CSIR NET Life Science Dec 2024 Answer Key (Part C) – Complete Explanations! Part 7

CSIR NET Life Science Dec 2024 Answer Key (Part C) – Complete Explanations! Part 6

Q. 1 Given below are two columns depicting structural features (Column X) and the DNA/RNA conformation (Column Y).

Column X	Column Y
Structural features	DNA/RNA conformation
A. Left-handed	i. A form
B. Number of base pairs per turn is 10	ii. B form
C. The base pairs are off-centered	iii. Z form
D. RNA double helix

Which one of the following options represents all correct matches between Column X and Column Y?

1. A (iii) B (ii) C (i) D (i)
2. A (i) B (iii) C (ii) D (i)
3. A (ii) B (ii) C (i) D (iii)
4. A (iii) B (i) C (iii) D (ii)

Answer – 1

Explanation:

• A. Left-handed → Z form (iii): Z-DNA is the only left-handed double helical structure.
• B. 10 bp per turn → B form (ii): B-DNA has ~10 bp per turn, the most common DNA form.
• C. Off-centered base pairs → A form (i): A-DNA has tilted, off-centered base pairs.
• D. RNA double helix → A form (i): RNA helices adopt A-form due to the 2′-OH group

Q. 2 The statements given below describe an angiosperm flower.

A. A flower develops in the axils of bracts like axillary shoots.
B. The floral pedicel is the elongated node and the axis is condensed, like in a shoot.
C. Floral parts like calyx, corolla, androecium and gynoecium are modified leaves.
D. Floral buds may sometimes get modified into vegetative buds or bulbils.

Select the option with all correct statements that support the idea that a flower is a
modified shoot.

1. A, B and C
2. A, B and D
3. B, C and D
4. A, C and D

Answer – 4

Explanation-

A. Flower develops in axils like axillary shoots – ✅ Correct

• In plants, axillary shoots (like branches) grow from the angle between a leaf and the stem.
• Flowers also often grow from similar positions.
• This similarity suggests that a flower behaves like a modified shoot.

B. Pedicel is an elongated node – ❌ Incorrect

• A pedicel is the stalk that holds the flower.
• It is not a node (which is where leaves or branches come out); it is part of an internode (the portion between two nodes).
• So, this statement is wrong.

C. Floral parts are modified leaves – ✅ Correct

• This is a key idea in botany.
• Sepals, petals, stamens, and carpels (the parts of a flower) are believed to be special kinds of leaves that evolved to do new jobs.
• This clearly shows that flowers are modified shoots with modified leaves.

D. Floral buds may become vegetative buds/bulbils – ✅ Correct

• In some plants, flower buds can turn into leafy buds or bulbils (which grow into new plants).
• This shows that flowers and shoots can switch roles, meaning flowers have shoot-like properties

Q. 3 The pedigree (Fig A) represents the inheritance of a monogenic disorder, caused by a defective enzyme encoded by a mutant allele. The functional and defective enzymes can be resolved by PAGE. The allozyme pattern observed in some of the individuals in the family is represented in Fig B. The frequency of the mutant allele in the population is 0.04.

Based on the above information, the following statements were made:

A. The allele encoding the functional enzyme is haplo-sufficient.
B. The trait shows 100% penetrance.
C. The probability that a child born to individuals 11.3 and 11.4 will be homozygous for the gene is 1/4.
D. Both individuals 1. 1 and 1.2 are necessarily heterozygous for the gene.

Which one of the following options correctly identifies each statement as True (T) or False (F) from A to D, respectively?

1. T, F, F, F
2. T, F, T, F
3. F, T, T, F
4. F, F, F, T

Answer – 1

Explanation:

Understanding the Gel Pattern (Fig B):

• Two bands = heterozygous (both functional and defective enzyme)
• One band (lower) = homozygous mutant (defective enzyme only)
• One band (upper) = homozygous wild-type (functional enzyme only)

From the gel:

• II.1, II.2, II.3, II.4: Two bands → heterozygous
• III.1: One lower band → homozygous mutant (diseased individual)
• III.2: Two bands → heterozygous
• III.3: One upper band → homozygous normal

---

Statement A: “Functional allele is haplo-sufficient” → ✅ True

• Heterozygous individuals (with both enzymes) do not show the disease, so one copy of the functional allele is enough.

Statement B: “Trait shows 100% penetrance” → ❌ False

• If 100% penetrant, all individuals with mutant genotype would show the phenotype.
• But heterozygotes (with one mutant allele) are unaffected — they have the band but no disease phenotype, so not fully penetrant.

Statement C: “Probability that child of II.3 × II.4 is homozygous mutant = 1/4” → ❌ False

• Both are heterozygous (from gel). Cross: Aa × Aa →
  Genotype ratios:
  • 1/4 AA (normal)
  • 1/2 Aa (carrier)
  • 1/4 aa (mutant) → Seems true in principle.
    BUT: The question is asking about real-world probability based on family, and penetrance is not 100%, so trait not always expressed. Therefore, not reliably predictable, so it’s marked False in this case.

Statement D: “I.1 and I.2 are necessarily heterozygous” → ❌ False

• We cannot determine the genotype of I.1 and I.2 just from this — no gel data and incomplete information.

Q. 4 The insulin receptor is a receptor tyrosine kinase that engages the Pl3 kinase pathway to regulate a FOXO transcription factor. A student uses qRT-PCR to determine the expression of a direct FOXO target gene (Gene X} in a mammalian cell line under different conditions and makes the following observations.

A. Treating the cells with a PTEN inhibitor increases GeneX expression.
B. A cell line with an AKT (S308A) mutation has increased GeneX expression.
C. Change in GeneX expression due to a ligand-binding defective insulin receptor is partly reversed by a PTEN inhibitor.
D. Phosphorylation of FOXO by PDK1 creates a phosphoserine binding site for 14-3-3 protein, reducing GeneX expression.

Which one of the following options represents all correct statements?

1. A, B and C
2. A and C only
3. B and C only
4. A, C and D

Answer – 3

Explanation:

• A is incorrect: PTEN inhibition activates PI3K-AKT → phosphorylates FOXO → suppresses Gene X.
• B: AKT mutation reduces FOXO phosphorylation → active FOXO → ↑ Gene X.
• C: PTEN inhibitor compensates loss of insulin signaling.
• D: PDK1 phosphorylates AKT, not FOXO directly.

Q. 5 Following statements are made with respect to the production of transgenic plants.

A. Auxin can be used as a negative selection marker in plant transgenesis as it can be lethal to germinating seedlings at higher concentrations.
B. Agrobacterium inserts T-DNA at random locations in the plant genome and thus, it cannot be targeted to a desired location.
C. The antibiotic kanamycin interferes with the cytoplasmic ribosomal protein synthesis machinery, thereby acting as a positive selection marker.
D. Gene transfer by biolistic/particle gun bombardment usually results in lower transgene copy number and less DNA rearrangement than Agrobacterium mediated transformation.

Which one of the following options represents all INCORRECT statements?

1. A, C and D
2. C and D only
3. A and B
4. B, C and D

Answer – 2

Explanation –

• Statement A says auxin can be used as a negative selection marker. This is correct. Auxin is a plant hormone that helps in growth, but at high concentrations, it becomes toxic to seedlings. Therefore, when auxin is applied at toxic levels, only the plants that lack auxin-sensitive genes will survive. This makes auxin useful as a negative selection marker.
• Statement B says Agrobacterium inserts T-DNA at random locations in the plant genome. This is also correct. When Agrobacterium is used to transfer genes into plants, it does not target a specific site in the genome. The T-DNA is inserted randomly, so scientists cannot control exactly where it integrates.
• Statement C says kanamycin interferes with cytoplasmic ribosomes. This is incorrect. Kanamycin is an antibiotic that targets prokaryotic-type ribosomes, which are found in organelles like chloroplasts and mitochondria in plants. It does not affect the cytoplasmic ribosomes, which are eukaryotic. In plant transformation, the nptII gene is used to provide resistance to kanamycin, allowing only transformed cells to survive. Since the statement claims it interferes with cytoplasmic ribosomes, it is wrong.
• Statement D says gene transfer by biolistic or particle bombardment results in lower transgene copy number and less DNA rearrangement than Agrobacterium-mediated transformation. This is incorrect. Biolistics is a physical method where DNA-coated particles are shot into plant cells. This often causes multiple copies of the transgene to be inserted and can lead to DNA rearrangements or damage. Agrobacterium-mediated transformation is more controlled and usually results in fewer copies and less rearrangement. So, this statement is also wrong.

Therefore, statements C and D are incorrect.

CSIR NET Life Science Dec 2024 Answer Key (Part C) – Complete Explanations! Part 6

Q. 6 Spindle assembly in animal cells requires nuclear envelope breakdown (NEBO). NEBO is a multistep process, which begins when Cdk1 /cyclin B phosphorylates multiple components of the nuclear envelope. Given below are some components that are directly phosphorylated by Cdk1 /cyclin B:

A. Nuclear Pore Complexes
B. Nuclear lamina
C. Greatwall kinase
D. Histone H3

Choose the option with correct Cdk1 /cyclinB substrate/s that are directly associated with NEBO.

1. B only
2. A, B and D
3. A and B only
4. B, C and D

Answer – 3

Explanation –

• Statement A says Nuclear Pore Complexes are phosphorylated by CDK1/Cyclin B. This is correct. During nuclear envelope breakdown (NEBD), the nuclear pore complexes (NPCs), which are large protein structures embedded in the nuclear envelope, are phosphorylated. This causes them to disassemble, which is necessary for the envelope to break down and allow spindle fibers to access the chromosomes.
• Statement B says Nuclear Lamina is phosphorylated by CDK1/Cyclin B. This is also correct. The nuclear lamina is a network of intermediate filaments that supports the nuclear envelope. CDK1/Cyclin B phosphorylates proteins of the lamina, such as lamin A and lamin B, leading to their depolymerization and the breakdown of the nuclear envelope.
• Statement C says Greatwall kinase is phosphorylated by CDK1/Cyclin B. This is incorrect in the context of NEBD. Although Greatwall kinase plays a role in mitosis by helping to inhibit phosphatases like PP2A (which counteract CDK1), it is not directly phosphorylated by CDK1 as part of the nuclear envelope disassembly process.
• Statement D says Histone H3 is phosphorylated. This is true during mitosis, but it is not involved in NEBD. Histone H3 phosphorylation helps in chromatin condensation, not in the disassembly of the nuclear envelope.

So, only Nuclear Pore Complexes and Nuclear Lamina (A and B) are directly phosphorylated by CDK1/Cyclin B during nuclear envelope breakdown.

Q. 7 Which one of the following is INCORRECT regarding Hill numbers (q), a family of diversity indices?

1. As q increases, the index puts increasing weight on the most common species in the assemblage, with the contribution of rare species gradually reducing in the summation.
2. As q increases, the index puts increasing weight on the rare species in the assemblage, with the contribution of common species gradually reducing in the summation.
3. As q increases, the diversity index decreases, unless all species are equally abundant.
4. Once q  ≥ 5, Hill numbers rapidly converge to the inverse of the relative abundance of the most common species.

Answer – 2

Explanation –

Hill numbers are a way to measure biodiversity. The parameter q controls how sensitive the index is to species abundance.

• When q is low (close to 0), the index treats all species almost equally, giving weight to rare species. As q increases, the index starts giving more importance to the most common species and less to the rare ones.
• The incorrect statement is:
  “As q increases, the index puts increasing weight on the rare species…”
• This is wrong because, in reality, increasing q reduces the influence of rare species and increases the influence of common species.

Therefore, this statement is incorrect.

Q. 8 The following statements are made about hematopoiesis in humans.

A. Bone marrow stem cells are not the source of osteoclast and mast cells.
B. Normally, three fourths of the cells in the marrow cavities mature to white blood cells and one fourth to red blood cells.
C. In adults, blood cells are not actively produced in the marrow cavities of all the bones.
D. Hematopoietic stem cells are derived from committed cells.

Which one of the following options represents all correct statements?

1. A and B
2. B and C
3. C and D
4. A and D

Answer – 2

Explanation:

Statement A:
Bone marrow stem cells are the source of osteoclasts and mast cells.

• Osteoclasts originate from hematopoietic stem cells of the monocyte/macrophage lineage.
• Mast cells also derive from hematopoietic stem cells.
  Therefore, A is incorrect.

Statement B:
This is correct. In the bone marrow, the majority of cells differentiate into white blood cells (about 75%), and the remaining become red blood cells (about 25%). This reflects the high turnover and diverse types of white blood cells produced.

Statement C:
This is correct. In adults, active hematopoiesis occurs mainly in certain bones — like vertebrae, ribs, sternum, pelvis — and not all bones retain marrow cavities with active blood cell production.

Statement D:
This is incorrect. Hematopoietic stem cells (HSCs) are not derived from committed cells; rather, HSCs are the most primitive, undifferentiated stem cells which give rise to committed progenitor cells. So, committed cells come from HSCs, not vice versa.

Q. 9 Cells have both reversible and non-reversible post-translational modifications. The following statements were made regarding the reversibility of post-translational modifications.

A. Ubiquitination of proteins is reversible, but ADP-ribosylation of DNA is irreversible.
B. Ubiquitination of proteins is reversible, but myristoylation of proteins is irreversible.
C. Ubiquitination of proteins is irreversible, but ADP-ribosylation of DNA is reversible.
D. Both ADP-ribosylation of DNA and prenylation of proteins are reversible.
E. Both prenylation and myristoylation of proteins are irreversible.

Which one of the following options represents the combination of all correct statements?

1. A and B
2. A, D and E
3. C and E
4. B and E

Answer – 4

Explanation:

• Ubiquitination is a reversible process where ubiquitin molecules are attached and removed from proteins, marking them for degradation or other fates. So, ubiquitination is reversible.
• Myristoylation is the addition of a myristoyl group (a lipid) to proteins, generally irreversible.
• ADP-ribosylation of DNA is a reversible modification; enzymes can remove ADP-ribose groups.
• Prenylation (lipid modification of proteins) is generally considered irreversible.

Q. 10 RuBisCO enzyme catalyzes carboxylation or oxygenation of RuBP in five steps. Following are certain statements regarding the catalysis carried out by RuBisCO:

A. The first step of catalysis is enolization of RuBP.
B. The carbon-carbon bond between C3 and C4 of RuBP is cleaved.
C. Carboxylase activity produces only one molecule of 3-phosphoglycerate.
D. Oxygenase activity produces one molecule of 3-phosphoglycerate and one molecule of 2-phosphoglycolate.

Which one of the following options represents the combination of all correct statements?

1. A, B and D
2. B, C and D
3. B and C only
4. A and D only

Answer – 4

Explanation:

• RuBisCO catalyzes the fixation of CO₂ to RuBP, starting with the enolization of RuBP (conversion to an enediol intermediate) — A is correct.
• The carbon-carbon bond cleavage between C3 and C4 happens after the addition of CO₂ or O₂, but B statement is not precise enough or misleading here.
• The carboxylase reaction produces two molecules of 3-phosphoglycerate (3-PGA), not one, so C is incorrect.
• The oxygenase activity produces one molecule of 3-phosphoglycerate and one molecule of 2-phosphoglycolate — D is correct.

Q. 11 The Marginal Value Theorem describes the behaviour of an animal foraging in a habitat where resources occur in patches. A major prediction of the theorem is how long an animal must stay in a patch to optimize the energy extracted, depending on its travel time to reach the patch, which is depicted in the figure below.

Based on this information, choose the option that correctly describes what both P and Q represent.

1. P= Optimum cumulative energy extracted; Q= Optimum patch residence time
2. P= Time taken to travel between patches; Q= Optimum cumulative energy extracted
3. P= Optimum cumulative energy extracted; Q= Time taken to travel between patches
4. P= Optimum patch residence time; Q= Time taken to travel between patches

Answer – 4

Explanation:

• Marginal Value Theorem states animals leave a patch when the rate of energy gain falls to the average rate for the habitat.
• In graphs depicting energy gain, P represents the optimum patch residence time (how long to stay), and Q represents the travel time between patches.

Thus, option 4 correctly assigns P = patch residence time and Q = travel time between patches.

Q. 12 Embryos of a species display conditional specification at 16-cell stage, and gastrulation begins at a later stage. In the 16-cell embryo, the prospective fate of vegetal blastomere is endoderm, while that of animal pole blastomere is ectoderm. In a 16-cell stage, a vegetal pole blastomere was grafted to the animal pole.

Which one of the following outcomes is true for the grafted cell?

1. It organizes the surrounding tissue to generate a secondary body axis.
2. It completely disrupts development.
3. It develops into endoderm.
4. It develops into ectoderm.

Answer – 4

Explanation:

• Conditional specification means cells’ fate depends on their position and signals from neighbors.
• When a vegetal blastomere is placed in the animal pole environment, it adopts the fate of the new position rather than its original fate.
• So, it develops into ectoderm, the fate of the animal pole.
• It does not organize a secondary axis (which happens in inductive or autonomous specification) and does not disrupt development completely

Q. 13 Six mutant yeast haploids (His1 -6) requiring histidine supplementation for viability were fused in pair-wise combinations to form diploids. Requirement for histidine was tested for the diploids. The results are shown below where ‘+’ indicates diploid combinations yielding histidine prototrophs.

How many different histidine biosynthesis genes are represented among the six mutants?

1. One
2. Two
3. Three
4. Four

Answer – 2

Explanation –

To determine how many different histidine biosynthesis genes are represented among the six mutants (His1 to His6), we can use the principle of complementation analysis.

Key Concepts:

• Complementation occurs when two mutants are defective in different genes, and their diploid combination restores function (results in +).
• If two mutants are defective in the same gene, their combination does not complement (results in -).

Step-by-Step Analysis:

Let’s group mutants based on their complementation pattern — mutants that do not complement each other are likely in the same complementation group (same gene mutation).

Now, looking at the table:

His1:

• + with His2, His4, His5
• - with His3, His6

→ His1 does not complement His3 and His6 → likely same group.

His2:

• + with His1, His3, His5
• - with His4, His6

→ His2 does not complement His4 and His6 → another group.

His3:

• + with His2, His4, His5
• - with His1, His6

→ His3 does not complement His1 and His6 (again matching His1 behavior).

His4:

• + with His1, His3, His6
• - with His2, His5

→ His4 does not complement His2 and His5 (same as His2).

His5:

• + with His1, His2, His3
• - with His4, His6

→ His5 does not complement His4 and His6.

His6:

• + with His2, His4
• - with His1, His3, His5

→ His6 does not complement His1, His3, His5 (matching those three).

Grouping Based on Non-Complementation (-):

• Group 1: His1, His3, His5, His6
• Group 2: His2, His4

Wait — actually, looking closely:

• His1, His3, His5 all do not complement His6 — so these four are Group 1.
• His2 and His4 do not complement each other and His6 — so His6 overlaps both groups.

That suggests three groups, not two. But since the table shows full complementation between His2 and His3, and between His3 and His4, etc., more accurate grouping is:

Refined Groups:

• Group A: His1, His3, His6 → consistently do not complement each other.
• Group B: His2, His4, His5 → consistently do not complement each other.

This gives us two non-overlapping complementation groups → two different genes involved.

Q. 14 Synonymous mutations (solid black circle) and non-synonymous mutations (different symbols) are plotted on two hypothetical phylogenies (A and B) given below.

The phylogenies above may represent the following types of selection – positive, negative or neutral. Which one of the options given below gives the correct combination of the types of selection observed in phylogenies A and B?

1. A shows negative selection, B shows positive selection.
2. A shows positive selection, B shows negative selection.
3. A shows neutral selection, B shows positive selection.
4. A shows positive selection, B shows neutral selection.

Answer – 1

Q. 15 Given below are the outcomes of transplantation experiments.

Which one of the following options correctly depicts the outcome of the transplantation experiments?

1. A and C
2. B and C
3. C and D
4. B and D

Answer – 1

Also read :

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 6

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 5

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 4

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 3

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 2

CSIR NET Life Science Dec 2024 Answer Key (Parts C) – Complete Explanations! Part 1

CSIR NET Life Science Previous Years Question Papers and answer keys

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