Question

question 35
in bacterial cells, when glucose is completely oxidized by all the pathways of aerobic cellular respiration, what is the
maximum number of atp generated?
a 38
b 36
c 2
d 3
e 24

Explanation:

In bacterial cells, during aerobic cellular respiration:

• Glycolysis produces 2 ATP (substrate - level phosphorylation) and 2 NADH. Each NADH can produce 3 ATP (in bacteria, the electron transport chain is more efficient than in eukaryotes, and there is no shuttle system loss as in eukaryotic cells). So from glycolysis - derived NADH: \(2\times3 = 6\) ATP. Total from glycolysis: \(2 + 6=8\) ATP.
• Pyruvate oxidation: 2 pyruvates are converted to 2 acetyl - CoA. Each pyruvate oxidation produces 1 NADH. So 2 NADH, which gives \(2\times3 = 6\) ATP.
• Krebs cycle: Each acetyl - CoA in the Krebs cycle produces 2 ATP (substrate - level phosphorylation), 6 NADH (\(6\times3 = 18\) ATP), and 2 FADH₂ (\(2\times2 = 4\) ATP). For 2 acetyl - CoA: \(2\times2=4\) ATP (substrate - level), \(6\times2\times3 = 36\) ATP (from NADH), \(2\times2\times2 = 8\) ATP (from FADH₂).

Sum up: \(8+6 + 4+36 + 8=38\) ATP.

Answer:

A. 38