AAMC CBT4 and 4R OFFICIAL Q&A

[Vihsadas]

This is the official Q&A thread for AAMC CBT4 and 4R.

Please post ONLY questions pertaining to AAMC CBT4 and 4R.
Out of respect for people who may not have completed the other exams, do not post questions or material from any other AAMC exam.

Please see this thread for the rules of order before you post.

Good luck on your MCAT!

---

Members don't see this ad.

 

[gettheleadout]

In the freeze-thaw frog passage, what does plasma hemoglobin level represent? Is it a measure of the hemolysis because normal hemoglobin shouldn't be found in the plasma? Is that why Figure 2 shows that survival and protection against hemolysis are promoted by exogenous glucose?

Yep. "Blood hemoglobin" would be different, but plasma excludes RBC's and we shouldn't find extracellular Hb unless cells had lysed.

 

[summergirl]

Hi, I guessed right on this question but I honestly did not understand it.I've attached a picture of it. What does the circle with the "S" in it mean? What role does the resistor play in this question? And which way does the electrons travel? As you can see, I just simple do not get this question. HELP!

 

[gettheleadout]

Sine wave (squiggly) in a circle means AC voltage source. The direction oscillates back and forth.

 

[pinklefluff]

Hey guys,

I am a little confused on this question:

Which of the following, if found true, would best refute the hypothesis that a membrain-bound enzyme is activated by a G protein?
a) the enzyme can be activated in the absence of bound GTP
d) the enzyme is always found in the activated state
which would you say is the better answer?

So the correct answer is A and TPR says that D would also refute the hypothesis but A is a better answer. From the way I see it, for A, just because the enzyme can be activated without a bound GTP (therefore without a G-protein) doesn't mean that a G protein wont activate the enzyme. Whereas in D, if the enzyme is always activated then it doesnt need the G-protein to do anything.

Does someone else have a more accurate way of reasoning this out?

Thanks!

 

[ashtonjam]

From the way I see it, for A, just because the enzyme can be activated without a bound GTP (therefore without a G-protein) doesn't mean that a G protein wont activate the enzyme.

I find that this is kind of a trend in logic/experiment problems in both BS and VR. The best evidence for something is the opposite case. By that, I mean that if an enzyme is hypothesized to be activated by G-protein with GTP, then the best evidence for that hypothesis is removing the GTP and seeing if it still works. If the enzyme is still activated, then the empty G-protein doesn't affect anything. G-proteins are defined as active when they have GTP bound in the passage info, IIRC.

 

[inasensegone]

I feel like there is an error in "Item 116". This is the Biology question concerning the E.Coli population in the colon:

"If the E. Coli population in the colon is severely reduced due to extended antibiotic treatment, which of the following effects on digestion and nutrition are most likely to occur?"

Answer: Deficiencies of specific vitamins.

OK so I get that the passage mentions that the bacterial species in the colon produces B12 and Vitamin K, however, I thought that we couldn't absorb these vitamins in the colon (mainly B12). These vitamins are released distal to the area of the small intestine where absorption occurs. The question specifically mentions the E. Coli population in the colon.

This is what Wikipedia has to say:
"Ultimately, animals must obtain vitamin B12 directly or indirectly from bacteria, and these bacteria may inhabit a section of the gut which is distal to the section where B12 is absorbed. Thus,herbivorous animals must either obtain B12 from bacteria in their rumens, or (if fermenting plant material in the hindgut) by reingestion of cecotrope feces"
Also:
"The human intestinal tract itself may contain B12 producing bacteria in the small intestine,[79] but it is unclear whether sufficient amounts of the vitamin could be produced to meet nutritional needs."

Sooo.... unless we eat our own poop, or unless the question was asking about E. Coli in the small intestine, then we really don't receive much Vitamin B 12 from our colonic E. Coli. Also, there is no intrinsic factor available. Now supposedly we do absorb a significant amount of Vitamin K from the E. Coli in our Colon, but I was unable to find any significant detail on this. Guyton & Hall's Physiology textbook just gives it a brief mention.

I guess I was mostly thrown off by The Berkeley Review's Passage: Chapter 6, Passage 8 in biology. Question 40. The answer explanation says that Colonic B12 is not absorbed because it is produced distally to the point of absorption (In the Ileum).

Regardless, I still feel like the better answer in this case was "Intestinal tract infections due to increased populations of other bacteria". I was thinking along the lines of C. Diff.

If anyone would like to share their thoughts, I would appreciate it.

 

[cookyjar]

Question 138 is not making sense to me. How is it that a hydrogen on a monoalkylated product is less acidic? If there are more carbon groups attached, due to inductive effect, the carbocation becomes more stable, allowing the hydrogen to become more acidic? Or is it that electron donating activity of the newly attached carbon group lead to instability because of the negative charge created when hydrogen is removed?

 

[sunshine02]

I have a question regarding PS #13
How does sound wave (even though it travels slower-->smaller v) increase the frequency shift? Based on the doppler shift equation, velocity of the wave in the medium is both on the top and bottom of the fraction. Wouldn't a decrease in v (which is what would happen if you use sound instead of radio waves) decrease the amount shifted?

Thanks!

 

[sunshine02]

Bump

 

[RiPSTONE]

I still don't understand this problem.

The above diagram represents the neural pathway that causes an individual to retract a stubbed toe.
If one were to modify this diagram to represent the pathway involved in feeling pain in the stubbed toe, where could additional neurons be placed?

A) At II and III
B) At II and IV
C) At III and IV
D) At I and IV

 

[BornToLead]

bump

 

[BornToLead]

I still don't understand this problem.
The above diagram represents the neural pathway that causes an individual to retract a stubbed toe.
If one were to modify this diagram to represent the pathway involved in feeling pain in the stubbed toe, where could additional neurons be placed?

A) At II and III
B) At II and IV
C) At III and IV
D) At I and IV

 

[DocKingh]

I still don't understand this problem.
The above diagram represents the neural pathway that causes an individual to retract a stubbed toe.
If one were to modify this diagram to represent the pathway involved in feeling pain in the stubbed toe, where could additional neurons be placed?

A) At II and III
B) At II and IV
C) At III and IV
D) At I and IV

In order to feel pain, you need to send neurons to the brain. These neurons are supposed to be around the interneuron.
Afferent Neuron --> Interneuron --> Efferent Neuron

 

[JAH360]

PS #23 explanation says the benzoate anion is a stronger base than hydroxide anion. Am I correct in saying this incorrect?

 

[justadream]

Just some comments on this old post:

1) internal resistance is normally little r

2) This can't be right because the explanation specifically says increasing circuit resistance would decrease the field.

This is referring to #41

"Which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount"

Why does increasing circuit resistance (of the resistor) decrease the field?

E = V / L

Why does V go down? I thought V should be constant since the voltage source's potential difference is always the same. I thought that increases in R should be reflected in decreases in I (V = IR). But since the electric field (E = V/L) only depends on V and L, then I thought the electric field strength should remain constant, regardless of the resistance of the circuit resistor.

 

[justadream]

BS #110

AAMC says we should separate based on molecular weight?

Seriously? The difference in molecular weight between 2 of the compounds of 5 grams (like 3%).

 

[redbird1133]

Still unsure if we're allowed to post questions word for word...on physical number 38, two objects with masses m1 and m2 collide with collinear velocities v1 and v2, so what's the final velocity.
Well my book specifically says that an inelastic collision equation is: m1v1 initial = (m1+m2)v final, so I got (m1v1)/(m1+m2), but the answer is (m1v1+m2v2)/(m1+m2). I guess I'm missing something here, can someone please explain?

 

[justadream]

Still unsure if we're allowed to post questions word for word...on physical number 38, two objects with masses m1 and m2 collide with collinear velocities v1 and v2, so what's the final velocity.
Well my book specifically says that an inelastic collision equation is: m1v1 initial = (m1+m2)v final, so I got (m1v1)/(m1+m2), but the answer is (m1v1+m2v2)/(m1+m2). I guess I'm missing something here, can someone please explain?

This is because the book example assumes that only 1 object in the collision is moving (while the other is stationary).

That is not the case here. Both objects have nonzero velocity initially.

 

[redbird1133]

This is because the book example assumes that only 1 object in the collision is moving (while the other is stationary).

That is not the case here. Both objects have nonzero velocity initially.

Ah I guess it's better to manipulate the primary momentum equation than try and memorize the elastic vs inelastic equations. Thanks!

 

[pacer]

I am confused with question 153 (AAMC 4R).

I know that the correct answer should have acidic and basic functional groups.

Choice B - Amine is basic, Amide is acidic

Choice D - Amine is Basic and Carboxylic acid is acidic

Why is choice B wrong??

 

[pacer]

Can someone explain question 157 about the functional groups (Old 4R test)? Please, thanks!

 

[justadream]

@pacer

Post those questions please.

 

[pacer]

Since this is a passage based Q, I will provide the details from the passage.

Passage states that Compound C reacts with dilute acid and base.

So, I thought that it must contain an acidic and basic functional groups.

 

[pacer]

 

[pacer]

#174 can someone explain this please? I don't have the explanation for this one in my solutions.

 

[pacer]

I am confused with this question. I looked at the explanation in the solutions and see how they are getting choice C.

But I was thinking that polling of blood increases blood volume, thereby increasing blood pressure (choice B states the opposite and would not be predicted).

 

[justadream]

Did anyone else find the biology section of the test very difficult?

I am confused with question 153 (AAMC 4R).

I know that the correct answer should have acidic and basic functional groups.

Choice B - Amine is basic, Amide is acidic

Choice D - Amine is Basic and Carboxylic acid is acidic

Why is choice B wrong??

Right, you need an acid and a base.

Why would you choose Choice B over D? Amides have pKa of 15. COOH's have pKa's of around 4-5. It's clear that COOH is a better acid.

 

[pacer]

Aren't amides also acidic?

 

[justadream]

View attachment 185492

I am confused with this question. I looked at the explanation in the solutions and see how they are getting choice C.

But I was thinking that polling of blood increases blood volume, thereby increasing blood pressure (choice B states the opposite and would not be predicted).

Not 100% sure but to me, it seems that if you have increased blood pressure, you wouldn't have pooling of blood in the first place. When I think of something "pooling", I think of low BP. High BP would tend to get rid of that excess blood.

 

[justadream]

Aren't amides also acidic?

I mean with a pKa of 15, I guess they are somewhat acidic.

Carboxylic ACIDS are clearly more acidic though. That makes COOH the best answer for the MCAT. Always go for the obvious answer.

 

[csx]

.

 

[Janus_Kinase]

#39 The decay goes from ^40K to ^40Ar. What particle is emitted?

Mass stayed the same so we know it could have not lost a proton or neutron. Can this be due to an electron emission? electrons dont weigh which satisfies the mass portion, and now without an electron it would look like Ar

 

[diroten]

In the BS section of the test, I'm confused by passage 8 question 189:

If (R)-styrene oxide (seen here http://i.imgur.com/WXce1bY.png) undergoes ring opening in the presence of NaOH(aq), what is the size of the ring that is formed as a result of intramolecular hydrogen bonding in the product? The mechanism that it undergoes is described in the passage as follows:

Mechanism 1:
A nucleophile attacks an unsymmetrical oxirane
predominantly at the least substituted carbon atom,
inverting its configuration. An alkoxide is produced,
which abstracts a proton from the solvent.

I thought it would be a 4-membered ring but the answer says it's a 5-membered ring. But doesn't the rxn result in a diol, since OH is the nucleophile? How would HO---C---C---OH make a 5 membered ring?

 

[Gauss44]

Can you tell for certain if someone or something like an ape is solving a problem "systematically" by looking at them? Or could that possibly be an invisible unobservable mental process?

Then whassup with this VR question:

Suppose researchers discover that only wild-born orangutans respond to IQ tests in the way described, whereas those born in captivity behave in the same way as chimps. Which of the following hypothesis is most compatible with passage information?

A. Orangutans have no need to solve problems systematically in their natural habitat.

 

[Gauss44]

on #89 on the verbal...

Suppose it is discovered that the first Zurvinates were influenced more by the Indian than the Babylonian conceptions of cyclical times. Does this discovert support the author's argument?

Answer B: No, it doesnt not affect it.

I thought that the babylonians had a very distinct idea of time and that was incorporated into the Zurvinate religion. At the same time it contrasted with what Zoroastrianism believed in. Are they sayin Indians also believed in cyclical time so there would be no difference?

#140 on bio...

Why is a different base often used for the second alkklaytion of acetoacetic ester?

A. Because the active hydrogen of the monoaalkylated product is less acidic, making a stronger base more effective.

The answer can be found through process of elimination but how is the hydrogren in the second step less acidic if its inbetween 2 carbonyls?

I think the explanation to your question about #89 could be that there are wiggle words in the paragraph about the Babylonians, and in the question stem.

"may well" have led them to....

influenced "more" by Indian than by Babylonians...

If there were no wiggle words, then the answer would shift to C in my opinion due to blow to credibility and not quite getting the story right

 

[yjj8817]

For passage 7 in the ps section, is the photon required to free the electron? Does the electric field between the plates free the electrons or does it only provide potential energy? Thanks!

 

[Cawolf]

@yjj8817 The photon ionizes an electron from the cathode when it strikes - this is the photoelectric effect.

The KE the electron has is the difference between the energy of the photon (hf) and the energy required to ionize the electron (work function).

The electric field does not free the electrons, but has potential energy that accelerates the electrons once they enter the field. The E field is between the plates, so the electron is accelerated qV.

 

[yjj8817]

Thank you! That makes more sense.

For #109 in the BS section passage 3, the explanation says that because compound A was soluble in dilute HCl, it has an amine group. Aren't there other functional groups that are soluble in acid? How are they making that conclusion?

It also says that compound C has carboxylic acid group because it is soluble in NaOH.

 

[Cawolf]

@yjj8817 Other bases are soluble in acid, but in this problem, that was the only choice.

You know it has an acid because it is soluble in base. It needs to have a basic group to be soluble in the strong acid. The only choices for the acid soluble group are amine, amide, and nitro. The neutral amine group readily accepts a proton to become charged, like in ammonia --> ammonium.

 

[yjj8817]

@yjj8817 Other bases are soluble in acid, but in this problem, that was the only choice.

You know it has an acid because it is soluble in base. It needs to have a basic group to be soluble in the strong acid. The only choices for the acid soluble group are amine, amide, and nitro. The neutral amine group readily accepts a proton to become charged, like in ammonia --> ammonium.

Thanks! Just out of curiosity, are carboxylic acids soluble in acidic solution? Like HCL

I thought it was since they are both protic or am I incorrect?

 

[Cawolf]

@yjj8817

It depends on the carboxylic acid. Small, organic acids and inorganic acids will likely be soluble in aqueous solutions.

You really look at the functional group characteristics when changes to them effect the solubility.

A larger acid like benzoic acid is not very soluble in water. The little solubility it has is due to the protic nature of the functional group. The bulk of the molecule is a non-polar benzene ring that is highly insoluble. If you place it in a basic aqueous solution, it can be deprotonated into the salt which has ionic character, and is therefore very soluble in water.

If you place it in an acidic aqueous solution, the salt is not readily formed. Therefore the solubility will be limited.

A general and likely too simple schematic that I follow when looking at solubility is just to see if the functional group is reactive. Generally this means acid groups are soluble in base and basic groups are soluble in acid.

 

[yjj8817]

@yjj8817

It depends on the carboxylic acid. Small, organic acids and inorganic acids will likely be soluble in aqueous solutions.

You really look at the functional group characteristics when changes to them effect the solubility.

A larger acid like benzoic acid is not very soluble in water. The little solubility it has is due to the protic nature of the functional group. The bulk of the molecule is a non-polar benzene ring that is highly insoluble. If you place it in a basic aqueous solution, it can be deprotonated into the salt which has ionic character, and is therefore very soluble in water.

If you place it in an acidic aqueous solution, the salt is not readily formed. Therefore the solubility will be limited.

A general and likely too simple schematic that I follow when looking at solubility is just to see if the functional group is reactive. Generally this means acid groups are soluble in base and basic groups are soluble in acid.

Thanks! But can the oh part in the carboxylic acid be protonated to gain charge and become soluble? Or maybe to carbonyl oxygen? But I guess that's not possible since it has Pka of 2?

 

[Cawolf]

That wouldn't be favorable - you are right to consider the pka.

You could also look up the pka of a protonated carboxylic acid and see it is -7. It isn't likely to occur. The same is for the protonated carbonyl oxygen. In general the acid functional group will not act as a base.

 

[CATSandKILOS]

Hi! Could anyone please explain these answers to me? Thank you!

 

[mauroz]

Hi! Could anyone please explain these answers to me? Thank you!

For the first one, it is asking for the fundamental wavelength (AKA the first harmonic). In a pipe open on both sides, the fundamental wavelength is 2x the length of the pipe, 2nd harmonic is the same length as the pipe, etc. Check out this picture http://www.ibiblio.org/kuphaldt/electricCircuits/AC/02377.png the situation is different in a pipe with one closed end.

For the second one, the buoyant force is force pushing up, so if an object loses 5g then that means the liquid displaced also weighs 5g. So if the original object weighed 15g, then it is 3 times denser than the fluid because of the 15:5 ratio of the forces with the same volume. Specific gravity is the ratio of densities then the mass will be 3 times denser than benzene. 3 * 0.7 = 2.1

 

[yjj8817]

That wouldn't be favorable - you are right to consider the pka.

You could also look up the pka of a protonated carboxylic acid and see it is -7. It isn't likely to occur. The same is for the protonated carbonyl oxygen. In general the acid functional group will not act as a base.

Thank you for your answers! Another question. It says in passage iii that compound A became alcohol and carboxylic acid after treatment with base. The passage also says that it was soluble in acid in the first paragraph. I thought ester can become carboxylic acid and alcohol after treatment with acid? Why didn't passage mention this?

 

[Cawolf]

@yjj8817

No problem.

I think what you are referring to is the reverse of Fischer Esterification - or an acid catalyzed ester hydrolysis.

I couldn't tell you why, it wasn't the point of the passage? We were dealing with saponification, and the acid information was given (I think) to show the presence of the amine group.

The experiment specifically talks about refluxing compound A in base, so it is likely there is no information to be gained from it's reaction in acid.

 

[yjj8817]

Can someone explain #119 in bs section the one about feeling pain in a stubbed toe?

 

[yjj8817]

Anyone? Thanks!

 

[Cawolf]

That is a map of a reflex arc involving three neurons.

1 to 2 is a sensory neuron from the toe to spinal cord.
2 to 3 is an interneuron that integrates the signal.
3 to 4 is a motor neuron that contracts a muscle.

Pain is determined in the brain, so we would need to introduce neurons that will take the signal to the brain, and then back down. These are called ascending and descending tracts.

The junction of 2 is in the CNS and so is the junction of 3. These are the best choices as to where the signal will go to and from the brain.