) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Outside Delhi 2016) with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

1. There are a total of 26 questions and five sections in the question paper. All questions are compulsory.
2. Section A contains questions number 1 to 5, very-short answer type questions of 1 mark each.
3. Section B contains questions number 6 to 10, short-answer type I questions of 2 marks each.
4. Section C contains questions number 11 to 22, short-answer type II questions of 3 marks each.
5. Section D contains question number 23, value based question of 4 marks.
6. Section E contains questions number 24 to 26, long-answer type questions of 5 marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of 2 marks, one question of 3 marks and all the three questions of 5 marks. In these questions, an examinee is to attempt any one of the two given alternatives.

Section A

Question 1.
A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Answer:
Male honeybees are formed by the process of parthenogenesis which involves development from unfertilized eggs. The unfertilised eggs carry only half number of chromosomes such as 16 chromosomes (haploid). While female honeybees are developed from unfertilised eggs and has 32 chromosomes.

Question 2.
Mention the role of genetic mother in MOET.
Answer:
MOET is Multiple Ovulation Embryo Transfer Technology. This programme is started for herd improvement. The genetic mother is available for another round of super ovulation.

Note
The generic mother or cow is administered with FSH (follicle stimulating hormone) in order to induce follicutar maturation and super ovulation.

Question 3.
What is biopiracy?
Answer:
Biopiracy is defined as the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment.

Question 4.
Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer:
The CNG is better than diesel because:

• CNG burns most efficiently than petrol and diesel in the automobiles and very little amount of it is left
  unburnt.
• CNG is cheaper than petrol or diesel.

Ñote
CNG is compressed Natural Gas and LPG is Liquified Petroleum Gas.

Question 5.
Write the probable differences in eating habits of (Homo habilis) and (Homo erectus).
Answer:
Difference between eating habits of Homo ha bilis and Homo erectus are as follows:

Characteristics	Homo habilis	Homo erectus
Eating habits	They did not eat meat.	They ate meat.

Note
The brain capccities of Homo habilis were in between 650-800cc while Horn erectus had larger brain capacities iround 900 cc.

Section B

Question 6.
A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain.
Answer:
The pea plant is monoecious as the male and female gametes are found on the same plant. So, self pollination takes place in such plants results in the production of seeds. Whereas the papaya plant is dioecious as male and female gametes are found located on the different plants. As only single parent (papaya) is involved according to question so cross pollination will not take place so there will be no seed production.

Question 7.
Following are the features of genetic codes. What does each one indicate? Stop codon; Unambiguous codon; Degenerate codon; Universal codon
Answer:
Stop codon: There are 64 codons out of which 61 codons codes for 20 amino acids and three are stop codon that terminates the process of protein synthesis (translation). Stop codons are UAA, UGA and UAG.

Unambiguous codon: One codon codes only for one amino acid and hence is called unambiguous and specific.

Degenerate codon: The codon is read in mRNA in a contiguous fashion as there are no punctuations.

Universal codon: The code is nearly universal. For example: From bacteria to human UUU would code for phenylalanine (Phe) amino acid.

Note
Codons are three nucleotide DNA or RNA sequences those codes for specific amino acid for the process of translation.

Question 8.
Suggest four important steps to produce a disease resistant plant through conventional plant breeding technology.
Answer:
The four important steps involved in the production of disease resistant plant through conventional plant breeding technology are:

1. Collection of variability: The entire collection of either plants or seeds having all diverse alleles for all genes in a given crop is called germplasm.
2. Evaluation and selection of parents: The germplasm is evaluated for the identification of plants having desirable combination of characters.
3. Cross hybridisation among the selected parents: The desired characters are combined from two different plants (parents). For example: high protein quality of one parent may required to be combined with disease resistance from another parents.
4. Selection and testing of superior recombinants: This step involves the selection among the progeny of the hybrids, those plants that have the desired character combination. The selection process is crucial to the success of the breeding objective and also requires careful scientific evaluation of the progeny.

Question 9.
Name a genus of baculovinis. Why are they considered good biocontrol agents?
Answer:
Baculoviruses are the pathogen that attack insects and other arthropods. The baculoviruses are used as a biological control agent. They belong the genus Nucleopolyhedrovirus. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications.

Question 10.
Explain the relationship between CFC’s and ozone is the stratosphere.
OR
Why are sacred groves highly protected?
Answer:
CFCs are widely used as a refrigerants and can be discharged in the lower part of the atmosphere that move upward and reach stratosphere. In stratosphere, UV rays act on them releasing Cl atoms. Cl degrades ozone releasing molecular oxygen with these atoms acting merely as catalysts, Cl atoms are not consumed in the reaction. When CFCs are added to the stratosphere, they have permanent and continuing affects on Ozone levels results in ozone layer depletion in the strastophere.

Note
Ozone layer depletion is marked over the Antarctic region results in the formation of a large area of thinned ozone layer which is commonly called as the ozone hole.

OR

The sacred groves are largely protected because they have large number of rare and threatened plants species. They are also protected because of the cultural and religious values of the communities. The sacred groves are found in Khasi and Jaintia Hills in Meghalaya, Aravalli Hills of Rajasthan, Western Ghat regions of Karnataka and Maharashtra and the Sarguja, Chanda and Bastar areas of Madhya Pradesh.

Section C

Question 11.
(a) Name the organic material exine of the pollen grain is made up of. How is this material advantageous to pollen grain?
(b) Still it is observed that it does not form a continuous layer around the pollen grain. Give reason.
(c) How are ‘pollen banks useftil?

OR

(a) Mention the problems that are taken care of by Reproduction and Child Health Care Programme.
(b) What is amniocentesis and why there is a statutoy ban on it?
Answer:
(a) The hard outer layer of the pollen grain is called the exine which is made up of sporopollenin. Sporopollenin is one of the most resistant organic material as it tolerate high temperatures and strong acids as well as alkali. It cannot be degraded by enzymatic degradation.

(b) The sporopollenin is not continuous because the exine of the pollen grain has prominent apertures called germ pores where the sporopollenin is absent.

(c) The pollen grain of a large number of species are stored for many years in liquid nitrogen at -196°C. The stored pollen grains are used as pollen banks and can be used in crop breeding programmes.

OR

(a) The “Reproductive and Child Health care (RCH) programmes” create awareness among people about the various reproduction related aspects and also providing facilities as well as support for building up a reproductively healthy society as the major goals of this programmes.

(b) Amniocentesis is a foetal sex determination test based on the chromosomal pattern in the amniotic fluid surrounding the developing embryo for determination of abnormalities in the foetus.
The statutory ban on amniocentesis is because of illegal sex determination that increases female foeticides, massive child immunisation.

Question 12.
What is a test cross? How can it decipher the heterozygosity of a plant?
Answer:
A test cross is a cross between the F₁ progeny and its homozygous recessive parents. Test cross is used for the determination of dominant character which is coming from the homozygous dominant genotype or heterozygous genotype. For example: TT and Tt for tallness.
Representation of genetic cross:

When heterozygous tall (Tt) plant is crossed with homozygous dwarf (tt) plant:

Result: 50% tall and 50% dwarf progenies are obtained in F2 generation.

Note
Test cross is used for the determination of the heterozygosity of the plant.

Question 13.
(a) What do Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project (HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGR
(c) Expand ‘SNPs’ identified by scientists in HGP.
Answer:
(a) BAC stands for Bacterial artificial chromosomes and YAC stands for Yeast artificial chromosomes.
BAC and YAC are commonly used host cells for cloning of DNA fragments using specialised vectors.

(b) Less than 2 percent of the genome codes for proteins and the function of over 50 percent of the discovered genes are unknown.

(c) SNPs are Single nucleotide polymorphism. Scientists have identified about 1.4 million locations where single base DNA differences occur in humans.

This information helps to revolutionise the processes of finding chromosomal locations for disease – associated sequences and also tracing the human history.

Question 14.
Differentiate between homology and analogy. Give one example of each.
Answer:
Difference between homology and analogy:

Homology	Analogy
(i) Homologous organs are those organs that are anatomically similar but perform different functions.	(i) Analogous organs are those that are anatomically not similar but perform same function.
(ii) This type of evolution is called divergent evolution.	(ii) This type of evolution is called convergent evolution.
(iii) Homology indicates common ancestry.	(iii) Analogy do not share common ancestry.
For example: Thom and tendrils of Bougainvillea and Cucurbita.	For example: Flippers of penguins and dolphins.

Question 15.
(a) It is generally observed that the children who had suffered from chicken-pox in their childhood may not contract the same disease in their adulthood. Explain giving reasons, the basis of such an immunity in an individual. Name this kind of immunity.
(b) What are interferons? Mention their role.
Answer:
(a) Children who had suffered from chicken pox in their childhood may not contract the same disease in their adulthood because they have developed antibody against chicken pox virus. The active memory initiates highly intense immune response during the second encounter. This kind of immunity is called active immunity which provides protection against the same disease as immune cells already produced antibodies against the disease causing antigens.

(b) Viral infected cells secrete proteins called interferons that protect non-infected cells from further viral infection.

Question 16.
(a) Write the two limitations of traditional breeding technique that led to promotion of micro propagation
(b) Mention two advantages of micro propagation
(c) Give two examples where it is commercially adopted.
Answer:
(a) The limitations related to traditional breeding techniques are as follows:

1. The traditional breeding techniques failed to keep pace with demand and to provide sufficiently fast and efficient systems for crop improvements.
2. It also takes a long time and not able to produce many offsprings in one time.

Note
The method of producing thousands of plants through tissue culture from a single explants is called micropropagation.

(b) The advantages of micropropagation are as follows:

1. Micropropagation is used for the production of many plants in a shorter time.
2. Disease free plants can be produced by this method.

(c) Micropropagation is commercially adopted for the production of banana and sugarcane.

Question 17.
(a) How do organic farmers control pests? Give two examples.
(b) State the difference in their approach from that of conventional pest control methods.
Answer:
(a) The organic farmers control pests by developing pest resistant in genetically modified plants. The desired gene of interest are introduced into the plant to make it resistant for pest. For example:

Bt cotton: The cotton plants were incorporated with Bt toxin gene of the bacterium Bacillus thur ingenes is. The gene which codes for toxic insecticidal proteins are called cry gene. Hence, crylAc and cryllAb control the cotton bollworm whereas crylAb controls com borer.

A nematode Meloidegyrte incognitia infects the roots of tobacco plant and causes reduction in yield. RNA interference (RNAi) involves silencing of a specific RNA. By using Agrobacterium vectors, nematode specific gene is introduced into host plant and the introduction of DNA produces both sense and antisense RNA into the host cells. The two RNAs are complementary to each other results in the formation of double-stranded RNA which initiates RNA interference mechanisms and provide protection against the pest.

(b) In conventional pest control method, pesticides are used for killing pest that causes harm to the crop. Pesticides are the toxic chemicals that cause harm to soil, crop as well as humans and animals. This results in reduction in crop yield and soil erosion.

Whereas organic farming does not involve the use of pesticides, hence causes less harm to nature.

Question 18.
(a) Name the selectable markers in the cloning vector pBR322? Mention the role they play.
(b) Why is the coding sequence of an enzyme galactosidase a preferred selectable marker in comparison to the one named above?
Answer:
(a) Selectable markers are used for the identification and elimination of non-transformants that selectively permits the growth of the transformants. The genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline and kanamycin are commonly used selectable markers.

Note
Transformation is a procedure through which a piece of DNA is introduced in a host bacterium.

(b) The selectable markers are developed to differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate.

In this, a recombinant DNA is inserted within the coding sequence of an enzyme, which is referred as insertional inactivation. The presence of a chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have insert.

The presence of insert results in insertional inactivation of the beta-galactosidase enzyme and the colonies does not produce any colour and are identified as recombinant colonies.

Question 19.
(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) It is necessary to make a cell competent in order to enhance the efficiency of the cell to take up the foreign DNA easily. As DNA being hydrophilic in nature, it cannot pass through the cell membranes. The cell can be made competent by treating it with a specific concentration of a divalent cation such as calcium which increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

(b) Biolistics or gene gun is suitable for plants, and in this method the cells are bombarded with high
velocity micro-particles of gold or tungsten coated with DNA.

Question 20.
Explain enzyme-replacement therapy to treat adenosine deaminase deficiency. Mention two disadvantages of this procedure.
Answer:
ADA (Adenosine deaminase deficiency) can be treated by enzyme replacement therapy in which functional ADA is given to the patient by injection. In this process:

• Lymphocytes from the blood of the patients are grown in a culture outside the body.
• A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes which are subsequently returned to the patient.

Disadvantages associated with enzyme-replacement therapy are:

• This method is not completely curative.
• The cells are immortal as the patients require periodic infusion of such genetically engineered lymphocytes.

Note
ADA deficiency is caused because of the deletion of the gene that codes for adenosine deaminase enzyme Severe combined Immunodeficieni v disorder (SCID) is caused because of the defect in gene which codes for adenosine deaminase enzyme.

Question 21.
Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.
Answer:
Mycorrhizae are association between fungi and the root nodules of higher plants. This type of population interaction is called mutualism. In this, the fungi help the plant in the absorption of essential nutrients from the soil whereas the plant in turn provides the fungi with energy-yielding carbohydrates.

The interaction that exist between cattle egret and cattle is known as commensalism. In this type of interaction, one species is benefitted whereas the other is neither benefitted nor harmed. The cattle, as they move, stir up and flush out insects from the vegetation that they graze, that otherwise might be difficult for the egrets to find and catch. The cattle receives neither harm nor benefit.

Note
Commensalism is a type of population interaction in which one species benefits and the other is neither harmed nor benefited. Mutualism is a type of population interaction in which both the interacting species are benefited from each other.

Question 22.
Differentiate between primary and secondary succession. Provide one example of each.
Answer:
Difference between Primary and Secondary succession:

Primary Succession	Secondary Succession
Primary succession occurs in an area where no living organisms ever existed.	Secondary succession occurs in a area where all the living organisms are lost that are existed there.
For example: Bare rock, ponds and deserts.	For example: The area affected by natural calamities, covered under deforestation.

Note
Ecological succession refers to the gradual and fairly predictable change in the species composition of a given area.

Section D

Question 23.
A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as a biology student can promote to check this social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with male partner.
Answer:
(a) Lack of proper education and unawareness about infertility are the main cause for blaming females in India.
People should be aware about several medical techniques that are available to overcome the infertility problems in both males and females. They must know not only females but also males can be affected by infertility

(b) Infertility is caused because of physical, congenital, diseases, drugs, immunological problems or even psychological problems.

(c) When infertility is caused because of inability of the male partner to inseminate the female or because of very low sperm count in the ejaculates is corrected by artificial insemination.
In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI-Intra-uterine insemination) of the female. (4 Marks)

Note
Infertility refers to the inability or failure of a couples to produce children in spite of unprotected sexual co-habitation.

Section E

Question 24.
(a) Explain the menstrual phase in a human female. State the levels of ovarian and pituitary hormones during this phase.
(b) Why is follicular phase in the menstrual cycle also referred as proliferative phase’? Explain.
(c) Explain the events that occur in a graafian follicle at the time of ovulation and thereafter.
(d) Draw a graafian follicle and label antrum and secondary oocyte.
OR
(a) As a senior biology student you have been asked to demonstrate to the students of secondary level in
your school, the procedure(s) that shall ensure cross-pollination in a hermaphrodite flower. List the different steps that you would suggest and provide reasons for each one of them.
(b) Draw a diagram of a section of a megasporangium of an angiosperm and label funiculus, micropyle, embryosac and nucellus.
Answer:
(a) The menstrual phase of the menstruation cycle starts when the menstrual flow occurs and it lasts for 3-5′ days. The menstrual flow results because of the breakdown of endometrial lining of the uterus and its blood vessels which forms liquid that comes out through vagina. Menstruation occurs if the released ovum is not fertilised.

Note
In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starting one menstruation till the next one is called the menstrual cycle.

(b) During the follicular phase, the primary follicles in the ovary grow to become a fully mature Graafian follicle. The endometrium of uterus regenerates through proliferation. This is reason that follicular phase is also called as proliferative phase.

The changes in the ovary and the uterus are induced by changes in the levels of pituitary and ovarian hormones. The secretion of gonadotropins (LH and FSH) increases gradually during this phase and also stimulates follicular development as well as secretion of estrogens by growing follicles. (2 Marks)

(c) The level of both LH and FSH increases and attain a peak level in the middle of cycle (about 14^th day). Rapid secretion of LH leads to its maximum level during the mid-cycle called LH surge induces rupture of Graafian follicle and thereby release of ovum (ovulation). The ovulatory phase or ovulation is followed by the luteal phase during which the remaining parts of the Graafian follicle transform as the corpus luteum.

The corpus luteum secretes a large amount of progesterone which is essential for maintenance of the endometrium. Endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy. During pregnancy all events of the menstrual cycle stop and there is no menstruation. In the absence of fertilisation, the corpus luteum degenerates and this causes disintegration of the endometrium results in menstruation. (1 Mark)

(d) Diagrammatic representation of Graafian follicle:

OR

(a) The procedure used to ensure cross-pollination in hermaphrodite flower are as follows:

• Emasculation: If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces by using a pair of forceps is called emasculation.
• Bagging: Emasculated flower is covered with a bag of suitable size generally made up of butter paper to prevent contamination of its stigma with unwanted pollen and this process is called bagging.

When the stigma of bagged flower attains receptivity, mature pollen grains collected from anthers of the male parent are dusted on the stigma and the flowers are rebagged and the fruits allowed to develop.

Note
The process of transfer of pollen grains to the stigma of a pistil is termed as pollination. The pollination can be divided into three types such as autogamy and geitonogamy are the type of self-pollination whereas xenogamy is a type of cross-pollination.

(b) Diagrammatic representation of megasporangium of an angiosperm:

Question 25.
Describe Meselson and Stahl’s experiment that was carried in 1958 on E. Coil. Write the conclusion they arrived after the experiment.
OR
(a) Describe the process of transcription in bacteria.
(b) Explain the processing the hnRNA needs to, undergo before becoming functional mRNA eukaryotes.
Answer:
Matthew Meselson and Franklin Stahl performed the following experiment in 1958 to prove that DNA replicates semiconservatively:

1. They grwo E. coli in medium containing ¹⁵NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. This result was that ¹⁵N was incorporated into newly synthesised DNA.
2. This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.
3. Then they transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at different time intervals as the cells multiplied and extracted the DNA that remained as double-stranded helices. The different samples were separated independently on CsCl gradients to measure the densities of DNA.
4. Thus, the DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid or intermediate density. DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of Tight’ DNA.

Note
Taylor and colleagues in 195S performed experiment by using radioactive thymidine to detect distribution of newly synthesised DNA in the chromosomes was performed on Vicia faba (faba beans). This experiment proved that the DNA in chromosomes also replicate semiconservatively.

OR

(a) Following are the steps depicting the process of transcription in bacteria.

1. Initiation: RNA polymerase binds to promoter and initiates the process of transcription. It uses nucleoside triphosphate as substrate and polymerises in a template depended manner.
2. Elongation: This is the second step involved in bacterial transcription that facilitates the opening of the helix and continues elongation of DNA duplex.
3. Termination: Once the RNA polymerase reaches the terminator region, the nascent RNA and RNA polymerase enzyme falls off, results in the termination of transcription.

The bacterial RNA polymerase requires initiation factor (sigma factor) to initiate the process of transcription and termination factor (Rho factor) to terminate the process of transcription.

Diagrammatic representation of transcription in bacteria:

(b) The hnRNA or heterogeneous nuclear RNA (hnRNA) is a precursor of mRNA transcribed by RNA polymerase II enzyme in eukaryotes.

The hnRNA is also called primary transcript and it contains both the exons and the introns. It is non
funtional.

So, the process of removal of introns from the exons in a hnRNA is called splicing and the exons are joined together by DNA ligase enzyme.

The process of capping involves addition of unusual nucleotide (methyl guanosine triphosphate) to the 5’-end of the hnRNA.

The process of tailing involves the addition of adenylate residues (200-300) are added at 3’-end in a
template independent manner.

The fully processed hnRNA is called mRNA which is transported Out of the nucleus for translation.

Note
In prokaryotes, the process of transcription and translation occurs simultaneously in the cytoplasm of the cell. Whereas in eukaryotes, the process of transcription takes place in the nucleus and translation takes place in the cytoplasm.

Question 26.
(a) Name the two growth models that represent population growth and draw the respective growth curves they represent.
(b) State the basis for the difference in the shape of these curves.
(c) Which one of the curves represent the human population growth at present? Do you think such a curve is sustainable? Give reason in support of your answer.

OR

(a) Taking an example of a small pond, explain how the four components of an ecosystem function as a unit.
(b) Name the type of food chain that exists in pond.
Answer:
(a) The two growth curves are exponential growth curves and logistic growth curves.
Diagrammatic representation of Growth curves:

Here, ‘a’ represents exponential growth curve whereas b represents logistic growth curve. (2 Marks)

(b) The exponential growth curve is formed when the availability of resources in the habitat are unlimited. A J-shaped curve is formed in exponential growth curve.
The exponential growth equation is N_t = N₀e^rt
where,
N_t = Population density at time t
N₀ = Population density at time zero
r = Intrinsic rate of natural increase
e = the base of natural logarithms

Logistic growth occurs when the available resources are limited results in competition between individuals for limited resources. So, the fittest individual will survive and reproduce.

A S-shaped or sigmoid growth curve is formed. The logistic growth equation is
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = rN\(\left[\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right]\)
where,
N = Population density at time(t)
r = Intrinsic rate of natural increase
K = Carrying capacity (2 Marks)

(c) The human population at present is represented by the logistic growth. No, the growth is not sustainable because with the growing population and the depleting natural resources, it would be difficult in the future to fulfill the demands of growing population.

Note
Carrying capacity (K) is defined as in a given habitat that has enough resources to support a maximum possible number, beyond which no further growth is possible.

OR

(a) The four components of an ecosystem are as follows:

• Productivity: The productivity is defined as the rate of biomass production. Productivity is represented by the autotrophic phytoplanktons, algae, floating and submerged plants.
• Decomposition: Decomposers such as fungi, bacteria, flagellates. The decomposers break down complex organic matter into inorganic substances such as carbon dioxide, water and nutrients.
• Energy flow: There is a unidirectional flow of energy from the sun to producers and then to consumers.
• Nutrient cycling: The movement of nutrient elements through the various components of an ecosystem. The pond ecosystem involves the process of conversion of inorganic substances into organic material with the help of solar energy.

(b) In the pond ecosystem, the aquatic food chain is the major food chain for the energy flow: