Chem/Phys MCAT missed problems

What types of interactions will occur between the biofunctionalized graphene oxide molecule from Figure 1 and the intended ssDNA molecules? Covalent bonding and ionic bonding Hydrogen bonding and covalent bonding Hydrogen bonding and Van der Waals forces Covalent bonding only

Choice C is correct. Hydrogen bonding is the primary type of attraction that facilitates complimentary base pair matching, but Van der Waals attractions will also occur. Answer choices A, B, and D are all incorrect because they include covalent bonding. Covalent bonding and ionic bonding are not involved in the interactions responsible for nucleotide base pair matching. Knowledge of the differences between hydrogen bonding, covalent bonding, ionic bonding, and Van der Waals forces is a general chemistry topic. In this case, it is integrated with conceptual understanding of the interactions that account for DNA base pairing, thereby combining chemistry and biology in a classic way often utilized by the AAMC authors.

The fourth step of glycolysis is functionally equivalent to a reversal of Scheme 1. This information suggests which relationship between the trials in Scheme 1 and biological metabolism? Trial 1 is similar to glycolysis and Trial 2 is similar to gluconeogenesis. Trial 2 is similar to glycolysis and Trial 1 is similar to gluconeogenesis. Trial 1 only is similar to gluconeogenesis. Trial 1 only is similar to glycolysis

Choice C is correct. The aldol addition in Trial 1 is quite similar to the step of gluconeogenesis which uses an aldolase enzyme to form fructose-1,6-bisphosphate out of two precursors. This makes C the best answer. Answers A and B are both incorrect because the biological sugars involved in the aldolase-catalyzed gluconeogenesis reaction feature hydroxyl groups and do not contain alkene groups as does Compound 4. The biological version is an aldol addition (or its reversal), which is similar to Trial 1, not a condensation, which would be more similar to Trial 2. Further, the temperature at physiological conditions, although above room temperature, never approaches 50 °Celsius (122 °Fahrenheit) as is seen in Trial 2. Such temperatures could denature enzymes and cause various other problems in a biological system. For these reasons, we can eliminate any answer that contains Trial 2. Answer D is false because Trial 1 is similar to gluconeogenesis and would be the reversal of glycolysis.

What is a possible reason for the increased yield produced by chicken egg white lysozyme? The product is stabilized by the enzyme's active site. The reactant is stabilized by the enzyme's active site. The reactants are correctly oriented by the enzyme's active site. The solvent is exposed to the enzyme's active site.

Choice C is correct. This question relies on a basic understanding of enzymes and how they act as catalysts. Choice C is correct because an enzyme's active site exhibits a specific morphology that allows it to arrange the reactants in the ideal spatial orientation for reaction. Answer A and B may sound tempting, but they are both impossible. If enzymes stabilized either the product or the reactant they would not increase reaction rate. To lower the energy of activation they must stabilize the transition state. Because the solvent molecules acted as reagents, it is obvious that the solvent was accessible to the active site.

The decomposition of H2O2 is an exothermic process. However, solutions of H2O2 in water are stable when stored in bottles over a long period of time. Which statement best explains these two observations? A. The decomposition of hydrogen peroxide to water is highly exothermic. B. Hydrogen peroxide contains two oxygen atoms per molecule, while H2O contains only one. C. The formation of water from hydrogen peroxide is associated with a high energy of activation. D. The O-O bond in hydrogen peroxide is stronger than the O-O bond in atmospheric oxygen.

Choice C is the correct answer. Consideration of Reaction 1 and the molecular structure of H2O2 shows that to generate H2O the O-O bond must be broken at some point along the reaction path. Breaking chemical bonds requires considerable activation energy, which is often the defining parameter affecting the reaction rate; in this case the kinetic stability of H2O2. Answer A is wrong because the question stem clearly stated that decomposition of hydrogen peroxide is an exothermic process. Answer B is wrong because the number of oxygen atoms has no direct relation to the thermodynamics of a reaction. Reactions involving more or less oxygen atoms could have a variety of thermodynamic properties. Answer D is wrong because the statement is incorrect. O2 has a bond order of 2 compared with its bond order of 1 in H2O2. Since bond strength roughly correlates with bond order it is O2 that has the stronger O-O bond, not H2O2.

Which change in solution composition would cause a protein to elute from a hydrophobic interaction column? Decreasing pH Increasing pH Decreasing salt concentration Increasing salt concentration

Choice C is the correct answer. Hydrophobic interaction chromatography (HIC) relies on high salt concentrations to enhance or strengthen hydrophobic interactions. Therefore, decreasing the salt concentration weakens these interactions, causing the protein to dissociate from the column. There is not a significant role for pH in HIC, making A and B incorrect. There could be a small effect if charges were to be neutralized but this is not the primary effect. Answer D is not correct since this would strengthen interactions with the column, not weaken them.

The two parallel plate electrodes give the solar cell an inherent capacitance. If the distance between the plates in Figure 1 were increased from 20 µm to 40 µm, the capacitance would: increase by a factor of 2. increase by a factor of 4. decrease by a factor of 2. decrease by a factor of 4.

Choice C is the correct answer. The capacitance of a parallel plate capacitor is given by C = ε0A/d, where ε0 is a constant, A is the area of the plates, and d is the separation of the plates. From this formula you can see that if the separation is doubled, C goes down by a factor of 2, which is Choice C. Choices A, B and D do not reflect the relationship shown in this formula.

Electrons flow through the electron transport chain from electron donors to electron acceptors. Which member of the electron transport chain has the greatest electron affinity? Cytochrome C NAD O2 Coenzyme Q

Choice C is the correct answer. The electrons in the electron transport chain of the mitochondria are always passed to carriers with greater electron affinity. O2is the final electron acceptor and therefore must have the greatest electron affinity. The electrons are donated from NADH or FADH2 originating from the Kreb's cycle. Then, in a multi-step pathway, those electrons are transferred to ubiquinone, later to cytochrome c, and eventually to oxygen. The ranking of the electron affinities of the other answer choices is in the order of electron transfer: NADH < ubiquinone < cytochrome c.

Suppose it is determined that the average distance between the device lens and the arteries in which blood flow is being measured is only 100 mm. If a lens with a focal length of 20 mm is used, how far from the lens should the sensor be placed so that the image will be focused on the surface of the sensor? 4 mm 20 mm 25 mm 80 mm

Choice C is the correct answer. This problem can be solved using the Thin Lens Equation: 1/di + 1/do = 1/f, where di is the image distance, do is the object distance, and f is the focal length of the lens. The value asked for by the stem is the image distance, the distance between the lens and the real image, which is where the surface of the sensor should be located to work properly. Plugging in the values given in the stem gives: 1/di + 1/100 = 1/20. This simplifies to: 1/di+ 0.01 = 0.05, which gives: di = 1/0.04 = 25. Choice C is therefore the correct answer. Without a calculator, a student may not recognize that 1/0.04 = 25, but scientific notation makes this calculation easy: (1.0 x 100)/(4 x 10-2) = 0.25 x 102 = 25. Answer A is related to 0.04 and might be chosen if a student did not complete the calculation by taking the inverse. Answer B is the focal length given in the stem. Students may choose Answer D if they simply subtract the focal length from the object distance (100-20).

Sound levels are measured in decibels according to the equation below, where I0 is the nominal threshold of human hearing, 10-12 W/m2.A sound source located 1.0 m from a microphone is measured at 80 dB. The same sound measured at a distance of 3.0 m from the microphone will have a decibel rating of: β = 10log10(I/I0) 50 dB 60 dB 70 dB 77 dB

Choice C is the correct answer. This problem requires understanding two important principles. The first principle is that typically sound waves spread out in all directions according to an inverse square law. One can see this by noting that if there is no loss of energy due to absorption/attenuation, the energy passing through the area of any imaginary sphere surrounding the source must be the same. Since the sphere area grows with the radius squared, the sound level must drop as the inverse of radius squared. The same would apply to light intensity from a light bulb. Therefore, doubling the distance from the source causes the intensity to go down by a factor of 4. Or, in this case, tripling the distance causes the intensity to go down by a factor of 9. The second principle one must understand is how the logarithmic decibel scale works: multiplying the intensity by 10 causes a +10 change to the decibel number. For example, 90 dB is 10 times more intense than 80 dB, and 110 dB is 1000 times more intense than 80 dB. In this problem, the intensity goes down by a factor of 9, which is close to 10, so the decibel number should have 10 subtracted from it. Thus 70 dB is the correct answer (Answer C). Choices A or D (reducing the decibels by 30 or by 3, respectively) might be tempting if you don't know how tripling distance actually affects the decrease in intensity. Choice B is similarly incorrect as the number must be close to a 10 dB decrease.

During Sprinter A's initial acceleration, how did the force of the track on the sprinter compare to the force of the sprinter on the track? Ftrack on sprinter > Fsprinter on track Ftrack on sprinter < Fsprinter on track Ftrack on sprinter = Fsprinter on track Ftrack on sprinter > Fsprinter on track from t = 0 s to t = 1.5 s and Ftrack on sprinter < Fsprinter on track from t = 1.5s to t = 10 s

Choice C is the correct answer. This question tests the student's understanding of Newton's Third Law, which says that the force of object 1 on object 2 is always equal and opposite to the force of object 2 on object 1 (sometimes called "partner forces" or "action-reaction" pairs). Thus, the force of the sprinter on the track is equal in magnitude to the force of the track on the sprinter. It doesn't matter whether the sprinter is accelerating at the time or not. The other answer choices all indicate an inequality of these two partner forces, and are therefore incorrect.

Which compound has a structural isomer with a boiling point that is significantly different than its own boiling point? Compound 1 Compound 2 CHCCH3. CH3OCH2CH3 Compound 1, because one of its isomers is a cyclic alkene. Compound 1, because triple bonds have a dramatic impact on boiling point. Compound 2, because one of its isomers is a cyclic ether Compound 2, because one of its isomers is capable of hydrogen bonding.

Choice D is correct. The isomer of Compound 2 which can accept and donate hydrogen bonds is propanol. Hydrogen bonding has a drastic effect on boiling point amongst molecules of the same molecular weight. Answer choice A pits the compound listed against cyclopropyne; these two compounds do not have significant differences in intermolecular forces, they both exhibit only London dispersion forces. Choice B is incorrect because the presence of triple bonds alone does not strongly affect boiling point. Choice C is wrong because there is no such cyclic ether that is an isomer of the listed compound; hydrogen atoms would have to be omitted to form a cyclic compound.

If the researchers had used 50 µL of the ADH1 solution instead of 5 µL, how would this change have affected their study of ADH1 activity? Increased production of acetaldehyde Decreased production of NADH Decreased pH of the solution Increased rate of acetaldehyde formation

Choice D is correct. The passage gives the reactant used by the researchers as EtOH. Based on Reaction 1, an ORE such as ADH1 will convert EtOH into acetaldehyde. Because ADH1 is an enzyme catalyst, the addition of more enzyme will have the effect of increasing the rate of the reaction. Answer A is incorrect because the addition of a catalyst does not change the position of an equilibrium, and therefore the overall yield of acetaldehyde would not be altered. Answer B is incorrect because the rate of production of NADH will increase, not decrease. Answer C is incorrect because the equilibrium concentration of H+ will not change. Answer D is correct because the addition of more catalyst will increase the rate of production of the acetaldehyde product.

The average height of an American adult is 1.70 m. When an individual of this height is standing, by how much will the blood pressure at the bottom of the feet differ from the blood pressure at the top of the head? (Densityblood = 1060 kg/m3) 2000 Pa 6000 Pa 10000 Pa 18000 Pa

Choice D is correct. The pressure in a static fluid varies with depth h according to P = Pair + ρgh, where the ρgh term arises from the weight of the fluid. In this case, the pressure at the top isn't from the air, but the same principle applies: Pfeet = Phead + ρgh. The pressure difference is thus ρgh = 1060×9.8×1.7 = 17660 Pa, which rounds to 18000 Pa. Since the answers are not particularly close to each other, the multiplication can easily be approximated as ρgh ≈ 1000×10×1.7 = 17000 Pa, which is closest to choice D. Choices A, B, and C are various options that could be obtained via multiplication errors or mistakes made in choosing which quantities to multiply, and are all incorrect.

For an acid-base equilibrium at constant temperature, which value remains constant? pH [conjugate base] [weak acid] pKa

Choice D is correct; pKa is the only constant among the available answers. The acid-base equilibrium constant, Ka, is a constant (as the name suggests) for a given reaction at a given temperature, and therefore the negative log of that constant would also be a constant. The pH of the solution is dependent on the concentration of acid and base in the equilibrium, making Answer A incorrect. The concentration of both the acid and the conjugate base c

Which experiment would be most useful in determining whether or not phenazonapthol glycosides could serve as effective prodrugs? Alter the chemical structure of the phenazonapthol and test its reactivity with β-glycosidases. Measure the activity of the phenazonapthol in vivo by administering them to live cell cultures. Measure the activity of lactose in vivo by administering it to live cell cultures containing β-glycosidases. Examine the reactivity of both the phenazonapthol glycosides and phenazonapthol in vivo using UV spectroscopy.

Choice D is the best answer. The phenazonapthol glycosides might be used as a prodrug if the drug could be combined with the phenazonapthol moiety and if—but only if—the drug-phenazonapthol compound were inactive when glycosylated and became active when deglycosylated. Passage information has already indicated that the sugar moiety can be cleaved in vivo. Scientists could also take advantage of the fact that the cleaved and uncleaved forms are distinguishable via UV spectroscopy (see the end of paragraph 3) to determine whether or not the prodrug has been successfully cleaved into its active form. Answer A can be eliminated because altering the chemical structure would produce a new compound with entirely new properties. Answers A and B both reference phenazonapthol being tested, when it would be the prodrug-like form, a phenazonapthol glycoside, which would be of interest when testing for a possible new prodrug. Testing the non-glycosylated precursor by itself would not provide useful information. Answer C can be eliminated because a) testing lactose alone would not provide information relative to the new prodrug, and b) giving the potential new prodrug to live cells would not directly tell us anything about lactose itself (as the answer suggests it would).

Device technicians are measuring blood flow in an artery that narrows from point A to point B. How will the speed (v) and pressure (P) of the blood compare between points A and B? vB < vA vB > vA PB < PA I only II only I and III only II and III only

Choice D is the correct answer. As a pipe or artery narrows, this causes an increase in the velocity of the fluid for the same reason that a garden hose sprays water more rapidly when you partially cover the end with your thumb. This is because the volume flow rate (Q in the continuity equation: Q = AV) remains the same for position A compared to position B. In order to maintain Q, as cross sectional area (A) decreases, velocity (V) must increase. When the speed increases in a flow at constant height, the Bernoulli effect says that the pressure will decrease, according to: P + ρgh + ½ rv2 = constant. If the ρghterm is zero (or does not change because both cases are at the same vertical height) and the right hand term increases due to the increased velocity of the fluid, then the P term must decrease. Thus vB > vA, and PB < PA, which is described by choice D. Statement I is false because it is the opposite of what was just described. Answers A and D cannot be true because they contain Statement I. Answer B is correct, but incomplete because Statement III is also true.

The Lelior pathway of galactose metabolism is shown. Galactosemia is a condition that results in the accumulation of galactose-1-phosphate. Which enzyme is defective in galactosemia? UDP-glucose + galactose-1-phosphate <--> glucose-1-phosphate + UDP-galactose Galactose-1,6-bisphosphokinase Hexokinase Galactokinase Galactose-1-uridylyltransferase

Choice D is the correct answer. Galactose-1-uridylyltransferase is the defective enzyme, leading to the accumulation of galactose-1-phosphate. This can be deduced by looking at the names of the substances in the reaction shown and comparing them to the names of the enzymes given as answer choices. With the exception of a few common names that have a long history of use, most enzyme names include a clear indication of the enzyme's substrate and the enzyme class. In order for galactose-1-phosphate to build up, the reaction must be impeded in the forward direction. Also, UDP is being transferred from glucose to galactose, telling us it is a transferase, not a kinase (which eliminates choices A, B and C). Answer B is an enzyme from glycolysis and not from the initial steps of metabolism of galactose (though galactose enters glycolysis later in the pathway). Answer A is not a known enzyme. It is a "fabricated distractor"—something the AAMC uses on occasion.

One theoretical model of enzyme-substrate interaction explains enzyme catalysis as a function of the stabilization of the transition state by the enzyme active site. This is an example of the: catalysis mechanism. lock and key mechanism. Michaelis-Menten mechanism. induced-fit mechanism.

Choice D is the correct answer. The accepted mechanism for enzyme function is the induced-fit mechanism, which proposes that the enzyme stabilizes the transition state of the substrate, causing a reduction in the activation energy of the transition state. Answer A is not correct because "catalytic mechanism" is not an accepted theory of enzyme-substrate interaction. Answer B is not correct because the lock and key mechanism suggests that the unchanged substrate is stabilized, not the transition state. Answer C is not correct since Michaelis-Menten is a kinetic equation and not a mechanism describing enzyme-substrate interaction.

Which concept best accounts for the pressure difference between upper and lower wing surfaces described in the passage? Archimedes' Principle, because the air is warmer above the wing than below it Archimedes' Principle, because the buoyant force is lower above the wing than below it Bernoulli effect, because the air is more dense above the wing than below it Bernoulli effect, because the air is moving faster above the wing than below it

Choice D is the correct answer. The pressure difference arises because the air is traveling at different speeds above and below the wing. The effect of fluid speed on pressure is called the Bernoulli effect, thus choice D is correct. Archimedes' Principle relates to static fluids, namely how a buoyant force is created by displacing a static fluid, thus Choice A and Choice B are incorrect. The statement in Choice C is likely untrue (if anything, the density would most likely be lower below the wing than above it), but even if it were true, that phenomenon would not be an example of the Bernoulli effect.

At maximum contraction, the length of the PAM decreases by 50% from 32 mm to 16 mm and its radius increases by 100% from 6 mm to 12 mm. The ratio of the number of moles of gas inside the PAM at max contraction to the initial moles of gas inside the PAM is closest to: A. 2:1 B. 4:1 C. 5:1 D. 10:1 Pressure: increases 5 fold

Choice D is the correct answer. This problem tests your ability to manipulate the ideal gas law in order to predict a ratio of moles between two scenarios. It also requires that you recall the five-fold increase in pressure noted in the caption for Table 1. There are many ways this problem could be solved. The more complex approach is to solve PV = nRT for moles and create a ratio for the two scenarios (the RT variables cancel, and are constant for both scenarios anyway): n2/n1 = P2V2/P1V1 =. The volume of the sleeve may be approximated as V = πr2L. This gives: [(500,000)(π(122)16)]/[(100,000)(π(62)32)], or 1,152,000,000/115,200,000 = 10. A more simple approach is to use the manipulating equations skills taught to you in the Altius Student Study Manual. Considering n = PV/RT, we see that if pressure increases by a factor of 5, moles must increase by a factor of (5). We also see that if volume goes up by a net factor of 2, this influence will increase moles by a factor of (2). Thus, moles increase by a factor (5)(2) = 10. The net volume increase is known as a result of the radius going up by a factor of 22 and the length decreasing by a factor of 2 (V = πr2L): 4/2 = 2. Answer A may be chosen if the student considers only the increase in moles due to volume. Answer B may be chosen if the student ignores the square on the radius, forgets the influence of the increased pressure, and erroneously treats length as being inversely related to volume. Answers C may be chosen if the student considers only the influence of the increased pressure.

A certain airplane has a cruising speed of 240 m/s and a body mass of 60,000 kg. If the wind resistance produces a backwards force of 240,000 N while the airplane is cruising at constant speed, how much power must the engines produce to keep the airplane moving forward? 4.8 × 102 W 1.0 × 103 W 1.4 × 107 W 5.8 × 107 W

Choice D is the correct answer. You must first realize that since the airplane is traveling at constant velocity, the forward thrust provided by the engines must be equal to the backwards force of wind resistance. Since power is equal to force times velocity (for constant velocity situations), the answer is P = 240,000 N × 240 m/s = 5.76 × 107 W. To avoid the full multiplication, notice that the numerical answer must be a little larger than (2 × 105) × (2 × 102) = 4 ×107. Choice D is the only answer fitting that requirement.

Hill coefficient

Cooperatively, slope

SN1 reaction

*unimolecular nucleophilic substitution reactions: 2 steps 1. Leaving group leaves forming a positively charged carbocation (rate limiting step) *The rate of rxn depends only on the concentration of the substrate rate = k [R-L] R-L is the alkyl group containing the leaving group *Anything that accelerates the formation of the carbocation increase the rate of rxn 2. Nucleophile attacks the carbocation (unstable) *results in substitution product

TLC Chromatography

-"Polar is Slower = lower Rf value" -Separation of mixtures using a Mobile and Stationary Phase

Suppose a researcher discovers a small molecule that inhibits CYPs once APAP is bound in the active site. This would illustrate which type of inhibition? A. Competitive B. Non-competitive C. Mixed D. Uncompetitive

D is the correct answer. The inhibitor inhibits the enzyme only after the substrate is bound. That means it is an uncompetitive inhibitor, and so D is the correct answer. Competitive inhibitors bind in the substrate pocket and mimic the substrate, so A is incorrect. A non-competitive inhibitor will bind the enzyme equally well whether or not the substrate is bound. Mixed inhibition is a combination of uncompetitive and non-competitive inhibition, in which the inhibitor can bind the enzyme with or without substrate bound, but the binding of the inhibitor affects the rate of binding of the substrate, and vice versa. Because the inhibitor can bind the enzyme whether or not the substrate is bound with both non-competitive and mixed inhibition, B and C are incorrect.

Newton's Third Law

Describes what happens when one object exerts a force on a DIFFERENT object

Suppose researchers need a purified sample of ᴅ-limonene with minimal contamination. To accomplish this, ᴅ-limonene could be doubly distilled: under a nitrogen atmosphere. below atmospheric pressure. under UV light. using filtered, purified air.

Distilling under nitrogen excludes air, and therefore oxygen, from reacting with the limonene. Because the main premise of the passage is that D-Limonene undergoes automatic oxidation in the presence of oxygen, excluding oxygen is a logical requirement if one is to obtain a pure sample. Answers B, C, and D are not applicable because they do not exclude air. Answer A would provide a distilled sample that is oxygen-free.

Prodrug

Drug that becomes more active after it is metabolized.

Urease uses nickel ions inside its active site. Chelating agents in high concentration inhibit the enzyme by forming coordination compounds with nickel. Which of the inhibitors from Figure 2 is most likely to act via nickel chelation?

From the information provided, it should be apparent that the substrate interacts with a nickel atom in the active site. Chelators will bind and "tie up" a metal ion, and would be expected to inactivate the enzyme. For an inhibitor to be effective as a chelator, it must interact with the active site, which is the definition of a competitive inhibitor. ACTIVE SITE=COMPETITIVE INHIBITOR

Aromaticity

1. Cyclic and conjugated 2. No sp3 atoms in ring 3. Planar 4. 4n + 2 pi electrons

Lineweaver burke

1/Vmax = y intercept 1/Km = x intercept

How many protons are translocated by NADH and by FADH2?

10 by NADH and 6 by FADH2

visible light spectrum

400-700 nm, 400=violet, 700=red

Based on its molecular structure, NAPQI most likely damages proteins through which process? Oxidation Phosphorylation Acetylation Cleavage of carbon-carbon bonds

A is the correct answer. NAPQI is damaging proteins by oxidizing them. This can be deduced by looking at the structure and noticing the carbonyl groups. Because of the unique nature of the ring with two double bonds (instead of three, as in an aromatic ring), the carbonyl oxygens are even more oxidizing. Answer B is incorrect because there is no phospho- group in NAPQI, and no reason to assume that it would participate in substrate level phosphorylation. Answer C is incorrect because while there is an acetyl group in NAPQI, there's no reason to assume that acetyl group transfer would damage a protein. Answer D is incorrect because there is nothing in the structure of NAPQI that would lend itself to cleaving carbon - carbon bonds.

Assuming viscous flow, water in the model vessel will flow: fastest in the top-center portion of the channel and more slowly along the sides and bottom. fastest along the sides and bottom of the channel, and more slowly at the center. fastest at the center of the channel and more slowly along the top and sides. at an equal rate throughout the channel, independent of location.

According to the physical properties of viscous flow there is friction between the fluid and container which creates drag. As the model is a long channel with sides and a bottom, there will be drag along the side walls and bottom of the channel. Water in the center of the vessel is further away from the side wall drag, and water at the very top is further away from the drag created by the bottom of the container. Thus, fluid at the top-center will flow the fastest. Answer A is correct, because it is the only one that describes a slow-down at the walls and bottom, with the fastest speed being in the top center. Answer C is incorrect; while this answer choice would explain the viscous fluid flow of an enclosed channel, the passage describes an open channel. Answer B is incorrect, fluid nearest the areas of drag will flow the slowest according to the physical properties of viscous flow. Answer D is incorrect, the fluid velocity will vary with distance from the areas providing drag. It would only be correct if there were not viscosity.

Figure 3 compares solar cell efficiency for light of various wavelengths. For four wavelengths, 400 nm, 525 nm, 700 nm, and 820 nm, which wavelength contains photons with the most energy? 400 nm 525 nm 700 nm 820 nm

According to the wave formula, v = fλ, frequency and wavelength are inversely related. All light travels at the speed of light, so velocity is a constant. Therefore, photons with a larger wavelength have a smaller frequency, and vice versa. From the formula for the energy of a photon, E = hf, it is plain to see that higher frequency equates to higher energy. At the extremes of the spectrum, x-rays (very short wavelength) are much more energetic than radio waves (very long wavelength). The shorter the wavelength is, the higher is its energy. The shortest wavelength given is 400 nm, making choice A correct. Choices B, C and D all suggest longer wavelengths, which contain less energetic photons.

Allosteric activator vs inhibitor

Activator shifts curve left, inhibitor shifts curve right

Suppose the native environment for a newly discovered bacterium is increasing in temperature. Which evolutionary change in the bacterium is most likely over time? Increased number of disulfide bonds Higher prevalence of beta sheets Increased number of hydrophobic patches Higher prevalence of premature stop codons

Answer A is the correct choice. The newly discovered bacterium is experiencing a shift in temperature, and as such, the genome is likely to be adapting through the natural selection of mutations that favor greater protein stability at higher temperatures. Answer A is the best choice because disulfide bonds are covalent, and as such they will not melt at high temperatures. Choice B is incorrect because the relative number of beta sheets has no particular bearing on whether a protein is more or less heat stable. Choice C is incorrect because, although increased hydrophobic interactions could increase protein stability, this is a much weaker interaction than disulfide bonds. Choice D is incorrect because mutations encoding early stop codons typically lead to non-functional proteins, not heat-stable proteins.

For which atom in levetiracetam is the most energy required to completely remove one valence electron? Oxygen Nitrogen Hydrogen Carbon

Answer B is correct. The amount of energy required to completely remove one valence electron from an isolated atom in the gaseous state is defined as the ionization energy. The periodic table trend for ionization energy increases as one move up and to the right. N and O are the exception to these trends, however, due to electron repulsion from oxygen's additional electron in the already half-filled 2p orbital. Therefore, the order of increasing ionization energy for the options listed is: N > O > C > H. This makes Answer B correct and eliminates the other options.

Ammonia is an example of a Bronsted-Lowry base. Through which process does ammonia raise the pH of biological solutions? Ammonia is a strong base, so small concentrations cause large increases in pH. Ammonia releases NH2-, which is analogous in size to the hydroxide ion. By abstracting protons from water, ammonia causes an increase in [OH-]. By abstracting protons from water, ammonia causes an increase in [H+].

Answer C is correct. Bronsted Lowry bases by definition accept protons. In aqueous solutions, they can accept a proton from water leaving behind hydroxide. If hydroxide increases, pH goes up. Choice A is incorrect, this describes strong bases such as KOH and NaOH. Choice B can be ruled out because the hydrogens of ammonia are not readily dissociable; therefore, this does not occur. Choice D states that by removing H+, the [H+] increases, which is impossible.

Stearic acid is found in animal fat and has the chemical formula C17H35CO2H. How many β-oxidation steps must stearic acid undergo until it is completely broken down into two-carbon fragments? 10 9 8 7

Answer C is correct. The chemical formula shows a total of 18 carbon atoms. Students may fail to count the carbon included in the CO2 portion of the formula. However, this is a frequent way of representing fatty acids and all carbons must be considered. Nine C2 fragments can be produced from 18 carbons. However, after 7 β-oxidations, a C4 fragment remains and only one cycle is required to achieve nine C2 fragments. Hence, 8 β-oxidations are required to complete the breakdown of stearic acid. Answer D is too few oxidations; it would be correct if the fatty acid had 16 carbon atoms. Answers A and B would be correct for 22- and 20-carbon fatty acids, respectively.

A new bacteria is discovered that utilizes a novel nucleotide, designated as H. The H nucleotide replaces guanine in the bacteria's genome and base pairs with cytosine. This bacteria is likely to be highly successful at evading host immune system responses because the: bacteria's genome will not be recognized by antibodies in the host's immune system. bacteria will produce new proteins which cannot be detected by the host's immune system. H nucleotide will disrupt restriction endonucleases that target non-self DNA molecules to degrade them. H nucleotide will make the bacterial surface proteins more similar to human surface proteins.

Answer C is correct. The newly discovered bacterium has a genome that uses ATCH instead of ATCG, but no other changes are described. We are told that H base pairs with C, so one can assume that it behaves in most respects just like G does. Answer C is the best answer because one way that organisms fight foreign invaders is through restriction endonucleases that recognize 'non-self" DNA sequences. These endonucleases have very specific restriction sites (It would be very dangerous to have endonucleases without specific sites, since they could cleave any DNA). Answer A is incorrect because antibodies are generated against proteins, not DNA molecules. Answer B is incorrect because there is nothing to indicate that new proteins are being generated. If H base pairs with C, tRNA should interact with codons the same way whether H or G is being used. Answer D is incorrect for the same reason, there is no reason to believe the bacteria will produce new proteins, or that those proteins will be more like host surface proteins because of the H for G transition.

Suppose a molecule composed of atoms A and B features a permanent dipole. This dipole is mostly likely the result of: an odd number of neutrons in the central atom. an incomplete octet for either atom, or both atoms. the presence of pi bonds between atoms A and B. the uneven distribution of electrons between atoms A and B.

Answer D is correct. A polar molecule has a positive pole and a negative pole, resulting from the uneven distribution of electrons. The negative pole has a surplus of electrons, while the positive pole has a deficiency of electrons. Choice A is incorrect because neutrons do not affect the polarity of a molecule. Choice B is incorrect; an incomplete octet does not necessarily indicate a permanent dipole. BF3 is an example of a nonpolar molecule with a central atom that has an incomplete octet. Choice C is incorrect because pi bonds can be present in both polar molecules (CO for example) and nonpolar molecules (C2H4 for example).

The light from an OLED device is produced by the: excitation of an electron to a higher energy level. absorption of a high-energy photon. emission of a positron associated with the relaxation of an electron. emission of a photon associated with the relaxation of an electron.

Answer D is correct. The exact process by which the electrical-to-light transition occurs in OLED devices is known, but not discussed in the passage, nor expected as prior knowledge. Therefore, reasoning about general principles of energy and light are required to answer this question. Answer A is impossible because light is energy and moving to a higher energy level requires energy—it does not release it. Answer B is false because it describes the absorption rather than the release of light. Answer C is false because it confuses a positron for a photon. It is photon emission, not positron emission that is associated with light emission. Answer D is the best choice. It describes the process responsible for most atomic light emission events: an electron is excited by absorption of energy and then releases light when it relaxes to its original energy state. Releasing light is caused by moving from high to low energy

Solutions A and B were placed in a flask separated by a selectively permeable membrane that mimics biological lipid bilayers and allowed to equilibrate overnight. What does this suggest about the original solutions A and B? A AND B EQUAL -> B HIGHER THAN A Solution A has a lower concentration of analytes. Solution B has a lower concentration of analytes. Both solutions have an equivalent concentration of analytes. The analyte in both solutions could freely cross the membrane.

Answer choice A is the correct choice. The volume on side B of the flask increases, which suggests that there is a higher concentration of the analyte on side B and water is diffusing to this side via osmosis. Answer B is not correct since the volume on side A would increase if Solution B had a lower concentration of analyte. Answers C and D are not correct since there would not be a change in volume in either of these scenarios. WATER MOVES LOW TO HIGH SOLUTE CONCENTRATION

During XPS, if the energy of the laser is sufficient to eject electrons, increasing the frequency of the laser will also increase the: binding energy of the ejected electrons. kinetic energy of the ejected electrons. number of photons ejected. wavelength of the laser light.

Answer choice B is correct. First, one must realize that increasing the frequency of the laser light will increase its energy, according to E = hf. If the laser source has more energy per photon, more energy will be available (in excess of the binding energy and work function) to be converted into the kinetic energy of the ejected electron. Solving Equation 1 for KEphoton makes this more clear: KEphoton= Elaser - (BE + j). This question is answered most easily if a student has a solid conceptual understanding of the photoelectric effect. Increasing the energy incident upon a photoelectric surface leaves more excess energy available to be converted into the kinetic energy of any electron that is ejected. Answer A is false because using a higher frequency laser will not alter the binding energy. Answer C is false because electrons are ejected, not photons. Answer D is false because the wavelength and frequency of light are inversely related; if the laser light is of higher frequency it will actually have a smaller wavelength. From a bigger-picture perspective, this question provides an excellent example of the kind of careful critical thinking required by the MCAT. At first glance, a student may assume that an increase in Elaser will increase BE (Answer A). However, this is impossible because BE is an innate characteristic of the atomic environment surrounding the electron prior to its ejection. Using a more energetic laser light source will clearly not CHANGE the pre-existing atomic characteristics of the sample being examined. When using equations to make predictions on the MCAT, it is always important to first determine if the prediction makes conceptual sense. For example, although it is true that F = mg, it is not logical to suggest that increasing the mass of an object (m) will decrease the magnitude of the gravitational field (g).

The lens shown in Figure 1 creates an image of the rotating and static diffusers on the sensor. If the focal length of the lens is 25 mm, the image formed at the sensor will be: virtual and upright. virtual and inverted. real and upright. real and inverted.

Answer is D. Positive images are Real and Inverted, and Negative images are Virtual and Upright. PRI NVU

At physiological conditions, the autoionization constant of water (Kw) is approximately 1 x 10-14. If temperature is increased by a factor of four, and pressure is increased to 25 MPa, Kw is increased by a factor of 100. How does the pH of water at this elevated temperature and pressure differ from pH under physiological conditions? It decreases by one pH unit. The pH is equal to 7 under both conditions. It increases by one pH unit. It increases by five pH units.

At room temperature, Kw = [H+][OH -] = 1 x 10-14. The stem uses this same figure for physiological conditions (pH ~ 7.4). We are told this value increases by a factor of 100, which would be 100 x 10-14, or 1 x 10-12. Thus, at the conditions stated in the problem, [H+][OH -] = 1 x 10-12 and [H+] = 10-6 M. From this we see that pH = 6 and the pH has decreased by one pH unit (Answer A). Answer B is incorrect because, as shown, the pH values are indeed different. Response C would be the pH of neutral water if the Kw was 100 times smaller than its room temperature value. Response D could result from correct conversion of Kw, but then taking -logKw instead of -log [H+].

Under anaerobic conditions, pyruvate is converted to lactate rather than being funneled into the PDH complex and the Citric Acid cycle. Which statement best explains the purpose of this condition-specific biochemical alteration? To generate additional energy from the pyruvate molecule. To regenerate NAD+ so that glycolysis can continue. To generate lactate, a required precursor for glycogen synthesis. To generate lactate, a precursor for subsequent energy-rich oxidation steps.

B is the correct answer. Glycolysis produces 2 NADH that generally donate their electrons to the electron transport chain. However, this only occurs when oxygen is present. Under anaerobic conditions, when oxygen is not present, the NADH has to be converted back to NAD+ or glycolysis stops. Answer A is not correct since more energy is produced if pyruvate is converted to Acetyl-CoA rather than lactate. Answer C is not correct since glucose-6-phosphate is the precursor for glycogen, not lactate. Answer D is not correct since lactate is a dead-end metabolically.

Carbon forms ethers similar to the silane ethers described in the passage. Reacting which two species will produce a carbon ether? tert-Butanol and an acid Ethyl alcohol and ethyl tosylate Ethyl alcohol and ethylamine tert-Butanol and tert-butyl chloride

B. ethylamine is a nucleophile and will not react with ethyl alcohol

Ethoxide

Base used as a catalyst

Capacitance equation

C=Q/V

Hyperventilation (rapid breathing) occurs in response to respiratory acidosis because hyperventilation: increases blood pH by decreasing CO2 concentrations. increases blood pH by decreasing HCO3- concentrations. decreases blood pH by decreasing CO2 concentrations. decreases blood pH by decreasing HCO3- concentrations.

Choice A is the correct answer. During hyperventilation, rapid breathing continually removes CO2 from the blood at a rate that is faster than the rate at which CO2 is being produced in the blood. Hyperventilation therefore reduces CO2, shifting the equilibrium in Equation 2 to the left (CO2 + H2O ↔ HCO3- + H+), leading to the conversion of HCO3- and H+ to CO2 and H2O to balance the system and return to equilibrium (Le Chatelier's Principles). Choice B is not correct since decreasing the concentration of HCO3- would lead to a decrease in blood pH and not an increase. Answer C is not correct because hyperventilation increases blood pH, it does not decrease it. Answer D is not correct since hyperventilation does not regulate HCO3- concentrations.

For most materials, the maximum static friction is: A. larger than the kinetic friction force. B. smaller than the kinetic friction force. C. equal to the kinetic friction force. D. equal to the gravitational force.

Choice A is the correct answer. Under normal situations, when a horizontal force is applied to an object the static friction force increases to its maximum limit, and then the object moves. The maximum limit is given by f = µsN. Compare this to the normal kinetic friction force experienced when the object is moving, which is f = µkN. The two expressions are very similar, but since µs > µk for essentially all known surfaces, the maximum static friction force will always be greater than the kinetic friction force. Choice A is therefore correct, and this eliminates choices B and C. Choice D could be a tempting option since the friction force is indeed related to the gravitational force mg. Under normal conditions, and if the object is on a horizontal surface, the maximum static friction force is equal to µN = µmg. However, µmg = mg only in the hypothetical circumstance where µ = 1, so choice D is also incorrect.

Which reagent from Figure 1 is an example of an ionic compound? MeOH NaOH HCl N2H4

Choice B is correct. NaOH is an ionic compound formed between a sodium cation (Na+) and a hydroxide anion (OH-). Answering this question requires general familiarity with ionic bonding and common ions, such as hydroxide. Answer A is incorrect because MeOH is methanol, which contains only covalent bonds. Answer C, HCl, may be tempting to some students because it is a strong acid that readily dissociates into ions in solution, but the H-Cl bond itself is covalent, not ionic (Caveat: It is worth noting that most covalent bonds have partial ionic character). Answer D is incorrect because hydrazine contains only covalent bonds. This is an example of an easy MCAT question that is answered correctly by nearly 90% of examinees. Your real MCAT will contain a broad mix of question difficulty. You should encounter some very easy questions such as this, several questions that are quite difficult, and most of the other questions will fall somewhere in-between.

If Zn2+ forms a quaternary coordination compound within the ADH1 active site, the bond angles between the coordinated ligands is closest to which of the following? 90° 110° 120° 180°

Choice B is correct. The question stem states that the zinc coenzyme is coordinated with the ADH1 active site in a quaternary format (i.e., it coordinate with four ligands). Four bound atoms or coordination members requires a tetrahedral arrangement. The bond angles of a tetrahedron are predicted to be 109.5°. Answer A is incorrect because this is the bond angle for octahedral molecular geometry. Answer B is the closest to 109.5° and is therefore correct. Answer C is incorrect because this is the bond angle for trigonal planar molecular geometry. Answer D is incorrect because this is the bond angle for linear molecular geometry.

When other biomolecules are unavailable, acetoacetic acid can be used as a fuel source in all of the following organs EXCEPT the: heart. liver. brain. muscles.

Choice B is the correct answer. Acetoacetic acid is a ketone body. Ketone bodies can be used as fuel in the brain, the heart, and muscles (so Answers A, C, and D are incorrect), but they cannot be used in the liver because it lacks the necessary enzymes to convert ketone bodies to ATP. This is because ketone bodies are produced in the liver when blood glucose is low.

Patients with Type I Diabetes can develop blood ketoacidosis due to the excessive breakdown of fatty acids. What effect does this increase in acid concentration have on blood pH during ketoacidosis? CO2 + H2O ↔ HCO3- + H+ Low pH, shifting Equation 2 to the right Low pH, shifting Equation 2 to the left High pH, shifting Equation 2 to the right High pH, shifting Equation 2 to the left

Choice B is the correct answer. High concentrations of ketone bodies leads to formation of keto acids and ultimately ketoacidosis. The high levels of keto acids results in decreased blood pH (i.e., more H+ ions in the blood), and shifts the plasma buffering system shown in Equation 2 to the left, in accordance with Le Chatelier's Principle. Ultimately, the increased acidity would result in increasing concentrations of CO2 as the system seeks to restore equilibrium. Answer A is incorrect, as a result of the previously mentioned shift. Answers C and D can be eliminated because blood is more acidic and therefore pH must be lower, not higher.

Activation of an enzyme by phosphorylation of a tyrosine residue is an example of what type of enzymatic regulation? Genetic control Covalent modification Allosteric modification Compartmentalization

Choice B is the correct answer. Phosphorylation is the addition of a phosphate group to a target molecule, such as the addition of phosphate to a tyrosine -OH group. This requires covalent modification, making Answer B correct. Answer A is incorrect; an example of genetic control would be the upregulation or downregulation of transcription through gene transcription factors. Answer C is also incorrect; allosteric modification describes the non-covalent interaction between a regulatory molecule and an enzyme away from the enzyme's active site. Finally, Answer D is incorrect. There is no information given in the stem that indicates the enzyme and the substrate are in different cellular compartments, so compartmentalization is not a logical choice. For phosphorylation to occur, the enzyme and the phosphorylating agent would need to be in the same cellular compartment.

The fluorescent light emitted from the β-isoindole fluorophore must: A. be infrared light. B. be ultraviolet light. C. have a wavelength greater than 420 nm. D. have a wavelength less than 420 nm.

Choice C is correct. Fluorescent light has to be of lower energy (longer wavelength) than the source light used to cause the excitation. The passage indicates that the light incident on the tube was 420 nm, so the light fluoresced must be of wavelength longer than 420 nm (lower energy). Answer A is unlikely to occur because infrared light is on the opposite end of the visible spectrum, in excess of 700 nm. Regardless, it is not true that it "must" occur as the stem requires; emission could occur at any wavelength greater than 420 nm. Answer B is much closer to the likely wavelength of the emitted light, but it is still impossible because 420 nm is already above the ultraviolet region and the wavelength is getting longer, not shorter. Answer D is false because it would indicate a higher energy wave of shorter wavelength. E=hc/wavelength Energy is lost when light is given off, wavelength and energy inversely proportional

Negative control

Group with no response expected

Radical

Has a single electron

highest priority

Highest atomic number

Which conclusion is most reasonable given the MO diagrams for benzene and cyclobutadiene? Electrons in antibonding MOs increase a molecule's energy. Nonbonding MOs are lower in energy than the atomic orbitals from which they form. In aromatic molecules, nonbonding MOs are lower in energy than bonding MOs. Partially filled antibonding MOs increase a molecule's stability.

Highest to lowest energy: Antibonding MO>Nonbonding MO (at baselevel, same energy as when they are formed> bonding MO

What happens to amides when exposed to water?

Hydrolysis

The ᴅ designation in ᴅ-limonene is an abbreviation for dextrorotatory. This indicates that ᴅ-limonene rotates polarized light: counterclockwise and could contain a chiral carbon with either the R or the S configuration. clockwise and could contain a chiral carbon with either the R or the S configuration. counterclockwise and must contain a chiral carbon with the S configuration. clockwise and must contain a chiral carbon with the R configuration.

In organic chemistry, ᴅ and ᴌ are used to indicate if an isomer is dextrorotatory or levorotatory, respectively. Dextrorotatory isomers rotate polarized light clockwise. These letters are not associated with any particular R or S absolute configuration, and thus answers C and D can be eliminated. It may be tempting to assign R/S configuration using the limonene structure shown in the passage--which includes the necessary stereochemistry to make that decision. However, the question stem only asks what the ᴅ designation indicates about ᴅ-limonene. That designation alone does not give any information about R/S. Answer A incorrectly uses counterclockwise for the rotation. Answer B is thus the best answer.

Anomeric carbon

Invertible epimer

Strong oxidizers

KMnO4, HClO4

Km definition

Km is a measure of the affinity of the enzyme for its substrate. Km is defined as the substrate concentration that will produce an initial velocity equal to ½ Vmax

Lactate dehydrogenase acts on

Lactate

Hepatic cells are located, drugs are metabolized in

Liver

Newton's Second Law can be used to relate force to mass and acceleration. In Trial 2, between seconds 8.5 and 33, the dynamometer does not accelerate because: The behavior of springs is not explained by Newton's Second Law. Newton's Second Law describes only non-variable forces. Newton's Second Law describes only unbalanced forces. Newton's Third Law predicts that the dynamometer pushes back on the subject's hand with an equal and opposite force.

Newton's Second Law explains that any net or unbalanced force on a mass will cause the mass to accelerate. There are two opposing forces on the dynamometer - one from the fingers and one from the palm and thumb. Basic physics is replete with examples of forces acting on objects at rest or in equilibrium that do not accelerate—because although a force is present, no net force is exerted on the object. This makes choice C the best answer. Answer A is false, as Newton's laws all apply to springs as well as to other objects. Answer B is false because Newton's Second Law applies to both variable and constant forces. Answer D can be eliminated because, although a true statement, it does not logically answer the query given by the stem. Newton's Third Law applies to equal forces on DIFFERENT objects (i.e., the earth pulls on YOU with the same magnitude as you pull on the EARTH).

Antibodies are generated against

Proteins

In an aromatic ring system, each carbon atom must be: sp hybridized. sp2 hybridized. sp3 hybridized. dsp2 hybridized.

SP2 - 3 BONDS

Ionization energy

The amount of energy required to remove an electron from an atom Low to high requires energy release

What is the empirical formula of the conjugate acid of the anion in strontium acetate, Sr(CH3COO)2? A. CH2O B. C2H4O2 C. SrC2H3O2 D. NaC2H3O2

The anion is acetate, the conjugate acid of which is acetic acid, formula C2H4O2. However, the question asks for the empirical formula, so this must be reduced to the lowest common denominator, which gives CH2O, or Answer A. Answer choice B cannot be an empirical formula, because it can be reduced to lower terms. Answer C is a salt, not an acid. Answer choice D is also a salt rather than an acid, so neither C or D can be the conjugate acid of acetate.

Students added HCl dropwise to a solution of phospholipids with an initial pH of 7.4. Upon reaching a pH of 4.0, what changes will have occurred to the [COO-] and [COOH] present on the phospholipid head? Increased [COO-], because water will abstract more protons. Increased [COOH], because water will donate more protons. Increased [COO-], because the [COOH] increased. Increased [COOH], because the [COO-] increased.

The correct answer is choice B. This problem is solved by understanding how Le Chatelier's principle applies to acid base chemistry. The dissociation for the carboxylic acid is RCOOH + H2O → RCOO- + H3O+. In this situation pH is DECREASING, which means the [H3O+] will be increasing. This will shift the dissociation equilibrium to the left, or reactant side, and increase the ratio of the protonated carboxylic acid. This proton could come from water or hydronium. Answers A and C are incorrect because the [COO-] will decrease not increase. Answer D is incorrect because both the acid and the base cannot increase simultaneously. Answer C makes this same claim, in reverse order, which provides a second reason to eliminate Answer C.

The high throughput screening assay was carried out by analyzing fluorescence emission spectra at 584 nm. What is the energy of photons at this wavelength? (Note: Planck's Constant = 6.6 × 10-34 m2kg/s) 3.4 × 10-19 J 1.3 × 10-48 J 2.7 × 1035 J 3.9 × 10-31 J

The energy of a photon can be calculated using the following equation E = (hc) / l. Therefore, E = (6.6 × 10-34 m2kg/s × 3.0 × 108 m/s) / 584 × 10-9 m = 3.4 × 10-19 J, the value of Answer A. Answer B is obtained if the equation is applied incorrectly: (6.6 × 10-34 m2kg/s × 584 × 10-9 m) / 3.0 × 108 m/s = 1.3 × 10-48J. Answer C is obtained if the equation is applied incorrectly: (3.0 × 108 m/s × 584 × 10-9 m) / 6.6 × 10-34 m2kg/s = 2.7 × 1035 J. Answer D is obtained by incorrectly multiplying the wave length in nm with Planck's Constant: 584 × 6.6 × 10-34 m2kg/s = 3.9 × 10-31 J. Although Planck's Constant was given in this case, the AAMC is about 50/50 on common constants of this type: sometimes they are given, and sometimes they are required to perform the calculation but are NOT given. Because they are often NOT given, you must know common values and constants such as the speed of light (c), Planck's constant (h), the gas constant (R), the charge on one electron (e-), etc.

Urease is required by colonizing bacteria as a way to obtain nitrogen from urea. As a result, an effective drug must inhibit urease continuously. What kind of inhibitor will best accomplish continuous inhibition? A noncompetitive inhibitor, because the inhibitor will reduce reaction velocity at any substrate concentration. A competitive inhibitor, because the inhibitor will compete with the urea for the active site. An uncompetitive inhibitor, because the inhibitor will reduce Km without affecting Vmax. A competitive inhibitor, because the buildup of product will inhibit the enzyme.

The question stem identifies the need for continuous inhibition to treat the given bacterial infection. A noncompetitive inhibitor interacts with the enzyme at a site other than the active site and the enzyme's function is at least partly compromised regardless of the substrate level; in other words, it will consistently inhibit the reaction even if the substrate (urea) concentration is high. This makes a noncompetitive inhibitor the best option for continuous inhibition. Answer choice A correctly identifies the needed inhibitor as noncompetitive and also provides the correct explanation for why a non-competitive inhibitor will provide constant inhibition, in that a non-competitive inhibitor will constantly inhibit the reaction regardless of substrate concentration. A competitive inhibitor is not ideal in this case, as the substrate will accumulate and overcome the inhibition, allowing the enzyme to reach its maximum velocity. The bacteria will, then, be able to utilize the substrate. This eliminates Choices B and D. The explanation for Choice B is true, but not relevant. The explanation in Choice D can also be true—products can inhibit enzymes, but again, it is not relevant. Choice C should be rejected as the explanation is incorrect. Uncompetitive inhibitors alter both Km and Vmax.

Uncompetitive inhibition vs noncompetitive inhibition

Uncompetitive binds to the enzyme substrate complex (parallel lines, both Km and vmax reduced), Noncompetitive binds to either e/s complex or just the enzyme (Km unaffected, vmax reduced)

Ohms law

V=IR (I and R inversely proportional)

Noncompetitive inhibitor

Vmax reduced, Km unchanged

Alcohol dehydrogenase

alcohol to aldehyde

mass spectrometry

an experimental method of determining the precise mass and relative abundance of isotopes in a given sample using an instrument called a mass spectrometer

Requirement to rotate polarized light

chiral carbon

Pyruvate dehydrogenase

converts pyruvate to acetyl-CoA (acetyl CoA enters citric acid cycle to make ATP and NADH)

Lewis acid

electron pair acceptor

Oligomer

multimer with just a few subunits, quaternary structure

Water is more polar than alcohol

nonpolar soluble in nonpolar and vice versa

Henderson-Hasselbalch equation

pH = pKa + log [A-]/[HA]

Carbocations

positively charged carbon atoms, cannot proceed under basic conditions

Bronsted lowry acid

proton donor

anhydrase

removal of water

pKa and acidity

smaller pKa = stronger acid

the slope of a force-displacement graph is

spring constant

What does an enzyme do to make the reaction go faster

stabilizes transition state

B-oxidation

the first step in fatty acid oxidation, in which fatty acids are broken into separate two carbon units of acetic acid, each of which is then converted to acetyl CoA (if even number of carbons in the fatty acid chain= only acetyl CoA formed, if odd number of carbons in the fatty acid chain=one succinyl coA formed)

Ammonia is (weak or strong) HCl is (weak or strong

weak; strong