CBSE Sample Papers for Class 12 Biology Set 2 with Solutions

Students can access the CBSE Sample Papers for Class 12 Biology with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Biology Set 2 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• The question paper has five sections and 33 questions. All questions are compulsory.
• Section – A has 16 questions of 1 mark each; Section – B has 5 questions of 2 marks each; Section – C has 7 questions of 3 marks each; Section – D has 2 case-based questions of 4 marks each; Section – E has 3 questions of 5 marks each.
• There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
• Wherever necessary, neat and properly labeled diagrams should be drawn.

Section – A(16 Marks)

Question 1.
Viola plants represent the largest genus of cleistogamous flowers. Which of the following is NOT a requirement for Viola plants to assure propagation?
(A) Flowering
(B) Fertilisation
(C) Bisexual flowers
(D) External pollinators
Answer:
(D) External pollinators

Explanation: Cleistogamy is a type of automatic self-pollination of certain plants that can propagate by using non-opening, self-pollinating flowers.

Question 2.
Which condition of gynoedums (pistil) is shown here:

(A) (i) Multicarpellary apocarpous, (ii) Multicarpellary syncarpous
(B) (i) Multicarpellary syncarpous, (ii) Multicarpellary apocarpous
(C) (i) Bicarpellary apocarpous, (ii) Bicarpellary syncarpous
(D) (i) Bicarpellary syncarpous, (ii) Bicarpellary apocarpous
Answer:
(B) (i) Multicarpellary syncarpous, (ii) Multicarpellary apocarpous

Explanation: The picture (i) is Papaver. It has many carpels (multicarpellary) and the carpels are fused. Such a condition is called syncarpous condition.
The picture (ii) is Michelia. It has many carpels (multicarpellary) and the carpels are free. Such a condition is called as apocarpous condition.

Question 3.
Codons UCA, UCG, UCU, UCC, AGC, and AGU code for the amino add serine.
What do these codons indicate?
(A) That the genetic code is universal
(B) That the genetic code is degenerate
(C) That UCA is the initiator codon
(D) That all amino adds are coded by six codons.
Answer:
(B) That the genetic code is degenerate

Explanation: Degenerate code means that some amino adds are coded by more than one codon.

Question 4.
Which of the following pairs is wrongly matched?
(A) Starch synthesis in pea: Multiple alleles
(B) ABO blood grouping: Co-dominance
(C) Flower colour in Snapdragon: Incomplete dominance
(D) T.H. Morgan: Linkage
Answer:
(A) Starch synthesis in pea: Multiple alleles

Explanation: Codominance is a phenomenon in which both alleles of a pair express themselves fully in the F, hybrid. A good example of codominance is human ABO blood grouping in which the four main blood groups A, B, AB and O are controlled by three alleles.
Linkage is a phenomenon of genetic inheritance in which genes of a particular chromosome show their tendency to inherit together. Linkage was discovered by T.H. Morgan in Drosophila melanogaster.
Flower colour in Mirabilis jalapa (Snapdragon) is an example of incomplete dominance.

Question 5.
In Hardy Weinberg equation, the frequency of heterozygous individual is represented by:
(A) q²
(B) p²
(C) 2pq
(D) pq
Answer:
(C) 2pq

Explanation: According to Hardy Weinberg principle, the gene pool remains constant, i.e., p² + 2pq +q² = 1. Here, p is the frequency of the A allele and q is the frequency of the ‘a’ allele in the population. Thus, 2pq represents the frequency of the heterozygous genotype Aa.

Question 6.
Excreta preserved as a fossil is known as:
(A) Coprolite
(B) Stromatolite
(C) Compressed fossil
(D) None of the above
Answer:
(A) Coprolite

Explanation: Coprolites are fossilised remains of the contents of the intestine and the fossilised faeces. It gives dues about an animal’s diet.

Question 7.
How does an anticodon on a tRNA molecule help?
(A) It signals the stop of translation.
(B) It helps to initiate a 5′ → 3′ translation.
(C) It ensures accurate reading of the code on the mRNA.
(D) It enables translation in the opposite polarity to the mRNA.
Answer:
(C) It ensures accurate reading of the code on the mRNA.

Explanation: Each time an amino add is added to a growing polypeptide during protein synthesis, a tRNA anticodon pairs with its complementary codon on the mRNA molecule, ensuring that the appropriate amino add is inserted into the polypeptide. It ensures accurate reading of the code on the mRNA.

Question 8.
Given below is a Karyotype of a human foetus obtained for screening to find any probable genetic disorder:

Based on the karyotype, the chromosomal disorder detected the child may suffer from:
(A) Turner’s syndrome: Sterile ovaries, short stature
(B) Down’s syndrome: Gynecomastia, overall masculine stature.
(C) Turner’s syndrome: Small round head, flat back of head
(D) Down’s syndrome: Furrowed tongue, short stature
Answer:
(A) Turner’s syndrome: Sterile ovaries, short stature

Explanation: Turner syndrome is a genetic disorder due to aneuploidy of the sex chromosome. One X chromosome is missing in all the cells or some cells. Only females are born with this condition.

Question 9.
Which of the following is derived from the hemp plant “Cannabis sativa”?
(A) Opium
(B) Marijuana
(C) MDMA
(D) Crack
Answer:
(B) Marijuana

Explanation: Marijuana is a natural drug derived from the hemp plant, Cannabis sativa.

Question 10.
The image below shows the DNA fingerprinting evidence obtained from a crime scene by the forensic department.

Who is most likely guilty of the crime?
(A) Suspect 1
(B) Suspect 2
(C) Suspect 3
(D) The DNA fingerprint of the criminal is not there.
Answer:
(B) Suspect 2

Explanation: Susped 2 may be involved in the crime as his DNA fragment matches the DNA segment from the crime scene.

Question 11.
The site of production of ADA in the body is
(A) Erythrocytes
(B) Lymphocytes
(C) Blood plasma
(D) Osteocytes
Answer:
(B) Lymphocytes

Explanation: ADA or adenosine deaminase is an enzyme produced by all cells in the body, however it is majorly produced by the lymphocyte cells that undergo maturation in the thymus gland and other lymph nodes. These cells defend the body against pathogen.

Question 12.
Productivity is the rate of production of biomass expressed in terms of:
(i) (kcal m^-3) yr^-1
(ii) g^-2yr^-1
(iii) g^-1yr^-1
(iv) (kcal m^-2) yr^-1
(A) (ii)
(B) (iii)
(C) (ii) and (iv)
(D) (i) and (iii)
Answer:
(C) (ii) and (iv)

Explanation: The rate of synthesis of energy containing organic matter by any trophic level per unit area in unit time is described as its productivity. It is measured as weight (e.g., g^-2yr^-1) or energy (e.g., kcalm^-2)yr^-1.

Question No. 13 to 16 consist of two statements- Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Question 13.
Assertion (A): Apomictic fruits are seedless.
Reason (R): Apomictic fruits are formed without fertilisation.
Answer:
(D) A is false but R is true.

Explanation: Parthenocarpy is the development of fruit without prior fertilisation and is seedless. This can be used to produce seedless fruit like grapes, bananas, etc. whereas apomixis is the development of seed without prior fertilisation that mimics sexual reproduction.

Question 14.
Assertion (A): The newly formed mRNA has the same sequence as the coding strand of transcriptional unit with uracil present in place of thymine.
Reason (R): The rule of complementarily guides the formation of DNA and RNA.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: The sequence of mRNA will be identical to the given sequence of coding strand except for the presence of uracil in place of thymine in mRNA. The rule of complementarity guides the formation of DNA and RNA.

Question 15.
Assertion (A): Rhizobium leguminosarum is a symbiotic bacterium found in the root nodules of leguminous plants.
Reason (R): They have the ability to fix atmospheric nitrogen.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: Rhizobium leguminosarum is symbiotic bacterium that serves as a biofertilizers. These bacteria fix atmospheric nitrogen into organic forms, which is used by the plant as nutrient.

Question 16.
Assertion (A): Genetic codes are commaless.
Reason (R): Genetic codes are overlapping.
Answer:
(C) A is true but R is false.

Explanation: A commaless genetic code means that no punctuations are needed between any two Words. The genetic code is non-overlapping. In actual practice six bases code for not more than two amino acids.

Section – B(10 Marks) 

Question 17.
A pregnant human female was advised to undergo MTE It was diagnosed that the fetus she was carrying had developed from a zygote having 45 chromosomes with only one X chromosome.
(a) What is this condition called and how does it arise?
(b) Why was she advised to undergo MTP?
Answer:
(a) The embryo has Turner’s syndrome due to aneuploidy of the sex chromosome. Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO.
(b) She was advised MTP as the child will have the following problems:
(i) Rudimentary ovaries
(ii) Poorly developed breasts
(iii) Lack of other secondary sexual characters
(iv) Delayed or no onset of the menstrual cycle and infertile.

Question 18.
Study the figure below and answer the following questions:
(a) Name the process depicted in the figure. Define it.
(b) Name the subunits of Ribosome

Answer:
(a) This process is known as translation. Translation involves decoding a messenger RNA (mRNA) to be translated into amino acids. Amino acids are linked together to build a polypeptide.
(b) Ribosomes consist of two major components: the small ribosomal subunit, which reads the RNA, and the large subunit, which joins amino acids to form a polypeptide chain.

Question 19.
Study the diagram showing replication of HIV in humans and answer the following questions accordingly.

(a) Write the chemical nature of the coat A’.
(b) Name the enzyme ‘B’ acting on ‘X’ to produce molecule ‘C’. Name ‘C’.
(c) Mention the name of the host cell ‘D’ the HTV attacks first when it enters into the human body.
(d) Name the two different cells the new viruses ‘E’ subsequently attacks.
Answer:
(a) The chemical nature of the coat: Viral protein coat.
(b) Enzyme B- Viral DNA is produced by reverse transcriptase.
X- Viral RNA is introduced into cell.
C- Viral DNA.
(c) Host cell (D)- Macrophage.
(d) Two different cells are macrophages and helper T-lymphocytes.

Question 20.
Given below is the diagram representing the observations made for separating DNA fragments by Gel electrophoresis technique? Observe the illustration and answer the questions that follow.

(a) Why are the DNA fragments seen to be moving in the direction A → B?
(b) Write the medium used on which DNA fragments separate.
(c) Mention how the separated DNA fragments can be visualised for further technical use.
Answer:
(a) Bistheanodeend. DNA fragments are negatively charged thereby moving towards the anode which is a positive rod, under the influence of an electric field during gel electrophoresis.
(b) Most commonly used matrix is agarose. Agarose is a natural polymer extracted from seaweeds.
(c) Ethidium bromide is used as a stain for DNA, which on exposure to UV-light appears as orange coloured bands.

Question 21.

An ecologist studied an area with population A, thriving on unlimited resources and showing exponential growth, and introduced population B and C to the same area. What will be the effect on the growth pattern of the population A, B and C when living together in the same habitat?
Answer:
This interaction will lead to competition between the individuals of population A, B and C for resources. Eventually, the ‘fittest’ individuals will survive and reproduce. The resources for growth will become finite and limiting, and population growth will become realistic.

OR

If in a population of size ‘N’ the birth rate is represented as ‘b’ and the death rate as ‘d’, the increase or decrease in ‘N’ during a unit time ‘t’ will be:
dN/dt = (b – d) × N
The equation given above can also be represented as:
dN/dt = r × N where r = (b – d)
What does V represent? Write any one significance of calculating V for any population.
Answer:
r = Intrinsic rate of natural increase; it is an important parameter for assessing the impacts of any biotic or abiotic factor on population growth.

Section – C(21 Marks)

Question 22.
Given below is a flow chart showing ovarian changes during the menstrual cycle. Fill in the spaces with the hormonal factor/s responsible for the events shown.

Answer:
a- FSH and Estrogen
b- LH
c- Progesterone.

Question 23.
During the reproductive cycle of a human female, when, where and how does a placenta develop? What is the function of the placenta during pregnancy and embryo development?
Answer:
During the reproductive cycle of a human female, placenta develops after the implantation of zygote in the uterus. After implantation, the cells of inner cell mass of blastocyst differentiate to form the embryo proper. The trophoblast differentiates into two layers, the outer layer secretes enzymes to dissolve the endometrium of uterus. The inner layer grows out as finger like projections called chorionic villi into the uterine stroma. They are surrounded by the uterine tissue and maternal blood vessels. The chorionic villi and the uterine tissue become interdigitated to form the structural functional unit called placenta.

Functions of placenta:
(i) The placenta secretes hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens and progesterone, that are necessary to maintain pregnancy.
(ii) Placenta also facilitates the supply of oxygen and provide is nutrients to the embryo through the umbilical cord.

Question 24.
Describe the three manners in which fertilisation of human ovum by a sperm can be prevented.
Answer:
In barrier methods, ovum and sperm are prevented from physically meeting. For examples:
(i) Condoms: They are made of thin rubber/latex sheath used to cover the penis in the male or vagina and cervix in the female just before coitus so that the ejaculated semen is not released in the female reproductive tract.
(ii) Diaphragms, cervical caps and vaults: These are also made of rubber inserted into the female reproductive tract to cover the cervix during coitus. They prevent fertilisation by blocking the entry of sperms through the cervix.
(iii) Intra Uterine Devices (IUDs): These devices are inserted by doctors in the uterus through vagina. These are available as non-medicated IUDs (Uppers loop), copper releasing IUDs (CuT Cu 7) and hormone releasing IUDs (LNG – 20). IUDs increase phagocytosis of sperms within the uterus and suppress sperm motility.

Question 25.
Study the figures given below and answer the following questions.

(a) Under the influence of which type of natural selection would graph (a) become like graph (b).
(b) What could be the likely reason of new variations arising in a population.
(c) Who suggested natural selection as mechanism of evolution?
Answer:
(a) Disruptive selection
(b) Because individuals at the extremes contribute more offspring compared to those in the centre and produces two peaks in distribution of a trait.
(c) Charles Darwin

Question 26.
Identify a, b, c, d, e and f in the table given below:

Answer:
a -Statins
b- Blood cholesterol lowering agent
c- Penicillium notatum
d- Penicillin
e- Trichoderma polysporum
f- Immuno-suppressive agent in organ transplantation.

Question 27.
You have created a recombinant DNA molecule by ligating a gene to a plasmid vector. By mistake your friend adds an exonuclease enzyme to the tube containing the recombinant DNA.
(a) How will your experiment get affected as you plan to go to the transformation now?
(b) What is EcoRI?
(c) How does EcoRI differ from an exonuclease?
Answer:
(a) The experiment is not likely to be affected as the recombinant DNA molecule is circular and closed, with no free ends. Hence, it will not be a substrate for exonudease enzyme, which removes nudeotides from fire free ends of DNA.
(b) EcoRI is a restriction endonudease enzyme.
(c) Exonudease removes nucleotides from the
ends of DNA.
EcoRI make cuts at a specific position within the DNA.

OR

Cloning vector is a small DNA molecule capable of self-replication inside the host cell. It is used for replicating donor DNA fragment within host cell.
(a) Name any two natural cloning vectors.
(b) Give reasons that make them act as cloning vectors.
(c) Write the two characteristics, the engineered vectors are made to possess.
Answer:
(a) Plasmids, bacteriophages
(b) Ability to replicate within bacterial cells, high copy number within the bacterial cells.
(c) Characteristics of engineered vectors: easy linking of foreign DNA, Selection of recombinants from non-recombinants / selectable marker.

Question 28.
(a) ‘The Evil Quartet’ describes the rate of species extinction due to human activities. Explain how the population of organisms is affected by fragmentation of the habitats.
(b) Introduction of alien species has led to environmental damage and decline of indigenous species. Give any one example of how it has affected the indigenous species?
(c) Could the extinction of Steller’s sea cow and passenger pigeon be saved by man? Give reasons to support your answer.
Answer:
(a) When a large habitat is broken into small fragments due to various activities, mammals and birds requiring large territories and certain animals with migratory habitats are badly affected, leading to population decline.
(b) Nile perch introduced in Lake Victoria eventually led to the extinction of an ecologically unique assemblage of more than 200 species of cichlid fish.
Ptirthenium/Lantana/water hyacinth caused environmental damage and threat to our native species.
African catfish-Clarias gariepinus introduced for aquaculture purposes is posing a threat to file indigenous catfish in our rivers.
(c) Yes, humans have overexploited natural resources for their ‘greed’ rather than “need” leading to extinction of these animals. Sustainable harvesting could have prevented extinction of these species.

Section – D(8 Marks)

Question 29,
A schematic representation of polymerase chain reaction (PCR) up to the extension stage is given below:
(a) Name the process ‘a’.
OR
Describe the role of heat in the process of PCR.
(b) Identify V.
(c) How many DNA duplex is obtained from one DNA duplex after 3 cycles of PCR?

Answer:
(a) Denaturation is the first step of PCR which involves the breaking of phosphate bond between the DNA base pairs at temperature 80°C.
OR
Role of Heat It helps in the denaturation process in PCR. The dsDNAs are heated in this process at a very high temperature 95°C so that both the strands separated.
(b) The primer binds on the DNA that initiates the polymerisation with the Ijelp of Taq polymerase enzyme at 72°C temperature.
(c) After each cycle, the number of duplexes doubles itself. Thus, after, the first cycle there are two DNA duplexes. A duplex has two DNA strands thus after second cycle there will be four duplexes, after third cycle there will be eight DNA duplex.

Question 30.
Population is a group of individuals of a single species living in a given area. It includes spatial arrangement, genetic variation and the dynamics including natality, mortality, biotic potential, carrying capacity, etc. Population structure includes features such as density, abundance, spacing, and genetic variations. The genetic structure reveals genetic variations and population size.
(a) Write any two characteristics, which only a population shows but an individual cannot.
(b) List any two ways of measuring the population density of a habitat.
(c) Mention the essential information that can be obtained by studying the population density of an organism.
OR
(c) Explain the death rate in a population.
Answer:
(a) Characteristics of population, not exhibited by individual are:
(i) Population size or density
(ii) Population interactions.
(b) (i) Percent cover for trees with larger canopy.
(ii) Number of fish caught per trap.
(c) The population density tells us about the status of a species, i.e., the outcome of competition, the impact of predation or effect of pesticides, etc.
OR
(c) Two types of the growth pattern of population are exponential and logistic growth.
When resources in the habitat are unlimited each species has the ability to realise fully its innate potential to grow in number. Then the population grows in exponential fashion.
Logistic growth pattern follow the sigmoid curve.

Section – E(15 Marks)

Question 31.
Apomixis (asexual seed formation) is the result of a plant gaining the ability to bypass the most fundamental aspects of sexual reproduction: meiosis and fertilisation. Without the need for male fertilisation, the resulting seed germinates a plant that develops as a maternal clone. This dramatic shift in the reproductive process has been documented in many flowering plant species, although no major seed crops have been shown to be capable of apomixis. The ability to generate maternal clones and therefore rapidly fix desirable genotypes in crop species could accelerate agricultural breeding strategies. The potential of apomixis as a next-generation breeding technology has contributed to increasing interest in the mechanisms controlling apomixis. In this review, we discuss the progress made toward understanding the genetic and molecular control of apomixis. Research is currently focused on two fronts. One aims to identify and characterise genes causing apomixis in apomictic species that have been developed as model species. The other aims to engineer or switch the sexual seed formation pathway in non-apomictic species, to one that mimics apomixis.

Study the above passage and answer the following questions.
(a) Explain any two ways by which apomictic seed can develop.
(b) List one advantage and one disadvantage of an apomictic crop.
(c) Why do farmers find the production of hybrid seeds costly?
Answer:
(a) In some species, the diploid egg cell is formed without reduction division and develops into the embryo without fertilisation. More often, as in many citrus and mango varieties, some of the nucellar cells surrounding the embryo sac start dividing, protrude into the embryo sac and develop into the embryos.

(b) Advantage: Apomixis is an effective means of rapid production of pure lines and provides an easy way of hybrid seed production.
Disadvantage: It reduces genetic diversity from parent to offspring plants due to a lack of variations in asexual reproduction.

(c) One of the problems of hybrids is that hybrid seeds have to be produced every year. If the seeds collected from hybrids are sown, the plants in the progeny will segregate and do not maintain hybrid characters. Production of hybrid seeds is costly and hence the cost of hybrid seeds becomes too expensive for the farmers.

OR

Angiosperms belong to one of the most diverse and largest extant groups of plants found in the universe. There are approximately 453 families of angiosperms that contain around 260,000 living species classified in them. Moreover, around 80 percent of all known green plants living on the earth are represented by angiosperms. It is very well elaborated that the vascular seed plants in which an egg is fertilised and developed into a seed in an enclosed hollow ovary are angiosperms. A unique evolutionary feature of angiosperms is double fertilization to yield embryo and endosperm through maternal and paternal contributions.
(a) Describe the process of double fertilisation in angiosperms.
(b) Trace the development of polyploid cell that is formed after double fertilisation in a non-albuminous seed and albuminous seed
Answer:
(a) Double fertilisation is a unique event to flowering plants. When the pollen grain falls on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of the synergids and releases two male gametes. Out of the two male gametes, one gametes fuses with the nucleus of the egg cell and forms the zygote. The process is known as syngamy. The other male gamete fuses with the two polar nuclei located in the central cell to form a triploid primary endosperm nucleus (PEN). Since, the process involves the fusion of three haploid nuclei, it is known as triple fusion. Since two kinds of fusions (syngamy and triple fusion) take place in an embryo sac it is known as double fertilisation.

(b) The triploid endosperm nucleus formed during double fertilisation leads to the formation of the triploid primary endosperm cell (PEC). This PEC undergoes repeated mitotic divisions to form a multicellular triploid endosperm. This serves as the nutritive tissue for the budding embryo within the seed.
In the case of albuminous seeds, the endosperm is retained and in many cases, a rudimentary cotyledon often termed as Scutellum is derived from it in case of monocots.
In case of non-albuminous seeds, this multicellular triploid endosperm is almost completely utilised or digested or absorbed to aid in the formation of fleshy cotyledon tissue as in case of most dicot seeds

Question 32.
Alfred Hersheyand Martha Chase started looking for the genetic material in organisms in 1952. Their experiments proved without a doubt that DNA is made up of genetic material.
Bacteriophages, which are viruses that kill bacteria, were the most important part of the experiment by Hershey and Chase. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer:
Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.
They grew some bacteriophages on a medium containing radioactive phosphorus (³²P) to identify DNA and some on a medium containing radioactive sulphur (³⁵S) to identify the protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated from the bacterial cell by blending and then subjected to the process of centrifugation.
Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.

OR

(a) Identify strands A’ and ‘B’ in the diagram of the transcription unit given above and write the basis on which you identified them.
(b) State the functions of the Sigma factor and Rho factor in the transcription process in a bacterium.
(c) Write the functions of RNA polymerase-I and RNA polymerase-in in eukaryotes.
Answer:
(a) A- Template strand
B- Coding strand
Template strand has polarity 3′ → 5′
Coding strand has polarity 5′ → 3′
On the basis of polarity with respect to promoter.

(b) Sigma factor associates with RNA polymerase to initiate transcription, Rho factor gets associated with RNA polymerase to terminate transcription.

(c) RNA polymerase I – Transcribes -RNAs
RNA polymerase III – Transcribes tRNA / 5srRNA / 5snRNA.

Question 33.
Describe the process of waste-water treatment under the following heads:
(a) Primary treatment.
(b) Secondary treatment.
Answer:
(a) Primary treatment

• Physical removal of particles like debris, soil,sand or silt through filtration, sedimentation in stages.
• Solids settle to form primary sludge, the supernatants form the primary effluent.

(b) Secondary Treatment
It is the biological treatment

• Effluent passed into aeration tanks.
• Vigorous growth of useful aerobic microbes into flocs.
• Significant reduction of BOD due to use of organic matter by microorganisms.
• After fall in the level of BOD, the effluent is Passed on to settling tanks where bacterial flocs settle to form activated sludge.
• Activated sludge is passed on to anaerobic sludge digester, where bacteria and fungi are aaerobically digested.

OR

(a) How does Bacillus thuringiensis act as a biocontrol agent for protecting Brassica and fruit trees? Explain.
(b) List the components of biogas.
(c) What makes methanogens a suitable source of biogas production?
Answer:
(a) Bacterium Bacillus thuringiensis (Bt) are available in sachets as dried spores, mixed with water and sprayed onto vulnerable plants, these are eaten up by the insect larvae, the toxins are released in the gut and larva gets killed.
(b) Methane, H₂ S, CO₂, H₂.
(c) Methanogens grow anaerobically on cellulosic material, produce large amount of methane, alongwith CO₂ & H₂.