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.

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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.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.Reduction of benzoquinone We see benzoquinone as our choice here and that contradicts our breakdown of the question.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.

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.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.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.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. 

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. 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.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.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 microjoulesEnergy2 = 2.2 mW * pulse duration of 2 ms = 4.4 microjoules.

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.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.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.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!

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.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.Hydrolysis. This matches our breakdown. The passage mentions we’re analyzing peptidases, and proteolytic cleavage involves hydrolysis.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.

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.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.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.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.

EmΔHuΔHnr0

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.

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Thiophenols: Similar to phenols, but sulfur-containing

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

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Phenols, thiophenols, and alkylbenzenes all have aromatic rings.

Aromatic ring that matches our prediction.Alkyl These are present in alkylbenzenes, but not all three compounds, so eliminate answer choice BCarboxylic acid This wasn’t present in any of our predicted figures.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.

be better focused laser type C. This isn’t relevant to our question, or explain the difference between the laser. Let’s keep comparing.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.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.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:

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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.

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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.

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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.

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 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.

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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.

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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.

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 carboxylateAdd 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. 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.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=IRVoltage=Current x resistance

Solving for resistance:R=V/I

R=80 mV/(400 x 10^-12 A) = 200 MΩThis 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

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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. 

Triacylglycerols. Triacylglycerols have three fatty acids linked to glycerol. No choline head.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.Phosphatides. This is consistent with our prediction. Phosphatides contain both phosphorus and choline head (nitrogen part). This is now the best answer choice.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.

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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.

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 choiceFluorescent 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.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.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.

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.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. 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. 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)

80 µg matches our prediction exactly8 mg 8 milligrams is equal to 8 x 10^-3 grams. Incorrect value80 mg 80 milligrams is equal to 8 x 10^-2 grams. Incorrect value.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 

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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 = /

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, =1.0M. 

Note: In spontaneous reactions, the sign for ∆G will be negativeSolve for / which equals e12This 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.

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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 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 4