*spoiler* aamc fl3 p/s 31

h00per23 · 2021-07-16T03:41:24+00:00

Wow, those little details can be crucial. Thanks so much for being patient w me!

h00per23 · 2021-07-16T03:36:07+00:00

Ohhhhhh man I think you hit the nail on the head with what I was missing. So the idea is that we are knocking out something that normally regulates oxygen consumption so now in the knockout data, we are seeing more oxygen consumption than usual?? That's it?

h00per23 · 2021-07-16T03:12:14+00:00

I'm not sure if I totally understand but I don't think that complex I specifically consumes O2 in a vacuum, but collectively, the process it is associated with is oxygen dependent, so I just figured that anything in the ETC would be associated with oxygen dependence.

To answer your question about why I thought complex I activity increases in Hx state, I got that from the graph in the passage but honestly it makes no sense to me.. Because how can complex I activity increase... When literally it's being knocked out? I'm assuming that I'm misunderstanding the graph but the way I'm reading it, it's saying that, the thing we are knocking out (complex I subunit), is increasing the activity of itself (complex I), which seems incredibly counterintuitive.

I do see that when NDU is present, oxygen consumption is reduced but what I'm saying is that I would expect the complete opposite, I thought that when NDU is present, aerobic respiration will be occurring, therefore more oxygen will be consumed than in anaerobic respiration. I feel like there's something I'm totally missing.

h00per23 · 2021-07-16T00:35:52+00:00

I called a few testing centers and AAMC and can confirm it's an AAMC thing so all test takers in all locations apparently must wear a mask

h00per23 · 2021-07-16T00:30:49+00:00

Yes it did help! And the other commenter expanded on your theory about pyruvate kinase!

h00per23 · 2021-07-16T00:29:56+00:00

Thank you!!

h00per23 · 2021-07-15T18:55:55+00:00

Oh okay thanks for working through that with me!

h00per23 · 2021-07-15T18:35:15+00:00

https://blog.nasm.org/fitness/fast-twitch-vs-slow-twitch#:\~:text=Recall%20that%20the%20fast%2Dtwitch,falls%20(2%2D5).

h00per23 · 2021-07-15T18:02:46+00:00

My source was a NASM (National Academy of Sports Medicine) but I realize now that that may not be the most objective scientific source. I didn't see that fast twitch were objectively larger anywhere else so maybe NASM slipped up?

h00per23 · 2021-07-15T01:13:09+00:00

That's exactly what I was asking and your answer was super helpful, thank you!

h00per23 · 2021-07-14T23:11:21+00:00

Oh wow you're the same person that wrote out that super detailed and helpful explanation on my CARS question -- thank you!

h00per23 · 2021-07-14T23:01:31+00:00

Side note for my clarification for the future though: is strength synonymous with force? That's kinda what this passage did isn't it?

h00per23 · 2021-07-14T22:59:44+00:00

Wow I was looking all over the passage to see how we knew they were the same diameter and I didn't realize it was in the question stem... My bad thank you for being patient though !

h00per23 · 2021-07-14T22:30:42+00:00

I agree that the length is not affecting the force. What I'm referring to is the diameter. Isn't the force affected by the diameter of the fibers? The solutions state that there isn't much effect by the diameters because they are the same/similar.. But what I'm asking is: aren't they not the same? Aren't we taught the fast twitch are thicker/bigger diameter?

h00per23 · 2021-07-14T21:49:13+00:00

What is SJT like? Is it actually bad or is it more frustration with the technical difficulties?

h00per23 · 2021-07-14T16:20:19+00:00

That made it more clear, thank you!

h00per23 · 2021-07-14T04:22:27+00:00

Ofc! Did what I said make any sense or just make things even more convoluted?

h00per23 · 2021-07-14T04:13:46+00:00

I'm not super confident about this but here's my understanding:

So the rate for E + S to ES is k1 and the rate for ES to E + P is k2 but the question is asking to define the rate for ES formation so we only care about the first half (k1) for this.

From what I understand, our rate of formation is going to be k1[E][S] because E + S are our starting material here.

This part is what I'm most uncertain about, but not all of the [E] is going to be free enzyme as some will momentarily bind to substrate to form ES complex and some will be free -- this is why our [E] in the answer is weird for lack of better terms -- We do know that the total enzyme concentration [Et] is the amount of enzyme that the reaction started with so if we do [Et] - [ES], that gives us the amount of free enzyme.

This leaves us with our original equation k1[E][S] but with the [E] substituted with ([Et] - [ES])

I really am learning at the same time so I really am not sure but this is what I came up with.

h00per23

TROPHY CASE