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MCAT Content / AAMC MCAT Practice Exam 1 Cp Solutions

AAMC FL1 CP [Web]

Exam 1 C/P Solutions: Passage 1

1) Despite being a passage-based question, we should be able to answer this question using external knowledge. To answer this question, we can do a quick overview of chromatography, then we’ll identify the principal factor determining the migration of individual components in the sample. Chromatography is a separation technique that takes advantage of the different products’ solubilities and relative affinities for the stationary phase used. All chromatography works by having a mobile phase (the part that moves), and a stationary phase (the part that stays still) which allow for separation of the different fractions of the original mixture. The higher the adsorption to the stationary phase, the slower the molecule will move through the column. We’re told in our question stem, water absorbed on cellulose (a very polar compound) functions as the stationary phase. Nonpolar molecules will not be attracted to the water molecules attached to the cellulose and will travel further along the paper, while polar molecules will have a high attraction for the molecules and not travel as far.

  1. Hydrogen bonding. Hydrogen bonding to the stationary phase will determine the migration of individual components in the sample. Nonpolar molecules will not be attracted to the water molecules attached to the cellulose and will travel further along the paper, while polar molecules will have a high attraction for the molecules and not travel as far
  2. Solute concentration. Solute concentration does not change the migration of individual components in the sample. The migration is dependent on affinities and hydrogen bonding. The most polar nonpolar molecules will travel furthest. 
  3. Stationary phase concentration. This is going to be similar to answer choice B. Stationary phase concentration is not going to affect the relative migration of individual components in the sample. That migration still depends on affinities. The most polar nonpolar molecules will still travel furthest. 
  4. Thickness of paper. This answer choice is similar to answer choice C. We expect thickness of paper is not going to affect the relative migration of individual components in the sample. The most polar nonpolar molecules will still travel furthest. We can stick with answer choice A as our best answer. 

2) The author goes through Scheme I in the passage to explain the formation of malic acid and subsequently fumaric/maleic acid.

The author then tells us ammonia reacted with fumaric, maleic, and malic acids to afford a prebiological synthesis of aspartic acid. Given this information, we have to decide what assumptions are being made if scientists conclude that aspartic acid was formed by the prebiological synthesis in the passage. 

  1. Aspartic acid is unstable at temperatures below 150°C. This is not an assumption that has to be made for aspartic acid to be formed by the prebiological synthesis in the passage. An increased temperature helps the reaction move forward and that temperature is not necessarily the temperature at which aspartic acid is stable.
  2. All of the malic acid underwent the dehydration reaction to form fumaric/maleic acid. We are told dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid. This does not necessarily mean the entirety of the malic acid underwent the dehydration reaction to form fumaric/maleic acid. We don’t know for sure whether a majority or all of the malic aicd underwent this reaction, but we do know fumaric acid and maleic acid are formed. 
  3. Compound A and cyanide were available on primitive Earth. This is an assumption that has to be made. Malic acid is formed from Compound A and cyanide, so it stands to reason both were available. If the two were not available, malic acid would not have been formed, and ultimately aspartic acid wouldn’t form. This is going to be the best answer choice so far.
  4. The reaction between ammonia and fumaric acid was catalyzed by the presence of water. There is no evidence in the passage that this reaction is catalyzed by the presence of water. Furthermore, we can see where the ammonia group is found in aspartic acid, and there is no hydroxyl group present in that spot in malic acid. We would expect that to be the case in a hydrolysis reaction (catalyzed by the presence of water). We can stick with answer choice C as the correct answer.

3) Although this question is part of the passage-related question set, we can answer it using only our external knowledge.  The retention factor (Rf) of a component is the distance travelled by the component over the distance traveled by the solvent. In this case, we can see the aspartic acid traveled a distance of two units, while the solvent front traveled 10 units. This is going to be a simple division problem:

Rf= 2 units / 10 units = 0.2

The approximate Rf value of aspartic acid is 0.2, or answer choice A.

4) To answer this question, we can go back to the middle of the passage and Scheme 1.

We’re told, “malic acid is thought to have formed via hydrolysis of a cyanohydrin intermediate… Subsequent dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid.” We have to be careful with the verbiage here. We want to know which of the four statements does NOT describe the dehydration of malic acid to fumaric acid and maleic acid which we see along the bottom of Scheme 1. Three of the statements will describe the reaction, while our correct answer will not. This can be answered by finding the answer choice that does not correctly describe the dehydration taking place.

  1. The reaction occurs most readily with tertiary alcohols. There is nothing in the passage that suggests this is not true. The dehydration of malic acid is an E1 reaction and we have a carbocation intermediate. Tertiary alcohols can be a great leaving group in this situation; there is nothing here that implies this is not a correct description.
  2. The reaction involves the loss of a water molecule. This is a dehydration reaction where there is a loss of a water molecule. There is nothing in the statement that implies this is not a correct description.
  3. The reaction has a carbocation intermediate. This is going to be consistent with what we said in our reasoning for answer choice A. Once again, there is nothing in the statement that implies this is not a correct description.
  4. The reaction is stereospecific. We are told in the passage “Subsequent dehydration of malic acid results in the formation of fumaric acid and its cis isomer, maleic acid.” The formation of two products in this case tells us the reaction is not stereospecific. A stereospecific reaction won’t produce a mixture. We can stick with answer choice D as our best answer.

5) This is another question in this same problem set that is going to rely predominantly on our external knowledge. Each amino acid is attached to its neighboring amino acid by a covalent bond, known as a peptide or amide bond. During the formation of this amide 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. The functional group formed is an amide group.

The only answer choice that is consistent with our breakdown here is going to be answer choice C, an amide group. 

 

Exam 1 C/P Solutions: Passage 2

6) This question relies on us revisiting the passage (specifically Figure 3) and using external knowledge about the rate of substitution of different reactions. 

Figure 3 is the SN2 Reaction of 1-pentanol and we have a 70%/30% product distribution as shown above. The author tells us later in the passage, “The rate of the reaction shown in Figure 3 increases as the concentration of alcohol, halide, or H+ increases.” We’re replacing 1-pentanol with 2-pentanol which is a secondary alcohol. We can visualize pentane with a hydroxyl group at position 2:

  1. the C–O bond in 2-pentanol is stronger than the C–O bond in 1-pentanol. The C-O bond is not going to be stronger in the different alcohols. Therefore, the rate of substitution is not going to differ because of this reason. 
  2. there is a competing elimination reaction that slows the rate of substitution. We’re told in the question stem we’re looking at the SN2 reaction in Figure 3, only with 2-pentanol. SN2 reactions favor primary alcohols more than they do secondary alcohols, so the rate of substitution will not slow.
  3. there is more steric hindrance at the oxygen atom in 2-pentanol than in 1-pentanol, making protonation less likely. There is not more steric hindrance at the oxygen atom itself in 2-pentanol; the 2-position of 2-pentanol has more steric hindrance, but not at the oxygen atom. We expect the likelihood of protonation will not change.
  4. there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol. This is going to tie into our reasoning for answer choice C. A secondary alcohol has been substituted for a primary alcohol. There is more steric hindrance at the 2-position, which we can see comparing Figure 3 and our 2-pentanol visual. Steric hindrance ultimately decreases the rate of substitution. Answer choice D is going to be our best option. 

7) To answer this question, we can pull up Reaction 2 from the passage. 

We’re told the reaction is repeated with HCl and the compound from the question stem. We want to go through our four answer choices and find the one that is NOT a direct product (without rearrangement). That means three of the answer choices will be direct products, while our correct answer will not.

  1. This answer choice is not going to be a direct product. I want you to pay attention to the verbiage in the question stem. We’re explicitly told “without arrangement.” How is answer choice A made? The carbocation shifts, or in other words, a rearrangement. This is going to be a strong answer choice.
  2. The way to get answer choice B is through an elimination reaction. Our hydroxyl group is protonated and chlorine ion deprotonates an adjacent hydrogen. 
  3. This is possible through an E1 and SN1 reaction. We stick with answer choice A as our best answer.
  4. This is possible through an E1 and SN1 reaction. The only answer choice that is not a direct product is going to be answer choice A. 

8) To answer this question, we can do an overview of gas chromatography, then we can decide the compound responsible for the first peak observed in the gc trace.

Gas chromatography involves samples being vaporized and passed through a liquid or solid stationary phase using a gaseous mobile phase. The stationary phase lets polar molecules elute more slowly. The molecules with the lowest boiling points come out of the column first. The molecules with the higher boiling points come out of the column last. The test-maker asks, “the first peak observed in the gc trace is attributable to which compound?” In other words, which substance will migrate the fastest?

  1. 2-Methyl-2-butanol. This is going to be the reactant in our reaction above. We have the ability to hydrogen bond given the hydroxyl group, and the polarity of the molecule compared to the alkene in the products is going to make this a weak answer choice. 
  2. 2-Methyl-2-butene. This is going to be the only alkene in the reaction above, and it is also nonpolar; alkenes have only carbon-carbon and carbon-hydrogen bonds. I mentioned the stationary phase lets polar molecules elute more slowly, so the nonpolar alkene would elute the fastest. This is looking like a strong answer choice. 
  3. 2-Chloro-2-methylbutane. Adding a chlorine to the hydrocarbon increases molecular weight and polarity as well. That rules out answer choice C.
  4. 2-Bromo-2-methylbutane. Reasoning here is going to be the same as answer choice C. Adding that bromine to the hydrocarbon increases molecular weight and polarity as well. We can stick with answer choice B as our best answer. 

9) To answer this question, we will once again reference the passage, and specifically Figure 3.

First thing we can do is figure out the configuration on Compound 1 by assigning priority to the different substituents. Hydroxyl group will be first as oxygen has the highest atomic number. The carbon chain will be second as carbon has the second highest atomic number. Third is the deuterium (more neutrons than hydrogen), and lastly hydrogen is priority 4. The original compound is in (R) configuration. We have an SN2 reaction and an inversion of this configuration; our answer is going to be (S) configuration. The bromine from HBr replaces our hydroxyl group on the first carbon. We get (S)-1-bromo-1-deuteriopentane, or answer choice C.

 

Exam 1 C/P Solutions: Questions 10-13

10) We have acetic acid on the left side of our reactants and we have ethanol on the right side of our reactants. To determine which reactant loses the -OH group, we would want to label either of the oxygens listed at the right side of each compound.

By doing this, we can determine which oxygen bonds to form the ester at the carbonyl carbon in our product, or leaves as a part of water. Glancing at our answer choices, we know replacing the hydrogens is not going to help best determine which reactant loses the -OH group. Answer choices A and B are not going to be correct. The carbonyl oxygen is not going to help us determine which reactant loses the -OH group. Instead, we want to focus on one of the oxygens in our visual. The only answer choice that matches our visual is going to be answer choice D. 

11) This is a standalone question and the author actually offers us a hint about how to approach this question. We’re told we should use the thin lens formula. We’re given a lens-to-retina distance and told the distance at which the person can clearly see objects. Let’s use the thin lens formula here and find the strength of the person’s eye lens:

As with any math or numbers-related question, we want to make sure we have the correct units. We can convert the lens-to-retina distance to meters using dimensional analysis:

2.0 cm x (1 m / 100 cm) = 0.02 m

We can plug in our values to our thin lens formula with O = 1.0 m and I = 0.02 m 

S= 1/1m + 1/0.02 m = 1 + 50 = 51 D

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

12) This is a standalone question that focuses on enzymes. We can use our external knowledge and what we know about enzymes to answer this question. 

The mechanisms of enzyme catalysis vary, but they are all similar in that they increase the reaction rate by reducing the activation energy. When an enzyme binds its substrate, it forms an enzyme-substrate complex. After binding between the enzyme and substrate takes place, one or more mechanisms of catalysis lowers the energy of the reaction’s transition state by providing an alternative chemical pathway for the reaction. 

The enzyme-substrate complex lowers the activation energy of the reaction and promotes its rapid progression by providing certain ions or chemical groups that actually form covalent bonds with molecules as a necessary step of the reaction process. Enzymes also promote chemical reactions by bringing substrates together in an optimal orientation and lining up the atoms and bonds of one molecule with the atoms and bonds of the other molecule. This can contort the substrate molecules and facilitate bond-breaking. The active site of an enzyme also creates an ideal environment, such as a slightly acidic or non-polar environment, for the reaction to occur. 

  1. the substrate changes the free energy of the reaction. This question is focusing on how enzymes affect chemical reactions. Enzymes function to lower activation energy; substrates do not change the free energy of the reaction.
  2. the transition state changes the free energy of the reaction. Enzymes will lower the energy of the reaction’s transition state so the reaction can proceed with less energy. The transition state itself is not changing the free energy of the reaction.
  3. the substrate changes the activation energy of the reaction. The enzyme functions to reduce the activation energy of the reaction; substrates do not change the activation energy of the reaction. 
  4. the transition state changes the activation energy of the reaction. After binding between the enzyme and substrate takes place, one or more mechanisms of catalysis lowers the energy of the reaction’s transition state by providing an alternative chemical pathway for the reaction. In other words, stabilization of the transition state provides the binding energy needed to reduce the activation energy of the reaction. We can stick with answer choice D as our best option.

13) This is a standalone question and the first thing we can do here is identify acetic acid as a weak acid. Ionization occurs when acetic acid yields a H+ ion and a conjugate base. In order to decrease the percentage ionization of acetic acid, the easiest method would be to add a strong acid. Why is that? Recall strong acids dissociate well, but weak acids do not. A strong acid will dissociate completely in solution, but acetic acid will not. What does the strong acid have to do with acetic acid? We can look at Le Chatelier’s principle. According to Le Chatelier’s principle, if we add an additional product to a system (like H+ ions), the equilibrium will shift to the left in order to produce more reactants. What strong acids might we introduce to dissociate better than acetic acid? We can consider our list of strong acids and bases:

Strong Acids Strong Bases
HCl- Hydrochloric acid LiOH- Lithium hydroxide
HBr- Hydrobromic acid NaOH- Sodium hydroxide
HI- Hydroiodic acid KOH- Potassium hydroxide
HNO3– Nitric acid RbOH- Rubidium hydroxide
HClO3– Chloric acid CsOH- Cesium hydroxide
HClO4– Perchloric acid Ca(OH)2– Calcium hydroxide
H2SO4– Sulfuric acid Sr(OH)2– Strontium hydroxide
  Ba(OH)2– Barium hydroxide
  1. Chlorinating the CH3 group. This can have the opposite effect where ionization actually increases with the addition of the chlorine. 
  2. Diluting the solution. Diluting the solution will decrease the concentration, but it shouldn’t change the percentage ionization of acetic acid. 
  3. Adding concentrated HCl(aq). This ties into our breakdown. Recall strong acids dissociate well, but weak acids do not. A strong acid will dissociate completely in solution, but acetic acid will not. If we have the strong acid dissociating, acetic acid will not ionize as much (common ion effect) like we mentioned in the breakdown of the question. This is a strong answer choice.
  4. Adding a drop of basic indicator. The key word here is indicator. An indicator can be helpful when experimenting or studying reactions, but the indicator isn’t going to do much in terms of changing the pH or the percentage ionization of acetic acid. We can stick with answer choice C as our best answer. 

 

Exam 1 C/P Solutions: Passage 3

14) This question involves knowing the relationship between variables. The MCAT is not a mathematics test. While you may have to do some level of addition, subtraction, division, and multiplication, it is far more important to know the relationships between the numbers with which you’re dealing. This question is a perfect example of that. We’re not told specific values for current or resistance, but we want to know how voltage is affected. How do we do that without knowing any numerical values? By understanding the relationship between these variables. When you see current, resistance, and voltage, you should be thinking Ohm’s law from your external knowledge:

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated by using the equation V=IR, where I is current in amps (A) and R is the resistance in ohms (Ω). 

V=IR. When we have an increase in current, I, and resistance, R remains constant, we should see an increase in voltage, V. If the right side of the equation increases, the left side also increases. The only answer choice that matches this is answer choice B: Increases.

15) This is going to be along the same reasoning as the first question within this question set. We’re focused on the relationship between variables which is ultimately going to help us use the given numerical values to solve the problem. We’re given the speed of light in a medium (vitreous humor) and the speed of light in a vacuum. We can use the relationship between these two values to solve for the index of refraction of the vitreous humor.

The index of refraction is n=c/v, where c is the speed of light in vacuum (3.0 x 108 m/s), v is the speed of light in the material (2.1 × 108 m/s in the vitreous humor),  and n is the index of refraction (which we’re solving for):

n = (3.0 x 108 m/s) / (2.1 × 108 m/s)  = 1.43

  1. 0.7 This answer choice is an incorrect value. Answer choice A would be the correct answer if you incorrectly rearranged the formula as n=v/c
  2. 1.4 This answer choice matches our calculated value. Answer choice B is our correct answer.
  3. 2.1 This answer choice incorrectly lists the integer in the speed of light in the vitreous humor.
  4. 3.0 This answer choice incorrectly lists the integer in the speed of light in a vacuum.

16) This question is testing whether we understand the information given to us in the passage, and specifically the research. The basis of this question is similar to the other questions in the set as we’re focused on relationships between variables. 

We’re asked about intensity of the radiation in this situation, which we should realize is going to be reliant on the number of photons emitted in a time frame. When we see intensity, we’re thinking a greater number. A higher intensity corresponds to a greater number of photons in that time, while a lower intensity corresponds to fewer photons. The energy of a photon is equal to the product of Planck’s constant and frequency of EM wave. The energy is quantized, so each photon is not going to be more intense, but rather each one described in our passage has a given frequency and energy. Instead, multiple photons emitted will increase the intensity. Answer choices A-C are not directly proportional to the intensity of the radiation emitted. The only answer choice that is consistent with our breakdown is going to be answer choice D.

17) A lot of the more physics-centric questions you see on the exam have to do with relationships between variables and understanding equations. Understanding equations means knowing proper units and being able to use dimensional analysis to go between units. That’s what we’re relying on here. We’re trying to solve for the energy of the photons emitted by the LED in the passage, and we’re give some numerical values.

The energy of a photon is equal to the product of Planck’s constant and frequency of EM Wave:

E = hf

We’re given a value for h, but we want to be careful about the units in frequency. We’re told we have a frequency of 610 THz. Terahertz is equivalent to 1012 Hz, so we have 610 x 1012 or 6.1 x 1014 Hz. Plug in our numbers: 

E = (6.6 × 10–34 J·s) x (6.1 x 1014 Hz) = 4.03 x 10-19

One thing I do want to point out here is that even if you didn’t know how to convert THz to Hz, you could still solve for the integer in the scientific notation, 4.0, by knowing the formula for energy and what each number in the question stem represents. This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D. 

 

Exam 1 C/P Solutions: Passage 4

18) This is a passage-related question, but it relies on our knowledge of algebra. Said differently, this question involves knowing the relationship between variables. The MCAT is not a mathematics test. While you may have to do some level of addition, subtraction, division, and multiplication, it is far more important to know the relationships between the numbers with which you’re dealing. This question is a perfect example of that. We’re not told specific values, but we do have to manipulate the two equations in the passage to find a relationship between PpPMPA, and Pd. Note, in all of our answer choices we have Pp isolated on one side of the equation. We can pull up Equation 1 and Equation 2.

First thing we want to note is the parentheses in Equation 1 are equivalent to the two sides of Equation 2. That means we can substitute Pp into the parentheses in Equation 1:

PMPA = Pd + (Pp)/3

Multiply everything by 3 to get rid of the 1/3 multiplying Pp:

3 PMPA = 3Pd + Pp

Isolate Pp:

Pp = 3 PMPA – 3Pd

The only answer choice that correctly lists the relationship between PpPMPA, and Pd is answer choice C. 

19) This question relies on our external knowledge. We’re told in the passage:

“The maximum and minimum blood pressures of a healthy adult (the systolic Pand the diastolic Ppressures, respectively) are about Ps  = 120 mmHg and Pd = 75 mm Hg.”

What is the pulse pressure given these values of systolic pressure and diastolic pressure? Pulse pressure is defined as the difference between systolic and diastolic blood pressure. We know Ps  = 120 mmHg and Pd = 75 mm Hg, so the difference between the two is 45 mmHg

Next, we want to relate that pulse pressure to standard atmospheric pressure. We know that from our content as 760 mmHg.

Dividing the pulse pressure by standard atmospheric pressure to get a percentage gives us:

45 mmHg / 760 mmHg = 5.92%. Glancing at our answer choices, this is only close to one answer choice, answer choice B: 6%. 

20) This question is going to rely on finding information about half-life in the passage, then using our external knowledge to determine the fraction of a sample remaining after 10 minutes.

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. We’re told in the passage the half-life of 15O is 2 minutes, therefore 10 minutes is 5 half-lives. We can multiply ½ x ½ x ½ x ½ x ½ or we can solve for (½)5. Both are going to tell you there is going to 1/32nd of the initial sample remaining. However, be careful with the verbiage here. The test-maker wants to know what fraction of the sample decays. We have 1/32nd of the initial sample remaining, meaning 31/32nd of the sample has decayed. We’re left with the only answer choice that matches this fraction: Answer choice D.

21) This question is related to our passage, but once again is going to rely on external knowledge and knowing the relationships between different variables. Figure 1 tells us the relationship between the brachial artery blood flow rate of a healthy adult and the power generated for body motion. First thing we’ll do is find the brachial artery blood flow rate so we can solve for power generated using Figure 1. We’ll subsequently take that power generated and multiply by the time frame to solve for our work.

Even if you didn’t know how to solve for blood flow rate, we can look at the units along the Y-axis of Figure 1 (mL/min). We’re give blood in liters in our question stem, so we can use dimensional analysis to convert to mL. 1000 mL in 1 L tells us the adult circulates 9000 mL of blood / 10 minutes. Simplifying here gives us 900 mL/min. If we find the 900 ml/min value along our Y-axis, we can find the corresponding value along the curve on the X-axis, or roughly 200 W.

Work is power x time, so we can multiply the power generated (200 W) by time (600s) to get 120,000 J or 120 kJ. This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D. 

 

Exam 1 C/P Solutions: Passage 5

22) This was a detail-heavy passage and this question is asking us to pick out some key details about Compound 1 and Compound 2 from the passage. We know molecules of Compound 1 spontaneously self-assemble into cylindrical micelles in water at pH 4. Oxidation of self-assembled, micellular Compound 1 with iodine resulted in a solid material (Compound 2). That should be our first clue about why Compound 1 is structurally more rigid.

Compound 2 is a solid and is formed by oxidation of self-assembled, micellular Compound 1 with iodine. We’re given the structure of Compound 1.

We see no shortage of cysteine residues here which we typically want to associate with covalent disulfide bonds on the MCAT. Cysteine residues get oxidized to form disulfide bonds, but if you forget that, ask yourself: How do we break disulfide bonds? Reducing agents. Now, the formation of these disulfide bonds is done using an oxidant. The oxidation of Compound 1 is what ultimately led to the structurally rigid Compound 2.  

  1. Intermolecular hydrogen bonding. We talked about the cysteine residues getting oxidized which caused the more rigid structure. This does not indicate stronger intermolecular hydrogen bonding.
  2. Intermolecular covalent bonding. This answer choice is consistent with our breakdown. We know we had the formation of disulfide bonds in the transition from Compound 1 to Compound 2 which is an example of intermolecular covalent bonding. 
  3. Intramolecular hydrogen bonding. Intramolecular forces hold atoms together in molecules, but intermolecular forces exist between molecules. Molecules of Compound 1 eventually form Compound 2, so we’re sticking with an answer choice that mentions intermolecular.
  4. Intramolecular covalent bonding. Reasoning here is the same as answer choice C. Intramolecular forces hold atoms together in molecules, but intermolecular forces exist between molecules. Molecules of Compound 1 eventually form Compound 2, so we’re sticking with an answer choice that mentions intermolecular.

23) This question relies on knowing a topic that is very high-yield, amino acids. Make sure you take the time to go through all of your amino acids and review their structures, properties, and abbreviations.

We can utilize the visual above that shows all of the necessary structures and we can break down Compound 1.

The author describes five regions:

  1. Hydrophobic at the left side of the compound
  2. Cross-linking is our cysteines which were touched on in a previous question as well
  3. Spacers are glycine
  4. Phosphorylation we can identify serine on the top side of Figure 1
  5. Residues with affinity for cells leaves asparagine, arginine, and glycine residues along the rightmost part of Figure 1. Glycine residues will not be useful for adhesion of cells on the scaffold surfaces. Ultimately, we can use the information from Figure 1 to pick our correct answer, answer choice D: Asp and Arg. Make sure you review your amino acids!

24) I love amino acid questions from AAMC because they should be easy points. Not because learning amino acids themselves is inherently easy, but because amino acids are so high-yield that there’s no downside to knowing them well. 

We have our visual above that shows the structures of the amino acids. We also want to pick out the pertinent information from the passage.

The author describes five regions:

  1. Hydrophobic at the left side of the compound
  2. Cross-linking is our cysteines which were touched on in a previous question as well
  3. Spacers are glycine
  4. Phosphorylation we can identify serine on the top side of Figure 1. This is going to be our correct answer.
  5. Residues with affinity for cells shows asparagine, arginine, and glycine residues along the rightmost part of Figure 1.

Reviewing the information from the passage and knowing our amino acid structures allows us to pick the only answer choice here that matches our breakdown, answer choice A: Serine.

25) To answer this question, we’ll first look at the key parts of the passage that clue us into the reasoning behind the pH-dependent self-assembly. 

We want to determine what causes that aggregation, or that self-assembly into cylindrical micelles. Biggest clue we have is the author mentions, “Molecules of Compound 1 spontaneously self-assemble into cylindrical micelles in water at pH 4 but freely redissolve at pH 8.” We’re thinking pH here. Big takeaway from Equation 1 above is that molecules of Compound 1 spontaneously self-assemble into cylindrical micelles in water at the lower pH, but freely redissolve at the higher pH. 

  1. Thiol side chains that can hydrogen bond. On the MCAT we reserve hydrogen bonding for the FON atoms, not sulfur. This is not going to influence pH like we’re looking for in our breakdown of the question.
  2. Long alkyl tail that exhibits predominantly London forces. The alkyl tail conveys hydrophobic character to the molecule, but in the grand scheme of things here the London forces are not responsible for the pH dependence of aggregation. Think back to the previous question in this set that talked about the different forces in play.
  3. Side chains whose net charge responds to pH. At a higher pH (8), side chains on cys and asp deprotonate and become negatively charged. Lowering the pH (4) has the opposite effect where these groups become protonated and we don’t have these negative, repulsive charges. At the lower pH we have aggregation. This is a solid answer choice.
  4. Covalent linkages that reversibly hydrolyze. We talked about covalent bonding earlier in the question set. In Question 22 we highlighted the formation of disulfide bonds in the transition from Compound 1 to Compound 2, which is an example of intermolecular covalent bonding. However, key to this question is the pH dependence. This answer choice is not as relevant as our best answer, answer choice C. 

 

Exam 1 C/P Solutions: Questions 26-29

26) This is a standalone question that is going to rely on us visualizing the question being asked and using external knowledge about flow of a fluid. Blood is incompressible, meaning the rate of flow into an area must equal the rate of flow out of an area. This is known as the equation of continuity. The equation of continuity can show how much the speed of a liquid increases if it is forced to flow through a smaller area. For example, if the area of a pipe is halved, the velocity of the fluid will double. As we’re answering, we’re going to make sure to keep blood circulation in mind as well.

  1. Capillary walls are more elastic than arterial walls. We know the opposite to be true. Capillaries are smaller than other blood vessels and have single-celled walls that are not elastic.
  2. Capillaries have less resistance to blood flow than arteries. Numerous factors can alter resistance, but the three most important are vessel length, vessel radius, and blood viscosity. With increasing length, increasing viscosity, and decreasing radius, resistance is increased. We know capillaries have the smallest radii of the blood vessels, so we see increased resistance. 
  3. The total cross-sectional area of capillaries exceeds that of arteries. Note this says total cross-sectional area of capillaries, and not a single capillary versus a single artery. I mentioned in our breakdown the equation of continuity can show how much the speed of blood increases if it’s forced to flow through a smaller area. The relatively higher number of capillaries in the body means a larger cross-sectional area and decreased velocity. The relatively lower number of arteries and cross-sectional area means the velocity of blood flow is faster in arteries. This is going to be our best answer.
  4. Blood pressure is higher in the capillaries than in arteries. We can use our blood circulation visual to determine this is false. Blood pressure is higher in arteries than it is in capillaries. We can stick with answer choice C as our best answer.

27) This question is one of my favorite types of AAMC questions because the answer is going to come from correctly reading the graph. We’re given a Pourbaix diagram and told we’re seeing the relationship between the predominant form of iron as a function of solution pH (changing along the x-axis) and applied potential (changing along the y-axis). Even if you’re unfamiliar with Pourbaix diagrams, AAMC wants to make sure you’re able to interpret data that is presented to you and find the true statement.

  1. At a potential of –0.4 V, as pH increases, Fe2+ is reduced and precipitates as Fe(OH)3. Go along the X-axis and find the potential of -0.4 v. From left to right as pH increases, we go from Fe2+ to Fe(OH)2, not Fe(OH)3
  2. At a potential of –0.44 V, the equilibrium between Fe and Fe2+ is independent of solution pH below pH 6. This is consistent with what we see in the diagram. Note at a potential of -0.44 V, we’re seeing our iron essentially on the line between Fe and Fe2+. That is independent of solution pH below pH 6.
  3. At pH = 1, as the potential is changed from –0.2 to +0.8, Fe3+ is reduced to Fe2+. We can look at pH=1 and potential of -0.2. As we go up along the pH=1 and potential increases to +0.8, we go from Fe2+ to Fe3+ and not the other way around. Answer choice B remains the best option.
  4. At pH = 8 and V = –0.1 V, Fe(OH)2 is the predominant form of iron. At pH=8 and V=-0.1 V, Fe(OH)3 is the predominant form of iron. The only answer choice that is consistent with the diagram is the correct answer, answer choice B. 

28) This is a standalone question that is going to involve visualizing compounds. To answer this question, we can consider octahedral geometry:

A central atom with six sigma bonds is octahedral in shape and is d2sp3 hybridized: 2d+1s+3p = 6 hybridized orbitals. Whether you have these hybridization geometries memorized or not (I recommend you do), you should be able to recall and contrast geometries. The only answer choice here that matches our breakdown is going to be answer choice D. 

29) This is a standalone question that relies entirely on external knowledge. AAMC will often ask question about fairly common topics we covered in our undergraduate courses and content review (in this case ideal gases), but make sure we really understand different parts of the topic. We have to be able to identify three incorrect assumptions, while our correct answer is going to be an assumption that is made about ideal gases. An ideal gas is a gas whose randomly-moving particles exhibit no attractive interactions; at high temperatures and low pressures, gases behave close to ideally. A gas is considered ideal if its particles are so far apart that they do not exert any attractive forces upon one another. In real life, there is no such thing as a truly ideal gas, but at high temperatures and low pressures (conditions in which individual particles will be moving very quickly and be very far apart from one another so that their interaction is almost zero), gases behave close to ideally.

  1. The law PV = nRT2 is strictly obeyed. This is incorrect, the law is PV = nRT. The small exponent is the difference between getting this answer correct vs incorrect, so make sure to read carefully.
  2. Intermolecular molecular forces are infinitely large. This is not an assumption for an ideal gas, but rather we there are no attractive forces upon one another. 
  3. Individual molecular volume and intermolecular forces are negligible. This is true of an ideal gas. Intermolecular forces are negligible, and individual molecular volume is also negligible.
  4. One gram-mole occupies a volume of 22.4 L at 25°C and one atmosphere pressure. STP means one mole of gas occupies a volume of 22.4L at 0°C and one atmosphere pressure, not 25°C. Answer choice C is going to be our correct answer choice.

 

Exam 1 C/P Solutions: Passage 6

30) This is a passage-based question that relies on external knowledge to explain something. We were told in the passage about the variety of opsins.

The question comes down explaining why there are additional opsin-based receptors expressed in specialized cone cells with different maximum absorptions. The one thing that stands out here is the mention of cones, and different wavelengths in the visible light range. That should steer your thinking toward different colors. 

  1. increase sensitivity to low light. We’re told by the author that the variety of opsins are expressed in specialized cone cells. If we continued to focus on rods in that part of the passage, this might be a viable answer, but for now, it’s not something that stands out.
  2. enable the detection of different colors. This is exactly what we mentioned in our breakdown. The author explicitly mentions cones and maximum absorptions that vary in the visible light range. That should point you towards an answer having to do with the detection of different colors. 
  3. ensure fast recovery of 11-cis-retinal after exposure. There is no evidence of the variety of opsins functioning to ensure recovery of 11-cis-retinal that has been exposed to light. Answer choice B remains superior.
  4. increase refractive index of the eye lens. This is out of scope. There is nothing in the passage that suggests an increase of the refractive index of the eye lens. We can stick with answer choice B as our best answer.

31) While this question is technically passage-related and part of a question set, this can likely be answered using the structural representation of 11-cis-retinal, and our external knowledge.

Recall we’re only focused on the covalently attached enzyme in this case. Based on the structure above, we need to determine the most likely environment of the retinal binding site. 

  1. hydrophilic. There is nothing in the Figure 1 structure that suggests the environment of the retinal binding structure is most likely hydrophilic. The excessive numbers of carbons and hydrogens suggest nonpolar bonds, so the opposite of answer choice A is true.
  2. positively charged. The compound shown in the passage is neutral, so there is nothing to suggest the environment of the retinal binding site is most likely positively charged. 
  3. negatively charged. Reasoning here is going to be similar to answer choice B. The compound shown in the passage is neutral, so there is nothing to suggest the environment of the retinal binding site is most likely negatively charged. Because the compound is neutral, that also makes it easier to eliminate both B and C together.
  4. hydrophobic. This answer choice is consistent with the structural representation of 11-cis-retinal and the reasoning we used in answer choice A. We have a hydrophobic molecule; the excessive numbers of carbons and hydrogens suggest nonpolar bonds. The environment of the retinal binding site is most likely answer choice D: hydrophobic. 

32) This is going to be a passage-based question where we have to dive into some details about the phosphate groups added to rhodopsin. The author mentions the following in the passage.

We associate kinases with phosphorylating a molecule, and rhodopsin kinase catalyzes the transfer of a phosphate group from ATP. The source of the phosphate group is therefore ATP.

  1. arrestin. We know that in the absence of phosphorylation, arrestin binding to wild-type rhodopsin decreased by 90%. Arrestin is not the one providing the phosphate groups that are added to rhodopsin.
  2. rhodopsin kinase. Rhodopsin kinase catalyzes the transfer of a phosphate group from ATP, but the kinase itself is not the source of the ATP. Note the distinction and be careful with the verbiage in the question stem.
  3. ATP. This is consistent with our breakdown of the question. We associate kinases with phosphorylating a molecule, and rhodopsin kinase catalyzes the transfer of a phosphate group from ATP. The source of the phosphate group is therefore ATP
  4. all-trans-retinol. We can reference Figure 1 in the passage and note that All-trans-retinol does not have a phosphate group. We can eliminate this answer choice. Answer choice C remains our best answer.

33) To answer this question, we can consider the carbon atoms involved in the conversion first. Once we find the carbon atom, we can determine the geometry we’re dealing with and the corresponding hybridization geometry.

The cis/trans carbon we’re focused on has 3 bonded atoms and no lone pairs, so the preferred geometry is trigonal planar. All three bonds are 120o apart and sp2 hybridized; there are three sigma and one pi bond. The only answer choice that matches our breakdown is answer choice B: sp2.

 

Exam 1 C/P Solutions: Passage 7

34) While this question is technically passage-related and part of a question set, this can likely be answered using external knowledge. LipA is lysosomal lipase. We should know from our general knowledge that lipases are hydrolase enzymes that catalyze the hydrolysis of fats. We can use our knowledge of different enzyme types to find the reaction that is most consistent with the hydrolysis of fats.

  1. ATP hydrolysis. This is out of scope. LipA is a lipase which catalyzes the hydrolysis of fats. ATP is not a fat.
  2. Peptide bond cleavage. Normally this is done by peptidases or proteases, not lipases. Lipases catalyze the hydrolysis of fats.
  3. Hydrolysis of triacylglycerides. This matches our breakdown exactly. Lipases are a subclass of esterases. We should know from our general knowledge that lipases are hydrolase enzymes that catalyze the hydrolysis of fats. This is going to be our best option.
  4. Transfer of carboxyl groups. Transferases catalyze reactions in which groups are transferred from one location to another. We can stick with answer choice C as our best option. 

35) To answer this question, we can pull up the necessary information from Table 1 that showed the different variants of LipA:

Variant Amino acid substitutions
XI R33G, K112D, M134D, Y139C, I157M

This question relies on knowing a topic that is very high-yield, amino acids. Make sure you take the time to go through all of your amino acids and review their structures, properties, and abbreviations.

We can note the net charge differences between the wild type and variant substitutions here, one at a time:

  • R33G -1 difference
  • K112D -2 difference
  • M134D -1 difference
  • Y139C No change
  • I157M No change

We have a combined change in net charge of -4. This was ultimately a basic math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D: -4.

36) This is a passage-based question that relies on our ability to analyze data. As I read through the passage, I like to note any trends and/or outliers (no need to memorize any specific details). When you go back to the passage to answer the question, you should already have a sense of what you’re looking at. The author says the enzymatic activity of the proteins was measured at 37°C, and then again after the protein was briefly exposed to higher temperatures and cooled down to 37°C. The results are in Figure 1.

Figure 1 shows enzymatic activity retention (% on y-axis) as a function of exposure temperature (oC on X-axis) for wild-type and variant LipA proteins. Wild-type and each variant are labeled for us in the legend. We have a pH 7 buffer in A, and a pH 4 buffer in B. Even if this sounds redundant, taking a few seconds to make sure you understand the visual before jumping into the answer choices ensures you’re giving yourself the best chance at identifying the correct answer.

  1. the unfolding of LipA proteins is a non-reversible process. If we look at the activity retention in graph B, we see activity approach or hit near 100% clearly. This is not a conclusion we can make from the data in Figure 1. 
  2. exposure to high temperature eliminates any variant protein activity. We can see from the Figure this is not true. Activity is retained to some extent in the variants, even at higher temperatures.
  3. variation in the flexible regions of LipA decreases the temperature sensitivity of activity. We can see this clearly in graph A. All three variants maintained higher activity at higher temperature. The variants are allowing LipA to not lose activity after exposure to the higher temperature like we see in the wild-type. This is going to be our correct answer. 
  4. wild-type LipA loses all activity when first heated to 50°C. This is an extreme answer. We still have activity at 50°C, so we can eliminate this answer choice. We can stick with answer choice C as our correct answer. 

37) We’re going to need to gather some key information from the passage about the different pHs in the experiments. Our answer is going to come down to the pH difference. We are told we have a pH 7 buffer in 1A, and a pH 4 buffer in 1B. pH is the negative logarithm to base 10 of the proton concentration. That means a pH change of 1 is actually a ten-times difference. In this case, the experiment in Figure 1B uses pH of 4 while the experiment in Figure 1A uses pH of 7. A pH difference of 3 means the proton concentration increases by a factor of 103, or 1000. This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D: 1000.

 

Exam 1 C/P Solutions: Passage 8

38) This is a passage-related question that requires us going back to the first paragraph and pulling out some key information.

We’re told glycogen will contain phosphate groups covalently linked to the C2 and C3 positions as monoesters. Now this question is as simple as finding a structure that shows phosphate groups at positions C2 and C3. The only answer choice that is consistent with the description in the passage is answer choice D.

39) To answer this question, we will need a few key pieces of information from the passage which we will combine with our external knowledge. The author begins the passage by saying glycogen synthase catalyzes the lengthening of glycogen polymers through the addition of glucose. Glycogen synthase stimulates glycogen synthesis. A glycosidic bond is a type of covalent bond that joins two carbohydrate molecules like in the case of glycogen. Linear chains of glycogen are bound by α-1,4-Glycosidic bonds (glycogen synthase), while branches have α-1,6-Glycosidic bond (glycogen branching enzyme). Glycogen works to incorporate glucose into glycogen via α-1,4-Glycosidic bonds and the release of UDP. That corresponds to answer choice A: α-1,4-Glycosidic bond

40) We can reference the first paragraph of the passage once again to answer this question. The author mentions UDP-glucose is formed from the enzymatically-catalyzed reaction between glucose 1-phosphate and UTP. 

We know glucose 1-phosphate has our first monosaccharide (glucose), so our other monosaccharide is going to come from UTP. The right side of Figure 1 shows the 5-carbon structure we’re looking to identify. 

  1. Xylose is an aldopentose, or a monosaccharide containing five carbon atoms which is what we’re looking for. We said our monosachharide is going to come from UTP which should mean a ribose sugar, so let’s keep comparing.
  2. Fructose is a ketose, meaning it has six carbon atoms. We can eliminate this answer choice for being inconsistent with Figure 1 from the passage. Answer choice A remains superior.
  3. Ribose is also an aldopentose, but it’s found in RNA. Why is that important in this case? Because we said our other monosaccharide comes from UTP which contains uracil linked to the 1’ carbon of a ribose sugar. This is going to be the best answer choice so far.
  4. Arabinose is also an aldopentose like xylose. It’s a monosaccharide containing five carbon atoms which is what we’re looking for. However, the reasoning here is going to be similar to the reasoning we used for answer choices A and C. the other monosaccharide comes from UTP which contains uracil linked to the 1’ carbon of a ribose sugar. Answer choice C is going to be our best answer.

41) To answer this question we can pull key information from the passage, and revisit Figure 2. When I read through the passage initially, I like to go through any figures and visuals and note key information (no need to memorize specific details). That way, when you go back to the passage to answer the question, you should already have a sense of what you’re looking at. In this case, we have a blot measuring the amounts of 14C and 32P in the samples. A thicker band means more of either is present.

What do we know about laforin from the passage? The author says the abundance of phosphate has been linked to a mutation in the gene for laforin, a phosphatase specific to carbohydrates. Phosphatases dephosphorylate a molecule, or remove a phosphate group from a molecule or compound. The presence of laforin should therefore decrease the intensity we see in the 32P band.

  1. a decrease in the intensity of the band representing WT without glucosidase when visualized with 14C decay. As we mentioned in our breakdown, laforin is a phosphatase which means it dephosphorylates a molecule. The presence of laforin should therefore decrease the intensity we see in the 32P band, not the 14C band
  2. an increase in the intensity of the band representing WT without glucosidase when visualized with 14C decay. Reasoning here is the same as answer choice A- laforin is a phosphatase which means it dephosphorylates a molecule. The presence of laforin should therefore decrease the intensity we see in the 32P band, not increase the intensity of the 14C band.
  3. a decrease in the intensity of the band representing WT without glucosidase when visualized with 32P decay. This matches our breakdown exactly. laforin is a phosphatase which means it dephosphorylates a molecule. The presence of laforin should therefore decrease the intensity we see in the 32P band. This is going to be the best answer choice so far.
  4. an increase in the intensity of the band representing WT without glucosidase when visualized with 32P decay. This answer choice goes against our breakdown and incorrectly says we will see an increase in the intensity of the 32P band. We can stick with answer choice C as our best answer.

42) The author mentions the reaction products were separated by SDS-PAGE and visualized by autoradiography that detects the products of β− decay. Beta radiation occurs when a neutron turns into a proton releasing an electron.

To calculate the atomic properties of an atom before and after radioactive decay, two important quantities are used – the atomic mass number A (number of protons + neutrons), and the atomic number Z (number of protons). We can see the equation for Beta- decay above. These numbers must balance before and after decay. We are going to add one to the atomic numbers of carbon (6) and phosphorus (15).

The radioactive decay described in the passage results in the formation of nitrogen (N) and sulfur (S). The only answer choice that matches our breakdown is answer choice A: N and S. The reason you may incorrectly pick answer choice B is if you incorrectly subtracted from the atomic numbers of carbon and phosphorus.

43) To answer this question, we’ll have to pinpoint some details the author brought up about Lafora bodies and their properties.

This is something we touched on in a previous question in this set as well. A glycosidic bond is a type of covalent bond that joins two carbohydrate molecules like in the case of glycogen. Linear chains of glycogen are bound by α-1,4-Glycosidic bonds (glycogen synthase), while branches have α-1,6-Glycosidic bond (glycogen branching enzyme). Being sparsely (small numbers) branched means we have a decrease in α-1,6-Glycosidic bonds. This corresponds to answer choice D. 

 

Exam 1 C/P Solutions: Questions 44-47

44) When we’re thinking about dissociation of acids in water, we want to think about strong acids first. Strong acids and strong bases are listed on AAMC’s content outline, so make sure you know these most common ones! And also you’ll want to know their properties. Strong bases will dissociate completely in an aqueous solution while yielding hydroxide ions. Strong acids will dissociate completely and yield protons.

Strong Acids Strong Bases
HCl- Hydrochloric acid LiOH- Lithium hydroxide
HBr- Hydrobromic acid NaOH- Sodium hydroxide
HI- Hydroiodic acid KOH- Potassium hydroxide
HNO3– Nitric acid RbOH- Rubidium hydroxide
HClO3– Chloric acid CsOH- Cesium hydroxide
HClO4– Perchloric acid Ca(OH)2– Calcium hydroxide
H2SO4– Sulfuric acid Sr(OH)2– Strontium hydroxide
  Ba(OH)2– Barium hydroxide
  1. HCl + H2O → H3O+ + Cl This answer choice contains one of the strong acids from our list so we can immediately eliminate this answer choice. Strong acids dissociate completely.
  2. HPO42− + H2O → H3O+ + PO43− This answer choice shows a negatively charged ion losing an additional proton. This is unlikely where we want a large negative charge on an ion. This is looking like a great answer choice.
  3. H2SO4 + H2O → H3O+ + HSO4 This answer choice contains one of the strong acids from our list so we can immediately eliminate this answer choice. Strong acids dissociate completely.
  4. H3PO4 + H2O → H3O+ + H2PO4 Phosphoric acid is a weak acid that doesn’t dissociate completely, however, compared to answer choice B which is already a negatively charged ion, our phosphoric acid is more likely to dissociate partially in water. That leaves answer choice B as our best answer choice.

45) This question is the classic setup for a thin lens equation problem and involves knowing the relationship between variables. The MCAT is not a mathematics test. While you may have to do some level of addition, subtraction, division, and multiplication, it is far more important to know the relationships between the numbers with which you’re dealing. This question is a perfect example of that. We’re not told specific values for distances, but rather we’re told the object O is at a distance three focal lengths from the center of a convex lens. Do we know exactly how far that is? We don’t have to know! We’re looking for a ratio in this case of the height of the image to the height of the object. Let’s set up our equation:

1/f = 1/o + 1/i, where f is focal length, o is object distance (triple focal length or 3f), and i is the image distance. We can Substitute o = 3f:

1/f = 1/3f + 1/i 

Isolating for i: i = 3f/2

We can use this value to find the ratio i/o:

i/o = (3f/2) / (3f) = ½

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

46) This is a standalone question that relies exclusively on our general knowledge. Gamma rays are waves of electromagnetic energy, or photons. The only answer choice here that is consistent with our definition is answer choice A. Make sure to study the decay types in case you get a simple definition question in the future!

47) First thing we want to not is the state of each of our reactants and products. We have solids and liquids in our reactants, but solids, liquids, and gases in our products. Entropy is the measure of the disorder of a system. The concept of disorder can best be described in terms of the states of matter. Solids, liquids, and gasses all have different degrees of disorder. We can think of the degrees of the number of different states the molecules in each can occupy. Because molecules in gases are able to change both volume and shape, they can be considered the most disordered. Liquids can be considered the second most disordered, and solids can be considered the least disordered. That means we’re looking for an answer with a positive ΔS°. The test-maker also mentions the reaction is spontaneous so we’re looking for an answer with negative ΔG°.

  1. negative ΔG° and positive ΔS°. Right away we have a strong answer choice. We said the reaction is spontaneous so we’re looking for an answer with negative ΔG°. We also said entropy increases as we have a gas in our products so we’re looking for an answer with a positive ΔS°. This matches our breakdown exactly.
  2. negative ΔG° and negative ΔS°. We mentioned an increase in entropy, so we prefer answer choice A with the positive ΔS°
  3. positive ΔG° and negative ΔS°. A positive ΔG° does not correspond to a spontaneous reaction. 
  4. positive ΔG° and positive ΔS°. A positive ΔG° does not correspond to a spontaneous reaction. Answer choice A is going to be our best option. 

 

Exam 1 C/P Solutions: Passage 9

48) First thing we want to do to answer this question is revisit reactions 3 and 4 from the passage. By doing so, we can see the best way to write out the overall reaction

In this case, our correct answer is going to be the net sum of reactions 3 and 4. Note reactants and products will cancel out as we sum the two reactions to find the overall reaction.

I have drawn a red line through the reactants and products that will “cancel” to get to our overall reaction.

  1. 2H2O2(aq) + IO(aq) → 2H2O(l) + I(aq) + O2(g) This incorrectly cancels certain reactants and products from one side of the equation and not the other.
  2. 2H2O2(aq) + 2I(aq) → 2H2O(l) + I2(g) + O2(g) This answer choice incorrectly leaves iodine ion in the reactants and generates diatomic iodine gas.
  3. H2O2(aq) → H2O(l) + O2(g). This answer choice incorrectly leaves out coefficients on some reactants and products.
  4. 2H2O2(aq) → 2H2O(l) + O2(g). This answer choice correctly lists the overall reaction for the decomposition of H2O2 that is shown in reactions 3 and 4.

49) To answer this question, we can revisit the part of the passage that talks about automobile catalytic converters, then we can go through the four diagrams to find the one not mentioned in the passage:

A.

This function was described in the passage and is not a good answer choice.

B.

This function was described in the passage and is not a good answer choice.

C.

The opposite reaction is mentioned in the passage. This is going to be our correct answer choice.

D.

This function was described in the passage and is not a good answer choice.

50) First thing we’ll do here is define a homogenous catalyst. Homogeneous catalysts are those that occupy the same phase as the reaction mixture (typically liquid or gas). Acid catalysis, organometallic catalysis, and enzymatic catalysis are examples of homogeneous catalysis. An advantage of homogeneous catalysis is that the catalyst mixes into the reaction mixture, allowing a very high degree of interaction between catalyst and reactant molecules. However, homogenous catalysts might not be able to be separated from the products at the end of the reaction. The reaction still happens, but we can’t get the final products we were looking for. We want to know the most likely outcome if that is the case. 

  1. The catalyst will become heterogeneous. Heterogeneous catalysts occupy a different phase. Generally, heterogeneous catalysts are solid compounds that are added to liquid or gas reaction mixtures. This is not what will happen to the catalyst.
  2. The products will be contaminated. This is consistent with what we said in our breakdown. Homogenous catalysts might not be able to be separated from the products at the end of the reaction. The reaction still happens, but we can’t get the final products we were looking for. This is a good answer choice. 
  3. The reaction will not occur. We talked about this in our breakdown and in answer choice B: The reaction still happens, but we can’t get the final products we were looking for. Answer choice B remains superior.
  4. The reaction rate will speed up. The reaction rate itself isn’t changing. The products are contaminated, but that is not the result of a sped-up reaction. We can stick with answer choice B as our best answer. 

51) This question actually ties into our last question and one of the advantages of using a homogenous catalyst. An advantage of homogeneous catalysis is that the catalyst mixes into the reaction mixture, allowing a very high degree of interaction between catalyst and reactant molecules. This question is not implying we use a homogenous catalyst, but it is mentioning the solid catalyst is finely ground, meaning a higher degree of interaction between the catalyst and reactant molecules. 

  1. The rate will be faster because a greater mass of catalyst will be present. This is incorrect as we expect the same mass of catalyst, only in a finely ground form and a more solid form.
  2. The rate will be faster because a greater surface area of catalyst will be exposed. This is consistent with our breakdown; the solid catalyst is finely ground, meaning a higher degree of interaction between the catalyst and reactant molecules. The reaction can proceed faster.
  3. The rate will be slower because the fine catalyst particles will interfere with product formation. This is the opposite of what we’d expect. Instead, a greater surface area of catalyst will be exposed and there is a higher degree of interaction between the catalyst and reactant molecules. The rate will be faster.
  4. The rate will remain the same because the mass of catalyst will be the same. While the mass of the catalyst will be the same, the rate will be faster. Recall we said a greater surface area of catalyst will be exposed and there is a higher degree of interaction between the catalyst and reactant molecules. Answer choice B is our best answer choice.

 

Exam 1 C/P Solutions: Passage 10

52) When you see current, resistance, and voltage, you should be thinking Ohm’s law from your external knowledge:

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated by using the equation V=IR, where I is current in amps (A) and R is the resistance in ohms (Ω). Voltage is not going to change, so to increase I, we would have to decrease R. Let’s see if this is a possible answer.

  1. Using electrical wire with a smaller diameter. This answer choice deals with resistivity. A decrease in diameter means increased resistance, which goes against what we said we’re looking for in our breakdown.
  2. Increasing the temperature of the electrical wire. An increase in temperature generally corresponds to an increase in resistance. That’s the opposite of what we need in this situation.
  3. Decreasing the concentrations of Xn+(aq) and Ym+(aq). This answer choice is out of scope. Changing the concentrations of the solutions will not increase current. 
  4. Replacing the lightbulb with one that has a resistance of 0.2 Ω. This answer choice is correct. By directly lowering the resistance from 0.5 Ω to 0.2 Ω, we’re decreasing resistance (R in V=IR). By decreasing resistance, we would have an increase in current. Answer choice D is going to be our best answer. 

53) First thing we’re told is that a strip of Zn is placed in a beaker with 0.1 M HCl and H2(g) evolves. What does that tell us? Zinc reduces H+ to H2(g). That’s all we know from the first part of the question stem. The second part of the question stem gets into a strip of Al and asked if we’d expect similar results as we did with Zn. We can look at Table 1 from the passage.

We want to focus on the metal Al(s) strip placed into the beaker containing Zn2+(aq). We see a new solid formed, so aluminum’s reduction potential is less than that of zinc. That means from highest to lowest we have H, Zn, and lastly Al. If we place Al(s) into a beaker containing 0.1 M HCl, we expect the same results as we got with the strip of Zn. The only answer choice consistent with our breakdown is going to be answer choice B:

Aluminum reduces H+ to H2(g) while aluminum is oxidized. 

54) First thing we can do to answer this question is pull up a more thoroughly labeled version of Figure 2. We have oxidation at the anode and reduction at the cathode. Note we’re dealing with aluminum and copper in this particular question:

We’re going to use Table 1 from the passage to determine which of our metals has the higher reduction potential.

We see the mixture of Al/Cu2+ yields a new solid, but the mixture of Cu/Al3+ yields no change. Cu2+ is more easily reduced than Al3+ in this situation, so we want our reactants to contain Cu2+. We want our products to be Cu(s) and Al3+. Let’s go through our answer choices and see if we can find a viable option consistent with this:

  1. 2Al3+(aq) + 3Cu(s) → 2Al(s) + 3Cu2+(aq) We mentioned in our breakdown we want our reactants to be Cu2+ and Al. This incorrectly shows them as products. 
  2. 3Al3+(aq) + 2Cu(s) → 3Al(s) + 2Cu2+(aq). Charges here do not balance like we need them to. Additionally, reactants and products are switched.
  3. 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s). This answer choice correctly lists Cu2+ and Al as reactants, and also lists Cu(s) and Al3+ as products. 
  4. 3Al(s) + 2Cu2+(aq) → 3Al3+(aq) + 2Cu(s). Reasoning here is going to be similar to answer choice B. Charges do not balance like we want them to. Answer choice C is going to be our best answer choice.

55) First thing we can do to answer this question is pull up a more thoroughly labeled version of Figure 2. We have oxidation at the anode and reduction at the cathode. Note we’re dealing with Pb and Cu in this particular question:

An oxidation-reduction reaction will occur spontaneously when we have a positive E°cell value. Is that possible given what we’re told in reactions 1 and 2?

If we flip Reaction 1 (need oxidation of Pb(s)) and combine with Reaction 2 the way Reaction 2 is currently written (reduction of Cu2+), we get an E°cell value of +0.466 V. We have oxidation of Pb(s) and combine with the reduction of Cu2+ to get a new E°cell value of +0.466 V which indicates a spontaneous reaction. This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D.

56) When you see current, resistance, and voltage, you should be thinking Ohm’s law from your external knowledge:

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated by using the equation V=IR, where I is current in amps (A) and R is the resistance in ohms (Ω). 

We have lightbulb resistance (R) in our passage as 0.5 Ω, and our question stem tells us potential (V) is 2.0 V. We can isolate current (I) in our V=IR equation and get:

V/R = I

Plugging in our values for V and R we get:

2.0 V / 0.5 Ω = 4.0 A

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

 

Exam 1 C/P Solutions: Questions 57-59

57) This is a classic example of a common ion effect problem. The common ion effect occurs when an ionic compound comes into contact with a substance sharing a common ion (CrO42- in this situation) and decreases the solubility of the ionic compound. We end up having a lot more CrO42- in our product and equilibrium has to shift left to counteract this effect (Le Chatelier’s Principle). 

  1. have no effect on the position of this equilibrium. We talked about the common ion effect and the shift in equilibrium. This is an incorrect answer choice. 
  2. shift this equilibrium left. This is consistent with our breakdown. We end up having a lot more CrO42- in our product and equilibrium has to shift left to counteract this effect. This is a great answer choice.
  3. shift this equilibrium right. This is the opposite of our breakdown. Note the equation in the question stem and CrO42- is in our product. We have to counteract the ions in the product through shifting equilibrium left. 
  4. shift this equilibrium first right and then left. Equilibrium shouldn’t fluctuate like described in this answer choice. We can stick with answer choice B as our best option. 

58) This question involves knowing the relationship between variables. The MCAT is not a mathematics test. While you may have to do some level of addition, subtraction, division, and multiplication, it is far more important to know the relationships between the numbers with which you’re dealing. This question is a perfect example of that. We’re relating molarity, moles, and volume together. In chemistry, the concentration of a solution is often measured in molarity (M), which is the number of moles of solute per liter of solution. In this instance we’re actually solving for liters of solution. Note our answers are given in mL, so we will have to be careful with our units!

Molarity = moles solute / L solution

0.120 M CaI2 = 0.078 mol solute / L solution

Solving for volume gives us 0.65 L of solution. This was a math problem where we did minimal rounding or approximating. We solved for an exact value that matches answer choice D. 

59) This is a standalone question that relies entirely on knowing content and being able to manipulate and compare units. Power is the amount of energy produced per unit time (Power = work/time). This is also equivalent to force x velocity. The unit of power is watt (W) which can also be measured as joules per second or Nm per second. Even if we don’t find an answer that matches these units exactly, we can still manipulate these units to find if the can express power or now.

  1. kg•m•s2 this is nearly equivalent to force x velocity which is kg•m/ s3 but it is not equal. Also note time is not in the denominator here. This is a viable answer to this question because these units cannot be used to express power. 
  2. J•s-1 We mentioned joules/second is a viable unit of power. Answer choice A remains superior.
  3. ft•lb •s-1 Foot pounds per second can be used to express power and can easily be converted to horsepower or watts. Answer choice A remains superior.
  4. W The unit of power is the Watt which is consistent with a way by which we can express power. The only measurement unit that cannot be used to express power is answer choice A.  


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