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.

 

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