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MCAT Content / AAMC MCAT Practice Exam 2 Cp Solutions

### AAMC FL2 CP [Web]

Exam 2 C/P Solutions: Passage 1

1) To answer this question, we can go back to the passage and gather any necessary information. The author tells us AMC is liberated from a peptide as a result of peptide bond hydrolysis.

AMC is attached to the peptide via a peptide bond. What do we know about peptide bonds? We usually see them in amino acids. Each amino acid is attached to its neighboring amino acid by a covalent bond, known as a peptide bond. During the formation of this peptide bond, the carboxyl (-COOH) group of one amino acid and the amino (-NH2) group of the other amino acid combine and release a molecule of water. Therefore, to break a peptide bond, it must be cleaved with the addition of a water, in a process called hydrolysis. In this specific question, we can find the amine in our structure: position I. Correct answer here is answer choice A.

2) This question is related to the passage, but ultimately, we’ll need to use external knowledge to deal with some math. First, what are the key points we need from the passage? Let’s revisit where the author talks about the excitation and emission events.

We have fluorescence emission at 440 nm and excitation at 360 nm.  The energy of a photon is equal to the product of Planck’s constant and frequency of EM Wave: E = hf. In this case, we have wavelengths so we can use the relationship between frequency and wavelength: f=c/λ. We can combine our equations and get:

E = h x c/λ where h = 6.62 × 10−34 J ∙ s and c = 3 × 108 m/s

We need to determine the energy converted based on the question stem, which will involve the two wavelengths I referenced from the passage.

E = h x c/λ

ΔE = h x c/λexcited – h x c/λemitted

We’ll plug the two wavelengths and our constants into our above equation to find the energy converted:

ΔE = h x c/λexcited – h x c/λemitted

ΔE = (6.62 × 10–34) × (3.0 × 108) × [1 / (360 × 10–9) – 1 / (440 × 10–9)] which corresponds to answer choice C.

3) This is a straightforward question in terms of how it’s worded, but we have to be careful to correctly identify the components mentioned in the question stem. We can pull up the relevant part of the passage to help us understand the question.

The proteasome is the enzyme introduced by the author in the first part of the passage. We’re told a porphyrin compound can inhibit proteasome activity. In this case, the substrate is going to be the peptide succinyl–LLVY–AMC. We want to note the factor difference in concentration of the substrate and the concentration of the proteasome. Note in the experimental procedure we’re told the proteasome concentration of 2 nM (2 x 10-9) while the peptide concentration is 100 μM (1 x 10-4). Dividing peptide concentration by proteasome concentration gives us

(1 x 10-4) / (2 x 10-9) = 5 x 104 or answer choice D.

4) First thing we’re going to do is identify the molecule in question. From reading the passage we know we’re looking at AMC:

sp2-hybridized means a set of hybrid orbitals produced when one s orbital and two p orbitals are combined mathematically to form three new equivalent orbitals. There are 10 total carbon atoms in AMC, with 9 of these carbons having a single double bond. That means our correct answer here is going to be 9 sp2-hybridized carbon atoms present in AMC. Answer choice C is correct.

Exam 2 C/P Solutions: Passage 2

5) To answer this question, we can revisit the passage.

We have the structure of the active site of a laccase above. We’re going to note which amino acid is most prevalent in the structure. We can pull up our amino acid chart:

Cross-referencing the amino acid chart and Figure 1 from the passage, histidine is found in greatest abundance in the active site of a laccase. Answer choice C is going to be our correct answer.

6) To answer this question, we’re going to use information from the passage. We’re ultimately looking for the product formed when Compound 3 is the substrate, but the key here is knowing what we’re looking for by studying Reaction 1.

We have an example of laccase catalyzing the oxidation of phenols. Phenols will have hydroxyl groups linked directly to a benzene ring. The hydroxyl groups in the products are now carbonyl groups. That means we’ll look at compound 3 and expect the hydroxyl groups to change to carbonyl groups in the correct answer choice.

We have compound 3 above. Like I mentioned, we can pinpoint the two hydroxyl groups on the left and right of the figure. I’ve outlined them in blue. We’re expecting carbonyl groups in our correct answer. Only answer choice that matches our breakdown is going to be answer choice D:

7) To answer this question, we can pull up Table 1 below. We’re looking at the results of laccase activity experiments.

We want to know: what effects do acid and chloride have on the enzymatic activity of laccases? We have 3 instances where pH is lower (acidic) and activity increases. 2 instances where pH is higher where activity is not as high. We can say acid increases activity.

We can look at the lower pH and see there are two rows without chloride, and one row with chloride. If we compare compound 3 at a pH of 4.5 with and without chloride, we can see chloride decreases the activity.

1. Both acid and chloride decrease the activity. While chloride does decrease activity, acid actually increases activity. This is only half correct
2. Both acid and chloride increase the activity. This is similar to answer choice A. Chloride actually decreases activity and acid increases activity.
3. Acid decreases the activity, whereas chloride increases the activity. This is the opposite of our breakdown. Acid actually increases activity and chloride decreases the activity.
4. Acid increases the activity, whereas chloride decreases the activity. This answer choice matches our breakdown. We note acid increases the enzymatic activity of laccases, while chloride decreases the activity.

8) First thing we’re going to do is pinpoint the information we need from the passage to answer this question. We’re dealing with Compound 3 at pH 4.5:

We can calculate kcat using the equation kcat=vmax/[E]. Plug in the numbers we know:

kcat= (125 nM/s) / 5.0 μM = 2.5 × 10–2 /s

This was a math problem where we did no rounding or approximation. We can stick with the answer that matches our breakdown exactly: Answer choice A.

Exam 2 C/P Solutions: Questions 9-12

9) This is a standalone question which means we’re relying on our external knowledge to answer the question. We’re dealing with the absorption of UV light.

1. Bond breaking. While this is possible, note the test-maker says always results in what process. We don’t know for a fact that absorption of UV light will result in bond breaking.
2. Excitation of bound electrons. When we have absorption of UV light, we have an increase in electron energy. Unlike answer choice A, we can say answer choice B is always correct.
3. Vibration of atoms in polar bonds. This refers to IR spectroscopy, not UV. Answer choice B remains superior.
4. Ejection of bound electrons. This is similar to answer choice A. the test-maker says always results in what process. We don’t know for a fact that absorption of UV light will result in ionization/bond breaking. We can stick with answer choice B as our best answer.

10) We’re given four organic compounds in our question stem, and all have similar sizes. Instead of using size-exclusion, we’ll have to consider the polarity of each molecule. Increased polarity will mean the molecule will increase affinity for the stationary phase. Decreased polarity will mean the opposite; the compound will elute more quickly. All we’re doing is deciding which compounds are the least to most polar. The least polar compound should be the first one listed in our correct answer choice.

First to elute of our answer choices should be the alkane: n-pentane. Second, we’re expecting an aldehydes or ketones: 2-butanone. At this point we should be able to narrow our answers down to A or B, just by knowing the alkane would elute first. In this case, we expect 2-butanone to elute before the alcohol that would elute third: n-butanol. Lastly, we’d have propanoic acid, the carboxylic acid. Answer choices C and D incorrectly list propanoic acid first and n-pentane last. We know the opposite to be true. We can stick with our correct answer, answer choice A.

11) The half-life is the time taken for the activity of a given amount of a radioactive substance to decay to half of its initial value. Highly radioactive substances rapidly transform to daughter nuclides, while those that radiate weakly take longer to transform. This question simply comes down to knowing the definition of half-life. We want to be very careful with the verbiage here.

1. half the time it takes for all of the radioactive nuclei to decay into radioactive nuclei. This sounds a bit confusing which might make it tempting, but this answer is incorrect. Half-life is not half the time nuclei take to decay into radioactive nuclei. Instead, it’s the time it takes for half of the radioactive nuclei to decay into their daughter nuclei. Wording here is very important.
2. half the time it takes for all of the radioactive nuclei to decay into their daughter nuclei. This is similar to answer choice A, but it correctly mentions daughter nuclei. The daughter nuclei are not necessarily radioactive. While this is a superior answer to answer choice A, we still want a better answer.
3. the time it takes for half of all the radioactive nuclei to decay into radioactive nuclei. First part of this definition matches our breakdown. I am liking this better than answer choices A and B. However, we mentioned the radioactive nuclei will decay into their daughter nuclei that don’t necessarily have to be radioactive.
4. the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei. This answer choice matches our breakdown completely. First part of the definition is correct, then it also correctly mentions daughter nuclei. Answer choice D is going to be our best answer.

12) The author provides us with a great visual here, and this is also something we’ve experienced. When a person stands up, they need to adjust to account for the location of their center of mass.

1. To increase the force required to stand up. If this were the case, it would be even more difficult to stand up. This is not what we’re looking for in this case.
2. To use the friction with the ground. There is friction with the ground in the position shown in the question stem. Adjusting your position will not be how friction is introduced into the equation.
3. To reduce the energy required to stand up. There’s a certain amount of energy required to go from the sitting position in the question stem to standing. The path taken is not going to affect the energy required. Energy in this case is a state function, meaning it’s a property that is defined only by the current state of the system rather than the changes that occurred between the initial and final state.
4. To keep the body in equilibrium while rising. The key to this question is maintaining rotational equilibrium. If we take away the chair, the feet are the pivot point and the center of mass is well away from this pivot point. There are two ways to correct this: by either moving your feet back and pushing that pivot point underneath your center of mass, or by leaning forward and pushing your center of mass forward on top of the pivot point. We can stick with answer choice D as our best answer.

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