Questions and Answers Archive - Page 60 of 87

True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body and, 6 short wings dark body. The mode of inheritance is (A) Typical Mendelian with curved wings and tan body being dominant (B) Typical non-Mendelian with curved wings and tan body not following any pattern (C) Mendelian with suppression of phenotypes (D) Mendelian with single crossover

True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 … Read more

Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 10⁸ cells / ml and the final acceptable contamination level is 10⁻³ cells, for how long should 1m³ medium be treated at temperature of 120° (thermal deactivation rate constant = 0.23 / min) to achieve acceptable load? (A) 48 min (B) 11 min (C) 110 min (D) 20 min

Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 10⁸ cells / ml and the final acceptable contamination level is 10⁻³ cells, for how long should 1m³ medium be treated at temperature of 120° (thermal deactivation rate constant = 0.23 / min) … Read more

Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non cms plant. (A) both (a) and (r) are true and (r) is the correct reason for (a) (B) both (a) and (r) are true and (r) is not the correct reason for (a) (C) (a) is false but (r) is true (D) (a) is true but (r) is false

Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non cms plant. (A) both (a) and (r) are true and (r) … Read more

Match the products in Group I with the microbial cultures in Group II used for their industrial production Group I: P. Gluconic acid Q. L-Lysine R. Dextran S. Cellulase Group II: 1. Leuconostoc mesenteroids 2. Aspergillus niger 3. Brevibacterium flavum 4. Trichoderma reesei (A) P-2, Q-1, R-3, S-4 (B) P-1, Q-3, R-4, S-2 (C) P-2, Q-3, R-1, S-4 (D) P-3, Q-2, R-4, S-1

Match the products in Group I with the microbial cultures in Group II used for their industrial production Group I: P. Gluconic acid Q. L-Lysine R. Dextran S. Cellulase Group II: 1. Leuconostoc mesenteroids 2. Aspergillus niger 3. Brevibacterium flavum 4. Trichoderma reesei (A) P-2, Q-1, R-3, S-4 (B) P-1, Q-3, R-4, S-2 (C) P-2, … Read more

A cell has five molecules of a rare mRNA. Each cell contains 4 × 10⁵ mRNA molecules. How many clones one will need to screen to have 99% probability of finding at least one recombinant cDNA of the rare mRNA, after making cDNA library from such cell? (A) 4.50 × 10⁵ (B) 3.50 × 10⁵ (C) 4.20 × 10⁵ (D) 4.05 × 10⁵

A cell has five molecules of a rare mRNA. Each cell contains 4 × 10⁵ mRNA molecules. How many clones one will need to screen to have 99% probability of finding at least one recombinant cDNA of the rare mRNA, after making cDNA library from such cell? (A) 4.50 × 10⁵ (B) 3.50 × 10⁵ … Read more

Receptor R is over expressed in CHO cells and analysed for expression. 6 × 10⁷ cells were incubated with its radioactive ligand (specific activity 100 counts per picomole). If the total counts present in cell pellet was 1000 cpm, the average number of receptors R per cell is (assume complete saturation of receptors with ligand and one ligand binds to one receptor) (A) 10⁴ (B) 10⁵ (C) 10⁶ (D) 10⁷

Receptor R is over expressed in CHO cells and analysed for expression. 6 × 10⁷ cells were incubated with its radioactive ligand (specific activity 100 counts per picomole). If the total counts present in cell pellet was 1000 cpm, the average number of receptors R per cell is (assume complete saturation of receptors with ligand … Read more

The dissociation constant Kd for ligand binding to the receptor is 10⁻⁷ M. The concentration of ligand required for occupying 10% of receptors is (A) 10⁻⁶ M (B) 10⁻⁷ M (C) 10⁻⁸ M (D) 10⁻⁹ M

The dissociation constant Kd for ligand binding to the receptor is 10⁻⁷ M. The concentration of ligand required for occupying 10% of receptors is (A) 10⁻⁶ M (B) 10⁻⁷ M (C) 10⁻⁸ M (D) 10⁻⁹ M

Dizygotic twins are connected to a single placenta during their embryonic development. These twins (A) have identical MHC haplotypes (B) have identical TH cells (C) have identical T cells (D) can accept grafts from each other (both (A) and (B))

Dizygotic twins are connected to a single placenta during their embryonic development. These twins (A) have identical MHC haplotypes (B) have identical TH cells (C) have identical T cells (D) can accept grafts from each other (both (A) and (B))

A mutant Gα protein with increased GTPase activity would (A) not bind to GTP (B) not bind to GDP (C) show increased signaling (D) show decreased signaling

Match Group I with Group II Group I: P. Staphylococcus aureus Q. Candida albicans R. Mycobacterium tuberculosis S. Lactobacillus lactis Group II: 1. Biofilms 2. Bacteriocins 3. Methicillin resistance 4. Isoniazid (A) P-1, Q-4, R-2, S-3 (B) P-2, Q-3, R-1, S-4 (C) P-3, Q-1, R-4, S-2 (D) P-1, Q-2, R-4, S-3

Match Group I with Group II Group I: P. Staphylococcus aureus Q. Candida albicans R. Mycobacterium tuberculosis S. Lactobacillus lactis Group II: 1. Biofilms 2. Bacteriocins 3. Methicillin resistance 4. Isoniazid (A) P-1, Q-4, R-2, S-3 (B) P-2, Q-3, R-1, S-4 (C) P-3, Q-1, R-4, S-2 (D) P-1, Q-2, R-4, S-3