Jack Westin's Cyber Month Extended - Valid through 12/05 11:59PM PST Learn More

Jack Westin's Cyber Month Extended - Valid through 12/05 11:59PM PST
Learn More

MCAT Content / AAMC Section Bank Cp Solutions

AAMC Section Bank CP [Web]

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 1

1) This question references the uniform electric field described in the passage and we’re asked its magnitude specifically. The question stem itself mentions this question is at least partially passage-based so we can reference the necessary part of the passage. I want you to keep in mind when you go back to the passage like this, I don’t want you re-reading the entire passage or even entire paragraphs. All I’m doing is walking you through the important parts here, just to be thorough. When you’re actually practicing or taking the exam, going back to the passage (unless it’s for specific details you don’t want to memorize) can be a waste of time. Also, this is a math problem that is asking for a quantitative value, magnitude, so make a mental note of the units of the answer choices.

We’re back at our passage here, and before we go into specific details, remember the units we saw in the answer choices: kilovolts over meters. Do we see any units here that correspond to kilovolts or meters? We have the 4.5 kilovolt electric voltage between the MALDI plate and MS detector. That’s the only reference of Volts. Travel distance is given as 0.5 meters, and previously we saw wavelength was given in nanometers. So, we’ve essentially isolated the source of our answer to a few options, before even having to dive into solving for the magnitude of the electric field. 

But now, we have to think to our physics content and how we can answer this question. The voltage across plates is given by the magnitude of our electric field times the distance between the plates. Said differently, the magnitude of the electric field is equal to voltage divided by distance. We’re given a voltage of 4.5 kV, and also told a travel distance of 0.5 m. Divide these two numbers and we have 9.0 kilovolts/meters. Now that’s convenient isn’t it?! We knew coming in, our answer choices were going to have these units. In fact, we thought we’d have to use the exact numbers we did to solve the problem before even thinking about any specifics.

Looking at the 4 options, every answer choice is in units where only voltage and travel distance can combine to give us a correct answer. And that’s what we combined in our breakdown of the question. 

  1. 1 kV/m doesn’t match our breakdown, but we’re keeping it regardless to compare with our additional answers.
  2. 3 kV/m closer to our predicted value, but still not exact. Remember with math problems, and especially when we don’t round or estimate any numbers, you’ll often find the exact answer you solved for. When you get into larger numbers or more complex equations, that can change.
  3. 6 kV/m once again, closer, but not 9 kilovolts per meter. Another observation I want to make: Even if we combined our numbers incorrectly and divided 0.5 meters by 4.5 kilovolts by accident here, we still haven’t stumbled across a potential answer. A lot of times the test maker will make that a potential answer choice, because they know students will make that mistake.
  4. 9 kV/m Units are good, our numeric value matches our predicted value exactly, and there’s no room for rounding errors here. We only dealt with whole number answers and we didn’t round during our calculation. That means D matches our breakdown and is our best answer.

2) DHB was the coating on our plate in the passage, so we want to know the reaction that leads to its formation. That means we’ll need to go back to the passage and see our molecule, and we’ll likely have to know some properties about the molecule. 

We have our structure here, and we’re expected to know how it is formed. Quick glance at our answer choices, we know we’re dealing with either hydroquinone or benzoquinione as our starting substances. Let’s quickly draw those:

First we have 1,4-benzenediol is also known as hydroquinone and has two hydroxyl groups attached to opposite ends of the benzene ring at carbons C1 and C4. Next, we have the simplest members of the quinone class. Benzoquinones, or you’ll ever hear them referred to as just quinones. Do you have to know these structures? Well, they’re found on AAMC’s content outline, so you should have these down. Our DHB looks much closer to hydroquinone, so I have a hunch our answer will be related to the hydroquionone structure more than the benzoquinone.

  1. Carboxylation of hydroquinone During carboxylation, a carboxylic acid group is produced by treating a substrate with carbon dioxide. Our DHB has two hydroxyl groups attached to opposite ends, but also a carboxyl group. This looks like a good choice for now.
  2. Oxidation of hydroquinone We do like answer choices with hydroquinone. But oxidation would be a loss of electrons by hydroquinone. That’s not what’s going on here. That contradicts the picture from the passage.
  3. Reduction of benzoquinone We see benzoquinone as our choice here and that contradicts our breakdown of the question.
  4. Hydroxylation of benzoquinone This answer choice is interesting. Hydroxylation would introduce the necessary hydroxyl groups to the central ring, the same way we see in hydroquinone. But our DHB has two hydroxyl groups and the carboxyl group. Correct answer is answer choice A, the carboxylation of hydroquinone.

3) During MALDI-MS, we have separation of ions that travel to the MS detector in the uniform electric field region. We’re asked specifically about the experimental feature that causes this. We’ll have to go back to our passage to revisit some specifics about the separation of the ions. But at its core, what is this question asking us to do? Identify the relationship between variables.

Here we see our diagram with the ions traveling toward the MS detector. The passage says “the velocity of the ions is inversely proportional to their mass-to-charge ratio.” So that would mean a high mass-to-charge ratio corresponds to a slower velocity. A low mass-to-charge ratio corresponds to a greatest velocity. We also know all ions travel a distance of 0.5 m to the detector, so we’ll likely have to explain the velocity difference along the way.

  1. Distance travelled by ions depends on the ion charge. We don’t think this is the case. All ions travel 0.5 meters. Distance is fixed, but we compare choice A with our other answer choices.
  2. Velocity of ions depends on the ion mass-to-charge ratio. We said a high mass-to-charge ratio corresponds to a slower velocity. A low mass-to-charge ratio corresponds to a greater velocity. That velocity difference could explain the separation of the ions. We’ll keep choice B for now.
  3. Time of travel is inversely proportional to the ion mass-to-charge ratio. This is the opposite of our breakdown. Slower molecules take more time because they have higher mass-to-charge ratio. Time of travel is directly proportional to the ion mass-to-charge ratio. Even when we don’t have exact numbers or quantitative values, we’re still having to keep track of the relationship between variables.
  4. Electric field between the MALDI plate and the MS analyzer is uniform. Electric field, just like distance traveled, is going to be the same for all ions. It’s not helping to separate the ions. Correct answer is answer choice B.

4) We’ve doubled frequency and we’re given four hypothetical lasers in our answer choices. We have to find the answer choice with statistics that are suitable for the MALDI technique. We’ll have to go back to our passage to look at our table that summarized characteristics of the radiation. Just like you did in the previous question, the relationship between our variables. Frequency multiplied by wavelength is a constant. So when one variable doubles the other is cut in half. If frequency is doubled, wavelength is halved and vice versa. Power is not related to our frequency or wavelength.

So here we have our table and again, these are suitable characteristics, so our potential answer should be related to one of these two sets of values. Our power is conserved, so it’s going to be 1.5 milliwatts or 2.2 milliwatts. We’ll also pay attention to the corresponding wavelengths as well. Our question stem says we need to find a laser suitable for MALDI technique after frequency is doubled. So, when we look at our answer choices, we’re looking for a wavelength that when halved corresponds to one of our values in our table. 

  1. Laser A: wavelength 826 nm, power 1.2 mW. If we double frequency and half 826 nanometers, we get 413 nanometers, and power of 1.2 milliwatts. 413 nanometers does not match the wavelengths in Table 1, and power of 1.2 milliwatts is not one of our options. 
  2. Laser B: wavelength 714 nm, power 1.2 mW.  If we double frequency and half 714 nanometers, we get 357 nanometers, and power of 1.2 milliwatts. Again, 357 nanometers does not match the wavelengths in Table 1, and that is not a viable power value.
  3. Laser C: wavelength 650 nm, power 1.5 mW. If we double frequency and half 650 nanometers, we get 325 nanometers, and power of 1.5 milliwatts. 325 nanometers matches Table 1. So far so good. But 325 nm also corresponds to power of 2.2 milliwatts, not 1.5 milliwatts. This answer choice is better than A and B, but still not fully there.
  4. Laser D: wavelength 532 nm, power 1.5 mW. If we double frequency and half 532 nanometers, we get 266 nanometers, and a power of 1.5 mW. This checks out with the table, and our breakdown of the question. What was the key in finding this answer? The relationship between our variables.

5) To answer this question, we’re going to look at Table 1 and see which of our answer choices corresponds to one of the values of electromagnetic energy delivered during a pulse. We’re going to be referencing the passage, we’ll be converting some units most likely, and converting between values based on some equations and relationships that are general knowledge. If necessary, we can round our numbers or change them to scientific notation, so note our answer choices go to one decimal place and units are microjoules. 

We have Table 1 here. We’re going to be doing some unit conversions to make sure everything is in the proper units and we’re answering the right question. Remember, our final unit is in microjoules, and we’re looking for a final answer that goes one place past the decimal point.

We have wavelength. Then power is defined as work/time. Work input is energy, so when we isolate for energy, Energy = Power * time. 

So, we two possible energy values then:
Energy1 = 1.5 mW * pulse duration of 5 ms = 7.5 microjoules
Energy2 = 2.2 mW * pulse duration of 2 ms = 4.4 microjoules.

  1. 2.0 µJ This was the length of one of the pulses in milliseconds, but not our predicted value. We’ll still keep A to compare.
  2. 3.5 µJ If we divide the pulse duration of 5 milliseconds by power of 1.5 milliwatts instead of multiplying, we get a value close to 3.5 microjoules, but we know that is not the correct way to solve for Energy here.
  3. 7.5 µJ We multiplied 1.5 milliwatts by a pulse duration of 5 milliseconds to get 7.5 microjoules. This matches one of our two predictions, so we can eliminate choices A and B.
  4. 8.0 µJ which again does not match our predictions. What we do want to pay attention to, is the other answer choices are similar to other values we have in the passage, but the units don’t always match up. What this means for you, is you can sometimes get by by just focusing on dimensional analysis, and lining up units. The MCAT doesn’t test super high-level physics, and like I mentioned, each calculation is often only a few steps. If you get stuck and have exhausted all of your knowledge and ideas, you still want to make sure you match up units and the corresponding numbers. This is a last resort. We always want our units to match up still, but we still do our calculations to make sure we have our proper number values.

6) To answer this question, we’ll have to think back to the passage and how the sample is prepared, and we’ll likely use external information to actually explain the reaction. 

The passage mentioned, ‘Proteins can be “fingerprinted” using MALDI if they are subjected to proteolytic cleavage before analysis.’ So, our focus here is going to be on proteolytic cleavage. Proteolytic cleavage is the hydrolysis of the peptide bonds between amino acids in proteins. This process is usually done by peptidases, which are enzymes. Enzymes are listed in both content Category 1A and Content category 5E, and other subtopics, so they’re critical on the MCAT!

  1. Oxidation. These are reactions resulting in the addition of oxygen and removal of hydrogen. We said proteolytic cleavage involves hydrolysis which contradicts this answer, but we still keep choice A to compare.
  2. Reduction. Reduction results in the addition of hydrogen and removal of oxygen. This is the opposite of answer choice A, but still contradicts our hydrolysis answer choice. We still keep choices A and B.
  3. Hydrolysis. This matches our breakdown. The passage mentions we’re analyzing peptidases, and proteolytic cleavage involves hydrolysis.
  4. Isomerization. Our molecule is undergoing proteolytic cleavage, it’s not just changing the arrangement of atoms. This answer is unreasonable, so we stick with our answer choice C, hydrolysis.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 2

7) It might be tempting to try and pick an answer based on differences we know about these techniques coming into the exam. But it explicitly says “based on the passage.” We’re going to breakdown what we know about calorimetry, and then compare to the setup given in the passage.

Calorimetry uses the energy released from whatever happens inside a chamber to heat water surrounding the chamber. The change in temperature of the water is measured to determine how much energy must have been released from the chamber. It takes some time to heat the water uniformly. The dissipation of the heat throughout the entire volume of the water makes it so that the exact location of the heat transfer can’t be traced. Meaning we’re unable to detect localized heat transfer.

The passage says “excess energy is absorbed by the solution, causing a local temperature increase. This fast and localized heating process…” that phrase should jump out to us. Fast and localized heating process creates a sound wave that can be recorded. Traditional calorimetry was anything but fast and localized, but PAC is.

  1. can be used on samples with specific heats larger than water’s. Neither the passage, nor the question stem talk about the specific heat of water. We don’t know which technique can be used on which specific heats. Let’s keep going through our answers and comparing.
  2. enables the measurement of fast and localized heat transfer processes. This matches our breakdown exactly. We touched on the steps involved with both experiments, and the PAC technique allowed for fast and localized heat transfer processes. We can keep answer choice B and eliminate answer choice A.
  3. is based on the second law of thermodynamics. 2nd law of thermodynamics is about entropy and this question doesn’t really concern itself with entropy. The entropy of a system, including the entire universe, is constantly increasing as long as nothing is hindering the increase. We can eliminate this answer choice, it’s out of scope.
  4. is useful on samples in the solid phase only. I mentioned breaking bonds of solute molecules in the breakdown. We can eliminate this answer choice. We’re left with our correct answer choice, answer choice B.

8) That means we’re using a specific laser to dissociate a specific chemical bond, and we want to know the energy meter reading. We can answer this question using content, but a quick glance at our answer choices shows that the answer could be in terms of the specific variables discussed in the passage. 

The passage say excess energy after the laser hits the sample is absorbed and temperature increases. But in this case, we’re using an appropriate laser, for a particular chemical bond. 

Everything works in a particular way, and the energy meter doesn’t pick up anything. It would only pick up if energy was to come off, and then you would know that there is an unexpected variable in your measurements.

Looking at the diagram, you can see that the basic flow of energy goes from laser (photon energy) –> to the lens (releases heat, causing sound waves) –> microphone –> and is picked up by the energy meter.

When the laser/photon energy is just right for a particular chemical bond, no excess heat is released from bond breakages inside the cell, and no sound waves are produced. The energy meter will read 0.

  1. Em
  2. ΔHu
  3. ΔHnr
  4. 0

Glancing at our answer choices, answer choice D matches our breakdown exactly. Answer choice B is the difference between the laser pulse energy, answer choice A, and the heat detected, answer choice C. Energy is conserved, as we’re using an appropriate laser. We’re going to eliminate answer choices A through C.

9) We’re going to go back to the passage to find our three compounds. We’ll use our general knowledge to identify the structure of all 3, and we’ll pick an answer based on similarities.

The passage mentions phenols, thiophenols and alkylbenzes. We can list them all out here. 

Phenols: hydroxyl group bonded to an aromatic hydrocarbon group.

Thiophenols: Similar to phenols, but sulfur-containing

Alkylbenzenes: Derivatives of benzene, in which one or more hydrogen atoms are replaced by alkyl groups. Shown is toluene, the simplest alkylbenzene

Phenols, thiophenols, and alkylbenzenes all have aromatic rings.

  1. Aromatic ring that matches our prediction.
  2. Alkyl These are present in alkylbenzenes, but not all three compounds, so eliminate answer choice B
  3. Carboxylic acid This wasn’t present in any of our predicted figures.
  4. Carbonyl Again, not present in any of our drawn-out structures. We can eliminate answer choices B-D and we’re left with our correct answer, answer choice A-aromatic ring.

10) In other words, what feature is going to allow laser A to be suitable to cleave the specific bond mentioned? Let’s go back to the passage to find Table 1 and the associated values.

We see our three types of bonds here and the type of laser associated with each. We’re focused on laser A and C. Enthalpy is higher for the -OH bond. Laser A is higher energy than laser C.

We can compare energy with frequency and wavelength.

Energy of a photon is proportional to frequency:

E=hf . h is Planck’s constant.

We also know c = λf, which means that wavelength and frequency are inversely proportional
 So, ↑ energy = ↑ frequency and ↓ wavelength 
and ↓ energy = ↓ frequency and ↑ wavelength.

Prediction is higher energy, higher frequency, and shorter wavelength for laser A.

  1. be better focused laser type C. This isn’t relevant to our question, or explain the difference between the laser. Let’s keep comparing.
  2. have a higher frequency than laser type C. This matches our prediction. We said laser A has higher energy, higher frequency, and shorter wavelength. Keep answer choice B, we can eliminate answer choice A.
  3. have a longer wavelength than laser type C. That’s the opposite of what we said. Laser A will have a shorter wavelength. We can eliminate answer choice C.
  4. emit fewer photons per unit time than laser type C. This would imply less energy in laser type A. That contradicts our prediction so we can eliminate answer choice D. We’re left with our correct answer, answer choice B. 

11) We can go back to the passage to reference Figure 1, but we’ll use the thin-lens formula to find focal length. Note the units in our answer choices are in centimeters.

We have our diagram here. Let’s write out our thin lens formula:

Object distance is 12 cm, image distance is 4 cm

1/12 + ¼ = 1/focal length

1/3 = 1/focal length

Focal length = 3 cm

This is a math problem where we did no rounding or approximation. We can look for our exact, calculated value of 3 centimeters. Our prediction matches answer choice A (3 cm) and we can eliminate answer choices B-D. 

12) We’re going to go back to the passage to find details, specifically the work function. We’re given two values in the question stem, frequency and h. We can solve this question using the passage and the equation for kinetic energy of photoelectrons.

We have part of the passage up above, we have the values for f and h below. 

The kinetic energy of ejected electrons (photoelectrons) is given by KE = hf – work function, where hf is the photon energy, 

The passage tells us the energy meter, based on the photoelectric effect, uses a detector with a work function of 3.4 eV

Multiply h and f. The seconds units cancel. The exponents on the 10s are opposites, so those cancel as well. We just multiply 5.0 x 4.1 to get 20.5 electron Volts. Plug in numbers to solve for kinetic energy. 

KE=20.5 eV-3.4 3V = 17.1 eV

This was a math problem where we did no rounding or approximation. We can look for our exact, calculated value of 17.1 electron volts; our prediction matches answer choice C.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 13-17

13) Let’s consider what we’re looking at in this question. We have a nucleophilic substitution, and the rate determining step is unimolecular. Or in other words, an SN1 reaction. We’re given 4 different alcohols, and told we have acidic conditions. Ultimately, which alcohol will most likely undergo substitution by an SN1 mechanism? 

A tertiary carbon or a secondary carbon is going to be favored by SN1, whereas a methyl carbon or a primary carbon goes to SN2. In this case, for SN1, we’re looking for a tertiary or a secondary carbon that would get attacked by the nucleophile. 

For SN2, remember the backside attack, and for SN1, remember the carbocation intermediate. In the SN1 reaction, the big barrier is carbocation stability. The first step of the SN1 reaction is loss of a leaving group to give a carbocation, the rate of the reaction will be proportional to the stability of the carbocation. Carbocation stability increases with increasing substitution of the carbon (tertiary > secondary >> primary). 

We have acidic conditions, meaning protonation and production of water

a.

Answer choice A is a primary alcohol. We said we’re looking for a tertiary or secondary carbon so we can have the most stable carbocation. Let’s keep this answer choice for now and compare with our other options

b.

 Answer choice B is a secondary alcohol. This matches our prediction. This carbocation is going to be more stable than the primary alcohol in answer choice A.

c.

Answer choice C is a primary alcohol. This is the opposite of what we wanted. We said primary alcohols are our worst option here. We’re still maintaining answer choice B is our best option so far.

d.

Answer choice D is a tertiary alcohol. This is the ideal answer choice we were looking for. The tertiary carbon is going to produce the most stable carbocation, so it’s going to most likely undergo substitution and produce a water molecule. We can eliminate answer choice B, we’re left with our best answer choice, answer choice D.

14) Let’s break this down a little bit at a time. We have a reaction between an amine and excess carboxylic acid anhydride. One of the products is going to be an amide, and we have to pick an extraction procedure that allows us to isolate this amide. 

First thing we want to do is to lay out the reaction happening. We have an amine and excess carboxylic acid anhydride that yields an amide and carboxylic acid. We had excess carboxylic acid anhydride, meaning all of our amine would be converted, and we’ll still have carboxylic acid anhydride that didn’t react to amine. We deal with the anhydride first by treating it with a strong base or a strong acid. That’s going to convert the anhydride to an unreactive form. So now we’ll have our amide, and two carboxylic acids. We have to separate the amide from the carboxylic acid. We’re using an extraction procedure, so we have to separate the two compounds.  We can use charge and solubility difference. The way we can separate by charge is to deprotonate the carboxylic acid. How can we do that? A strong base. The strong base can deprotonate the carboxylic acid, while amide is unaffected because of its high pKa. So we can separate our charged carboxylate and uncharged amide using an organic solvent. The amide can then be filtered out.

  1. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. This is consistent with our breakdown of the question. Strong base will quench the unreacted anhydride. Meaning it’s now unreactive. The strong base deprotonates the anhydride and we have a charge difference. We add diethyl ester, which is analogous to our organic solvent. The charge difference allows us to dissolve the amide, but the carboxylate won’t dissolved. So, evaporating the solvent separates the amide and leaving only the carboxylate
  2. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. The first part of the process is consistent with our process. But using a strong acid instead of a strong base doesn’t allow for the deprotonation of the anhydride and give us that charge difference. We can still maintain that A is our best option. 
  3. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl(aq). Our amide is in the organic layer and we have to filter it to get our amide. This contradicts what I said in the breakdown of the question.
  4. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with NaOH(aq). This is similar to answer choice C, but also, we’re using a strong acid instead of a strong base. This contradicts our breakdown as well. We’re left with our correct answer, answer choice A.

15) Voltage is constant, so we need to relate resistance and current. We’ll do that using Ohm’s Law. Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points. We can write out Ohm’s law as:

V=IR. Voltage equals current times resistance. Voltage is constant in this question. Minimum resistance will be when current is at a maximum (since they are inversely proportional). The current is highest at about 7 to 8ms. Value of the current is roughly 400 x 10^-12 Amps according to the graph. We can plug in the values we’re given to solve for minimum resistance.

V=IR
Voltage=Current x resistance

Solving for resistance:
R=V/I

R=80 mV/(400 x 10^-12 A) = 200 MΩ
This was a math problem that didn’t involve much rounding. Instead, we solved for the minimum resistance as 200 MΩ. We can compare all 4 of our answer choices at once. Our correct answer is going to be answer choice C.

16) We have a graph showing the relationship between frequency and intensity. We’re focusing on the y-axis and want to know the ratio between the maximum and minimum sound intensities. Looking at the chart, the max intensity level is 80 dB. The minimum intensity level is 30 dB. Decibels are logarithmic. meaning every 10-dB increase is more powerful by a 10 multiplier. So, a 20-dB increase is an increase by 10^2. A 30-dB increase is more powerful by 10^3. 50-dB increase is more powerful by 105

This is a fairly straightforward question that just involved reading the graph given to us. We solved for an exact value, so we can compare all of our answer choices at the same time. We’re going to pick the answer choice that’s the same as our prediction, the ratio is 105.

17) We’re using the image in the question stem and external knowledge to answer this question. We’re going to know about Ohm’s law, currents, and resistors in series. First thing we want to point out, is when the voltmeter reads zero, the voltage DIFFERENCE between those points is zero. So that means they are at the same voltage. 

Let’s look at our resistors: R1 and R2 are in series. R3 and R are in a series. Then R1 and R2 are in parallel with R and R3. Current through R1 and R2 is equal, and the current through R3 and R is equal. We have a current through R1 and R2 (I) and a current through R and R3 (I’).

Ohm’s law is V=IR. We know that the voltage is the same for the segment containing R1 and the segment containing R; we can set up the relationship:
V= I*R1 and that is also equal to I’*R

as well as the relationship:
V= I*R2 and that is equal to I’*R3 as the voltage must be same in these portions as well.

Now, if we do some math and isolate R in the first equation, we get I/I’ * R1= R. But nothing in our answer choices contains current. So, we need to find a way to get rid of current.

If we solve from I from the second equation, we get I = I’*R3/R2.

Now, we plug the quantity of I into the equation isolated for R, and see that I’ cancels out, and we are left with:
(R3*R1)/R2=R

We solved for the exact relationship. We said R equals R3*R1/R2. We can pick our correct answer, answer choice D.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 3

18) To answer this question, we’re going to reference the figure from the passage, but then we’ll use our general knowledge to actually classify the lipid molecule.

Compound 1 is a phosphatidylcholine. Phosphatidylcholines contain both phosphorus and choline head (nitrogen part), and are found in cell membranes. Compound 1 is going to be used to synthesize liposomes. How can we connect this structure to liposomes? Liposomes are spherical vesicles and they mimic cell membranes (phospholipid bilayer). Keep these key points of our structure handy, and go back to our answer choices. 

  1. Triacylglycerols. Triacylglycerols have three fatty acids linked to glycerol. No choline head.
  2. Pyrophosphates. Pyrophosphates have two phosphates in Phosphorus-oxygen-phosphorus linkage. Not the case here-we have our phosphorus and choline head, but it’s not a pyrophosphate.
  3. Phosphatides. This is consistent with our prediction. Phosphatides contain both phosphorus and choline head (nitrogen part). This is now the best answer choice.
  4. Phosphonic acids. These consist of a single pentavalent phosphorus covalently bound via single bonds to a single hydrogen and two hydroxy groups and via a double bond to an oxygen. Our phosphorus here is double bonded to an oxygen, but we don’t have the two hydroxyl groups here. Answer choice C remains our best option. 

19) To answer this question, we’re going to reference the experiment in the passage, but then we’ll use our general knowledge to explain the properties of liposomes and how size-exclusion chromatography work.

In size-exclusion chromatography, molecules are separated based on their size. Separation is achieved through differential exclusion where the biggest molecules elute the fastest, while the smallest molecules elute the slowest.

Liposomes have a spherical structure while allows them to be used as a vehicle to transport drugs and nutrients. The experiment involves synthesizing liposomes through a reaction between Compound 1 and in the presence of dye. As the liposomes are formed, they will be filled with the dye-containing solution. Size exclusion chromatography is uses to separate things. As the liposomes are filtered through size-exclusion chromatography, the liposome will break up and release the dye, meaning the solution fluoresces.

  1. The macromolecule had extensive conjugation. When you look at the molecule there is no conjugation. Our big indicator for conjugation is alternating double bonds. This is an unreasonable answer choice
  2. Fluorescent dye was trapped inside. This matches our breakdown of the question. The liposomes break up and release the dye that was trapped inside. This answer choice is superior to answer choice A.
  3. Intermolecular interactions lower the energy of the excited state. Decreasing an energy state would release a color, not fluorescence.  And if this was the case the experimenters wouldn’t need to have to deal with dye and fluorescence in the first place. Also, keep in mind that a color would only be released if a covalent/ionic bond happens. In this case, the liposomes are creating a bilayer that is held together by Van der Waals forces, which doesn’t really decrease the energy state. There’s no color, and that’s also why they needed fluorescent dye.
  4. Light reflects from the surface of the sphere. This isn’t how light works, and the fluorescence is from the dye being physically released from the liposome. We’re left with our correct answer choice, answer choice B fluorescent dye was trapped inside.

20) In the passage we found there were differences when we mixed the compounds. Compound 1 was stable to mixing, but Compound 2 formed new liposomes. We want to know why exactly this difference existed. That means we’re going to reference the experiment in the passage, but then we’ll use our general knowledge to explain the behavioral differences. Quick glance at our answer choices shows we’ll have to distinguish between kinetic and thermodynamic control. 

The passage says “Liposomes formed from Compound 1 were stable to mixing, but mixing those from Compound 2 formed new liposomes with an average size expected for the effective final lipid concentration.” Compound 1 doesn’t change; Compound 2 quickly results in the formation of new liposomes. 

We have to distinguish between kinetic control and thermodynamic control, so let’s do some quick background. Short reaction times favor kinetic control, whereas longer reaction times favor thermodynamic reaction control.

Example: product A forms faster than product B because the activation energy for product A is lower than that for product B, but product B is more stable. In such a case A is the kinetic product and is favored under kinetic control and B is the thermodynamic product and is favored under thermodynamic control. The pathway that’s favored depends on the condition of the reaction. 

If a reaction is under kinetic control, it is said that the preferred product/major output product of the reaction is the one that is formed most quickly and has the lowest activation energy barrier. It’s usually associated with irreversible reactions. It explains why Compound 1, after forming the liposomes, doesn’t form more energetically stable liposomes upon mixing.

Thermodynamic control, is preference for a more stable product (lower free energy). No matter which product forms first, if there is a lower energy state product that can be formed, then it will be formed. This is associated with reversibility. That explains why Compound 2 is able to reverse the liposome forming reaction, and form more energetically stable liposomes after mixing. Our answer is kinetic control for Compound 1 and thermodynamic control for Compound 2.

  1. both Compound 1 and Compound 2 are under kinetic control. Remember, we said only compound 1 is under kinetic control. Compound 2 is under thermodynamic control. This contradicts our breakdown.
  2. Compound 1 are under kinetic control, but those prepared from Compound 2 are under thermodynamic control. This answer is saying exactly what I said in our breakdown of the question. 
  3. Compound 1 are under thermodynamic control, but those prepared from Compound 2 are under kinetic control. This answer choice is the opposite of our breakdown, so answer choice B remains the best option. 
  4. both Compound 1 and Compound 2 are under thermodynamic control. Only Compound 2 is under thermodynamic control, not Compound 1. Compound 1 is under kinetic control. We can stick with our correct answer choice, answer choice B.

21) We’re shown the above table that gives us diameter vs elution volume. Diameter goes from 50 nanometers to 200 nanometers in the table. We want to know the lipid concentration necessary to prepare liposomes with a diameter of 250 nm. We can solve this question by using information given in the passage. Our goal is to relate the visual in the question stem with the one from the passage.

The chart in the question stem shows us an inverse relationship with diameter and elution volume. As diameter increases, elution volume decreases. With this information, we can predict that a diameter of 250 nanometers will correspond to an elution volume of less than 10mL. 

Now, look at Figure 1 on the right- we have another inverse relationship here. As elution volume increases along the X axis, we have a decrease in concentration.

Let’s combine this information: 250 nanometers means we have a larger diameter, a smaller elution volume. That means an increase in concentration. The concentration must be bigger than 0.20mM because that elutes at ~10mL. So that’s our expected answer: greater than 0.2 millimolar.

Quick glance at our answer choices and we can actually compare all of them at once. The concentrations are all in the same units and go from the smallest concentration to largest concentration from choices A to D. The only answer choice greater than 0.2 millimolar is answer choice D: 0.30 millimolar.

22) We can solve this question by using information given in the passage. We’ll use Figure 1 in the passage, and we’ll be using data analysis.

For liposomes that elute at 20 mL, the solution concentration is 0.1 mM. The solution used to synthesize liposomes was 1 mL of varying concentrations, so we can solve for numbers of moles using volume and molarity. 

We can use dimensional analysis. Let’s write out our calculations to make sure units and prefixes are in order.

We have molarity (0.1 x 10^-3 mol/L) X our factor (1 mL x 1 L/1000mL) = 0.1 x 10^-6 moles

0.1x 10^-6 moles x (800 g/mol) = 8 x 10^-5 grams OR 80 micrograms (μg)

  1. 80 µg matches our prediction exactly
  2. 8 mg 8 milligrams is equal to 8 x 10^-3 grams. Incorrect value
  3. 80 mg 80 milligrams is equal to 8 x 10^-2 grams. Incorrect value.
  4. 8 g is another incorrect value, so our correct answer choice is answer choice A, 80 micrograms.

23) We can solve this question by using information given in the passage, specifically Figure 1. Compound 2 is able to reverse the liposome forming reaction, and form more energetically stable liposomes after mixing. We’ll analyze our figure with this in mind. 

Mixing equal volumes of the 0.10 mM and 0.2mM will yield a suspension equal to a concentration of 0.15mM. 

The combined concentration of 0.15mM, has an elution volume of about 16 or 17 mL. So that 16mL is the only place we would see the fluorescence bump. 

And just for comparison purposes: Individual elution volumes for 0.10 mM and 0.2mM are at about 20 mL and ~12 mL 

Answer choice A has two bumps at roughly 12 ml and 20 ml. We just said these are individual elution volumes for 0.1 millimolar and 0.2 millimolar concentration. 

Answer choice B has the same two bumps at 12 mL and 20 mL. We’re getting the same bumps as answer choice A, so there’s a chance we end up eliminating answer choice along with answer A for saying the same thing.

Answer choice C has a single bump at roughly 25 mL. This would be the case at a very low concentration, not the roughly 0.15 millimolar concentration in this question.

Answer choice D has a single bump at roughly 16 mL, which is consistent with our breakdown. We can pick our correct answer, answer choice D, and eliminate all our other answer choices.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 24-28

24) We’re given the free energy of the hydrolysis of ATP. We’re asked about the ratio of ADP to ATP at equilibrium given certain conditions. To answer this question, we’ll need to set up the equilibrium constant expression and find how it relates to the free energy of the reaction.

First thing we can do is find the equilibrium constant expression. 

For reactions involving a solid or a liquid: the amounts of the solid or liquid will change during a reaction. But their concentrations won’t change. Instead, the values for solids and liquids remain constant. And because they’re constant, their values are not included in the equilibrium constant expression. Water is a liquid so it’s not included. Let’s write out our expression:

Keq = [ADP][Pi] / [ATP]

Now we can relate our expression with the free energy of the reaction:
∆G° = -RT ln K 

We can substitute K from above into this equation, so plug in given values: 
∆G=-30 kJ/mol, RT= 2.5 kJ/mol, [Pi]=1.0M. 

Note: In spontaneous reactions, the sign for ∆G will be negative
Solve for [ADP]/[ATP] which equals e12
This was a math problem that didn’t involve rounding. Instead, we solved for the ratio of ADP to ATP at equilibrium. Our calculated value is e12. We can compare all 4 of our answer choices at once. Our correct answer is going to be answer choice D.

25) We’ll have to use the periodic table and subsequently find the electron configuration of cobalt ion.

Elemental cobalt is element number 27, meaning it has 27 protons and 27 total electrons typically. In this case, we’re dealing with cobalt ion. Cobalt loses two electrons to get a +2 charge. Elemental cobalt has 9 valence electrons, but cobalt cation has 7 valence electrons.

Elemental cobalt has an electron configuration [Ar] 3d7 4s2

We have to be careful with some transition metals. For cobalt, we need to determine whether the 2 electrons are lost from the 4S or the 3d orbital. In this case, when electrons are lost from Co atom, they are lost from the 4𝑠 orbital first, because those electrons have a higher reactivity when both 3𝑑 and 4𝑠 are filled with electrons.

[Ar] 3d7 instead of [Ar]4s2 3d5.

Every answer choice listed starts with Argon first, that’s the closest noble gas and it has an octet.

  1. [Ar] 3d7 This matches our breakdown. Cobalt has 7 valence electrons, and it loses 2 valence electrons from the 4s orbital in its cation form.
  2. [Ar] 4s2 3d5 This would have been our only other potential option. But we said electrons are lost from the 4𝑠 orbital first, because those electrons have a higher reactivity when both 3𝑑 and 4𝑠 are filled with electrons. Answer choice A remains superior.
  3. [Ar] 4s2 3d7 This is the electron configuration of elemental copper, not the cation. Answer choice A still the best option. 
  4. [Ar] 4s2 4d5 This is an unreasonable answer. That isn’t how the valence electrons of cobalt fill up, so we can stick with answer choice A.

26) We’re given a rate in micromoles over seconds. We have a volume in liters, and a time in minutes. Let’s do some conversions and get to the number of moles that’re in our correct answer.

First thing, write out the rate in scientific notation. We can convert micro molar to molar by multiplying by 10-6. We know to do this from our prefixes:
(2×10^-6 mol/L /s) 

Next, convert time to seconds: 1 minute * (60 seconds/1 minute) =60 seconds.

We can multiply our molarity by seconds and by (.001L) to solve for moles. This gives us 1.2×10-7 mol

This was a math problem, and there was no rounding or approximating. Instead, we solved for the number of ATP molecules that hydrolyzed after 1 minute. Our calculated value was 1.2×10-7 mol. We can compare all 4 of our answer choices at once. Our correct answer is going to be answer choice A.

27) We want to know which of the four cations listed can replace iron in the square planar binding hemoglobin; we want something with similar properties. We’re dealing with ions and specific elements, so we’ll be using our periodic table.

We’ve got our periodic table on the left, and our answer choices on the right. We can find iron on the periodic table; iron’s atomic number is 26. Next, we can find our 4 answer choices. Magnesium is a metal in group 2, lithium is a metal in group 1, cobalt is a transition metal next to iron on the periodic table, and sodium is a metal in group 1.

What more can we say about our answers and iron? Iron is also a transition metal. The transition metals are found in the middle of the periodic table. Lithium and sodium are both alkali metals, while magnesium is an alkaline earth metal. Typically, elements that are found in the same group have the most similar properties. Even though iron and cobalt are not in the same group, they are both transition metals and share properties that are not shared with representative elements. Presumably cobalt and iron will have the most similar properties. We’re going to keep our correct answer, answer choice C.

28) We just need to know about the different classes of enzymes in the answer choices. And in particular, which one catalyzes ATP-dependent phosphorylation of a protein target. Phosphorylation is done by phosphotransferases. They catalyze the transfer of phosphoryl groups. Let’s see which of the answer choices is the best option here.

  1. Oxidoreductase. Oxidoreductases are enzymes that will catalyze a transfer of electrons from one molecule to another molecule in an oxidation-reduction reaction. This doesn’t match what we’re looking for, so let’s keep comparing.
  2. Transferase. Transferases are enzymes that will transfer specific functional groups from one molecule to another molecule. Transferases are used in a variety of reactions that take place through the body and in many organisms. Transferases are further classified based on the functional group transferred. Kinases are phosphotransferases-they catalyze the transfer of phosphate groups from one molecule to another, and kinases often transfer phosphate groups from ATP. This sounds like our ideal answer
  3. Hydrolase. Hydrolase describes the class of enzymes that use water to break chemical bonds. This doesn’t match what we’re looking for. Answer choice B is still superior.
  4. Ligase. Ligases are enzymes that catalyze the junction of molecules and can also be accompanied by a hydrolysis reaction. Not what we’re looking for either, so we’re left with our correct answer, answer choice B: Transferase.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 4

29) What is this question referencing? It’s asking about the experimental results specifically, so we can reference our passage. 

The passage mentions Compound 1 was utilized because it produced two compounds: Compound 2a and Compound 2b. One product is intensely colored, so we can say that compound was clearly detected if the students made this observation. We’re looking for the colored compound, but we also want to know the reasoning behind the intense color. What causes the color to appear? Extensive conjugation causes color to appear in organic compounds. In transition metals, it’s the d orbital that gives color. Let’s look at our two compounds closely.

Only B has a conjugated system, meaning Compound B will be the one that’s colored, and it’s the one that was detected. Having a “conjugated system” for our sake simply means consecutive alternating double bonds. Double bonds indicate there are overlapping pi orbitals and this delocalizes the electrons. Highly delocalized compounds (or highly conjugated compounds) are better at absorbing and emitting light. At its core, really all this question is asking is which product is emitting light, and why. We’re looking for an answer like Compound 2b, because of the conjugation.

  1. Compound 2a; it has three aromatic rings. That goes against our prediction of Compound 2b first of all. Secondly, we want the greater number of conjugated double bonds. The fewer aromatic rings works against answer choice A
  2. Compound 2a; it is a diphenol. Does that affect our answer? It sounds like it could be a better choice than A, but still not the compound we predicted.
  3. Compound 2b; it has extensive delocalization of electrons. This sounds like our breakdown, just in different words. Double bonds indicate there are overlapping pi orbitals and this delocalizes the electrons. We like answer choice C the best so far.
  4. Compound 2b; it is a mixed phenol and quinone. Much like answer choice B, this doesn’t exactly answer our question, even if it’s technically a true statement. We’re sticking with option C as the best answer. 

30) In the passage we have Table 1. We can compare the Km value for all of the esterases, but the Km for the wildtype wasn’t listed. This question is asking us to solve for the missing Km, that’s our Michaelis constant.

We’re looking at Figure 1 and Table 1 in the passage. We need to solve for Km of the wildtype esterase. What do we know about Km first of all? The Michaelis constant, Km is equal to substrate concentration when reaction rate is ½vmax. vmax is the maximum enzymatic reaction rate. Km, helps compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate. 

So first, let’s use the graph from the passage. We can look at the graph to see if we can use the relationships between variables to solve our problem.

Vmax looks to be slightly above 1.5. Call it 1.6. At ½ of vmax (so approximately 0.8), we can look at our graph. The fluorogenic substrate value looks to be about 1.25 at ½ vmax. 

  1. 0.67. This value is much lower than our estimated value. We can likely eliminate it for being an incorrect value, but let’s compare with our additional answer choices
  2. 1.31. This is very close to our prediction of 1.25. Looking again at our line graph, 1.31 looks to correspond to 1/2vmax, so we’ll hold on to this answer choice; answer choice B is superior to answer choice A.
  3. 1.73. If we look at the graph and the associated formula, This value is closer to Vmax that it is to ½ vmax. We can eliminate it for being unreasonable.
  4. 3.04. This is again unreasonable. Our Vmax looks like it’s somewhere between 1.5 and 1.75, so it makes no sense for 1/2vmax to be greater than that. We stick with our correct answer choice, answer choice C. Km is 1.31 for the wildtype enzyme

31) The passage listed five different esterases: there was a wildtype, and four variants. There were amino acid substitutions in each variant and we have to explain the substitutions that took place.

Look at substitutions in the table to understand what’s happening, and we’ll use our knowledge of different amino acids to make observations.

Table 1 it lists the various esterases. What is being substituted is given by the first letter in the “esterase” column. For example V86A means the 86th residue was a valine, and the substitution was an alanine. V86A means valine substituted to alanine. M87A means methionine substituted to alanine. L209A means leucine substituted to alanine. I276A means isoleucine substituted to alanine. So that’s 4 substitutions from amino acids to alanine. You want to reference your amino acid chart to answer this question. Just as an aside, if you’re shaky about your amino acids, make sure to spend time reviewing this chart. It ends up being easy points on test day and it will make your life a lot easier. 

Alanine is classified as hydrophobic (It’s the 2nd smallest nonpolar amino acid. And it only has CH3 as its side group). The previous esterases are all hydrophobic already, so hydrophobicity and hydrophilicity do not change. Alanine isn’t acidic or basic. 

  1. Increased hydrophobicity. Even though alanine is hydrophobic, so were the initial amino acids. We’ll hold on to answer choice A and compare with additional answer choices to see if we can find a better option.
  2. Decreased basicity. The alanine substitution won’t affect acidity or basicity in these situations. This contradicts what we know from our content. Answer choice A remains superior for the time being.
  3. Increased hydrophilicity. This is the opposite of answer choice A, but it’s similar. Neither answer choice explains what is taking place. Alanine is hydrophobic, but so were the amino acids it was replacing. We’ll hold on to both answer choices A and C for now.
  4. Decreased steric constraints. We mentioned alanine is the 2nd smallest nonpolar amino acid. There are decreased steric constraints when valine, methionine, leucine, and isoleucine are substituted. The best answer here is going to be answer D. The substitutions changed the side chains via decreased steric constraints.

32) The students used 0.1 mg/ml bovine serum albumin in their setup. This question wants to know the purpose of using this BSA. The passage tells us BSA mobilizes proteins and lipids, but doesn’t add any additional information. BSA is used to stabilize some enzymes during the digestion of DNA, and to prevent adhesion of the enzyme to reaction tubes, pipette tips, and other vessels-hence the mobilization of proteins and lipids mentioned in the passage.

  1. BSA acts as a co-catalyst for the reaction. BSA does not bring about the catalysis in the reaction. Co-catalysts would do that, while BSA only mobilizes proteins and lipids. This contradicts information given in the passage.
  2. prevents the esterase from adhering to the walls of the vessel. Again, we said BSA mobilizes proteins and lipids. That mobilization can prevent the esterase from adhering to the walls of the vessel.
  3. helps buffer the solution from changes in pH. There was no mention of BSA buffering the solution from changes in pH. This contradicts information given in the passage.
  4. is a non-specific target for protease contaminants. Again, not a purpose of this experimental setup. This contradicts my initial breakdown of the question. We can stick with answer choice B-BSA mobilizes proteins and lipids, and it prevents the esterase from adhering to the walls of the vessel.

33) The passage listed five different esterases: there was a wildtype, and four variants. We want to know which esterase showed the lowest catalytic efficiency, so we can pull up Table 1.

Catalytic efficiency is given by Kcat/Km and measures how well an enzyme is able to “grab” its substrate and turn it into product. This means you want the greatest number of reactions happening, for the least amount of input of time/substrate. That’s basically what kcat/km says. 

Kcat, the catalytic constant, measures the number of molecules of substrate turned over into product per unit time (usually seconds). Km is our Michaelis constant. It’s used to determine the affinity between an enzyme and substrate. 

The higher your kcat is, the more reactions happen per unit time (i.e. Vmax) 

All of this is taking place when conditions are optimal, and you have enough substrate. So, Kcat is positively proportional to efficiency; 

The higher km is, the more concentration of substrate you need before you reach optimal conditions for kcat, so km is inversely proportional. 

We have all of our experimental data organized in the table, so we can find the lowest ratio of Kcat to Km to see which variant showed the lowest catalytic efficiency. The I276A esterase has one of the smallest catalytic constants listed, and has by far the largest Michaelis constant. I276A esterase has the lowest catalytic efficiency. You’re not expected to do this math in your head. You want to get a sense of the magnitude of the numbers, and estimate. Do not try and find an exact numerical value for each.

Looking at our answer choices, our correct answer for this question is D) I276A.

34) We’re told the properties of the PBS in the passage, but now we want to know the mass of NaCl that was present in 100 microliters of the solution. Let’s look at the passage to get the specific details about our PBS solution.

pH is 7.4, and 0.1 mM NaCl. First thing we’ll do is look at units and find information about PBS solution and its NaCl composition. PBS given as 0.1mM NaCl. We can use dimensional analysis to convert this number into molarity. 

0.1 mM is multiplied by (1M/1000mMol) to give (0.0001 mol/L). Next, we’re given the molar mass of sodium chloride as 58.5 g/mol. We can multiply: (0.0001 mol/L) x (58.5 g/mol) which gives us 0.00585 grams/L sodium chloride. Next, we multiply by the 100 microliters (or 0.0001 L) of PBS solution.

Final answer (grams): 0.000000585, or 5.85 x 10-7grams which is 585 nanograms or 0.585 micrograms.

Let’s go through our answer choices:

  1. 585 ng. Converted to scientific notation, that is 585 x 10^-9 grams, or 5.85 x 10^-7 grams, which matches our calculation.
  2. 5.85 µg. 5.85 micrograms is 5.85 x 10^-6 grams. That’s an incorrect value. 
  3. 585 µg. 585 micrograms is 585 x 10^-6 grams, or 5.85 x 10^-8 grams. That’s an incorrect value.
  4. 5.85 mg. 5.85 mg is 5.85 x 10^-4 grams. That’s an incorrect value as well. We’re left with the final, correct answer choice. Answer choice A, 585 nanograms.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 5

35) To answer this question, we’re explaining the reasoning for the deprotonation prior to reaction from the passage. The question explicitly mentions a specific instance in the passage. This doesn’t always mean we go back and revisit the passage, but in this case we will. Quick glance at the answer choices shows we’ll have to know about nucleophiles and electrophiles, so we’ll ultimately need our content to answer the question as well.

The passage says “The proposed mechanism of action by hAcyl involves binding of the substrate, deprotonation of the bound water by a glutamate at position 147, and subsequent reaction with substrate.” We’re explaining this deprotonation of bound water. What else do we know about this bound water from the passage?

For hAcyl, the resting state has H2O instead of the glycine (so at the bottom of the figure, replace the glycine with your water). By deprotonating this water (which is done by the glutamate at position 147) this creates a hydroxide, which is a great nucleophile. This hydroxide is able to attack an amide’s electrophilic, carbonyl carbon in a nucleophilic addition reaction. 

  1. amine group to be more nucleophilic, we said the water, and subsequently the hydroxide is created and is a great nucleophile. We’ll still keep this answer choice to compare to our other options. 
  2. carbonyl group to be more electrophilic. While the reaction does include the electrophilic carbonyl group, the deprotonation isn’t what causes this. This is why we always do step 1 and summarize the question, because we have to make sure we’re answering the question being asked. This is a better answer than option A.
  3. water molecule to be more nucleophilic. This is exactly what we described in our breakdown. The deprotonation process creates a great nucleophile. We can keep answer choice C as our superior answer. 
  4. zinc ion to be more electrophilic. This isn’t consistent with what is going on during the deprotonation and contradicts our breakdown. We can eliminate this answer choice and keep our best answer: answer choice C.

36) The researchers performed several amino acid substitutions in their study. This question wants to know why the substituted amino acid was amino acid “A”, or alanine, for every site variant. To answer this question, we’ll go back to the passage to reference the substitutions taking place. We’ll use our external knowledge about amino acids to explain the reasoning for the alanine substitution and the effect it has.

We have our table from the passage that shows all of our variants. Like we mentioned, there is an alanine substituted in every variant. H80A means the 80th residue was histidine, but has now been replaced by alanine. D113A means the 113th residue was aspartate, but has now been replaced by alanine. And amino acid E is glutamate and that’s replaced by alanine.

So that’s 6 substitutions from amino acids to alanine. We can pull up the amino acid chart here for reference:

What do we know about alanine? Alanine is classified as hydrophobic (It’s the 2nd smallest nonpolar amino acid. And it only has an uncharged CH3 as its side group). The previous esterases are all electrically charged. The acidic asparate and glutamate both have negative charges on their side groups. Basic histidine has a positive charge on its side group.

I’ve listed quite a few parts to our breakdown so far, but biggest change here is going from larger, charged side groups to an uncharged and much smaller side group. Alanine is more unique for being a smaller side group. There are quite a few other amino acids that are uncharged. Let’s look at our answer choices and see if our breakdown can help us narrow them down or pick one.

  1. reduces the net charge on the bimetallic center. Like we just mentioned, the researchers could have used a variety of different, uncharged amino acids. Using alanine specifically isn’t consistent with answer choice A. 
  2. increases strength of substrate binding to the active site. Again, not the case here. The Michaelis constant decreases with the substitution. This answer choice contradicts what’s happening in the passage; we’re expecting weaker binding.
  3. increases the conformational rigidity of the active site and the enzyme. This is not true; the substitution should do the opposite of this. 
  4. reduces the interaction of the side chain with other active site components. This sounds plausible. We have one of the smallest side chains of any amino acid on alanine. The side chain is also uncharged like we mentioned, so that would decrease the interactions even more.

37) We need to explain the role of the metal centers in the catalytic center, so our two Zinc ions. Their role in catalysis (so we’ll use our two constants in table 1) and in conformation stability (so we’ll use melting point). Each metal was given a distinct letter in the figure in the passage, so we’ll find the role of each. We’re going to use the table and information in the passage, and external knowledge to see how the different substitutions in Table 1 affected each of our Zinc ions.

Here we have Table 1 and our Figure. Quick recap of the information in our Table that we looked at in our previous questions. What is our wild type? Wild type (WT) refers to the prototypical form of enzyme, as it occurs in nature. What do we see in the experimental data? We touched on this a bit already. Each variant that differs from the wild type has a decreased melting temperature, Tm, meaning both ions provide conformational stability that corresponds with a higher melting temperature.

Next, the catalytic constant, Kcat, measures the number of molecules of substrate turned over into product per unit time. Kcat is positively proportional to efficiency. So, the higher your kcat is, the more reactions happen per unit time when you have enough substrate. The fact that kcat is so much higher for the wild type compared to the other variants means both ions are involved in catalysis also. Final answer is going to involve: both ZnA and ZnB provide confirmation stability and are involved with catalysis.

  1. ZnA is responsible for catalysis, and ZnB provides conformational stability. These are both true statements, but not necessarily exclusively true. Zn-B is also responsible for catalysis, and Zn-A is also responsible for conformational stability.
  2. ZnB is responsible for catalysis, and ZnA provides conformational stability. This is the same answer as answer choice A. Both statements are true again, but the answers are only partially complete.
  3. Both ions are involved in catalysis, and both provide conformational stability. This matches our breakdown exactly. Answer choices A and B both said essentially the same thing, so answer choice C is our best answer so far.
  4. Both ions are involved in catalysis, but neither provides conformational stability. The second part of the answer choice contradicts our breakdown. We said the melting temperature proved both ions provided conformational stability as well. We can eliminate answer choice D and we’re left with our correct answer, answer choice C.

38) We want to explain the result of a specific substitution and its effect on the enzyme as a whole. E147D variant means glutamate is substituted for by aspartate. The passage says the melting point following this substitution was below 37 degrees C when this variant was observed. This means conformational stability decreased quite a bit from the wild type and enzyme activity also decreased. What do we know about the specific substitution? Both amino acids are negatively charged & acidic, meaning charge doesn’t change. The substituted variant won’t act as more of a base or acid. There’s not a ton of size difference either-side chain on aspartate is actually smaller than the side chain on glutamate. If you need to, review the amino acid chart in Question 36 to help you out here.

  1. a repulsive interaction that reduces conformational stability and thus enzymatic activity. We know for sure there is reduced conformational stability. Melting point is below 37 degrees, while the wild type and other variants were all higher. 
  2. a change in net charge that changes the reactivity of the nearby Zn ion. There was no change in net charge. Both amino acids are negatively charged. Answer choice A remains the superior answer choice so far. 
  3. increased steric hindrance of the active site and reduced access by the substrate. Like we mentioned, there is no increased steric hindrance. The side chain on aspartate is actually slightly smaller. Once again, answer choice A remains superior.
  4. increased capacity of the site to act as a general base which eliminates the proposed reaction mechanism. Another contradiction. We said both amino acids have the same electric charge, and both are acidic. We’re left with our correct answer choice, answer choice A.

39) We’re using the table in the passage to compare the differences between the wild types of each and the variants of each. 

Right away we can tell both substrate binding and catalytic turnover are affected because of the differences in Kcat and Km. We want to see which one is affected more relative to the other. The Kcat differs by roughly 10^-3 times. Substrate binding can change by about a factor of 6 or even 8. Both are affected, but Kcat is affected significantly more.

  1. substrate binding more than catalytic turnover. Both variables are affected, but this is the opposite of our prediction. We said catalytic turnover is affected more. We’ll still compare to our other options. 
  2. catalytic turnover more than substrate binding. This matches our prediction exactly. Both are affected, but catalytic turnover is affected more than substrate binding. 
  3. substrate binding, but not catalytic turnover.
  4. catalytic turnover, but not substrate binding. Answer choices C and D both contradict our breakdown. We said both variables are affected, but these both say only one. We can eliminate both answer choices C and D. We’re left with our correct answer choice, answer B

40) We’re going to have to use specifics about the enzyme from the passage, but external knowledge to actually list the state changes during the reaction. 

The passage says the enzyme is catalyzing hydrolysis of terminal peptide bonds. The enzyme does this by adding water to the carbonyl carbon of the terminal peptide bond. The mechanism says that the glutamate residue deprotonates water molecule to create a hydroxide (we mentioned this is a good nucleophile in a previous question). The hydroxide then attacks the carbonyl carbon in a nucleophilic addition reaction. The carbonyl carbon is now a tetrahedral intermediate with four substituents bonded to it. So, it went from sp2 to sp3. To complete hydrolysis, the double bond between the oxygen and carbon reforms. and the carbon-nitrogen bonds breaks. So, the hybridization of the carbon atom changes back to sp2.

So, our predicted state changes are sp2  sp3  sp2. We can compare all of the answer choices to find the one that matches our breakdown. Answer choice A matches our train of thought. The other answer choices all begin with sp3, and can be eliminated.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 6

41) The passage started out with lysozymes and the mechanism by which it works. We have to categorize the type of bond that is cleaved by lysozymes. 

The passage said lysozymes kill gram-positive bacteria by cleaving N-acetylglucosamine oligosaccharides. While we were reading the passage, we said N-acetylglucosamine is an amide derivative of glucose, and we know oligosaccharides are saccharide polymers with a small number of monosaccharides. We’re dealing with carbohydrates.

  1. Answer choice A says peptide bond. Peptide bonds typically happen between amino acids. That’s not what we said is happening here. We’ll still compare with our other answer choices.
  2. Answer choice B says Phosphate esters which are made of phosphoric acids and alcohols. Again, not what’s going on here. We’re dealing with saccharides. This contradicts our breakdown.
  3. Answer choice C says glycoside. Glycosidic bonds join carbohydrates together, or to other compounds. That sounds like a keeper! We’re cleaving N-acetylglucosamine oligosaccharides, so expecting glycosidic bonds between monosaccharides that will be cleaved. We can hold on to answer choice C and eliminate answer choices A and B.
  4. Answer choice D says pyrophosphate. Pyrophosphates contain two phosphorus and an oxygen in a P-O-P linkage. Not what’s happening here. We’re cleaving bonds that link monosaccharides, not pyrophosphates. We can eliminate answer choice D for contradicting our breakdown; we’re left with our correct answer, answer choice C.

42) We want to know the approximate change in entropy for binding of NAG3 to HEW at 27°C. We’re going to need to find specific values from the passage, but we’ll also need our general knowledge of changes in state functions-we’re dealing with change in Gibbs free energy. Note, we’re dealing with whole numbers, and the answers are all given in J/k. 

We’re given values of ΔH, the change in enthalpy and delta G, for Gibbs free energy. Let’s write out our equation: ΔG=ΔH–TΔS where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is temperature (kelvin), and ΔS is the change in entropy.

Remember, we need to convert our values to J and Kelvin.

Delta H is now =-50,000 J/mol
Delta G is now =-30,000 J/mol
T=27 degrees Celsius=300K (add 273 degrees for Kelvin)

Let’s plug in values and we’ll isolate for our entropy. 

ΔG = ΔH – TΔS
-30,000 J/mol = -50,000 J/mol – 300KΔS

ΔS=-67 J/K. We said our final answer is a whole number with joules and kelvin as the units. 

This was a math problem, and there was no rounding or approximation. The values in the answers are all fairly far apart, so we can compare all four at once. Our correct answer corresponds to answer choice B. All of the other answers can be eliminated for being incorrect values.

Another reminder about the importance of units. The math in the problem was not difficult. AAMC makes sure you can do every math problem in a couple of steps and round numbers to approximate. None of our values are even close to each other, but if you make a mistake in how you calculate the answer, it’s easy to get a different answer choice. That means it’s extra important to keep track of your numbers, and always manipulate your values properly.

43) We’re going to need to find specific details from the passage about the amino acids listed, but we’ll also need our general knowledge about hydrogen bonding.

Which amino acids are found at each number in the answer choices? Let’s work through each one.

  1. 62=W -tryptophan- that’s an aromatic amino and has a nitrogen in its side chain.
  2. 101=D -aspartate- is a negatively charged amino acid. It’s got that negative oxygen charge in its side chain
  3. 103=N -asparagine- is a polar amino acid and also has a nitrogen in its side chain.
  4. 107=A -alanine. That’s a nonpolar amino acid. It’s got the 2nd smallest and most basic side chain, it’s a methyl group. 

Before moving on, I’ve included the amino acids here. If you haven’t already, make sure you commit these to memory.

Quick comparison of our four answer choices. Alanine does not allow for potential hydrogen bonding. Amino acids typically will have a hydrogen atom attached to an oxygen or a nitrogen. Both atoms that are more electronegative than carbon, and can potentially form hydrogen bonds. We can eliminate answer choices A-C because those can form hydrogen bonds. Answer choice D corresponds to alanine’s methyl side chain. That’s going to be our correct answer choice here.

44) We’re thinking back to the passage and HEW lysozyme was titrated with injections of NAG3. How much NAG3 was needed to reach the equivalence point, and our answer is given in different units of moles. We’ll need to visit the figures in the passage and get specific details. But we’ll also have to know about titrations, and specifically equivalence points from our general knowledge. Let’s solve for our quantity in either nanomoles or micromoles

The number of mol of NAG3 must equal number of mol of HEW at the equivalence point.

The information about NAG3(25 injections, 10 micro L and 2.5mM) in the passage is about the TOTAL amount of NAG3 added in the titration but we don’t actually know what amount of NAG3 was needed to reach equivalence. 

The only thing the titration graphs tell you is that to reach equivalence you need a 1:1 mol ratio of HEW to NAG3. However, if you look back at the passage info about HEW, it tells us that there is 1 mL (1×10^-3 L) and .1mM (1×10^-4 Molar) of HEW. We can convert this to moles. Multiplying, we get 100nmol or 0.1 micromoles (1×10^-7 mol) of HEW. 

We said there are an equal number of moles. So, we’d also need 100nmol or 0.1 micromoles of NAG3 to reach the equivalence point.

We can compare all four of our answer choices at once. Our correct answer is answer choice B. We can eliminate answer choices A, C, and D for being incorrect values.

45) Our answer is going to come from the passage, and specifically the figures. We’re given units of M-1

The passage says that plotting KNAG3/Kapp vs [NAG] will allow you to determine KNAG. This is shown in Figure 2, and we’re told that’s what’s happening in the description of Figure 2 as well. We can find KNAG by finding the slope of the line comparing the two variables. Find the change in Y, divide by the change in X:

ΔY/ΔX = (2.0 – 1.0)/(0.020m – 0.00M) = 50/M

We can compare all four of our answer choices at once. Our correct answer is answer choice D. We can eliminate answer choices A, B, and C for being incorrect values.

46) We have the same heat curve and an apparent affinity constant equal to association constant in two different compounds. How can we change the conditions to have heat differences and have the association constant for Compound 1 be greater than the apparent affinity constant?

The affinity constant (KNAG3) essentially tells us the affinity of an enzyme for substrate. A higher affinity constant means more of the enzyme substrate complex is formed from the starting concentrations of enzymes and substrate.

Apparent affinity constant considers a second substrate. So in our passage we had enzume+NAG3, and our second substrate is NAG. So we now have a KAPP, not KNAG3. So our KNAG3 doesn’t change with the 2nd substrate. Only the Kapp does.

So in our graph, while we’re increasing [NAG], KNAG3/Kapp is also increasing. But we said KNAG3 isn’t changing. That means KAPP is decreasing. 

Adding more NAG will decrease KAPP to compensate and keep the slope the same. KNAG3 doesn’t change. 

  1. Dilute all of the solutions. The net effect here would cancel out. 
  2. Increase the concentration of Compound 1 in both titrations. We’re trying to decrease the apparent affinity constant between Protein A and Compound 1 here. 
  3. Increase the concentration of Protein A in both titrations. Same basic reasoning as answer choice B. We’re ultimately trying to decrease apparent affinity between Protein A and Compound 1.
  4. Increase the concentration of Compound 2 in the second titration. This is what we mentioned we’re trying to do in our breakdown of the question. We increase concentration 2, our NAG, and this ultimately decreases Kapp. We’re left with our correct answer, answer choice D: Increase the concentration of Compound 2 in the second titration

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 47-51

47) Like most standalone questions, this is going to rely on you knowing your content, and knowing the definition of different enzyme models. Right below you can see how a visual showing how enzymes and substrate fit together. Note what it says above “Enzyme/substrate complex.” It says the enzyme changes shape slightly as substrate binds. We’re looking for an answer that’s consistent with this. Typically, this is called the induced-fit model. We want an answer choice that’s similar or the same.

  1. transition-state model. Transition-state model is not something we typically hear on the MCAT or from AAMC’s content outline. Transition state is the state corresponding to the highest potential energy along this reaction coordinate. Not what we’re looking for in the question stem, so let’s keep looking for a better answer.
  2. active-site model. The active site describes the location on the enzyme that binds with the substrate. The question stem is not describing an “active-site model” as this is not something we’re expected to know from our MCAT content outline.
  3. lock-and-key model. This is model that asserts an enzyme and substrate fit together perfectly in one instantaneous step. This was the primary model used by scientists for many years, but current research supports the induced fit model. While the definition in the question stem doesn’t match this answer specifically, we can keep this as a viable option for the time being. It’s a more relevant answer than options A and B.
  4. induced-fit model. The induced-fit model proposes that the initial interaction between enzyme and substrate is relatively weak, but that these weak interactions rapidly induce conformational changes in the enzyme that strengthen binding. In other words, the question stem is describing the induced-fit model. 

48) Kcat, the catalytic constant, measures the number of molecules of substrate turned over into product per unit time (usually seconds). The higher your kcat is, the more reactions happen per unit time (i.e. Vmax). All of this is taking place when conditions are optimal, and you have enough substrate. So, Kcat is positively proportional to efficiency. Let’s see if we can find an answer choice consistent with what we’ve laid out for Kcat.

  1. limiting. I mentioned in the brief breakdown of the question and Kcat, we want conditions to be optimal, and we need sufficient substrate. Limiting substrate goes against what I said.
  2. equal to 1/2 KM. The Michaelis constant, Km, can help compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate. The higher km is, the more concentration of substrate you need before you reach optimal conditions for Kcat, so km is inversely proportional. When measuring Kcat, substrate concentration does not have to be equal to ½ Km
  3. equal to KM. This is similar to answer choice B in that we’re going to say that when measuring Kcat, substrate concentration does not have to be equal to Km. We’re still hoping for a better answer choice. 
  4. saturating. This matches what we said in the breakdown of this question; we want conditions to be optimal, and we need sufficient substrate. This is the best of the four answer choices listed. 

49) Let’s consider the following graph and put a name to the shape. We even have our equations up top, this is a hyperbolic relationship. 

  1. A hyperbolic dependence on [S]. This matches exactly what we said in the breakdown of the question and what we see in the image above. This is likely going to be our best answer. Let’s compare to be thorough.
  2. A linear dependence on [S]. Note the equations and the actual graph above. The relationship here is not linear. We can still stick with our best answer choice, answer choice A.
  3. A sigmoidal dependence on [S]. This is similar to answer choice B in that it’s factually incorrect. A remains the best option.
  4. A parabolic dependence on [S]. Just like B and C, this does not describe what we’re seeing in the above graph, nor is it consistent with our equations. We’ll stick with our best answer here, answer choice A. 

50) Uncompetitive inhibitors bind at a site other than the active site:

We can already see in the graphic that we have a decrease in both Km and Vmax. We can also look at a graph to illustrate the effect of this uncompetitive inhibitor: 

a. the KM increases and Vmax decreases. This would be possible if we were looking at mix inhibitors, which is not the case. We said we expect both Km and Vmax to decrease here.

b. the KM decreases and Vmax increases. This answer choice also contradicts what we’ve shown in our visuals. Let’s keep looking for a superior answer. 

c. both the KM and Vmax decrease. This answer choice matches what we see in both of our visuals. If you don’t know the characteristics of enzyme inhibitors, make sure to review:

d. both the KM and Vmax increase. This is also an incorrect answer based on our breakdown of the question. We can stick with our correct answer, answer choice C: Both Km and Vmax decrease. 

51) An anion-exchange column will bind to negatively charged molecules. We’re going to find an answer choice that has a net negative charge so it can actually bind. We’re told we want to find the peptide that requires the lowest concentration of NaCl for elution. That means we want a net negative charge for binding, but we also want the peptide to elute at a lower salt concentration. In other words, we want a negative charge, but the least negative possible. We’ll have to know our amino acids and their abbreviations:

  1. AVDEKMSTRGHKNPG We have a net positive charge here, so this is not a great option.
  2. YPGRSMHEWDIKAQP We have a net neutral charge here. Slightly better than option A, but still not ideal.
  3. HIPAGEATEKALRGD. This is going to be hard to beat. We have a net charge of -1 which is exactly what we’re looking for. 
  4. EAPDTSEGDLIPEVS. If answer choice C didn’t exist, this would be a viable option. However, the -5 net charge is too high. Answer choice C is our best option because it’s the least negative anion.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 7

52) The students had to determine the rate of reaction before plotting Figure 1. This question wants to know how the students were able to do this. We’re going to need the passage to determine which of the experimental steps allowed them to determine the rate of reaction. We’ll need external knowledge about laboratory techniques to analyze the steps and the data we’re given. Quick glance at our answer choices show we’re going to have an answer dealing with absorbance at different wavelengths.

We’re told compound 1 produced beta-D-galactose a yellow Compound 2 with lactase. Visible light is found at wavelengths in the high 300s to mid 700s. Yellow is in the middle, at around 550 nanometers. If yellow is being observed in Compound 2, that means there is absorption of the complementary color to yellow at the same time. 

Yellow and purple are complementary colors. Purple is in the high 300 nm, and at that wavelength, light is absorbed and yellow is reflected. So, there’s more absorbance of purple light in the yellow compound. An increase in production of Compound 2, the yellow product, would mean that the absorption of purple light would increase. We have to know which colors are complementary to answer this. It’s not mandatory to be able to draw out the entire color wheel, but you should know complements.

  1. Monitor the increase in absorbance of the solutions at 200 nm. The wavelength here is a little low based on what we said in our breakdown, but we can still keep this answer choice to compare to our other options.
  2. Monitor the increase in absorbance of the solutions at 360 nm. This is closer to our prediction. We expected an increase in absorbance of the complementary color to yellow. Purple is found at the high 300s and low 400 nm wavelength. So we’ll keep this answer choice. It’s better than answer choice A, so we can eliminate that answer choice.
  3. Monitor the decrease in absorbance of the solutions at 200 nm. Any decrease in absorbance could be at a higher wavelength. In this case, we’re expecting that increase in purple absorbance at 360. This contradicts our breakdown, so we’re sticking with answer choice B for now as our superior option. 
  4. Monitor the decrease in absorbance of the solutions at 360 nm. Another decrease in absorbance which we said contradicts our breakdown. We’re left with our correct answer choice, answer choice B.

53) We’re going to need the passage for details about lactose being hydrolyzed by lactase, and the role of water. We’re also going to use external knowledge to know about the results of the cleavage reaction.

First a little background. Lactose is a disaccharide that is made up of two monosaccharide residues: glucose and galactose. Lactose is formed through the condensation of galactose and glucose; hydrolysis is the breakdown of a compound due to the addition of water. Glutamate deprotonates water. Then the remaining hydroxide ion, that was derived from this deprotonated water, nucleophilically attacks the bound galactose to free it up. That means the oxygen is going to appear on the galactose product.

  1. neither the galactose nor the glucose products. We know this isn’t the case for the galactose, even if it is for the glucose
  2. the glucose product only. This is the opposite of our prediction. The glucose doesn’t react with the oxygen provided by water, that’s what happens with galactose. The glucose only has a new proton, not a new oxygen.
  3. the galactose product only. This is consistent with our prediction. We can keep this answer choice for now and eliminate answer choices A and B. 
  4. both the galactose and the glucose products. This is only partially correct. We confirmed this is the case for galactose, but not the case for glucose. Our correct answer here is going to be answer choice C.

54) To clarify, we want to know why glutamate 1 of lactase acts as a proton donor and donates a proton to the oxygen atom. We’re going to need the passage for details about this process. But we’re also going to need external knowledge about Bronsted acids, and nucleophilic substitutions. We need to think of this question in terms of nucleophiles and leaving groups.

First thing we want to note, is the passage says Glutamate 2 acts as a nucleophile in this step to liberate D-glucose. Our answer is going to have to do with either G2 being a better nucleophile, or glucose being a better leaving group. The question is asking in terms of G1 being a Bronsted acid though. G1 protonates the glycosidic oxygen so glucose becomes a better leaving group. Even though G2 is the nucleophile in this situation, it’s not as relevant to the mechanistic reasoning of G1 acting as a Bronsted acid.

This question was a bit open-ended, but in this case, our breakdown ends up matching answer choice C. Based on our breakdown, A and B are out of scope because they don’t answer the specific question being asked, and D is a contradiction. We can eliminate those answer choices and stick with answer choice C as our best option.

55) In other words, we need to find the Michaelis constant of compound 1 with respect to lactase. We’re going to need the passage for the Lineweaver-Burk plot. We can use external knowledge to actually read the plot. Quick glance at our answer choice for units. We’re trying to find an answer choice in millimolar.

X-intercept in a Lineweaver-Burk plot is -1/Km and that the y-intercept is 1/Vmax. You can label both on the graph. Remember, it’s important to understand any visuals presented in the passage, and what our experimental data means.

X-intercept here is at -0.05 millimolar. Solving for Km gives 20mM

That means we’ve calculated the value of our Michaelis constant as 20 millimolar. We can compare all four of our answer choices at once. Our correct answer is answer choice D. We can eliminate answer choices A-C for being incorrect values.

56) Just like the previous question, we’re going to need the passage for the Lineweaver-Burk plot. We can use external knowledge to actually read the plot and find Vmax.

We went over this question while we were just answering our last question. Vmax is the reaction rate when the enzyme is fully saturated by substrate. We said the X-intercept in a Lineweaver-Burk plot is -1/Km and that the y-intercept is 1/Vmax. We can label both on the graph. In this case, we didn’t have to find the specific value for Vmax, but rather we want to know how we can solve for Vmax (using the y-intercept of the graph). We can pick our correct answer choice, answer choice D. Making these connections while reading the passage can really pay off. If you took some time to look at the main parts of that graph, you would’ve been able to answer questions 55 and 56 within seconds. We all know time is extremely valuable on the MCAT. A few seconds extra on the passage can shave entire minutes off your question-answering time!

57) We need to know about the concentrations of our two samples so we can write an expression that relates the two; we’ll get that from our passage.

As always, carefully look at the units and find the concentrations for each step. Stock solution was made by diluting 0.100 mL of the commercial preparation to 25.0 mL in the buffer solution. We can divide the two. This means the commercial preparation was diluted 1250. 

Next, 1.0 mL of substrate solution was mixed with 1.0 mL of the enzyme solution to further dilute it. It’s now half as concentrated so 12 here.

Final solution means we multiply: 1/250 x ½  to get 1/500th of the commercial preparation. So, the concentration of lactase in the commercial preparation of the enzyme is [E]T x 500. This was a calculation problem and we can compare all four of our answer choices at once. Our correct answer is answer choice D. We can eliminate answer choices A-C for being incorrect values.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 8

58) The passage gave specific details about the makeup of DNMT3a, so we can go back and get these details.

Our enzyme in question is classified as a homotetramer. That’s a 4-subnit complex that’s made of four identical subunits.

SDS-PAGE showed a single band at 35 kDa. 

SDS-Page denatures proteins and breaks down quaternary structure, meaning the band on the SDS-Page will only be from ¼ of the homotetramer weight after it is denatured. That means the weight of each subunit was 35 kDa. We can multiply the weight of each subunit by the total number of subunits in our homotetramer to get 140 Kda.

This is a math problem where we did no rounding, and our answer choices are relatively far apart from one another. We can pick the answer that matches our calculation, answer choice D. Eliminate answer choices A-C for being incorrect values.

59) Recall from the passage that was a modification at the C5 position by our methyltransferase. We have to know about the structure of cytosine and how methylation works. 

Cytosine, thymine, and uracil are all pyrimidines which are single-ringed. All answer choices given to us are single rings as well. We have to make sure we pick an answer that’s cytosine.

Passage said methylation occurs at the C5 position of the cytosine base. Number your rings so that the nitrogens end up with the lowest number combination. The nitrogen on the bottom of every answer choice is going to be numbered as 1 in the ring, the carbonyl is #2, so we’re numbering counterclockwise from the bottom-most nitrogen in each base. We’re looking for a methyl group at position 5.

Biggest difference between answers A&B and C&D is position 4 in the ring. A and B have an amino group, while C and D have another carbonyl. Answer choices A and B are both cytosine, while answer choices C and D represent thymine. We can eliminate answer choices C and D.

Answer choice A has a methyl group at position 6 on the ring, but the passage said methylation occurs at C5. That’s consistent with answer choice B. We can eliminate answer choice A for contradicting the passage. We’re left with our correct answer, Answer B.

60) Tritium is the hydrogen isotope from the passage. We’re told the half-life of tritium, and now we have to determine how many half-lives have passed.

The half-life of our tritium is 4500 days according to the passage. So, every 4500 days, the concentration of our tritium falls to half of its previous value. 13,500 days is the equivalent of 3 half-lives of 4,500 days each. If the initial concentration is X, we need to solve for the concentration after 3 half-lives. There’re two ways to do this. We’re taking ½ of the concentration 3 times. So X times (1/2)3 = 1/8 of our initial concentration.

Alternatively, we can do it one step at a time. After 4500 days, ½ of our concentration remains. After another 4500 days, ½ of ½ remains, or ¼. And after the last 4500 days, ½ of ¼ remains, or 1/8.

This is a math problem where we did no rounding. We solved for the fraction of our starting amount (two different ways) and got answer choice A, 1/8 of the starting amount. We can eliminate answer choices B-D for being incorrect values.

61) Chromatography is used to separate components of a mixture, so we want to know the type of chromatography that was used to purify DNMT3a. 

We have the excerpt from our passage here. It says the DNMT3a was purified with nicel-nitrilotriacetic acid-agarose chromatography. 

In chromatography, a mixture is separated as it passes through a medium. The components of the mixture will move at different rates based on their affinity for the medium through which they travel. Nickel columns are used to purify molecules in this experiment. The key component of protein purification through polyhistidine tags is its affinity purification. It exploits the molecular properties and the selective interactions of macromolecules (in this case the protein).

  1. affinity. This matches our breakdown. I said affinity chromatography can exploit selective interactions and molecular properties, like was done in the passage. Polyhistidine tages are used for affinity purification. I’m liking this answer choice for the time being, let’s keep comparing.
  2. anion-exchange. Anion-exchange chromatography is a process that separates substances based on their charges using an ion-exchange resin. The resin contains positively charged groups and has an affinity for molecules with net negative surface charges. Not what’s taking place in this experiment, we can eliminate this answer choice. 
  3. cation-exchange. Cation exchange chromatography separates molecules based on their net surface charge. It uses a negatively charged ion exchange resin with an affinity for molecules having net positive surface charges. This is the opposite of answer choice B, and neither is indicative of what’s going on here. 
  4. size-exclusion. In size-exclusion chromatography (SEC), molecules are separated based on their size. Separation is achieved through differential exclusion where the biggest molecules elute the fastest, while the smallest molecules elute the slowest. We’re not separating based on size, but rather affinity. That means we’re left with our correct answer choice, answer A.

62) We have to decide between 3 options, and with these Roman numeral questions, we can have multiple reasons be correct. We’re going to need to reference the passage to see where it talks about the wash step in the experiment, and any details about its purpose. We’re going to use our knowledge of biochemical laboratory techniques to reason out the purpose.

Passage states samples were loaded onto plates containing immobilized avidin, which strongly binds biotin. The wash would therefore not be designed to remove the biotinylated substrate which binds the immobilized avidin. That eliminates choice I, we can’t wash away the immobilized and bound substrate. We now know II is a correct answer choice. Non-biotinylated substrate isn’t bound to the avidin, so it’s not immobilized. Next, we have to decide about III (methylated cytosine). We’re ultimately measuring methylation rates, meaning III wouldn’t be washed out. If it was, we’d have no experiment. 

Option II is a correct option and comparing all 4 of our answer choices, we see answer choice B says II only. That is going to be our correct answer, we can eliminate answer choices A, C and D. The wash is designed to remove non-biotinylated substrate.

63) To answer this question, we want to know which statement of our four answer choices is consistent with the data in Figure 1. That means we’re going to need to reference the passage to see Figure 1. We’ll analyze the figure and use what we know about the experiment in the passage to answer this question.

Above we have Figure 1 from the passage. We’re observing relative methylation rate of biotinylated substrate. We say biotinylated, because any non-biotinylated substrate has been removed from the plate by the washing process (this is something we covered in the previous question). In theory, the longer the methyltransferase is bound to the DNA, we should see more methylation by sheer volume because there is more cytosine to be methylated. 

  1. Biotinylation prevents DNA strands from binding to Dnmt3a. We see a large amount of methylation in condition 1 of the 509mer. Biotinylation does the opposite of answer choice A.
  2. The stability of the protein/DNA fibers formed on the 30mer and 509mer DNA strands was the same. We see differences between methylation in the 509mer and 30mer. Even with our error bars. It’s not likely the stability of the fibers formed was the same. Our methylation rate would be identical if that were the case.
  3. The stability of the protein/DNA fibers formed on the 509mer DNA strand was greater than those formed on the 30mer DNA strand. We have an increase in methylation rate in Condition 1 compared to our control, Condition 3. Another interesting thing we note is non-biotinylated DNA’s methylation rate is decreased relative to Condition 3.
    We don’t have that same discrepancy in the 30mer. Biotinylated vs non-biotinylated doesn’t make a difference, even though biotinylation should increase methylation. The methyltransferase isn’t able to methylate cytosine on such a short strand, and that adversely affects the stability of the protein & DNA fibers.
  4. Biotinylated DNA strands are methylated at a higher rate than non-biotinylated DNA strands. This looks correct from the perspective of our 509mer, but we’re only going by the data in Figure 1. The biotinylated DNA is methylated at the same rate as the non-biotinylated DNA strands. So, comparing all of the 4 options. Option C seems like the best option based on Figure 1.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 64-67

64) For this question, we need some background on SDS-PAGE gels. We have to know proteins can be chemically denatured before running them on a gel and proteins are separated exclusively by size. The chemical SDS denatures proteins, and so denaturing SDS-PAGE gels are used for these experiments to separate proteins by size. We’re told there are no disulfide interactions unless we’re explicitly told there are. Disulfide bonds between cysteine residues are not fully broken by SDS alone. The key here is finding the smallest subunits that will be broken down. The protein with the smallest subunits has the highest electrophoretic mobility in SDS-PAGE. Let’s go through our 4 options and consider size.

  1. Protein 1. Protein 1 is a monomer, and at first glance through our answer choices this would be the smallest protein of the 4 options listed. However, we don’t just want the smallest protein, we want the smallest subunits. We’ll keep the 32 kDa size in mind and keep comparing. 
  2. Protein 2. We want to make sure to not get tricked here. In the question stem, we’re told there are no disulfide interactions unless we’re explicitly told there are. Disulfide bonds between cysteine residues are not fully broken by SDS alone. That means the size here is going to be 38 kDa and Protein 1 is still our superior answer. 
  3. Protein 3. This homotrimer is comprised of 25 kDa monomers, meaning when it’s broken down, we’ll have 25 kDa subunits. Protein 3 will have the highest electrophoretic mobility of the first three answer choices we compared.
  4. Protein 4. Protein 4 is similar to protein 3 in that we’ll have subunits that are 38 kDa, despite the homodimer itself being larger. We’re still going to stick with the smallest subunits in this situation: Protein 3.

65) To answer this question we have to think about the movement of electrons within the context of breaking of disulfide bonds. This is an open-ended a very broad question, so glancing at our answer choices, we can see we’re going to be considering NADH and NAD+. Think of which of the two we’d need. NADH can split into NAD+, a proton, and a pair of electrons. NAD+ can accept two electrons and a proton to form NADH. In this case we want the former. One mole of NADH can reduce one of the disulfide bonds through the donation of two electrons. However, we have a protein with four disulfide bonds. That means we’re going to need four moles of NADH as well to break these four bonds. This is not a subjective answer, so we can pick the answer choice we came up with in our breakdown, answer choice B: 4 moles of NADH for each mole of protein.

66) This is a fairly wordy question, so we can pick out the important bits. We’ll likely have to know the properties of steroid hormones versus peptide hormones. We’ll have to know how each reacts to the mixture, and the extraction method. 

Steroid hormones are a type lipid hormone and are usually ketones or alcohols. Steroid hormones are insoluble in water, so transport proteins carry them in the blood. As a result, steroid hormones remain in circulation longer than peptide hormones. Peptide hormones are water-soluble and act on the surface of target cells via secondary messengers. 

  1. Estrogen would be in the hexane phase; insulin would be in the aqueous phase. This answer choice is consistent with the idea that the steroid hormone, estrogen, would be in the hexane, or hydrophobic phase. We said steroid hormones are insoluble in water. It also makes sense that insulin would be in the aqueous phase. That’s because peptide hormones are water-soluble. This answer choice hits everything we were looking for in our breakdown. We can quickly go through our other options to be thorough.
  2. Both estrogen and insulin would be in the aqueous phase. This answer choice is only half correct. We expect insulin to be in the aqueous phase, but we expect estrogen to be in the hexane phase. We already went through the reasoning in our breakdown and in answer choice A. Answer choice A remains the best answer.
  3. Insulin would be in the hexane phase; estrogen would be in the aqueous phase. This answer choice is the opposite of what we expect. Steroid hormones are insoluble in water and peptide hormones are water-soluble. We can eliminate this answer choice as well.
  4. Both estrogen and insulin would be in the hexane phase. This is similar to answer choice B. It’s half correct, but we expect insulin to be in the aqueous phase, not the hexane phase. We can eliminate answer choice D. We’re left with our best answer, answer choice A. 

67) pH of 7 is neutral, so we’re looking for a protein that will have the largest negative net charge at this pH. We might be looking at the amino acid profile of this protein, or we might be looking at what the protein binds. A lot of options here because this question is so open-ended, so we can dive into the answer choices.

  1. A protein that binds to an anion-exchange column at pH 7 and requires a high concentration of NaCl for elution. In an anion-exchange column we expect negatively charged molecules (hence the name anion-exchange) to bind to the positively charged column. What does it mean if it requires a high concentration of NaCl for elution? The bound molecule has a very negative charge relative to the positively charged column. This sounds like a great option for now. Let’s keep comparing with our additional answers to be thorough.
  2. A protein that binds to an anion-exchange column at pH 7 and requires a moderate concentration of NaCl for elution. This is similar to answer choice A, but now we have a moderate concentration of NaCl for elution. That means there’s less of a negative charge on the protein. The protein is not bound as tightly, so less NaCl is required for elution. Answer choice A is going to be a better option because the protein requited a high concentration of NaCl for elution. We can eliminate answer choice B.
  3. A protein that binds to a cation-exchange column at pH 7 and requires a high concentration of NaCl for elution. A protein that binds to a cation-exchange column would be positively charged. Right away we can eliminate this answer choice.
  4. A protein that binds to a cation-exchange column at pH 7 and requires a moderate concentration of NaCl for elution. This is similar to answer choice C. A protein that fits this description will have a positive charge, not a negative charge. Answer choice A fit our criteria the best and is our best answer.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 9

68) To answer this question, we’re simply going to count the number of purines and pyrimidines in Figure 1 from the passage. 

Which of our nucleobases are purines and which are pyrimidines? 

Thymine, uracil, and cytosine are classified as pyrimidines. Adenine and guanine are purines. If you ever forget which nucleobases are classified as which, I’ve always used a few short phrases. 

Pure As Gold (purines = adenine, guanine) -2 rings

CUT The Pie (pyrimidine = cytosine, thymine, uracil) -1 ringed.

Make sure to commit this visual to memory because it can mean easy points on test day:

All we have to do is count the number of purines and pyrimidines here. There are 17 purines and 9 pyrimidines. 

Purines are listed first for every answer choice, and the only answer choice that matches our prediction is answer choice A, there are 17 purines. We also said there are 9 pyrimidines, so our answer choice checks out. We can eliminate answer choices B-D.

69) To answer this question, we’re going to need Figure 2 in the passage to see the experimental results. Then we’re using general knowledge to actually interpret the results. 

Gel electrophoresis can be used to separate macromolecules based on size and charge-smaller molecules will migrate across the gel more quickly than larger molecules. 

The native gel shows the HPmutant band further down the gel than the wild type, or the MBmutant bands. This means the folded structure of HPmutant is more compact than the other two. 

When all three samples are denatured, they all migrate the same distance along the denaturing gel, meaning the number of nucleotides is identical. If you are unsure about gel electrophoresis, make sure to reference the following figure: 

  1. The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is less compact. The first part of the answer choice is correct, and we saw this from the native gel. But the second part of the answer choice says HPmutant M1 is less compact. The HPmutant band migrated more quickly, meaning it is more compact than the wild type and MBmutant.
  2. The folded structures of wild-type and MBmutant M1 mRNA are similar, but the structure of HPmutant M1 mRNA is more compact. Again, first part of the answer choice is correct. Second part correctly says the structure of HPMutant M1 mRNA is more compact, just like we said. Answer choice B is our best answer choice so far.
  3. Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has fewer base pairs. We expect all 3 to have the same number of base pairs based on the results of the denaturing gel electrophoresis, so answer choice B remains superior.
  4. Wild-type and MBmutant M1 mRNA have the same number of base pairs, but HPmutant M1 mRNA has more base pairs. This is the same issue as answer choice C. All 3 variants have the same number of base pairs; our correct answer is answer choice B.

70) To answer this question, we’re trying to do base pair substitutions to increase the stability of wild-type M1-mRNA. We have to pick one of the 4 options that would increase stability the most. 

There are three hydrogen bonds between cytosine and guanine. There are two hydrogen bonds between adenine and uracil. More hydrogen bonds means more stability. 

We can do a quick example in DNA also: Guanine forms three hydrogen bonds with cytosine to form a stronger bond than is found between adenine and thymine. There are only two hydrogen bonds found between adenine and thymine. DNA that has additional guanine and cytosine bases will have the most hydrogen bonds between strands. Samples with the highest adenine and thymine content would be the easiest to break apart.

  1. Changing position 132 to a G and position 151 to a T. We’re dealing with mRNA, so we’re not going to pick an answer that gives us a thymine.
  2. Changing position 134 to an A and position 149 to a U. Both adenine and uracil form two hydrogen bonds, so this won’t change the stability.
  3. Changing position 129 to an A and position 154 to a U. Same as answer choice B, both adenine and uracil form two hydrogen bonds each.
  4. Changing position 133 to a G and position 150 to a C. In this answer choice, we replace one adenine and one uracil with a guanine and cytosine. So, we go from two hydrogen bonds to three, and increasing stability. We can eliminate answer choices A-C and our correct answer choice is answer choice D.

71) We’re going to need Figure 3 from the passage, but we’re also going to have to be able to understand the data and associate the graph to melting temperature. 

The question is asking about the solution containing 100 mM potassium chloride, or the most solid line in the figure above. That was the one that has the most unfolding at the lowest temperature. 

For nucleotides, Temperature of Melting (Tm) is defined as the temperature at which 50% of double-stranded DNA is changed to single-standard DNA. When does that happen? When the fraction unfolded is at 0.5, or 50% on the graph. We can estimate the temperature at that point to be slightly above 50 degrees Celsius, we can call it roughly 53 or 55 degrees. 

Looking at the answer choices, none of our options are within more than 8 degrees Celsius of each other, so our approximation should be good enough here. The only answer choice that matches our low 50s range is answer choice B.

72) To answer this question, we’re going to need Figure 3 from the passage, but we’re also going to have to be able to understand the data and the influence on stability by each ion.

We have Figure 3 here, and the four ions in question are on the far left. We simply need to interpret the graph and the results of the experiment. We want the ion corresponding to the least amount of denaturation or unfolding:

100 mM potassium chloride: most unfolding (mRNA denaturation)

100 mM potassium chloride + 10 mM MgCl2: less unfolding than the 100 mM potassium chloride. MgCl2 must have helped to stabilize the mRNA. Since Cl- is in the potassion chloride salt also, the stabilization must be from Mg2+.

300 mM KCl + 10 mM MgCl2: only slightly more unfolding, but MgCl2 still helps to stabilize the mRNA.

1 M NaCl: also low unfolding, but we had to add a lot more salt to get the same stabilizing effect.

Answer choices: K+ and Cl- both correspond with more denaturing so we can eliminate these answer choices. Mg2+ helped. Choice D was Na+ which also helped, but less effectively given the amount needed.

Ultimately, a lot of NaCl had the same effect as a tiny bit of MgCl2. 

Think of it this way, if you study using Jack Westin materials for 2 weeks, and then a different course for 20 weeks, but you still get the same MCAT score. Then you would tell all your friends to just use Jack Westin materials because they’re the best and most effective. Same with our magnesium ion here. Our correct answer choice is answer choice C-we can eliminate answer choices A, B, and D.

73) We’re going to need the passage to get the concentration of magnesium chloride used. We’ll convert our mass to moles using the molar mass of magnesium chloride. Once we have our number of moles, we can solve for approximate volume. Quick glance at our answer choices shows our answer will be in microliters or milliliters. 

Above we have Figure 3 here once again, and we have information for elemental magnesium and chloride on the left side. 

First step is to solve for number of moles in 0.5 mg of MgCl2. We can find molar mass using individual masses and get 95.3 g/mol. 

Use dimensional analysis. 0.5 mg can be converted to grams, multiply by a factor (1 g/1000mg) to get 0.0005 grams. Divide grams by molar mass to get the number of moles of magnesium chloride to be 5 x 10-6 moles MgCl2

Looking at our graph, concentration =10mM, meaning 10-2 moles MgCl2 / Liter solution

Divide our number of moles by molarity to isolate our volume in liters:

5 x 10-6 moles MgCl2 divided by 10-2 moles MgCl2 / 1 Liter

Volume is =5×10-4 L. We said our answer will be in microliters or milliliters. 

Volume is 0.5 mL or 500 microliters. Only answer choice that matches our calculation is answer choice A, 500 microliters. We can eliminate answer choices B-D. 

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 10

74) To answer this question, we’re going to need to reference a few things from the passage: the list of amino acids in the substrate binding pocket, and the bonds that occur at the substrate binding pocket. 

The passage mentioned that the amino acid residues found in the substrate binding pocket engaged in hydrogen bonds. The four amino acids mentioned were the four given here in the answer choices. 

We’ve got 4 different amino acids in the glucokinase active site: arginine, glutamate, aspartate, and glycine. We’ll have to know the properties of these amino acids. This is vital information that you’ll need to know for the MCAT, so make sure you don’t miss points by not knowing your amino acids!

Arginine, glutamate, and aspartate are all electrically charged amino acids. Glycine is a nonpolar amino acid and has the smallest side chain of all the amino acids, a single hydrogen. Aspartate and glutamate both have negatively charged oxygen in their side group, while arginine has a positive charged nitrogen.

Amino acids can engage in hydrogen bonding through their side chains when they have electronegative atoms like oxygen or nitrogen. Both atoms that are more electronegative than carbon, and can potentially form hydrogen bonds. Remember our three main elements that form hydrogen bonds for the purposes of the MCAT: it’s our FON elements: fluorine, oxygen, nitrogen. Glycine’s side group is the simple hydrogen atom, so we’re not going to see the same electronegativity difference or hydrogen bonding. Correct answer here is going to be answer choice D.

75) To answer this question, we’re going to need to reference the table in the passage for specific values, but we’re ultimately using general knowledge to interpret the data and solve for catalytic efficiency.

Catalytic efficiency measures how well an enzyme is able to “grab” its substrate and turn it into product. this means you want the greatest number of reactions happening, for the least amount of input of time/substrate. Catalytic efficiency is given by: Kcat/km. 

Kcat measures the number of molecules of substrate turned over into product per unit time (usually seconds). kcat is positively proportional to efficiency. The higher your Kcat is, the more reactions happen per unit time.

The Michaelis constant, Km, can help compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate.

We want the highest ratio of Kcat to Km. Highest Kcat is the wild type. Km ATP is much lower than Km galactose. Highest ratio is going to be wildtype with respect to ATP. That’s going to be our highest ratio, and highest catalytic efficiency possible.

  1. wild type with respect to galactose. The Km for galactose is much higher than that of ATP.
  2. wild type with respect to ATP. This matches our breakdown; it’s going to be the highest ratio. We can eliminate answer choice A and hold on to answer choice B as the superior answer.
  3. D45A with respect to galactose. This is one of our lower ratios. Km is highest in this situation, so our ratio is going to be smaller. We can also eliminate this answer choice. 
  4. D45A with respect to ATP. This was our only other plausible answer choice. But looking at Kcat, answer choice B is still the better answer choice. We can eliminate answer choice D and we’re left with our correct answer choice: the highest catalytic efficiency results from wild-type with respect to ATP, or answer choice B. 

76) Note the natural substrate of GalK is galactose, not glucose. We need to describe galactose’s structure. This is almost like a standalone, or discrete, question.

Galactose is a monosaccharide and six-carbon aldose. Even if you did not remember the structure of galactose, recall it is an epimer of glucose. Galactose and glucose are both epimers at C-4. Epimers are stereoisomers that differ in their configuration at a single stereocenter. The position of the -OH group on carbon C-4 is in opposite directions in galactose and glucose.

Answer choice C matches our breakdown exactly, galactose is a six-carbon aldose.

77) To answer this question, we’re going to have to reference the figure in the passage, and any additional information given about this variant. We’ll need to use general knowledge to reason out why Km was smaller for this variant.

We have Table 1 here. We have to explain the smaller Km values for the D45G galactokinase variant. The amino acid switch is from aspartate to glycine in this case. Let’s jump into Km, and we touched on Km a bit earlier in our question set. Km is equal to substrate concentration when reaction rate is ½vmax. It’s the Michaelis constant, and it can help us compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate. 

The rate of a reaction is labeled as v and vmax is the maximum enzymatic reaction rat, so a lower Km corresponds to a lower vmax. 

  1. the substrate binding pocket is too crowded. Think about the amino acid substitution taking place in this variant. From aspartate to glycine will reduce crowding in the pocket. Glycine has the smallest amino acid side chain. If you need to, review the amino acid chart provided for question 74.
  2. a key hydrogen bond between the enzyme and substrate is lost. This could be a plausible explanation, but the substitution from aspartate to alanine in the other variant also means it lost potential hydrogen bonding, but Km was not affected the same way. This is still a better option than answer choice A
  3. the enzyme is unfolded as a result of the substitution. We still see some activity from the enzyme, so it’s not unfolded. We can eliminate this answer choice for contradicting the information in Table 1.
  4. the Vmax is much lower, which means less substrate is needed to reach it. This matches our breakdown and what we said about Vmax. A lower Vmax means a lower Km. Answer choice D is our best answer choice.

78) To answer this question, we can reference the passage for the specific aspects of our reactions. But we’re going to need general biochemistry knowledge to identify the full reactions taking place.

We have an excerpt from our passage here. Galactokinase transfers a phosphate group from ATP to galactose, producing ADP. This ADP can be converted back to ATP by pyruvate kinase. PEP + ADP in the presence of pyruvate kinase yields pyruvate and ATP. 

Pyruvate is converted to lactate by lactate dehydrogenase and NADH is converted to NAD+ in the process. Considering the four answer choices: lactate, NAD+, and ATP are all end products. ADP is a reactant that is part of the conversion from PEP to pyruvate. ADP is going to have the lowest concentration of the answer choices listed. That means our correct answer choice is answer choice B and we can eliminate answer choices A, C, and D.

79) This is going to be similar to our last question. We can reference the passage for the specific aspects of our reactions. But we’re going to need general biochemistry knowledge to identify the full reactions taking place and what is being oxidized and what is being reduced. Quick glance at our answer choices shows our answer is going to be in terms of NADH or NAD+ and pyruvate or PEP.

We have an excerpt from our passage and we’re told the reaction was monitored by the decrease in concentration of NADH.

This happens in the conversion of pyruvate to lactate. Pyruvate is our reactant, lactate is our product. Reaction happens via lactate dehydrogenase and NADH is oxidized to NAD+. If NADH is being oxidized, our reactant pyruvate is being reduced. 

NADH is oxidized (loses electrons) to form NAD+. Pyruvate is reduced (gains electrons) and forms lactate. 

     a.       NADH is oxidized and pyruvate is reduced. This matches our breakdown exactly, so this is an early frontrunner for our best/correct answer.
      b.       NAD+ is reduced and pyruvate is oxidized. NAD+ could be reduced in the reverse reaction, but we said concentration of NADH decreases in the passage. We can eliminate answer choice B.
      c&d.  PEP is oxidized or reduced. The formation of pyruvate from PEP isn’t coupled with NADH to NAD+. This isn’t a redox reaction, so we can eliminate answer choices C and D for contradicting our breakdown of the question. We’re left with our correct answer choice, answer choice A.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 80-84

80) Like most standalone questions, this question relies completely on knowing your content. To answer this question, we can consider DNA base pairing:

  1. A. Adenine contains 1 hydrogen bond acceptor
  2. C. Cytosine contains 2 hydrogen bond acceptors, so it’s going to be our best answer choice.
  3. G. Guanine contains 1 hydrogen bond acceptor
  4. T. Adenine contains 1 hydrogen bond acceptor. We can eliminate answer choices A, C, and D. Answer choice B, cytosine, has N and O hydrogen bond acceptors, so cytosine is our correct answer.

81) Something I want to point out here. If we are able to eliminate any of the three options listed, we can get to our correct answer. What do I mean by that? Every one of the three options is listed in three answer choices. For example, if we eliminate option I, that would allow us to eliminate answer choices A, B, and D, and pick C as the correct answer. Let’s go through the three choices and see if they affect the Tm of dsDNA in the experiment. 

  1. pH of solution. This could mean an increase or decrease in pH. In acidic solution, we would have protonation of bases, so that disrupts the usual base-pairing and leads to errors. This is a viable option. 
  2. Ionic strength of solution. If we have a solution with a lot of positive ions, we have the phosphate groups stabilize. If we gave increased stability, we have a higher melting point and it’s more difficult to denature. This is also a viable option.
  3. Length of DNA strands. The length of a DNA strand determines how many hydrogen bonds are in the strand. A longer DNA strand would be more stable because there are more hydrogen bonds. That means this is also a viable option.

We went through all three choices here and determined each one was a viable option and could affect the Tm of dsDNA in the experiment. The only answer choice that’s consistent with that is answer choice D: I, II, and III.

82) We can look at the structure of the four options listed to determine which has the largest molecular weight. This is exclusively a content question.

We have adenine on the left, and guanine on the right. Right away we can say the additional oxygen atom means guanine has the higher molecular weight. What’s the difference between guanosine and deoxyguanosine? In deoxyguanosine we replace a hydroxyl group with a hydrogen, meaning deoxyguanosine will have a lower molecular weight than its RNA counterpart. We’re left with our correct answer, answer choice B: Guanosine. 

83) To answer this question, we’ll have to know the structure of the four amino acids listed in the answer choice. This is general knowledge that you should have memorized for test day. Amino acid information is high-yield, and potentially easy points! 

We’re asked to decide between our four options and determine which one can form a bond that is similar to a peptide bond. Glycine, phenylalanine, and alanine all have nonpolar side chains, while lysine is the only option that has a charged R group so it’s the odd one out. Lysine’s side chain has an amino group, and it can form peptide bonds by reacting with a carboxylic acid group:

That sounds like exactly the criteria we’re looking for. That means our correct answer is going to be answer choice C: Lysine. 

84) To answer this question, we’re going to have to consider which is the most basic amino acid listed and would, therefore, increase pI. 

  1. Lys. Lysine has the positively charged side chain and has one of the highest pKa values listed in our visual above. This is a strong answer choice to start.
  2. Glu. Glutamate has the negatively charged side chain, and we can also see it has a much lower pKa value than lysine. Lysine remains our best answer.
  3. Gln. This is a neutral amino acid, so the positively charge lysine is still our best option.
  4. Val. Once again, this is a neutral amino acid, so the positively charge lysine is going to be our correct answer.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 11

85) We can answer this question using external information, or general knowledge. It’s a very broad question, so quick glance at our answer choices shows we have to determine if it’s reducing or nonreducing, and which anomeric carbons are involved in the glycosidic bond.

Maltose is a disaccharide that is composed of two linked glucose units and can be a product in the breakdown of starch. The two glucose units that make up maltose are joined by an alpha 1-4’ linkage. 

In the cyclic form of a sugar, the anomeric carbon is the carbon that was part of the carbonyl group in the straight-chain structure. Maltose’s two glucose molecules are linked a way to leave one anomeric carbon that can open to form an aldehyde group. C1 is the anomeric carbon, so only one anomeric carbon is involved in the glycosidic bond. The anomeric carbon of our left glucose here. The anomeric carbon on the right side can open up to form an aldehyde group, so we can classify maltose as a reducing sugar. A reducing sugar is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group. All monosaccharides are reducing sugars, along with some disaccharides. 

  1. It is a reducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. This sounds like the breakdown we just went through. One anomeric carbon involved in the glycosidic bond, the other can open up to form an aldehyde group, so we classify maltose as a reducing disaccharide.
  2. It is a reducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. Both anomeric carbons aren’t involved in the glycosidic bond, we can eliminate this answer choice for contradicting our breakdown of the question.
  3. It is a nonreducing disaccharide in which only one anomeric carbon is involved in the glycosidic bond. Answer choices C and D both say maltose is a nonreducing disaccharide, which we determined isn’t true.
  4. It is a nonreducing disaccharide in which both anomeric carbons are involved in the glycosidic bond. Answer choices C and D both say maltose is a nonreducing disaccharide, which we determined isn’t true during our breakdown of the question. We’re left with our correct answer, answer choice A.

86) We’re going to answer this question as a percent of our initial supernatant. We’re going to need our data from the passage, and we’re going to use dimensional analysis to calculate the number of units in each step, and ultimately the percentage purification yield.

Supernatant step: 3000 mg x 0.1 units/mg = 300 units

Ammonium sulfate precipitation: 1000 mg x 0.2 units/mg = 200 units

Ion-exchange chromatography: 30 mg x 4.0 units/mg = 120 units

Size-exclusion chromatography: 8 mg x 10.0 units/mg = 80 units

Ion-exchange chromatography: 3 mg x 20.0 units/mg = 60 units

We started with 300 units in our supernatant and are left with 60 units in our final step. We divide: 

60 units final/300 units starting = 20% purification yield

This was a math problem with no rounding, so we can use our calculated value to compare all of our answer choices at once. Answer choice B is going to be our correct value, and we can eliminate answer choices A, C, and D.

87) To answer this question, we’re going to use the figure from the passage. We’ll need to know about amino acids and different types of molecular interactions.

We want to know the primary means of stabilization of phosphate bound to maltose phosphorylase. We’ve got the side chains of two different amino acids here surrounding phosphate. Two serine side chains on the left side, we have a tyrosine side chain on the bottom right. 

The negatively charged oxygen atoms of the phosphate can electrostatically interact with the partial positive charges on the hydrogen atoms. What type of bonding is this? Hydrogen bonding. We have three different atoms that we always look out for when we’re thinking hydrogen bonding: fluorine, oxygen, and nitrogen (FON atoms).

And note, the phosphorus atom itself isn’t participating in hydrogen bonding. It’s the negatively charged oxygen atoms.

Upon first glance, only a single answer choice here matches our breakdown of the question: Answer choice D, hydrogen bonding.

Note that while the amino acids can form hydrophobic interactions, the phosphate itself doesn’t. That means answer choice C doesn’t address the interaction between the residues and the phosphate.

88) We have Figure 1 shown here and we actually went over this in our previous question as well. We’ve got the side chains of two different amino acids here surrounding phosphate. Two serine side chains on the left side, we have a tyrosine side chain on the bottom right. This answer is as simple as knowing the amino acids and knowing your content.

Only answer choice that matches our breakdown is answer choice C. If you are still shaky with the amino acids, make sure to take some time to memorize them. It will make your life a lot easier:

89) To answer this question, we want to know the change in state, and the pH at which this happens. We’re going to need to revisit Figure 2 from the passage, and we’re combining that with our external knowledge of amino acids. If you need to, make sure to review the amino acid chart I added to question 88. 

Figure 2 shows that at very high and very low pH, relative activity is decreased. In terms of protonation, at low pH, functional groups are more likely to be protonated. Low pH means acidic. At high pH, functional groups are more likely to be deprotonated. High pH means more basic. 

The passage points out that catalytic activity is driven by Glutamate protonating the substrate in the active site. Glutamate itself is an acidic amino acid. If glutamate is protonated at low pH, that wouldn’t affect the catalytic activity. If its de-protonated (at high pH), then it won’t be able to protonate the substrate and we’d have a decrease in catalytic activity.

  1. low pH is due to the protonation of Glu487. We said at low pH we would be expecting protonation, so that checks out. But we also said protonating glutamate at low pH wouldn’t affect catalytic activity.
  2. low pH is due to the deprotonation of Glu487. This answer not only contradicts our breakdown, but it also contradicts itself. At low pH we expect protonation, not deprotonation. We can automatically eliminate answer choice B
  3. high pH is due to the protonation of Glu487. Another answer choice that contradicts our breakdown. At high pH we expect deprotonation.
  4. high pH is due to the deprotonation of Glu487. This is consistent with our breakdown. Deprotonating glutamate would mean it can no longer act as a general acid during catalysis. We can stick with our correct answer choice, answer D.

90) In other words, we’re trying to find a ratio of hydronium ion concentrations in two different situations: at the pH with the highest enzyme activity, and pH at the lowest enzyme activity. We’re going to need to need Figure 2 from the passage, and we’re combining that with our external knowledge of pH.

Figure 2 shows the highest MP activity occurs at slightly above pH 6, while the lowest maltose phosphorylase activity occurs at pH 4.

The pH scale is logarithmic, so each single unit increase in pH correlates with a tenfold decrease in hydronium ions. Going from a pH of 4 to a pH of 5 will decrease hydrogen ion concentration by 10x. Going from a pH of 5 to 6 will decrease hydrogen ion concentration another 10x. Ultimately, the hydronium ion concentration will decrease 10*10, a total decrease of 100x.

This is a math problem with no rounding, so we can use our calculated value of 100x to compare all of our answer choices at once. Answer choice D is going to be our correct value.

Note the only approximation we did was when we read the X-axis on the chart. However, our answer choices here aren’t relatively close to one another. While both are important, the MCAT cares more about knowing that you can solve the problem, rather than knowing you can approximate.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Passage 12

91) In the passage we focused on the enzyme glycerol kinase. We want to know the product of the reaction catalyzed by this enzyme. We can reference the passage for specific details, but ultimately, we’re going to use our general knowledge to come up with the structure of the product. We’ll need to know our main biochemical pathways to do so.

From the passage: Glycerol kinase (GK) catalyzes transfer of the gamma-phosphoryl group from ATP to glycerol to produce glycerol-3-phosphate and ADP. Our question wants to know the structure of glycerol-3-phosphate.

Glycerol kinase catalyzes the reaction to form glycerol-3-phosphate which can eventually feed into glycolysis or gluconeogenesis, so it’s a fairly important structure to know. It’s drawn out for you above. We’ve still got our hydroxyl groups on carbons 1 and 2 (same as glycerol). Carbon 3 has a phosphate.

a. 

this looks like our breakdown. Hydroxyl groups on our first two carbons, and phosphate on 3.

b.

this is missing our hydroxyl group on carbon two, so we can eliminate this answer choice.

c.

this has our substituents on the wrong carbons.

d.

Carbon 1 is attached to a hydroxyl group, it doesn’t have this carbonyl. Best answer here is going to be answer choice A.

92) We can reference the passage for specific details, but we’re going to need to know about electrophoresis and the GK enzyme to answer the question. 

Glycerol kinase functions as a homodimer-a protein homodimer is formed by two identical proteins. So, are we going to see one band, or two in both of our electrophoresis runs?

In native PAGE, proteins maintain their secondary structure and charge density during analysis. When analyzing, we see a single band. 

In SDS PAGE, the SDS detergent is going to disrupt secondary, tertiary, and quaternary structure of a protein, but we’d still see all of the protein subunits show up at the same band because they are all the same molecular weight. The key here is to know a homodimer is formed by two identical proteins, and that includes molecular weight.

  1. Two bands in both the native PAGE and the SDS-PAGE. We said we’d see a single band in both. This answer choice contradicts our breakdown.
  2. One band in the native PAGE and 2 bands in the SDS-PAGE. This answer is slightly better than answer choice A. We’re expecting 1 band in the native PAGE, but also 1 in the SDS page.
  3. Two bands in the native PAGE and 1 band in the SDS-PAGE. This is similar to answer choice B in that it’s not fully correct, but better than answer choice A. We’re expecting 1 band in SDS PAGE, but also 1 in the native PAGE
  4. One band in both the native PAGE and the SDS-PAGE. This sounds like our breakdown and we can eliminate answer choices B and C now. Those both incorrectly have one of the PAGEs having two bands. Answer choice D is our best answer.

93) To answer this question, we’ll draw out our glycerol structure, and then choose which of the 4 types of interactions is most likely to contribute to stabilization. We have 3 hydroxyl groups which usually points to hydrogen bonding. We’re going to have the electronegativity difference with oxygen:

  1. hydrophobic interactions. Hydrophobic molecules usually contain long carbon chains and are nonpolar. We know glycerol is part of the hydrophilic head of fatty acids. It contains polar hydroxide groups.
  2. Aromatic interactions. There are no aromatic structures, so not a great option here. Let’s see if we can find a better option.
  3. hydrogen bonding interactions. Hydrogen bonds are formed when hydrogen bonds with a very electronegative atom. The hydrogen gains a partial positive charge, and the atom gains a partial negative charge. What are our 3 main hydrogen bonding atoms? Our FON elements: fluorine, oxygen, nitrogen. The hydroxide functional groups contain an oxygen that is capable of forming hydrogen bonds. This is now our best option.
  4. disulfide bonding interactions. Disulfide bonds are between thiol groups so we can eliminate answer choice D. We can stick with our correct answer, answer choice C: the stabilization is predominantly due to hydrogen bonding interactions.

94) This question is asking us to look at and interpret the information in Table 1, and we want to explain the data for the H232R variant. We’re going to need to go back to the passage to revisit Table 1. We’ll analyze the numbers using our general knowledge, particularly about enzyme activity. 

Look at Table 1 and see how activity is affected by substitutions:

H232A is substituting histidine for alanine

H232E is substituting histidine for glutamic acid

H232R is substituting histidine for arginine

Activity is actually higher following H232R substitution. So, what are the key differences? We’ll have to know about the different amino acids.

Arginine is an basic amino acid with a positively charged side chain. Histidine is a basic amino acid with a positively charged side chain.

Substituting out something at the same location as the histidine likely caused a structural change. Since H232R caused an increase in activity, it more than likely caused a change similar to phosphorylated glycerol kinase, which also has a similar spike in activity.

  1. making similar interactions to those made by His within GK-P. This is unlikely. If arginine in H232R GK produced similar interactions to histidine in WT GK, the structure would resemble WT GK, not GK-P. The question is asking why H232R produces activity resembling GK-P, instead of activity resembling WT GK. This can’t be the answer.
  2. causing similar changes in GK structure as are observed in GK-P. This sounds similar to our breakdown. We have a similar structural change, which causes a similar activity change. This is superior to answer choice A.
  3. being directly involved in catalysis. The passage explicitly states that the activation loop is 25 angstroms away from the catalytic cleft. There’s no real evidence in the passage that this is happening.
  4. causing dimeric GK to dissociate into monomers. Another option that assumes something that isn’t consistent with what was told to us in the passage. We try not to make assumptions or connections like this unless they’re in the passage. We can keep our correct answer choice, answer choice B. 

95) To answer this question, we’re looking at 4 different lineweaver-burk plots in our answer choices and we have to pick the one that represents glycerol kinase, and the one that represents the phosphorylated glycerol kinase. We’re going to need to go back to the passage to revisit specific details about kinetic data. We’re going to need external knowledge to actually analyze the plots.

The passage says it was noted after phosphorylation, the Km of glycerol kinase did not change with respect to either substrate, but the Kcat increased 10-fold.

Quick overview first. The Michaelis constant, Km, compares the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate. In this case, there was no change after phosphorylation. So remember, Km will be the same. 

Kcat measures the number of molecules of substrate turned over into product per unit time (usually seconds). kcat is proportional to efficiency. The higher your kcat is, the more reactions happen per unit time. So, the phosphorylation increased efficiency quite a bit.

X-intercept in Lineweaver-Burk plots is -1/Km and that the y-intercept is 1/Vmax. 

The passage states Km does not change with respect to either substrate following phosphorylation. Meaning the X-intercept will remain the same. Alternatively, Kcat increases 10-fold. 

The higher kcat is, the more reactions happen per unit time, when conditions are optimal and you have enough substrate. Kcat is proportional to Vmax. The Y-intercept is proportional to 1/Vmax, and therefore proportional to 1/Kcat as well. A higher Kcat means a lower Y-intercept. So, our phosphorylated glycerol kinase will have a lower Y-intercept.

To find our correct answer: we’re expecting the same X-intercept for both glycerol kinase and phosphorylated glycerol kinase. We also predicted glycerol kinase will have a higher Y-intercept because it has a lower catalytic constant.

Answer choices A and B right away contradict our breakdown. Both lines don’t show the same X-intercept, despite Km being the same. 

Answer choice C says shows the same x-intercept, but our phosphorylated glycerol kinase should have a lower Y-intercept, not higher.

Answer choice D matches our breakdown. Both lines have the same x-intercept. Phosphorylated glycerol kinase has a lower y-intercept because it has a higher catalytic constant:

96) This is a very open-ended question, but glancing at our answer choices we see that we need to distinguish between an ordered and random order mechanism. We also have to establish if a ternary complex is formed. We can revisit the passage to get specific details about the catalysis mechanism.

The passage says “ATP and glycerol both occupy the catalytic cleft prior to catalysis, and it has been determined that glycerol binds first.” The glycerol always binding first means it’s an ordered mechanism, not random.

We have both ATP and glycerol occupying the catalytic cleft. A ternary complex is composed of three parts, so we can confirm a ternary complex is formed. 

  1. an ordered mechanism in which a ternary complex is formed. This is exactly the answer we’re looking for based on my breakdown. Let’s go through our additional answers, just to be thorough.
  2. an ordered mechanism in which no ternary complex is formed. First part of this answer is correct, but a ternary complex is formed. Answer choice A remains superior.
  3. a random order mechanism in which a ternary complex is formed. This answer choice mentions a random order mechanism, but we know that’s not the case. Glycerol always binds first, so it’s ordered. 
  4. a random order mechanism in which no ternary complex is formed. This answer choice also mentions a random order mechanism. Glycerol always binds first, so it’s ordered. Best answer choice here is going to remain answer choice A.

 

Section Bank: Chemical and Physical Foundations of Biological Systems: Questions 97-100

97) There are two pieces of information we’re going to need to answer this question. First, we have to know which amino acids allow for formation of a covalently bonded dimer? That means we’re looking for a peptide with a cysteine residue because those can form covalent bonds and disulfide-linked dimers. 

The next piece of information is something I always mention to students: you have to know your amino acids. That means knowing the properties of each amino acid, the structures, and the one and three-letter abbreviations. Cysteine is Cys or C. Only one of the four answer choices here actually list cysteine, so this is going to be a simple, straightforward answer. We can pick answer choice B as our correct answer because of the presence of cysteine. 

98) This is a fairly wordy question, but we can break down what it’s asking. There is chemical denaturation. The fraction of folded protein is just how much of the protein is still folded and not broken down by the breaking of tertiary and secondary structures. As denaturant concentration increases, the fraction of folded protein will decrease (as it’s broken down). Once part of the protein become unstable, that part is no longer stabilizing other parts of the protein, so the entire protein becomes more unstable. Segments of the protein will cooperate to unfold the protein. Why did I use that exact verbiage? Because we can think of how cooperative processes are shown on a graph-they’re sigmoidal. Initially the denaturant concentration won’t have as obious of an effect on the fraction of folded protein. However, as some unfolding begins, the entire protein quickly becomes more unstable. We can stick with our best answer, answer choice C. 

99) Catalytic efficiency measures how well an enzyme is able to “grab” its substrate and turn it into product. This means you want the greatest number of reactions happening, for the least amount of input of time/substrate. Catalytic efficiency is given by: Kcat/km. 

Kcat measures the number of molecules of substrate turned over into product per unit time (usually seconds). kcat is positively proportional to efficiency. The higher your Kcat, the more reactions happen per unit time.

The Michaelis constant, Km, can help compare the affinity of an enzyme for a substrate. A lower Km means a higher affinity of an enzyme for its substrate.

For this specific question, we want the lowst ratio of Kcat to Km because that would mean the lowest catalytic efficiency possible.

  1. A high KM and a low kcat. Catalytic efficiency is Kcat/Km, so this answer choice is exactly what we’re looking for. Remember the question stem as for lowest catalytic efficiency. This is going to be the lowest ratio possible. 
  2. A high KM and a high kcat. This answer choice is going to yield higher catalytic efficiency than answer choice A because of the larger denominator. A high Kcat is going to lead to more efficiency based on our breakdown. We can eliminate this answer choice.
  3. A low KM and a high kcat. If we were looking for the highest catalytic efficiency, this would be our best option. However, we were asked to consider the lowest catalytic efficiency, which means this is the opposite of what we’re looking for. We can eliminate answer choice C as well. 
  4. A low KM and a low kcat. This is similar to answer choice B. This answer choice is going to yield higher catalytic efficiency than answer choice A because of the smaller denominator. A low Km is going to lead to more efficiency based on our breakdown. We can eliminate this answer choice. Our best answer is going to be answer choice A.

100) Let’s not get lost in the verbiage here since this is a longer question. Let’s break down what we know. We know more erythrocyte interactions with walls yield greater viscosity. When blood enters small vessels (smaller diameter), there are less erythrocyte interactions with the wall (they’re instead flowing in the center of the vessel). Fewer interactions (and smaller diameter) means lower viscosity. More interactions with vessel walls (and larger diameter) mean higher coefficient of viscosity. Let’s find an answer choice that demonstrates this relationship.

a. 

This answer choice is showing no change in the coefficient of viscosity as diameter increases. In reality, we said a larger diameter should mean a higher coefficient of viscosity. Let’s look for something that better represents that.

b. 

This answer choice shows an initial increase in diameter and the coefficient of viscosity which we expect. However, this trend reverses about halfway across the x-axis. There’s no information in our question stem that tells us this should happen. We can eliminate this answer choice.

c. 

This answer choice matches what we’re looking for. What were our key points? Fewer interactions (and smaller diameter) means lower viscosity. More interactions with vessel walls (and larger diameter) mean higher coefficient of viscosity. That’s exactly what we see in this answer choice, so this is our best answer for the time being.

d. 

This is the opposite of what we’d expect in a graphic demonstrating what we learned in the question stem. As diameter increases, we expect the coefficient of viscosity to also increase, not decrease. We can also eliminate this answer choice. We’re left with our best answer, answer choice C. 



Billing Information
We had trouble validating your card. It's possible your card provider is preventing us from charging the card. Please contact your card provider or customer support.
{{ cardForm.errors.get('number') }}
{{ registerForm.errors.get('zip') }}
{{ registerForm.errors.get('coupon') }}
Tax: {{ taxAmount(selectedPlan) | currency spark.currencySymbol }}

Total Price Including Tax: {{ priceWithTax(selectedPlan) | currency spark.currencySymbol }} / {{ selectedPlan.interval | capitalize }}