How's MCAT Studying Going?

Hey everyone -

I recognize that many of you are now getting to the final week before your MCAT, and I recognize that your nerves may be getting to you.  I wanted to check in to see how everyone's doing and give you a few reminders.

I want to remind you all that I am in clinics currently (I'm at Penn Medicine at Radnor in their Internal Medicine practice as I write this), but that doesn't mean that I am inaccessible.  Many of you have still been writing me with questions, and -- while it takes me a bit more time to respond than in the past -- I am making sure to answer them all as best I can.

Please do me a favor and send me back a note -- even if it's very brief -- on how you're doing with studying for the MCAT and when your Test Date is.  I recognize that many people have changed their test dates since we finished MSCT III back in December.  I am eager to hear about your successes, but also want to make sure that nobody is suffering "silently"!

As a few reminders:

-The class blog (mchv10702.blogspot.com; username mchv10702; password letsgo45T) is still there!  I recently wrote a pretty extensive entry about what to do if you're thinking about changing your test date.

-End-of-class surveys were just recently sent out.  If you didn't see it in your mailbox, please check your Spam folder.  These surveys are very important to Kaplan and (more importantly) to me personally!  I have gotten five responses so far and I really hope to hear from more of you.

-If you need a wake-up call on Test Day, let me know.  Since I have to be at the hospitals by 8 am, I'm usually up around 6:30.

-I really want to hear back from you after your MCAT.  It means a lot to me to get feedback from students -- both positive and negative -- and it's definitely helpful for me to know what actually showed up on each administration of the MCAT.  I want to hear about what was easy or difficult for you, what surprised you on the test, how you scored, and how you're feeling.  I may just be your MCAT teacher, but my goal is to help you acquire the best application portfolio you can.  It's very important to me for you to succeed both locally on the MCAT and in getting into the school(s) of your dreams.

It has truly been an honor teaching you all.

Deciding to take the MCAT on Jan. 29th

Hey Alex,

I am currently planning on taking the MCAT on Jan 29th. My highest cumulative score is currently 4 points below my "worst case scenario" score and my highest in each section adds up to one point below. I plan on taking 6 more full lengths before test date. I guess I am just looking for advice as to whether or not I should actually go through with taking the MCAT when I am scheduled to.Thanks!
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This is, of course, just as hard a question to answer as it is to ask.  While I don't think there is any definitive answer, here's some advice and what I've seen in the past.


Let me also preface this by reminding you all what AAMC's policy is regarding changing test dates and registration costs.  This info can also be found in the MCAT Essentials guide:
https://www.aamc.org/students/download/63060/data/mcatessentials.pdf
If it is at least two weeks from your test date, you can change your test date for $60.  If it is within two weeks of your test date, you forfeit the ability to change your test date.  You can register for another MCAT for $235.  If you cancel your original test date, you can register right away.  If you still go to that test date, you will have to wait until the day after that MCAT.  My advice is not to cancel your original test date.  You should still go and feel out the test.  It will give you a chance to take the test in an actual testing center, and see exactly what procedures are like.  Hopefully you end up feeling like you did well and keep that score.  If you decide the test really didn't go well, you do have the option to void your score the day of your test.  In any case, you pay the same amount of $235 if you take the test again.


So in summary, if it's >2 weeks before your test, choose to either keep the original test date or change it for $60.  If it's <2 weeks, definitely keep the original test date.  Wait until after the test to decide if you want to register again for $235.


(This is going to be a long response -- but I'm sure I will be referring other students to it in the future, too.)


So let me answer your original question.  I'll stratify it by how far you are from your target score.  Also, to remind you, the average MCAT score is right around 25, the average applicant is around 28, and the average matriculant is near 31.  This is school-dependent, of course.  Top schools tend to average around 37.


If you are >5 points from your target score, this is difficult to accomplish in two weeks (but possible).  You should strongly consider taking a later test date.
If you are 3-4 points from your target score, this is very doable but it will require you to work hard.  I describe how to do that below.
If you're 0-2 points from your target score, I expect you to get there.  Don't rest on your laurels -- keep taking practice tests and reviewing them, and you should do solid.


So, how do you get from the few points away to your target score?  Full-lengths, of course; however, active review is really key to getting that score up.  What that means is when you finish a full-length, don't just look at the numbers.  Look at the breakdown by topics and question types on SmartReports.  Make those Why-I-Missed-It Sheets.  Use the test to gain more than just practice.  Allow it to guide what you need to know.  Learn everything you didn't know on that test for next time.  Analyze your timing and use of the methods.


When do you take them?  Well, I suggest doing Mini-Diagnosis cycles.  That  means taking a full-length (or three to four section tests, if there's one section in particular that needs working) on one day, and alternating it with a day of active review, reading and lighter passage work (maybe a topical test or two).  Do three cycles per week, making sure to take a day off.  (No matter how stressed you are, you need to take a day off or -- I promise you -- you will burn out.  I have taught over 1000 students, and any of my students who tried to study every single day for multiple weeks were too exhausted come Test Day.)


My experiences with students in the past generally shows an improvement in the last two weeks.  I should note that a number of people have a syndrome of going down a few points at the very end (on the last full-length or two they took).  This should not cause you to panic; the vast majority of people who had this occur did just fine on the test, and jumped back up to where they were.


I have also noted that people tend to go up 1-2 points on the actual MCAT from their top scores on the full-lengths.


You also need to think about how this fits into your med school application cycle.  Remember that the application opens in May, and you can submit in June.  You do not want to hesitate on it.  That means that you should try to take your MCAT in April or earlier.  May and June are still fine, but July and August start making it a bit more challenging to get in (they've already considered a number of applicants, so you need to be that much more competitive to make it).


One other important point.  If you do decide to push back your test date, or register again, do not stop studying.  You can certainly relax some (you shouldn't do alternating full-lengths and study days for another three months!), but don't allow yourself to say "Well, now I have a few months.  I'll take a week or two off and then start up again."  You're running a marathon.  You can run a bit slower, but you don't just go sit on the sidelines for an hour to cool off, right?


I've come to realize I'm quite verbose.  But I hope you all find this helpful!

full-length 4, ps #35

I just don't understand number 35 at all, could you explain it? Thanks!
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For this question, it asks us to find the length of the “sweet zone” on a 81 cm bat.  Let’s first consult our passage for the definition of “sweet zone”:

“The sweet spot is the node of the second harmonic nearest the thicker end of the bat…there is a ‘sweet zone’ that extends from the old sweet spot to the node of the third harmonic which is closest to the thick end of the bat.”

Alright, so we’re dealing with standing waves (based on the words “node” and “harmonic”).  Earlier in the passage, it informs us that the bat resonates much like an open pipe.  Let’s take a look at a diagram of the second and third harmonics (n = 2 and n = 3).  Remember that the a harmonic for an “open-at-both-ends” or “closed-at-both-ends” tube is the number of half-wavelengths.

n = 2 

n = 3 

Considering that our bat is 81 cm long, let’s locate the node in the second harmonic. Each node is ¼ the distance of the bat from each end; divide the bat into four equal lengths and you’ll see that you have either a node or antinode at each division.  So the node for the n = 2 bat is 81/4 = 20.25 cm from the end.

For the third harmonic, let’s do a similar process.  The node closest to the end is 1/6 the length from the end of the bat – again, look at the third harmonic diagram above, and you’ll see that dividing the bat into six equal lengths will give you a node or antinode at each end.  The node closest to the end is 1/6 of the way over, which is 81/6 = 13.5 cm.

So the distance between them is 20.25 cm – 13.5 cm = 6.75 cm.  This, thus, represents the length of the “sweet zone”.

Physical sciences - FL7

Hi Alex,
My question is about number 4 on the physical sciences section. I guess I just don't understand the concept behind setting the torque equal to the gravitational potential energy in that I can't visualize/understand why they would be equal.

Also, for number 49 on the biological science section of Kaplan Full Length 8, wouldn't the flow be reversed and be from the pulmonary artery to the aorta?

Thanks!
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For the first question, let's start by considering the forces acting on each of the small balls (m).  They each have a force of gravity created by Earth acting downwards (F = mg), and a force of gravity created by the large balls (F = GMm/r²).  Let’s then translate these forces into torques.  We have forces in two directions, so let’s consider rotation caused by torques in those two directions.  The two forces of gravity from Earth are acting in opposite directions by rotation, so the two torques cancel out.  That is to say that we have rotational equilibrium, looking only at the forces of gravity from Earth.

For the other direction, we have a torque on each ball pulling counterclockwise.  Since they’re pulling the same rotational direction, we can just add them up.  Torque = (lever arm)(force), so torque from each is (L/2)(GMm/r²).  Adding the torques together, we have 2(L/2)(GMm/r²) = GMm/r².  That’s answer (D).

Overall, we’re not setting equal to gravitational potential energy; we’re calculating the torque by multiplying the lever arm (L/2) by the force between the two masses – a gravitational force equal to GMm/r².

For the other question, remember that this question says after birth.  The ductus arteriosus connects the pulmonary artery and the aorta.  After birth, when an individual has the adult-style circulation, the pressure in the aorta is MUCH higher than the pulmonary artery.  Thus, we would expect blood to travel through the shunt – if it’s still open – from the area of higher pressure to lower pressure.  Therefore, there will be flow from the aorta to the pulmonary artery.

PS Test 2, question 30

Hi Alex, could you help explain question 30 from the physical sciences
section test 2. I found that question to be really challenging and I am
not quite understanding the explanation.
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This question reads:  “If Step 2 above were the rate-determining step of Reaction 1, which of the following equations would correctly define the rate?”

Remember that the rate-limiting step can be related to the rate by simply taking each of the reactants from that step and raising them to their stoichiometric coefficients.  For this question, that means that rate = k₂[B][D].  There’s a problem here, though.  When we write rate laws, all of the terms must be reactants of the overall reaction, meaning that we’ll have to have our rate dependent on A and D (based on the reaction above the mechanism).  So having [B] in our answer is a problem.  How do we fix this?

The key is then looking at Step 1, where [B] is manufactured.  To find out how to rewrite our concentration of B, let’s use the fact that it tells us this reaction is “slow.”  In reaction kinetics, “slow” means that the reaction is at equilibrium.  If it’s at equilibrium, then the forward and reverse reactions of Step 1 have the same rate.  Mathematically, that means:

rate_f = rate_r

k₁[A] = k_-1[B][C]

The “-1” on the rate constant is indicating that it is the rate constant for the reverse reaction of Step 1.  This is a common convention in kinetics.  Thus, to find [B], we can rearrange:

[B] = k₁[A]/k_-1[C]

Plugging into our rate law from before, we get:

rate = k₂[B][D] = k₂(k₁[A]/k_-1[C])[D] or k₁k₂[A][D]/k_-1[C].  This matches with answer choice C.  We could get picky and point out that C is in our rate law, and it’s not a reactant of the overall reaction.  That’s true, but you’ll see that we actually cannot get any further with [C] than this step.  Ideally, we’d know more information and be able to remove [C] from our rate law, but we can’t do that in this particular scenario.

Le chatelier's

This makes sense- but I just want to be sure I am understanding this correctly. Adding or removing a substance that is NOT in the equilibrium expression, such as a solid, has no effect on the state of the rxn at equilibrium- so le chatlier's pricinple does not apply?
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Yes.  Raising or lowering the amount of a solid or liquid will not disturb the equilibrium because these terms are not even part of the Keq of the reaction.

Nomeclature and functional groups, #5

Hey Alex,

Just wondering if you could explain to me why the alkene isn't a site
of electrophilic attack.  I know it has a cloud rich in electron
density, but couldn't the extra electrons move to the neighboring bond
to bump off the Br?  So basically why the answer isn't D.

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This question reads:

Which of the functional groups on the following molecule are susceptible to nucleophilic attack?

Looking first at (a) - the Br attached to the secondary carbon is electron-withdrawing, so it renders the carbon at a electrophilic (thus, suscpetible to nucleophilic attack).  For the carbonyl carbon at (c), there is similar logic since the oxygen is electron-withdrawing.  As a side note, carbonyl carbons are the most common electrophile on the MCAT.

For (b) - the main problem with the logic above is that there is no driving force for this sort of reaction to occur.  All reactions in organic chemistry need to have reason to run.  Here, an incoming nucleophile would find it very difficult to be attracted to the reaction site; it has a large electron-rich cloud that will strongly repel any negatively-charged (or even neutral) nucleophile coming in.  Even if the nucleophile were attracted, there is an energy barrier to overcome here.  This reaction would require breaking the double bond, and then undergoing some E2-like reaction mechanism.  Still, any incoming nucleophile would be very strongly repelled from the sp2-hybridized carbons at (b).

Thus, a and c are the reaction sites susceptilbe to nucleophilic attack.

FL 9 - 1, 21, 41

I had a couple questions from the physical sciences section of Full-Length 9. The first one is actually #1. I chose answer c, and I still can't understand why the answer is b. Wouldn't the elastic part be easier to stretch? My next question is number 21, I just can't put together an equation-based relation that includes velocity. Also, could you explain number 41? Thank you!!
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Question 1:

For this question, we need to analyze the graph.  The graph is stretch versus time (NOT force), for a system with a viscous component and an elastic component in parallel.  Well, if we give a certain CONSTANT force to a spring over time, what happens?  It stretches until it reaches a certain deformation from equilibrium, and then just stays there.  Think about hanging a mass from a spring – that’s a constant force on the spring which causes deformation until the mass is at translational equilibrium.

What would happen to a viscous object without an elastic force?  Well, it would stretch and stretch and stretch… and keep stretching until (probably) it eventually broke.  But either way, it doesn’t reach equilibrium because it’s not elastic.  So it just keeps stretching.

Why is it asymptotic, then?  Well the spring will only stretch until it’s in equilibrium with the force.  At that point, it won’t stretch anymore.  Thus, the stretch of the viscous component (which should be infinite) is limited by the stretch of the spring.

Question 21:

We’re supposed to relate the force created by the spring with this scenario of a pilot landing a jet.  Well, anytime you see force, think acceleration!  F_net = ma, but F also represents the force of the spring here, and thus F_spring = kx.  So, we can say F_spring = ma.  We’re given one more key piece of information:  the final velocity is zero.  Acceleration is simply change in velocity per time or Δv/t.  Since v_f = 0, we can say that a = (v_f – v_i)/t = (0 - v_i)/t = -v_i/t.  Plugging into our equation, we get F_spring = m(-v_i/t) = -mv_i/t.  Thus, the force is directly proportional to the starting velocity.

Question 41:

The first thing we see when looking at the answers is the fact that it says that one has higher or lower energy than the other, and differences in principle quantum number (5sto 5d, etc.).  So, let’s start with energy.  E = hf, so the higher the frequency, the higher the energy.  Red light has the lowest frequency while violet has the highest, so that means red light has lower energy than violet.  Here, we’re dealing with red (Sr) and green (Ba).  So that means that strontium has a lower energy jump than barium does.

Well, for principle quantum number, where is strontium?  It’s in the 5s grouping.  Barium, however, is 6s.  Where are these electrons jumping to?  Well, it should jump to the next-highest-energy orbital.  If we start at an s orbital, we’ll just jump to p or d.  So we’ll want to see that electrons jumping from 5s to 5p or d have less energy than those jumping from 6s to 6p or d.  And that matches with answer choice (B).  

AAMC #3, BS 114

I'm having a hard time understanding why developing a leak in the apparatus increases the surface pressure and thus increases the BP of both substances. How do we know the pressure in the appartus is different  from atmospheric pressure to begin with?

When the leak occurs does the vapor pressure of the liquids also decrease because temperature is going to be decreasing? This wouldnt necessarily mean the BP will be different, just that it will take longer to get there right?
Thanks!
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This question reads:  "If a leak develops in the vacuum distillation apparatus, the boiling points of the two components of caraway seed oil will:"

The key, then, is the fact that it says "vacuum distillation."  Remember that vacuum distillation will lower the pressure above the two liquids being distilled, so that it lowers their boiling point.  This is because liquids will boil when their vapor pressure equals the ambient pressure.  By lowering the ambient pressure, liquids will boil at a lower temperature because they don't need as high of a vapor pressure.  Vacuum distillation is used for liquids that have boiling points >125 degrees Celsius.

Thus, if a leak develops, we will expect the pressure inside the apparatus to go up.  It starts lower than atmospheric pressure, because that's the whole goal of vacuum distillation.  When a leak develops, the pressure will go up because air comes in from the surrounding atmosphere.  By increasing the ambient pressure, the boiling point starts going up again.

The vapor pressure of the liquids shouldn't change here just because of the leak.  Indirectly, it will cause them to go up in the end.  Remember that boiling is an isothermal process.  By increasing the boiling point, we increase the vapor pressure needed to cause that boiling.  Thus, assuming that we're supplying heat to the apparatus, the temperature will go up because the boiling point has gone up.  Before, it would hit the boiling point and stop at that temperature since the boiling process is isothermal.  Now, it adds heat until the boiling point -- at a higher temperature and higher vapor pressure -- and then boils at this temp.

Scaled scores

Does scaling the scores according to the percentage correct vary from exam to exam? I've just been noticing differences with the percent I got correct and my scaled score- like on FL 5, my scaled score for physical sciences was 2 points higher than what the percentage I got correct would have said.
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I've said this a few times before -- please be careful with the rubric I gave you in the MSCT III packet.  It gives you an idea of where your score falls, but it is not wholly accurate for the MCAT.  The MCAT varies from test administration to test administration; likewise, our scaling varies from Full-Length to Full-Length.  The scaled scores you are given by the Kaplan tests are quite accurate; it wouldn't be fair for us to either give you a false sense of confidence or to make you feel like you're not doing as well as you actually are.

The scaled score conversion chart should be used from section tests, and remember that it predicts usually +/- 1.  In your case, it was off by 2, but usually it should be accurate -- give or take 1 scaled score point.