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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

(2 x 10-9) / (1 x 10-4) = 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.

Exam 2 C/P Solutions: Passage 3

13) To answer this question, we’ll go back to the passage to pick up on some key points mentioned by the author. We want to know about the largest CD signal in the near UV region which the author brings up in the second paragraph:

1. present as a free amino acid. This is not a great answer because we’re told in the passage asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region of peptides. A single amino acid is not going to contribute to this tertiary structure we’re looking for.
2. part of an α-helix. The passage says “The asymmetry of the α-carbon atom does not impact the CD signal of the aromatic side chain nor do elements of secondary structure.” Part of an alpha helix can be ruled out.
3. part of a β-sheet. Reasoning here is going to be the same as answer choice B. We can rule out part of a beta sheet.
4. part of a fully folded protein. This matches what the author tells us in the passage. If tryptophan is part of a fully folded protein, it will give rise to the largest CD signal in the near UV region.

14) This is going to tie into our previous question.

The author mentions side chains of aromatic amino acid residues absorb in this region. To find an amino acid that will not contribute to the CD signal in the near UV region, we can look for any non-aromatic amino acids.

1. Trp Look at the bottom right of the visual and note tryptophan is aromatic.
2. Phe Phenylalanine is aromatic
3. Ala. This is the only answer choice so far that is not aromatic. This is going to be our best answer and we can eliminate answer choices A and B.
4. Tyr Tyrosine is aromatic. We can stick with answer choice C as our best answer.

15) To answer this question, we’ll determine between a peptide bond or an aromatic side chain exhibiting an electronic excited state that’s closer in energy to the ground state. The ground state is when we have energy as low as possible. What I want to note is we’re given two options: peptide bond or aromatic side chain. We don’t even have to know the exact reasoning, but if we can decide between one or the other, we can eliminate two answer choices right away. Let’s think about what we know about energy:

Using the above equations, we have E= h x c/λ. That means as wavelength lessens, we have higher energy. A higher wavelength means lower energy. Let’s use this and apply it to the wavelengths given in the passage.

Aromatic side chains absorb in the 250-290 nm range, while peptide bond absorbs in the 190-250 nm range. Based on our breakdown, we know as wavelength increases we have an electronic excited state that is closer in energy to the ground state. That means we can say our correct answer is going to be aromatic side chain. We can eliminate answer choices C and D. We also mentioned we’re looking for the ground state, or lower energy. That means answer choice B is going to be our best option.

16) This question just boils down to knowing what to look for in the passage and using the necessary information in the question stem. We can look at Figure 1 in the passage that shows ellipticity at different wavelengths.

What do we notice here? Graph A from our question stem matches the alpha-helix line. Graph B from our question stem matches the beta-sheet line. That means the change that most likely took place is from alpha helix to beta sheet. Answer choice D is our best answer.

Exam 2 C/P Solutions: Passage 4

17) To answer this question, we can go back to the passage and get more information about the sT-loop. We are told the amino acid sequence: KTFCGPEYLA and we can determine which charged amino acids are in this sequence. What does that entail? Knowing your amino acid abbreviations and knowing your amino acid properties. Not knowing these can cost you several points on test day.

The only charged amino acids in the sequence are one lysine and one glutamic acid. That yields a net charge of zero. Our correct answer is going to be answer choice C.

18) To answer this question, we have to consider what is happening in the passage. The sT-loop was incubated with 32P-labeled ATP in the presence of PDK1 for different time periods at 37 ° C and pH 7.2, and the amount of radioactivity incorporated into sT-loop was measured by detection of β decay.

1. α32P-ATP. We’re thinking about the ATP molecule with three phosphates: alpha, beta, and gamma. The alpha phosphate is the one attached to the ribose sugar, but the gamma phosphate (furthest away from the ribose sugar) is actually the one that will attach to the protein. We’re looking for the gamma phosphate, not alpha.
2. β32P-ATP. Reasoning here is going to be the same as answer choice A. We’re looking for the gamma phosphate.
3. γ32P-ATP. This answer choice matches what we’re looking for. Key here is knowing the alpha phosphate is the one attached to the ribose sugar, next is the beta phosphate, and lastly, we have the gamma phosphate.
4. δ32P-ATP. There is no delta phosphate we’re dealing with in our question. We can eliminate this answer choice as well. We’ll stick with answer choice C as our best option.

19) This question is going to tie into some of the basic information we needed to answer Question 17. Knowing your amino acid abbreviations and knowing your amino acid properties is crucial! Note we’re asked about spHM, the phosphorylated sHM, and not sHM here. If we want to achieve the same experimental results, we will need to find an amino acid with similar properties to phosphorylated threonine to replace T in the FLGFTY sequence. We will need a negatively charged amino acid which means aspartic of glutamic acid. The only answer choice substituting phosphorylated threonine with one of these options is answer choice D. We have amino acid E, glutamic acid, substituting for phosphorylated threonine.

20) This question actually ties into our previous question and just knowing your amino acids. I can’t understate how important that is! PDK1 phosphorylates specific Ser or Thr residues, which means we want to know the functional group in common between these two amino acids. Once again, we’ll look at our amino acid chart.

Serine and threonine both have a hydroxyl group, which corresponds to answer choice A. Reactions involving serine and threonine involve their hydroxyl side chain.

Exam 2 C/P Solutions: Passage 5

21) This question comes down to knowing how to identify cooperativity in the form of a graph. We’re going to look at figures 2 and 3 and determine if we see cooperativity or not. How do we know if there is cooperativity? We can either look at the Hill coefficient, or see if you have a sigmoidal graph. A Hill coefficient greater than 1 will typically mean cooperativity, as will a sigmoidal graph.

Looking at figures 2 and 3, we can say activation and folding are both cooperative.

1. Both activation and folding are cooperative. This answer choice is consistent with our breakdown. We can consider the figures in the passage and see both have a sigmoidal shape. That tells us we have cooperativity (in both activation and folding).
2. Activation is cooperative, but folding is not. This answer choice is only half correct. Folding is cooperative as well.
3. Folding is cooperative, but activation is not. This answer choice is only half correct. Activation is cooperative as well.
4. Neither activation nor folding is cooperative. This goes against what we said in our breakdown. Both activation and folding are cooperative. We can stick with answer choice A as our best answer.

22) Similar to Question 21, this comes down to analyzing the data presented to us in the question stem. We’re asked which variant is most stable based on Figure 3. Figure 3 shows us percentage unfolded at different temperatures. As temperature increases, the less unfolding we see, the more stable the variant. Think about additional temperature needed to unfold as extra energy. The most stable proteins won’t unfold until a high temperature is reached.

Looking closely at the Figure shows us that L203A, the solid gray line that unfolds at the highest temperature, is the most stable variant. We can stick with answer choice A as our correct answer.

23) This ties into something we looked at in our previous question. Figure 3 shows us percentage unfolded at different temperatures. As temperature increases, the less unfolding we see, the more stable the variant. When we look at Y229A, we see a decrease in stability. What does that tell us? T229 is important for stability. What do we know about cAMP activation? The passage says “L203A and Y229A show the same activation as WT.” Because we have the same activation as WT, we can say Y229 is not essential for cAMP activation.

1. Y229 is important for protein stability but not critical for cAMP activation. This answer choice matches our breakdown. We looked at figures 2 and 3 and found Y229 is important for protein stability, but not for cAMP activation.
2. Y229 is important for cAMP activation but not critical for protein stability. This answer choice is the opposite of our breakdown. Y229 is important for protein stability, but not for cAMP activation.
3. Y229 is important for protein stability and critical for cAMP activation. This answer is only half right. Y229 is not critical for cAMP activation.
4. Y229 is not important for protein stability and not critical for cAMP activation. This answer choice is only half right. Y229 is important for protein stability. We can stick with answer choice A as our best option.

24) This is a content question that relies on knowing what’s happening in Figure 1: the reactions in the enzymatic assay. We can pull up Figure 1 here.

First clue is going to be lactate dehydrogenase. Dehydrogenase enzymes are oxidoreductases, meaning they are involved in oxidation and reduction reactions. In this specific case, we have the carbon attached to the oxygen on the left side of our compounds. That carbon is reduced (with NADH oxidized). Just by identifying the enzyme in the reaction, we were able to narrow down to either answer choice A or B. We know Compound 2 undergoes reduction in order to get to Compound 3. Answer choice B is our best answer.

Exam 2 C/P Solutions: Questions 25-28

25)

1. Doppler effect. The Doppler effect explains the perceived increase (or decrease) in the frequency of sound, light, or other waves as the source and observer move toward (or away from) each other. Not relevant in this case.
2. Venturi effect. The Venturi effect is the reduction in fluid pressure when a fluid flows through a constricted section. In other words, if a gas is moving more quickly, we have a reduction in pressure. In our example, the person inhales oxygen from the tank into the mask and creates a low-pressure system. Static air from outside the mask is higher pressure and moves into the mask, which is lower pressure.
3. Diffusion. Diffusion involves the movement of gases from regions of high concentration to regions of low concentration.
4. Dispersion. Dispersion involves suspension of bubbles of gas into a liquid. This is not relevant in this case. We can stick with answer choice B as our best option.

26) In other words, if we had additional G+C, why do we expect duplex DNA to melt at a higher temperature? Typically, this has to do with additional hydrogen bonds between G+C.

1. Stronger van der Waals forces of pyrimidines. We will have an equal number of purines and pyrimidines in both DNA samples.
2. Stronger van der Waals forces of purines We will have an equal number of purines and pyrimidines in both DNA samples.
3. Increased π- stacking strength. While this is not explicitly talking about the additional hydrogen bonds between G+C, that extra hydrogen bond does mean less rotation and strain, and this increases π- stacking strength.
4. Reduced electrostatic repulsion of phosphates. This is out of scope. We don’t have phosphates interacting or repulsion of phosphates in this situation. We can stick with answer choice C as our best answer.

27) This question might seem simple at first glance, but we have to consider a few things. The boiling point of a liquid is when vapor pressure is equal to atmospheric pressure. However, if we change different factors like molecular weight or bonding, we can also affect vapor pressure. This question wants to know the property best used to estimate relative vapor pressure.

1. Melting point. If deciding between melting and boiling point, we stick with boiling point based on our breakdown of the question.
2. Boiling point. This answer choice is our best option for the time being.
3. Molecular weight. As I mentioned, molecular weight can affect vapor pressure, but it would also affect boiling point. What does that tell us? Boiling point and vapor pressure are going to be related closely.
4. Dipole moment. This is going to be similar to answer choice C. Boiling point is going to be the best answer for this question because it provides the best estimate of relative vapor pressure.

28) Fats provide energy, insulation, and storage of fatty acids for many organisms. They consist of three fatty acids and glycerol; they are also called triacylglycerols or triglycerides.

1. Two fatty acids ester-linked to a single glycerol plus a charged head group This answer choice describes a phospholipid. Phospholipids consist of a glycerol molecule, two fatty acids, and a phosphate group that is modified by an alcohol. The phosphate group is the negatively-charged polar head.
2. Three fatty acids ester-linked to a single glycerol. This answer choice describes a triglyceride or triacylglycerol. This is going to be our correct answer.
3. Two fatty acids ester-linked to a single sphingosine plus a charged head group. This answer choice describes a sphingolipid

Exam 2 C/P Solutions: Passage 6

29) This is obviously a very broad question, but we can look at an example of a zinc sulfate and nickel sulfate visual to get a sense of the movement of electrons.

We have oxidation at the anode and reduction at the cathode. Right away that means we can eliminate answer choices C and D which mention oxidation at the cathode and reduction at the anode. In our passage, we can see what happens with nitrogen and hydrogen.

Nitrogen goes from a 0 oxidation state to -1, while hydrogen goes from 0 to +1. This means nitrogen is reduced and hydrogen is oxidized. Answer choice B is our correct answer.

30)

1. Boyle’s law. Boyle’s law states that the absolute pressure and volume of a given mass of confined gas are inversely proportional, provided the temperature remains unchanged within a closed system.
2. Charles’s law. Charles’s law describes the relationship between the volume and temperature of a gas. This law states that at constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature (in Kelvin); in other words, temperature and volume are directly proportional.
3. Heisenberg’s principle. The Heisenberg Uncertainty Principle states that it is impossible to determine both the position and the velocity of an object simultaneously
4. Le Châtelier’s principle. Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state. We’re considering the removal of the ammonia from the reaction mixture: N2(g) + H2(g) ⇄ NH3(g). If we remove ammonia, more product is formed to get back to equilibrium. Answer choice D is our best answer.

31) This is almost like a standalone question. We did just read a passage about ammonia, so this is tangentially related, but ultimately, we can answer this question without focusing on the passage.

1. assume a planar structure. The lone pair on ammonia makes it trigonal pyramidal.
2. act as an oxidizing agent. The oxidizing agent is a substance that causes oxidation by accepting electrons. Ammonia might donate electrons in water, but not accept electrons.
3. act as a Lewis acid in water. This is similar to answer choice B. A Lewis acid is any substance that can accept a pair of electrons. Not what happens with ammonia.
4. act as a Lewis base in water. A Lewis base is any compound capable of donating an electron pair. In this specific scenario, ammonia is more basic and water is amphoteric, so ammonia acts as a Lewis base.

32) We were told in the passage commercial production requires a catalyst.

We can go over and define the role of a catalyst to answer this question. Where do we see catalysts the most? Enzymes! In this specific question, we have to know what a catalyst is, and why it is used in the Haber process. Enzymes are catalysts that are able to lower activation energy without being used up or being destroyed in the reaction. Catalysts work to increase the rate of a reaction by lowering activation energy.

1. It increases the amount of ammonia produced per unit time. This is consistent with our definition of a catalyst. By increases the rate of the reaction, more ammonia is produced per unit time.
2. It increases the total amount of ammonia produced. The total amount of ammonia produced is not going to change, but ammonia is produced in less time.
3. It decreases the amount of ammonia that decomposes per unit time. More ammonia is produced, but the catalyst does not affect the amount that decomposes per unit time.
4. It decreases the total amount of ammonia produced. This is similar to answer choice B. The total amount of ammonia produced is not going to change, but ammonia is produced in less time. Answer choice A remains our best option.

33) To answer this question, we’ll consider what we know from the passage about the effect of different temperatures.

Proton conductivities increase substantially with temperature, but we also don’t want temperature too high where we have decomposition of ammonia. We see the importance of half-reaction 3 in producing NH3, so we want a high enough temperature to make this reaction go quickly, but not so high that ammonia decomposes. That means we want SCY temperature higher than electrode temperature: Answer choice A is our best answer.

Exam 2 C/P Solutions: Passage 7

34) We can use the equation ΔG°=-RTlnK.

We have a positive T according to the question stem and a positive K according to the passage. That means we have a negative ΔG°. A negative ΔG° means a spontaneous process. Answer choice B is going to be our correct answer.

35) We’re told in the passage “The formation of [Cu(NH3)4]2+ takes place in a stepwise manner with one water molecule being replaced by an ammonia molecule in each step.” We’re starting with [Cu(H2O)4]2+. We’re determining the product of the second step. Answer choice C would be the product of the first step. After replacing another water molecule with an ammonia molecule, we get answer choice A, which is our correct answer. Answer choices B and D are unreasonable as they list more than 4 combined water and ammonia molecules.

36) This question comes down to knowing some periodic trends and how different atoms will bond.

Copper and nitrogen have a small electronegativity difference so we’re likely only looking at answer choices B and D. For an ionic bond, we need much larger electronegativity differences. Furthermore, when copper and nitrogen bond, nitrogen will actually donate an electron pair for the bond. This is going to be a coordinate covalent bond. We can stick with our correct answer, answer choice D.

37) We can pull up Equation 1 here:

[Cu(H2O)4]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]2+(aq) + 4H2O(l)

We’re asked about adding hydrochloric acid and the effect on the product [Cu(NH3)4]2+. By adding a strong acid (HCl), we’re going to react with NH3 (one of the reactants in Equation 1). By having increased reaction of HCl with NH3, there is now a decrease in reactants and Le Chatelier’s principle comes into play. A decrease in reactants means we have a shift to the left in our Equation 1. That means the concentration of our reactants will decrease. Answer choice B is our best answer choice here.

38)

1. The oxidation number of Cu only. The oxidation number describes explicitly the degree to which an element can be oxidized (lose electrons) or reduced (gain electrons). The 4 is not telling us the effective charge on the copper.
2. The coordination number of Cu2+ only. The coordination number is the number of atoms, molecules, or ions bonded to a central atom in a molecule. In this case, the copper is bonded to the four ammonia.
3. Both the oxidation number of Cu and the coordination number of Cu2+. While the 4 is the coordination number of copper, it does not also represent the coordination number.
4. Neither the oxidation number of Cu nor the coordination number of Cu2+. The 4 does represent the coordination number of copper. We can stick with answer choice B as our best option.

39) We can answer this question by going through all three of our options.

• First option mentions ammonia containing more lone pairs than water. Ammonia has a single lone pair while water has two. That means option I is not correct. We could technically stop here and eliminate answer choices A, B, and D because they contain option I. We want to be thorough here.
• Second option talks about ammonia being a Lewis base. A Lewis base is any compound capable of donating an electron pair. In this specific scenario, ammonia is more basic and water is amphoteric, so ammonia can act as a Lewis base.
• This ties into our previous option. Ammonia can donate a lone pair of electrons more readily than water. That means answer choice C is going to be our best option.

Exam 2 C/P Solutions: Passage 8

40) Ionization energy is the energy needed to remove and electron from an atom, and second ionization energy is the energy needed to remove a second electron from an atom. Typically, alkaline earth metals like magnesium and calcium will have the smallest second ionization energy. For example, after losing the first electron, magnesium would only have a single valence electron remaining, so it would be more willing to lose this remaining electron to achieve a valence octet.

1. Na sodium would be willing to lose one electron to achieve a valence octet, but it would not want to give up that second electron once it has a valence octet.
2. C carbon is a group 4 element so its second ionization energy would likely be somewhere in the middle of these answer choices.
3. O be careful to not get tricked by oxygen needing 2 electrons to achieve a valence octet. Ionization energy involves removing electrons from an atom.
4. Ca calcium is an alkaline earth metal which means it has a small second ionization energy. After losing the first electron, calcium would only have a single valence electron remaining, so it would be more willing to lose this remaining electron to achieve a valence octet. We can stick with answer choice D as our best option.

41) While this is a passage-related question, we can almost treat this like a standalone question. To find the mass percent of oxygen, we can find the mass of oxygen in the species and divide by the total mass of the species. We’ll use the periodic table:

1. H2O (16/18) = 89% This is the largest mass percent of oxygen
2. CaCO3 (48/100) = 48%
3. CO2 (32/44) = 73%
4. HCO3 (48/61) = 79%

42) This comes down to knowing your content. When you’re given multiple pieces of information like this, you should know in a matter of seconds which information is necessary, which information is superfluous, and which additional information you may need to answer the question. To answer this question, we can use the Henderson-Hasselbach equation:

We’re given pKa and the two concentrations we need, so we can plug in numbers to solve for pH:

pH= 6.37 + log (0.2/2) = 5.37

This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice B.

43) This is another question in this set that doesn’t necessarily rely on information we learned from the passage to answer. We’re given the solubility product constant Ksp for CaCO3. We can write out the expression as:

Ksp= [Ca2+] [CO32-]

When calcium carbonate is dissolved, we get equal moles of Ca2+ and CO32-. That means we can write out the following expression. Because we have equal quantities, we can denote both as “x”:

Ksp= [Ca2+] [CO32-] = [x][x] = x2 = 4.9 × 10–9

Solving for x gives us 7.0 x 10-5, or answer choice C.

Exam 2 C/P Solutions: Questions 44-47

44) This is one of my favorite types of questions because it’s so easy to see its application in medicine. As you’re practicing these concepts, never lose sight of the end goal!

1. 85 mmHg was the diastolic pressure. This sounds like a true statement. Blood pressure is noted as systolic/diastolic pressure. However, we’re looking for the statement that is LEAST likely true. This is not a good answer choice.
2. Blood flow was heard when the pressure of the cuff was greater than 130 mmHg. Blood flow was not heard when the pressure of the cuff was greater than 130 mmHg. If systolic pressure is 130 mmHg, that means the maximum pressure exerted by the blood cannot overcome the pressure being exerted by the cuff. The cuff has to be less than 130 mmHg.
3. 130 mmHg was the systolic pressure. This also sounds like a true statement. Reasoning is going to be the same as answer choice A. Blood pressure is noted as systolic/diastolic pressure. However, we’re looking for the statement that is LEAST likely true. This is not a good answer choice.
4. Blood flow was heard when the pressure of the cuff was 90 mmHg. Blood flow would be heard when the pressure of the cuff was 90 mmHg. This is the opposite of the reasoning for answer choice B. Maximum pressure exerted by the blood against the artery wall is greater than 90 mmHg. That means only answer choice B is unlikely to be true. We can pick answer choice B as our best answer.

45) The structure of DNA is called a double helix, which looks like a twisted staircase.

The sugar and phosphate make up the backbone of the nucleotides and are bonded together by phosphodiester bonds. The nitrogenous bases that are bonded to the pentose sugars pair to each other by weak hydrogen bonds. Adenine (A) hydrogen bonds to Thymine (T) and Guanine (G) hydrogen bonds to Cytosine (C) – the bases pair in this way due to the relative sizes and functional groups of the bases. This is known as complementary base pairing and causes the two separate strands of nucleotides to be joined together forming a double strand. The double-strand coils up forming a long double helix molecule of genetic material.

Nucleotide strands are complementary to each other and run in opposite directions. This ensures the base pairs match up. Due to the unique complementary nature of a DNA double helix it is called antiparallel as both strands are in the opposite sequence of base pairs to each other.

A. Nitrogenous bases pair with other bases in the same purine or pyrimidine groups. Purines and pyrimidines do not pair with members of the same group. Instead, purines pair with their corresponding pyrimidines and vice versa.

B. The two DNA strands of the double helix are oriented in the same direction. The two strands are complementary and antiparallel.

C. The amount of guanine will equal the amount of cytosine in a DNA sequence. This is correct as guanine bonds to cytosine. For every guanine we have a similar amount of cytosine. This is a good option so far.

D. Sugar-phosphate backbones form the interior of the double helix. Sugar-phosphate backbones make up the exterior of the double helix. That’s why it’s called a backbone! Answer choice C is the best answer.

46) We’re going to have to visualize to answer this question. First thing we want to remember is that opposites attract. If we only had C and B, then C would head directly toward B in a straight line. Alternatively, same charges will repel one another. If we only had A and C, then C would be repelled by A. In this case, we’re going to have the push from and the pull from B. If we were to draw out the vectors, we’d get a net movement right. The vertical component of the attraction and repulsion will cancel out. Ion C will only move right. This is consistent with answer choice D.

47) First thing we want to note is that we can’t move protons or that would change the identity of any elements you’re dealing with. Removing electrons will lead to different ion which we’re okay with. We need to quantify the number of electrons that have been removed, and we’re told the magnitude of electron charges is 1.6 x 10-19 C. We can divide the total charge on the rod by this magnitude to get the total number of electrons:

3.2 × 10–9 C / 1.6 x 10-19 C = 2.0 x 1010 which corresponds to answer choice D.

Exam 2 C/P Solutions: Passage 9

48) We can look at the structure of nitroglycerin.

1. C–H Bond length corresponds roughly to atomic radius. In this case, hydrogen has the shortest atomic radius so this will be a short single bond.
2. C–O Unlike answer choice A, the single bond between carbon and oxygen will be between non-hydrogen atoms that have a larger atomic radius. Answer choice A remains the shortest bond.
3. C–C Reasoning here is similar to answer choice B. The hydrogen having the shortest radius is still going to be the biggest factor in having the shortest single bond.
4. O–N Reasoning here is similar to answer choices B and C. The hydrogen having the shortest radius is still going to be the biggest factor in having the shortest single bond. We can stick with answer choice A as our best option.

49) Hess’s law states that, if an overall reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions, at the same temperature. To answer this specific question, we can find use Table 1 to find ΔH° for 2 moles nitroglycerin and the products that are formed.

4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

We have 2 moles of nitroglycerin, which means we also have half as many moles as are shown in Reaction 1 above. We have ΔH°f values for CO2 and H2O, so we will include 6 moles carbon dioxide and 5 moles of water in our calculation:

ΔH°f 2 moles nitroglycerin, 6 moles carbon dioxide, 6 moles water:

2(364.0 kJ/mol) – 6(393.5 kJ/mol) – 5(241.8 kJ/mol) = –2842 kJ/mol

We came up with a rounded answer, but the question stem explicitly asks for an answer that is closest to our calculated value. We can pick answer choice D as our best answer.

50) This question just comes down to dimensional analysis and balancing equations. We have reaction 1:

4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

For every 4 moles of nitroglycerin, we have 6 moles N2. Therefore, for every 1 mole of nitroglycerin, we have 1.5 moles N2.

At STP, 1 mole of an ideal gas will occupy 22.4 L/mol, but we’re interested in the volume of N2 produced in this situation. That means we can multiply our 22.4 L/mol by 1.5 moles of N2 to get a volume of 33.6 L. Once again, the test-maker asks for an answer closest to our calculated value. We can pick answer choice D as our best answer.

51) This is going to be similar to another question in this set. We previously figured out the shortest single bond in nitroglycerin, now we’re looking for the bond that is the least polar. We can look at the structure of nitroglycerin.

Bond polarity comes into play when atoms from different elements are covalently bonded. The shared pair of electrons will be attracted more strongly to the atom with the higher electronegativity. As a result, the electrons will not be shared equally, and such bonds are said to be ‘polar’. However, if we have a bond between identical atoms, that bond is not likely to be polar because there is no electronegativity difference.

1. C–H Electronegativity difference here is minor on the Pauling scale, so this is actually a good choice right off the bat.
2. C–O Similar to answer choice A here. We don’t have a huge electronegativity difference.
3. C–C As I mentioned in the breakdown of the question, if we have a bond between identical atoms, that bond is not polar because there is no electronegativity difference. Answer choice C is going to be our best option.
4. O–N Reasoning here is the same as answer choices A and B. Not a huge electronegativity difference, but answer choice C remains the best option because it listed identical atoms bonded to one another.

Exam 2 C/P Solutions: Passage 10

52) We’re going to use some of the data from the passage to help us answer this question, but ultimately, we will need some outside information to get to our final answer. We need to use the equation for power and manipulate it so we can use the information we’re given. Keep your units straight, and rely on dimensional analysis whenever possible to make sure your math is clean.

Power = Work/ time

Work=Force x distance (Force can be given as mass x acceleration, or acceleration due to gravity in this case. Distance is height, h)

Work = mg x g which we can substitute back into our power equation:

Power = mgh/time where m is the mass of our subject, g is 10 m/s2 acceleration from gravity, and h is height of 0.15 m steps x 30 steps, and time is found in Table 2 for a 64-year old woman climbing 30 steps as 27 seconds.

Power = (54 kg) (10 m/s2) (0.15 m x 30 steps) / 27 seconds = 90 W, or answer choice D.

53) This question is similar to the last question in this question set. We previously set power equal to mgh/t. Now we’re dealing with stretching a rubber band, so we’ll consider potential energy. Work done on an elastic object can be given by ½ kx2.

We can pick out some key numbers from the passage. Maximum stretching of an elastic band by the 83-year old is going to be 20 cm or 0.2 m. We’re told the elastic band has an elastic constant of k = 200 N/m. Let’s plug this into our ½ kx2

Work = ½ (200 N/m) (0.2)2 = 4.0 J, or answer choice A.

54) First thing we’ll do is find the minimum sound intensity heard by each. For the 64-year-old male that is 20 dB, while for the 74-year-old female that is 40 dB. This comes directly from Table 1 and Table 2. We’re dealing with decibels here, which means we’re on the log scale. Recall the equation dB = 10log(I/Io)

Solving for the ration (I/Io) with a 20 dB difference is going to get us a 102 difference in sound intensity. Our correct answer is answer choice D.

55) For this question we can check Table 1 once again for the focal length of the 68-year-old male’s glasses.

We have a focal length of -0.5 m and an object is 2 m away. A negative focal length implies a diverging lens, meaning it will produce a virtual image. That narrows our options down to answer choices C or D.

We can use our lens equation here:

1/f = 1/o + 1/i where f is our focal length (-0.5), o is the distance of the object (2m) and we’re solving for the image. Plugging in our numbers gets us i= -0.4

Now, we plug into our magnification formula: M= -i/o where i is -0.4 and o is 2

M=-(-0.4)/2 = 0.2 which shows we have a reduced image. That means we can stick with answer choice D: the image is virtual and reduced.

56) To answer this question, we can first find the time it took the 75-year-old male to walk one lap around the gym.

If five laps take 200 seconds, we can approximate one lap as 40 seconds. We know one half-life is 20 seconds, so 40 seconds is two half-lives. The amount remaining after a certain number of half-lives, n, is 1/2n. After two half-lives we have 1/22 or 25%. Answer choice C is going to be our best answer.

Exam 2 C/P Solutions: Questions 57-59

57) This is going to come down to a few steps and knowing our content. A redox reaction is taking place in a battery (which is chemical energy), but is converted to electric current, while current is running, heat is going to dissipate as we have resistance (thermal). We can pick answer choice D as our best answer here. Unfortunately, there aren’t many ways around knowing how to solve this problem besides committing the content to memory. I highly recommend reviewing our content outline online if you are not comfortable with circuit elements specifically. jackwestin.com/resources/mcat-content/circuit-elements

58) The structure of DNA is called a double helix, which looks like a twisted staircase.

The sugar and phosphate make up the backbone of the nucleotides and are bonded together by phosphodiester bonds. Secondary structure like alpha helices and beta pleated sheets are held together by hydrogen bonds between the backbones of amino acids. We have to be careful here. Tertiary and quaternary structure are stabilized by side-chain interactions, so we’re ruling out answer choices C and D for a question about protein secondary structure. Secondary structure we associate with hydrogen bonding between the backbone, or answer choice A.

59) To answer this question, we can use the equation for resistors in parallel:

1/R = 1/R1 + 1/R2 + ….. but in this case we only have two resistors. We can plug in our values for R1 and R2

1/Rtotal = 1/60-Ω + 1/20-Ω

Simplify and get 1/ Rtotal = 1/15 so resistance is 15 Ω. We did no approximating and no rounding here. We can stick with our correct answer, answer choice C.

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