AAMC Sample Test Cp Solutions - MCAT Content (2023)

Sample AAMC CP [Web]

Part chemistry/physics exam example passage 1

1)

1. an agonist. – The passage states that Compound 1 is a “analog transition state' for the HIV protease enzyme, meaning that it mimics the product of substrate binding. Howeverhinderso that the enzyme can perform its function, which we know because it is used toget over HIV, not increase the infection. An agonist would activate the enzyme, not inhibit it.
2. an antagonist. - An antagonist would inactivate the enzyme, which as mentioned above appears to be the effect of compound 1.
3. a placebo. - A placebo serves as a negative control; is given to determine the magnitude of the effect caused by a person's therapeutic valuebelieveThey are treated but have no clear biological mechanism affecting outcomes. When Compound 1 is used to treat HIV infection, it ishe musthave a biological mechanism for treatment - and we know they do, because it binds to the HIV protease. This means that connection 1it's not a placebo.
4. a catalyst. – Catalysts increase the rate of a reaction, but in the case of compound 1, the reaction is stoppedstopped, not accelerated.Compound 1 is an antagonist, which means that answer alternative B is correct.

2) The number of stereoisomers is equal to 2^Nwhere n = number of chiral centers. Recall the criteria for a chiral center—a carbon bonded to four unique substituents. Let's see how many chiral centers there are in compound 1:

1. 8– This would be true if there were only 3 chiral centers (2³=8), but there are 5 chiral centers.
2. 16– This would be true if there were only 3 chiral centers (2³=16), but there are 5 chiral centers.
3. 32– There are 5 chiral centers and 2⁵=32, making this answer option the correct answer.
4. 64– This assumes that there are 6 chiral centers (2⁶=64), but there are only 5, which overestimates and doubles the number of stereoisomers. We must confine ourselves to answering alternative C.

3) For the formation of kidney stones thatdirect reactionthat produces the stones would have to bemore accommodatingas the reverse reaction. For the formation of a precipitate or solid this means that the favoring of the forward reaction must be greater than for the dissolution reaction and Kprecipitate > Ksp.

1. [Ca²⁺] + [C₂Ö₄^2–] > K_sp– are kidney stonesconsisting of calcium oxalate CaC₂Ö₄then the solubilityProductsthe two reagents calcium and oxalate must showmultipliedenot added. This answer option shows both added and incorrect.
2. [Ca²⁺][C₂Ö₄^2–] > K_sp– The solubility product needs to be multiplied by the reactants, so this answer option looks good so far. For the second half we see that the solubility product is larger than the solubility product constant Ksp. This would mean that the direct reaction, the formation of the precipitate, is favored. Here we go, this must be the right answer!
3. [Ca²⁺] + [C₂Ö₄^2–] <K_sp– Like answer option A, this shows that the reagents are added and not multiplied. Next.
4. [Ca²⁺][C₂Ö₄^2–] <K_sp– During the expressionhe doesshow the multiplication of the reactants, shows that their product isless thanthe Ksp, which would mean that theA precipitate is less likely to formand rather dissolve. Answer alternative is B.

4) If you passed the sample test, you know that amino acids arevery importantfor the test day and that you should know the one-letter abbreviations, three-letter abbreviations, side chain structures, properties, and pKas. Cowardly. Figure 1 shows HIV protease cleavage, so let's jump back to Figure 1 (below) before continuing with the possible answers.

1. Phe e Ala- The amino acid to the left of the dashed line (representing where the enzyme cleaves the peptide chain) is phenylalanine, but the amino acid to the right of the dashed line is not alanine, which would have a single methyl group as a side chain.
2. Pro and Val- The amino acid to the right of the dashed line is proline, but the amino acid to the left is not valine, which would have a 3-carbon "V"-shaped R group.
3. Val e Ala– Neither valine nor alanine are immediately to the left or right of the dashed line.
4. Phe e Pro- The amino acid to the left of the dashed line has a phenyl group and is phenylalanine, and the amino acid to the right has a ring containing the backbone and is proline. That's the correct answer.

An amino acid table is provided below for reference; You should be able to replicate it on test day.

Part chemistry/physics exam example passage 2

5) To answer this question, it is important to remember that by convention, electric field lines represent the motion of apositiveTest charge, so they should point away from the plus and towards the minus. We also know from Fig. 2 that the inside of the neuron has aNegative voltageat baseline before stimulation:

Figure 2A typical nerve impulse, also called an action potential

A.

This picture would mean that is the inside of the neuronpositiveand not negative because the electric field lines are pointingabsentof the positive area.

B.

This image correctly shows the electric field lines pointing away from the relatively positive extracellular solution and into the axon at ~-70 mV.

C.

The voltage potential is between the outside of the neuron and the inside of the neuron, so the electric field lines must beParallel to the difference, and perpendicular to the axon–extracellular fluid boundary. This answer option shows the field linesparallel to the borderwhat is wrong.

D.

Like the answer option above, this answer option incorrectly shows the voltage difference (represented graphically by the electric field lines) parallel to the axon-extracellular fluid boundary instead of perpendicular to it. Answer option B is the correct answer.

6)

A.Isolate the axon from the surrounding area.- The first answer sounds good... The myelin sheath serves to increase the speed at which the action potential travels down the axon, inducing saltatoric conduction through the Ranvier nodes.

B.Reduce axon radius. -As noted in Table 1, axon radius does not change between myelinated and unmyelinated axons.

C.produce Schwann cells. - Schwan's cellsinventthe myelin sheath, not the other way around. This choice of response is the opposite of one of the features of the myelin sheath.

D.Increase in axon capacity. – According to Table 1, the myelinated axon haskleinerCapacity,is not biggercapacity (10^–7<10^–4). Answer A is the only one that accurately describes the function of the myelin sheath.

7) If you're like me, one of your eyebrows went up when you read the question:Where does channel X come from? I don't remember the passage where an X-channel was mentioned.Well, you would be right. This question is pseudo-discrete and only asks us to identify the smallest molecule among the possible answers. Many of us forget that in the MCAT testing interface we have access to a periodic table. This is one of the problems that bringing up the periodic table can help a lot.

1. Protein– Proteins are quite large compared to a single ion; Think how many atoms are in just one amino acid!
2. sodium ions- Already⁺the ions would have the electronic configuration of Ne and be slightly smaller. It has the same number of electrons as Ne, but it has an extra proton that pulls those electrons closer to the nucleus, reducing the radius of the electron cloud.
3. potassium ions–K⁺Ions would have the electronic configuration of Ar and as Na⁺in terms of Ne, it would be slightly smaller than its noble gas equivalent. k⁺the ions are also one period below Na⁺and it would begreateras Na⁺cations.
4. chloride ions– Kl^–Anions would have the electronic configuration of Ar, but unlike K⁺, there is anot lessProton than Ar and would be larger than the other answer choices given (wait for proteins). N / A⁺Ions would be the smallest substance listed and would be the substance carried through the X channel.

8)

A.0,01 m– 0.01 m is too short and would require even less travel time than the time for an action potential. See answer option B for a visual representation.

B.0,1 m– The minimum distance at which electrodes can be placed is determined by the time it takes for an action potential to be signaled and the speed at which the impulse can propagate. Multiplying the action potential time, 1 ms, by the pulse rate given in the last paragraph, 100 m/s, we get a distance of 0.1 m.

C.1,0 m– This is the length given at the passage of the longest axon, which would be much longer thanrequired for the nearest distance.

D.10m– 10m is even longer than the length of the longest axon and would even be longer than necessary. OcloserDistance to attach the electrode is 0.1 m, answer option B.

9) Current is the charge moved per unit time and can be calculated using Ohm's law: V=IR. To calculate the current, we need the voltage, also called potential, and the resistance.

1. conductivity, resistance and length- This answer option does not take into account a key variable in Ohm's law, voltage/potential.
2. potential, conductivity and radius– The conductivity is theturning backvonResistance, and the radius is perpendicular to the stream, so it wouldn't help us to calculate the stream.
3. Potential, Resistance and Radius– Although resistivity can be used to calculate resistance, radius is perpendicular to current and would not be useful for calculating current directly.
4. potential, resistance per unit length and length– Here the potential is taken into account, and if we multiply the resistance per unit length by the length over which this resistance occurs, we get the total resistance. I=V/R and both variables needed to calculate the current are taken into account. This is our golden ticket to another place in the field of chemistry/physics.

Part chemistry/physics exam example passage 3

10) Time for some math! The pH of a solution can be determined by pH = -log[H⁺]. The last sentence of the passage reads: "The hydrogen ion concentration of the unknown aqueous solution was1 × 10^–5M.” Substituting this into the pH equation, we get the following:

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

1. 4– This would be the pH if the hydrogen ion concentration was 1*10^-4. This answer option is very acidic.
2. 5– As mentioned above, pH = -log[1*10^-5] = 5; That's the right pH.
3. 9– This would be the pH if the hydrogen ion concentration was 1*10^-9. This answer option is very simple.
4. 10– This would be the pH if the hydrogen ion concentration was 1*10^-10. This answer option is very simple. Choice C is the correct pH.

11) The passage gives us limited information about the unknown compound: "The compound completely dissolves in water and conducts electricity weakly. The hydrogen ion concentration of the unknown aqueous solution was1 × 10^–5M.Unfortunately, dissolving in water and weakly conducting electricity doesn't help us much in distinguishing between the options given below. However, we just calculated the pH for the question above and know that a pH < 7 is acidic and generally less than about 2 is strongly acidic. This isn't a hard and fast rule, just a quick and dirty way of estimating; Remember that a strong acid is one that completely dissociates. A pH of 5, as calculated above, is slightly acidic.

1. weak base.– A weak base would have a pH greater than but slightly closer to 7, not <7.
2. strong base.– A strong base would have a pH much higher than 7, not lower than 7.
3. weak acid.- This is a solid answer option. The pH < 7 but not low enough to be a strong acid.
4. strong acid.– We would expect the pH to be much lower if the unknown compound was a strong acid because it would completely dissociate. A weak acid is a better characterization of the unknown compound.

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

NO₂(OH) dissolved in water creating an acidic solution-> Well, if it produced an acidic solution, it must be aoxyacid.Ni(OH)₂only dissolved in an acidic solution–> oppositeNO₂(OH),Ni(OH)₂notto producean acid solution,dissolved in itan acidic solution. We know that acids dissolve in bases and bases because of the joint ionic effect in acids, so ifNi(OH)₂dissolved in an acid it is aBase.

1. Both were oxyacids. - Just no₂(OH) is an oxy acid, Ni(OH)₂it is a base.
2. Both were bases. – Ni(OH)₂It's a base, but NO₂(OH) is an oxy acid.
3. NO₂(OH) was a base and Ni(OH)₂it was an oxyacid. – This is the opposite of the correct answer and mislabeled the molecules. Ni(OH)₂is a base and NOT₂(OH) is an oxy acid.
4. NO₂(OH) was an oxy acid and Ni(OH)₂it was a foundation.- Is it right! NO₂(OH) is an oxy acid and Ni(OH)₂it is a base.

13)

1. an alkali or alkaline earth metal. - The second paragraph reads: "It can be theorized that if the central atom A is an alkali or alkaline earth metal, the compound is basic.” The pH of the unknown compound indicates that it is aacidand no base, then the central atom shouldNObe an alkali or alkaline earth metal.
   
2. a transition metal. – The passage does not address what type of molecule is formed when the central atom is a transition metal. There has to be a better answerbased on the passage.
3. a nonmetal. – In the second paragraph the author says: “But when A is a nonmetal, the compound is acidic.” The unknown compound produces aacidsolution, then it is safe to say that the central atom A is a nonmetal.
4. a noble gas.– Noble gases are “happy” with their 8 valence electrons and relatively uninterested in bonding with other atoms, which is whymajorityhave no listed electronegativity. Answer C is the best answer.

discrete questions

14)

1. Triacylglycerine– Triacycloglycerols have three long carbon chains, specifically fatty acids, attached to a glycerol backbone. In squalene, several long carbon chains are attached to a glycerol backbone.
2. Phospholipide– Phospholipids have a polar head containing a phosphate and two long carbon chains forming the “tails”. Squalene doesn't fit that description.
3. steroids– Steroid hormones, like the cholesterol from which they are derived, have a basic structure of four fused rings. We could easily draw a line between eachalmost closedRings to form four fused rings, allowing them to be easily prepared from a squalene precursor.
4. Prostaglandins– Prostaglandins have a central ring with chains coming out from these rings. We don't see much evidence for this structure. Squalene is similar to cholesterol and steroid hormones, so C is the best answer.

15) Calcium chloride is written as CaCl₂; There are two chloride ions per calcium ion in the molecule. If CaCl₂is dissolved in water to make a 0.1 M solution, CaCl₂will dissociate into a Ca²⁺etwoKl^–ions.

1. 0,02 Mio- This answer choice is one decimal place apart - twice 0.1 is 0.2, not 0.02.
2. 0,05 M- That's ithalfof concentration, noGutConcentration as it should be.
3. 0,10 M- This is thatsame concentrationsuch as CaCl₂that was resolved, but we know there will betwoKl^–for each CaCl₂dissolved.
4. 0,20 M– The concentration of Cl^–will be twice as much CaCl₂, 2*0.01 M = 0.02 M. Mental arithmetic can give us the right choice of answer if we configure it carefully.

16) I've never been a big fan of carbs in the MCAT because they always seemed unnecessarily complicated to me. Let's see if we can help you feel more confident about carbs!

The solution mentioned in the question stem is silver oxide and is used to test reducing sugars. In order for a sugar to reduce another compound, it must be able to do sobe rusty. In practice, this occurs when the sugar ring is opened to aHalbacetalto form an intermediate aldehyde. To find a hemiacetal, we need to look for an oxygenated ring with an alcohol attached directly to a carbon next to the oxygen, the anomeric carbon. The disaccharide underneathdoes not contain hemiacetalornot be a reducing sugarand willnot producethe silver mirror when reacted with silver oxide. This is the disaccharide you're looking for.

A.

Cellobiose has a hemiacetal outlined in yellow.

B.

Lactose contains a hemiacetal, shown in yellow.

C.

Maltose contains a hemiacetal, shown in yellow.

D.

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

17) Amino acids, you can't get far without them popping up again! For this question it is sufficient to add the charges of each of the side chains at a pH very close to physiological pH. The +1 from the N-terminus and the -1 from the C-terminus cancel each other out. Arginine is a basic amino acid with a +1 charge, alanine is non-polar and neutral, phenylalanine is non-polar and neutral, and leucine is non-polar and neutral. If we add all the charges, we get 1 (N terminal) + -1 (C terminal) + 1 (arg) + 0 (ala) + 0 (phe) + 0 (leu) = +1

1. –1– The sign is wrong; the net fee is +1.
2. 0- If you tried this answer option, you may have forgotten to add the +1 for arginine.
3. +1- That's the right answer! 1 (N-terminus) + -1 (C-terminus) + 1 (arg) + 0 (ala) + 0 (phe) + 0 (leu) = +1
4. +2- If you tried this answer option, you may have forgotten to add the -1 from the C terminal.

Part chemistry/physics exam example passage 4

18)

1. Increasing the polarity of the mobile phase decreases the retention time of the compound1related to the connection2.– Increasing affinity for a mobile phase results in faster elution and shorter retention time, and increasing affinity for the stationary phase slows elution and lengthens retention times. Connection 1 isless polaras compound 2 because it lacks the alcohol in its methyl group. If the polar mobile phase became even more polar, compound 1 would have beenfewerAffinity to the mobile phase and its retention timeincrease, not decrease.
2. composed1is eluted first because it is more polar than compound2.— That is wrong and the opposite; Connection 1 isless polaras connection 2.
3. reducing the affinity of the compound1because the stationary phase increases its retention time compared to the compound2.– If compound 1 had less affinity for the stationary phase, it would elute faster and would have ashorterretention time,no longerretention time. Remember that retention time refers to how long the compound stays on the instrument before it elutes.
4. composed2will elute first because it does not interact as favorably with the stationary phase as does the compound1.– That's right – compound 2 is more polar than compound 1 due to the alcohol attached to the upper left carbon and interacts more favorably with the polar mobile phase than with the non-polar stationary phase, which means thatelute faster. Compound 1, which is less polar, interacts more favorably with the non-polar stationary phase and elutesafterconnection 1

19)

1. composed1is still eliminated from the body of patients expressing the CYP2C9*3 allele. – While this is true, the fact that it is cleared from the body doesn't tell us much about the affinity or how tightly and easily compound 1 binds to CYP2C9 enzymes.
2. OK_MValues ​​for variant enzymes are not significantly different from wild-type enzyme. - This is trueedeals with the binding affinity of the compound with the enzyme. Km is the dissociation constant, making it inversely related to binding affinity. If binding affinity changed, there would be a change in Km. Table 1 shows no significant changes in Km, meaning that there are no significant changes in the binding affinity of the CYP2C9 enzymes relative to compound 1. That's what we're looking for because the question is aNOAsk.
3. Amino acid substitutions at positions 144 and 359 do not alter the enzyme binding pocket of the variant."We can't say for sure. If amino acids 144 and 359 are in the binding compartment, swapping the amino acids would alter the binding compartment. Let's see what option D says.
4. The side chains of the amino acid residues at positions 144 and 359 are charged at physiological pH. – Considering the four amino acids at these two positions (arginine and isoleucine in WT and cysteine ​​and leucine in mutants), only arginine is charged at physiological pH. Let's stick with answer option B because it's true.eaddresses the root of the question.

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

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

Looking at the graphs above, we see that the y-intercept is 1/Vmax. Decreasing Vmax causes the y-axis intercept to move in the opposite direction and the y-axis to move up.

1. is the same for all three variants. - That is not true. The Vmax is different for each enzyme, so the y-trap must change.
2. it is the same for CYP2C9*1 and CYP2C9*2 but different for CYP2C9*3.- CYP2C9*1 and CYP2C9*2 have different Vmax and are outside each other's confidence intervals. They must have different Vmaxs.
3. it is the same for CYP2C9*2 and CYP2C9*3 but different for CYP2C*1.- CYP2C9*2 and CYP2C9*3 have different Vmax and are outside the respective confidence intervals. They must have different Vmaxs.
4. is different for all three variants.- Is it right. Each of the enzyme variants has a different Vmax, and since the y-intercept in a Lineweaver-Burk plot is 1/Vmax, they must all have different y-intercepts.

21)

1. an oxidizing agent. - an oxidizing agentAgentIt isreducedin a reaction; NADPH does not accept electrons, which it would if reduced and acting as an oxidizing agent. It does not look good.
2. a reducing agent. - a cutAgentreduces your goal andis oxidizedincluded. Oxidation is the loss of electrons and NADPH loses electrons, evidenced by the creation of a positive chargeNADP⁺; that looks good.
3. a catalyst.– The catalytic converters mustnot be consumedduring a reaction. NADPH is consumed and not regenerated, so it's not a catalyst here.
4. an electrophile. – An electrophile “wants” electrons; NADPHloseselectrons and charges itself positively in the course of the reaction. It is oxidized and serves as a reducing agent, so answer B.

Part chemistry/physics exam example passage 5

22)

2HCl(aq) + Mg(S) → MgCl₂(aq) +H₂(G)

Reaction 1

1. oxidation reduction– Reaction 1 involves changes in oxidation states, making it a redox reaction. Magnesium begins in its elemental state and is then oxidized to a +2 oxidation state. The hydrogen in HCl is converted to its elemental form and its oxidation state ranges from +1 to 0.
2. Lewis acid/Lewis base"This is so, so incredibly tempting!" HCl is an acid! Unfortunately, this does not necessarily mean that the reaction isbest describedas a Lewis acid/base reaction. The oxidation states have changed, and a Lewis acid/base reaction involves thesplitof donated electrons, but MgCl₂it is ionic. If in doubt, calculate the oxidation states.A change in oxidation states means the reaction is a redox reaction.
3. double substitution– The reaction would theoretically be a single substitution, not a double substitution. Mg exists in its elemental form as a reactant and is not bound to anything else that can be displaced.
4. Ionisation– An ionization reaction would produce a free electron, which does not exist. Answer option A is the best answer.

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

1. lower. - Is it right. As mentioned above, the pressure of the small column of air is inside the buretteless thanAmbient pressure, causing the liquid level in the burette to rise and reach equilibrium.
2. the same.– The air/gas pressure in the burette wasless thanthe ambient pressure, not the same or the same.
3. 2 times larger.– The air/gas pressure in the burette wasless thanAmbient pressure, not higher. There's no need to quantify if you're going in the wrong direction.
4. 26 times larger. – The air/gas pressure in the burette wasless thanAmbient pressure, not higher. A quantification is not yet necessary; Let's choose A and move on.

24)

A.PV = nRT– Although this is the classic version of the ideal gas law, the question arises, what is the simplest way to calculate the gas constant, which is what we needisolate R. This equation has no isolated R.

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

C.Nr=PV/T– Like answer option A, this answer option R does not show in isolation or alone.

D.R=PV/nT– D is the correct answer because it isolates the gas constant R and is a precise transformation of the ideal gas equation.

25) In the second paragraph of the passage it says: “Die ideally GaskonstanteRcan be found by measuring the volume occupied by a given amount of gas at a given temperature and pressure. One technique to achieve this is to react a known amount of magnesium with acid to produce hydrogen gas.The section then goes on to describe the experiment that this excerpt just told us is one way to calculate the gas constant R.

1. To determine the molar mass of H₂Gas– The hydrogen gas produced in reaction 1 is not the focus of the experiment. The passage tells us that the experiment is a means of calculating the ideal gas constant R.
2. Investigate the mechanism of reaction 1– The experiment uses Response 1 as the baseline, but the goal is to calculate R by measuring other related variables. The specific mechanism of Reaction 1 is not investigated.
3. For determining the measurands of the ideal gas law– This is an appropriate answer option. In the second paragraph, experiment was introduced as a way to calculate the ideal gas constant, R, using other variables measured in the ideal gas law (volume, temperature, and pressure).
4. Investigation of the reactivity of magnesium with acids– Again, the focus is on the ideal gas law constant, not on a specific part of Reaction 1, including magnesium reactivity. Answer C is the best answer.

discrete questions

26)

1. 0,212 kg– This answer option is too small and results from the use of 2 instead of 5 liters.
2. 0,530kg– This answer option is a decimal place; Check your decimals and moves! Keeping scientific notation in mind is important for chemistry/physics calculations.
3. 5,30kg– As shown above, this is the correct blood mass.
4. 10,6kg- This answer option is double the correct mass and is likely the result of rounding from 5 to 10. While rounding can help on test day, we don't want to round too much. The closer the possible answers are to each other, the less you should round. Option C is the correct answer.

27) This question asks us to remember the Doppler Effect equation, which allows us to create images with ultrasound and determine the variables and minimum values ​​that would be required for imaging. There are four variables in the Doppler equation, that is, theMinimumwould be necessarythree out of fourVariables.

1. The speeds of sound and the moving object.– These two variables are very few andless than the minimumwe would need If an equation has four variables, it would take three to calculate the fourth.
2. The speed of sound and the frequencies of the emitted and observed sound waves.– This answer option includes three of the four variables above, namely theMinimumInformation required to use ultrasound becomes the correct answer.
3. The sound velocities of the moving object and the frequencies of the emitted and observed sound waves.– This answer option includes all four variables and deliversmore than the minimumInformation.
4. The speeds of sound and the moving object, as well as the frequencies and wavelengths of the emitted and observed sound waves. – Wavelengths are not directly part of the Doppler equation and this answer option provides a lot of information. The “B” answer option means the “B”est answer.

28) Which factors influence the boiling point? Intermolecular forces—mainly hydrogen bonding, but also dipole-dipole interactions and van der Waals forces to a lesser extent—as well as ionic bonding, branching, and molecular size. When you think about these factors, what can fluorine or can that other halogens can or can't?

1. The H-F bond is much less polar than the bonds between H and the other halogens.– Fluorine is the most electronegative element, meaning it creates bonds that arepolarsthan any other element that binds to the same atom.
2. HF has the lowest molecular weight of the Group 7A hydrides. - While this is true, is the boiling pointpositively connectedto the molecular mass, i.e. with alowerMass would mean alowerBoiling point. There has to be something else at play for HF to have that abnormalAltBoiling point compared to the other Group 7A hydrides.
3. HF is influenced by hydrogen bonding to a much greater extent than other Group 7A hydrides. – In contrast to the other halogens, fluorine is excellently suited for hydrogen bonds. Hydrogen bonding increases boiling point - yields H₂Also its particularly high boiling point. This is a good answer.
4. HF has the highest vapor pressure among the Group 7A hydrides. – Vapor pressure and boiling point arethe oppositerelated to each other such that a higher vapor pressure would correspond to a lower boiling point. We stay with C

29)

1. Approximate radial size of an electron cloud– The principal quantum number describes the energy of a specific orbital or electron cloud in which an electron is located. We know that the energy is affected by the distance between two charges, so the radial size of the cloud would affect how much energy is in each electron cloud. As the radius and size of the electron cloud change, so does the energy.
2. Approximate shape of an electron cloud- ÖDrehimpulsThe quantum number is associated with the shape of a cloud of electrons.
3. Number of valence electrons orbiting a nucleus– Quantum numbers can be used to describe the orbitals in which certain electrons reside andare not limited to valenceelectrons.
4. Number of protons and neutrons in the nucleus of an atom. – Quantum numbers are used to describe the orbitals containedelectrons,not protons and neutrons. Answer option A is the best answer option.

For an overview of quantum numbers, seehttps://jackwestin.com/resources/mcat-content/electronic-structure/orbital-structure-of-hydrogen-atom-principal-quantum-number-n-number-of-electrons-per-orbital

Part chemistry/physics exam example passage 6

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

1. Increase the value of the CPFX connected to the BSA.– In order to answer this question, we have to read the character description carefully. The MCAT test writers are under no obligation to bring any specific information to your attention - you are expected to critically evaluate any information they provide to you, taking into account what you should know. However, be extra careful when they draw your attention to specific information by adding a "Note:". The “Note” to Table 2 reads as follows: “Note: k₀is the binding constant of CPFX-BSA in the absence of a metal ion.” If Ka is the binding constant with the metal ions present and K₀is the binding constant in the absence of metal ions, i.e. Ka/K₀will tell us the binding with metal ions compared to without. Ifk_A/k₀>1, metal ionsimprovebond if it =1 they haveNo effectwhen calling, and if it's <1, shereduceBinding. We see that all values ​​are in the last columnk_A/k₀<1 then the metal ionsreduceCPFX and BSA mandatory, do not increase.
2. Decrease the amount of CPFX associated with the BSA.– As mentioned in the explanation above, the presence of metal ions produces ak_A/k₀<1 and decreases the amount of CPFX associated with the BSA.
3. decrease the amount of free CPFX in plasma.– If there was a drop in free CPFX, there should beIncrease in linked CPFX. The decrease in binding observed in the table shows that the opposite is true.
4. have little impact on the level of linked CPFX.- That is not true. As noted in the explanations above, there is a reduction in the amount of CPFX associated with the BSA, Answer Alternative B.

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

A.Glucose– We need to be able to recognize a glucose molecule on the day of the test. Glucose is a monosaccharide with a six-membered ring and an oxygen inside the ring. Ibuprofen has a benzene ring but no glucose molecule.

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

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

D.palmitoleic acid– Palmitoleic acid is a fatty acid, meaning it has a hydrocarbon tail (the “fat” component) and a carboxylic acid (ie, an acid). Although tails are usually linear, this answer option is the best.closerto what we see and what we must do. Next question!

32)

1. mainly in Local II.-According to the paragraph, "Warfarin and ibuprofen have been shown to bind specifically to separate sites (sites I and II, respectively).', so in the table the results for warfarin are in site I and the results for ibuprofen are in site II. Looking at binding at site II with ibuprofen we see ak_A/k₀<1 hence a decrease in binding for ibuprofen. So far things are looking good, but we have yet to see where warfarin binds. Here,k_A/k₀It isEven smaller,0.46 < 0.87. That means there ismore competitionat the warfarin binding site, site I... When there is less binding and more competition, CPFX and warfarin are more likely to compete for the same site, making the I site more likely to be the CPFX binding site.
2. split equally between Local I and Local II.- Ök_A/k₀because warfarin and ibuprofen are not the same thing.
3. mainly in Local I.- And that. As mentioned above, the presence of warfarin has a greater impact on CPFX binding, making it more likely that the two share a binding site.
4. at a location other than Site I or Site II.- The presence of warfarin reduces CPFX binding by more than halfk_A/k₀<0.5), making site I a likely binding site for CPFX, implying that C is the best response option.

33)

The first paragraph tells us that "Ligands tend to be primarily hydrophobic with anionic or electronegative properties’ and since opposites attract, the side chains of the amino acids at the binding site must be positively charged.

1. R e L– Arginine is positive and leucine is neutral, creating a net positive binding site. Let's evaluate the other possible answers.
2. E e Y– Glutamate is negatively charged and tyrosine is neutral, resulting in a net repulsive negative charge.
3. D e E– Aspartate and glutamate are negatively charged, which would repel ligands even more than the previous response option.
4. D e H– Aspartate is negatively charged while histidine is positively charged, creating a neutral binding site; This is a poor answer option compared to the attractive web site with net positive links in answer option A.

Part chemistry/physics exam example passage 7

34) This question is quite simple. Density is mass divided by volume, the author gives the mass of the yarn as 4 × 10^-3kg and the question stem says the volume is 5×10^–7M³.

Density = mass/volume = (4 × 10^-3kg)/( 5 × 10^–7M³) = 0,8 × 10⁴kg/m³= 8000 kg/m³

1. 2000kg/m³– This answer choice is ¼ of the correct answer choice and probably results from misjudging 4/5 as 0.2 instead of 0.8.
2. 3600kg/m³– This answer selection is too small and is the result of an arithmetic error or incorrect equation construction.
3. 6400kg/m³– This answer selection is too small and is the result of an arithmetic error or incorrect equation construction.
4. 8000kg/m³– This is the answer we calculated above, and it is the correct answer.

35) The wording of the question does not explicitly direct us to Table 1, but in it we can find the relationship between T and R:

As T increases, R increases. The increase of neither T nor R is exponential. Changes are evenly distributed; if T increases by 100 K, R increases by about 2–3 Ω.

A.

This is a positive linear relationship. It seems good.

B.

This incorrectly assumes that R does not change as T increases.

C.

While this is positive, it shows an exponential rather than a linear relationship. The response of R in this diagram is very strong relative to T.

D.

This shows a negative relationship and goes in the wrong direction; we stay with a

36) This question is almost a pseudo-discreet question. In order to calculate the heat required to bring a substance to a specific temperature, we must use the equation of thermodynamics q=mCT. We get the temperature change and specific heat capacity in the rod in question and the mass in the passage.

Q=mCT = (4 × 10^-3kg)(460 J/kg*K)(200 K) = (8 × 10^–1kg*K)(460 J/kg*K)

= 0,8 x 460J = 368J

For this question, you could also round from 460 to 500 on test day, taking into account the distance between answer choices.

Q=MCDT = (4 × 10^-3kg)(500 J/kg*K)(200 K) = (8 × 10^–1kg*K)(500 J/kg*K)

= 0.8 × 500J = 400J, which is an overestimate since we have rounded and is closer to 368J than 550J.

1. 368J– We calculated that.
2. 550J– Check your arithmetic, this is too big.
3. 1840J– That is also very big. Check your math.
4. 3680J– This answer option is incorrect by a power of ten. Check the math for your exponents. Answer option A is the correct answer.

37) This question is multifaceted. We need to calculate the proportional increase in resistance with respect to temperature and then multiply the original current by the new ratio. According to Ohm's law, resistance and current are inversely proportional. So if resistance were to increase at the same rate as temperature (which it should be because we've found that the two have a positive linear relationship), current should decrease at the same rate. equal share.

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

We can do the math and calculate that 2.3 * 2A = 4.6A but if we look at the answer options we can see that the two values ​​used are the current of 2A and a larger current of 4.6A are, so no further calculations are required.

1. It stayed constant at 2 A.– We know that the currentis loyalto 2A, so did not remain constant.
2. It stayed constant at 4.6 A.- We may remove this answer option for the same reason as the previous one.
3. Increased from 2A to 4.6A.- Oh indeedreduced, has not increased.
4. Decreased from 4.6A to 2A. - This is the correct answer and shows a reasonable decrease in current as temperature and resistance increase.

Part chemistry/physics exam example passage 8

38) The best way to approach this question is to follow the marked carbons along the illustrations, highlighted in purple below.

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

39)

40) On test day you should be aware of some of the most common absorbance values ​​in the IR spectrum including but not limited to alcohol and carboxylic acid as seen in mevalonate.

1. II only– 1700–1750 cm^–1corresponds to a carboxylic acid, which actually isGiftin mevalonate. We need to evaluate the other Roman numerals and answer choices.
2. I and II only– 3200–3500 cm^–1is a reference to an alcohol thatit's availableat the opposite end of mevalonate as the carboxylic acid. We also know that the Roman numeral II is correct. We will continue reading and evaluating.
3. II and III only– We know that the Roman numeral I is correct, so this answer choice is wrong. Also 1580-1610cm^–1corresponds to an aromatic C=C bond, that isnot presentin Mevalonat.
4. I, II and III– Roman numerals I and II are correct, but Roman numeral III is incorrect, meaning B is the correct answer.

41) This can be a difficult question. We are very used to chiral reactants producing chiral products, but none of the reactants in the above reaction are actually chiral...

1. One of the reagents is chiral. – As already mentioned, none of the reactants are chiral.
2. Both reagents are chiral.– Like option A, this answer can be ruled out since none of the reactants are chiral.
3. The solvent medium is chiral.– At first glance, this answer option looks good—a chiral solvent would promote chiral reactions. However, given the context of the reaction, an in vivo reaction within a living organism, this choice of reaction carries less weight. Perhaps answer option D can provide a more convincing mechanism.
4. The enzyme is chiral. – This is a great answer because not only would it produce a chiral product, but it makes sensecontexthuman metabolism and organic chemistry. There is precedent for chiral enzymes - remember how in living organisms we only use L-amino acids - making this the most robust and correct answer option.

42) Squalene is similar to the four fused rings of cholesterol and its steroid hormone derivatives.

1. Glucose– Glucose is a single ring and not a steroid hormone. There is no clear-cut relationship with squalene, making it a good answer to this question.NOAsk.
2. Testosterone– Testosterone is a steroid hormone and has a structure resembling squalene; it could very well be the product of squalene metabolism.
3. cholesterol– Cholesterol is structurally similar to squalene.
4. Cortisone– Like the two answers directly above, cortisone is a steroid hormone structurally similar to squalene. Glucose is the only option that doesn't resemble squalene, so it should beNObe a product of squalene metabolism.

discrete questions

43)

1. cylindrical. – A cylindrical lens focuses the light in a line that does not change where the image is formed. Next.
2. convergent. – A converging lens would cause light rays to be focused further in front of and farther from the retina. On the day of the test, we should know that converging lenses are used to treat long-sightedness or long-sightedness. People who are farsighted have an image that is focused above/behind the retina, not in front of the retina.
3. deviating. – Negative lenses are used to treat nearsightedness or short-sightedness and cause light rays to focus more backwards and onto the retina. That's itseekexcellent;verwhat did we do there? Okay, let's stop the puns!
4. spherical. – A spherical lens focuses light into a point. This does not change the distance at which the image is created. Alternative answer C is!

To learn more about lenses:https://jackwestin.com/resources/mcat-content/geometrical-optics/thin-lenses

44)

1. Sublimation under reduced pressure– Sublimation is the conversion of a solid into a gas without first going through the liquid phase. This would be a useful procedure as there would be no melting or heating and the proteins would remain in their current/frozen conformation. It would also successfully remove water when it turns to gas.
2. steam distillation– We want to freeze the sample and keep the protein in its native state. Distillation involves heating the sample, which would defeat the purpose of freezing.
3. Extraction with organic solvents– An organic solvent is hydrophobic, which means thatNOreadily react with water and would be a poor strategy to remove water from the sample.
4. Addition of magnesium sulfate– Magnesium sulfate, usually more recognizable as Epsom salts, would not help remove moisture from the solution if it was frozen. We don't want to melt the sample we just froze, so this isn't a viable option. The answer option "A" means "A" great, so we'll keep it.

45) This question is intimidating. We are asked to identify a mysterious structure with three different classes of substituents that would result from acid reflux. Before we look at the possible answers, let's take a short break. the watery H₂THEN₄would promote acid catalyzed hydrolysis, so we are looking for a structure that could be hydrolyzed to produce HCN, benzaldehyde C₆H₅CHO and two glucose molecules. Go to the answer options.

A.

That's promising. Working from right to left we have a cyanide that can be released as HCN, then a benzene with an oxygen based bond to the two glucose molecules such that hydrolysis of the bond would release an aldehyde and two glucose molecules.

B.

Here the benzene is placed between the two glucose molecules and the cyanide is attached to the same carbon where the putative benzaldehyde is attached. Hydrolysis of this site would not result in HCN and benzaldehyde, and further hydrolysis would not produce lower right glucose - we woulda short carbon.

C.

As with the molecule in answer option B, the position and order of the structures does not lend itself to hydrolysis into the products of interest. Although HCN could be produced, benzaldehyde would not be a product. Indeed, if one of the rings attached to the extra carbon in benzene were hydrolyzed after removal of the HCN, there would be aextra coalfrom oxygen and would not lead to benzaldehyde.

D.

Again, we can make HCN, but not benzaldehyde - here we aremissing the extra carbonbound to benzene before oxygen.

46) The best way to solve this question is to look at Archimedes' principle with the information we got in the question and then determine which answer option corresponds to the expression found.

1. C_ar/(C_ar–C_C). – This expression corresponds to what we calculated above and is the correct answer.
2. (C_ar–C_C)/C_ar. - This is thatturning backthe right answer.
3. (C_ar–C_C)/C_C. - This is the oppositeeincorrectly swaps body weight in air for body weight in water for what the meter should be.
4. C_C/(C_ar–C_C). – Again, the counter must be the weight of the body in the air and not the weight of the body in the water.

Examination part chemistry/physics example passage 9

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

1. 0 for pantothenate and 0 for phosphopantothenate– Both pantothenate and phosphopantothenate contain a carboxylic acid and not a basic functional group, so they must have a net charge ofat least-1.
2. -1 for pantothenate and -1 for phosphopantothenate– Phosphopantothenate loses two additional hydrogen atoms from its phosphate group for a net charge of -3.
3. -1 for pantothenate and -3 for phosphopantothenate– Pantothenate loses one hydrogen from its carboxyl group and phosphopantothenate loses two hydrogens from its phosphate and one from its carboxylic acid. This is the correct pair of net charges.
4. -3 for pantothenate and -4 for phosphopantothenate– Pantothenate loses only one hydrogen from its carboxylic acid and phosphopantothenate loses 3 hydrogens in total, so the charges should be -1 and -3 respectively, like answer option C.

48)

1. β-mercaptoethylamine, phosphopantothenate, dAMP with added 3'-phosphate– All three components are wrong here, coenzyme A does not contain any of them.
2. β-mercaptopropylamine, pantothenate, AMP with added 5' phosphate– Although Coenzyme A contains pantothenate, it contains ADP and not AMP and must be β-mercaptopÄthylnon-propylamine.
3. β-mercaptopropylamine, phosphopantothenate, dAMP with added 5' phosphate– Like answer option A, coenzyme A does not contain any of these groups.
4. β-mercaptoethylamine, pantothenate, ADP with added 3'-phosphate– Coenzyme A contains all three of these structures (see above), so this is the correct answer.

49) In the last paragraph the passage reads “Bonding is stabilized by interactions between C2'is C4'Hydroxyl groups of pantothenate and a carboxylate group of PanK3', meaning that PanK3 requires an amino acid with a carboxylic acid that is deprotonated to form a carboxylate.

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

50) In order to stabilize negatively charged phosphates in ADP, amino acids must be presentpositiveFees.

1. Asn9 and Thr10– Although polar, both amino acids have no charge.
2. His11 and Arg27– Both amino acids are positively charged, so they would stabilize the negative charges of ADP.
3. Asp6 is Arg27– Aspartate is negatively charged and would cancel out the positive charge of arginine, so the negatively charged ADP would not be stabilized.
4. Asp6 and His11– Aspartate is negatively charged and would neutralize the positive charge of histidine, so the negatively charged ADP would not be stabilized. Only B would stabilize the negative charges.

51) The second paragraph indicates the fate of NADH: “Pyruvate is then converted to lactate by lactate dehydrogenase using NADH at the same time."It is notneeded; is converted to NAD⁺and gives up electrons so that it is oxidized. If you're still unsure, take a look at the lactic acid fermentation picture below.

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

Examination part chemistry/physics example passage 10

52)

Using the conversion factor given in the question, the energy of the photons is approximately 224*10^-16J. When connected to E=h/f, the frequency is 3.38*10¹⁹Hz, slightly less if we round.

1. 2,11 × 10³⁵Hertz– The orders of magnitude are very large in this answer. We could have lost the 10^-16initially no counter.
2. 3,38 × 10¹⁹Hertz- That's roughly what we calculated.
3. 3,01 × 10^–20Hertz– The minus sign in the exponent should tell us that it is wrong, even if the 3 in front of it misleads us.
4. 1,45 × 10^–47Hertz– We probably forgot to change the sign of the power in the denominator when we put it on the fraction bar, if this answer tempts us. Option B is the correct answer.

53) From Figure 2 we can see that half of the original sample at 500 MBq dropped to 250 MBq after a little over 5 hours but 10 hours ago. The half-life must be > 5 hours and < 10 hours.

1. 6h– 6 hours is within the range determined by the figure.
2. 12h– 12 hours is longer than 10 hours and too long.
3. 6 p.m– In 18 hours more than 75% of the sample has decayed. This is too long.
4. 24 hours–Almost the entire sample decayed at this point. Option A is the correct half-life.

54) The disintegration of⁹⁹Mo is described in the second paragraph: "Ö⁹⁹Mo falls apart^99mTc releasing a beta particleB^–and an antineutrinoῡ.”

1. a photon.– A charged particle is released,beta particleB^–, is an electron. Photon is not the correct answer.
2. a neutrino. - SameAntiNeutrino was emitted and isn't charged, so doesn't answer the question about thatloadedParticle.
3. an electron. – That's right, the released negatively charged beta particle is an electron.
4. a positron. – A positron is positive, not negative.

55)

1. Distinguishing between liquids and tissues. – A standard ultrasound is capable of this, so no advantage ofDopplerUltrasonic.
2. measure blood flow. – The Doppler effect describes the change in perceived frequency, when there is onerelative movementbetween a source and an observer. This allows movement to be measured, in this case movement/blood flow in the heart. It seems good!
3. Measure tissue density. – The stock density is independent of the relative movement. This is not an advantage of Doppler ultrasound.
4. Measurement of the thickness of the heart wall. – Like tissue density, heart wall thickness does not describe relative motion and measuring heart wall thickness is not an advantage of aDopplerUltrasonic. Answer option B is the only option that describes an additional advantage or an additional measurement by using the Doppler shift.

56) In the last paragraph, the author mentions that the stationary bike “a 30W load', which gives us the power of the bike (the unit of power is watts). Rearranging the power equation: power=work/time -> work=power*time

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

1. 5400J- That's the correct answer.
2. 90J– In this answer, we probably forgot to convert minutes to seconds. Track your units.
3. 6J– This answer option is too short.
4. 0,16 J- This answer is also very small; The correct answer is 5400 J.

discrete questions

57) A buffer ideally has a pKa value that is +/- 1 pH unit of the desired pH value.

1. 2.14. – This pKa value is too acidic for a reaction taking place at pH 5.3.
2. 4.75.– This pKa is 0.55 pH units below the desired pH value and thus in the range of 1 pH unit.
3. 6.5. - Close, but not quite. A pKa of 6.5 is 1.2 pH units above the desired range, so B is a better answer.
4. 7.0.– This pKa means that the buffer is even more basic than the previous one; a pKa of 4.75 is the only pKa that meets the threshold of 1 pH unit.

58) The specific heat can be found by rearranging q=mCΔT equation.

Q=mCΔT –> C = q/ mΔT = 250J/(0,1kg*5°C) = 250/0,5 J/(kg·°C) = 500 J/(kg·°C)

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

59) On test day it is unlikely that theFirstThe step you take for a question like this is to write down and solve each of the half reactions. That is, we will provide the math in theFilmexplanation for this question.

Let's say you're short on time or exactly where you want to be. How can you quickly and accurately solve this question? First check if the NADH is to NAD⁺Proportions are balanced. Then check that the net charges balance out on both sides of the equationeneutral. If you still don't have the answer, make sure each side has the same number as each atom (so if there are 5 carbons on the left, there are 5 carbons on the right).

1. CH₃C(=O)CO₂^–+ 2NADH → CH₃CH(OH)CO₂^–+2NAD⁺– There is a NAD⁺for each NADH, but the net charges on each side do not match. On the left there is a net charge of -1, but thetwoNAD positive⁺s on the right, in addition to the negative charge of lactate, means that the net charge to the right of the arrow is +1, not -1.
2. CH₃C(=O)CO₂^–+ 2NADH + 2H⁺→ CH₃CH(OH)CO₂^–+2NAD⁺– O NADH para NAD⁺relationship is correct and the net charges are the same (+1 and +1). However, the net chargepositive. Something must have gone wrong with adding and balancing electrons; they areshould have canceledwhen the half reactions have been added. Let's see if there is a better answer choice.
3. CH₃C(=O)CO₂^–+ NADH + 2H⁺→ CH₃CH(OH)CO₂^–+ YOU⁺– O NADH para NAD⁺Relationship is correct, but net charges are unequal (+1 and 0 on the left and right, respectively). Next.
4. CH₃C(=O)CO₂^–+NADH+H⁺→ CH₃CH(OH)CO₂^–+ YOU⁺– We discarded the other possible answers, so this must be the correct answer. It is indeed - the NADH to NAD⁺The relationship is correct, the net charges are equal and neutral, and each side has the same total number of each atom. See below if you want to walk through the whole process of balancing reactions.