NS FL3 BB Q33 and 35

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Amber9
Posts: 29
Joined: Thu Jul 04, 2019 11:08 am

NS FL3 BB Q33 and 35

Post by Amber9 » Tue Jul 23, 2019 7:36 pm

Q33 I thought the answer was D because the ATP production was reduced by 67% and that would involve oxidative phosphorylation. I'm not sure about what it means that there is not enough oligomycin molecules.

Q35 I thought if the parent chain was 2 carbons shorter forming the acetyl-CoA molecule then that would be isotopically-labeled only

Thanks!
NS_Tutor_Mathias
Posts: 280
Joined: Sat Mar 30, 2019 8:39 pm

Re: NS FL3 BB Q33 and 35

Post by NS_Tutor_Mathias » Wed Jul 24, 2019 3:09 pm

#33:
A few problems with substrate-level phosphorylation:
1. "Researchers studying the energy yield of fatty acid metabolism estimated the amount of ATP produced in the mitochondria" - glycolysis happens in the cytosol, so apparently they are already accounting for this to get an accurate estimate of mitochondrial production.

2. If true, the oxphos to substrate level ratio should resemble 16:1, not 3:1.

3. Glycolysis is also not invovled in the breakdown of fatty acids (you can assume any cell will be undergoing some gylcolysis, but again: this speaks for how the researchers would likely control for baseline glycolysis, or just use a separate inhibitor of a glycolysis enzyme, for example PFK).

This one seems to catch a lot of people, but ultimately it's still a fair question. And in general, some of the trickiest MCAT questions are those in which a true statement is made, but it does not answer the question.

#35.
What you said in your post is correct, the chain will be two carbons shorter. However, that means the acetyl-CoA generated will contain carbons #1 and #2 (#2 being labeled in this case). At the same time, the first carbon of the acyl-CoA we generated (the big remainder of our chain) will be the previously labeled carbon #3. So both molecules end up radiolabeled!
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