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Home » GATE BT 2010

GATE BT 2010

Total MCQ: 47

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1 Q46 True-breeding Drosophila with curved wings & dark bodies × true-breeding short wings & tan bodies; F₁: curved wings & tan body; F₂: 45 curved/tan, 15 short/tan, 16 curved/dark, 6 short/dark. The mode of inheritance is

  • A) (A) Typical Mendelian; curved & dark dominant
  • B) (B) Typical non-Mendelian; no pattern
  • C) (C) Mendelian with suppression of phenotypes
  • D) (D) Mendelian with single crossover
View Answer & Discuss

2 Q47 Match Group I with Group II: P. Real-Time PCR; Q. 2-D Electrophoresis; R. Affinity Chromatography; S. Microarray. Group II: 1. Biochips; 2. SYBR Green; 3. Antibody‐linked Sepharose beads; 4. Ampholytes.

  • A) (A) P-1, Q-2, R-4, S-3
  • B) (B) P-2, Q-3, R-4, S-1
  • C) (C) P-2, Q-4, R-3, S-1
  • D) (D) P-3, Q-2, R-1, S-4
View Answer & Discuss

3 Q45 Thermal death of microorganisms follows first‐order kinetics (k = 0.23 min⁻¹ at 120 °C). Initial concentration 10⁴ cells/mL; acceptable level 10³. Time to reach acceptable load?

  • A) (A) 48 min
  • B) (B) 11 min
  • C) (C) 110 min
  • D) (D) 20 min
View Answer & Discuss

4 Q43 Match products in Group I with microbial cultures in Group II: P. Gluconic acid, Q. Lysine, R. Dextran, S. Cellulase. Group II: 1. Leuconostoc mesenteroides; 2. Aspergillus niger; 3. Brevibacterium flavum; 4. Trichoderma reesei.

  • A) (A) P-2, Q-1, R-3, S-4
  • B) (B) P-1, Q-3, R-4, S-2
  • C) (C) P-2, Q-3, R-1, S-4
  • D) (D) P-3, Q-2, R-4, S-1
View Answer & Discuss

5 Q44 Assertion (A): Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason (R): cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non-cms plant.

  • A) (A) both (A) and (R) true; (R) correct reason for (A)
  • B) (B) both (A) and (R) true; (R) not correct
  • C) (C) (A) true; (R) false
  • D) (D) (A) false; (R) true
View Answer & Discuss

6 Q42 A cell has five molecules of a rare mRNA amid 4×10⁵ total mRNAs. How many clones must be screened for 99% probability of finding at least one recombinant cDNA of the rare mRNA?

  • A) 4.50×10⁵
  • B) 3.50×10⁵
  • C) 4.20×10⁵
  • D) 4.05×10⁵
View Answer & Discuss

7 Q40 The dissociation constant Kₙ 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
View Answer & Discuss

8 Q41 Receptor R is overexpressed in CHO cells (6×10⁷ cells). Incubated with radioligand (100 cpm/pmol), pellet counts = 1000 cpm. Assuming one ligand binds one receptor, average receptors per cell is

  • A) 10⁴
  • B) 10³
  • C) 10⁶
  • D) 10⁷
View Answer & Discuss

9 Q38 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
View Answer & Discuss

10 Q39 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
View Answer & Discuss
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