aamcas q bank

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Squish miss

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i would have never thought to use units to calculate this. anyone else knew how to do it/?

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Good question! The key with this one is to first isolate exactly what we're supposed to be comparing. In an extremely general sense, whenever we're dealing with yield in chemistry, we want to take some value divided by another value, then multiply by 100. For example, the equation for % yield is (actual yield) / (theoretical yield) x 100.

Here, though, that can't be what we're supposed to do, since we have no way to know what the "theoretical yield" is. Instead, we must be comparing what we have after the ion-exchange chromatography step (our "final" value) with what we had initially (the supernatant). Here, we finally get to your question - how do we know we have to use the units, and not something else? The answer is basically that nothing else works. We've already figured out that we need to divide some amount corresponding to the ion-exchange chromatography step by some amount corresponding to the supernatant, then multiply by 100. We can't just use the specific activity values, since that will give us a yield of something like 20000%, which doesn't make any sense for a purification, where the yield should be less than 100%. And we can't just use total protein values, because that includes all the protein present, not just the product of interest.

Since we can't use either value alone, we need to combine them in some way. Since one value is in mg and the other in units/mg, it makes sense to multiply them together, giving us simply "units." (Any other manipulation would give us weirdly complicated units and wouldn't fit any of the answer choices.)

This strategy - quickly narrowing down options to find what they must want us to do - is very helpful on the MCAT. Most students wouldn't see this question and automatically think "we need to compare activity units from the last step with activity units in the first step!" But they might be able to work it out as I described above.

Good luck with your prep :)
 
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thank you. can you also please explain why acetyl co a will not work. in my notes, it does. thank you :) @Next Step Tutor @NextStepTutor_2 @NextStepTutor_3

Absolutely! This question is basically asking which of the options can be used as precursors in gluconeogenesis. We know this by going back to the part of the passage where they mention that "the glucose pool must be replenished" and then reference glucogenic amino acids.

In humans, acetyl-CoA CANNOT be used as a precursor in gluconeogenesis. You mentioned your notes - are those that first image you attached? If so, take a look at acetyl-CoA (in the bottom right corner). From this image, we see that acetyl-CoA is already in the mitochondria. From here, acetyl-CoA can be used in the citric acid cycle, but it is NOT considered a gluconeogenic substrate because it cannot feasibly be used to produce pyruvate in the cytosol.

If this is still unclear, let me know - some online sources do a great job explaining it. And regarding your second comment - what question were you talking about where the answer is glucagon?

Again, good luck with your MCAT studying :)
 
Absolutely! This question is basically asking which of the options can be used as precursors in gluconeogenesis. We know this by going back to the part of the passage where they mention that "the glucose pool must be replenished" and then reference glucogenic amino acids.

In humans, acetyl-CoA CANNOT be used as a precursor in gluconeogenesis. You mentioned your notes - are those that first image you attached? If so, take a look at acetyl-CoA (in the bottom right corner). From this image, we see that acetyl-CoA is already in the mitochondria. From here, acetyl-CoA can be used in the citric acid cycle, but it is NOT considered a gluconeogenic substrate because it cannot feasibly be used to produce pyruvate in the cytosol.

If this is still unclear, let me know - some online sources do a great job explaining it. And regarding your second comment - what question were you talking about where the answer is glucagon?

Again, good luck with your MCAT studying :)


thank you. do you get this last concept.

The Kinetic Molecular Description of Liquids and Solids

if you look at the bottom, decreasing temperature and increasing pressure have the same effect. but aren't they directly proportional as in pv=nrt and should have the same effect?
 
Good question!

First of all, note that PV = nRT is the ideal gas law and only applies to gases assumed to be ideal. One assumption of the ideal gas law is that intermolecular forces between gas particles are negligible. The source you referenced is discussing intermolecular forces, which means that they must be describing real - not ideal - gases. (It also looks like they're talking about liquids as well, which wouldn't be expected to adhere to the ideal gas law.)

Second, look at the diagram on the bottom of that page. Do you see how it looks like the molecules are becoming closer together - in other words, like the volume is decreasing? Your point makes sense - we would expect pressure and temperature to be proportional assuming that the other variables remain constant. But imagine a case in which pressure is doubled, while volume is decreased by a factor of 4. If you plug these values in to the ideal gas law, you'll see that (assuming n remains the same) temperature should be cut in half. Thus, this is a case where pressure increased at the same time that temperature decreased.

Finally, the reasoning I like to use for why decreasing temperature and increasing pressure have the same effect has to do with, again, deviations from the ideal gas law. Decreased temperature and increased pressure are the conditions that promote condensation, or the formation of a liquid from a gas. This is a great example of those two sets of conditions - decreased T and increased P - working together to promote the same effect. This also relates to why gases behave LEAST ideally under low temperature and high pressure - because those are the conditions that promote formation of a liquid, which is very much NOT an ideal gas.
 
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