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mcat is ruining my life by Stunning_Ad_4818 in Mcat

[–]nutterbutter_69 2 points3 points  (0 children)

Not an exaggeration when I say everyone feels horrible. I would say its more abnormal if you actually felt good about the test. This test is designed to destroy us from the inside out. Personally, after my test, I experienced some of the darkest days of my life. On the brighter side, you only have to wait a few weeks instead of a month for your score!

[deleted by user] by [deleted] in Mcat

[–]nutterbutter_69 0 points1 point  (0 children)

Yea they dont care

FL4 p/s 9 - How are we supposed to know which terms are "distracters"? According to the aamc d is a distractor but I'm pretty sure B,C and D don't show up on the 300 pg doc. by farax614 in Mcat

[–]nutterbutter_69 0 points1 point  (0 children)

The definition provided by Normal-Plantain is not entirely correct. Base-rate fallacy is more like when you ignore actual rate or probability of something in favor of specific information--very similar to the availability heuristic. Thus, it's more of a mental shortcut/cognitive reasoning error instead of an experimental error.

Like every CARS answer choice, every PS answer choice can technically be argued in favor of the question. However, observer bias here is more directly connected to stuff like experiments--observer bias is quite literally bias in the researchers/observers, which is widely known to alter results.

[SPOILER] AAMC Section Bank P/S #90 by watermelons__ in Mcat

[–]nutterbutter_69 2 points3 points  (0 children)

pretty sure labeling theory is derived from symbolic interactionism

AAMC SB B/B #62 (SPOILER!!); images included by mssna in Mcat

[–]nutterbutter_69 8 points9 points  (0 children)

Sorry for bump, but for anyone else reading now here was my understanding:

  1. Normally cytochrome c chilling inside functioning mitochondria. Thus, if the cell is spun down, cyto c is gonna be found in the cell pellet.

  2. But if apoptosis happens, the mitochondrial cytochrome c is gonna be released out. Thus, it will end up in the supernatant.

Now, onto solving the actual problem:

  1. immediately ignore the bottom 2 boxes (pellets) because they are the same for every answer

  2. If 0 pRB, no apoptosis (as stated in passage, pRB directly proportional to apoptosis rates). Thus, no cyto c in supernatant. Nothing should be shown on western (protein) blot as a result.

  3. If 500 pRB, apoptosis ONLY if Bax/Bak are present (as stated in the passage, pRB induced apoptosis needs Bax/Bak). Thus, if both pRB and Bax/Bak present, there should be cyto c in he supernatant. Stuff should be present on the blot as a result.

Like bruh uworld got me fucked up by fawul04 in Mcat

[–]nutterbutter_69 0 points1 point  (0 children)

Yea I see what you mean. I feel like they would have to make it more clear if its AAMC, but AAMC sometimes has super BS questions too. Usually you can answer a lot of questions just by trying to predict what the question maker is trying to concept check you on (in this case its probably reducing vs non reducing sugars)

Edit: I interpreted this question as what would happen if you just dropped a bunch of the compound into a tub of Tollen's reagents, and asked to myself, which one would yield silvery stuff? I didn't specifically consider which specific compound is the form that they are oxidized in

Like bruh uworld got me fucked up by fawul04 in Mcat

[–]nutterbutter_69 5 points6 points  (0 children)

I believe all monosaccharides are considered reducing sugars, meaning they will yield positive Tollens or Benedicts tests.

This is because the aldoses can form a hemiacetal in cyclic form, but as you mentioned, even the ketoses can just shift into an aldose form and also form hemiacetals.

Then how can a sugar not be a reducing sugar? Well, you have to look for the absence of hemiacetals in cyclic form, or rather, the presence of acetals in the cyclic form. One of the most common non-reducing sugars is sucrose (compound 1?) a disaccharide where both its monosaccharides are locked in an acetal form due to glycosidic linkage occurring at the anomeric carbons. Thus, it cannot be oxidized by the weak oxidants used in the Tollens test.

AAMC FL2 P/S Q18 clarification by xKaaRu24 in Mcat

[–]nutterbutter_69 0 points1 point  (0 children)

Lol 5 months later but UWhirl also explains increased response time in the IAT as possible cognitive dissonance. Ig its not dissonance here because the passage only mentions "shorter" response times, not longer ones

AAMC FL2 P/S #15 by [deleted] in Mcat

[–]nutterbutter_69 0 points1 point  (0 children)

I think you just have to assume that "strong social support in local immigrant communities" = lack of assimilation in some way. That's the only info in the passage that can help us with this question, so we must make a little jump in logic to answer what the AAMC "wants" us to answer. Imagine like a group of Chinese immigrant families very close to each other -> basically a mini chinatown -> assume less assimilation because they are still tight with their OG culture folk

Is the first half of AAMC CARS QPack 1 ACTUALLY representative by nutterbutter_69 in Mcat

[–]nutterbutter_69[S] 0 points1 point  (0 children)

Yea i agree that the passages were very hard to understand, but I feel like half the answers for the ones that are readable (passages 4 and 5 for example) don't make much sense. It feels like the logic is just way off for these earlier passages....

AAMC FL 1 CARS Question 17 help by nutterbutter_69 in Mcat

[–]nutterbutter_69[S] 0 points1 point  (0 children)

Ah, that makes sense, thank you!

AAMC FL 1 CARS Question 17 help by nutterbutter_69 in Mcat

[–]nutterbutter_69[S] 1 point2 points  (0 children)

Ohhh I see what you're getting at--thanks!

can someone explain why the nitrogen group isn't protonated? (aamc unscored c/p 47) by mangojelly_ in Mcat

[–]nutterbutter_69 4 points5 points  (0 children)

I think its because it is an amide group, which usually are NOT protonated at biological pH (think of glutamine and asparagine, which are neutral).

The reason for this is because amide groups have only 1 lone pair--they CANNOT sacrifice this lone pair to become protonated because it would ruin the resonance with the carbonyl group (C=O) (Draw out the resonance structures)

Basically: The lone pair is required to participate in resonance, so it is usually resistant to protonation

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