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MCAT Content / AAMC Official Guide Cp Solutions

AAMC Official Guide CP [Web]

Official Guide Chemistry/Physics Section Passage 1

1) Amino acids are a high yield topic and we should be very comfortable with them. Come test day, we should know that naturally occurring amino acids used by all living organisms are L amino acids, and most chiral ones are S amino acids except for cysteine. Let’s look at the answer choices and see how else we can reach the correct answer. 

To the left you’ll see the structure of aspartic acid with the chiral carbon of interest starred. Since the hydrogen is not explicitly shown and we have an amine group pointing out of the page while the other two substituents are within the plane of the page, the hydrogen must be pointing into the page (dashed line). Whatever R/S configuration we settle upon will be the final configuration. 

The nitrogen of the amine group has the greatest atomic number compared to the two yellow carbons so it will be the highest priority. We now have two carbons (yellow) so we look to what each of these carbons are immediately attached to. The carbon on the left is attached to another carbon and two hydrogens, while the carbon on the right is attached to the equivalent of 3 oxygens. Thus the carbon on the right is attached to substituents that are higher priority and the carbon on the left is lowest priority. Drawing a line from 1 -> 2 -> 3, we have a counterclockwise arc and an S configuration. 

  1. S – Based on what we know, as well as the configuration shown above, we can say that naturally occurring aspartic acid is S configuration.
  2. Z – Z refers to an alkene with each carbon’s highest priority substituents on the same side (think cis- and ze zame zide). The only double bonds we see are carbonyls, and we don’t apply Z and E designations to carbonyls. Moving on.
  3. D – D and L designations are usually mentioned in the context of carbohydrates on the MCAT. They can also be used to describe amino acids. However, the aspartic acid we see above, like all other amino acids humans use, is an L amino acid. 
  4. E – as mentioned above, Z and E designations are not appropriate for amino acids as they lack a C=C bond. Answer choice A is then our best answer.

2) This question can seem difficult at first glance. Yet, there are a few different ways to reach the same, correct answer. First, the question stem asks specifically about transaminases. We know that the “-ase” ending refers to an enzyme, so the “amin” will have to tell us what functional group the enzyme uses. “Amin” looks a whole lot like “amine”, so we’ll look for the amino acid that has an amine group in its side chain. 

We can also go back to the passage. The second paragraph reads “All transaminases appear to follow a common reaction mechanism illustrated in Figure 2 and utilize the same prosthetic group, pyridoxal phosphate (Compound 5). Pyridoxal phosphate acts as a carrier of the amino acid NH2 group during the reaction.” So if the transaminase uses pyridoxal phosphate, and pyridoxal phosphate carries an NH2 group, then it follows that the transaminase uses an NH2 group.

Let’s say we didn’t pick up on the enzyme name or are looking for a visual representation of the amino acid. We can look at the first step in Fig. 2 and look at what is attached to the enzyme.

We see that attached at the end of the enzyme, there is an NH2 group. While here we’re working through three different routes to reach the same answer, you would only use one of these on test day to save time. Now that we know that we’re looking for an amino acid with an amine group in its side chain, let’s look at the answer choices.

  1. Val – valine is a hydrophobic amino acid with a three carbon side chain in “V” formation.
  2. Asp – aspartic acid is an acidic amino acid, meaning its side chain should be one that readily loses a proton. Aspartate has a carboxylic attached to another carbon as its R group.
  3. Phe – phenylalanine is easily recognizable due to its benzene ring attached to a carbon within the side chain. No amine group here either.
  4. Lys – lysine is a basic amino acid that can readily accept a proton. In particular, lysine has a single amine group following a four carbon chain as its R group. This is the answer we’ve been eagerly awaiting!

Below are the amino acids and their side chains for reference.

3) This question is testing our knowledge of Le Chatelier’s principle. Before moving on to the answer choices, we need to recognize that compounds 1 and 2 are the reactants, and compounds 3 and 4 are the products. The question is then asking what would cause an increase in the production of products.

  1. Adding Compound 3 – compound 3 is a product. According to Le Chatelier’s principle, adding a product would cause a leftward shift to restore equilibrium. This is the opposite of what we want.
  2. Increasing the temperature – temperature can be tricky. To know the effect that temperature would have on a reaction, we would need to know whether the reaction is endothermic or exothermic. If it were an endothermic reaction, then this would be an appropriate answer choice because more product would be formed. However, because we’re not given this information, we need to keep going.
  3. Adding more catalyst and pyridoxal phosphate – adding more catalyst will not change the amount of product made, but instead will increase the rate or speed at which it’s made. They almost got us on this one.
  4. Adding Compound 2 – compound 2 is a reactant. According to Le Chatelier’s principle, adding more reactants causes the reaction to proceed forward and increase the amount of product made to restore equilibrium. This is the best answer choice.

4) The question stem is sending us to Fig. 1. The only component of the image that might give us a hint as to the thermodynamic properties of the reaction would be the equal magnitude double-headed arrows, indicating equilibrium. 

While the question stem sends us to Fig. 1, reading each of the figure descriptions before moving on to the questions also indicates that Fig. 2 is just the intermediate steps within the same reaction. The Keq~1.0 mentioned with Fig. 2 must then also apply to Fig. 1. A Keq of 1, like the bidirectional arrows of equal magnitude seen in Fig. 1, points us to thermodynamic equilibrium. 

  1. Since ΔG < 0, ATP is produced. –  As noted above, the reaction is in thermodynamic equilibrium. ATP is neither used nor produced. 
  2. Since ΔG = 0, no ATP is required or produced. – ΔG=-RTlnKeq and ln(1)=0 so given that Keq~1 for this reaction, ΔG=0. At equilibrium, ΔG= 0 and no ATP is used or produced. This is exactly what we were looking for. 
  3. Since ΔG > 0, ATP is required. – Because the reaction is at equilibrium, ΔG must be 0 and ATP is not involved.
  4. Since ΔG > 0, ATP is produced. – This statement is factually incorrect. If  ΔG>0, energy/ATP would be required for the reaction to proceed forward as it would be nonspontaneous. We’ll stick with answer choice B.

5) Here’s another Le Chatelier’s Principle question, this time in disguise. The question is asking us what would react with Compound 11, now a reactant, to regenerate Compound 5, now a product. In order to form more products, we can add more reactants. Compound 12 is shown below so we can compare its structure to each of the answer choices.

  1. ATP – We know this reaction is thermodynamically at equilibrium so ATP is not needed, and Compound 12 does not resemble ATP.
  2. Oxaloacetate – The figure description for Fig. 2 tells us that it is the intermediate steps for the reaction in Fig. 1. When we take a look at Fig. 1, we see that we are given the structure for oxaloacetate. The region that is similar to compound 12 is highlighted.  eWZ563kTBHxqroVGIKruGQCSpZC1P55amabsd_yPCEvFEMLEKOsowOz6XFF7MRJph5yNrMtBpNKDyGIfMuYioHHSnzlZL8v7HeJbM30qa7olW7p4OIBgdrdJW4M3ow
  3. NADH – NADH is involved in redox reactions and does not structurally resemble Compound 12. Answer choice B is a better answer.
  4. FAD+ – FAD+ is involved in redox reactions and does not structurally resemble Compound 12. We’re sticking with answer choice B.


Official Guide Chemistry/Physics Section Passage 2

6) The passage says The deflection of the cantilever was measured using visible laser light…” The correct answer choice will have to fall within the visible light spectrum, approximately 400nm-700nm. 

  1. 226 nm – This wavelength falls below the visible light spectrum.
  2. 633 nm – 633nm is well within the visible light spectrum.
  3. 1.26 μm – 1.26 μm = 1.26*10-6 m=1260*10-9 m=1260nm, this is outside of the 400-700nm range.
  4. 3.17 μm – 3.17 μm = 3.17*10-6 m=3170*10-9 m=3170nm, this is outside of the 400-700nm range. Answer choice B was the only choice that fell within the visible light spectrum and as such is the correct answer.

7) The question stem does not explicitly send us to a given figure, however from our preliminary figure review before moving on to the questions, we know that Fig. 2 related force and extension distance. We also know that Work=F*d. To calculate the mechanical work done over the range in the question stem, we can simply take the area under the curve since the area under the curve is the distance (x-axis) * force (y-axis). Of note, we cannot simply take the peak force and extension distance and multiply them directly because the force is variable. The variability of the force means that we need to answer this question graphically. We could also use the spring equation for work, W= (1/2)*kx2. We’re not given k but since k has units of N/m, we can equate it to W=(1/2)(F/x)x2=(1/2)Fx , the same as the area of a triangle/the area under the curve. 

  1. 2.00 × 10–22 J – this answer choice is orders of magnitude off. 10–22 << 10–19 
  2. 4.70 × 10–20 J – while not as far from the answer we calculated above, it’s not quite right either.
  3. 2.50 × 10–19 J – this is exactly what we calculated.
  4. 5.00 × 10–18 J – this answer choice is larger than what we calculated so we’ll stick with answer choice C and move on.

8) Once again, the question stem does not send us to a specific figure, however Fig. 3 can help us answer this question. This figure has velocity on the x-axis and force on the y-axis. Well, Power=F*v so we can use the graph to determine the force when the speed is 1,000 nm/s and then multiply that speed by the force to get power.

  1. 1.5 × 10–18 W – The power here is off. Surely we can find an answer choice that is closer to what we calculated.
  2. 7.5 × 10–17 W – This is exactly what we calculated, love to see it!
  3. 3.5 × 10–6 W – We might arrive at this answer choice if we forgot to convert pN to N and just plugged it into the power equation, and possibly made an arithmetic error as well. Remember to convert force units to N.
  4. 5.5 × 106 W – We might arrive at this answer if we forgot to convert both sets of units in addition to making an arithmetic error. When plugging values into physics equations, SI units are generally our friend. We’ll stick with answer choice B.

9) Let’s take a moment and acknowledge that this can be a really difficult question. Now that we’ve done that, we’ll brush ourselves off and get right back to it! The question is asking for units for the rate constant. But first, what’s the rate? Well, the concentration of unfolded protein is increasing by some amount per unit of time, so out rate units are M/s. Now we have to ask ourselves, is this a zero, first, or even second order mechanism? The protein isn’t spontaneously unfolding so it’s not zero order. The figures showing us the unfolding have also been linear, so it should be a unimolecular mechanism, that is, first order.

M/s = (units for k) * (M) => units for k = (M/s)*(1/M) = (1/s)

  1. M–1•s–1 – this answer choice would represent a second order rate.
  2. M•s  – a rate constant has to have time in the denominator so this is incorrect regardless of the rate order.
  3. M•s–1 –  this answer choice would be correct if the unfolding were zero order
  4. s–1 – this answer choice represents a first order rate and is exactly what we’ve been looking for. 


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