CBSE Class 12 Biology Set 3 Question Paper PDF (57/5/3​)​ is now available for download. CBSE conducted the Class 12 Biology examination on March 19, 2024, from 10:30 AM to 1:30 PM. The question paper consists of 33 questions carrying a total of 70 marks. Section A includes 16 MCQs for 1 mark each, Section B contains 5 very short-answer questions for 2 marks each, Section C comprises 7 short-answer questions for 3 marks each, Section D comprises 2 Case-based questions carries 4 marks each and Section E comprises 3 long-answer questions carries 5 marks each. 

Candidates can use the link below to download the CBSE Class 12 Biology Set 3 Question Paper with detailed solutions.

CBSE Class 12 Biology Question Paper 2024 (Set 3- 57/5/3) with Answer Key

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

CBSE Class 12 2024 Biology Questions with Solutions


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



A person with trisomy of 21st chromosome will be afflicted with Down's syndrome. This condition is characterized by a small round head, furrowed tongue, partially open mouth, and characteristic palm crease. Quick Tip: For conditions like Down's syndrome, remember that trisomy 21 leads to physical and developmental traits such as furrowed tongue and palm crease.


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 impact of chromosomal events on genetic variation.
- Independent assortment occurs during meiosis, leading to the random distribution of maternal and paternal chromosomes, which increases genetic variation.
- Crossing over involves the exchange of genetic material between homologous chromosomes, creating new genetic combinations.
- Mutation introduces new genetic variations in the DNA sequence.
- Linkage, however, refers to genes located close together on the same chromosome being inherited together, reducing recombination and, therefore, genetic variation.

Step 2: Conclusion.
The correct answer is \( \textbf{(C)} \), as linkage does not contribute to genetic variation but rather conserves specific gene combinations. Quick Tip: Linkage reduces the chance of recombination between genes, whereas independent assortment and crossing over actively promote genetic diversity.


Question 3:

Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below:

diagram

  1. A – zygote, B – degenerating synergids, C – degenerating antipodals, D – PEN
  2. A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals
  3. A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote
  4. 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



The correct labellings in the figure are:

- A represents degenerating synergids as they are no longer needed after fertilization.

- B is the zygote formed after the fusion of the male and female gametes.

- C is the primary endosperm nucleus (PEN) formed by the fusion of a sperm with two polar nuclei.

- D represents degenerating antipodals as their function is complete. Quick Tip: In a fertilised embryo sac: - Zygote forms at the center of the sac. - Degenerating synergids and antipodals are found at opposite poles. - PEN results from the fusion of a sperm with two polar nuclei.


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:

[Diagram of Pedigree Chart]

  1. Dominant X-linked
  2. Recessive X-linked
  3. Autosomal dominant
  4. Autosomal recessive
Correct Answer: (4) Autosomal recessive
View Solution

Solution: The disorder depicted in the pedigree chart follows an autosomal recessive inheritance pattern. This can be inferred from the following observations:
1. The trait skips generations, which is typical for recessive traits.
2. Both males and females are equally affected, indicating an autosomal pattern.
3. The trait is expressed only when an individual inherits two copies of the recessive allele, one from each parent.


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



In agarose gel electrophoresis, DNA is separated based on size. Staining with ethidium bromide allows the DNA bands to be visualized under UV light. Ethidium bromide binds to DNA and fluoresces under UV light, making the bands visible. Quick Tip: Always use UV light to observe ethidium bromide-stained DNA during gel electrophoresis.


Question 6:

A phenomenon where a male insect mistakenly identifies the patterns of an orchid 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 the phenomenon.
Pseudocopulation occurs when a male insect is deceived by the appearance of a flower, mistaking it for a female insect partner, and attempts to copulate with it. This behavior leads to pollination of the flower.
- The other options:
- Pseudopollination: Not a valid term in this context.
- Pseudoparthenocarpy: Refers to fruit development without fertilization, unrelated here.
- Pseudofertilisation: Does not describe this phenomenon.

Step 2: Conclusion.
The correct answer is \( \textbf{(A)} \), as pseudocopulation is the phenomenon described. Quick Tip: Pseudocopulation is a strategy used by certain flowers, such as orchids, to mimic female insects and achieve pollination without offering nectar or other rewards.


Question 7:

Observe the schematic representation of assisted reproductive technology given below:

the schematic representation


Identify the most appropriate technique depicted in the above diagram.

(A) IUT
(B) IUI
(C) ICSI
(D) ZIFT

Correct Answer: (C) ICSI
View Solution



The diagram represents Intracytoplasmic Sperm Injection (ICSI), where a single sperm is directly injected into the cytoplasm of the egg using a fine needle. This technique is used in cases of severe male infertility or failed fertilization in conventional IVF. Quick Tip: Remember: In ICSI, a single sperm is injected directly into the egg's cytoplasm using a fine needle.


Question 8:

The source of ‘Smack’ is:

(A) Leaves of Cannabis sativa
(B) Flowers of Datura
(C) Fruits of Erythroxylum coca
(D) Latex of Papaver somniferum

Correct Answer: (D) Latex of \textit{Papaver somniferum}
View Solution



Step 1: Identifying the source of ‘Smack’.
- Smack is a street name for heroin, which is derived from the latex of Papaver somniferum (the opium poppy).
- The other options:
- \textit{Cannabis sativa: Source of marijuana and hashish.
- \textit{Datura: Contains toxic alkaloids but not heroin.
- \textit{Erythroxylum coca: Source of cocaine, unrelated to heroin.

Step 2: Conclusion.
The correct answer is \( \textbf{(D) \), as the opium poppy produces the raw material for heroin. Quick Tip: Heroin is synthesized from morphine, which is extracted from the opium latex of Papaver somniferum.


Question 9:

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

(A) A – Waksman; B – Streptococcus
(B) A – Fleming; B – Penicillium notatum
(C) A – Waksman; B – Bacillus brevis
(D) A – Fleming; B – Staphylococci

Correct Answer: (D) A – Fleming; B – \textit{Staphylococci}
View Solution



Step 1: Understanding the discovery of the first antibiotic.
- Alexander Fleming discovered penicillin accidentally in 1928 while studying the bacterium Staphylococci.
- The mold \textit{Penicillium notatum inhibited bacterial growth in his cultures.

Step 2: Analyzing the options.
- Options (A) and (C) refer to discoveries by Selman Waksman, not Fleming.
- Option (B) incorrectly pairs Fleming with \textit{Penicillium notatum as the working organism. Fleming observed \textit{Staphylococci.

Step 3: Conclusion.
The correct answer is \( \textbf{(D) \), as Fleming’s work involved Staphylococci. Quick Tip: Penicillin’s discovery marked the start of the antibiotic era, with Fleming’s observation of inhibited \textit{Staphylococci growth.


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’ – GUAAGUA – 3’

Correct Answer: (A) 5' – AUGAAUG – 3'
View Solution



The RNA transcript is complementary to the template strand and identical to the coding strand (except that uracil (U) replaces thymine (T) in RNA). Therefore, the RNA transcript of the sequence 5' – ATGAATG – 3' is 5' – AUGAAUG – 3'. Quick Tip: Remember: RNA is complementary to the template strand and identical to the coding strand, except that U replaces T.


Question 11:

Match the following genes of the lac operon listed in column ‘A’ with their respective products listed in column ‘B’:

A (Gene) B (Products)
a. ‘i’ gene (i) β-galactosidase
b. ‘z’ gene (ii) lac permease
c. ‘a’ gene (iii) repressor
d. ‘y’ gene (iv) transacetylase

Select the correct option:

  1. (i), (iii), (ii), (iv)
  2. (iii), (i), (iv), (ii)
  3. (iii), (i), (ii), (iv)
  4. (iii), (iv), (i), (ii)
Correct Answer: (3) (iii), (i), (ii), (iv)
View Solution

Solution: In the lac operon:
- The ‘i’ gene codes for the repressor protein, which inhibits the operon when lactose is absent.
- The ‘z’ gene produces β-galactosidase, which breaks down lactose into glucose and galactose.
- The ‘a’ gene encodes transacetylase, which helps in lactose metabolism.
- The ‘y’ gene produces lac permease, which facilitates the entry of lactose into the cell.


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

Correct Answer: (C) Chromosome 1 and Y
View Solution



Step 1: Understanding the gene distribution among chromosomes.
- Chromosome 1 has the highest number of genes, as it is the largest human chromosome.
- Chromosome Y has the least number of genes, as it primarily carries genes related to male sex determination and reproduction.

Step 2: Conclusion.
The correct answer is \( \textbf{(C)} \), as Chromosome 1 has the highest and Chromosome Y has the least number of genes. Quick Tip: Remember: Chromosome 1 is the largest and most gene-rich, while Chromosome Y is the smallest and least gene-dense.


Question 13:

Assertion (A): In birds, the sex of the offspring is determined by males.

Reason (R): Males are homogametic while females are heterogametic.

(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.
(D) (A) is false, but (R) is true.

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



Step 1: Understanding sex determination in birds.
- In birds, the sex of the offspring is determined by females, not males, as females are heterogametic (ZW), while males are homogametic (ZZ).
- Thus, the assertion (A) is incorrect.

Step 2: Analyzing the reason.
- The reason (R) correctly states that males are homogametic while females are heterogametic in birds.

Step 3: Conclusion.
Since the assertion is false and the reason is true, the correct answer is \( \textbf{(D)} \). Quick Tip: In birds: - Males (ZZ) are homogametic. - Females (ZW) are heterogametic and determine the offspring's sex.


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.

(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.
(D) (A) is false, but (R) is true.

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



Step 1: Analyzing the assertion (A).
- Biodiversity hotspots are indeed regions of high species richness and endemism, so the assertion is true.

Step 2: Analyzing the reason (R).
- The total number of biodiversity hotspots worldwide is 36, not 22. India is home to 4 biodiversity hotspots (the Himalayas, Indo-Burma, Western Ghats-Sri Lanka, and Sundaland). Hence, the reason is false.

Step 3: Conclusion.
Since the assertion is true but the reason is false, the correct answer is \( \textbf{(C)} \). Quick Tip: Currently, there are 36 biodiversity hotspots in the world, with four located in India: Himalayas, Indo-Burma, Western Ghats-Sri Lanka, and Sundaland.


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.

(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.
(D) (A) is false, but (R) is true.

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



Step 1: Analyzing the assertion (A).
- AIDS (Acquired Immunodeficiency Syndrome) is indeed caused by the HIV (Human Immunodeficiency Virus). Hence, the assertion is true.

Step 2: Analyzing the reason (R).
- HIV is a retrovirus with RNA, not DNA, as its genetic material. Therefore, the reason is false.

Step 3: Conclusion.
Since the assertion is true but the reason is false, the correct answer is \( \textbf{(C)} \). Quick Tip: HIV is a retrovirus with RNA as its genetic material, not DNA.


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.

(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.
(D) (A) is false, but (R) is true.

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



Step 1: Analyzing the assertion (A).
- Single-stranded DNA or RNA tagged with a radioactive molecule is indeed called a probe. Probes are used in molecular diagnosis to detect specific DNA sequences. Hence, the assertion is true.

Step 2: Analyzing the reason (R).
- A probe searches for and hybridizes with its complementary DNA sequence within a sample, confirming the assertion's purpose. The reason is true and correctly explains the assertion.

Step 3: Conclusion.
Both the assertion and the reason are true, and the reason explains the assertion. The correct answer is \( \textbf{(A)} \). Quick Tip: Probes in molecular biology: - Single-stranded DNA or RNA tagged with a radioactive or fluorescent molecule. - Used to detect specific DNA/RNA sequences by hybridization.


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, which states:

The percentage of adenine (A) equals the percentage of thymine (T), and the percentage of cytosine (C) equals the percentage of guanine (G).
Since adenine constitutes 31%, thymine also constitutes 31%. Together, A + T = 62%.
The remaining 38% is divided equally between cytosine and guanine. Thus, cytosine constitutes 19%. Quick Tip: Chargaff’s Rule: A = T and C = G. Use this to calculate the percentage of any base in DNA when the percentage of one base is known.


Question 18:

Observe the population growth curve and answer the questions given below:

curve

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

(b) Mention what does ‘K’ in the graph represent.

Correct Answer:
View Solution



(a)

Growth curve ‘A’ represents exponential growth, which occurs when resources are unlimited and environmental constraints are absent.
Growth curve ‘B’ represents logistic growth, which occurs when resources are limited, leading to population stabilization at carrying capacity.


(b) ‘K’ in the graph represents the carrying capacity of the environment, which is the maximum population size that can be sustained indefinitely with available resources. Quick Tip: - Exponential growth: Unlimited resources, no constraints.
- Logistic growth: Limited resources, stabilizes at carrying capacity (K).
- Carrying capacity (K): Maximum sustainable population size.


Question 19:

19

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

(b) What are the arrows in the given figure indicating? Write the result obtained thereafter.

Correct Answer:
View Solution

N/A Quick Tip: Remember: - EcoRI recognizes the sequence 5' – GAATTC – 3'. - Palindromic sequences are the same on complementary strands when read in opposite directions. - Sticky ends are crucial for recombination in genetic engineering.


Question 20:

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

Correct Answer:
View Solution



The following events occur during sewage treatment to reduce the Biochemical Oxygen Demand (BOD) of primary effluent:

The primary effluent is passed into a large aeration tank.
The effluent is constantly agitated mechanically, and air is pumped into it.
Vigorous growth of useful aerobic microbes (such as fungi and bacteria) occurs, forming flocs.
These microbes consume the major part of the organic matter present in the effluent, reducing the BOD. Quick Tip: Remember: BOD reduction is achieved in the aeration tank through the action of aerobic microbes that degrade organic matter in the primary effluent.


Question 21:

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

Correct Answer:
View Solution



Farmers prefer apomictic seeds because:

They do not require repeated hybridization, saving time and cost.
They maintain the hybrid vigor (heterosis) over generations, as apomixis produces genetically identical offspring. Quick Tip: Remember: Apomictic seeds are cost-effective and maintain hybrid vigor across generations, making them favorable for farmers.


OR
Question 21:

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

Correct Answer:
View Solution



Advantage:

Amniocentesis can detect genetic abnormalities and chromosomal disorders in the fetus, aiding in early diagnosis and management.

Disadvantage:

It carries a small risk of complications such as miscarriage or infection. Quick Tip: Amniocentesis: Think of "genetic diagnosis" (advantage) vs. "risk of miscarriage" (disadvantage).


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 7-methylguanosine cap is added to the 5' end of hnRNA to protect it from degradation and assist in ribosome binding during translation.
Tailing: A poly-A tail (a string of adenine nucleotides) is added to the 3' end of the hnRNA to enhance stability and facilitate export from the nucleus.
Splicing: Introns (non-coding regions) are removed, and exons (coding regions) are joined together to form a continuous sequence that codes for a protein.

This processing occurs in the nucleus of the eukaryotic cell. Quick Tip: Remember the three main steps in hnRNA processing: Capping, Tailing, and Splicing. These modifications occur in the nucleus and are essential for mRNA stability and functionality.


Question 23:

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

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

Correct Answer:
View Solution



(a) Importance of Mother’s Milk:

Mother’s milk provides all essential nutrients required for the growth and development of the newborn.
It contains antibodies (such as IgA) that boost the immune system and protect the baby from infections.
Breast milk promotes healthy gut flora and reduces the risk of allergies and illnesses.


(b) Vaccine and Its Principle:

A vaccine is a biological preparation containing dead or weakened pathogens or their components.
The principle of vaccination is to stimulate the immune system to produce memory cells against a specific pathogen, enabling a faster and stronger response upon subsequent exposure. Quick Tip: Breastfeeding provides both nutritional and immunological benefits. Vaccines work on the principle of immunological memory.


Question 24:

(a) Tropical regions harbor more species than temperate regions. How have biologists tried to explain this in their own ways? Explain.

OR

(i) What does an ecological pyramid represent?

(ii) The Ecological pyramids may have an ‘upright’ or an ‘inverted’ shape. Justify with the help of suitable examples.

Correct Answer:
View Solution



(a) Reasons for Higher Biodiversity in Tropical Regions:

Favorable Climate: Tropical regions have a warm and stable climate throughout the year, which promotes the growth and reproduction of diverse organisms.
Higher Productivity: Abundant sunlight and rainfall increase primary productivity, supporting a variety of life forms.
Long Evolutionary Time: Tropical regions have been relatively undisturbed over geological timescales, allowing species to evolve and diversify.


(b) (i):
An ecological pyramid represents the flow of energy and the number of organisms at each trophic level in an ecosystem. It is a graphical representation that shows the relative amount of energy or biomass at each level, starting with producers at the base, followed by herbivores, carnivores, and top predators. The pyramid illustrates the decrease in energy and number of organisms as we move up through the trophic levels.




(b) (ii):
Ecological pyramids may take two forms:


Upright Ecological Pyramid: In this type, the number of organisms, biomass, or energy decreases as we move up from the producers to the apex consumers. It is the most common shape of ecological pyramids, where producers form the base and each subsequent level contains fewer organisms or less energy.
Example: A forest ecosystem is an example of an upright pyramid, where plants (producers) form the base, followed by herbivores (primary consumers), then carnivores (secondary consumers), and finally apex predators.

Inverted Ecological Pyramid: This occurs when the number of organisms or the biomass increases at higher trophic levels. This happens typically in aquatic ecosystems, where the biomass of primary producers (e.g., phytoplankton) is less than that of primary consumers (e.g., zooplankton). In these ecosystems, producers are consumed rapidly, and a large number of consumers are supported.
Example: In a pond or lake, the pyramid of biomass may be inverted, with more zooplankton than phytoplankton. Quick Tip: - Tropical regions have high biodiversity due to favorable conditions and longer evolutionary times. - In species-area relationships, larger areas generally support more species, but the rate of increase diminishes.


Question 25:

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

Correct Answer:
View Solution



Reasons for Interest in Artificial Hybridisation:

It allows breeders to combine desirable traits from two different plants, such as disease resistance and high yield.
It enhances genetic variation in crops, increasing the potential for better adaptability and productivity.
It helps in developing new crop varieties with improved quality and performance.


Process of Artificial Hybridisation:

The process involves emasculation, where the anthers of a flower are removed to prevent self-pollination.
The emasculated flower is then covered with a bag (bagging) to protect it from unwanted pollen.
Pollen from the desired parent plant is collected and dusted onto the stigma of the emasculated flower, ensuring cross-pollination. Quick Tip: Artificial hybridisation is widely used to develop high-yielding, disease-resistant, and stress-tolerant crop varieties.


Question 26:

(a) What are transgenic animals?

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

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

Correct Answer:
View Solution



(a) Definition of Transgenic Animals:

Transgenic animals are those that have had their DNA modified by the introduction of a foreign gene or genes, making them capable of expressing traits that are not naturally found in their species.

(b) Transgenic Animal with the Largest Population:

Mice are the transgenic animals with the largest number among all existing transgenic animals. They are extensively used in genetic and biomedical research.

(c) Reasons for Producing Transgenic Animals:

To study the function of specific genes and their role in development and diseases.
For producing pharmaceutical proteins and drugs (e.g., recombinant insulin).
To create animal models for human diseases, enabling research on treatments and therapies. Quick Tip: Transgenic animals like mice and sheep are instrumental in biomedical research, drug production, and understanding genetic functions.


Question 27:

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

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

Correct Answer:
View Solution

27

(a) Pyramid of Biomass in the Sea:

The pyramid of biomass in the sea is typically inverted. This is because:

The biomass of phytoplankton (producers) is much smaller than the biomass of the fishes (primary and secondary consumers).
Phytoplankton have a high turnover rate due to rapid reproduction and consumption by herbivorous fishes, maintaining their population despite low biomass.



Inverted Pyramid of Biomass (Marine Ecosystem) \[ Fishes (Consumers) - Large Biomass at the Top
Phytoplankton (Producers) - Small Biomass at the Bottom \]



(b) Conditions for an Upright Pyramid:

The pyramid of biomass will always remain upright in terrestrial ecosystems where:

The biomass of producers is always higher than the biomass of consumers.
Energy transfer follows the 10% rule, with significant energy loss at each trophic level. Quick Tip: - In marine ecosystems, the pyramid of biomass is often inverted due to the high productivity and turnover of phytoplankton. - Terrestrial ecosystems generally have upright pyramids due to greater producer biomass.


Question 28:

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

(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



(a) Reason for Replication within the Replication Fork:

DNA replication occurs within the replication fork because DNA unwinds progressively at the fork to expose the two strands for replication.
The entire length of DNA cannot replicate simultaneously as it would require the entire molecule to be unwound at once, which is not feasible due to its length and complexity.
The replication fork allows for localized unwinding and replication, ensuring accuracy and efficiency.


(b) Continuous and Discontinuous Replication:

DNA polymerase synthesizes DNA only in the 5' to 3' direction.
On the leading strand, replication is continuous because it proceeds in the same direction as the unwinding of the helix.
On the lagging strand, replication is discontinuous, as it proceeds in short segments (Okazaki fragments) opposite to the direction of unwinding.
These fragments are later joined by DNA ligase to form a continuous strand.


Schematic Representation:
28
Quick Tip: - Leading strand: Continuous synthesis (5' to 3'). - Lagging strand: Discontinuous synthesis via Okazaki fragments. - DNA replication is semi-discontinuous due to the opposite orientation of strands.


SECTION D

Question 29:

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

29
(a) Name the hormones and their source organ, which are responsible for the menstrual cycle at puberty.

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

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

OR

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

Correct Answer:
View Solution



(a) The hormones responsible for the menstrual cycle at puberty are:

FSH (Follicle Stimulating Hormone): Secreted by the anterior pituitary gland, stimulates follicle development.
LH (Luteinizing Hormone): Secreted by the anterior pituitary gland, triggers ovulation and corpus luteum formation.
Estrogen and Progesterone: Secreted by the ovaries, regulate the uterine lining during the cycle.


(b) Implantation should occur during the secretory phase (days 15-28) of the menstrual cycle. During this phase, the uterine lining (endometrium) is thick and vascularized due to the action of progesterone, creating a suitable environment for embryo implantation.

(c) The hormone responsible for the proliferative phase is Estrogen, secreted by the developing ovarian follicles. It stimulates the regeneration and thickening of the uterine lining (endometrium) after menstruation.

OR

If the released ovum is not fertilized, the corpus luteum degenerates, leading to a drop in progesterone levels. This causes the uterine lining (endometrium) to break down and shed, resulting in menstruation. Quick Tip: - Menstrual cycle phases: Menses, Proliferative, Secretory. - Implantation occurs in the secretory phase due to high progesterone levels. - Progesterone = Maintains endometrium; Estrogen = Builds endometrium.


Question 30:

Read the following passage and answer the questions that follow:

``Mosquitoes are drastically affecting 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.

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

(OR)

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

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

Correct Answer:
View Solution



(a)

(i) In humans, Plasmodium enters in the form of sporozoites, which are injected by the female \textit{Anopheles mosquito during a bite.
(ii) In the female \textit{Anopheles body, \textit{Plasmodium enters in the form of gametocytes, which are taken up with the blood meal from an infected person.


(b) The symptoms of malaria do not appear immediately after the bite because the parasite undergoes an incubation period inside the human liver and blood cells. Symptoms appear 10–14 days later when the parasite multiplies and ruptures red blood cells, releasing toxins. This results in fever, chills, and sweating.

OR

In the female \textit{Anopheles mosquito, after sucking blood from an infected person:

The gametocytes of \textit{Plasmodium develop into sporozoites in the mosquito's gut.
The sporozoites migrate to the salivary glands of the mosquito, making it ready to infect another human during its next bite.


(c) A mosquito species other than \textit{Anopheles is \textit{Aedes aegypti, which carries the pathogen for dengue fever. Quick Tip:Plasmodium stages: Sporozoites enter humans; gametocytes enter mosquitoes.
- Aedes aegypti: Carries pathogens for dengue and chikungunya. - Malaria symptoms appear after the incubation period (10–14 days).


SECTION E

Question 31:

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

(ii) In a human female, the probability of an ovum getting fertilized by more than one sperm is impossible. Give a reason.



OR

(b)
(i) With the help of a labelled diagram only, show the different stages of embryo development in a dicot plant.

(ii) Endosperm development precedes embryo development. Justify.

 

View Solution

31 Functions of labelled parts:
• Head: Contains the nucleus with genetic material and acrosome for penetrating the egg.
• Middle Piece: Contains mitochondria that provide energy for motility.
• Tail: Facilitates movement of the sperm towards the egg.
• Plasma Membrane: Encloses the sperm and maintains its structural integrity.

(ii) The probability of an ovum being fertilized by more than one sperm is impossible due
to a phenomenon called cortical reaction, where the egg membrane changes after the first
sperm enters, preventing entry of additional sperms.
Solution (b):
(i) Diagram showing stages of embryo development in a dicot plant:

31
(ii) Endosperm development precedes embryo development because the endosperm pro-
vides nutrients to the developing embryo. It supports early stages of growth until the plant
can photosynthesize independently.
Quick Tip
- Human sperm: Acrosome helps penetrate the egg; mitochondria in the middle piece
generate energy.
- Cortical reaction prevents polyspermy in humans.
- Dicot embryos develop after endosperm formation, ensuring nutrient availability.


Question 32(a):

Draw a schematic diagram of the cloning vector pBR 322 and label (1) Bam HI site (2) gene for ampicillin resistance (3) ‘ori' (4) ‘rop' gene.

State the role of ‘rop' gene.

A cloning vector does not have a selectable marker. How will it affect the process of cloning?

Why is insertional inactivation preferred over the use of selectable markers in cloning vectors?

OR

Question 32(b):

Name the nematode (scientific name) that infects the roots of tobacco plant and reduces its yield.

Name the vector that is used to introduce nematode-specific genes into the host plant (tobacco).

How do sense and anti-sense RNAs function?

Why could the parasite not survive in a transgenic tobacco plant?

View Solution

Solution (a):
(i) A schematic diagram of the cloning vector pBR 322 is shown below:


(ii) The 'rop' gene in pBR 322 encodes proteins involved in maintaining the copy number of the plasmid.
(iii) If a cloning vector does not have a selectable marker, it would be difficult to differentiate between transformed and non-transformed cells, complicating the identification of successful clones.
(iv) Insertional inactivation is preferred because it allows direct identification of recombinant colonies. It inactivates a selectable marker gene (e.g., antibiotic resistance), so only recombinant cells are easily identified.

Solution (b):
(i) The nematode that infects the roots of tobacco plants is *Meloidogyne incognita*.
(ii) The vector used to introduce nematode-specific genes into the host plant is *Agrobacterium tumefaciens*.
(iii) Sense and anti-sense RNAs are complementary RNA strands. When expressed together, they form a double-stranded RNA that silences the target nematode gene through RNA interference (RNAi).
(iv) The parasite could not survive in the transgenic tobacco plant because the RNAi mechanism silences the nematode-specific genes, preventing its growth and reproduction.


Question 33:

Work out a dihybrid cross up to \( F_2 \) generation between pea plants bearing violet-coloured axial flowers and white-coloured terminal flowers using Punnett’s square. Give their \( F_2 \) phenotypic ratio. State the Mendel’s law of inheritance that was derived from such a cross.

OR

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

Correct Answer:
View Solution




(a) Dihybrid Cross:


Let:

Violet axial flowers = ( VVAA )
White terminal flowers = ( vvaa )


33

Mendel’s Law Derived: Mendel’s law of Independent Assortment states that the inheritance of one trait is independent of the inheritance of another.

% Solution for part (b)
(b) Transcription in Prokaryotes:

Transcription in prokaryotes involves the synthesis of mRNA from DNA using RNA polymerase. The process includes:

Initiation: RNA polymerase binds to the promoter region of DNA, aided by sigma factors.
Elongation: RNA polymerase synthesizes RNA in the 5’ to 3’ direction using the DNA template strand.
Termination: Transcription ends when RNA polymerase reaches a terminator sequence, releasing the mRNA.


Differences from Eukaryotic Transcription:

In eukaryotes, transcription occurs in the nucleus, whereas in prokaryotes, it occurs in the cytoplasm.
Eukaryotic transcription requires multiple RNA polymerases and transcription factors, whereas prokaryotes use a single RNA polymerase.
Eukaryotic mRNA undergoes capping, tailing, and splicing, which are absent in prokaryotic transcription. Quick Tip: - In a dihybrid cross, the \( F_2 \) phenotypic ratio is always \( 9:3:3:1 \). - Prokaryotic transcription is simpler, whereas eukaryotic transcription involves additional post-transcriptional modifications.