AAMC CBT7 and 7R OFFICIAL Q&A

Can somebody please explain number 69 in the verbal section? Its under passage 3 question number 4. I don't understand why choice A cannot be the correct answer. Even based on the explanation they provide it still seems that "A" is a suitable answer.

can anyone explain #46 on the physical sciences. If a rupture is moving from south to north how does Doppler effect influence this? Thanks.

TypeSH07 said:

can anyone explain #46 on the physical sciences. If a rupture is moving from south to north how does Doppler effect influence this? Thanks.

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The Doppler shift can impact any wave. In this case, the wave is an impact wave travelling through the ground away from the rupture. The source of the wave is the rupture, and given that the rupture is moving, the wave it emits will be Doppler shifted. You can thus treat it as a source moving north emitting a wave, much like a car travelling north emitting a sound.

The explanation from the BR answer sheet is:

• The primary rupture is moving from south to north, so the Doppler effect is associated with north-south movement and not east-west movement. This eliminates choices A and B. Seismic waves propagating from this rupture are Doppler shifted by the motion of the rupture. Seismic waves emitted to the north are compressed by the motion of the rupture while seismic waves moving south are elongated by the motion of the rupture. Compression results in a shorter wavelength, so the wavelength of seismic waves moving to the north is decreased while the wavelength of seismic waves moving to the south is increased. The best answer is choice C.

For #130 on why can it not be C? I was trying to decided between A and C and chose C since it can "elude" host defenses. Was I jumping too far in thinking that it can elude defenses by suppressing immune activity? Thanks.

I have a question about this test. It's the #7 AAMC test from Kaplan. The very last question on biological section is about an experiment on the phases of cell cycle. They incorporate radioactive 2-deoxythimidine in these dividing cells and monitor the # of radioactive signal, then graph it. The choices are:
A/ mitosis
B/ meiosis
C/ DNA synthesis
D/ RNA synthesis

I don't understand why it cannot be A/ mitosis. I understand that DNA synthesis will incorporate those radioactive deoxythymidine into new DNAs, and that will shows up on the graph. However, in mitosis, cells that are dividing also carry these radioactive T in their DNA. Technically, these cells will also provide the answer for those sharp increase of radioactive level on the graph. Am I wrong?

Gpan said:

I have a question about this test. It's the #7 AAMC test from Kaplan. The very last question on biological section is about an experiment on the phases of cell cycle. They incorporate radioactive 2-deoxythimidine in these dividing cells and monitor the # of radioactive signal, then graph it. The choices are:
A/ mitosis
B/ meiosis
C/ DNA synthesis
D/ RNA synthesis

I don't understand why it cannot be A/ mitosis. I understand that DNA synthesis will incorporate those radioactive deoxythymidine into new DNAs, and that will shows up on the graph. However, in mitosis, cells that are dividing also carry these radioactive T in their DNA. Technically, these cells will also provide the answer for those sharp increase of radioactive level on the graph. Am I wrong?

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If it is a linear shape, it would be mitosis, curvelinear meiosis...

Now, assuming I'm correct (and I don't have access to the graph) why would I assume that?

Because exponential functions don't graph linearly, ruling out Mitosis.

JWLuiza said:

If it is a linear shape, it would be mitosis, curvelinear meiosis...

Now, assuming I'm correct (and I don't have access to the graph) why would I assume that?

Because exponential functions don't graph linearly, ruling out Mitosis.

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wait, what??? I think you got the wrong question. The correct answer is DNA synthesis, not those. I just wanna know why can't it be mitosis.

It's the last question on the biological section in AAMC #7 test provided by kaplan. If you have access to this test, please chime in

Gpan said:

wait, what??? I think you got the wrong question. The correct answer is DNA synthesis, not those. I just wanna know why can't it be mitosis.

It's the last question on the biological section in AAMC #7 test provided by kaplan. If you have access to this test, please chime in

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Sorry misread. I'll take a look later today if I still have access as a teacher...

These questions are supposed to be posted in the Study Q & A forum; there will be a thread that is specifically for AAMC 7. That's there so people don't get questions spoiled

A cell goes through interphase before it goes through mitosis. At the S phase there is DNA replication which also incorporates the radioactive thimidine (but is not part of mitosis) whereas the mitosis just splits the DNA and cytoplasmic contents to the daughter cell. Therefore it is more accurate to say DNA synthesis.

Anyways, even if mitosis is correct the MCAT is always looking for the best answer. And how can an answer be better than the actual process that incorporates the Thymidine to make DNA?

TypeSH07 said:

For #130 on why can it not be C? I was trying to decided between A and C and chose C since it can "elude" host defenses. Was I jumping too far in thinking that it can elude defenses by suppressing immune activity? Thanks.

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yea i would say eluding the immune system is more evasive means of infection, while suppressing is more active and aggressive means of infection.

can someone explain #27 on the PS section?? i dont understand why the rate is decreased OR how they got to (1.0 x 10-2 M/s) x (1 x 10-2) ???

thanks!

maluskeeter said:

can someone explain #27 on the PS section?? i dont understand why the rate is decreased OR how they got to (1.0 x 10-2 M/s) x (1 x 10-2) ???

thanks!

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Are you talking about 27 PS on AAMC 7? That's the one about the electromagnetic spectrum. I think you're talking about 33.

So they're telling us that everything is constant except for the H+ concentration. If the pH is 1, then the H+ concentration is 10^-1. For the rate to be 10^-2 then we have to square the H+ concentration:

[H+]^2 = rate

so now if the pH becomes 2, then the H+ concentration is 10^-2, and plugging that into the equation that we figured out, the rate will be [10^-2]^2 = 10^-4

Does my explanation make sense? If not, I can clarify.

mdhopefully23 said:

Are you talking about 27 PS on AAMC 7? That's the one about the electromagnetic spectrum. I think you're talking about 33.

So they're telling us that everything is constant except for the H+ concentration. If the pH is 1, then the H+ concentration is 10^-1. For the rate to be 10^-2 then we have to square the H+ concentration:

[H+]^2 = rate

so now if the pH becomes 2, then the H+ concentration is 10^-2, and plugging that into the equation that we figured out, the rate will be [10^-2]^2 = 10^-4

Does my explanation make sense? If not, I can clarify.

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I get what you are saying, but isn't the question asking for the rate of formation for Cl-, not H+??

TawMus said:

I get what you are saying, but isn't the question asking for the rate of formation for Cl-, not H+??

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They give you the rate of formation equation in the passage (going off memory). H+ is second order and if you take the ratio of the rate with [H+] at pH=1 vs pH=2 you will see that there is a difference of the order of 1x10^-4 by the explanation above.

This is the passage on Landers earthquakes and I had a real hard time understanding the passage, which ended up screwing me over on the questions...

First of all, the waves that are described in this passage.. are they simple harmonic waves or transverse waves? I thought they were transverse waves because simple harmonic waves are energy and momentum conserved because they have no outside forces acting on them and continue to propagate at a continuous frequency and period (F=1/T). The waves described in the passage dissapate over time..

I don't remember how exactly the passage went but earthquakes produce two different types of waves (actually 3 but i wont go there). In seismology transverse waves are called secondary, or s waves because they arrive later than the primary, or p waves from an earthquake, which are longitudinal. So an earthquake can produce different types of waves. Also, only p waves can travel completely through the earth because of the liquid properties of our Earth's core. This should help you understand all the earthquake passages now.

Also seismograph are just those weird spike lines on a graph you see (similar to an EKG) that receives the waves the earthquake makes from the opposite side of the Earth.

BloodySurgeon said:

I don't remember how exactly the passage went but earthquakes produce two different types of waves (actually 3 but i wont go there). In seismology transverse waves are called secondary, or s waves because they arrive later than the primary, or p waves from an earthquake, which are longitudinal. So an earthquake can produce different types of waves. Also, only p waves can travel completely through the earth because of the liquid properties of our Earth's core. This should help you understand all the earthquake passages now.

Also seismograph are just those weird spike lines on a graph you see (similar to an EKG) that receives the waves the earthquake makes from the opposite side of the Earth.

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In one of the questions based on the passage, frequency is converted to 1/T. Isn't that something you can only do with simple harmonic motion??

this was the first practice test that I have done, so I was wondering how its compares to the others if anyone could tell me???

Why is 3-ketocyclohexene less polar than phenol??

TawMus said:

Why is 3-ketocyclohexene less polar than phenol??

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alcohols are more polar than ketone's because they have a hydrogen on the oxygen which allows them to make hydrogen bonds with water's oxygen while the water's hydrogen can make a hydrogen bond with alcohol's oxygen. That and hydroxyl groups are more dipole(ish).

(1) O-H ---- OH2
(2) H-O ---- H2O


(1) C=O ---- H2O

If the compound was a benzene instead of cyclohexene, would that make a difference in the polarities???

I have a question about #44 on the PS section. The answer is A but I picked B. My reasoning for that is that if the deformations from the quake are less than the daily periodic distortions due to tidal forces then wouldnt these forces set off random earthquakes before the initial quake would thus making it a poor choice. C and D are also poor choices so by process of elimination B is the best answer.

TawMus said:

If the compound was a benzene instead of cyclohexene, would that make a difference in the polarities???

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no. phenol is still more polar than benzene with any keto substituent that does not contain a hydroxyl group

So polarity is related to stability and hydrogen bonding?

i have a similar question about Retention factors; from what i could learn off you guys--ketones are less polar than alcohols

...yet the answer in the question is D.


then here comes this question:

the answer is B.
I thought hydroxyls were 3300 and carbonyls were 1700???

MrMattOglesby said:

i have a similar question about Retention factors; from what i could learn off you guys--ketones are less polar than alcohols

...yet the answer in the question is D.


then here comes this question:

the answer is B.
I thought hydroxyls were 3300 and carbonyls were 1700???

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to the first question. D makes sense because since the OH is more polar, it will be more attracted to the polar film. therefore, it won't travel as far and will have a smaller Rf value.

for the 2nd question, you are correct the answer is D. check again to make sure your eyes didn't play a trick on you. even my AAMC test tells me D is the right one...maybe yours had a weird glitch or something? OH is broad stretch around 3400 and carbonyl is a peak around 1700

damn...one point away from getting an 11 on that bio section!
woops...i messed up the math again. I DID get an 11!!!
by far the highest score on any section done so far.

so my scores for each part of the paper test for AAMC 7 were:
PS: 8
VR: 7
BS: 11

composite = 26
thats good enough for my DO schools =]

argh...i can't figure out #15 on the PS..the electron configuration for a ground state silicon atom? how can it be A? it doesn't even have the right number of orbitals in the energy level..does anyone know what i'm missing?

If you printed the exam I think that is the problem. The paper version only shows 2 p orbitals but on the online version there are three like there should be.

In everyone honest opinion how do you think this test ranks in terms of difficulty?

Can someone please clarify these:

Q 19) Independent question on electric potential of copper ion and water reaction.

My answer: C, 1.57V. But the solution key says A.

Q20) A gas that occupies...

My answer: C; Solution key has no answer!

Q23) Question on absorpance of Bromine.

My answer: B; cannot find answer in solution key

Thank you!

Premedico said:

Can someone please clarify these:

Q 19) Independent question on electric potential of copper ion and water reaction.

My answer: C, 1.57V. But the solution key says A.

Q20) A gas that occupies...

My answer: C; Solution key has no answer!

Q23) Question on absorpance of Bromine.

My answer: B; cannot find answer in solution key

Thank you!

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Q 19) Answer is A: -0.89V, you just have to sum up the potential for the half rxns ... looks like you forgot to put in the "-" sign -1.23+0.34=-0.89.

Q 20) B, here is there explanation (I got this wrong): The ideal gas law makes the assumption that molecules have no volume. This assumption is adequate when the gas is at 1 atm, but when the pressure is increased to 500 atm the volume of the gas molecules is no longer negligible.

Q 23) B, you were right. This one was simple math (which is usually a little more difficult under timed conditions with no calculator).

Premedico said:

Can someone please clarify these:

Q 19) Independent question on electric potential of copper ion and water reaction.

My answer: C, 1.57V. But the solution key says A.

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I had the same question also. Isn't it Ecathode - Eanode?

This was how I did it, can someone please correct my flaw?

E for Cu was +0.34
E for the water reaction was -1.23

Based on the above values, I concluded that water was being oxidized and Cu was being reduced. If that is the case, then Ecell = 0.34 - (-1.23) = 1.59.

If the two were reversed (i.e., Cu is oxidized and water is reduced), then it would be - 1.59. I don't understand how or why you can get -0.89. I mean, I don't understand the reasoning behind adding the two potentials.

Any help is much appreciated.

Thank you

travelbug73 said:

I had the same question also. Isn't it Ecathode - Eanode?

This was how I did it, can someone please correct my flaw?

E for Cu was +0.34
E for the water reaction was -1.23

Based on the above values, I concluded that water was being oxidized and Cu was being reduced. If that is the case, then Ecell = 0.34 - (-1.23) = 1.59.

If the two were reversed (i.e., Cu is oxidized and water is reduced), then it would be - 1.59. I don't understand how or why you can get -0.89. I mean, I don't understand the reasoning behind adding the two potentials.

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This is what the reaction looked like:
2 Cu2+ + 2 H2O → 2 Cu(s) + O2 + 4 H+
The proper equation for determining Ecell is to ADD the potentials, not to subtract them (Ecell=Eanode+Ecathode). +0.34V represents copper being reduced while -1.23V represents water being oxidized; this is exactly what is occurring in the given formula for which we need to determine Ecell. Therefore, we only need to add the potentials: -1.23V + 0.34V = -0.89V = Ecell.


Does this help?

Was I the only one that felt unfairly treated by #95?

According to the Cahn-Ingold-Prelog priority rules, which group bonded to the chiral carbon atom (x) in Figure 1 has the highest priority?

I got it right by using a different method of assigning priorities. But what the hell? Cahn-Ingold-Prelog rules?

And I know you've heard it, Vihsadas, but come on.

Kaustikos said:

Was I the only one that felt unfairly treated by #95?

According to the Cahn-Ingold-Prelog priority rules, which group bonded to the chiral carbon atom (x) in Figure 1 has the highest priority?

I got it right by using a different method of assigning priorities. But what the hell? Cahn-Ingold-Prelog rules?

And I know you've heard it, Vihsadas, but come on.

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The highest priority would be the atom with the highest atomic number. If two bonded atoms have the same atomic number than we go to the next atom and so on. This is actually expected knowledge for the mcat. Also for assigning R or S configuration, the lowest priority group should be facing away from you. Therefore if it is towards you and you get a clockwise priority configuration for R the real answer is the opposite and is then S. Even though you did not know what Cahn-Ingold-Prelog was, you should still have been able to answer this question.

BloodySurgeon said:

The highest priority would be the atom with the highest atomic number. If two bonded atoms have the same atomic number than we go to the next atom and so on. This is actually expected knowledge for the mcat. Also for assigning R or S configuration, the lowest priority group should be facing away from you. Therefore if it is towards you and you get a clockwise priority configuration for R the real answer is the opposite and is then S. Hope this helps.

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Sorry, I meant I've never EVER heard it called that. THAT is what threw me off.

A lot of times with the mcat, they will use names and rules that nobody has ever heard of. Your job as a test-taker is to use your common knowledge to try and reason out the problem. In this case, you did not know what Cahn-Ingold-Prelog was. Could you have solved the problem without it? Sure. This question was just quizzing you if you know how to assign highest priority. The mcat will try to confuse you as much as possible, but in the end it is just testing you on basic knowledge.

BloodySurgeon said:

The highest priority would be the atom with the highest atomic number.

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I want to add a small caveat here. You consider atomic mass in the event you have isotopes of the same atom. For instance, deuterium takes priority over hydrogen. This came up once upon a time on a question about SN2 and inversion of a chiral, primary carbon. You had to realize that the primary carbon can only be chiral if the two H atoms are unique, which is accomplished by isotopic labeling.

LovelyBRass said:

This is what the reaction looked like:
2 Cu2+ + 2 H2O → 2 Cu(s) + O2 + 4 H+
The proper equation for determining Ecell is to ADD the potentials, not to subtract them (Ecell=Eanode+Ecathode). +0.34V represents copper being reduced while -1.23V represents water being oxidized; this is exactly what is occurring in the given formula for which we need to determine Ecell. Therefore, we only need to add the potentials: -1.23V + 0.34V = -0.89V = Ecell.


Does this help?

Click to expand...

Ok!! I get it now. The mistake I made was that I thought that both values were REDUCTION potentials. No, one was the reduction potential and the other was the oxidation potential. In this case, we add them up like you said.

Thank you for your help!

96. is there another way to figure this out without knowing the Wittig reaction?

98. how is it determined a hydrogen is more acidic? and what does that have to do with a base closing the ring?

99. i guess on this one and got it right, but can someone point out where the ester group is?

102. how do i approach this problem?

133. i picked that the immune system defeats initial cancers, but the answer is that people tolerate the infection without developing tumors. i don't know what concepts to use to arrive at the right answer or what's being tested here.

sorry thats lots of questions. but any help is appreciated!

wooki said:

96. is there another way to figure this out without knowing the Wittig reaction?

98. how is it determined a hydrogen is more acidic? and what does that have to do with a base closing the ring?

99. i guess on this one and got it right, but can someone point out where the ester group is?

102. how do i approach this problem?

133. i picked that the immune system defeats initial cancers, but the answer is that people tolerate the infection without developing tumors. i don't know what concepts to use to arrive at the right answer or what's being tested here.

sorry thats lots of questions. but any help is appreciated!

Click to expand...

96. This just something that should be known. The reduction of a ketone to an alkene occurs via wittig reactions. I actually loved seeing this question because I memorized this reaction and haven't had the opportunity to use this "knowledge". I had to do a double take.

98. Alpha hydrogens for ketones/carboxylic acids tend to be acidic because of the ability of the carboxyl group to stabilize the deprotonation (electron withdrawing capabilities - basically the same thing dealing with aromatics and activating/deactivating ordeal you remembered. The carboxylic acids/esters/ketones/aldehydes tend to have an electron withdrawing affect on carbons alpha to them, stabilizing the deprotonation of that carbon.
If taht doesn't help, look back in your books to the enolization reaction in your review books. I ddin't see it at first and missed this question but saw it after. The carbonyl group above the alkene wasn't something I saw which would've been a clear-cut enolization reaction. And the initial step in this is the base-catalyzed removal of alpha hydrogens of aldehydes/ketones, which then attacks the unsaturated carbons.

102. I cannot offer any real advice than to know that cyclin proteins are necessary proteins for initiating mitosis. These work in conjunction with Cyclin-dependent kinases in a cascade event for starting mitosis. If you want to further solidify this knowledge, just know that the main target/aim of cancer agents nowadays are cyclin-dependent kinases. Tumor/cancer is mitosis that goes on forever. You stop cyclin dependent kinases, you have the ability of inhibiting mitosis. Cyclin-dependent. Looking at the graph you see a high amount of the cyclins right before mitosis - thus, you know they're making cyclins. A and B are wrong, thus, because those don't cause in the making of Cyclins. D is wrong because you don't translate in mitosis, or even if you did, the cyclins wouldn't cause the regulated initiation of mitosis.

133. I'll get to that shortly.

Sorry. Massive thunderstorms and week wifi made me post the above in haste.

Anyways.

133.
A - Bacteria don't give us chromosomes, those are viruses that incorporate genomes into our cells.
B - Look at D
C - This I just took as wrong because our immune system doesn't initially stop the infection, thus, the bacteria still have a chance of "infecting". They call the individuals "infected". You can't eradicate the infection immediately and then call the individual infected. THere has to be chance for a certain amount of bacterial growth in order to classify an individual as infected. Thus, C is eliminated because that just wouldn't realistically happen. As long as you have 1 bacteria, you have a chance of the cancer.
D - I had to struggle between this and B, honestly. But there was a 1% chance of picking B. But I just cannot see a reason to explain why B or not B. B just looked wrong. It seemed like D was the best answer overall.

wow thanks a lot Kaustikos. much appreciated! just #99 left. I cant seem locate an ester, which i know is

.....O
.....||
..../...\
...R.....OR

For #133

From the passage:

'Infected individuals have a two-fold increased risk of gastric cancer, although >75% of patients with active infections do not develop cancer'

Hence, answer D.

wooki said:

wow thanks a lot Kaustikos. much appreciated! just #99 left. I cant seem locate an ester, which i know is

.....O
.....||
..../...\
...R.....OR

Click to expand...

You probably mistook the Et as Ethanol as I first did initially. But if you look above the Carbon labeled "a", you'll see the CO2Et (et meaning ethyl). There's your ester. Seemed rather lazily/purposely put in there to trick.

And no problem. Long as I can help.

I guess I dont even understand what they are asking in #141. How do you read figure 2 to answer this question.

Thanks