Zollege is here for to help you!!
Need Counselling
Simran Zutshi's profile photo

Simran Zutshi

Content Strategist|Tech-innovator|National Hackathon Winner | Updated On - Jan 28, 2025

CBSE Class 12 2024 Biology Set 3 Question Paper with Solutions (Q.P. Code: 57/5/3) 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 students’ initial reactions, the CBSE Class 12 2024 Biology Set 3 Question Paper was reported as Moderate. The Ecology section was considered Easy to Moderate, the Genetics and Evolution section as Challenging, and the Human Physiology section as Moderate. 

CBSE Class 12 2024 Biology Set 3 57/5/3 Question Paper with Answer Key PDF

Candidates can download the CBSE Class 12 Biology Question Paper with Solution and Answer Key PDFs for Set 3 Question Paper (Code: 57/5/3) 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:

A person with trisomy of 21st chromosome shows:
(i) Furrowed tongue
(ii) Characteristic palm crease
(iii) Rudimentary ovaries
(iv) Gynaecomastia
Select the correct option from the choices given below:

  • (A) (ii) and (iv)
  • (B) (i), (ii), and (iv)
  • (C) (ii) and (iii)
  • (D) (i) and (ii)
Correct Answer: (D) (i) and (ii)
View Solution

Step 1: Understanding the condition of trisomy 21. Trisomy of chromosome 21 leads to Down syndrome, a genetic disorder caused by an extra copy of chromosome 21. Step 2: Symptoms of Down syndrome. Key features include: - Furrowed tongue (i) - Characteristic palm crease (ii) Other traits, such as developmental delays, may also occur. Quick Tip: Down syndrome is caused by the presence of an extra copy of chromosome 21, leading to distinct physical and cognitive traits.


Question 2:

Which one of the following chromosomal events will not result in genetic variation amongst the offspring?

  • (A) Independent assortment
  • (B) Crossing over
  • (C) Linkage
  • (D) Mutation
Correct Answer: (C) Linkage
View Solution

Step 1: Understanding the role of linkage. - Linkage refers to the inheritance of genes located close together on the same chromosome. - Since linked genes are inherited together, they do not contribute to genetic variation. Step 2: Contrasting linkage with other mechanisms. - Independent assortment, crossing over, and mutations result in genetic diversity by introducing new combinations or changes in genetic material. - Linkage conserves the gene arrangement, thereby reducing variation. Quick Tip: Linkage conserves the arrangement of genes on a chromosome, while mechanisms like crossing over, mutation, and independent assortment promote genetic diversity.


Question 3:

Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below:
Labeled structure with Primary Endosperm Cell (PEC).

  • (A) A – zygote, B – degenerating synergids, C – degenerating antipodals, D – PEN
  • (B) A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals
  • (C) A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote
  • (D) A – degenerating synergids, B – zygote, C – degenerating antipodals, D – PEN
Correct Answer: (B) A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals
View Solution

Step 1: Identifying the labeled parts of the fertilized embryo sac. - A represents degenerating synergids, which are no longer functional after fertilization. - B is the zygote, formed by the fusion of male and female gametes. - C is the Primary Endosperm Nucleus (PEN), which develops into endosperm to nourish the embryo. - D represents degenerating antipodal cells, which have no role post-fertilization. Quick Tip: In a fertilized embryo sac, unused cells like synergids and antipodals degenerate, while the zygote and PEN develop further.


Question 4:

Study the pedigree chart of a family showing the inheritance pattern of a certain disorder. Select the option that correctly identifies the nature of the trait depicted in the pedigree chart:
Pedigree chart showing inheritance pattern.

  • (A) Dominant X-linked
  • (B) Recessive X-linked
  • (C) Autosomal dominant
  • (D) Autosomal recessive
Correct Answer: (D) Autosomal recessive
View Solution

Step 1: Identifying the pattern of inheritance. - The disorder skips generations, which is characteristic of recessive inheritance. - Both males and females are equally likely to inherit the trait, indicating autosomal inheritance. Step 2: Analyzing the pedigree. - Affected individuals have parents who are carriers but not affected themselves, confirming an autosomal recessive trait. - X-linked recessive traits would primarily affect males, which is not evident in this pedigree. Quick Tip: Autosomal recessive traits affect both genders equally and typically skip generations, requiring both parents to be carriers for an offspring to be affected.


Question 5:

Which one of the following statements is correct in the context of observing DNA separation by agarose gel electrophoresis?

  • (A) DNA can be seen in visible light.
  • (B) DNA can be seen without staining in visible light.
  • (C) Ethidium bromide stained DNA can be seen in visible light.
  • (D) Ethidium bromide stained DNA can be seen under UV light.
Correct Answer: (D) Ethidium bromide stained DNA can be seen under UV light
View Solution

Step 1: Role of ethidium bromide. - Ethidium bromide is a dye used to stain DNA in agarose gel electrophoresis. - It binds to DNA and fluoresces under UV light, making DNA bands visible. Quick Tip: Ethidium bromide is widely used in molecular biology for visualizing DNA under UV light during gel electrophoresis.


Question 6:

A phenomenon where a male insect mistakenly identified the patterns of a flower as the female insect partner, and tries to copulate and thereby pollinates the flower is said to be:

  • (A) Pseudocopulation
  • (B) Pseudopollination
  • (C) Pseudoparthenocarpy
  • (D) Pseudofertilisation
Correct Answer: (A) Pseudocopulation
View Solution

Step 1: Understanding pseudocopulation. - Pseudocopulation is a process in which a male insect is attracted to a flower that mimics the appearance or scent of a female insect. - During its attempt to copulate, the insect facilitates pollination by transferring pollen. Step 2: Mimicry in flowers. - This is an example of mimicry used by flowers to attract specific pollinators, ensuring effective reproduction. Quick Tip: Pseudocopulation is a specialized pollination strategy seen in some orchids to ensure effective reproduction.


Question 7:

Observe the schematic representation of assisted reproductive technology given below:
Illustration of sperm being injected into the cytoplasm of the egg using a fine needle.
Identify the most appropriate technique depicted in the above diagram.

  • (A) IUT
  • (B) IUI
  • (C) ICSI
  • (D) ZIFT
Correct Answer: (C) ICSI
View Solution

Step 1: Understanding ICSI. - The diagram represents Intracytoplasmic Sperm Injection (ICSI). - In this technique, a single sperm is directly injected into the cytoplasm of an egg using a fine needle. Step 2: Use of ICSI. - ICSI is used in cases of male infertility, such as low sperm count or motility. Quick Tip: ICSI is a specialized form of in-vitro fertilization (IVF) that overcomes severe male infertility.


Question 8:

The source of ‘Smack’ is:

  • (A) Leaves of \textit{Cannabis sativa}
  • (B) Flowers of \textit{Datura}
  • (C) Fruits of \textit{Erythroxylum coca}
  • (D) Latex of \textit{Papaver somniferum}
Correct Answer: (D) Latex of \textit{Papaver somniferum}
View Solution

Step 1: Source of Smack. - Smack is a common name for heroin, a derivative of morphine. - Morphine is extracted from the latex of \textit{Papaver somniferum, commonly known as the opium poppy. Quick Tip: Heroin is an opioid drug derived from the latex of \textit{Papaver somniferum}, often used illicitly for its euphoric effects.


Question 9:

The first antibiotic was discovered accidentally by A while working on B. ‘A’ and ‘B’ are:

  • (A) A – Waksman; B – \textit{Streptococcus}
  • (B) A – Fleming; B – \textit{Penicillium notatum}
  • (C) A – Waksman; B – \textit{Bacillus brevis}
  • (D) A – Fleming; B – \textit{Staphylococci}
Correct Answer: (D) A – Fleming; B – \textit{Staphylococci}
View Solution

Step 1: Discovery of the first antibiotic. - Alexander Fleming discovered the first antibiotic, penicillin, in 1928. Step 2: Observation during research. - He observed that a mold, \textit{Penicillium notatum, inhibited the growth of \textit{Staphylococci bacteria. Quick Tip: The discovery of penicillin marked the beginning of the antibiotic era, revolutionizing medicine.


Question 10:

If the sequence of nitrogen bases of the coding strand in a transcription unit is 5’ – ATGAATG – 3’, the sequence of bases in its RNA transcript would be:

  • (A) 5’ – AUGAAUG – 3’
  • (B) 5’ – UACUUAC – 3’
  • (C) 5’ – CAUUCAU – 3’
  • (D) 5’ – GUAAUGA – 3’
Correct Answer: (A) 5’ – AUGAAUG – 3’
View Solution

Step 1: Relationship between the coding strand and RNA transcript. - The RNA transcript is complementary to the template strand and identical to the coding strand, except uracil (U) replaces thymine (T). Step 2: Base sequence of the transcript. - Coding strand: 5’ – ATGAATG – 3’ - mRNA transcript: 5’ – AUGAAUG – 3’ Quick Tip: During transcription, the RNA sequence is complementary to the DNA template strand and matches the coding strand (with U in place of T).


Question 11:

Match the following genes of the lac operon listed in column ‘A’ with their respective products listed in column ‘B’:
\begin{tabular{|c|c| \hline A: Gene & B: Products
\hline a. ‘i’ gene & (i) $\beta$-galactosidase
b. ‘z’ gene & (ii) lac permease
c. ‘a’ gene & (iii) repressor
d. ‘y’ gene & (iv) transacetylase
\hline \end{tabular Select the correct option:

  • (A) (i) (iii) (ii) (iv)
  • (B) (iii) (i) (ii) (iv)
  • (C) (iii) (i) (iv) (ii)
  • (D) (iii) (iv) (i) (ii)
Correct Answer: (C) (iii) (i) (iv) (ii)
View Solution

Step 1: Functions of the lac operon genes. - 'i' gene: Produces the repressor protein (iii). - 'z' gene: Produces $\beta$-galactosidase (i), which breaks down lactose. - 'a' gene: Produces transacetylase (iv). - 'y' gene: Produces lac permease (ii), which facilitates lactose entry into the cell. Quick Tip: The lac operon is an inducible operon that regulates lactose metabolism in prokaryotes.


Question 12:

The human chromosome with the highest and least number of genes in them are respectively:

  • (A) Chromosome 21 and Y.
  • (B) Chromosome 1 and X.
  • (C) Chromosome 1 and Y.
  • (D) Chromosome X and Y.
  • (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
  • (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  • (C) (A) is true, but (R) is false.
Correct Answer: (C) Chromosome 1 and Y
View Solution

Step 1: Chromosome with the highest number of genes. - Chromosome 1 has the highest number of genes, with approximately 2,000–3,000 genes. Step 2: Chromosome with the least number of genes. - Chromosome Y has the least number of genes, with around 50–60 functional genes. Quick Tip: The human genome has 23 pairs of chromosomes, with gene distribution varying significantly across them.


Question 13:

Assertion (A): In birds, the sex of the offspring is determined by males.
Reason (R): Males are homogametic while females are heterogametic.

Correct Answer: (D) (A) is false, but (R) is true
View Solution

- In birds, the sex of the offspring is determined by females, not males. - Female birds are heterogametic (ZW), while males are homogametic (ZZ). - Therefore, (A) is false, but (R) is true. Quick Tip: In birds, females determine the sex of the offspring because they have two different sex chromosomes (ZW), unlike males (ZZ).


Question 14:

Assertion (A): “Biodiversity hotspots” are the regions which possess high levels of species richness, high degree of endemism.
Reason (R): Total number of biodiversity hotspots in the world is 22 with two of these hotspots found in India.

Correct Answer: (C) (A) is true, but (R) is false
View Solution

- Biodiversity hotspots are regions with high species richness and a high degree of endemism, making (A) true. - However, there are 36 biodiversity hotspots worldwide, not 22, making (R) false. Quick Tip: India has two biodiversity hotspots: the Western Ghats and the Himalayas, contributing significantly to global biodiversity.


Question 15:

Assertion (A): AIDS is a syndrome caused by HIV.
Reason (R): HIV is a virus that damages the immune system with DNA as its genetic material.

Correct Answer: (C) (A) is true, but (R) is false
View Solution

- AIDS (Acquired Immunodeficiency Syndrome) is caused by the Human Immunodeficiency Virus (HIV), making (A) true. - However, HIV is a retrovirus with RNA as its genetic material, not DNA, making (R) false. Quick Tip: HIV is a retrovirus, meaning it uses RNA as its genetic material and converts it into DNA within host cells.


Question 16:

Assertion (A): In molecular diagnosis, single stranded DNA or RNA tagged with radioactive molecule is called a probe.
Reason (R): A probe always searches and hybridises with its complementary DNA in a clone of cells.

Correct Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)
View Solution

- Probes are single-stranded DNA or RNA molecules tagged with radioactive or fluorescent labels used in molecular diagnostics. - They specifically hybridise with complementary sequences in the target DNA, enabling detection in molecular diagnostic techniques. Quick Tip: Probes are used in techniques like Southern blotting to detect specific DNA sequences.


SECTION - B

Question 17:

If the base adenine constitutes 31% of an isolated DNA fragment, then write what will be the expected percentage of the base cytosine in it. Explain how did you arrive at the answer given.

Correct Answer:
View Solution

- DNA follows Chargaff’s rule: A = T and C = G. - If Adenine (A) constitutes 31%, then Thymine (T) also constitutes 31%. - The remaining percentage of the bases is: \[ 100% - (A + T) = 100% - 62% = 38%. \] - Since Cytosine (C) = Guanine (G), the remaining 38% is equally divided: \[ C = G = \frac{38%}{2} = 19%. \] - Thus, the percentage of cytosine in the DNA fragment is 19%. Quick Tip: Chargaff’s rule ensures that purines (A, G) and pyrimidines (T, C) are present in equal proportions in double-stranded DNA.


Question 18:

Observe the population growth curve and answer the questions given below:
Population growth curves showing exponential growth

 

(a) State the conditions under which growth curve ‘A’ and growth curve ‘B’ plotted in the graph are possible.

Correct Answer:
View Solution

- Growth curve ‘A’ (Exponential growth): This type of growth occurs under \textit{ideal conditions, where there are no limitations on resources such as food, space, or other environmental factors. - Growth curve ‘B’ (Logistic growth): Logistic growth takes place under \textit{natural conditions where resources are limited. Environmental resistance, such as competition and predation, slows down the growth and stabilizes the population at the carrying capacity. Quick Tip: Exponential growth is unsustainable in natural conditions due to resource limitations, while logistic growth reflects a realistic population pattern that stabilizes at the carrying capacity.


(b) Mention what does ‘K’ in the graph represent.
Correct Answer:
View Solution

‘K’ represents the carrying capacity, which is the maximum population size that the environment can sustain indefinitely with the resources available. It reflects the balance between resource availability and population growth. Quick Tip: Exponential growth occurs when resources are abundant, while logistic growth occurs when resources become limited, leading to stabilization at carrying capacity.


Question 19:

The given DNA sequence:
5’ – G$\uparrow$A A T T C – 3’
3’ – C T T A A$\uparrow$G – 5’

 

(a) Name the restriction enzyme that recognises the given specific sequence of bases. What are such sequences of bases referred to as?

Correct Answer:
View Solution

- The restriction enzyme that recognizes this sequence is EcoRI. - Such sequences of bases are known as palindromic sequences, as they read the same in both directions on complementary strands. Quick Tip: Sticky ends created by restriction enzymes enhance the precision and efficiency of ligation during recombinant DNA technology.


(b) What are the arrows in the given figure indicating? Write the result obtained thereafter.
Correct Answer:
View Solution

- The arrows indicate the sites where the DNA strand is cleaved by EcoRI. - The cleavage results in two DNA fragments with sticky ends. These sticky ends can anneal to complementary sequences, facilitating the insertion of desired DNA fragments into vectors during recombinant DNA technology. Quick Tip: Sticky ends created by restriction enzymes enhance the precision and efficiency of ligation during recombinant DNA technology.


Question 20:

List the events that reduce the Biochemical Oxygen Demand (BOD) of a primary effluent during sewage treatment.

Correct Answer:
View Solution

During secondary treatment of sewage, the BOD of the effluent is reduced through the following processes: - Consumption of organic matter: Aerobic microorganisms in the effluent consume the organic matter. - Formation of flocs: Microbial activity leads to the formation of flocs, which are aggregations of bacteria and fungi. - Decomposition of organic matter: Organic matter is broken down into simpler substances like carbon dioxide and water. - Separation of activated sludge: The flocs settle down as activated sludge, further reducing the organic load. Quick Tip: A decrease in BOD during sewage treatment reflects the removal of organic pollutants, ensuring cleaner water for reuse or discharge.


Question 21:

(a) “Farmers prefer apomictic seeds to hybrid seeds.” Justify giving two reasons.

Correct Answer:
View Solution

Farmers prefer apomictic seeds over hybrid seeds for the following reasons: - Genetic Uniformity: Apomictic seeds are genetically identical to the parent plant, ensuring uniform traits in the crop, such as yield and quality. - Cost-effectiveness: Farmers can reuse apomictic seeds without purchasing new seeds every season, reducing overall costs. Quick Tip: Apomictic seeds eliminate genetic variability, providing consistency in crop quality and reducing dependency on external seed suppliers.


(b) Mention one advantage and one disadvantage of amniocentesis.

Correct Answer:
View Solution

Amniocentesis has the following advantage and disadvantage: - Advantage: It helps in the early detection of genetic disorders in the fetus, enabling timely medical interventions. - Disadvantage: Amniocentesis can be misused for prenatal sex determination, which may lead to unethical practices like female feticide. Quick Tip: Amniocentesis is a valuable diagnostic tool but must be used ethically to prevent misuse and uphold societal balance.


SECTION - C

Question 22:

Explain the processing of heterogeneous nuclear RNA (hnRNA) into a fully functional mRNA in eukaryotes. Where does this processing occur in the cell?

Correct Answer:
View Solution

The processing of hnRNA into mRNA in eukaryotes involves the following steps: - Capping: A methyl guanosine cap is added to the 5’ end of hnRNA, ensuring stability and aiding in ribosome recognition. - Tailing: A poly-A tail is added to the 3’ end of hnRNA to prevent degradation and enhance translation. - Splicing: Introns (non-coding regions) are removed, and exons (coding regions) are joined together to form a functional mRNA. This processing occurs in the nucleus of eukaryotic cells. Quick Tip: mRNA processing ensures the stability and proper translation of genetic information in eukaryotic cells.


Question 23:

(a) “Mother’s milk is considered very essential for the new born infant.” Justify.

Correct Answer:
View Solution

Mother’s milk is essential for a newborn because: - It provides all the necessary nutrients required for the infant's growth and development. - It contains colostrum, which is rich in immunoglobulins that build the infant’s immunity and protect against infections. Quick Tip: Colostrum in mother’s milk acts as the first vaccine for the infant, strengthening immunity and protecting against infections.


(b) What is a ‘vaccine’? Explain the principle on which it works.

Correct Answer:
View Solution

- Definition: A vaccine is a biological preparation that provides active immunity against specific diseases. - Principle: Vaccines work by stimulating the immune system to produce antibodies against the disease-causing agent without causing the disease itself. This enables the body to fight the actual pathogen in the future. Quick Tip: Vaccines mimic infections, enabling the immune system to recognize and combat real pathogens effectively in the future.


Question 24:

Tropical regions harbour more species than the temperate regions. How have biologists tried to explain this in their own ways? Explain.

Correct Answer:
View Solution

Biologists explain the higher species richness in tropical regions through the following reasons: - Stable Climate: Tropical regions have a stable climate over long periods, which supports the survival and evolution of diverse species. - Higher Solar Energy: These regions receive abundant solar energy, increasing productivity and resource availability, which supports a higher number of species. - Absence of Severe Climatic Changes: Unlike temperate regions, tropical areas experience minimal climatic disruptions, allowing uninterrupted species evolution. Quick Tip: Biodiversity thrives in tropical regions due to their consistent climate, allowing uninterrupted species adaptation and survival.


Question 24:

How does latitude influence the species richness of an ecosystem?

Correct Answer:
View Solution

Latitude influences species richness in the following way: - Higher Richness Near the Equator: Tropical regions near the equator have higher species richness due to stable climates, abundant solar energy, and higher productivity. - Decrease with Distance from Equator: As latitude increases (moving toward the poles), species richness decreases due to harsher climatic conditions and limited resources. Quick Tip: Latitude impacts biodiversity, with regions closer to the equator housing more species due to favorable climatic conditions and higher productivity.


Question 25:

State why plant breeders are interested in artificial hybridisation programme. How do they carry out this process?

Correct Answer:
View Solution

Plant breeders are interested in artificial hybridisation because it allows them to combine desirable traits from two different plant varieties into a single variety. This helps improve crop yield, resistance to diseases, and adaptability to environmental conditions. It is also a method to develop new plant varieties with enhanced traits. The process of artificial hybridisation involves the following steps: - Emasculation: Anthers are removed from the flower of the female parent to prevent self-pollination. - Bagging: The emasculated flower is covered to avoid contamination by unwanted pollen. - Pollination: Pollen from the desired male parent is manually transferred to the stigma of the emasculated flower. - Re-bagging: The pollinated flower is covered again to ensure successful fertilisation. Quick Tip: Artificial hybridisation is a key technique in plant breeding to produce superior crop varieties with desired traits.


Question 26:

(a) What are transgenic animals?

Correct Answer:
View Solution

Transgenic animals are animals whose genomes have been altered through the insertion of a foreign gene. This enables them to express specific traits, such as producing therapeutic proteins, serving as model organisms for disease research, or improving agricultural products. Quick Tip: Transgenic animals are key tools in biotechnology, enabling advancements in research, medicine, and agriculture.


(b) Name the transgenic animal having the largest number amongst all the existing transgenic animals.

Correct Answer:
View Solution

Mice are the transgenic animals with the largest numbers among all existing transgenic animals. They are widely used due to their short generation time and genetic similarity to humans. Quick Tip: Mice are ideal transgenic animals for studying gene functions and human diseases due to their compatibility with genetic modifications.


(c) State any 3 reasons for which these types of animals are being produced.

Correct Answer:
View Solution

Transgenic animals are produced for the following reasons: - To study genetic diseases and understand gene functions: Transgenic animals are used as model organisms to study the role of specific genes and genetic diseases. - To produce pharmaceutical proteins: Transgenic animals help in the production of therapeutic proteins like insulin or clotting factors. - To enhance agricultural products: They are used to improve the nutritional value of milk, meat, or other animal-derived products. Quick Tip: Transgenic animals contribute significantly to scientific research, therapeutic advancements, and agricultural improvements.


Question 27:

(a) Construct a pyramid of biomass in sea with phytoplankton and fishes. Explain giving reasons about the characteristic of the constructed pyramid.

Correct Answer:
View Solution

In aquatic ecosystems, the pyramid of biomass is inverted. This means that the biomass of primary producers (phytoplankton) is smaller than the biomass of primary and secondary consumers (fishes). This happens because phytoplankton reproduce rapidly, compensating for their lower biomass with a high turnover rate. Pyramid of Biomass in a Marine Ecosystem. Quick Tip: Inverted biomass pyramids in aquatic ecosystems highlight the rapid reproduction and energy efficiency of primary producers like phytoplankton.


(b) In which condition will the pyramid remain always upright?

Correct Answer:
View Solution

The biomass pyramid will always remain upright in terrestrial ecosystems, where primary producers (plants) have the highest biomass. This occurs because plants form the largest biomass base, which supports consumers at higher trophic levels. Quick Tip: Terrestrial ecosystems maintain upright biomass pyramids due to the abundance and slower turnover of plant biomass.


Question 28:

(a) Why does DNA replication occur within a replication fork and not in its entire length simultaneously?

Correct Answer:
View Solution

DNA replication occurs within a replication fork because: - Directionality of DNA polymerase: The enzyme DNA polymerase can add nucleotides only in the 5’ to 3’ direction. - Anti-parallel nature of DNA: Due to the anti-parallel arrangement of DNA strands, replication occurs continuously on one strand (leading strand) and discontinuously on the other strand (lagging strand) within the fork. Quick Tip: The replication fork ensures accurate DNA synthesis by coordinating the replication of both strands in complementary directions.


(b) “DNA replication is continuous and discontinuous on the two strands within the replication fork.” Explain with the help of a schematic representation.
Correct Answer:
View Solution

- Leading strand: Synthesized continuously in the 5’ to 3’ direction towards the replication fork. - Lagging strand: Synthesized in short Okazaki fragments in the 5’ to 3’ direction away from the replication fork. These fragments are later joined by DNA ligase, ensuring the continuity of the strand. Quick Tip: The semi-discontinuous replication mechanism ensures efficient and accurate duplication of genetic material.


SECTION - D

Question 29:

In a human female, the reproductive phase starts on the onset of puberty and ceases around middle age of the female. Study the graph given below regarding menstrual cycle and answer the questions that follow:
hormonal regulation and events during the menstrual cycle.

(a) Name the hormones and their source organ, which are responsible for menstrual cycle at puberty.

Correct Answer:
View Solution

- Hormones: Follicle Stimulating Hormone (FSH) and Luteinising Hormone (LH). - Source organ: Pituitary gland. Quick Tip: FSH and LH play a pivotal role in regulating ovarian cycles, triggering ovulation, and preparing the body for potential pregnancy.


(b) For successful pregnancy, at what phase of the menstrual cycle an early embryo (up to 3 blastomeres) should be implanted in the uterus (IUT) of a human female who has opted for Assisted Reproductive Technology (ART)? Support your answer with a reason.

Correct Answer:
View Solution

- Phase: Luteal phase (Day 15–28). - Reason: During this phase, the endometrium is thick and rich in blood supply, suitable for implantation and nourishment of the embryo. Quick Tip: The luteal phase provides an optimal environment for embryo implantation and development due to high progesterone levels.


(c) Name the hormone and its source organ responsible for the events occurring during the proliferative phase of menstrual cycle. Explain the event.

Correct Answer:
View Solution

- Hormone: Estrogen. - Source organ: Ovaries (developing follicles). - Event: Estrogen helps repair and thicken the uterine lining (endometrium) that was shed during menstruation, preparing it for potential implantation. Quick Tip: The proliferative phase ensures the regeneration of the uterine lining, creating a receptive surface for potential implantation.


Question 29:

In a normal human female, why does menstruation only occur if the released ovum is not fertilised? Explain.

Correct Answer:
View Solution

- If the ovum is not fertilised, the corpus luteum degenerates, leading to a drop in progesterone levels. - This drop causes the breakdown of the thickened uterine lining, resulting in its shedding as menstrual blood. Quick Tip: The degeneration of the corpus luteum after unfertilized cycles leads to menstruation, resetting the reproductive system.


Question 30:

Read the following passage and answer the questions that follow:
“Mosquitoes are drastically affecting the human health in almost all the developing tropical countries. Different species of mosquitoes cause very fatal diseases so much so that many humans lose their life and if they survive, are unable to put in productive hours to sustain their life. With the result the health index of the country goes down.”

 

(a) Name the form in which Plasmodium gains entry into (i) human body (ii) the female Anopheles body.

Correct Answer:
View Solution

- (i) Human body: Sporozoite – The infective stage that enters the human body. - (ii) Female \textit{Anopheles} body: Gametocytes – The stage that enters the mosquito during a blood meal. Quick Tip: The sporozoite stage of \textit{Plasmodium} is crucial for initiating infection in humans, while gametocytes ensure transmission to mosquitoes.


(b) Why do the symptoms of malaria not appear in a person immediately after being bitten by an infected female \textit{Anopheles}? Give one reason. Explain when and how do the symptoms of the disease would appear.

Correct Answer:
View Solution

- The sporozoites take time to multiply in the liver cells before infecting red blood cells, which delays the onset of symptoms. - Symptoms like fever, chills, and sweating appear when the parasite ruptures red blood cells and releases toxins into the bloodstream. Quick Tip: The incubation period of malaria allows the parasite to multiply in liver cells before invading red blood cells, causing symptoms.


(b) Explain the events which occur within a female \textit{Anopheles} mosquito after it has sucked blood from a malaria patient.

Correct Answer:
View Solution

- The gametocytes of \textit{Plasmodium enter the mosquito’s gut along with the blood meal. - In the mosquito's stomach: - The gametocytes develop into gametes. - The gametes fuse to form zygotes. - The zygotes develop into sporozoites. - The sporozoites migrate to the mosquito’s salivary glands. - The mosquito is now ready to infect another host. Quick Tip: The development of \textit{Plasmodium} in the mosquito’s gut and migration of sporozoites to the salivary glands are key for disease transmission.


(c) Name a species of mosquito other than female \textit{Anopheles} and the disease, for which it carries the pathogen.

Correct Answer:
View Solution

- Mosquito: \textit{Aedes aegypti. - Disease: Dengue or Zika virus. Quick Tip: Different mosquito species are vectors for specific diseases; for example, \textit{Aedes aegypti} spreads Dengue and Zika, while \textit{Anopheles} spreads Malaria.


SECTION - E

Question 31:

(a) Draw a diagram of a human sperm. Label any four parts and write their functions.

Correct Answer:
View Solution

A diagram of a human sperm should include the following parts:
a human sperm cell showing its structure and parts.
- Head: Contains the nucleus with genetic material.
- Acrosome: Contains enzymes that help in penetrating the egg.
- Midpiece: Contains mitochondria to provide energy for movement.
- Tail: Propels the sperm forward for fertilization.


(b) In a human female, probability of an ovum to get fertilized by more than one sperm is impossible. Give reason.

Correct Answer:
View Solution

- Once a sperm penetrates the ovum, a chemical reaction occurs, forming a fertilization membrane. - This fertilization membrane prevents the entry of additional sperms, ensuring monospermy. Quick Tip: The fertilization membrane is a key defense mechanism to ensure only one sperm fertilizes the ovum, preventing polyspermy.


(b) (i) With the help of a labeled diagram, show the different stages of embryo development in a dicot plant.
Correct Answer:
View Solution

A labeled diagram should show:
Stages of Embryo Development in a Dicot Plant.
- Zygote formation.
- Embryo development stages:
- Globular stage. - Heart-shaped stage. - Mature embryo. Quick Tip: Embryo development in dicot plants progresses through organized stages, supported by endosperm nutrition during early growth.


(ii) Endosperm development precedes embryo development. Justify.

Correct Answer:
View Solution

- The endosperm provides nutrients to the developing embryo. - It is formed by the fusion of a male gamete with two polar nuclei, creating a triploid cell that develops into the endosperm. Quick Tip: The early formation of endosperm ensures that the embryo has a consistent supply of nutrients during critical development stages.


Question 32:

(a) Draw a schematic diagram of the cloning vector pBR322 and label (1) Bam HI site (2) gene for ampicillin resistance (3) ‘ori’ (4) ‘rop’ gene.

Correct Answer:
View Solution

A diagram of pBR322 should include the following labels: Schematic diagram of the pBR322 cloning vector. Bam HI site: A restriction site for cloning. - Ampicillin resistance gene: A selectable marker. - ‘Ori’: Origin of replication, essential for plasmid replication. - ‘Rop’ gene: Regulates plasmid copy number. Quick Tip: The pBR322 vector is one of the most widely used plasmids in genetic engineering due to its multiple cloning sites and selectable markers.


(ii) State the role of ‘rop’ gene.

Correct Answer:
View Solution

- The ‘rop’ gene regulates the replication of plasmid copies, ensuring a low copy number of the plasmid. Quick Tip: Regulating plasmid copy number is critical for maintaining genetic stability in recombinant DNA experiments.


(iii) A cloning vector does not have a selectable marker. How will it affect the process of cloning?
Correct Answer:
View Solution

- Without a selectable marker, it becomes difficult to distinguish between recombinant and non-recombinant cells. - Additional screening methods, such as PCR or DNA sequencing, are needed, which increases time and cost. Quick Tip: Selectable markers simplify recombinant identification, making cloning faster and more efficient.


(iv) Why is insertional inactivation preferred over the use of selectable markers in cloning vectors?
Correct Answer:
View Solution

- Insertional inactivation allows for direct identification of recombinant colonies, as the disrupted gene leads to observable changes (e.g., loss of color or enzyme activity). - It avoids the use of antibiotics, promoting biosafety. Quick Tip: Insertional inactivation is a biosafe method for identifying recombinants without relying on antibiotics.


Question 33:

(a) Work out a dihybrid cross up to F\textsubscript{2} generation between pea plants bearing violet coloured axial flowers and white coloured terminal flowers using Punnett’s square. Give their F\textsubscript{2} phenotypic ratio. State the Mendel’s law of inheritance that was derived from such a cross.

Correct Answer:
View Solution

- Parents: Violet axial (VVAA) × White terminal (vvaa). - Gametes: Parent 1 (VVAA): VA. Parent 2 (vvaa): va. - F\textsubscript{1} Generation: All offspring are heterozygous (VvAa), bearing violet axial flowers. - F\textsubscript{2} Generation (Punnett Square): Dihybrid Cross: Punnett Square showing the F2 generation of pea plants. Phenotypic Ratio: - 9 Violet axial: Plants with dominant traits for flower colour (V) and position (A). - 3 Violet terminal: Plants with dominant colour (V) and recessive position (aa). - 3 White axial: Plants with recessive colour (vv) and dominant position (A). - 1 White terminal: Plants with recessive traits for both colour (vv) and position (aa). - Phenotypic Ratio: 9:3:3:1. - Mendel’s Law of Inheritance Derived: Law of Independent Assortment: Alleles of different genes assort independently during gamete formation, leading to new combinations of traits. Quick Tip: The dihybrid cross demonstrates the Law of Independent Assortment, showcasing how genes for different traits segregate independently during gamete formation.


(b) Explain the process of transcription in prokaryotes. How is it different from transcription in eukaryotes?

Correct Answer:

View Solution

Transcription in Prokaryotes: - Initiation: RNA polymerase binds to the promoter region of DNA, initiating the process of transcription. - Elongation: RNA polymerase synthesizes RNA in the 5’ to 3’ direction, creating an RNA sequence complementary to the DNA template strand. - Termination: Transcription stops when RNA polymerase reaches the terminator sequence. The newly formed RNA transcript is released. Differences between Prokaryotic and Eukaryotic Transcription: - RNA Polymerase: - Prokaryotes use a single RNA polymerase for transcription. - Eukaryotes have three distinct RNA polymerases (RNA Polymerase I, II, and III), each responsible for transcribing specific types of RNA. - Location: - In prokaryotes, transcription occurs in the cytoplasm, and translation begins immediately after transcription. - In eukaryotes, transcription occurs in the nucleus, and the RNA transcript must undergo processing before it is transported to the cytoplasm for translation. - RNA Processing: - Prokaryotic RNA does not require processing and is directly functional as mRNA. - Eukaryotic RNA undergoes capping (addition of a methylated guanosine cap at the 5’ end), tailing (addition of a poly-A tail at the 3’ end), and splicing (removal of non-coding introns and joining of coding exons) to become a mature mRNA. Quick Tip: Transcription in prokaryotes is simpler and faster due to the absence of a nucleus and RNA processing, unlike the complex and compartmentalized transcription in eukaryotes.

 

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

Ask your question

Subscribe To Our News Letter

Get Latest Notification Of Colleges, Exams and News

© 2026 Patronum Web Private Limited