CSIR NET Life Sciences Mock Test - 2 - UGC NET MCQ

Calculation:
Difference in two store of A = 19% ×12600 ~ 15% × 14500 = 2394 - 2175 = 219
The difference in two stores of B = 21% × 12600 ~ 14% × 14500 = 2646 - 2030 = 616
The difference in two stores of C = 8% × 12600 ~ 18% × 14500 = 1008 - 2610 = 1602
The difference in two stores of D = 16% × 12600 ~ 12% × 14500 = 2016 - 1740 = 276
The difference in the two stores of E = 11% × 12600 ~ 14% × 14500 = 1386 - 2030 = 644
The difference in two stores of F = 25% × 12600 ~ 27% × 14500 = 3150 - 3915 = 765
∴ Garment type A has the minimum difference between the two stores.

Calculation:
The total number of students in school = 1000
Students who play chess = 300
Students who play football = 600
Students who play both chess and football = 50

Hence the total number of students playing sports = 250 + 50 + 550 = 850
∴ The total number of students not playing any of the sports = 1000 – 850 = 150

Let EF = x units. Then, DF = 2x units.

AB = 2AE = 6 units.
∴ Area of rect. ABCD = AB × AD
= (6 × 4) sq. units = 24 sq. units.

The ways of placing the balls would be

5, 1, 1, 1 ( = 4 w a y s ) ;

4,2,1, 1 ( = 12 w a y s ) ;

3, 3, 1, 1 ( × 2 ! = 6 w a y s ) ;

3, 2, 2, 1 ( = 12 w a y s ) and

2, 2, 2, 2 (1 way).

Total number of ways = 4 + 12 + 6 + 12 + 1 = 35 ways.

Calculation:

Consumption (km/liter) means distance covered by the car in 1 liter of fuel

⇒ Fuel consumption per liter =

From the above table, fuel consumption per km was least during the lap Q.

∴ The correct answer is Q.

The correct answer is Prerna
Let's denote the positions of the girls as follows:

1. _

2. _

3. _

4. _

5. _

From the given information:

1. Leela has exactly one girl to her left. So, Leela must be in the second position.

2. Alice is just right of Prerna. So, Alice must be immediately to the right of Prerna.

3. There are at least two girls between Radha and Zarina.

Let's place Leela in the second position:

1. _

2. Leela

3. _

4. _

5. _

Since Alice is just right of Prerna, they must occupy two consecutive positions. Let's consider the possible positions for Prerna and Alice:

If Prerna is in position 3, Alice will be in position 4.

If Prerna is in position 1, Alice will be in position 2, but this is not possible since Leela is already in position 2.

Therefore, Prerna must be in position 3 and Alice in position 4:

1. _

2. Leela

3. Prerna

4. Alice

5. _

Now, there are at least two girls between Radha and Zarina. The only possible positions for Radha and Zarina that satisfy this condition are positions 1 and 5:

1. Radha

2. Leela

3. Prerna

4. Alice

5. Zarina

The girl in the middle position (position 3) is Prerna.

The correct answer is Phage display

Concept:

• Phage display is a molecular technique where a library of peptides or proteins is expressed on the surface of bacteriophages. This allows for the identification of peptides or proteins that bind to a specific target, such as a protein or DNA.
• This technique can be used to study protein-peptide interactions as well as protein-DNA interactions, making it highly versatile in molecular biology research.
• Phage display involves the fusion of a peptide or protein of interest to a coat protein of a bacteriophage, which then displays the peptide or protein on its surface. The displayed molecules can then be screened against various targets to identify binding interactions.

Explanation:

• DNA footprinting: This technique is used to identify the specific binding sites of DNA-binding proteins. It involves treating the DNA-protein complex with nucleases and then analyzing the protected DNA fragments. This technique is specific to protein-DNA interactions and does not study protein-peptide interactions.
• 2D-gel electrophoresis: This technique is used to separate proteins based on their isoelectric point and molecular weight. It is primarily used for protein analysis and does not provide information on protein-peptide or protein-DNA interactions.
• Phage display: This technique can be used to study both protein-peptide and protein-DNA interactions. It allows for the identification of peptides or proteins that bind to a specific target, making it a versatile tool in molecular biology research.
• ChIP-on-chip assay: This technique combines chromatin immunoprecipitation (ChIP) with microarray technology to identify DNA sequences bound by specific proteins in vivo. It is specific to protein-DNA interactions and does not study protein-peptide interactions.

The correct answer is Xa21

Concept:

• Bacterial blight is a devastating disease in rice, caused by the bacterium Xanthomonas oryzae pv. oryzae. It significantly reduces crop yield and quality.
• To combat this, researchers have identified several genes in rice that confer resistance to bacterial blight, known as Xa genes.
• Among these, the Xa21 gene is particularly well-known for its strong resistance to multiple strains of the pathogen.

Explanation:

• PiB: This is a gene associated with resistance to blast disease in rice, not bacterial blight. Blast disease is caused by the fungus Magnaporthe oryzae.
• Rar: This is not a recognized gene associated with bacterial blight resistance in rice. It may be a misnomer or an incorrect option.
• Xa21: This gene is well-documented to provide resistance against bacterial blight in rice. It was one of the first resistance genes to be cloned and has been widely studied for its effectiveness in combating Xanthomonas oryzae pv. oryzae.
• Pi9: Similar to PiB, this gene is related to resistance against blast disease in rice, not bacterial blight.

The correct answer is Option 1 i.e. 5

Concept:

• Trypsin, chymotrypsin are endopeptidases and will cleave only internal bonds.
• Carboxypeptidases are exopeptidases cleaves amino acid at C-terminus.

Enzymes specificity:

• Trypsin cleaves at the C-terminal side of lysine (K) and arginine (R), but not if followed by proline (P).
• Chymotrypsin cleaves at the C-terminal side of aromatic amino acids (W, Y, F), but not if followed by proline (P).
• Carboxypeptidase-A cleaves at the C-terminal end, removing amino acids one at a time, preferentially removing aliphatic and aromatic side chains except Lys(K) and Arg (R)

Explanation:

Trypsin Action

• Trypsin cleaves at the C-terminal side of lysine (K) and arginine (R).
• A-K → AK | VR | FWGKT
• Trypsin is an endopeptidase that cleaves at the C-terminal side of lysine (K) and arginine (R) except when these residues are proximal to the C-terminal where it cannot act as an exopeptidase.

Chymotrypsin Action

• Chymotrypsin cleaves at the C-terminal side of phenylalanine (F), tryptophan (W), and tyrosine (Y).
• After F → AK | VR | F | WGKT
• After W → AK | VR | F | W | GKT

After action by trypsin and chymotrypsin,the generated fragments are:-

• AK
• VR
• F
• W
• GKT

This means trypsin and chymotrypsin action results in 5 distinct fragments apart from the original peptide: AK, VR, F, W, GKT.

DNA typically follows Chargaff's rules, which state that in double-stranded DNA, the amount of adenine (A) equals the amount of thymine (T), and the amount of guanine (G) equals the amount of cytosine (C). This means A ≈ T and G ≈ C.

Given the base composition:

• A = 40
• T = 22
• G = 21
• C = 17

The amounts of adenine (40) and thymine (22) are not equal, and the amounts of guanine (21) and cytosine (17) are also not equal. This indicates that the sample does not follow Chargaff's rules and therefore, it is unlikely to be double-stranded DNA, either linear or circular.

Therefore, the given base composition suggests that the DNA is not a typical double-stranded molecule (duplex).

So, the correct answer is DNA is single-stranded

The correct answer is UDP glucuronosyl transferase.

Key Points

• Enzyme Function: UDP-glucuronosyltransferase is a critical enzyme in the process of detoxifying bilirubin, a by-product of the breakdown of hemoglobin in red blood cells.
• This enzyme catalyzes the attachment of glucuronic acid to bilirubin, making it water-soluble.
• Location: This enzyme is primarily found in the endoplasmic reticulum of liver cells, reflecting its central role in processing and excreting bilirubin from the body.
• Biological Importance: The water-soluble form of bilirubin (bilirubin diglucuronide) can be readily excreted in the bile and eventually eliminated in the stool.
• This process is essential for preventing the buildup of bilirubin in the body.
• Bilirubin Metabolism: Bilirubin conjugation is a key step in bilirubin metabolism.
• Initially, bilirubin is insoluble (unconjugated) and bound to albumin in the bloodstream.
• The liver then takes up bilirubin, where UDP-GT catalyzes its conversion to a soluble form.
• Clinical Significance: Impairment in the activity of UDP-glucuronosyltransferase can lead to various forms of jaundice, characterized by a yellowing of the skin and eyes due to elevated levels of bilirubin.
• One common condition related to this enzyme's deficiency is Gilbert's syndrome, a mild, chronic form of unconjugated hyperbilirubinemia.
• Genetic Variations: The activity of UDP-glucuronosyltransferase can vary greatly among individuals, influenced by genetic factors.
• Mutations in the gene encoding this enzyme can lead to disorders affecting bilirubin conjugation and metabolism.
• Drug Interaction: UDP-glucuronosyltransferase also plays a significant role in drug metabolism by catalyzing the glucuronidation of various drugs, which can impact drug effectiveness and toxicity.
• Understanding this enzyme's activity is crucial in pharmacology for assessing drug interactions and responses.

The correct answer is Option 1 i.e. Cyclin B-CDK1

Concept:

• Cell cycle is a highly regulated and ordered series of events. The engines that derive the progression from one step of the cell cycle to the next are cyclin-CDK complexes.
• These complexes are composed of two subunits- cyclin and cyclin-dependent protein kinase. Cyclin is a regulatory protein whereas CDK is a catalytic protein and acts as serine/threonine protein kinase.
• Cyclins are so named as they undergo a cycle of synthesis and degradation in each cycle.
• Humans contain four cyclins- G1 cyclins, G1/S cyclins, S cyclins, and M cyclins.

Diagram:

Explanation:

The G2/M checkpoint is a critical control mechanism that ensures the cell's DNA is fully replicated and any damage is repaired before the cell proceeds into mitosis. This checkpoint thereby prevents the propagation of damaged or incomplete genetic material to daughter cells.

A) Cyclin B-CDK1: This is the correct answer. The Cyclin B-CDK1 complex, also known as the maturation-promoting factor (MPF), plays a pivotal role in controlling the transition from G2 to M phase.

• Accumulation of Cyclin B and activation of CDK1 are necessary for the cell to enter mitosis.
• The activation of Cyclin B-CDK1 complex promotes the initiation of multiple mitotic processes including chromosome condensation, nuclear envelope breakdown, and spindle formation.
• The G2/M checkpoint works by inhibiting the activation of Cyclin B-CDK1 until DNA replication is complete and any DNA damage is repaired.

B) Cyclin D-CDK4: This complex is mainly involved in the early G1 phase of the cell cycle and plays a crucial role in the transition from G0 to G1 and the progression through G1, helping the cell respond to extracellular growth signals. It is not directly involved in the G2/M checkpoint regulation.

C) Cyclin E-CDK2: This complex is important for the G1/S transition, promoting the progression into S phase where DNA replication occurs. While crucial for the initiation of the S phase, it is not the complex directly responsible for regulating the G2/M transition.

D) Cyclin A-CDK2: While Cyclin A-CDK2 has a role in facilitating the cell's progression through S phase and preparation for mitosis, and it is involved in the regulation of the G2 phase, it is specifically the Cyclin B-CDK1 complex that is most directly responsible for the actual checkpoint control at the G2/M transition.

Conclusion:-

Therefore, the Cyclin B-CDK1 complex is the most relevant answer in the context of the G2/M checkpoint.

• Reed beds (also known as constructed wetlands) are engineered systems designed to treat wastewater using natural processes involving wetland vegetation, soils, and their associated microbial assemblages.
• They are effective in removing pollutants through filtration, sedimentation, and biological uptake.
• Specifically, in the context of nitrifying secondary effluent, reed beds support the growth of nitrifying bacteria, which convert ammonia and ammonium from the effluent into nitrates a less harmful form of nitrogen.
• This makes them suitable for treating crude sewage and for nitrifying secondary effluent as part of a more comprehensive wastewater treatment process.

A. CRISPR-Cas9 genome editing is a versatile tool that allows for precise modification of the genome, including the introduction of foreign DNA into specific locations. It can be used to create transgenic Drosophila by targeting and modifying their genomes in a controlled manner.

B. RNA interference (RNAi) is a molecular technique used to suppress gene expression. While it is not used to add new genes to the Drosophila genome, RNAi can be utilized to create transgenic strains where the expression of specific endogenous genes is knocked down, leading to interesting phenotypes for research.

C. P-element transposition is a classic method of generating transgenic Drosophila, harnessing naturally occurring mobile DNA elements (P-elements) to insert foreign DNA into the fly's genome. This technique has been widely used for many years to create various transgenic lines for research purposes.

Therefore, all the mentioned techniques (A, B, and C) can be used to manipulate the Drosophila genome and generate transgenic strains, each through different mechanisms and with different outcome

Concept:

• The extracellular matrix (ECM), also called intercellular matrix, is a three-dimensional network consisting of extracellular macromolecules and minerals, such as collagen, enzymes, glycoproteins, and hydroxyapatite that provide structural and biochemical support to surrounding cells.
• Because multicellularity evolved independently in different multicellular lineages, the composition of ECM varies between multicellular structures; however, cell adhesion, cell-to-cell communication, and differentiation are common functions of the ECM.
• The primary structural protein in the extracellular matrix found in the different connective tissues of the body is collagen.
• The collagen helix, also known as the collagen molecule, is made up of three amino acids bound together to create an elongated fibril. It is primarily present in connective tissue, which includes the epidermis, tendons, ligaments, cartilage, and bones.
• It is an important part of the endomysium in muscular tissue. One to two percent of muscle tissue and 6% of the weight of skeletal muscle tissue are made up of collagen.
• The cell that produces collagen most frequently is called a fibroblast.

Explanation:

• All multicellular animals contain a class of fibrous proteins known as collagens. Collagen is the rigid protein that is most prevalent in connective tissue.
• Deformable sites also contain rubber-like elastin strands that can be stretched and relaxed (e.g., skin, tendons, heart).
• They are produced in significant amounts by connective-tissue cells and in smaller amounts by a wide variety of other cell types. Collagens are the most prevalent proteins in animals and makeup 25% of the total protein mass. They are a crucial part of skin and bone. A normal collagen molecule is distinguished by its long, stiff, triple-stranded helical structure.
• collagen superhelix is made of three chains of collagen polypeptides known as α chains wrapped around one another.
• The creation of the triple-stranded helix depends on the amino acids proline and glycine, both of which are abundant in collagen.
• One-third of the amino acids are made up of glycine. Repeatedly contain Gly-X-Y repetitions (where X is proline and Y is either hydroxyproline or hydroxylysine).

The correct answer is α - Ketoglutarate

• α-Ketoglutarate is a vital precursor for the synthesis of certain amino acids. It is directly involved in the biosynthesis of glutamate, an amino acid that can subsequently be converted into other amino acids like proline and glutamine. This conversion process essentially links the energy metabolism pathways with the synthesis of these amino acids, showcasing the metabolic flexibility within the cell.
• α-Ketoglutarate is an intermediate in the citric acid cycle (Krebs cycle) and serves as a starting point for the synthesis of the amino acids proline, glutamine, and also glutamate.

Concept:

Amino acid metabolism encompasses a wide range of biochemical processes that allow for the synthesis, breakdown, and use of amino acids in the body. Amino acids play crucial roles in various biological processes, including protein synthesis, serving as precursors for neurotransmitters and hormones, and participating in metabolic pathways for energy production.

Synthesis of Amino Acids

• Non-essential Amino Acids: These amino acids can be synthesized by the human body and do not need to be obtained through the diet. The pathways for their synthesis involve intermediates from glycolysis, the pentose phosphate pathway, and the citric acid cycle. For example, alanine is synthesized through transamination involving pyruvate, while serine is derived from 3-phosphoglycerate, an intermediate of glycolysis.
• Essential Amino Acids: These amino acids cannot be synthesized in the body at a rate sufficient to meet physiological needs and must be obtained from the diet. Examples include lysine, methionine, and tryptophan.

Breakdown and Catabolism of Amino Acids
Amino acid catabolism involves the removal of nitrogen-containing amino groups and the breakdown of the remaining carbon skeletons. The catabolic process can be divided into two main phases:-

• Transamination and Deamination: These processes remove the amino group from amino acids. Transamination transfers the amino group to a keto acid, usually forming glutamate, while deamination removes the amino group as ammonia. Glutamate can then undergo oxidative deamination in the liver, releasing ammonia, which is converted to urea in the urea cycle for excretion.
• Carbon Skeleton Metabolism: The carbon skeletons, known as keto acids, are directed into various metabolic pathways, depending on their structure. They can enter:

1. Glycolysis: For example, the carbon skeletons of alanine and serine can be converted into pyruvate.
2. Citric Acid Cycle: For instance, the breakdown products of isoleucine, valine, and methionine can be directed into the citric acid cycle as acetyl-CoA or other intermediates such as succinyl-CoA.
3. Ketogenesis and Gluconeogenesis: Certain amino acids can also serve as precursors for glucose synthesis (gluconeogenesis) or ketone bodies (ketogenesis) under specific physiological conditions, like fasting or intensive exercise.

Amino Acid Derivatives :-
Some amino acids serve as precursors to important biological molecules:-

• Neurotransmitters: For example, tryptophan is a precursor to serotonin, and tyrosine can lead to the synthesis of dopamine, norepinephrine, and epinephrine.
• Nitric oxide: Derived from arginine.
• Porphyrins: Including heme, synthesized partly from glycine.

Fig:The metabolic relationship of amino acids to the citric acid (Krebs) cycle determines whether they are glucogenic or not.

Conclusion:

In summary, amino acid metabolism is a complex interplay of biochemical pathways that manage the synthesis, degradation, and conversion of amino acids. It integrates with other metabolic pathways, reflecting the central role of amino acids in cellular physiology and human health.

The correct answer is In turbulent flow, mixing time is independent of N_Re

Concept:

• The Reynolds Number (N_Re) is a dimensionless number that helps predict flow patterns in different fluid flow situations. It is defined as the ratio of inertial forces to viscous forces and is used to characterize the flow regime as laminar or turbulent.
• In a stirred tank bioreactor, the Reynolds Number plays a crucial role in determining the mixing efficiency and flow regime of the medium being stirred.
• The Reynolds Number is given by the formula: N_Re = (ρ * D * N) / μ,

where

• ρ is the fluid density
• D is the impeller diameter
• N is the impeller speed
• μ is the fluid viscosity.

Explanation:

• Option 1: N_Re is independent of the viscosity of the medium - This statement is incorrect because the Reynolds Number is directly dependent on the viscosity of the medium (μ). An increase in viscosity would lead to a decrease in N_Re, assuming all other factors remain constant.
• Option 2: In laminar flow, mixing time increases with an increase in N_Re - This statement is incorrect. In laminar flow, the mixing time generally decreases with an increase in N_Re because higher N_Re signifies a more vigorous mixing action due to increased inertial forces relative to viscous forces.
• Option 3: N_Re is inversely proportional to the impeller speed - This statement is incorrect. The Reynolds Number is directly proportional to the impeller speed (N). An increase in impeller speed increases N_Re, assuming all other factors remain constant.
• Option 4: In turbulent flow, mixing time is independent of N_Re - This statement is correct. In turbulent flow, mixing is generally very efficient, and the mixing time becomes largely independent of N_Re. This is because turbulent flow ensures rapid mixing and homogeneity regardless of further increases in N_Re.

The correct answer is Option 3

• Nitrogenase enzyme converts nitrogen into ammonia. This is a reduction process where nitrogen gains electrons and is reduced to ammonia.
• Nitrogenase enzyme is highly sensitive to oxygen and may get irreversibly inactivated. Symbiotic nitrogen fixes are present in the root nodules of legumes where leghemoglobin is present in large amounts. It has high affinity for oxygen and it can react with nitrogenase enzyme to inactivate it. In order to protect nitogenase, these N₂-fixers such as Rhizobium have special cells called heterocysts in which N₂-fixation occurs. Heterocytes possess only photosystem I which only generates ATP but not oxygen.

The correct answer is Trastuzumab

Concept:

• Immune checkpoint inhibitors are a class of drugs that help to activate the immune system to recognize and attack cancer cells. They work by blocking checkpoint proteins from binding with their partner proteins, which prevents the "off" signal from being sent, allowing T cells to kill cancer cells.
• Common immune checkpoint inhibitors target proteins such as CTLA-4, PD-1, and PD-L1.

Explanation:

• Ipilimumab: This drug is an immune checkpoint inhibitor that targets CTLA-4 (Cytotoxic T-Lymphocyte-Associated protein 4). By blocking CTLA-4, Ipilimumab enhances the immune system's response against cancer cells. It is commonly used in the treatment of melanoma.
• Pembrolizumab: This is another immune checkpoint inhibitor, but it targets PD-1 (Programmed Cell Death protein 1). Pembrolizumab blocks PD-1, thereby boosting the immune system's ability to attack cancer cells. It is used in treating various cancers including melanoma, lung cancer, and head and neck cancer.
• Nivolumab: Similar to Pembrolizumab, Nivolumab is an immune checkpoint inhibitor targeting PD-1. By inhibiting PD-1, Nivolumab enhances T cell response against cancer cells. It is utilized in the treatment of multiple cancer types, such as melanoma, non-small cell lung cancer, and renal cell carcinoma.
• Trastuzumab: The correct answer is Trastuzumab, which is not an immune checkpoint inhibitor. Instead, Trastuzumab is a monoclonal antibody that targets the HER2 (Human Epidermal growth factor Receptor 2) protein. It is primarily used in the treatment of HER2-positive breast cancer and gastric cancer. Trastuzumab works by inhibiting the growth of cancer cells that overexpress the HER2 protein.

The correct answer is G2 phase nucleus will wait for the S phase nucleus to complete the replication and both the nuclei simultaneously enter into M phase.

Concept:

• In the cell cycle of mammalian cells, the S phase (Synthesis phase) is when DNA replication occurs, and the G2 phase (Gap 2) is the period following DNA replication, leading up to mitosis (M phase).
• Cell cycle phases are tightly regulated by specific molecules and checkpoints to ensure proper progression and replication of DNA.
• When cells at different stages of the cell cycle are fused, the regulation mechanisms interact, sometimes causing synchronization of the cell cycle phases.

Explanation:

• Option 1: G2 phase nucleus will wait for the S phase nucleus to complete the replication and both the nuclei simultaneously enter into M phase.
  • This is the correct answer. The S phase promoting factors (SPFs) present in the S phase nucleus can influence the G2 phase nucleus, causing it to pause until DNA replication is completed in the S phase nucleus.
  • After the S phase completes, both nuclei can proceed together into the M phase, synchronizing their cell cycles.
• Option 2: S phase nucleus would immediately enter into G2 phase without completing the replication phase.
  • This is incorrect. The S phase nucleus cannot skip DNA replication; it must complete the replication process before progressing to the G2 phase.
  • The cell cycle checkpoints ensure that DNA replication is complete before moving forward.
• Option 3: Both the nuclei would follow their corresponding cell cycle without influencing each other.
  • This is incorrect. When fused, the regulatory factors from one nucleus can influence the other, leading to synchronization or adjustments in the cell cycle phases.
• Option 4: Due to the influence of S phase promoting factor, G2 phase nucleus will enter into S phase.
  • This is incorrect. The G2 phase nucleus has already completed DNA replication and does not revert to the S phase.
  • Instead, it waits until the S phase nucleus completes its replication before both can enter the M phase together.

The correct answer is 9.4 and 75

• Protein concentration and enzyme activity are key metrics in biochemical purification processes.
• Fold purification is calculated by dividing the specific activity of the purified fraction by the specific activity of the initial extract.
• Percentage recovery is calculated by dividing the total activity of the purified fraction by the total activity of the initial extract and multiplying by 100.

Initial Extract:

• Protein concentration: 5 mg/mL
• Enzyme activity: 2 units/mL
• Total volume: 100 mL
• Total protein: 5 mg/mL x 100 mL = 500 mg
• Total activity: 2 units/mL x 100 mL = 200 units
• Specific activity: 2 units/mL / 5 mg/mL = 0.4 units/mg

Purified Fraction:

• Protein concentration: 4 mg/mL
• Enzyme activity: 15 units/mL
• Total volume: 10 mL
• Total protein: 4 mg/mL x 10 mL = 40 mg
• Total activity: 15 units/mL x 10 mL = 150 units
• Specific activity: 15 units/mL / 4 mg/mL = 3.75 units/mg

Fold Purification:

• Fold purification = Specific activity of purified fraction / Specific activity of initial extract
• Fold purification = 3.75 units/mg / 0.4 units/mg = 9.375 ≈ 9.4

Percentage Recovery:

• Percentage recovery = (Total activity of purified fraction / Total activity of initial extract) x 100
• Percentage recovery = (150 units / 200 units) x 100 = 75%

The correct answer is Using CaMV35S promoter-specific forward and GAPDH-specific reverse primer.

Concept:

To confirm the presence of a transgene in transgenic plants, you typically want to verify both the promoter and the gene of interest. In this case, the transgenic rice has the GAPDH gene under the control of the CaMV35S promoter.

• Using CaMV35S promoter-specific forward and GAPDH-specific reverse primer: This approach allows you to amplify a fragment that includes both the promoter and the gene, confirming that the transgene is correctly integrated and being expressed in the rice plant. The forward primer will bind to the promoter region, and the reverse primer will bind to the GAPDH coding region, creating a PCR product only if both components are present.

Explanation:

• 1) Using exon-specific primers of the GAPDH gene: While this can confirm the presence of the GAPDH gene, it won't confirm that it is under the control of the CaMV35S promoter.

• 2) Using intron-specific primers of the GAPDH gene: If the GAPDH gene has introns and you use intron-specific primers, it might not work in a transgenic line where the gene is expressed as a cDNA without introns, leading to a lack of amplification.

• 3) Amplification of the promoter region of the GAPDH gene: This would only verify the presence of the promoter but wouldn’t confirm whether the GAPDH gene is present and functioning downstream of that promoter.

Thus, the best approach to confirm transgenic lines in this scenario is Using CaMV35S promoter-specific forward and GAPDH-specific reverse primer as it checks for both the promoter and the gene together.

The correct answer is 25.

• CMS line: This line has cytoplasmic male sterility due to mitochondrial mutations, resulting in male-sterile flowers.
• Rf line: This is a homozygous restorer of fertility line with the dominant Rf gene, which restores fertility in the presence of CMS.

Cross Details:

1. Cross the CMS line (cc) with the homozygous Rf line (RR):
   • The CMS line (cc) cannot produce functional pollen.
   • The Rf line (RR) can restore fertility when crossed.

F1 Generation:

• The F1 progeny will be heterozygous for the Rf gene: Rr.
• All F1 plants will be male fertile since they possess the dominant Rf allele (the Rf gene restores fertility).

Self-Pollination of F1:
When the F1 progeny (Rr) is self-pollinated, the genotypic ratio in the F2 generation will be as follows:

• Rr x Rr leads to:
  • RR (fertile): 25%
  • Rr (fertile): 50%
  • rr (male sterile): 25%

F2 Progeny:

• The male sterile phenotype corresponds to the rr genotype.
• The percentage of male sterile plants in the F2 generation is 25%.

Conclusion: Thus, 25% of the F2 progeny will be male sterile.

Gene therapy is a technique for correcting defective genes responsible for disease development. It can be classified into two main types: somatic gene therapy and germline gene therapy.

Germline gene therapy targets the germ cells, which are the sperm and eggs, or the embryos at a very early stage. The primary goal here is to make genetic changes that will be carried forward into every cell of the resulting individual, including their germ cells. Therefore, these modifications are not restricted to the individual who receives the therapy but are also passed down to their descendants. This kind of therapy has the potential to eradicate certain genetic diseases from a family line.

Somatic gene therapy targets non-reproductive cells and is designed to treat or cure a genetic condition within an individual by modifying the genes in some of their body (somatic) cells. Since these genetic changes are confined to the somatic cells of the individual and do not affect their germ cells (sperm or eggs), the modifications are not inherited by any offspring.

Key Points

[P] Affected individuals, but not their progeny, can be cured through germline gene therapy: This statement is incorrect. Germline gene therapy targets the reproductive cells, meaning changes would be passed on to future generations. So, if successful, it can potentially cure affected individuals and their progeny.

[Q] Affected individuals, as well as their progeny, can be cured through germline gene therapy: This statement is correct. Germline gene therapy involves modifying the genes in a person's gametes (egg or sperm) or the genes in the early embryos. These changes would be heritable, affecting not only the individual who received the therapy but also their descendants.

[R] Affected individuals, but not their progeny, can be cured through somatic gene therapy: This statement is correct. Somatic gene therapy targets only the non-reproductive cells, meaning the modifications are not passed on to the individual's offspring. This type of therapy is aimed at treating or curing disease within the individual without affecting the genetic makeup of future generations.

[S] Affected individuals, as well as their progeny, can be cured through somatic gene therapy: This statement is incorrect. Since somatic gene therapy does not alter the reproductive cells, any changes made would not be inherited by the individual's progeny.

Conclusion:

Therefore, the correct statements are Q and R

The correct answer is Option 1 i.e.All statements are true and correctly describe the mechanisms of protein sorting to the respective organelles.

Concept:

Signal Recognition Particle (SRP) and the ER
Signal Recognition Particle (SRP):

• The SRP is a ribonucleoprotein complex that plays a critical role in the targeting of proteins to the ER.
• When a ribosome synthesizes a protein destined for the ER, the emerging polypeptide chain presents an ER signal sequence (also known as a signal peptide) at its N-terminus.
• The SRP recognizes and binds to this signal sequence, temporarily halting translation (a process termed translation arrest).
• The SRP-ribosome-nascent chain complex is then targeted to the ER membrane, where it interacts with the SRP receptor.
• Upon docking, the signal sequence is inserted into a translocon complex in the ER membrane, SRP is released, and translation resumes, threading the polypeptide into the ER lumen or membrane.
• Within the ER, signal peptidases often cleave off the signal sequence, and the protein can undergo folding and post-translational modifications.

Nuclear Pore Complexes and Nuclear Localization Signal (NLS)
Nuclear Transport:

• Proteins destined for the nucleus contain a nuclear localization signal (NLS), which is a specific sequence of amino acids that serves as a tag.
• The NLS is recognized by nuclear transport receptors (importins), which facilitate the protein's binding to the nuclear pore complex (NPC).
• The NPC is a massive structure that spans the nuclear envelope, allowing controlled transport into and out of the nucleus.
• Through a process of facilitated diffusion and active transport, the importin-protein complex passes through the NPC into the nucleus.
• Once inside, the importin is recycled back to the cytoplasm, and the protein is released to fulfill its nuclear functions.

Mitochondrial Targeting Sequence
Mitochondrial Protein Import:

• Proteins destined for the mitochondria are usually synthesized in the cytosol with a mitochondrial targeting sequence (MTS) at their N-terminal end.
• This MTS is a positively charged sequence that directs the protein to the mitochondrial surface.
• At the outer mitochondrial membrane, the MTS is recognized by translocase receptors.
• The protein is then threaded through translocases in the outer and inner mitochondrial membranes, often with the aid of chaperones.
• Upon successful translocation into the mitochondrial matrix or membrane, the MTS is typically cleaved off by specific peptidases.

Peroxisomal Targeting Signal (PTS1) and Pex5p
Peroxisomal Import:

• Peroxisomal proteins often harbor a specific peroxisomal targeting signal (PTS1) at their C-terminus, which consists of a short sequence of amino acids, typically ending in SKL (serine-lysine-leucine) or a variant thereof.
• Pex5p is a cytosolic receptor that specifically recognizes the PTS1, binds to it, and directs the protein to the peroxisomal membrane.
• At the peroxisome, Pex5p interacts with components of the docking complex, facilitating the translocation of the cargo protein into the peroxisome.
• Following protein delivery, Pex5p is recycled back into the cytosol for further rounds of transport.

Explanation:

• Statement 1: The Signal Recognition Particle (SRP) guides the ribosome-nascent polypeptide complex to the ER membrane, where the polypeptide is co-translationally translocated into the lumen or inserted into the membrane of the ER, depending on the presence and nature of other signaling sequences.
• Statement 2: The presence of a Nuclear Localization Signal (NLS) is a key feature for the identification and transport of proteins into the nucleus through the nuclear pore complexes. This process is highly regulated and ensures that only proteins with the NLS can enter the nucleus.
• Statement 3: Mitochondrial proteins often have a mitochondrial targeting sequence (MTS) at their N-terminus, which directs the protein to the mitochondria. These sequences are recognized by receptors on the surface of the mitochondria, facilitating the translocation of these proteins across the mitochondrial membranes.
• Statement 4: The peroxisomal targeting signal 1 (PTS1) is indeed recognized by the cytosolic receptor protein Pex5p. This interaction is critical for the targeting and import of proteins into the peroxisomes, where Pex5p functions as a shuttle to transport these proteins across the peroxisomal membrane.

Conclusion:

Therefore, the correct answer is Option 1

Concept:

• The temperature of the environment affects the metabolic rate of individual organisms and it also determines its geographical distribution.
• Based on how an organism obtains its body heat, organisms are classified as follows:
  • Ectotherm: Organisms depend on the environment to obtain their heat
  • Endotherm: Organisms obtain heat from metabolic interaction and do not depend on the environment to obtain heat.
• Based on how the animal maintains its body temperature, organisms are classified as follows:
  • Poikilotherm: These organisms regulate their body temperature primarily by environment. They cannot maintain a constant body temperature
  • homeotherm: These organisms regulate their body temperature primarily by their body metabolism. They can maintain their constant body temperature.
• The temperature of the organism can vary over time (temporal heterothermy) and from one region of its body to another (regional heterothermy).

Explanation:

• Regional heterothermy refers to the phenomenon where organisms are able to maintain different temperatures in different parts or regions of their body.
• Bumblebees are another example of an organism having regional heterothermy. They show counter-current heat exchange where heat is retained in the thorax and lost from the abdomen.
• Moths also show regional heterothermy as they generate heat mainly in the thorax before they take a flight but their abdomen remains relatively cool.
• Hence, the correct answer is option 3.

A cross between two organisms with contrasting phenotypes for two traits. One parent has violet flowers and green seeds (genotype VVGG), and the other has white flowers and yellow seeds (vvgg). Given that these parents are pure-breeding (homozygous) for both traits, the F1 generation would all be heterozygous (VvGg) for both traits, exhibiting the dominant phenotypes (violet flowers and green seeds).

• When these F1 individuals (VvGg) are self-crossed to produce the F2 generation, the phenotypic distribution follows a pattern that we would expect from a dihybrid cross, with expected phenotypic ratios of 9:3:3:1 for the respective traits if the genes assort independently (according to Mendel's law of independent assortment).
• However, the observed ratio in the F2 generation significantly deviates from the 9:3:3:1 expected ratio, especially evident in the low numbers of plants with combination phenotypes (violet flowers and yellow seeds, and white flowers and green seeds).

Key Points

• Same genes control both flower and seed colors: This is not supported by the data, as we observe four distinct phenotypes, indicating that the traits are controlled by separate genes.
• Genes for flower and seed colors are genetically interacting: Genetic interaction implies that the expression of one gene affects the expression of another. While this is possible, the distribution suggests more about how the genes are inherited rather than their interaction in determining the phenotype.
• Genes for flower and seed colors are present on the same chromosome: This is the most plausible explanation for the observed phenotypic distribution. When genes are linked (on the same chromosome), they do not assort independently and are usually inherited together during meiosis, leading to a deviation from the expected 9:3:3:1 ratio in a dihybrid cross. Recombination through crossing-over during meiosis can explain the appearance of the non-parental phenotypes (violet flowers and yellow seeds, white flowers and green seeds), but in smaller numbers compared to the parental phenotypes, which is consistent with the observed numbers.
• Flower color in this plant species is a polygenic trait: Polygenic inheritance involves multiple genes contributing to a single trait, typically resulting in a continuous variation (like skin color in humans). This does not directly explain the specific phenotypic ratio observed in the progeny.

Thus, the observed distribution suggests that the genes for flower and seed color are not assorting independently as expected according to Mendel's second law. The most likely explanation for the observed deviation from the expected Mendelian ratio is that the genes for flower and seed colors are present on the same chromosome (linked genes), leading to fewer recombinant phenotypes than would be expected with independent assortment.

Conclusion:

Therefore, the correct interpretation is Genes for flower and seed colors are present on the same chromosome

Concept:

Equilibrium theory of Island biogeography

• Biogeography is the study of the geographical distribution of species and the reason for their pattern of distribution.
• Ecologists Robert MacArthur and E.O. Wilson, proposed the equilibrium theory of Island biogeography.
• By 'islands' we mean not only oceanic islands, but also habitat islands on land, such as the peaks of mountains and isolated springs in the desert.
• Any isolated habitat, totally separated from others of its kind, can be considered as an island.
• The oceanic islands can be formed in two basic ways: by the disappearance of a connecting land from the mainland either the sea rises and floods an area in between or if joining land is eroded or sinks or by land rising from the sea usually form by volcanic activity on the ocean floor gradually building up igneous material until it reaches above sea level.
• The theory proposes that the number of species on any island is determined by a balance between the rate at which new species colonize it i.e. the rate of immigration of species from the mainland to the island and the rate at which populations of established species become extinct i.e. the rate of extinction of species on the island.
• The rate at which species might become extinct on the island would be related to the number that has become residents
• Two physical features of the island further affect immigration and extinction rates: its size and its distance from the mainland.
• The mainland is that place where new immigrant species originally inhabited.

Effect of size:

• It is well established that the number of species on islands decreases as island area decreases.
• Small islands have higher extinction rates, as they generally contain fewer resources and less diverse habitats for colonizing species.
• Larger islands reduces the probability of extinction, as they contain larger habitat areas and more resources for colonizing species.
• The lower rate of extinction on larger islands results in a higher equilibrium number of species as compared to smaller islands.

Effect of distance:

• Islands that are more isolated are less likely to receive immigrants than islands that are less isolated.
• The greater the distance between the island and the mainland, the less the likelihood that many immigrants will successfully complete the journey.
• The result is a decrease in the equilibrium number of species.
• Thus, for two islands of equal size, an island close to mainland generally has a higher immigration rate than one farther away.
• When an island is nearly empty, the extinction rate is necessarily low because few species are available to become extinct.
• As more species inhabit an island, extinction rates on the island increase because of the greater likelihood of competitive exclusion.
• At the same time, as the number of species on the island increases, the immigration rate of new species decreases.
• Hence, the rate at which additional species will establish populations will be high when the island is relatively empty, and the rate at which resident populations go extinct will be high when the island is relatively full.
• Over time, the countervailing forces of extinction and immigration result in an equilibrium level of species richness.
• The equilibrium species richness of an island at which immigration balances extinction and which remains roughly constant.

Explanation:

Statement A is correct

• The theory proposes that the number of species on any island is determined by a balance between the rate at which new species colonize it i.e. the rate of immigration of species from the mainland to the island and the rate at which populations of established species become extinct i.e. the rate of extinction of species on the island.

Statement C is correct

• Islands that are more isolated are less likely to receive immigrants than islands that are less isolated.
• The greater the distance between the island and the mainland, the less the likelihood that many immigrants will successfully complete the journey.
• The result is a decrease in the equilibrium number of species.

Hence the correct answer is option 3

Key Points

• Polyadenylation is a post-transcriptional modification process that involves the addition of a polyadenylate (polyA) tail to the 3' end of most eukaryotic mRNA molecules.
• This process is essential for mRNA stability, nuclear export, and translation efficiency. The addition of the polyA tail is carried out by the enzyme Poly(A) polymerase.
• During polyadenylation, the pre-mRNA undergoes cleavage at a specific site downstream of the coding region.
• This site is called the polyadenylation site or poly(A) site.
• After cleavage, Poly(A) polymerase adds a stretch of adenosine residues (AMP) to the 3' end of the mRNA molecule, resulting in the formation of the poly(A) tail.
  ​

Explanation:

Statement 1: Poly(A) polymerase is a template independent enzyme:

• This statement is true. Poly(A) polymerase does not require a template to add the poly(A) tail to mRNA.
• It adds adenine nucleotides (AMP) sequentially to the 3' end of the mRNA molecule.

Statement 2: Poly(A) polymerase catalyzes the addition of AMP from dATP to the 3' end of mRNA:

• This statement is incorrect. Poly(A) polymerase utilizes ATP as the source of adenosine monophosphate (AMP) during polyadenylation.
• It catalyzes the addition of AMP residues from ATP to the 3' end of mRNA.

Statement 3: Poly(A) polymerase is an RNA-template dependent enzyme:

• This statement is incorrect. Poly(A) polymerase does not require an RNA template for its activity.
• It adds AMP residues to the mRNA in a template-independent manner.

Statement 4: Poly(A) polymerase catalyzes the addition of ADP from ATP to the 3' end of mRNA:

• This statement is incorrect.
• Poly(A) polymerase adds AMP residues (adenosine monophosphate) to the 3' end of mRNA, not ADP (adenosine diphosphate).

Statement 5: Poly(A) polymerase catalyzes the addition of AMP from ATP to the 3' end of mRNA:

• This statement is true.
• Poly(A) polymerase utilizes ATP (adenosine triphosphate) as the source of AMP and catalyzes the addition of AMP residues to the 3' end of mRNA during polyadenylation.

Statement 6: Poly(A) polymerase catalyzes the addition of AMP from dADP to the 3' end of mRNA:

• This statement is incorrect. Poly(A) polymerase uses ATP as the source of AMP, not dADP (deoxyadenosine diphosphate).

• Therefore, the correct combination is: (1,5)
• Poly(A) polymerase is a template independent enzyme.
• Poly(A) polymerase catalyzes the addition of AMP from ATP to the 3' end of mRNA.

Antigen processing and presentation, a fundamental aspect of the adaptive immune response.

P. Tc cells recognize antigen with class I MHC molecules on target cells.

• Tc cells, also known as cytotoxic T lymphocytes (CTLs), primarily recognize antigens presented by Class I Major Histocompatibility Complex (MHC) molecules on the surface of target cells. These cells are essential for the immune system's ability to eliminate virus-infected cells, certain types of tumor cells, and cells infected with intracellular pathogens. The interaction between the T-cell receptor (TCR) on Tc cells and the peptide-MHC Class I complex on target cells is critical for the activation of CTLs.

Q. Endogenous antigens are degraded into peptides within the cytosol by proteasomes.

• Endogenous antigens, which typically originate from within the cell (e.g., viral proteins or abnormal cellular proteins), are degraded into peptide fragments by proteasomes in the cytosol. These peptides are subsequently transported into the endoplasmic reticulum (ER) where they are loaded onto Class I MHC molecules for presentation on the cell surface. This pathway is crucial for the immune surveillance of cells by CTLs.

R. Exogenous antigens are internalized and degraded within the acidic endocytic compartments.

• Exogenous antigens, which come from outside the cell (e.g., pathogens or foreign proteins), are internalized by antigen-presenting cells (APCs) such as dendritic cells, macrophages, and B cells. These antigens are then processed in acidic endocytic compartments (such as endosomes and lysosomes) where they are degraded into peptide fragments. These peptides are loaded onto Class II MHC molecules for presentation to CD4⁺ T-helper cells, a critical step for initiating immune responses including antibody production and activation of macrophages.

S. Presentation of nonpeptide antigens derived from bacteria involves the class I-like CD1 molecules.

• CD1 molecules are MHC class I-like molecules that present lipid and glycolipid antigens, rather than peptides, to T cells. This includes nonpeptide antigens derived from the cell walls of mycobacteria and other bacteria. CD1 molecules thus play a unique role in the immune response against a variety of pathogens, expanding the repertoire of antigens that can be recognized by the immune system beyond peptides.

Conclusion:

Therefore, all statements are correct, each describing a different aspect of antigen processing and presentation critical to the adaptive immune response. So, Option 4 is correct.