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Simran Zutshi

Content Strategist|Tech-innovator|National Hackathon Winner | Updated On - Jan 23, 2025

CBSE Class 12 2024 Biology Set 1 Question Paper (Paper Code: 57/4/1) is available for download. The exam was successfully conducted by CBSE on March 19 in the morning session from 10:30 AM to 1:30 PM. As per the student’s initial reactions, the CBSE Class 12 2024 Biology Set 1 Question Paper was reported as Moderate. The Ecology section was reported as Easy to Moderate, the Genetics and Evolution section as Challenging, and the Human Physiology section as Moderate.

CBSE Class 12 2024 Biology (Set 1- 57/4/1) 2024 Answer Key With Solution

Candidates can download the CBSE Class 12 Biology Question Paper with Solution and Answer Key PDFs for Set 1 Question Paper (Code: 57/4/1) using the link below.

CBSE Class 12 2024 Biology​ Question Paper with Answer Key download iconDownload Check Solution

CBSE Class 12 2024 Biology Questions with Solutions

SECTION - A
 

Question 1:

In a fertilized ovule of an angiosperm, the cells in which n, 2n, and 3n conditions respectively occur are:

  1. antipodal, zygote, and endosperm
  2. zygote, nucellus, and endosperm
  3. endosperm, nucellus, and zygote
  4. antipodals, synergids, and integuments
Correct Answer: 1
View Solution

• In a fertilized ovule of an angiosperm:
– n (haploid): Found in antipodal cells.
– 2n (diploid): Found in the zygote formed by fertilization.
– 3n (triploid): Found in the endosperm formed by double fertilization.
• This distribution of ploidy levels is characteristic of the angiosperm reproductive process.

Question 2:

Select the option that gives the correct identification of ovum, morula, and blastocyst in a human female reproduction system as shown in the following diagram:
Human Female Reproductive System

  1. Ovum – B, Morula – D, Blastocyst – F
  2. Ovum – A, Morula – B, Blastocyst – G
  3. Ovum – A, Morula – E, Blastocyst – G
  4. Ovum – B, Morula – D, Blastocyst – G
Correct Answer: 4
View Solution

Ovum: The unfertilized egg cell (B) as shown in the diagram.
Morula: A solid ball of cells (D) resulting from the early cleavage stages post-fertilization.
Blastocyst: A hollow structure (G) formed in the early development of mammals, which implants in the uterine wall.

Question 3:

Study the table given below:
Contraceptive/Contraceptive Method
Select the option where contraceptive/contraceptive method are correctly matched with their mode of action.

  1. A – III, B – II, C – I, D – IV
  2. A – II, B – III, C – I, D – IV
  3. A – III, B – I, C – IV, D – II
  4. A – III, B – I, C – II, D – IV
Correct Answer: 3
View Solution

A – The pill: Works by inhibiting ovulation (III).
B – Condom: Prevents sperm from reaching the cervix (I).
C – Vasectomy: Involves cutting/tying vas deferens, resulting in semen containing no sperm (IV).
D – Copper-T: Prevents implantation of the fertilized egg in the uterus (II).

Question 4:

Identify the category of genetic disorder depicted in the pedigree chart given below:
genetic disorder depicted in the pedigree chart

  1. X-Linked recessive
  2. X-Linked dominant
  3. Autosomal recessive
  4. Autosomal dominant
Correct Answer: 3
View Solution

Step 1: Analyze the Pedigree Chart.
- Both males and females are affected, indicating autosomal inheritance.
- The disorder appears in siblings but not in the parents, which is characteristic of a recessive trait.
Final Answer: The genetic disorder depicted is autosomal recessive.

Question 5:

Which was the last of the 24 human chromosomes to be completely sequenced?

  1. Chromosome – 1
  2. Chromosome – 11
  3. Chromosome – 21
  4. Chromosome – X
Correct Answer: 1
View Solution

Step 1: Recall the Human Genome Project.
- Chromosome 1 was the last human chromosome to be fully sequenced due to its size and complexity. It contains the largest number of genes and a high number of repetitive sequences, making sequencing particularly challenging.
Final Answer: The last chromosome to be completely sequenced was Chromosome 1.

Question 6:

Study the following diagram of Transverse Section of a young anther of an angiosperm:
Transverse Section of a young anther of an  angiosperm

  1. A – Connective, B – Endothecium, C – Pollen grain.
  2. A – Endothecium, B – Connective, C – Pollen grain.
  3. A – Pollen grain, B – Connective, C – Endothecium.
  4. A – Endothecium, B – Pollen grain, C – Connective.
Correct Answer: 1
View Solution

Step 1: Interpret the Anther Diagram.
- Connective tissue joins the two lobes of the anther.
- Endothecium forms the outer wall of the pollen sac.
- Pollen grains are present inside the pollen sac.
Final Answer: The correct identification is A – Connective, B – Endothecium, C – Pollen grain.

Question 7:

Turner’s syndrome in humans occurs due to:

  1. Aneuploidy
  2. Euploidy
  3. Polyploidy
  4. Autosomal abnormality
Correct Answer: 1
View Solution

Step 1: Understand Turner’s Syndrome.
- Turner’s syndrome is a chromosomal disorder caused by the presence of a single X chromosome (45, XO) instead of two sex chromosomes.
- This is a form of aneuploidy, which refers to the loss or gain of chromosomes.
Final Answer: Turner’s syndrome is caused by aneuploidy.

Question 8:

Which of the options has correct identification of ‘P’, ‘Q’ and ‘R’ in the illustration of ‘Central Dogma’ given below?
Central Dogma

  1. P – Replication, Q – rRNA, R – Transcription.
  2. P – Translation, Q – mRNA, R – Transcription.
  3. P – Replication, Q – mRNA, R – Translation.
  4. P – Transcription, Q – mRNA, R – Translation.
Correct Answer: 4
View Solution

Step 1: Recall the Central Dogma.
- Transcription (P): DNA to mRNA.
- mRNA (Q): Carries genetic information.
- Translation (R): mRNA to protein.
Final Answer: P – Transcription, Q – mRNA, R – Translation.

Question 9:

Who proposed the mutation theory in favour of organic evolution?

  1. Weismann
  2. Louis Pasteur
  3. Darwin
  4. Hugo de Vries
Correct Answer: 4
View Solution

Step 1: Recall Evolutionary Theories.
- Hugo de Vries proposed the mutation theory, which states that evolution occurs through sudden, heritable changes in an organism’s genetic material.
- These mutations serve as raw material for natural selection.
Final Answer: The mutation theory was proposed by Hugo de Vries.

Question 10:

Study the following list of bioactive substances and their action:
Bioactive Substance

  1. A – II, B – III, C – I, D – IV
  2. A – III, B – IV, C – II, D – I
  3. A – IV, B – I, C – II, D – III
  4. A – IV, B – III, C – I, D – II
Correct Answer: 2
View Solution

Step 1: Match Bioactive Substances to Their Actions.
- Statin: Lowers blood cholesterol (III).
- Cyclosporin A: Acts as an immuno-suppressive agent (IV).
- Streptokinase: Removes clots from blood vessels (II).
- Lipase: Removes oil stains (I).
Final Answer: A – III, B – IV, C – II, D – I.

Question 11:

The ‘molecular scissors’ fall in the category of:

  1. Cleaving enzyme
  2. Endonuclease
  3. Exonuclease
  4. Restriction enzymes
Correct Answer: 4
View Solution

Step 1: Understand Restriction Enzymes.
- Restriction enzymes, also known as molecular scissors, cut DNA at specific sequences.
- These enzymes are extensively used in genetic engineering and molecular biology.
Final Answer: Restriction enzymes are referred to as molecular scissors.

Question 12:

ELISA technique is based on the principle of:

  1. DNA replication
  2. Antigen-antibody interaction
  3. Pathogen–antigen interaction
  4. Antigen–protein interaction
Correct Answer: 2
View Solution

• ELISA (Enzyme-Linked Immunosorbent Assay) works on the principle of antigen-antibody binding to detect specific proteins or antibodies in a sample.
• The reaction between an enzyme-labeled antibody and its antigen is measured through color development, indicating the presence of the antigen.

Question 13:

Assertion (A): A given fig species can be pollinated only by its partner wasp.
Reason (R): The wasp pollinates the fig inflorescence while searching for suitable egg-laying sites.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true and (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Correct Answer: 1
View Solution

• Fig and wasp share a mutualistic relationship where each depends on the other for survival and reproduction.
• Wasps pollinate fig flowers while laying eggs, ensuring mutual benefit.

Question 14:

Assertion (A): Plasmids are autonomously replicating circular extra-chromosomal DNA.
Reason (R): Plasmids are usually present in eukaryotic cells.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true and (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Correct Answer: 3
View Solution

• Plasmids are indeed circular DNA molecules capable of autonomous replication, but they are commonly found in prokaryotic cells like bacteria, not eukaryotes.

Question 15:

Assertion (A): Patents are granted by governments to an inventor.
Reason (R): Patents prevent others from commercial use of an invention.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true and (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Correct Answer: 1
View Solution

• Patents are legal rights granted to inventors for exclusive use of their inventions.
• The purpose of patents is to prevent unauthorized commercial use and ensure the protection of intellectual property, which directly explains the reason provided.

Question 16:

Assertion (A): Some aquatic ecosystems have inverted biomass pyramids.
Reason (R): More energy is required by organisms occupying higher trophic levels.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true and (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Correct Answer: 2
View Solution

• Inverted biomass pyramids occur in aquatic ecosystems where the biomass of primary consumers exceeds that of producers.
• The energy transfer efficiency in trophic levels determines biomass, not the amount of energy required by higher trophic levels, which makes the reason valid but not explanatory for the assertion.

SECTION – B

Question 17:

Study the graph given below that represents the changes in the thickening of the uterine wall in women ‘X’ and women ‘Y’ over a period of one month:
graph with respect to woman ‘X’ and woman ‘Y’ indicate

What does the graph with respect to women ‘X’ and women ‘Y’ indicate? Give a suitable reason.

View Solution

• The graph shows:
Woman ‘X’: Normal thickening of the uterine wall due to regular hormonal activity.
Woman ‘Y’: Lack of thickening indicates a hormonal imbalance or insufficient estrogen/progesterone secretion.

Question 18:

(a) Intensely lactating mothers generally do not conceive. Why?

View Solution

• Lactational amenorrhea prevents ovulation during intense lactation.
• Prolactin, the hormone responsible for milk production, suppresses gonadotropin release.

(b) Why has our government intentionally imposed strict conditions for MTP (Medical Termination of Pregnancy)?

View Solution

• To prevent misuse and ensure MTP is conducted only for genuine medical or legal reasons.
• Protects the health and rights of women.

Question 19:

(a) Name the source from which insulin was extracted in earlier times. Why is this insulin no more in use by the diabetic patients?

View Solution

Source: Pancreas of pigs and cows.
Reason: Animal insulin caused allergic reactions in some patients due to structural differences from human insulin.

(b) Why does the insulin synthesized in the human body undergo processing whereas the insulin produced by Eli Lilly company does not need to undergo any processing? Explain.

View Solution

• Human insulin is produced as proinsulin and requires processing to remove C-peptide.
• Eli Lilly produces ready-to-use recombinant insulin without requiring further processing.

Question 20:

(a) Differentiate between grazing food chain and detritus food chain.

View Solution

Grazing food chain: Begins with producers consumed by herbivores. (e.g., Grass → Deer → Lion)
Detritus food chain: Begins with decomposers breaking down organic matter. (e.g., Dead leaves → Fungi → Earthworm)

OR

(b) Explain Brood parasitism with the help of a suitable example.

View Solution

Brood parasitism: A reproductive strategy where one species lays eggs in the nest of another species.
Example: Cuckoo lays eggs in crow nests; cuckoo chicks outcompete crow chicks for food.

Question 21:

(a) Biodiversity hotspots cover less than 2% of Earth’s land area. Strict protection of these areas can reduce the rate of ongoing extinctions. Explain.

View Solution

• Biodiversity hotspots are regions rich in endemic species but threatened by habitat loss.
• Protecting hotspots helps conserve numerous species, ensuring ecosystem stability and reducing extinction rates.

(b) Name any two hotspots in India.

View Solution

• Western Ghats.
• Indo-Burma region.

SECTION – C

Question 22:

Draw a well-labelled diagram of sectional view of male gametophyte/microspore of an angiosperm and write the functions of any two parts labelled. (Any four labels).

View Solution Sectional view of a male gametophyte of an angiosperm. Labelled parts include: Exine, Intine, Vegetative Cell, and Generative Cell.

Functions:
Generative cell: Divides to form two male gametes.
Tube cell: Grows into a pollen tube to facilitate fertilization.

Question 23:

(a) A man with blood group ‘A’ marries a woman with blood group ‘AB’. The first child born to them has blood group ‘B’. Work out a cross to find the genotype of the father. Give the possible blood groups and their genotypes of the children that could be born to this couple. (Use a Punnet square).

View Solution

Father’s Genotype: IAi
Mother’s Genotype: IAIB
Punnett Square:
[Insert Punnett Square Image/Text Here - Example below]
| | IA | IB |
|---|---|---|
| IA | IAIA | IAIB |
| i | IAi | IBi |
Children’s Blood Groups:
Blood Group A: IAIA, IAi
Blood Group B: IBi
Blood Group AB: IAIB

(b) State the basis of ‘ABO’ blood grouping in humans.

View Solution

The ABO blood grouping is based on:
• The presence or absence of antigens (A and B) on the surface of red blood cells.
• The presence of antibodies (anti-A and anti-B) in the plasma.

Question 24:

(a) Whose skulls ‘A’, ‘B’, and ‘C’ are shown below? Which of the two are more similar to each other?
skull

View Solution

Skull A: Homo sapiens
Skull B: Neanderthals
Skull C: Australopithecus
• Skulls A and B are more similar to each other.

(b) Name the (i) ape-like (ii) man-like primates that existed 1.5 million years ago.

View Solution

• (i) Australopithecus (ape-like)
• (ii) Homo erectus (man-like)

Question 25:

(a) (i) Name the group of drugs whose skeletal molecule is shown below:
Drug

View Solution

Steroids.

(ii) How are such drugs consumed?

View Solution

These drugs are consumed orally, injected, or applied topically.

(iii) Name the human body organ affected by the consumption of these drugs.

View Solution

Liver.

OR

(b) Draw a schematic diagram of an antibody molecule and label any 4 parts. Mention their chemical nature. Name the cells which produce them.

View Solution

[Insert Antibody Diagram Here]
• Antibodies are Y-shaped molecules produced by B-lymphocytes.
Chemical Nature: Proteins made of light and heavy polypeptide chains.

Question 26:

Explain the role of the following during the sewage treatment:
(a) Flocs
(b) Anaerobic sludge digester

View Solution

Flocs: Aggregates of bacteria and fungi used in the secondary treatment to degrade organic matter in sewage.
Anaerobic sludge digester: Breaks down organic matter into methane, CO2, and water during the anaerobic digestion of sludge.

Question 27:

Study the steps shown below, that are carried during a specific technique:
Steps

(a) Identify the steps ‘A’ and ‘D’ in the diagram.

View Solution

A: Denaturation of DNA at high temperatures.
D: Amplification of DNA after 30 cycles.

(b) What does ‘B’ represent?

View Solution

Primer annealing.

(c) Write what is ‘C’ ? Name its source organism.

View Solution

C: Taq DNA polymerase.
Source: Thermus aquaticus.

(d) Mention the use of this technique in molecular diagnostics.

View Solution

• Used in Polymerase Chain Reaction (PCR) to amplify DNA for genetic testing, disease diagnosis, and forensic analysis.

Question 28:

Explain the role of transgenic animals in:

(a) Production of Biological Products

View Solution

• Transgenic animals are engineered to produce important biological products like proteins and hormones.
• Example: Transgenic cows producing human lactoferrin, which has therapeutic applications.

(b) Studying Diseases

View Solution

• Transgenic animals are used as models to study human diseases and their progression.
• Example: Transgenic mice are commonly used to study cancer, diabetes, and Alzheimer’s disease.

(c) Chemical Safety Testing

View Solution

• Transgenic animals are used to test the safety of chemicals and drugs before they are approved for human use.
• Example: Transgenic mice are used to assess the toxicity of new pharmaceutical compounds.

SECTION D

Question 29:

Populations evolve to maximise their reproductive fitness in the habitat in which they live. Study the population growth curves shown in the given graph and answer the questions that follow:

(a) Identify the growth curves ‘A’ and ‘B’.
Population Growth Curves

View Solution

Curve ‘A’: Exponential growth curve.
Curve ‘B’: Logistic growth curve.
Explanation: Exponential growth occurs under unlimited resources, showing a J-shaped curve. Logistic growth accounts for environmental constraints and levels off at the carrying capacity (K), forming an S-shaped curve.

(b) Mention what the dotted line in the graph indicates and state its importance also.

View Solution

• The dotted line represents the carrying capacity (K).
Importance: Carrying capacity is the maximum population size that an environment can sustain indefinitely. It is determined by resource availability and environmental factors.

OR

Growth curve ‘B’ shows a different pattern from that of growth curve ‘A’. Justify giving one reason.

View Solution

• Curve ‘B’ (logistic growth) differs from curve ‘A’ (exponential growth) because it incorporates the concept of limited resources and environmental resistance.
• Logistic growth considers population stabilization when reaching the carrying capacity, making it more realistic.

(c)(i) Which one of the two curves is more ‘realistic’ and why?

View Solution

• Curve ‘B’ (logistic growth) is more realistic because populations are subject to resource limitations, predation, and environmental constraints.
• Curve ‘A’ (exponential growth) assumes unlimited resources, which is rarely the case in natural ecosystems.

(ii) Which one of the two curves is relevant in present days with respect to human population in our country and why?

View Solution

• The exponential growth curve (Curve ‘A’) is more relevant because the human population continues to grow at a rapid rate, often exceeding the carrying capacity temporarily.
• However, environmental resistance like resource depletion and climate change will eventually necessitate stabilization, transitioning to a logistic growth model.

Question 30:

Generally, in eukaryotic cells the average length of a transcription unit along a DNA molecule is about 8,000 nucleotides, so the RNA product of the transcription is also that long. But it only takes about 1200 nucleotides from the above RNA product to translate average sized polypeptide of 400 Amino acids.

(a) Name this RNA product transcribed from the DNA that subsequently translates into a polypeptide of 400 amino acids. Mention the enzyme responsible for transcribing this type of RNA from the DNA.

View Solution

• The RNA product is mRNA (messenger RNA).
Enzyme responsible: RNA polymerase II.

(b) Name and explain the process the RNA molecule transcribed from 8000 nucleotide long DNA undergoes to be able to translate a polypeptide of 400 amino acids.

View Solution

Process: Splicing.
Explanation: During splicing, non-coding regions (introns) are removed from the pre-mRNA, leaving only coding regions (exons) to form mature mRNA. This ensures the correct sequence for translating into a 400 amino acid polypeptide.

(c) Write the number of RNA polymerases involved in the transcription of DNA in a prokaryote and eukaryotes.

View Solution

Prokaryotes: One RNA polymerase is involved in transcription.
Eukaryotes: Three RNA polymerases (RNA polymerase I, II, and III) are involved in transcription.

OR

(c) Mention the difference in the site of transcription in a prokaryote and eukaryote cell.

View Solution

Prokaryotes: Transcription occurs in the cytoplasm.
Eukaryotes: Transcription occurs in the nucleus.

SECTION E

Question 31:

(a) The given diagram shows the sectional view of a seminiferous tubule of Human testis.
Testis

(i) Name and describe the process depicted in the diagram which results in the development of spermatozoa.

View Solution

Process: Spermatogenesis.
Description: Spermatogenesis is the process of formation of sperm cells (spermatozoa) from the spermatogonial cells in the seminiferous tubules of the testes. It includes the following stages:
Multiplication Phase: Mitotic division of spermatogonia to produce primary spermatocytes.
Growth Phase: Enlargement of primary spermatocytes.
Maturation Phase: Meiosis to produce haploid spermatids from secondary spermatocytes.
Differentiation Phase: Spermatids transform into spermatozoa.

(ii) Identify the cell where you are seeing a cluster of spermatozoa attached in the diagram. Write the function of the cell.

View Solution

• The cell is the Sertoli cell.
Function: Sertoli cells provide nourishment and structural support to the developing sperm cells. They also secrete inhibin, which regulates spermatogenesis.

OR

(b) Observe the picture of Commelina plant bearing two types of flowers:
Commelina

(i) Identify the two types of flowers labelled ‘A’ and ‘B’ in the picture.

View Solution

• ‘A’: Chasmogamous flower.
• ‘B’: Cleistogamous flower.

Question 31:

(ii) Compare the two types of flowers with reference to:
(1) Characteristic feature
(2) Modes of pollination

View Solution

Characteristic feature:
Chasmogamous flowers: Open flowers, reproductive parts exposed.
Cleistogamous flowers: Closed flowers, reproductive parts not exposed.
Modes of pollination:
Chasmogamous flowers: Cross-pollination facilitated by agents like wind, insects, or water.
Cleistogamous flowers: Self-pollination occurs as flowers never open.

(iii) List any two ‘outbreeding devices’ in flowering plants. Explain why do plants develop such devices.

View Solution

Outbreeding devices:
– Self-incompatibility.
– Production of unisexual flowers.
Explanation: These devices prevent self-pollination and promote cross-pollination, which enhances genetic diversity and adaptation.

Question 32:

(a) Study the schematic diagram given below and answer the questions:
polarity from ‘X’ to ‘Y’ in the mRNA

(i) Identify the polarity from ‘X’ to ‘Y’ in the mRNA segment shown. Mention how many more amino acids can be added to the polypeptide that is being translated and why.

View Solution

Polarity: 5’ to 3’ direction.
Number of additional amino acids: Depends on the number of codons left untranslated on the mRNA.
Reason: Each codon corresponds to one amino acid, and translation stops when a stop codon is encountered.

(ii) Write the initiating codon for translation, its anticodon, and the amino acid it codes for.

View Solution

Initiating codon: AUG.
Anticodon: UAC.
Amino acid: Methionine.

(iii) Explain the charging of an adaptor molecule. Why does this molecule need to be charged?

View Solution

Charging of tRNA: Aminoacyl-tRNA synthetase attaches a specific amino acid to its corresponding tRNA, forming aminoacyl-tRNA.
Importance: Charged tRNA ensures accurate addition of amino acids to the growing polypeptide chain during translation.

OR

(b) Answer the following questions on sickle-cell anaemia:

(i) Why is sickle-cell anaemia, a human blood disorder, so named?

View Solution

• The disorder causes red blood cells to assume a sickle shape under low oxygen conditions, leading to its name.

(ii) Explain the genetic basis that results in the expression of this disorder.

View Solution

• It is caused by a point mutation in the β-globin gene, where glutamic acid is replaced by valine at the sixth position of the haemoglobin protein.
• The mutated haemoglobin (HbS) polymerizes under low oxygen conditions, leading to cell deformation.

(iii) Work out a cross to explain how normal parents may have a sickle-cell anaemic child.

View Solution

[Insert Punnett Square Here]
• When both parents are carriers (HbA HbS), the offspring have a 25% chance of inheriting two sickle-cell alleles (HbS HbS), causing the disorder.

Question 33:

Describe the following:

(a) Describe the life cycle of HIV from the time of its entry into the human body till full blown AIDS sets in.

View Solution

Entry: HIV enters host cells (CD4+ T-cells) via surface receptors.
Reverse transcription: Viral RNA is converted into DNA by reverse transcriptase.
Integration: Viral DNA integrates into the host genome via integrase enzyme.
Replication: Host machinery synthesizes viral components.
Assembly and release: New viral particles are assembled and released, infecting other cells.
Progression to AIDS: Gradual destruction of CD4+ cells leads to weakened immunity and opportunistic infections, culminating in AIDS.

OR

(b) Write the symptoms of malaria in humans and explain what causes these symptoms.

View Solution

Symptoms:
– High fever with chills.
– Sweating and headache.
– Nausea and vomiting.
Cause: The symptoms are caused by the rupture of red blood cells during the release of merozoites and the immune response to parasitic toxins.

(ii) Describe the different steps in the sexual mode of reproduction in the life cycle of a malarial parasite.

View Solution

• Gametocytes develop in human blood and are taken up by a mosquito during a blood meal.
• Fertilization occurs in the mosquito’s gut, forming a zygote.
• The zygote develops into an ookinete and then an oocyst.
• Sporozoites are released from the oocyst and migrate to the salivary glands, ready to infect another human.

 

*The article might have information for the previous academic years, please refer the official website of the exam.

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