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MCAT Content / AAMC Sample Test Cp Solutions

AAMC Sample CP [Web]

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+] + [C2O42–] > Ksp – 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+][C2O42–] > Ksp – 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+] + [C2O42–] < Ksp – Like answer choice A, this shows the reactants being added, and not multiplied. Next.
  4. [Ca2+][C2O42–] < Ksp – 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.

 

Sample Test Chemistry/Physics Section Passage 3

10) Time for some math! The pH of a solution can be determined by pH = -log[H+]. The last sentence of the passage states “The hydrogen ion concentration of the unknown aqueous solution was 1 × 10–5 M.” Plugging this into the pH equation we get the following:

pH = -log[H+] = -log[1*10-5] = -log[10-5] = -(-5) = 5 

  1. 4 – This would be the pH if the hydrogen ion concentration were 1*10-4. This answer choice is too acidic.
  2. 5 – As noted above, pH = -log[1*10-5] = 5; this is the correct pH.
  3. 9 – This would be the pH if the hydrogen ion concentration were 1*10-9. This answer choice is too basic.
  4. 10 – This would be the pH if the hydrogen ion concentration were 1*10-10. This answer choice is too basic. Answer choice C is the correct pH.

11) The passage gives us limited information on the unknown compound: “The compound completely dissolved in water and weakly conducted electricity. The hydrogen ion concentration of the unknown aqueous solution was 1 × 10–5 M.” Unfortunately, dissolving in water and weakly conducting electricity won’t help us much with differentiating between the options given below. However, we just calculated the pH for the question above, and we know that a pH < 7 is acidic, and usually less than about 2 is strongly acidic. This is not a hard and fast rule, just a quick and dirty way to estimate; remember that a strong acid is one that fully dissociates. A pH of 5 as calculated above is weakly acidic.

  1. weak base. – A weak base would have a pH greater than but somewhat close to 7, not <7.
  2. strong base. – A strong base would have a pH much greater than 7, not less than 7.
  3. weak acid. – This is a solid answer choice. The pH < 7 but is not low enough to be a strong acid.
  4. strong acid. – We would expect the pH to be much lower if the unknown compound were a strong acid because it would fully dissociate. A weak acid is a better characterization of the unknown compound.

12) The second paragraph of the passage says that “It can be theorized that if the central atom, A, is an alkali or alkaline earth metal, the compound is basic. But if A is a nonmetal, the compound is acidic.” We have two options, either produce an acidic solution or a basic solution. The first paragraph says the molecules can be “either oxyacids or bases.” This means that if the molecule is a base, it will produce a basic solution, and if it is an oxyacid, it will produce an acidic solution. Let’s use that information and look at these compounds one at a time.

NO2(OH) dissolved in water and produced an acidic solution –> well, if it produced an acidic solution, it should be an oxyacid. Ni(OH)2 dissolved only in an acidic solution –> unlike the NO2(OH), Ni(OH)2 did not produce an acidic solution, it dissolved in an acidic solution. We know that acids dissolve in bases, and bases dissolve in acids due to the common ion effect, so if Ni(OH)2 dissolved in an acid, it is a base.

  1. Both were oxyacids. – Only NO2(OH) is an oxyacid, Ni(OH)2 is a base.
  2. Both were bases. – Ni(OH)2 is a base, but NO2(OH) is an oxyacid.
  3. NO2(OH) was a base and Ni(OH)2 was an oxyacid. – This is the opposite of the correct answer and has the molecules incorrectly labeled. Ni(OH)2 is a base and NO2(OH) is an oxyacid.
  4. NO2(OH) was an oxyacid and Ni(OH)2 was a base. – This is correct! NO2(OH) is an oxyacid and Ni(OH)2 is a base.

13)

  1. an alkali or alkaline earth metal. – The second paragraph says that “It can be theorized that if the central atom, A, is an alkali or alkaline earth metal, the compound is basic.” The pH of the unknown compound indicates that it is an acid and not a base so the central atom should not be an alkali or alkaline earth metal.
  2. a transition metal. – The passage does not address what type of molecule is produced if the central atom is a transition metal. There must be a better answer choice based on the passage.
  3. a nonmetal. – In the second paragraph the author says, “But if A is a nonmetal, the compound is acidic.” The unknown compound produced an acidic solution, so it’s safe to say that the central atom A is a nonmetal.
  4. a noble gas. – Noble gases are “happy” with their 8 valence electrons and are relatively uninterested in bonding with other atoms which is why most do not have an electronegativity listed. Answer C is the best answer. 

 

Discrete Questions

14)

  1. Triacyl glycerols – Triacyclglycerols have three long carbon chains, specifically fatty acids, attached to a glycerol backbone. There aren’t multiple long carbon chains attached to a glycerol backbone in squalene.
  2. Phospholipids – Phospholipids have a polar head that contains a phosphate and two long carbon chains that make up the “tails.” Squalene does not fit this description.
  3. Steroid hormones – Steroid hormones, like the cholesterol they are derived from, have a four fused ring base structure. We could easily draw a line between each of the almost closed rings to make four fused rings so they could easily be produced from a squalene precursor. 
  4. Prostaglandins – Prostaglandins have one central ring with chains coming off of those rings. We don’t see much evidence of that structure. Squalene resembles cholesterol and steroid hormones, so C is the best answer. 

15) Calcium chloride is written as CaCl2; there are two chloride ions per calcium ion in the molecule. When CaCl2 is dissolved in water to make a 0.1M solution, the CaCl2 will dissociate into one Ca2+ and two Cl ions. 

  1. 0.02 M – This answer choice is one decimal place off—double of 0.1 is 0.2, not 0.02.
  2. 0.05 M – This is half of the concentration, not double the concentration as it should be.
  3. 0.10 M – This is the same concentration as the CaCl2 that was dissolved, but we know that there will be two Clfor every CaCl2 dissolved. 
  4. 0.20 M – The concentration of Cl will be double that of CaCl2, 2*0.01M=0.02M. Mental math can get us the correct answer choice if we set it up carefully.

16) I was never a huge fan of carbohydrates on the MCAT because they always felt unnecessarily complicated. Let’s see if we can help you feel more confident with carbohydrates!

The solution mentioned in the question stem is silver oxide, and it is used to test for reducing sugars. For a sugar to be able to reduce another compound, it should be able to be oxidized. Practically, this will occur when the ring of the sugar is opened at a hemiacetal to form an aldehyde intermediate. To find a hemiacetal, we need to look for an oxygen-containing ring with an alcohol directly attached to a carbon that is adjacent to the oxygen, the anomeric carbon. The disaccharide below that does not have a hemiacetal will not be a reducing sugar and will not produce the silver mirror when reacted with silver oxide. This is the disaccharide we are looking for.

A.

Cellobiose has a hemiacetal, outlined in yellow.

B.

Lactose contains a hemiacetal, shown in yellow.

C.

Maltose contains a hemiacetal, shown in yellow.

D.

Unlike the others, sucrose does not contain a hemiacetal and is not a reducing sugar. It would not produce a silver film when reacted with silver oxide and is the correct answer.

17) Amino acids, can’t go too far without them coming up again! For this question, we just need to add the charges of each of the side chains at a pH that is very close to physiological pH. The +1 from the N-terminus and the -1 from the C-terminus cancel one another out. Arginine is a basic amino acid with a +1 charge, alanine is nonpolar and neutral, phenylalanine is nonpolar and neutral, and leucine is nonpolar and neutral. Adding all of the charges together we get 1 (N terminus) + -1 (C terminus) + 1 (arg) + 0 (ala) + 0 (phe) + 0 (leu) = +1

  1. –1 – The sign is incorrect; the net charge is +1.
  2. 0 – If you were tempted by this answer choice, you might have forgotten to add the +1 of arginine.
  3. +1 – This is the correct answer! 1 (N terminus) + -1 (C terminus) + 1 (arg) + 0 (ala) + 0 (phe) + 0 (leu) = +1
  4. +2 – If you were tempted by this answer choice, you might have forgotten to add the -1 of C terminus.

 

Sample Test Chemistry/Physics Section Passage 4

18)

  1. Increasing the polarity of the mobile phase will decrease the retention time of Compound 1 relative to Compound 2. – Increased affinity for a mobile phase causes faster elution and shorter retention time, and increased affinity for the stationary phase slows down elution and increases retention times. Compound 1 is less polar than Compound 2 because it lacks the alcohol on its methyl group. If the polar mobile phase became even more polar, Compound 1 would have even less affinity for the mobile phase and its retention time would increase, not decrease. 
  2. Compound 1 will elute first because it is more polar than Compound 2. – This is incorrect and opposite; Compound 1 is less polar than Compound 2.
  3. Decreasing the affinity of Compound 1 for the stationary phase will increase its retention time relative to Compound 2. – If Compound 1 has less affinity for the stationary phase, it would elute faster and have a shorter retention time, not a longer retention time. Remember that retention time refers to how long the compound remains in the apparatus before eluting.
  4. Compound 2 will elute first because it does not interact as favorably with the stationary phase as Compound 1. – This is true—Compound 2 is more polar than Compound 1 due to the alcohol attached to the upper leftmost carbon and interacts more favorably with the polar mobile phase than the nonpolar stationary phase, meaning it will elute faster. Compound 1, which is less polar, will interact more favorably with the nonpolar stationary phase and will elute after Compound 1.

19)

  1. Compound 1 is still eliminated from the body of patients expressing the CYP2C9*3 allele. – While this is true, the fact that it is eliminated from the body does not tell us much about the affinity, or how tightly and readily Compound 1 binds to the CYP2C9 enzymes.
  2. The KM values for the variant enzymes do not differ significantly from the wild-type enzyme. – This is true and addresses the binding affinity of the compound with the enzyme. Km is the dissociation constant, making it inversely related to the binding affinity. If the binding affinity changed, there would be a change in the Km. Table 1 shows no significant changes in Km, meaning no meaningful changes to the binding affinity of the CYP2C9 enzymes toward Compound 1. This is what we are looking for because the question is a NOT question.
  3. The amino acid substitutions at positions 144 and 359 do not change the binding pocket of the variant enzymes.– We cannot say this with any degree of certainty. If the amino acids 144 and 359 are in the binding pocket, changing the amino acids would change the binding pocket. Let’s see what answer choice D says.
  4. The side chains of the amino acid residues at positions 144 and 359 are charged at physiological pH. – Looking at the four amino acids at those two positions (arginine and isoleucine in the WT and cysteine and leucine in the mutants), only arginine is charged at physiological pH. We’ll stick with answer choice B because it is true and addresses the question stem.

20) For this question, we need the information from Table 1 and we need to remember the anatomy of a basic Lineweaver-Burk plot. 

All three variants have roughly the same Km, the WT enzyme has the greatest Vmax, and the CYP2C9*3 enzyme has the lowest Vmax.

Looking at the plots above, we see that the y-intercept is 1/Vmax. Decreasing Vmax will cause the y-intercept to move in the opposite direction and move upwards on the y-axis.

  1. is the same for all three variants. – This is not true. The Vmax is different for each enzyme, so the y-intercept should change.
  2. is the same for CYP2C9*1 and CYP2C9*2 but different for CYP2C9*3. – CYP2C9*1 and CYP2C9*2 each have a different Vmax and are outside of each other’s confidence intervals. They should have different Vmaxs. 
  3. is the same for CYP2C9*2 and CYP2C9*3 but different for CYP2C*1. – CYP2C9*2 and CYP2C9*3 each have a different Vmax and are outside of each other’s confidence intervals. They should have different Vmaxs. 
  4. is different for all three variants. – This is correct. Each of the enzyme variants has a different Vmax and because the y-intercept on a Lineweaver-Burk plot is 1/Vmax, they should all have different y-intercepts.

21)

  1. an oxidizing agent. – An oxidizing agent is reduced in a reaction; NADPH does not gain electrons which would be the case if it were reduced and served as an oxidizing agent. Not looking good.
  2. a reducing agent. – A reducing agent reduces its target and is oxidized in the process. Oxidation is the loss of electrons and NADPH loses electrons as evidenced by the production of the positively charged NADP+; this is looking good.
  3. a catalyst. – Catalysts should not be consumed during a reaction. NADPH is consumed and not regenerated so it is not a catalyst here.
  4. an electrophile. – An electrophile “wants” electrons; NADPH loses electrons and becomes positively charged in the process of the reaction. It is oxidized and serves as a reducing agent, so answer B.

 

Sample Test Chemistry/Physics Section Passage 5

22)

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Reaction 1

  1. Oxidation/reduction – Reaction 1 involves changes in oxidation states, making it a redox reaction. The magnesium begins in its elemental state and then becomes oxidized to a +2 oxidation state. The hydrogen in HCl is converted to its elemental form and its oxidation state goes from +1 to 0.
  2. Lewis acid/Lewis base –This is so, so incredibly tempting! HCl is an acid! Alas, that doesn’t necessarily mean the reaction is best described as a Lewis acid/base reaction. The oxidation states have changed, and a Lewis acid/base reaction involves the sharing of donated electrons, but the MgCl2 is ionic. When in doubt, calculate the oxidation states. A change in oxidation states means the reaction is a redox reaction.
  3. Double replacement – The reaction would theoretically be a single replacement, not a double replacement. The Mg is in its elemental form as a reactant and is not attached to anything else that could be displaced.
  4. Ionization – An ionization reaction would produce a free electron which does not occur. Answer choice A is the best answer.

23) If the level of the solution is higher inside the buret than outside, it means that at this fluid level, the sum of the pressure of the gas and the pressure of the fluid is not enough to balance the pressure of the ambient pressure and the pressure of the fluid; at this point the fluid pressures are equal and cancel out. The fluid level inside the buret rises to increase the contribution of the fluid pressure until a new equilibrium is reached. The air pressure in the buret must have been smaller than the ambient pressure.

  1. lower. – This is correct. As noted above, the pressure from the small column of air inside the buret is less than the ambient pressure, causing the fluid level inside the buret to rise and reach equilibrium. 
  2. the same. – The air/gas pressure inside the buret was less than the ambient pressure, not the same or equal.
  3. 2 times greater. – The air/gas pressure inside the buret was less than the ambient pressure, not greater. No need to quantify if it’s in the wrong direction.
  4. 26 times greater. – The air/gas pressure inside the buret was less than the ambient pressure, not greater. Still no need to quantify; we’ll choose A and move on. 

24)

A. PV = nRT – While this is the classic version of the ideal gas law, the question is asking for the easiest way to calculate the gas constant, which means we need to isolate R. This equation does not have R isolated.

B. R = nPT/V – This equation is not a rearrangement of the ideal gas equation; n and T should be in the denominator, and V should be in the numerator. 

C. nR = PV/T – Like answer choice A, this answer choice does not show R isolated or by itself. 

D. R = PV/nT – D is the correct answer because it isolates the gas constant R and is an accurate rearrangement of the ideal gas equation.

25) The second paragraph in the passage states “The ideal gas constant R can be found by measuring the volume that a given amount of gas occupies at a certain temperature and pressure. One technique for accomplishing this is to react a known amount of magnesium with acid to produce hydrogen gas.” The passage then goes on to describe the experiment, which this excerpt has just told us is a way to calculate the gas constant R. 

  1. To evaluate the molar mass of H2 gas – The hydrogen gas produced in reaction 1 is not the focus of the experiment. The passage tells us that the experiment is a means to calculate the ideal gas constant R.
  2. To study the mechanism of Reaction 1 – The experiment does use reaction 1 as its basis, but the goal is to calculate R by measuring other related variables. The specific mechanism of reaction 1 is not studied.
  3. To determine the measured variables of the Ideal Gas Law – This is an appropriate answer choice. The second paragraph introduced the experiment as a way to calculate the ideal gas constant R using other measured variables in the ideal gas law (volume, temperature and pressure).
  4. To study the reactivity of magnesium with acids – Again, the focus is the ideal gas constant from the ideal gas law, not any specific part of reaction 1, including the reactivity of magnesium. Answer C is the best answer.

 

Discrete Questions

26)

  1. 0.212 kg – This answer choice is too small and is the result of using 2 instead of 5 liters.
  2. 0.530 kg – This answer choice is one decimal place off; check your powers and decimal movements! Keeping track of scientific notation is important for chem/phys section calculations.
  3. 5.30 kg – As shown above, this is the correct mass of blood.
  4. 10.6 kg – This answer choice is twice the correct mass and is likely the result of rounding 5 to 10. While rounding can be helpful on test day, we don’t want to round too far. The closer the answer choices are to one another, the less you should be rounding. Answer choice C is the correct answer. 

27) This question is asking us to recall the Doppler effect equation which is what allows us to image using ultrasound, and to determine the minimum variables and values that would be needed for the imaging. There are four variables in the Doppler equation, so the minimum required would be three of the four variables. 

  1. The speeds of the sound and of the moving object. – These two variables are too few and less than the minimum we would need. If an equation has four variables, three would be required to calculate the fourth.
  2. The speed of the sound, and the frequencies of the sound waves emitted and observed. – This answer choice includes three of the four variables above which would be the minimum information needed to make use of the ultrasound making it the correct answer.
  3. The speeds of the sound and of the moving object, and the frequencies of the sound waves emitted and observed. – This answer choice includes all four variables and provides more than the minimal information.
  4. The speeds of the sound and of the moving object, and the frequencies and wavelengths of the sound waves emitted and observed. – Wavelengths are not directly a part of the Doppler equation and this answer choice provides too much information. Answer choice “B” stands for “B”est answer.

28) What factors affect boiling point? Intermolecular forces—mostly hydrogen bonding but also dipole-dipole interactions and van der Waals forces to a lesser extent— as well as ionic bonds, branching and size of the molecule. Thinking about these factors, what does fluorine have or is fluorine capable of that the other halogens do not have or cannot do?

  1. The H–F bond is much less polar than the bonds between H and the other halogens. – Fluorine is the most electronegative element, meaning it will produce bonds that are more polar than any other element binding to the same atom.
  2. HF has the lowest molecular mass of the Group 7A hydrides. – While this is true, boiling point is positively related to molecular mass, so having a lower mass would mean a lower boiling point. Something else must be at play for HF to have the abnormally high boiling point compared to the other group 7A hydrides.
  3. HF is affected by hydrogen-bonding interactions to a much greater degree than the other Group 7A hydrides. – Fluorine, unlike the other halogens, is great at hydrogen bonding. Hydrogen bonding increases boiling point—it gives H2O its particularly high boiling point as well. This is a good answer.
  4. HF has the highest vapor pressure among the Group 7A hydrides. – Vapor pressure and boiling point are inversely related to one another, so a higher vapor pressure would correspond to a lower boiling point. We’ll stick to C.

29)

  1. Approximate radial size of an electron cloud – The principal quantum number describes the energy of a given orbital or electron cloud in which an electron resides. We know that energy is influenced by the distance between two charges, so the radial size of the cloud would influence how much energy is in each electron cloud. As the radius and size of the electron cloud changes, so does the energy.
  2. Approximate shape of an electron cloud – The angular momentum quantum number is associated with the shape of an electron cloud.
  3. Number of valence electrons that orbit a nucleus – The quantum numbers can be used to describe the orbitals in which specific electrons reside and are not limited to valence electrons.
  4. Number of protons and neutrons found in the nucleus of an atom. – The quantum numbers are used to describe the orbitals containing electrons, not protons and neutrons. Answer choice A is a better answer choice. 

For a review of quantum numbers, go take a look at https://jackwestin.com/resources/mcat-content/electronic-structure/orbital-structure-of-hydrogen-atom-principal-quantum-number-n-number-of-electrons-per-orbital

 

Sample Test Chemistry/Physics Section Passage 6

30) Many of us see the mention of metal ions in the question stem and know to go to Table 2 in the passage but aren’t sure where to go from there. Let’s go through the answer choices and interpret the table as we go along.

  1. increase the amount of CPFX bound to BSA. – To answer this question, we need to read the figure description carefully. The MCAT test writers are under no obligation to draw your eye to any information in particular—you are expected to critically evaluate all of the information they provide you given the content you are expected to know. That said, when they draw your eye to specific information by adding a “Note:” pay extra attention. The “note” with Table 2 is as follows: “Note: K0 is the binding constant of CPFX-BSA in the absence of a metal ion.” If Ka is the binding constant with the metal ions present, and K0 is the binding constant in the absence of the metal ions, then Ka/K0 will tell us the binding with metal ions compared to without. If Ka/K0 >1, metal ions enhance binding, if it’s =1 they have no effect on binding, and if it is <1, they decrease binding. We see that all of the values in the last column are Ka/K0<1 so the metal ions decrease CPFX and BSA binding, not increase it.
  2. decrease the amount of CPFX bound to BSA. – As noted in the explanation above, the presence of the metal ions produces a Ka/K0 <1 and decreases the amount of CPFX bound to BSA.
  3. decrease the amount of free CPFX found in plasma. – If there were a decrease in free CPFX, there would need to be an increase in bound CPFX. The decrease in binding seen in the table shows the opposite to be true.
  4. have little effect on the amount of bound CPFX. – This is not true. As noted in the explanations above, there is a decrease in the amount of CPFX bound to BSA, answer choice B.

31) To answer this question, we will need to review the structure of ibuprofen, which is included below for reference. The molecule most likely to bind at the same site should have a similar structure or functional groups compared to ibuprofen.

A. Glucose – We should be able to recognize a glucose molecule on test day. Glucose is a monosaccharide with a six-member ring and an oxygen within the ring. Ibuprofen has a benzene ring, but not a glucose molecule.

B. ATP – There are no phosphates or adenine in ibuprofen. See AMP below.

C. Glycerol – Glycerol consists of a three-carbon backbone, each with an alcohol attached to the carbon. The only singular alcohol is a part of a carboxylic acid. Hopefully answer choice D is a better answer choice because we’re running out of options.

D. Palmitoleic acid – Palmitoleic acid is a fatty acid, which means it has a hydrocarbon tail (the “fatty” component) and a carboxylic acid (hence it is an acid). While the tails are usually linear, this answer choice is the closest to what we see and will have to do. Next question!

32)

  1. primarily at Site II. – According to the passage, “warfarin and ibuprofen have been shown to specifically bind to separate sites (sites I and II, respectively)” so the results in the table for warfarin are at site I and the results for ibuprofen are at site II. Looking at the binding at site II with ibuprofen, we see a Ka/K0<1, so a decrease in binding when ibuprofen. This is looking good so far, but we still need to check site I where warfarin binds. Here, Ka/K0 is even smaller, 0.46<0.87. This means there is more competition at the warfarin binding site, site I… If there is less binding and more competition, it is more likely that CPFX and warfarin are competing for the same site, making site I more likely to be the CPFX binding site.
  2. equally partitioned between Site I and Site II. – The Ka/K0 for warfarin and ibuprofen are not equal.
  3. primarily at Site I. – This is it. As noted above, the presence of warfarin has a greater impact on the binding of CPFX making the two more likely to share a binding site.
  4. at a site distinct from either Site I or Site II. – The presence of warfarin cuts the binding of CPFX by more than half Ka/K0<0.5) making site I a likely binding site for CPFX meaning C is the best answer choice.

33)

The first paragraph tells us that “the ligands tend to be mainly hydrophobic with anionic or electronegative features” and since opposites attract, the side chains of the amino acids in the binding site should be positively charged.

  1. R and L – Arginine is positive and leucine is neutral making for a net positive binding site. Let’s assess the other answer choices. 
  2. E and Y – Glutamate is negatively charged and tyrosine is neutral making for a repulsive net negative charge.
  3. D and E – Aspartate and glutamate are both negatively charged which would repel the ligands even more than the previous answer choice.
  4. D and H – Aspartate is negatively charged while histidine is positively charged making for a net neutral binding site; this a poor answer choice compared to the attractive net positive binding site in answer choice A.

 

Sample Test Chemistry/Physics Section Passage 7

34) This question is fairly straightforward. Density is mass divided by volume, the author gives the mass of the wire as 4 × 10–3 kg and the question stem says the volume is 5 × 10–7 m3.

Density = mass/volume = (4 × 10–3 kg)/( 5 × 10–7 m3) = 0.8 × 104 kg/m3 = 8000 kg/m3

  1. 2000 kg/m3 – This answer choice is ¼ the correct answer choice and likely results from miscalculating 4/5 as 0.2 instead of 0.8.
  2. 3600 kg/m3 – This answer choice is too small and the result of an arithmetic error or improper setup of the equation.
  3. 6400 kg/m3 – This answer choice is too small and the result of an arithmetic error or improper setup of the equation.
  4. 8000 kg/m3 – This is the answer we calculated above and is the correct answer.

35) The question stem does not explicitly send us to Table 1, but that is where we can find the relationship between T and R:

As T increases, R increases. The increase of neither T nor R is exponential. The changes are evenly spaced out; as T increases by 100 K, R increases by approximately 2-3 Ω.

A.

This is a positively linear relationship. Looking good.

B.

This incorrectly assumes R doesn’t change as T increases.

C.

While positive, this shows an exponential relationship, not linear. The response of R on this graph is too strong relative to T.

D.

This shows a negative relationship and is in the wrong direction; we’ll stick with A.

36) This question is almost a pseudodiscrete question. To calculate the heat required to increase a substance by a given temperature, we need to use the thermodynamics equation q = mCΔT. We are given the temperature change and specific heat capacity in the question stem, and the mass in the passage.

q = mCΔT = (4 × 10–3 kg)(460 J/kg*K)(200 K) = (8 × 10–1 kg*K)(460 J/kg*K)

                   = 0.8 × 460 J = 368 J

For this question, you could also round 460 to 500 on test day given how spaced out the answer choices are. 

q = mCΔT = (4 × 10–3 kg)(500 J/kg*K)(200 K) = (8 × 10–1 kg*K)(500 J/kg*K)

                   = 0.8 × 500 J = 400 J which is an overestimation because we rounded up and is closer to 368 J than to 550 J.

  1. 368 J – This is what we calculated.
  2. 550 J – Double check your arithmetic, this is too large.
  3. 1840 J – This is also too large. Double check your math.
  4. 3680 J – This answer choice is off by an order of 10. Verify the math for your exponents. Answer choice A is the correct answer.

37) This question is multifaceted. We’ll need to calculate the proportional increase in resistance relative to temperature and then multiply the original current by the new proportion. According to Ohm’s law, resistance and current are inversely related so if the resistance increased by the same proportion as the temperature (which it should because we established that the two have a positive linear relationship), then the current should decrease by the same proportion.

Estimating the proportional increase in T we get 673/293 ≅ 700/300 = 7/3 or 2.3 so the R should increase by a factor of 2.3 and the current should decrease by a factor of 2.3.

We can do the math and calculate that 2.3*2A=4.6A, but looking at the answer choices we can see that the two values used are the 2A current and a larger 4.6A current so no more calculations are actually needed.

  1. It remained constant at 2 A. – We know that the current dropped down to 2A so it did not remain constant.
  2. It remained constant at 4.6 A. – We can get rid of this answer choice for the same reason as the one above.
  3. It increased from 2 A to 4.6 A. – The current decreased, it did not increase.
  4. It decreased from 4.6 A to 2 A. – This is the right answer and shows an appropriate decrease in the current as the temperature and resistance increased.

 

Sample Test Chemistry/Physics Section Passage 8

38) The best way to approach this question is to follow the labeled carbons down the figures, shown below highlighted in purple.

  1. C1, C3, and C4. – C1 and C3 are not labeled with carbon-14.
  2. C1, C3, and C5. – C1 and C3 are not labeled with carbon-14.
  3. C2, C3, and C5. – C3 is not labeled with carbon-14.
  4. C2, C4, and C5. – All three of these carbons are labeled with carbon-14.

39)

A.

This is the same structure as we saw above.
B.

This answer has incorrectly reduced the carbonyl; the carbonyl is not involved in the ring formation during equilibrium.

This molecule has too many oxygens and still has a negatively charged oxygen which will want to continue reacting.
D. 

This ring has a lot more ring strain than the correct answer, A.

40) On test day, you should know some of the more common IR spectrum absorption values, including but not limited to alcohol and carboxylic acid as seen on mevalonate.

  1. II only – 1700–1750 cm–1 corresponds to a carboxylic acid which is indeed present on mevalonate. We need to evaluate the other roman numerals and answer choices.
  2. I and II only – 3200–3500 cm–1 is indicative of an alcohol which is present on the opposite end of mevalonate as the carboxylic acid. We also know that roman numeral II is correct. We’ll keep reading and evaluating.
  3. II and III only – We know that roman numeral I is correct so this answer choice is incorrect. Also, 1580–1610 cm–1 corresponds to an aromatic C=C bond which is not present on mevalonate.
  4. I, II, and III – Roman numerals I and II are correct, but roman numeral III does not apply meaning B is the right answer.

41) This can be a tough question. We’re very used to chiral reactants producing chiral products but neither of the reactants in the above reaction are actually chiral…

  1. One of the reactants is chiral. – As noted, neither of the reactants are chiral.
  2. Both reactants are chiral. – Like answer choice A, this answer can be ruled out because neither reactant is chiral. 
  3. The solvent medium is chiral. – At first glance, this answer choice sounds good—a chiral solvent would promote chiral reactions. However, keeping in mind the context of the reaction, an in vivo reaction inside a living organism, this answer choice holds less weight. Maybe answer choice D can provide a more convincing mechanism.
  4. The enzyme is chiral. – This is a great answer because not only would it produce a chiral product, but it also makes sense in the context of human metabolism and organic chemistry. There is a precedent for chiral enzymes—think about how we only use L amino acids in living organisms—making it the most robust and correct answer choice. 

42) Squalene resembles the four fused rings of cholesterol and its steroid hormone derivatives.

  1. Glucose – Glucose is a single ring and is not a steroid hormone. There is no clear relationship with squalene making it a good answer choice for this NOT question.
  2. Testosterone – Testosterone is a steroid hormone and has a structure reminiscent of squalene; it could very well be the product of squalene metabolism.
  3. Cholesterol – Cholesterol structurally resembles squalene.
  4. Cortisone – Like the two answers immediately above, cortisone is a steroid hormone that is structurally similar to squalene. Glucose is the only choice that does not resemble squalene so it should not be a product of squalene metabolism.

 

Discrete Questions

43)

  1. cylindrical. – A cylindrical lens focuses light into a line which does not change the location at which the image is formed. Next.
  2. converging. – A converging lens would make the light rays focus more to the front and further from the retina. On test day, we should know that converging lenses are used to treat farsightedness, or hyperopia. People who are farsighted have an image that is focused past/behind the retina, not in front of the retina.
  3. diverging. – Diverging lenses are used to treat myopia, or nearsightedness and will cause the light rays to focus further back and onto the retina. This is looking great; see what we did there? Okay, we’ll stop with the puns!
  4. spherical. – A spherical lens focuses light into a point. This will not change the distance at which the image is formed. Answer choice C it is!

For more on lenses: https://jackwestin.com/resources/mcat-content/geometrical-optics/thin-lenses

44)

  1. Sublimation under reduced pressure – Sublimation is the conversion of a solid into a gas without first passing through the liquid stage. This would be a useful procedure because no melting or heating would be involved, and the proteins would stay in their current/frozen conformation. It would also successfully remove the water as it turns into a gas.
  2. Distillation using steam – We want to keep the sample otherwise frozen and the protein in its native state. Distillation involves heating the sample which would defeat the purpose of having frozen it.
  3. Extraction with organic solvent – An organic solvent is hydrophobic which means it will not readily interact with water and would be a poor strategy for removing the water from the sample.
  4. Addition of magnesium sulfate – Magnesium sulfate, usually more recognizable as Epsom salt, would not help remove moisture from the solution if it is frozen. We don’t want to have to melt the sample we just froze so this is not a viable option. Answer choice “A” stands for “A”wesome so we’ll keep it.

45) This question is intimidating. We’re being asked to identify a mystery structure with three different classes of substituents that would result from a reflux with an acid. Before looking at the answer choices, let’s pause for a quick moment. The aqueous H2SO4 would promote an acid-catalyzed hydrolysis, so we are looking for a structure that could be hydrolyzed to produce HCN, benzaldehyde C6H5CHO and two molecules of glucose. On to the answer choices.

A.

This is promising. Working from right to left we have a cyanide that could be released as HCN, then a benzene with an oxygen-based bond to the two glucose molecules so hydrolyzing the bond would release an aldehyde and the two glucose molecules.

B.

Here the benzene is sandwiched between the two glucose molecules and the cyanide is attached to the same carbon as where the would-be benzaldehyde is attached. Hydrolysis of this site would not result in HCN and benzaldehyde, and further hydrolysis would not produce a glucose from the bottom right moiety—we would be one carbon short.

C.

Like the molecule in answer choice B, the placement and order of the structures does not lend itself to hydrolysis into the products of interest. While HCN could be produced, benzaldehyde would not be a product. If one of the rings attached to the benzene’s additional carbon were hydrolyzed after the removal of HCN, there would actually be an extra carbon before the oxygen and would not result in benzaldehyde.

D.

Again, we might be able to make HCN but not benzaldehyde—here we are missing the additional carbon attached to the benzene before the oxygen.

46) The best way to work through this question is to consider Archimedes’ principle with the information we are given in the question stem and then determine which answer choice matches the expression we come up with.

  1. Wair/(WairWw). – This expression matches what we calculated above and is the correct answer.
  2. (WairWw)/Wair. – This is the inverse of the correct answer. 
  3. (WairWw)/Ww. – This is the inverse and incorrectly exchanges the weight of the body in air with the weight of the body in water for what should be the numerator.
  4. Ww/(WairWw). – Again, the numerator should be the weight of the body in the air not the weight of the body in the water.

 

Sample Test Chemistry/Physics Section Passage 9

47) The best way to go about this question is to determine the sum of the charges of each of the functional groups on pantothenate and phosphopantothenate.

  1. 0 for pantothenate and 0 for phosphopantothenate – Both pantothenate and phosphopantothenate contain a carboxylic acid and no basic functional groups so they should have a net charge of at least -1.
  2. −1 for pantothenate and −1 for phosphopantothenate – Phosphopantothenate will lose an additional two hydrogens from its phosphate group for a net charge of -3.
  3. −1 for pantothenate and −3 for phosphopantothenate – Pantothenate loses a hydrogen from its carboxylic group and phosphopantothenate loses two hydrogens from its phosphate and one from its carboxylic acid. This is the correct pair of net charges.
  4. −3 for pantothenate and −4 for phosphopantothenate – Pantothenate only loses one hydrogen from its carboxylic acid and phosphopantothenate loses 3 hydrogens total so the charges should be -1 and -3 respectively, like answer choice C.

48)

  1. β-Mercaptoethylamine, phosphopantothenate, dAMP with additional 3′–phosphate – All three components are incorrect here, Coenzyme A does not contain any of these.
  2. β-Mercaptopropylamine, pantothenate, AMP with additional 5′–phosphate – While Coenzyme A does contain pantothenate, it contains ADP not AMP and it should be β-Mercaptopethylamine not -propyl-.
  3. β-Mercaptopropylamine, phosphopantothenate, dAMP with additional 5′–phosphate – Like answer choice A, Coenzyme A does not contain any of these groups.
  4. β-Mercaptoethylamine, pantothenate, ADP with additional 3′–phosphate – Coenzyme A contains all three of these structures (see above) so this is the right answer.

49) In the last paragraph, the passage says “binding is stabilized by interactions between C2 and C4 hydroxyl groups of pantothenate and a carboxylate group of PanK3” which means that PanK3 needs an amino acid with a carboxylic acid that is deprotonated to form a carboxylate. 

  1. arginine. – Arginine is basic and does not contain a carboxylic acid in its side chain.
  2. asparagine. – Asparagine is polar but uncharged and does not contain a carboxylic acid in its R group.
  3. aspartate. – Aspartate contains a carboxylic acid within its side chain, making this the best answer choice.
  4. glutamine. – Not to be confused with glutamate, glutamine is a polar, uncharged amino acid with no carboxylic acid in its R group.

50) To stabilize the negatively charged phosphates in ADP, the amino acids should have positive charges.

  1. Asn9 and Thr10 – While polar, both of these amino acids are uncharged.
  2. His11 and Arg27 – Both of these amino acids are positively charged so they would stabilize the negative charges on ADP.
  3. Asp6 and Arg27 – Aspartate is negatively charged and would cancel out the positive charge of the arginine so the negatively charged ADP would not be stabilized.
  4. Asp6 and His11 – Aspartate is negatively charged and would cancel out the positive charge of the histidine so the negatively charged ADP would not be stabilized. Only B would stabilize the negative charges.

51) The second paragraph gives a hint as to the fate of the NADH: “The pyruvate is then converted to lactate by lactate dehydrogenase with simultaneous use of NADH.” NADH is used; it is converted to NAD+ and loses electrons so it is oxidized. If you’re still unsure, take a look at the lactic acid fermentation image below.

  1. Oxidation of NADH to NAD+ – This is correct; the NADH donates its electrons to pyruvate and is oxidized to NAD+.
  2. Reduction of NADH to NAD+ – The loss of electrons is oxidation, the gain of electrons is reduction. NADH lost its electrons and was oxidized; it was not reduced.
  3. Oxidation of NAD+ to NADH – NADH is being used, not produced. Also, going from NAD+ involves the gain of electrons (reduction), not the loss of electrons (oxidation).
  4. Reduction of NAD+ to NADH – NADH is being used, not produced. Answer A is the correct answer.

 

Sample Test Chemistry/Physics Section Passage 10

52)

Using the conversion factor given in the question stem, the energy of the photons is approximately 224 * 10-16 J. When plugged into E=h/f, the frequency is 3.38*1019 Hz, a little less if we round.

  1. 2.11 × 1035 Hz – The orders of magnitude are much too large in this answer. We might have lost the 10-16 initially in the numerator.
  2. 3.38 × 1019 Hz – This is about what we calculated.
  3. 3.01 × 10–20 Hz – The negative sign in the exponent should tip us off that this is incorrect, even if we are tempted by the 3 out front.
  4. 1.45 × 10–47 Hz – We likely forgot to change the sign of the power in the denominator when we brought it up over the fraction line if we’re tempted by this answer. Choice B is the correct answer.

53) Using figure 2, we can see that half of the original sample at 500 MBq has decayed to 250 MBq after just over 5 hours but before 10 hours. The half-life should be >5hrs and <10hrs.

  1. 6 h – 6 hours falls between the range determined from the figure.
  2. 12 h – 12 hours is greater than 10 hours and is too long. 
  3. 18 h – By 18 hours, more than 75% of the sample has decayed. This is too long.
  4. 24 h –Almost all of the sample has decayed at this point. Choice A is the correct half life.

54) The decay of 99Mo is described in the second paragraph: “The 99Mo decays to 99mTc, releasing a beta particle β and an antineutrino .” 

  1. a photon. – The charged particle released, beta particle β, is an electron. Photon is not the correct answer.
  2. a neutrino. – An antineutrino was emitted and is not charged so it does not answer the question which asks about the charged particle.
  3. an electron. – This is correct, the negatively charged beta particle released is an electron.
  4. a positron. – A positron is positive, not negative.

55)

  1. distinguishing between fluids and tissue. – A standard ultrasound is able to do this so it is not an advantage of the Doppler ultrasound.
  2. measuring the blood flow. – The Doppler effect describes the change in perceived frequency when there is relative motion between a source and an observer. This allows for the measurement of movement, in this case the movement/flow of blood in the heart. Looks good! 
  3. measuring the tissue density. – Tissue density is not dependent on relative motion. This is not an advantage of a Doppler ultrasound.
  4. measuring the heart wall thickness. – Like tissue density, heart wall thickness does not describe relative motion and measuring heart wall thickness is not an advantage of a Doppler ultrasound. Answer choice B is the only option that describes an advantage or additional measurement when taking advantage of the Doppler effect.

56) In the last paragraph, the author mentions that the stationary bike has “a load of 30 W” which gives us the power of the bike (unit of power is Watts). Rearranging the power equation: Power=Work/time –> Work=Power*time

Work = 30W * 3min = 30W * 180sec = 5400 J

  1. 5400 J – This is the correct answer.
  2. 90 J – With this answer, we likely forgot to convert minutes to seconds. Keep track of your units.
  3. 6 J – This answer choice is much too small.
  4. 0.16 J – This answer is also way too small; the correct answer is 5400 J.

 

Discrete Questions

57) A buffer will ideally have a pKa that is +/- 1 pH unit from desired pH.

  1. 2.14. – This pKa is much too acidic for a reaction occurring at a pH of 5.3.
  2. 4.75. – This pKa is 0.55 pH units below the desired pH and so it falls within the 1 pH unit range.
  3. 6.5. – Close, but not quite. A pKa of 6.5 is 1.2 pH units above the desired range so B is a better answer.
  4. 7.0. – This pKa means the buffer is even more than basic than the one before; a pKa of 4.75 is the only pKa that meets the 1 pH unit cutoff.

58) Specific heat can be found by rearranging the q = mCΔT equation. 

q = mCΔT –> C = q/ mΔT = 250J/(0.1kg*5°C) = 250/0.5 J/(kg·°C) = 500 J/(kg·°C)

  1. 125 J/(kg·°C) – This is half of the heat absorbed, not the specific heat.
  2. 250 J/(kg·°C) – This is simply the heat absorbed, not the specific heat.
  3. 375 J/(kg·°C) – This is less than the actual specific heat calculated. Make sure you plugged in the correct values from the question stem.
  4. 500 J/(kg·°C) – Here we have it, a quick calculation and on to the next question.

59) On test day, unless you’re doing well with time and can afford to systematically go through all of the steps for balancing redox reactions, it’s unlikely that the first step you’ll take for a question like this will be to write out and resolve each of the half reactions. That said, we’ll provide the math at the end of the explanation for this question. 

Let’s say you’re short on time or just where you want to be, how can you solve this question quickly and accurately? First, check to make sure the NADH to NAD+ ratios are balanced. Next, make sure the net charges on each side of the equation are balanced and neutral. Lastly, if you still don’t have the answer, make sure that each side has the same number of each atom (so if there are 5 carbons on the left, there are 5 carbons on the right).

  1. CH3C(=O)CO2 + 2NADH → CH3CH(OH)CO2 + 2NAD+ – There is one NAD+ for every NADH, but the net charges on each side do not match. On the left, there is a -1 net charge, but the two positive NAD+s on the right in addition to the negative charge on lactate mean the net charge to the right of the arrow is +1, not -1. 
  2. CH3C(=O)CO2 + 2NADH + 2H+ → CH3CH(OH)CO2 + 2NAD+ – The NADH to NAD+ ratio is correct and the net charges are equal (+1 and +1). However, the net charge is positive. Something must have gone wrong with our addition and balancing of electrons; they should have canceled out when the half-reactions were added. Let’s see if there’s a better answer choice.
  3. CH3C(=O)CO2 + NADH + 2H+ → CH3CH(OH)CO2 + NAD+ – The NADH to NAD+ ratio is correct but the net charges are unequal (+1 and 0 on the left and right respectively). Next.
  4. CH3C(=O)CO2 + NADH + H+ → CH3CH(OH)CO2 + NAD+ – We’ve ruled out the other answer choices so this should be the correct answer. Indeed it is—the NADH to NAD+ ratio is correct, the net charges are equal and neutral, and each side has the same total number of each atom. See below if you’d like to go through the entire process of balancing the reactions. 



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