CBSE Class 12 Biology Set 2 Question Paper PDF (57/3/2)​ 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 2 Question Paper with detailed solutions.

CBSE Class 12 Biology Question Paper 2024 (Set 2- 57/3/2) 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:

The ploidy of the apomictic embryos developing from the integument cells and synergids respectively would be:

  • (A) \( n, 2n \)
  • (B) \( 2n, n \)
  • (C) \( 3n, 2n \)
  • (D) \( 2n, 3n \)
Correct Answer: (B) \( 2n, n \)
View Solution

Apomictic embryos arise without fertilization. When formed from integument
cells, they retain the somatic ploidy level, which is 2n, whereas synergids, being haploid
structures, remain n.
This aligns with option (B) 2n, n.
Quick Tip: Apomixis leads to clonal reproduction, maintaining genetic uniformity, as the offspring are genetically identical to the parent.


Question 2:

A DNA fragment has 2500 nucleotides, out of which 240 are Guanine. How many bases having double hydrogen bonds between them does this DNA fragment possess?

  • (A) \( 480 \)
  • (B) \( 720 \)
  • (C) \( 1010 \)
  • (D) \( 2020 \)
Correct Answer: (D) \( 2020 \)
View Solution

In DNA, Guanine (G) pairs with Cytosine (C) via triple hydrogen bonds, while
Adenine (A) pairs with Thymine (T) via double hydrogen bonds.
Since there are 240 Guanine bases, there must be 240 Cytosine bases. This totals 480
nucleotides, leaving 2500 − 480 = 2020 bases as Adenine and Thymine.
Since A-T pairs form double hydrogen bonds, the number of such bases forming double
bonds is 2020. Quick Tip: Adenine pairs with Thymine (A-T) through double hydrogen bonds, while Guanine pairs with Cytosine (G-C) through triple hydrogen bonds.


Question 3:

The mechanism to produce seeds without fertilization has evolved in the given family of the flowering plants:

  • (A) Asteraceae
  • (B) Solanaceae
  • (C) Malvaceae
  • (D) Liliaceae
Correct Answer: (A) Asteraceae
View Solution

Apomixis is a special type of asexual reproduction in which plants can produce
seeds without fertilization. It results in the formation of clonal progeny, ensuring genetic
uniformity across generations.
This phenomenon is common in the Asteraceae family, which includes species such as
Taraxacum (dandelions) and Hieracium.
Apomixis provides advantages such as bypassing meiotic recombination, producing
genetically identical offspring, and ensuring stability in hybrid varieties.
Quick Tip: Apomixis is an evolutionary advantage that allows plants to reproduce without genetic variation, helping to maintain desirable traits in agricultural crops.


Question 4:

Louis Pasteur dismissed the theory of spontaneous generation by his experiments using pre-sterilized flasks and:

  • (A) Live yeast
  • (B) Killed yeast
  • (C) Live bacteria
  • (D) Killed bacteria
Correct Answer: (B) Killed yeast
View Solution

The theory of spontaneous generation proposed that living organisms could arise
from non-living matter. Louis Pasteur refuted this idea by designing an experiment using
pre-sterilized swan-neck flasks containing nutrient broth.
When the flasks remained sealed, no microbial growth was observed, proving that
microorganisms do not arise spontaneously.
However, when the flasks were exposed to air, microbial contamination occurred, confirming
that microorganisms originate from pre-existing life forms in the environment. Quick Tip: Louis Pasteur’s experiment provided strong evidence for the biogenesis theory, which states that all life originates from existing life.


Question 5:

After the 1850s in the post-industrialization era in England, the expected effect of natural selection on the number of white-winged moths as compared to the dark-winged moths was:

  • (A) Less in number
  • (B) More in number
  • (C) Both were less in number
  • (D) Both were more in number
Correct Answer: (A) Less in number
View Solution

Before industrialization, white-winged moths were more common because they
camouflaged well against lichen-covered tree trunks.
Industrial pollution darkened the tree bark due to soot deposition, making dark-winged
moths less visible to predators and increasing their survival rate.
White-winged moths, now more visible, were predated more frequently, leading to a decline
in their population. This is a classic example of natural selection known as industrial
melanism.
Quick Tip: Natural selection favors traits that enhance survival in response to environmental changes.


Question 6:

A Snapdragon plant bearing red color flowers was crossed with a Snapdragon plant bearing white color flowers. The \( F_1 \) progeny on selfing produced the progeny in the ratio of:

  • (A) \( 50% \) Red : \( 50% \) White
  • (B) \( 25% \) Pink : \( 50% \) White : \( 25% \) Red
  • (C) \( 50% \) Red : \( 50% \) Pink
  • (D) \( 25% \) Red : \( 50% \) Pink : \( 25% \) White
Correct Answer: (D) \( 25% \) Red : \( 50% \) Pink : \( 25% \) White
View Solution

Snapdragon flower color follows incomplete dominance, where the heterozygous
condition results in a pink phenotype instead of the dominant red.
Crossing a red flower (RR) with a white flower (rr) results in all pink (Rr) flowers in the F1
generation.
Selfing the F1 generation (Rr × Rr) produces a 1:2:1 ratio in the F2 generation: 25% Red
(RR) - 50% Pink (Rr) 25% White (rr)Quick Tip: Incomplete dominance results in intermediate traits, where the heterozygous condition shows a blend of both alleles rather than the expression of one dominant trait.


Question 7:

In which of the following chromosomal disorders do individuals have short stature, underdeveloped feminine character, and sterile ovaries?

  • (A) Down’s syndrome
  • (B) Turner’s syndrome
  • (C) Klinefelter’s syndrome
  • (D) Patau’s syndrome
Correct Answer: (B) Turner’s syndrome
View Solution

Turner’s syndrome is caused by monosomy of the X chromosome (45, X).
Individuals with this syndrome exhibit characteristics such as short stature, webbed neck,
lack of secondary sexual development, and sterility due to ovarian dysgenesis.
Unlike Klinefelter’s syndrome (which affects males), Turner’s syndrome only affects
females.
Quick Tip: Turner’s syndrome is a chromosomal disorder caused by the absence of one X chromosome in females.


Question 8:

Which one of the following represents the correct annealing of primers to the DNA to be amplified in the PCR?

  • (A)
  • (B)
  • (C)
  • (D)
Correct Answer: (B)
View Solution

Polymerase Chain Reaction (PCR) is a method used to amplify specific DNA sequences by using primers that bind to the template strand.

The primers must anneal in a complementary fashion, with one primer binding to the 3' end of one strand and the other binding to the 3' end of the opposite strand.

Option (B) correctly illustrates the primers binding in an antiparallel orientation, which is essential for successful amplification. Quick Tip: In PCR, primers must be complementary to the template DNA and bind in an antiparallel direction to allow DNA polymerase to extend in the 5' to 3' direction.


Question 9:

A patient is suffering from fever, chills, cough, headache, and bluish fingernails and lips. These symptoms are of the disease:

  • (A) Malaria
  • (B) Typhoid
  • (C) Common cold
  • (D) Pneumonia
Correct Answer: (D) Pneumonia
View Solution

Pneumonia is a lung infection that leads to symptoms such as cough, chills, fever, and difficulty breathing.

The bluish tint seen in the fingernails and lips is a sign of oxygen deficiency, a prominent symptom in severe cases of pneumonia.

Other conditions, such as malaria and typhoid, do not primarily cause respiratory issues or oxygen shortage. Quick Tip: Pneumonia is often caused by bacterial infections like *Streptococcus pneumoniae* or viral infections, and it requires prompt medical treatment.


Question 10:

In an experiment, E. coli is grown in a medium containing \( ^{14}NH_4Cl \) (\( ^{14}N \) is the light isotope of nitrogen) followed by growing it for six generations in a medium having heavy isotope of nitrogen (\( ^{15}N \)). After six generations, their DNA was extracted and subjected to CsCl density gradient centrifugation. Identify the correct density (Light/Hybrid/Heavy) and ratio of the bands of DNA in CsCl density gradient centrifugation.

  • (A) Hybrid : Heavy, 1 : 16
  • (B) Light : Heavy, 1 : 31
  • (C) Hybrid : Heavy, 1 : 31
  • (D) Light : Heavy, 1 : 05
Correct Answer: (C) Hybrid : Heavy, 1 : 31
View Solution

Meselson and Stahl's experiment provided evidence for semi-conservative replication, where the parental DNA strands act as templates for the synthesis of new strands.

Initially, *E. coli* grown in \( ^{14}N \) medium contains only light DNA. When transferred to \( ^{15}N \) medium, the newly synthesized DNA incorporates \( ^{15}N \), resulting in a hybrid DNA fraction after the first round of replication.

After six generations in \( ^{15}N \) medium, most of the DNA will be heavy (completely incorporated with \( ^{15}N \)), but a small fraction (1 in 31) will still be in a hybrid form. Quick Tip: In semi-conservative DNA replication, each daughter DNA strand consists of one original parental strand and one newly synthesized strand.


Question 11:

For the replication of the first recombinant DNA, Stanley Cohen and Herbert Boyer used the DNA polymerase of:

  • (A) Thermus aquaticus
  • (B) Salmonella typhimurium
  • (C) Escherichia coli
  • (D) Haemophilus influenzae
Correct Answer: (C) Escherichia coli
View Solution

Stanley Cohen and Herbert Boyer were the pioneers in creating the first recombinant DNA by combining DNA from two different organisms.

They utilized DNA polymerase from Escherichia coli (E. coli), which played an essential role in amplifying the DNA during recombinant DNA technology.

The use of E. coli was crucial due to its status as a model organism for genetic manipulation, along with its ability to grow rapidly, making it easier to clone recombinant DNA. Quick Tip: Escherichia coli is a commonly used organism in molecular biology for cloning and recombinant DNA work due to its quick replication and genetic flexibility.


Question 12:

The population growth curve applicable for a population growing in nature with limited resources available to them will be:

  • (A)
  • (B)
  • (C)
  • (D)
Correct Answer: (C)
View Solution

A population growing in the wild with limited resources follows a logistic growth model, where the population initially grows rapidly but slows down as it approaches the environment's carrying capacity.

This growth pattern is illustrated by a sigmoid curve (S-shaped curve), where the population density begins low, grows exponentially, and then levels off once resources become scarce.

Graph 3 represents this logistic growth, showing the population density stabilizing at the carrying capacity. Quick Tip: Logistic growth is a common model for natural population growth, where resource limitations prevent continuous increases in population size.


Question 13:

Assertion (A): The zygote gives rise to a heart-shaped embryo and subsequently proembryo in most angiosperms.

Reason (R): The zygote is present at the micropylar end of the embryo sac and develops into an embryo.

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Assertion (A) is false, but Reason (R) is true.
Correct Answer: (D) Assertion (A) is false, but Reason (R) is true
View Solution

The assertion about the zygote developing into a heart-shaped embryo is true, as it forms the proembryo in angiosperms.

However, the reason incorrectly states that the zygote develops from the micropylar end. The zygote actually forms from fertilization, and the assertion explains the general process rather than being directly related to the specific micropylar location. Quick Tip: In angiosperms, the zygote develops into a heart-shaped proembryo, but the micropylar end refers to the fertilization site, not the development origin.


Question 14:

Assertion (A): Primary transcripts in eukaryotes are subjected to splicing to remove the introns.

Reason (R): Primary transcripts contain both exons and introns and the introns are non-functional in eukaryotes.

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Assertion (A) is false, but Reason (R) is true.
Correct Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)
View Solution

In eukaryotes, primary RNA transcripts indeed contain both exons (coding regions) and introns (non-coding regions).

The assertion that primary transcripts undergo splicing to remove the introns is true, and the reason correctly explains that the introns are removed because they are non-functional for protein coding.

Both the assertion and reason are true and the reason directly explains the assertion. Quick Tip: In eukaryotic cells, splicing removes non-functional introns to ensure the correct formation of mature mRNA for translation.


Question 15:

Assertion (A): The stirrer facilitates the even mixing of oxygen availability in a bioreactor.
Reason (R): Stirred-tank bioreactors generally have a flat base.

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Assertion (A) is false, but Reason (R) is true.
Correct Answer: (C) Assertion (A) is true, but Reason (R) is false
View Solution

The assertion is correct as the stirrer in a bioreactor is used to mix the medium, ensuring an even distribution of oxygen for microbial growth.

However, the reason is incorrect as stirred-tank bioreactors typically have a cylindrical base rather than a flat one. The base shape doesn't directly explain the stirrer's function.

The assertion is true, but the reason does not properly explain the assertion. Quick Tip: In bioreactors, efficient mixing is essential for uniform oxygen distribution, which supports optimal microbial growth.


Question 16:

Assertion (A): The chronic use of alcohol by a person leads to cirrhosis.

Reason (R): Alcohol addiction at times becomes the cause of mental and financial distress to the entire family of the addicted person.

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Assertion (A) is false, but Reason (R) is true.
Correct Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A)
View Solution

The assertion is correct because chronic alcohol consumption can result in cirrhosis, which is liver scarring due to prolonged damage.

The reason is also valid, as alcohol addiction can significantly affect the family, causing both mental and financial strain.

However, the reason does not directly explain cirrhosis; it pertains to the impact on the family, not the liver itself. Therefore, the reason does not provide an explanation for the assertion.
Quick Tip: Chronic alcohol consumption can lead to irreversible liver damage, causing cirrhosis, while also affecting the emotional and financial stability of the family.


Question 17:

(a) Write important features of 'humus' formed during the decomposition cycle in a terrestrial ecosystem.

Correct Answer:
View Solution

Humus is the dark, organic material formed during the decomposition process in soil. The key features of humus are:


1. Humus is a rich source of essential nutrients such as nitrogen, phosphorus, and sulfur, which are crucial for plant growth.

2. It enhances the soil's water retention and aeration, which supports healthy root development.

3. Humus improves soil structure by making it more porous, aiding in better drainage.

4. It plays a vital role in maintaining soil fertility and pH stability, acting as a buffer against extreme fluctuations.

5. The process of humus decomposition is slow but important for recycling organic matter and facilitating nutrient cycling.

6. Humus increases biological activity in the soil, fostering a thriving community of microorganisms and beneficial soil fauna.
Quick Tip: Humus is vital for maintaining soil health and fertility, and it supports the long-term sustainability of ecosystems by promoting nutrient cycling.


Question 17:

(i) Graphically represent the relationship between species richness and area on a log-log scale for bats and fishes.

Correct Answer:
View Solution

On a log-log scale, the relationship between species richness and area can be represented by the following graph:


1. The x-axis represents the area (on a log scale), while the y-axis represents species richness (also on a log scale).

2. The general trend shows that as the area increases, species richness tends to increase as well, following a power law.

3. The curve typically demonstrates a positive correlation, with a steep slope at the beginning of the area increase, which then levels off as the area continues to grow.



Quick Tip: On a log-log scale, the relationship between area and species richness typically follows a power law, indicating that larger areas tend to support more species.


Question 17:

(ii) Write the equation for the relationship as on a logarithmic scale.

Correct Answer:
View Solution

The species-area relationship on a logarithmic scale is expressed by the following equation:
\[ \log(S) = z \cdot \log(A) + c \]
where:

1. \(S\) represents species richness,

2. \(A\) represents the area,

3. \(z\) is the slope of the curve (the species-area exponent), and

4. \(c\) is a constant.
Quick Tip: The species-area relationship is key to understanding biodiversity patterns in ecosystems, and this equation helps quantify how species richness varies with area.


Question 18:

What is a vaccine? Write the basis on which it acts when administered in the body.

Correct Answer:
View Solution

A vaccine is a biological substance that provides active acquired immunity against a specific infectious disease. It generally contains weakened or inactivated forms of the microorganism, its toxins, or one of its surface proteins.


Basis of Action:

1. When administered, a vaccine prompts the immune system to recognize and prepare to combat the pathogen if encountered in the future.

2. In response to the vaccine, the immune system produces antibodies that "remember" the pathogen, allowing for a faster immune response if the body is exposed to the same pathogen later.

3. This process, called immunization, helps prevent future infections by providing "memory" immunity.
Quick Tip: Vaccines are a powerful public health tool for preventing the spread of infectious diseases by "training" the immune system to recognize harmful pathogens.


Question 19:

State the procedure followed in the technique of amniocentesis. List any two advantages of this technique.

Correct Answer:
View Solution

Amniocentesis is a procedure used to collect a sample of amniotic fluid for testing during pregnancy. The steps involved are:

1. A needle is inserted through the abdomen and into the uterus to collect a small amount of amniotic fluid, which contains fetal cells.

2. The fluid is then sent to a laboratory for analysis to identify genetic conditions or other potential abnormalities in the fetus.


Advantages:

1. It aids in identifying genetic disorders such as Down's syndrome, cystic fibrosis, and other chromosomal abnormalities.

2. It can also detect infections, neural tube defects, and other conditions that could impact the health of the fetus. Quick Tip: Amniocentesis is an important diagnostic tool in prenatal care, though it does carry some risks, such as miscarriage, and is usually performed when there are medical indications.


Question 20:

Consider the given data of a hypothetical small portion of mRNA that codes for a functional polypeptide chain and answer the questions that follow:

mRNA: 5' UCAUUACACCGAUUCUUUUAAAAGA 3'

(a) How many amino acids will be formed from the given codons, if substitution of 'U' by 'C' takes place at the 5th codon? Explain your answer.

 
View Solution

The given mRNA sequence has the following codons:

UCA UUA CAC GAA UCU UUU AAA GAA

If 'U' is substituted by 'C' at the 5th codon, the new sequence will be:

UCA UUA CAC GAA C(U)U UUU AAA GAA

From the above sequence, the number of amino acids will be 8 since there are 8 codons that code for amino acids.


Question 20:

(b) Write the number of amino acids that would be in the polypeptide synthesized by a similar mRNA as above, where in the fourth codon instead of 'C' there is 'U'. Justify your answer.

 
View Solution

If in the fourth codon, 'C' is replaced with 'U', the new sequence becomes:

UCA UUA UUA GAA UCU UUU AAA GAA

The new codon is 'UUA', which still codes for an amino acid. The sequence will still produce 8 amino acids.


Question 21:

With reference to the set-ups (A, B, and C) given below, of the electrophoretic separation of a mixture of DNA fragments of varied lengths, answer the questions that follow:

(a) In which one of the two Set-ups, A or B, would you see the DNA fragments separated and why? Justify your answer.


 

Correct Answer:
View Solution

In Set-up A, the DNA fragments will be separated.

The gel matrix, along with the applied current, generates an electric field that separates DNA fragments according to their size. Smaller fragments move more quickly towards the anode, while larger fragments move at a slower pace. Quick Tip: Electrophoresis separates charged particles based on their size and charge using an electric field. Smaller DNA fragments travel faster through the gel.


Question 21:

(b) In Set-up C, which one of the two, I or II, are the bands of longer fragments of DNA? Justify your answer.

Correct Answer:
View Solution

In Set-up C, the bands at position I correspond to longer DNA fragments, while the bands at position II correspond to shorter DNA fragments.

Longer DNA fragments move more slowly through the gel matrix because of their larger size, so they form bands closer to the well at position I. Quick Tip: In gel electrophoresis, longer DNA fragments travel more slowly and therefore appear closer to the well, while shorter fragments migrate farther.


Question 22:

(a) Draw a diagram of a human sperm. Label the following parts:

(i) Which helps in the motility of the sperm.

(ii) Which helps in penetrating through the corona radiata and zona pellucida.

(iii) Which enters into the cytoplasm of the human egg.

Correct Answer:
View Solution




Explanation of labeled parts:


(i) Tail (Flagellum) - Responsible for movement

The tail, also known as the flagellum, enables the sperm to propel forward.

It moves in a whip-like motion, allowing the sperm to navigate through the female reproductive tract to reach the egg.


(ii) Acrosome - Aids in penetration

The acrosome is situated at the tip of the sperm’s head.

It contains digestive enzymes like hyaluronidase and acrosin, which help break down the egg’s protective layers, such as the corona radiata and zona pellucida, facilitating fertilization.


(iii) Head - Enters the egg’s cytoplasm

The sperm head contains the nucleus, which holds the genetic material (haploid chromosomes).

During fertilization, the head fuses with the egg's plasma membrane and releases its genetic material, resulting in the formation of a zygote. Quick Tip: The acrosome contains enzymes like hyaluronidase, which help the sperm penetrate the egg's protective layers during fertilization.


Question 22:

(b) Mention how polyspermy is prevented during the course of fertilization in humans.

Correct Answer:
View Solution

Polyspermy, the fertilization of an egg by multiple sperm, is prevented by the following mechanisms:


1. After the first sperm fuses with the egg, the egg's membrane undergoes a change that prevents other sperm from fusing with it.

2. A cortical reaction occurs, which causes the release of enzymes that alter the zona pellucida, making it impenetrable to additional sperm.
Quick Tip: Polyspermy is prevented through mechanisms that modify the egg's membrane and zona pellucida to ensure only one sperm fertilizes the egg.


Question 23:

(a) (i) List two major reasons for using cow-dung in a biogas plant instead of using domestic sewage.

Correct Answer:
View Solution

1. Cow dung is rich in organic matter and microorganisms, making it a more efficient source for biogas production.

2. It is easier to handle and process cow dung compared to domestic sewage, which may contain chemicals and non-organic substances that could interfere with the biogas production process.
Quick Tip: Cow dung is a preferred substrate for biogas production due to its high organic content and the presence of microorganisms that aid in the fermentation process.


Question 23:

(a) (ii) Mention one use of the unspent slurry of the biogas plant.

Correct Answer:
View Solution

The unspent slurry produced by a biogas plant, often referred to as digestate, contains valuable organic matter and nutrients that make it an excellent fertilizer for crops. This slurry is the residual solid material left after biogas production, and it is rich in essential nutrients such as nitrogen, phosphorus, and potassium, which are crucial for plant growth.


1. Organic Matter Content:

The slurry is primarily composed of decomposed organic material from the input waste, which helps improve soil structure, enhancing its ability to retain moisture and improve aeration. These properties support healthier root development and increase soil productivity.


2. Nutrient Enrichment:

The presence of nitrogen, phosphorus, and potassium, as well as trace elements, in the unspent slurry makes it an ideal fertilizer. These nutrients are vital for the growth of crops and are often required in large quantities, making the slurry a sustainable alternative to synthetic fertilizers.


3. Environmental Benefits:

Using the unspent slurry as fertilizer reduces the need for chemical fertilizers, which can have adverse environmental impacts such as nutrient runoff, soil degradation, and pollution. Additionally, it helps in recycling organic waste, promoting a circular economy approach.


Thus, the unspent slurry from a biogas plant not only serves as an effective fertilizer but also contributes to sustainable agricultural practices by enhancing soil health and reducing reliance on chemical fertilizers. Quick Tip: Biogas plants not only generate renewable energy but also provide valuable by-products like fertilizer, which plays a key role in promoting sustainable agriculture.


Question 23:

(b) Name the bioactive molecule and its microbial source generally used by physicians to treat the patients for:

(i) Myocardial infarction

(ii) High blood cholesterol level

(iii) Organ transplantation

Correct Answer:
View Solution

(i) Streptokinase is used to treat myocardial infarction (heart attack) and is derived from Streptococcus bacteria.
(ii) Statins are used to lower high blood cholesterol levels and are produced by fungi like Aspergillus terreus.
(iii) Cyclosporine is used in organ transplantation to prevent rejection and is derived from the fungus Tolypocladium inflatum. Quick Tip: Bioactive molecules like Streptokinase, Statins, and Cyclosporine are important in treating critical conditions and have revolutionized medical treatments.


Question 24:

Three crosses were carried out in pea plants with respect to flower colour violet/white (V/v) and flower position axial/terminal (A/a). Study in the table the crosses 'a', 'b' and 'c' where parental phenotypes and their \( F_1 \) progeny phenotypes are given.

Find the genotypes of each of the parental pairs of crosses 'a', 'b' and 'c'.



 

Correct Answer:
View Solution

For cross (a):
The offspring show both violet and white flowers, which indicates that the violet parent is heterozygous (Vv) and the white parent is homozygous recessive (vv).

The axial flower phenotype suggests that both parents must have at least one dominant allele for flower position, meaning both are either homozygous (AA) or heterozygous (Aa). Therefore, the genotypes of the parents are:

Violet, axial: VvAa

White, axial: vvAA


For cross (b):

The presence of both axial and terminal flowers in the offspring implies that the violet, axial parent is heterozygous for both traits (VvAa).

The white, terminal parent must be homozygous recessive for both traits (vvaa). Therefore, the parental genotypes are:

Violet, axial: VvAa

White, terminal: vvaa


For cross (c):

The offspring consist of both violet axial and white axial plants, suggesting that the violet, axial parents are heterozygous for both traits (VvAa).

The genotypes of the parents in this cross are:

Violet, axial: VvAa

Violet, axial: VvAa
Quick Tip: In Mendelian genetics, analyzing phenotypic ratios and their genetic origins helps determine the genotypes of the parents.


Question 35:

(a) Mention the advantages of 'ART'.

Correct Answer:
View Solution

Assisted Reproductive Technology (ART) offers several benefits:

It assists couples facing infertility issues by utilizing techniques like in-vitro fertilization (IVF) and intrauterine insemination (IUI) to help them conceive.

ART enables genetic testing of embryos, increasing the likelihood of healthy pregnancies.

It supports fertility preservation for individuals with medical conditions, such as cancer or age-related fertility decline, that may impact their reproductive health.

ART provides opportunities for single women and same-sex couples to have biological children.

It helps overcome reproductive challenges caused by male or female infertility.
Quick Tip: ART has transformed reproductive medicine by offering practical solutions to infertility, enabling many individuals and couples to achieve parenthood.


Question 25:

(b) Explain the 'ICSI' and 'AI' methods of assisted reproductive technologies.

Correct Answer:
View Solution

ICSI (Intracytoplasmic Sperm Injection):


ICSI is a type of in-vitro fertilization (IVF) in which a single sperm is injected directly into an egg to achieve fertilization.

This method is commonly used for male infertility, especially in cases of low sperm count or poor sperm motility.

AI (Artificial Insemination):

Artificial insemination involves introducing sperm into a woman's reproductive system by means other than sexual intercourse.

It is used when a male has low sperm count, motility issues, or other fertility problems. Intrauterine insemination (IUI) is a common type of AI, where sperm is directly placed into the uterus.
Quick Tip: ICSI is particularly beneficial for men with severe male factor infertility, while AI helps couples with unexplained infertility or male fertility issues.


Question 26:

(a) Give the scientific name of the bacteria widely used in biotechnology to create a GM cotton crop resistant to bollworm attacks.

Correct Answer:
View Solution

The bacterium commonly used in biotechnology to develop genetically modified (GM) cotton resistant to bollworm infestations is Bacillus thuringiensis (Bt).

Bacillus thuringiensis produces a protein known as Cry protein, which is toxic to certain pests, such as the cotton bollworm, but is harmless to humans, animals, and beneficial insects.

The incorporation of Bt in GM crops like cotton reduces the need for chemical pesticides, benefiting both the environment and farmers by lowering pesticide-related expenses. Quick Tip: *Bacillus thuringiensis* (Bt) acts as a natural insecticide in genetically modified crops, offering sustainable pest control without the harmful effects of chemical pesticides.


Question 26:

(b) Explain how GM cotton crop is able to resist insect attacks.

Correct Answer:
View Solution

GM cotton is engineered to produce Bacillus thuringiensis (Bt) proteins, specifically the Cry proteins, which are toxic to specific pests like the cotton bollworm. Here's how the mechanism works:

1. The cotton plants express the Bt protein in their tissues. When a bollworm feeds on the cotton, it ingests the Cry protein.

2. The Cry protein is activated in the alkaline environment of the insect’s gut and binds to receptors in the gut wall, causing a disruption in the digestive system.

3. This leads to the insect's inability to digest food, causing it to stop feeding and ultimately die.

4. This built-in protection reduces the need for chemical insecticides, providing an environmentally friendly and cost-effective solution to pest control.
Quick Tip: Bt cotton is an example of an environmentally sustainable solution to pest control, as it uses the plant's own defenses to protect against harmful insects, reducing pesticide use.


Question 27:

Describe the mutualistic relationship that exists between the Mediterranean orchid \textbf{Ophrys} and a bee species.

Correct Answer:
View Solution

The Mediterranean orchid Ophrys forms a mutualistic relationship with male bees of certain species. Here's how the interaction works:

The Ophrys flower closely resembles the appearance and scent of a female bee, attracting male bees that mistake it for a potential mate.

As the male bee attempts to mate with the flower, it inadvertently collects pollen from the flower's anthers.

When the bee tries to mate with another flower, it transfers the pollen, thereby pollinating the orchid.

While the bee does not receive any direct benefit, its behavior aids in its reproductive strategy, and the orchid benefits from successful pollination.


Thus, the relationship is mutualistic, as the orchid receives pollination, while the bee's mating behavior plays a role in its reproductive cycle. Quick Tip: Mutualistic relationships, like that between orchids and bees, can lead to evolutionary adaptations, where plants evolve traits to attract specific pollinators for effective reproduction.


Question 28:

A population of snakes lived in a desert with brown sand. Study the drawings given below showing the change in the population from 'one' to 'two' over time and answer the question that follows. Brown snakes and Grey snakes are represented by alleles A/a (Dominant/recessive).

(a) If the frequency of the recessive trait is 9% in population-one, work out the frequency of homozygous dominant and heterozygous dominant snakes.



 

Correct Answer:
View Solution

Given that the frequency of the recessive trait (aa) in population-one is 9%, we can use Hardy-Weinberg equilibrium to calculate the frequencies of the homozygous dominant (AA) and heterozygous dominant (Aa) genotypes.

First, we find the frequency of the recessive allele (a) using \(q = \sqrt{0.09} = 0.3\).

The frequency of the dominant allele (A) is then calculated as \(p = 1 - q = 1 - 0.3 = 0.7\).

The frequency of homozygous dominant (AA) is \(p^2 = (0.7)^2 = 0.49\).

The frequency of heterozygous dominant (Aa) is \(2pq = 2(0.7)(0.3) = 0.42\).


Therefore, the genotype frequencies are:
Homozygous dominant (AA): 49%
Heterozygous dominant (Aa): 42% Quick Tip: The Hardy-Weinberg equilibrium principle allows the calculation of genotype frequencies based on allele frequencies, providing insights into genetic variation in populations.


Question 28:

(b) Name the mechanism of evolution that must have operated so that population-two evolved from population-one.

Correct Answer:
View Solution

The mechanism of evolution that must have operated is natural selection. The change in the snake population can be attributed to selective pressures in the environment. The brown sand likely provided camouflage for brown-colored snakes, offering them a survival advantage over grey snakes, leading to an increase in the frequency of brown-colored (dominant) snakes in the population. Additionally, migration and genetic drift might have contributed to the change in allele frequencies. Quick Tip: Natural selection leads to the adaptation of populations over time as individuals with traits that are favorable for survival are more likely to reproduce and pass on those traits.


Question 29:

Study the figures given below that depict the comparative age distribution of human populations in Sweden and Rwanda. (International Data Base 2003) and answer the questions that follow :

(a) What can be inferred from the very broad base of Rwanda’s age pyramid? Support your answer with the data provided in the figure.


 

Correct Answer:
View Solution

Rwanda’s age pyramid features a very wide base, indicating a high birth rate, which is characteristic of developing countries. This suggests that a significant portion of the population is in the younger age groups (0-14 years), with the numbers gradually decreasing as age increases.

The pyramid data shows that the proportion of people in younger age groups, such as 0-4 years and 5-9 years, is considerably higher compared to older age groups. For instance, in Rwanda, the age group 0-4 years comprises 8.2% of the male population and 8.2% of the female population.

This wide base points to a youthful population with a high fertility rate. Quick Tip: A broad base in an age pyramid signifies a high birth rate and a large proportion of the population in younger age groups, typically indicating rapid population growth.


Question 29:

(b) Sweden has an age distribution that is approximately the same width near its base as at the apex. What does this indicate?

Correct Answer:
View Solution

The similar width of Sweden’s age pyramid at both the base and the apex suggests that the country has a stable population with slow growth. This is typical of developed nations with low birth and death rates, where the population is more evenly spread across various age groups.

In Sweden, the proportion of individuals in younger age groups (e.g., 0-4 years, 5-9 years) is relatively smaller compared to countries like Rwanda, and the older age groups (60-64 years and 65+) make up a comparable percentage of the population.

This distribution indicates that Sweden’s population is gradually aging, with a steady rate of younger individuals replacing the older generation, contributing to a stable, non-expanding population.
Quick Tip: An age pyramid with a similar width at both the base and the apex indicates a stable population, which is characteristic of developed countries with low birth and death rates.


Question 29:

(c) Name the type of age pyramid shown above for Sweden.

Correct Answer:
View Solution

The age pyramid for Sweden is classified as a stationary pyramid. In this type of pyramid, the population is evenly distributed across different age groups, which indicates a stable or zero-growth population. This pattern suggests that birth and death rates are in balance, and the population is neither expanding nor shrinking significantly.

The stationary pyramid shape is characterized by similar proportions of people in each age group, from the youngest to the oldest. This reflects a demographic equilibrium where the number of births is roughly equal to the number of deaths, leading to a population that remains stable over time.

Such a population typically has low fertility rates and high life expectancy, which are common characteristics of developed countries like Sweden. In these countries, healthcare improvements, family planning, and social support systems contribute to lower birth rates and higher survival rates across all age groups.
Quick Tip: A stationary age pyramid reflects a stable population with low fertility rates and high life expectancy, commonly found in developed nations.


OR
Question 29:

(c) Name the type of age pyramid shown above for Rwanda.

Correct Answer:
View Solution

The type of age pyramid shown for Rwanda is a expansive pyramid. This pyramid has a very broad base, indicating a high birth rate and a youthful population, which is typical of developing countries with high fertility rates and lower life expectancy. Quick Tip: An expansive age pyramid is characteristic of developing countries with high birth rates and a younger population, often associated with rapid population growth.


Question 30:

Read the passage given below and answer the questions that follow.

In recombinant DNA technology, restriction enzymes are used as they recognize and cut DNA within a specific recognition sequence. BamH I is one such restriction enzyme which binds at the recognition sequence

5' \, G-G-A-T-T-C-C \, 3' and cleaves this sequence between G and G on each strand, whereas Alu I binds at the recognition sequence

5' \, A-G-C-T \, 3' and cleaves these sequences between G and C on each strand.

(a) If Alu I is used to cut the given DNA strand, how many DNA fragments would be formed? Write the sequence of each fragment with its polarity.



Correct Answer:
View Solution

The given DNA strand is:



When Alu I cuts this sequence at the recognition sequence \(\textbf{5' A-G-C-T 3'}\), the DNA strand is cleaved between G and C on each strand. The cuts will result in two fragments. The sequence of each fragment is as follows:

First fragment: \[ 5' \, \textbf{C-C-G-T-A-G-C-T} \, 3' \]

Second fragment: \[ 5' \, \textbf{A-G-C-T-A-T-C-G-A-C-T-G-C-T-G-G} \, 3' \]

Thus, 2 DNA fragments would be formed. Quick Tip: Restriction enzymes like Alu I cut the DNA at specific sequences, creating fragments with sticky ends that can be used for cloning or recombinant DNA technologies.


Question 30:

(b) Which one of the two restriction enzymes BamH I or Alu I will preferably be used on the same given DNA strand to make a recombinant DNA molecule and why?

Correct Answer:
View Solution

BamH I would preferably be used on the given DNA strand to make a recombinant DNA molecule. The reason is that BamH I creates sticky ends by cutting between G and G, which can facilitate the insertion of foreign DNA into a vector.

Sticky ends are useful in recombinant DNA technology as they allow for the ligation of complementary DNA fragments with high specificity.

In contrast, Alu I produces blunt ends, which are less efficient for ligating DNA into vectors compared to sticky ends.
Quick Tip: Sticky ends created by BamH I are ideal for the ligation of foreign DNA into vectors, while blunt ends created by Alu I are less efficient for recombination.


Question 30:

(c) After binding to the two strands of the double helix DNA, where specifically does the restriction enzyme act to cut the two strands of DNA? Write the specific term used for the specific nucleotide sequences of DNA recognized by a restriction endonuclease.

Correct Answer:
View Solution

Restriction enzymes act at specific sequences of nucleotides called restriction sites. The restriction enzyme binds to these sequences and cuts both strands of the DNA, usually between specific nucleotides.
For example, the restriction enzyme BamH I recognizes the sequence \[ 5' \, \textbf{G-G-A-T-T-C-C} \, 3' \]
and cuts between the G and G on each strand.
Similarly, Alu I recognizes the sequence \[ 5' \, \textbf{A-G-C-T} \, 3' \]
and cuts between G and C on each strand.

The process is called restriction digestion or restriction enzyme cleavage. Quick Tip: Restriction enzymes recognize specific DNA sequences, and by cutting the DNA at these sites, they allow for precise manipulation of DNA in molecular biology experiments.


OR

Question 30:

(c) Write the specific sequence of DNA segment recognized by the restriction endonuclease EcoRI.

Correct Answer:
View Solution

EcoRI is a commonly used restriction endonuclease that recognizes the specific palindromic DNA sequence: \[ 5' \, \textbf{G-A-A-T-T-C} \, 3' \]
It cuts between the G and A on each strand, resulting in sticky ends with overhanging bases. These overhangs allow the DNA fragments to easily pair with complementary sequences, making EcoRI invaluable in cloning and recombinant DNA technology.


The EcoRI recognition site is highly specific, meaning only DNA fragments containing this sequence will be cleaved by the enzyme.

The sticky ends generated by EcoRI are particularly advantageous in molecular cloning, as they facilitate the ligation of DNA fragments with matching sticky ends. Quick Tip: EcoRI is one of the most widely used restriction enzymes in molecular biology. Its ability to generate sticky ends makes it essential for the efficient recombination of DNA fragments in genetic engineering.


Question 31:

(a) (i) Describe the life cycle of Plasmodium from the time it enters the human body till a female Anopheles mosquito bites an infected person.

Correct Answer:
View Solution

The life cycle of Plasmodium involves both humans and Anopheles mosquitoes. The key stages in the life cycle are as follows:


1. Infection of the human host: When a female Anopheles mosquito bites an infected person, it injects Plasmodium sporozoites into the bloodstream. These sporozoites travel to the liver, where they mature and reproduce.

2. Liver stage: Inside the liver, the sporozoites undergo asexual reproduction, forming merozoites. These merozoites are released into the bloodstream.

3. Blood stage: The merozoites invade red blood cells, where they continue to reproduce asexually. This causes the red blood cells to burst, releasing more merozoites into the bloodstream, which can infect additional red blood cells.

4. Transmission to the mosquito: Some merozoites transform into gametocytes, which are then ingested by another Anopheles mosquito when it bites an infected person. Within the mosquito's gut, the gametocytes fuse to form zygotes, which mature into sporozoites that travel to the salivary glands, ready to be transmitted to a new human host.
Quick Tip: The Plasmodium life cycle alternates between human and mosquito hosts, involving both asexual and sexual reproduction stages, which makes its transmission complex.


Question 31:

(a) (ii) Mention the two events of Plasmodium life cycle that occur within the female Anopheles body.

Correct Answer:
View Solution

The two key events of the Plasmodium life cycle that occur within the female Anopheles body are:


1. The fusion of male and female gametocytes to form a zygote, which develops into an ookinete.

2. The ookinete penetrates the mosquito's gut lining and forms an oocyst, where it undergoes division to produce sporozoites. These sporozoites travel to the salivary glands of the mosquito, ready to infect a new human host. Quick Tip: The sexual cycle of \textbf{Plasmodium} occurs in the mosquito host, leading to the production of sporozoites that are essential for transmission to the human host.


Question 31:

(b) (i) Write two differences between malignant tumor and benign tumor.

Correct Answer:
View Solution

The two main differences between malignant and benign tumors are:


1. Malignant tumors are cancerous and can invade nearby tissues and spread to other parts of the body through the blood or lymphatic system, whereas benign tumors are non-cancerous and do not spread to other parts of the body.

2. Malignant tumors grow uncontrollably and can metastasize, while benign tumors grow slowly and are generally encapsulated, making them easier to remove surgically.
Quick Tip: Malignant tumors are more dangerous because they spread and invade other tissues, while benign tumors are typically localized and non-threatening.


Question 31:

(b) (ii) Explain any three diagnostic techniques for the detection of cancer.

Correct Answer:
View Solution

Three common diagnostic techniques for detecting cancer are:


1. Biopsy: A sample of tissue is removed from the suspected tumor and examined under a microscope to check for cancer cells.

2. Imaging techniques: Methods such as X-rays, CT scans, and MRIs are used to visualize tumors inside the body.

3. Blood tests: Some cancers produce specific markers that can be detected in the blood, helping in diagnosis and monitoring the progress of treatment. Quick Tip: Early detection through various diagnostic techniques improves the chances of successful cancer treatment.


Question 32:

(a) (i) Explain any four devices that flowering plants have developed to encourage cross-pollination.

Correct Answer:
View Solution

Four devices that flowering plants have developed to encourage cross-pollination include:


1. Colorful flowers: Brightly colored flowers attract pollinators such as bees, butterflies, and birds.

2. Scent: Flowers often release scents that attract pollinators, ensuring they visit and transfer pollen between flowers.

3. Nectar: The presence of nectar in flowers encourages pollinators to visit flowers and aids in the transfer of pollen.

4. Specialized floral structures: Some plants have specialized mechanisms, like long tubes or sticky pollen, which ensure that pollen is transferred effectively from one flower to another. Quick Tip: Cross-pollination helps maintain genetic diversity, which is important for the health and survival of plant species.


Question 32:

(a) (ii) Why do plants discourage self-pollination? State any one reason.

Correct Answer:
View Solution

Plants discourage self-pollination to promote genetic diversity. Cross-pollination allows for the mixing of genetic material, which helps in producing offspring with a broader genetic pool, making them more resilient to diseases and environmental changes. Quick Tip: Self-pollination can limit genetic diversity, while cross-pollination promotes healthier and more adaptable offspring.


Question 32:

(b) Explain the ovarian and uterine events taking place along with the role of pituitary and ovarian hormones, during menstrual cycle in a normal human female under the following phases:

(i) Follicular phase/proliferative phase

(ii) Luteal phase/secretory phase

(iii) Menstrual phase

Correct Answer:
View Solution

The menstrual cycle involves several phases that are regulated by hormones from the pituitary and ovaries. The phases are:

1. Follicular phase (Proliferative phase):

During this phase, the pituitary hormone FSH (follicle-stimulating hormone) stimulates the growth of ovarian follicles.

The growing follicles release estrogen, which thickens the endometrium in preparation for implantation.


2. Luteal phase (Secretory phase):

After ovulation, the ruptured follicle forms the corpus luteum, which secretes progesterone.

Progesterone further thickens the endometrium and prepares it for potential pregnancy.


3. Menstrual phase:

If fertilization does not occur, the corpus luteum degenerates, leading to a drop in progesterone levels.

This causes the endometrial lining to break down and shed, leading to menstruation. Quick Tip: The menstrual cycle is regulated by the interplay between pituitary and ovarian hormones, ensuring proper reproductive function.


Question 33:

(a) “The influence of both the alleles in a heterozygous state is clearly expressed in codominance.” Explain with the help of inheritance of ABO blood group in humans.

Correct Answer:
View Solution

In the case of ABO blood groups, the alleles A and B exhibit codominance. This means that when both alleles are present in a heterozygous individual, both are fully expressed.


For example:

Genotype AB results in a person having both A and B antigens on the surface of their red blood cells.

This is in contrast to dominant-recessive inheritance, where one allele is completely dominant over the other.


The ABO blood group system is controlled by three alleles: A, B, and O.

A and B alleles are dominant, while O is recessive.

Individuals with genotype AO or AA have blood group A, BO or BB have blood group B, AB has blood group AB, and OO has blood group O. Quick Tip: Codominance results in both alleles being expressed simultaneously in a heterozygous individual, as seen in the ABO blood group inheritance pattern.


Question 33:

(b) A group of genes are regulated and expressed together as a unit in lac operon.
(i) Explain the mechanism of switching 'on' of the structural genes of lac operon.

Correct Answer:
View Solution

The lac operon is regulated by the presence of lactose. When lactose is available, it binds to the repressor protein, causing a conformational change that prevents the repressor from binding to the operator. This allows RNA polymerase to bind to the promoter region and transcribe the structural genes, which are involved in the breakdown of lactose. Quick Tip: The presence of lactose induces the transcription of the \textbf{lac operon} genes by inactivating the repressor protein, allowing the production of enzymes for lactose metabolism.


Question 33:

(b) (ii) “Regulation of lac operon is referred to be negatively regulated.” Justify giving a reason.

Correct Answer:
View Solution

The regulation of the lac operon is considered negatively regulated because the repressor protein normally binds to the operator region of the operon and prevents transcription. The presence of lactose inactivates the repressor, thus allowing transcription to occur. Without lactose, the repressor remains bound to the operator, blocking the operon. Quick Tip: In negative regulation, the default state of the operon is "off," and the presence of an inducer like lactose removes the repression to allow gene expression.