Microbiology

Step 1: Understanding Beggiatoa's Metabolism

- Beggiatoa is a chemoautotrophic bacterium that derives energy from sulfur oxidation.
- Therefore, Assertion (A) is incorrect.
- Reason (R) is correct because chemoautotrophs derive energy from inorganic compounds.

Step 1: Understanding Microbial Products

- Aspergillus produces Citric acid (W).
- Streptococcus produces Streptokinase (X).
- Trichoderma produces Cyclosporin A (Y).
- Monascus produces Statins (Z).
Thus, the correct match is option (D).

Step 1: Stoichiometry of the Calvin Cycle:
The standard Calvin Cycle equation for the production of PGAL (G3P) is: Out of these 6 PGAL:

• 1 PGAL is the Net Gain (used for glucose synthesis).
• 5 PGAL are used for the **Regeneration** of 3 RuBP molecules.

Step 2: Calculate for 24 PGAL Formed:
The question states that **24 PGAL molecules are formed** (Gross production).

• Scale Factor: .
• Used: .
• G3P (PGAL) for Regeneration: .

Step 3: Final Answer:
12 and 20 G3P. This matches Option (C).

Step 1: Analyze ATP yield via ETS for each stage:
The question asks specifically for ATP formed through ETS alone (excluding substrate-level phosphorylation). A) Glycolysis:

• Glycolysis produces 2 NADH molecules in the cytoplasm.
• These NADH must be shuttled into the mitochondria to enter ETS.
• If the Glycerol-3-Phosphate shuttle is used (common in skeletal muscle/brain), each NADH yields 2 ATP.
• Total ATP via ETS = .
• (Note: If Malate-Aspartate shuttle is used, yield is 6 ATP, but Option A starts with 4, suggesting the G-3-P shuttle assumption).

B) Oxidative Decarboxylation (Link Reaction):

• 2 Pyruvic acid 2 Acetyl CoA.
• Produces 2 NADH (mitochondrial matrix).
• Each mitochondrial NADH yields 3 ATP in ETS.
• Total ATP via ETS = .

C) Krebs Cycle (2 Acetyl CoA molecules):

• Per Acetyl CoA: 3 NADH and 1 FADH are produced.
• For 2 Acetyl CoA: and .
• ATP from NADH: .
• ATP from FADH : .
• Total ATP via ETS = .
• (Substrate-level phosphorylation [GTP] is excluded as per the "ETS alone" condition).

Step 2: Match with Options:
Values are: A=4, B=6, C=22. This matches Option (A).

A: Molecular oxygen participates in final electron acceptance (so connected to electron transport, option may vary; verify in context). B: Electron acceptor = Cytochrome C. C: Connecting link = Acetyl CoA (links glycolysis to TCA cycle). D: Decarboxylation = α-Ketoglutaric acid undergoes decarboxylation in TCA cycle.

1. In lac operon, A = Repressor protein, which binds operator to prevent transcription.
2. B = Inducer (allolactose) inactivates repressor and allows transcription.
3. C = Permease, facilitates lactose entry into the cell.
4. D = β-galactosidase, hydrolyzes lactose to glucose and galactose.
5. Hence the correct assignment is (1) A-Repressor, B-Inducer, C-Permease, D-B galactosidase.

1. Fixing one CO2 in C3 cycle requires 3 ATP & 2 NADPH → A → IV.
2. One glucose formation requires 6 CO2, 18 ATP & 12 NADPH → B → III.
3. ATP & NADPH required for one glucose = 18 ATP & 12 NADPH → C → II.
4. Number of CO2 molecules per glucose = 6 → D → I.
5. Correct matching is (1) A-II, B-III, C-IV, D-I.

1. In lac operon, A = Repressor protein, which binds operator to prevent transcription.
2. B = Inducer (allolactose) inactivates repressor and allows transcription.
3. C = Permease, facilitates lactose entry into the cell.
4. D = β-galactosidase, hydrolyzes lactose to glucose and galactose.
5. Hence the correct assignment is (1) A-Repressor, B-Inducer, C-Permease, D-B galactosidase.

In bacteria, transcription termination can be Rho-dependent. - A = newly synthesized RNA transcript. - B = RNA polymerase synthesizing RNA. - C = Rho factor, which is a helicase that moves along RNA to unwind the DNA-RNA hybrid and terminate transcription. Sigma factor is only required for initiation, not termination.

1. Fixing one CO2 in C3 cycle requires 3 ATP & 2 NADPH → A → IV.
2. One glucose formation requires 6 CO2, 18 ATP & 12 NADPH → B → III.
3. ATP & NADPH required for one glucose = 18 ATP & 12 NADPH → C → II.
4. Number of CO2 molecules per glucose = 6 → D → I.
5. Correct matching is (1) A-II, B-III, C-IV, D-I.