Question

Let's use 32 as the number of ATP produced per glucose's aerobic catabolism. (See the last sentence in the legend to Table 21.1 for the rationale for this number.)

Now, what would the ATP yield be, in aerobic catabolism, for the processing of dihydroxyacetone phosphate?

32

16

17

18

Answer 1

#A. Glucose -------> 2 Pyruvate        ; net gain 2 ATP (1 ATP per pyruvate)

[Note: Preparatory phase consumes 2 ATPS and Payoff phase releases 4 ATP from 1 glucose molecule. Thus, net gain = 2 ATP.]

#B. 2 Pyruvate ------> 2 Acetyl CoA --TCA--->         , net gain 30 ATP (15 ATP per pyruvate)        Total ATP gain = A + B = 32 ATP            [As mentioned in question]   

Now,

#C. 1 DHAP -----------> 1 Pyruvate , net gain 2 ATP gain

Since, this step does not include ATP consumumption during “preparatory step”, yield of this step = 2 ATP.

Total ATP yield using 1 DHAP = ATP yield from #C + ATP yield from #B per 1 Pyruvate

                                                = 2 ATP + 15 ATP = 17 ATP

Correct option: C. 17 ATP.