Sample Test Chemistry/Physics Section Passage 1

1)

1. an agonist. – The passage says that Compound 1 is a “transition state analog” for the HIV protease enzyme, which means that it mimics the product of substrate binding. However, it prevents the enzyme from performing its function which we know because it’s used to treat HIV, not amplify the infection. An agonist would activate the enzyme, not inhibit it.
2. an antagonist. – An antagonist would inactivate the enzyme which, as noted above, seems to be the effect of Compound 1.
3. a placebo. – A placebo serves as a negative control; it is given to determine the amount of the effect that is caused by the therapeutic value of someone believing they are being treated, but does not have a clear biological mechanism that would affect the results. If Compound 1 is being used to treat HIV infection, it should have some biological mechanism for treatment—and we know that it does because it binds the HIV protease. This means that Compound 1 is not a placebo.
4. a catalyst. – Catalysts increase the rate of a reaction, but in the case of Compound 1, the reaction is being stopped, not sped up. Compound 1 is an antagonist, meaning answer choice B is correct. 

2) The number of stereoisomers is equal to 2ⁿ where n = number of chiral centers. Recall the criteria for a chiral center—a carbon attached to four unique substituents. Let’s see how many chiral centers are present in compound 1:

1. 8 – This would be true if there were only 3 chiral centers (2³=8) but there are 5 chiral centers.
2. 16 – This would be true if there were only 3 chiral centers (2³=16) but there are 5 chiral centers. 
3. 32 – There are 5 chiral centers, and 2⁵=32, making this answer choice the correct answer.
4. 64 – This assumes there are 6 chiral centers (2⁶=64), but there are only 5, making this an overestimation and double the number of stereoisomers. We should stick to answer choice C.

3) For kidney stones to form, the forward reaction that produces the stones would need to be more favorable than the reverse reaction. For the formation of a precipitate or solid, this means that the favorability of the forward reaction must be greater than for the reaction for dissolving and Kprecipitate > Ksp.

1. [Ca²⁺] + [C₂O₄^2–] > K_sp – The kidney stones are “composed of calcium oxalate CaC₂O₄” so the solubility product will need to show the two reactants, calcium and oxalate, multiplied and not added. This answer choice shows the two added and is incorrect.
2. [Ca²⁺][C₂O₄^2–] > K_sp – The solubility product should have the reactants multiplied so this answer choice looks good so far. For the second half, we see that the solubility product is greater than the solubility product constant Ksp. This would mean that the forward reaction, the formation of the precipitate, is favored. Here we have it, this should be the correct answer!
3. [Ca²⁺] + [C₂O₄^2–] < K_sp – Like answer choice A, this shows the reactants being added, and not multiplied. Next.
4. [Ca²⁺][C₂O₄^2–] < K_sp – While the expression does show the multiplication of the reactants, it shows that their product is less than the Ksp, which would mean that the precipitate is less likely to form and more likely to dissolve. Answer choice B it is.

4) If you’ve made it to the sample test, you know that amino acids are very important for test day, and that you should know the one letter abbreviations, three letter abbreviations, structures, properties and side chain pKas. Fig. 1 depicts the HIV protease cleavage, so we’ll jump back to Fig. 1 (shown below) before moving on to the answer choices. 

1. Phe and Ala – The amino acid to the left of the dashed line (representing where the enzyme cleaves the peptide chain) is phenylalanine, but the amino acid to the right of the dashed line is not alanine which would have a single methyl group as the side chain.
2. Pro and Val – The amino acid to the right of the dashed line is proline, but the amino acid to the left is not valine which would have a “V” shaped 3-carbon R group.
3. Val and Ala – Neither valine nor alanine are immediately to the left or to the right of the dashed line.
4. Phe and Pro – The amino acid to the left of the dashed line has a phenyl group and is phenylalanine, and the amino acid to the right has a ring that incorporates the backbone and is proline. This is the correct answer.

An amino acid chart is provided below for reference; you should be able to replicate it by test day.

 

Sample Test Chemistry/Physics Section Passage 2

5) To answer this question, it’s important to remember that by convention, electric field lines represent the movement of a positive test charge, so they should point away from positive and to negative. We also know from Fig. 2 that the inside of the neuron has a negative voltage at baseline prior to stimulation:

Figure 2 A typical nerve impulse, also known as the action potential

A.

This image would mean that the inside of the neuron is positive and not negative because the electric field lines point away from the positive region.

B.

This image correctly shows the electric field lines pointing away from the relatively positive extracellular solution and towards the ~ -70mV axon interior.

C.

The voltage potential is between the outside of the neuron and the inside of the neuron, so the electric field lines should be parallel to the difference, and perpendicular to the axon-extracellular fluid boundary. This answer choice shows the field lines parallel to the boundary which is incorrect.

D.

Like the answer choice above it, this answer choice incorrectly shows the voltage difference (graphically represented by the electric field lines) as parallel to the axon-extracellular fluid boundary instead of perpendicular to it. Answer choice B is the right answer.

6)

A. insulate the axon from the surroundings. – The first answer choice is looking good already… The myelin sheath serves to increase the speed at which the action potential travels down the axon by inducing saltatory conduction down the nodes of Ranvier. 

B. decrease the radius of the axon. –As noted in Table 1, the radius of the axon does not change between the myelinated and nonmyelinated axons.

C. produce Schwann cells. – Schwan cells make up the myelin sheath, not the other way around. This answer choice is opposite one of the characteristics of the myelin sheath.

D. increase the capacitance of the axon. – According to Table 1, the myelinated axon has a smaller capacitance, not a larger capacitance (10^–7<10^–4). Answer A is the only one to accurately describe the function of the myelin sheath.

7) If you’re anything like me, one of your eyebrows went up when you read the question stem—where did channel X come from? I don’t remember the passage mentioning a channel X. Well, you would be correct. This question is a pseudodiscrete and is just asking us to identify the smallest molecule from the answer choices. Many of us forget that we have access to a periodic table within the MCAT testing interface. This is one of the questions where pulling up the periodic table could be very useful.

1. Proteins – Proteins are quite large compared to a single ion; think of how many atoms are in a single amino acid alone!
2. Sodium ions – Na⁺ ions would have the electron configuration of Ne and would be a bit smaller. It has the same number of electrons as Ne but has an additional proton that pulls those electrons closer to the nucleus and decreases the electron cloud radius.
3. Potassium ions – K⁺ ions would have the electron configuration of Ar, and like Na⁺ relative to Ne, would be slightly smaller than its noble gas equivalent. K⁺ ions are also one period below Na⁺ and would be larger than Na⁺ cations. 
4. Chloride ions – Cl^– anions would have the electron configuration of Ar, but unlike K⁺, it has one less proton than Ar and would be larger than the other answer choices given (expect for proteins). Na⁺ ions would be the smallest substance listed and would be the substance transmitted by channel X.

8)

A. 0.01 m – 0.01m is too short and would require a travel time even shorter than the time for an action potential. See answer choice B for a visual representation.

B. 0.1 m – The minimum distance at which the electrodes can be placed is determined by the time for the signaling of an action potential, and the speed at which the pulse can travel. Multiplying the time for the action potential, 1ms, by the pulse velocity given in the last paragraph, 100m/s, we get a distance of 0.1m.

C. 1.0 m – This is the length given in the passage of the longest axon and would be much longer than necessary for the closest distance.

D. 10 m – 10m is even longer than the length of the longest axon and would be even longer than needed. The closest distance for electrode placement is 0.1m, answer choice B.

9) Current is the charge moved per unit time, and can be calculated using Ohm’s law: V=IR. To calculate current, we need the voltage, also known as the potential, and the resistance.

1. Conductivity, resistivity, and length – This answer choice does not account for a key variable in Ohm’s law, the voltage/potential.
2. Potential, conductivity, and radius – Conductivity is the inverse of resistivity, and the radius is perpendicular to the current so it would not help us calculate the current.
3. Potential, resistivity, and radius – While the resistivity could be used to calculate the resistance, the radius is perpendicular to the current and would not be useful in directly calculating the current.
4. Potential, resistance per unit length, and length – The potential is accounted for here, and if we multiply the resistance per unit length by the length over which that resistance occurs, we would have the total resistance. I=V/R and both variables needed to calculate current are accounted for. This is our golden ticket to another point on the chem/phys section.

 

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