CBSE Sample Papers for Class 12 Biology Set 5 with Solutions

Students can access the CBSE Sample Papers for Class 12 Biology with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Biology Set 5 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.
The aquatic plant having long and ribbon like pollen grains is:
(A) Vallisneria
(B) Hydrilla
(C) Eichhomia
(D) Zostera
Answer:
(D) Zostera

Explanation: In Zostera (sea grasses), the female flowers remain submerged in water and the long, ribbon like pollen grains are carried inside the water to reach the stigma.

Question 2.
In some species of flowering plants, fruits develop without the process of fertilisation.
Which of these can be an identifying factor of such fruits?
(A) They are not developed from the ovary.
(B) They are always composite.
(C) They are always seedless.
(D) They are not juicy.
Answer:
(C) They are always seedless.

Explanation: The fruits formed without fertilisation of the ovary are called parthenocarpic fruits. These fruits are seedless such as banana.

Question 3.
Which one of the following diagrams correctly represents DNA replication in eukaryotes?

Answer:

Explanation: DNA replication take place in the 5′ to 3′ direction because DNA polymerase acts on the 3′-OH of the existing strand for adding free nucleotides.

Question 4.
Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group, respectively. What would be the genotype of both mother and father?
(A) Mother is homozygous for A’ blood group and father is heterozygous for ‘B’.
(B) Mother is heterozygous for A’ blood group and father is homozygous for ‘B’.
(C) Both mother and father are heterozygous for A and ‘B’ blood group, respectively.
(D) Both mother and father are homozygous for A and ‘B’ blood group, respectively.
Answer:
(C) Both mother and father are heterozygous for A and ‘B’ blood group, respectively.

Explanation: In this question the mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group, respectively. This can be possible only when both the mother (I^A i) and the father (I^B i) are heterozygous.

Question 5.
The probability of all possible genotypes of offsprings in a genetic cross can be obtained with the help of
(A) Test cross
(B) Back cross
(C) Punnett square
(D) Linkage cross
Answer:
(C) Punnett square

Explanation: Punnett square is a graphic representation of the probabilities of all the possible genotypes and phenotypes of offsprings in a cross.

Question 6.
The correct statement with respect to thalassemia in human is:
(A) α-Thalassemia is controlled by a single gene HBB.
(B) The gene for α-Thalassemia is located on chromosome-16.
(C) β-Thalassemia is controlled by two closely linked genes HBA-1 and HBA-2.
(D) In β-Thalassemia the production of alpha globin chain is affected.
Answer:
(B) The gene for α-Thalassemia is located on chromosome-16.

Explanation: α-Thalassemia is controlled by two dosely linked genes HBA1 and HBA2 on chromosome 16. In α-thalassemia, production of α globin chain is affected while in β-thalassemia, the production of β globin chain is affected. β-thalassemia is controlled by a single gene HBB on chromosome 11.

Question 7.
A single gene is usually seen to exhibit a single phenotypic expression. Pleiotropic genes, however, can impact multiple phenotypic expressions.
Which of these genetic disorders have multiple phenotypic expressions being impacted by a change in a single gene?
(A) thalassemia
(B) haemophilia
(C) phenylketonuria
(D) Down’s syndrome
Answer:
(C) phenylketonuria

Explanation: A pleiotropic gene is a single gene which exhibits multiple phenotypic expression. An example of this is die disease phenylketonuria, which occurs in humans. The disease is caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase (single gene mutation).

Question 8.
Study the pedigree analysis of humans given here and identify the type of inheritance along with an example:

(A) Sex-linked recessive, Haemophilia
(B) Sex-linked dominant, Vitamin D resistant rickets
(C) Autosomal recessive, Sickle-cell anaemia
(D) Autosomal dominant, Myotonic dystrophy
Answer:
(D) Autosomal dominant, Myotonic dystrophy

Explanation: Autosomal dominant are the traits whose encoding gene is present on any one of the autosomes, and the wild type allele is recessive to its mutant allele, which means the mutant allele is dominant. For example, myotonic dystrophy.

Question 9.
The decrease in the T-lymphocytes count in human blood will result in:
(A) decrease in antigens
(B) decrease in antibodies
(C) increase in antibodies
(D) increase in antigens
Answer:
(B) decrease in antibodies

Explanation: T-lymphocytes help B cells to secrete antibodies. Hence, the decrease in the T-lymphocytes count in the human body will result in decrease in antibodies.

Question 10.
Name the enzymes ‘P’ and ‘Q’ that are involved in the processes given below.

(A) Enzyme P- Exonudease and Enzyme Q-Fermease.
(B) Enzyme P- Exonudease and Enzyme Q- Ligase.
(C) Enzyme P- Endonuclease and Enzyme Q- Permease.
(D) Enzyme P- Restriction endonuclease and Enzyme Q- Ligase.
Answer:
(D) Enzyme P- Restriction endonuclease and Enzyme Q- Ligase.

Explanation: EnzymePisa restriction endonudease that cuts die DNA into fragments while enzyme Q is ligase that joins the two fragments.

Question 11.
Given below are the list of the commercially important products and their source organisms, Select the option that gives the correct matches.
Table
Options:
(A) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
(B) (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
(C) (A)-(iv), (B)-(iii), (C)-(ii), P)-(i)
(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
Answer:
(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Explanation: Cydosporin A, a bioactive molecule is produced by the fungus Trichoderma polysporum. Statins are by the yeast Monascus purpureas, Streptokinase is produced by the baderium Streptococcus and Penicillin, an antibiotic is produced by Penicillium notatum.

Question 12.
Given below, js a food web representative of the Arctic region.

Increasing temperatures have been causing changes in the ocean ecosystem. These changes have caused the population of Arctic cod to decline rapidly.
Which of the following statements is/are most likely to be TRUE based on this information?
P. The population of arctic birds will increase.
Q. The ringed seal will slowly become extinct.
R. The harbour seal will be dependent on capelins alone.
(A) only P
(B) only R
(C) only Q and R
(D) all – P Q and R
Answer:
(C) only Q and R

Explanation: As there is a decrease in the population of Arctic cod due to increased temperature, the ringed seal which feeds only on it will slowly become extinct. The decline in the population of Arctic cod will also affect Harbour seal which now will be dependent only on capelins.

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): Tapetum is formed dining the process of the formation of microsporangium.
Reason (R): It plays an important role in guiding the pollen tube into the synergid.
Answer:
(C) A is true but R is false.

Explanation: Tapetum plays an important role in nourishing pollen mother cells (FMCs) or microspores.

Question 14.
Assertion (A): RNA polymerases are able to catalyse all three steps of translation.
Reason (R): RNA polymerases contain an initiation factor and a termination factor in them.
Answer:
(D) A is false but R is true

Explanation: DNA-dependent RNA polymerase is the “only enzyme” that has the “capability” to catalyse initiation, elongation, and termination in the process of transcription in prokaryotes. Initiation factor sigma helps in the process of initiation while the termination factor Rho help in the termination of the process.

Question 15.
Assertion(A): The colostrum provides passive immunity to the new-born baby.
Reason(R): In this, the ready-made antibodies are directly given to protect the body.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: Colostrum (mother’s first milk) contains good amount of readymade antibodies like IgA, which provide passive immunity to the new bom and protect it from various infections.

Question 16.
Assertion (A): Gene therapy is a method of treating a disorder but cannot cure it.
Reason (R): Cells are drawn from a patient and the functional gene is introduced into these cells and transferred back to the patient.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

Explanation: Gene therapy is the insertion of genes into an individual’s cells and tissues to treat diseases especially hereditary diseases. It does so by replacing a defective mutant allele with a functional one or gene targeting which involves gene amplification.

Section – B(10 Marks)

Question 17.
The graph given below shows the number of primordial follicles per ovary in women at different ages. Study the graph and answer the questions that follow.

(a) What is the average age of the women at the onset of menopause?
(b) At what age are maximum primordial follicles present in the ovary according to the given graph?
Answer:
(a) The average age of women at the on set of menopause is 45 years.
(b) According to the given graph, at the age of 10, maximum primordial follicles are present in the ovary.

Question 18.
Study the schematic representation of the genes involved in the lac operon given below and answer the questions that follow:

p

i

p

o

z

y

a

(a) Identify and name the regulatory gene in this operon. Explain its role in ‘switching off’ the operon.
(b) Why is lac operon’s regulation referred to as negative regulation?
Answer:
(a) T gene is the regulatory gene.
Role of the regulatory gene in switching off operon is:
(i) Regulatory gene codes for the repressor protein ofthe operon,which is synthesised in constant amounts.
(ii) The repressor has an affinity for the operator gene. It binds to the operator and prevents the RNA polymerase from transcribing the structural genes.
(b) Regulation by lac operon is referred to as negative regulation because the repressor binds to the operator for ‘switching off’ the operon.

Question 19.
In the given flow diagram, the replication of retrovirus in a host is shown.

Observe and answer the following questions.
(a) Fill in 1 and 2.
(b) Why is the virus called retrovirus?
Answer:
(a) 1: Viral DNA is produced; 2: Viral RNA is produced by host cell.
(b) The virus is called retrovirus because it does not follow the central dogma (DNA → RNA → Protein). Its genetic material is RNA which is transcribed to DNA by using the enzyme reverse transcriptase.

Question 20.
Nidhi performed gel electrophoresis after treating one vector with restriction enzymes. She added one mixture in well Q and another mixture in well R. Given below is an image of the results.

(a) What can be concluded about the mixtures loaded in wells P and Q?
(b) What is the likely reason that the fragments in wells Q and R are different?
Answer:
(a) Well P contains the uncut vector whereas well Q contains the vector cut by a restriction enzyme.
(b) The vectorin well Q has beencutbya restriction enzyme that has two sites whereas the vector in well R has been cut either by different enzymes or by one enzyme that has more than two sites.

Question 21.

(a) Given here is a pyramid of biomass in an ecosystem where each bar represents the standing crop available in the trophic level. With the help of an example explain the conditions where this kind of pyramid is possible in nature?
(b) Will the pyramid of energy be also of the same shape in this situation? Give a reason for your response.
OR
Construct a pyramid ofbiomass starting with phytoplankton, Label its three trophic levels. Is the pyramid upright or inverted? Justify your answer.
Answer:
(a) Inverted pyramids of biomass are seen:
(i) In aquatic conditions where a small standing crop of phytoplankton supports a large standing crop of zooplankton/fish.
(ii) In terrestrial ecosystem where a large number of insects are feeding on the leaves of a tree.
(b) No, the pyramid of energy is always upright, and can never be inverted because when energy flows from one trophic level to the next trophic level, some amount of energy is always lost as heat at each step.

OR

The pyramid is inverted because the biomass of fishe is much more than that of the zooplankton and phytoplankton.

Section – C(21 Marks)

Question 22.
Study the transverse section of the human ovary given below and answer the questions that follow:

(a) Name the hormone that helps in the growth of A → B → C.
(b) Name the hormone secreted by A and B.
(c) State the role of the hormone produced by D.
Answer:
(a) Follicle stipulating hormone (FSH) secreted under the influence of releasing hormone from the hypothalamus stimulates the development of primary follicles into Graafian follicles. LH leads to the growth of follicle and the secretion of estrogen.
(b) Hormone secreted is estrogen.
(c) Progesterone hormone helps in the maintenance and preparation of the endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of LH and FSH, therefore inhibiting further ovulation

Question 23.
Human Genome Project (HGP) was a mega project launched in the year 1990 with some important goals.
(a) Enlist any four prime goals of HGP
(b) Name any one common non-human animal model organism which has also been sequenced thereafter.
Answer:
(a) Prime goals of HGP are:
(i) Identify all the approximately 20,000-25,000 genes in the human DNA.
(ii) Determining the sequences of the 3 billion chemical base pairs that make up human DNA.
(iii) Store this information in databases.
(iv) Improve tools for data analysis.
(b) Drosophila

Question 24.
List any three types of IUDs that are available for human females and state their mode of action.
Answer:
(a) Non-medicatfid lUDs, increase phagocytosis of sperms within the uterus.
(b) Copper releasinglUDs, Cu ions suppress sperm motility and fertilising capacity of sperms.
(c) Hormone releasing lUDs, make the uterus unsuitable for implantation/makes cervix hostile to sperms.

Question 25.
With the help of an algebraic equation, how did Hardy-Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations?
Answer:
Hardy-Weinberg principle states that allele frequencies in a population are stable and is constant from generation to generation unless disturbances such as mutations, genetic drift, natural selection, etc., are introduced.
The sum total of all the allelic frequencies is one.
In a diploid, p and q represent the frequency of allele A and allele a. The frequency of AA individuals in a population is simply p², that of aa is q² and that of Aa is 2pq.
Hence, p² + 2pq + q² = 1.
When the frequency measured is different from that expected, it is indicative of evolutionary change.

Question 26.
The primary effluent in the treatment of sewage is sent to tanks for secondary treatment in the presence of aerobic bacteria.
(a) How would the BOD of the effluent be affected if anaerobic bacteria are used for secondary treatment?
(b) Name one condition that should be maintained in a sludge digester where biogas is produced.
(c) The slurry formed after biogas production is recommended as manure for plants. Which nutrients will the slurry be rich in and why?
Answer:
(a) Anaerobic bacteria would not use the oxygen in the wastewater, thus the BOD would not reduce.
(b) Anaerobic conditions need to be maintained as biogas-producing bacteria are strict anaerobes.
(c) The slurry will mainly be rich in nitrogen and phosphorus as the carbon in the sludge would be used in the formation of methane and carbon dioxide.

Question 27.
Selectable markers are the genes present in the cloning vectors. It helps in eliminating the non-transformants. It primarily aid in the detection and eradication of transformants as well as enabling their proliferation. A selectable marker is used in the section of recombinants on the basis of their ability to produce colour in presence of chromogenic substrate.
(a) Mention the name of mechanism involved.
(b) Which enzyme is involved in production of colour?
(c) How is it advantageous over using antibiotic resistant gene as a selectable marker?
Answer:
(a) Insertional inactivation.
(b) p-Galactosidase
(c) Selection of recombinants due to inactivation of antibiotics requires simultaneous platingon two plates having different antibiotics.

OR

On spraying Bacillus thuringiensis on an infected cotton crop field the pests are killed by the toxin, however the toxin although produced by the bacteria does not affect it. Explain by giving reason.
Answer:
Bacillus thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. These proteins are present in inactive protoxin form, but become active toxin in the alkaline pH of the insect gut.
The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis and eventually cause death of the insect. Hence, Bt toxin does not kill the bacterium that produces it, but kills the insect that ingests it.

Question 28.

(a) The given image shown below is of a sacred grove found in India. Explain how human involvement has helped in the preservation of these biodiversity-rich regions.
(b) Value of Z (regression coefficient) is considered for measuring the species richness of an area. If the value of Z is 0.7 for area A, and 0.15 for area B, which area has higher species richness and a steeper slope?
Answer:

(a) India’s history of religious and cultural traditions emphasised the protection of nature. In many cultures, tracts of forest are set aside; all the trees and wildlife within are venerated and given total protection. Sacred groves in many states are the last refuges for a large number of rare and threatened plants.
(b) Area A will have more species richness and a steeper slope.

Section – D(8 Marks)

Question 29.
Carefully observe the given picture. A mixture of DNA with fragments ranging from 200 base pairs to 2500 base pairs was electrophoresed on agarose gel with the following arrangement.

(a) What result will be obtained on staining with ethidium bromide? Explain with reason.
(b) The above set-up was modified and a band with 250 base pairs was obtained at X.

(i) What change(s) were made to the previous design to obtain a band at X?
(ii) Why did the band appear at position X?
Answer:
(a) No bands will be obtained as all DNA will be seen in the well only;
DNA fragments being negatively charged will not move towards negative end/ cathode.
DNA being negatively charged will remain stationed at the positive end/ anode end of the agar block.
(b) (i) Position of the positive terminal/ end/anode and the negative terminal/ end/ cathode was inter-changed.
(ii) The fragment with the least base pairs will get separated faster and move faster to the anode end.

OR

(b) How are DNA fragments visualised during gel electrophoresis? What is elution?
(c) Why ethidium bromide is used in gel electrophoresis?
Answer:
(b) Separated DNA fragments stained with ethidium bromide, followed by exposure to UV radiations, removal of DNA bands from agarose gel and its extraction from the gel is elution.
(c) Ethidium bromide is sometimes added to running buffer during the separation of DNA fragments by agarose gel electrophoresis. It is a fluorescent dye which is used in gel electrophoresis for visualizing nucleic adds, such as DNA and RNA

Question 30.
When a microorganism invades a host, a definite sequence of events usually occurs leading to infection and disease, causing suffering to the host. This process is called pathogenesis. Once a microorganism overcomes the defence system of the host, the development of the disease follows a certain sequence of events as shown in the graph. Study the graph given below for the sequence of events leading to the appearance of a disease and answer the questions that follow:

(a) In which period, according to the graph there are maximum chances of a person transmitting a disease/infection and why?
(b) Study the graph and write what is an incubation period. Name, a sexually transmitted disease that can be easily transmitted during this period. Name the specific type of lymphocytes that are attacked by the pathogen of this disease.
OR
(b) Draw a schematic labelled diagram of an antibody.
(c) In which period, the number of immune cells forming antibodies will be the highest in a person suffering from pneumonia? Name the immune cells that produce antibodies.
Answer:
(a) There are maximum chances of a person transmitting a disease/infection during the period of illness because the number of microorganisms in the body is very high at that time.
(b) Incubation period refers to the period between exposure to an infection and the appearance of the first symptoms.
AIDS is a sexually transmitted disease that can be easily transmitted during this period.
T helper cells are a specific type of lymphocytes that are attacked by the pathogen of this disease.

OR

(b) Labelled diagram of an antibody:

An antibody

(c) During ‘Period of decline’, the infected person will have maximum number of cells producing antibodies.
Immune cell that produces antibodies is B lymphocytes.

Section – E(15 Marks)

Question 31.
(a) IVF is a very popular method these days that is helping childless couples to bear a child. Describe the different steps that are carried out in this technique.
(b) Would you consider Gamete Intra-fallopian Transfer (GIFT) as an IVF? Give a reason in support of your answer.
Answer:
(a) In in vitro fertilisation method, popularly known as test tube baby programme, ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory.
The zygote or early embryos (with upto eight blastomeres) could then be transferred into the fallopian tube (ZlFT-zygote intra fallopian transfer) and embryos with more than eight blastomeres, into the uterus (IUT- intra uterine transfer), to complete its further development.
(b) No, GIFT cannot be considered as IVF technique because fertilisation takes place in the female body/ in vivo.

OR

Given is the diagram of a human ovum surrounded by a few sperms. Study the diagram and answer the following questions:

(a) Which one of the sperms would reach that ovum earlier?
(b) Identify ‘D’ and ‘E’. Mention the role of ‘E’.
(c) Mention what helps the entry of sperm into the ovum and write the changes occurring in the ovum during the process.
(d) Name the specific region in the female reproductive system where the event represented in the diagram takes place.
Answer:
(a) Sperm A will reach the ovum fastest.
(b) D – Cells of the corona radiata
E – Zona pellucida
The zona pellucida layers play several roles such as –

• it supports the communication between oocytes and follicle cells during oogenesis
• it protects the oocytes, eggs, and embryos during development
• it regulates interactions between ovulated eggs and free-swimming sperm during and following fertilization

(c) The anterior portion of the head of the sperm is covered by a cap-like structure called the acrosome. This acrosome is filled with enzymes that help in fertilization of the ovum. As a sperm comes in contact with the zona pellucida layer, it induces changes in the layer and blocks the entry of other sperms. The enzymes of the acrosome help the entry of the sperm into the cytoplasm of the ovum.
(d) This event of fertilization occurs in the ampullary region of the fallopian tube.

Question 32.
Study the pedigree chart given below and answer the questions that follow:

(a) On the basis of the inheritance pattern exhibited in this pedigree chart, what conclusion can you draw about the pattern of inheritance?
(b) If the female is homozygous for the affected trait in this pedigree chart, then what percentage of her sons will be affected?
(c) Give the genotype of offspring 1, 2, 3 and 4 in third generation.
(d) In this type of inheritance pattern, out of male and female children which one has less probability of receiving the trait from the parents? Give a reason.
Answer:
(a) X- linked, Recessive trait
(b) 100%
(c) 1. XY 2. XX, 3. XY, 4. XX
(d) The possibility of the female getting the trait is less. The female will get the trait only if the mother is at least a carrier and the father is affected.

OR

(a) How and why is charging of tRNA essential in the process of translation?
(b) State the function of the ribosome as a catalyst in bacteria during the process of translation.
(c) Explain the process of binding of ribosomal units to mRNA during protein synthesis.
Answer:
(a) Charging of tRNA (Aminoacylation of tRNA): Here, amino adds are activated (amino add + ATP) and linked to their cognate tRNA in the presence of aminoacyl tRNA synthetase. This process is commonly known as the charging of tRNA or aminoacylation of tRNA. If two such charged tRNAs are brought dose enough, the formation of peptide bonds between them would be favoured energetically. This is an essential step as only activated amino acids are carried to the site of protein synthesis by their respective tRNA.

(b) In bacteria, the ribosome acts as a catalyst (23S rRNA) for the formation of peptide bonds.

(c) Translation is initiated by the formation of an initiation complex consisting of 30S ribosomal subunit, formyl-methionyl (fMet) tRNA, and mRNA. It begins at the 5′-end of mRNA in the presence of an initiation factor.
The mRNA binds to the small subunit of ribosome. AUG is recognised by the initiator tRNA. The initiation codon for methionine is AUG. So methionyltRNA complex would have UAC at the Anticodon site. Now, the large subunit (50S) binds to the small subunit to complete the initiation complex.
Large subunit (70S) has two binding sites to which tRNA-carrying amino acids can bind. One is called aminoacyl tRNA binding site (A site) and the other is called peptidyl site (P-site).

Question 33.
(a) Farmers are often suggested to use the following organisms in their crop land so as to improve the soil fertility, (i) Rhizobium, (ii) Anabaena. Explain.
(b) Organic farmers use Trichoderma and Baculovirus as biological control agents. Explain.
Answer:
(a) (i) Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants that have the ability to fix atmospheric nitrogen.
(ii) Anabaena has the ability to fix the atmospheric nitrogen. They are free-living nitrogen fixing bacteria. Hence, they are used as biofertilisers in the field of rice.
(b) Trichoderma sp. is free living fungi. They live in the roots of higher plants and protect them from various pathogens. They are effective biocontrol agents of several plant pathogens. Baculoviruses (especially genus Nucleopolyhedrovirus) attack insects and other arthropods. These are suitable for spedes-spedfic, narrow spectrum insectiddal applications. This is desirable in IPM (Integrated pest management) programe to conserve beneficial insects.

OR

(a) Explain the life cycle of Plasmodium starting from its entry into the body of female Anopheles till the completion of its life cycle in humans.
(b) Explain the cause of periodic recurrence of chill and high fever during malarial attack in humans.
Answer:
(a) When a female Anopheles mosquito bites an infected person, the parasites enter the mosquito’s body as gametocytes.
It leads to fertilisation and development in the gut of the mosquito and undergoes further development to form sporozoites that are stored in salivary glands until their transfer to the human body. In the human body – the sporozoites reach the liver and reproduce asexually, bursting the cells and releasing them into the RBCs as gametocytes.

(b) The rupture of RBCs releases a toxic substance called haemozoin.