MCAT AAMC practice exam Chem Phys

On a graph of distance vs force, how is work calculated?

Since work is defined as distance times force, the area under the line can be calculated to find the work done by the force.

What is the ratio of cation (0.150M) to enzyme (0.75mM) in the spectroelectrochemical experiments described in the passage? A. 1:2 B. 2:1 C. 20:1 D. 200:1

0.150M=150mM 150/0.75= 200: 1 (D)

How many moles of NaCl were contained in 500 mL of the buffer solution used to elute hMPRα? 300mM NaCl A. 7.5 x 10-3 B. 3.0 x 10-3 C. 3.0 x 10-2 D. 1.5 x10-1

1.5x10-1 300 mmol/1L x 1/1000mL x 500mL=150 mmol= 1.5x10-1 mol

Which classification of amino acids applies to the Trp residues after photochemical modification by CCl3CO2H? (adds carboxylic acid) A. Acidic B. Basic C. Hydrophobic D. Polar neutral

Acidic

Samples from various time points of the proteolysis of TPMTwt were subjected to SDS-PAGE under reducing conditions. Which figure best depicts the expected appearance of the gel? (Note: The arrow indicates the movement of the protein through the gel.)

As befits proteolysis, the number of lower molecular weight bands with time increases and the original protein band at the highest molecular weight diminishes with time.

Approximately how many moles of Kr+ are contained in the laser tube at 0°C and 1 atm? V= 11cm^3 A. 3 x 10-7 B. 2 x 10-6 C. 4 x 10-5 D. 5 x 10-4

D 11cm^3= 11x10^-3L 1 mole of gas at STP is 22.4 L 11/22.4= 0.5 (x10^-3), therefore 5e-4 is the answer

In μM•s-1 and μM, what should the approximate values of kcat/KM and Ki be, respectively, when [I] = 180 μM?

Neither is affected by a change in I, therefore they will stay the same (150 and 60.3). Ki is the equilibrium constant and is never affected by increasing I.

Consider the fully protonated amino acid shown. As the pH of a solution of this amino acid is raised, which group deprotonates first? A. I B. II C. III D. IV

Solution: The correct answer is A. A. As the pH is raised, the most acidic group deprotonates first. Of the choices, two are carboxylic acids, one is a hydroxyl group, and one is a protonated amine. The hydroxyl group is least acidic, while the protonated amine is less acidic (pKa ~ 10) than a carboxylic acid group (pKa ~ 5). Of the two carboxylic acid groups, the one next to the chlorine atoms will be more acidic since its conjugate base (an anion) will be stabilized by an inductive effect which dissipates negative charge building up. B. A proton on alcohol is much less acidic than one in a carboxylic acid because the negative charge cannot be delocalized to several electronegative atoms. C. This carboxylic acid proton is less acidic than the one on the left because it is not near two electronegative chlorine atoms. D. Protonated amines are less acidic than carboxylic acids (pKa of 10 versus 5).

If all else is held constant, which of the following changes would NOT double the volume of a gas? A. Doubling the pressure B. Doubling the absolute temperature C. Halving the pressure D. Doubling the number of gas molecules

Solution: The correct answer is A. A. Based on Boyle's Law, P is inversely proportional to volume, thus, doubling the pressure of a gas sample will decrease, not increase, the volume. B. Doubling the absolute temperature of a gas sample will double the volume because T is directly proportional to V for an ideal gas according to Charles' Law. C. Halving the pressure of a gas sample will double its volume because P is inversely proportional to V for an ideal gas. D. Doubling the number of gas molecules present will double the volume according to Avogadro's Law.

Which chromatographic technique would most likely separate a mixture of native carbonic anhydrase from carbonic anhydrase photochemically modified by CCl3CO2H? A. Anion-exchange chromatography B. Cation-exchange chromatography C. Gas-liquid chromatography D. Size-exclusion chromatography

Solution: The correct answer is A. A. Because the passage states that native carbonic anhydrase has a net charge of -2.9 and the modified enzyme would have greater negative charge, anion-exchange chromatography can separate them as this technique separates proteins with different negative charges. B. The native and modified Enzymes both have net negative charge. In cation-exchange chromatography, both Enzymes would elute together in the void volume. C. Proteins degrade before they would vaporize. D. The molecular weights of the native and modified proteins are too close (< 1 kDa difference) to allow separation by size-exclusion chromatography.

Which of the following species has an electron configuration equivalent to that of a noble gas? A. Ca2+ B. Cu2+ C. O D. H

Solution: The correct answer is A. A. Ca2+ has lost its valence electrons and thus attains the electronic configuration of the previous noble gas (Ar). B. Copper has 11 outer electrons. Loss of 2 electrons leaves the cation with a [Ar] 3d10 configuration. C. An oxygen atom is not a noble gas. It has a partially filled 2p subshell. D. A hydrogen atom is not a noble gas. It has a half-filled 1s subshell.

The second purification step (His tags) is which type of chromatographic separation? A. Affinity B. Size exclusion C. Cation exchange D. Anion exchange

Solution: The correct answer is A. A. Displacement of the protein from the column in this step involved disrupting the binding of the (His)6 tag to the column. This is a classic example of affinity chromatography. B. Size-exclusion chromatography separates proteins by molecular weight, not selective column binding. C. Cation-exchange chromatography separates proteins with different positive charges (or positive versus negative/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. D. Anion-exchange chromatography separates proteins with different negative charges (or negative versus positive/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction.

The radiation of wavelength 605 nm CANNOT be used to produce the fluorescence radiations depicted in Figure 3 because: A. the energy of the absorbed radiation must be larger than the energy of the fluorescence radiation. B. the energy of the absorbed radiation must be smaller than the energy of the fluorescence radiation. C. the 605-nm radiation has more energy than the 407-nm radiation. D. the 605-nm radiation is not visible.

Solution: The correct answer is A. A. Fluorescence can occur when the absorbed radiation has a photon energy larger than the photon energy of the radiation emitted through fluorescence. The photon energy is inversely proportional to the radiation wavelength, thus the 605-nm wavelength radiation cannot produce the entire fluorescence radiation spectrum shown in Figure 3 as its photon energy is below that of the fluorescence radiation of wavelength 604 nm. B. The photon energy of the absorbed radiation must exceed the photon energy of the fluorescence radiation because the energy difference cannot be created from nothing, per the energy conservation principles. C. The photon energy is inversely proportional to the radiation wavelength according to the formula Energy = Planck's constant multiplied by the speed of light divided by the wavelength, thus the 605-nm wavelength radiation has less photon energy than the 407 nm wavelength radiation. D. The 605 nm wavelength radiation is visible, but this feature is unrelated to fluorescence.

If the red line in the Balmer series has a wavelength of 656 nm, which of the following is closest to its frequency? A. 4.6 × 1014 Hz B. 4.6 × 10−​14 Hz C. 2.1 × 1015 Hz D. 2.1 × 10−​15 Hz

Solution: The correct answer is A. A. For light that travels in vacuum, wavelength and frequency are related according to wavelength × frequency = 3.0 × 108 m/s. Solving for frequency yields (3.0 × 108 m/s)/(656 × 10-9 m) = 4.6 × 1014 Hz. B. A frequency of 4.6 × 10-14 Hz corresponds to a wavelength of 6.56 × 1021 m , not 656 × 10-9 m. C. A frequency of 2.1 × 1015 Hz corresponds to a wavelength of 143 nm, not 656 nm. D. A frequency of 2.1 × 10-15 Hz corresponds to a wavelength of 1.43 × 1023 m, not 656 × 10-9 m.

The CD spectroscopy signal that was used to generate the data in Figure 1 arises from the chirality of the: A. α carbon. B. amide nitrogen. C. carbonyl carbon. D. β carbon.

Solution: The correct answer is A. A. In a protein, each amino acid residue, except Gly, has a chiral α carbon. B. Amide nitrogen is achiral. C. Carbonyl carbon is achiral. D. With the exception of Ile and Thr,the β carbon in amino acid residues is achiral. CD spectra arise from the massively greater presence of α carbon chirality.

Which structural change to Compound 1 would make it more water soluble? A. Replacing benzene CH with N in the ring B. Replacing C=O with C=CH2 C. Replacing N-N=N with CH-CH=CH D. Replacing NH with NCH3

Solution: The correct answer is A. A. Nitrogen in the benzene ring would have a lone pair that could accept a hydrogen bond from water, thus increasing the solubility of the compound. B. Replacing C=O with C=CH2 would decrease the polarity of the compound and make it less water soluble. C. Replacing N−N=N with CH−CH=CH would decrease the polarity of the compound and make it less water soluble. D. Replacing NH with NCH3 removes a hydrogen bond donor, thus decreasing water solubility.

If the defibrillator described in the passage were fully charged and the entire charge were discharged through a patient in 10 ms, which of the following is closest to the average electrical current that would flow through the paddles? (3000V and 25 microfarads) A. 7.5 A B. 15 A C. 22.5 A D. 30 A

Solution: The correct answer is A. A. The full charge of the capacitor is 25 μF × 3000 V = 75 mC. The average discharge current is (75 mC)/(10 ms) = 7.5 A. B. If the average discharge current is 15 A, it implies the full charge of the capacitor should have been 10 ms × 15 A = 150 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC. C. If the average discharge current is 22.5 A, it implies the full charge of the capacitor should have been 10 ms × 22.5 A = 225 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC. D. If the average discharge current is 30 A, it implies the full charge of the capacitor should have been 10 ms × 30 A = 300 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC

Which of the following substances is polar? A. NF3 B. CCl4 C. CO2 D. Li2

Solution: The correct answer is A. A. The geometry of trifluoroamine is impacted by the lone pair on nitrogen, making it trigonal pyramidal. No bond dipoles cancel; this results in a polar molecule. B. While each C-Cl bond in carbon tetrachloride is polar, the sum of the dipole moments cancel as a result of its tetrahedral molecular geometry. C. Carbon dioxide is a linear molecule. The bond dipole moments of each C=O bond cancel as they are in opposite directions. D. This molecule is necessarily non-polar as it is comprised of two identical atoms.

Two open flasks I and II contain different volumes of the same liquid. Suppose that the pressure is measured at a point 10 cm below the surface of the liquid in each container. How will the pressures compare? A. The pressures will be equal. B. Pressure in flask I will be less. C. Pressure in flask II will be less. D. The pressures cannot be compared from the information given.

Solution: The correct answer is A. A. The pressure at a point 10 cm below the surface of the liquid is the same in both flasks because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. B. The pressure at a point 10 cm below the surface of the liquid in flask I is the same as the pressure in flask II at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. C. The pressure at a point 10 cm below the surface of the liquid in flask II is the same as the pressure in flask I at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. D. The pressures can be compared because both pressures are calculated according to the hydrostatic pressure formula p = ρgd, where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured.

Based on the reported Hill coefficient (2.56), in what way do the MCS oligomers affect inhibition? A. As one MCS oligomer binds to the ATPase, it makes it easier for the others to bind, leading to inhibition. B. As one MCS oligomer binds to the ATPase, it makes it more difficult for the others to bind, leading to inhibition. C. A single MCS oligomer binds to the ATPase, leading to inhibition. D. MCS oligomers randomly bind to the ATPase, leading to inhibition.

Solution: The correct answer is A. A. The reported Hill coefficient of 2.56 indicates positive cooperativity because the value is greater than 1.0. Therefore, when one MCS oligomer binds to the ATPase, it makes it easier for others to bind. When there are sufficient MCS oligomers bound, they lead to inhibition of the ATPase. B. This would be negative cooperativity, requiring a fractional Hill coefficient. C. This would indicate no cooperativity, and the Hill coefficient would be 1.0. D. The Hill coefficient does not indicate whether binding is random.

Which type of heterocycle is found on two amino acid residues blocking access to W15? (blocked by two His residues and one Lys residue) A. Imidazole B. Indole C. Pyrimidine D. Pyrrole

Solution: The correct answer is A. A. The two amino acid residues blocking access to W15 are histidines, each with an imidazole ring-containing side chain. B. The blocking residues are His, not Trp. C. Pyrimidine is found in DNA and RNA, not in the side chain of natural amino acids. D. Pyrrole is found in heme, not in the side chain of natural amino acids.

What functional group transformation occurs in the product of the reaction catalyzed by Na+-NQR (NADH and ubiquinone)? A. RC(=O)R → RCH(OH)R B. ROPO32- → ROH + Pi C. RC(=O)NHR'→ RCOOH + R'NH2 D. RC(=O)OR'→ RCOOH + R'OH

Solution: The correct answer is A. A. This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. A. This is the reaction catalyzed by a phosphatase. A. This is the reaction catalyzed by a protease or amidase. A. This is the reaction catalyzed by an esterase.

An ester is prepared by the method of direct esterification using an esterase enzyme as a catalyst. Which of the following modifications will NOT appreciably increase the final yield of ester? RCOOH+RCH2OH-->RCOOCH2R +H2O A. Using 2 times as much enzyme B. Using 2 moles of RCOOH instead of 1 mole C. Using 2 moles of RCH2OH instead of 1 mole D. Removing RCOOCH2R from the reaction mixture as it is formed

Solution: The correct answer is A. A. This modification will increase the rate of formation, not the final yield of ester. B. Doubling the concentration of RCOOH will increase the yield of ester by Le Châtelier's principle. C. Doubling the concentration of RCH2OH will increase the yield of ester by Le Châtelier's principle. D. By Le Châtelier's principle, removing the ester product as it forms will drive the reaction to form more ester.

Which of the following is closest to the wavelength of a photon whose energy is 2 eV? h= 4.1e-15 eV/s and c=3e8m/s A. 740 nm B. 620 nm C. 450 nm D. 310 nm

Solution: The correct answer is B. A. A wavelength of 740 nm corresponds most closely to a photon with energy E = 1.66 eV, not 2 eV. B. Using hc = 4.1 × 10-15 eV•s × 3 × 108 m/s = 1230 eV•nm, and λ = hc/E, then E = 2 eV corresponds to λ = (1230 eV•nm)/(2 eV) = 615 nm, which is closest to 620 nm. C. A wavelength of 450 nm corresponds most closely to a photon with energy E = 2.75 eV, not 2 eV. D. A wavelength of 310 nm corresponds most closely to a photon with energy E = 4 eV, not 2 eV.

In which phase(s) will the lipophilic MCS precursor be predominantly found after the extraction step? The MCS precursor will: A. be found in the aqueous layer. B. be found in the tert-butyl methyl ether layer. C. be distributed equally between the aqueous layer and the tert-butyl methyl ether layer. D. form a precipitate between the aqueous and tert-butyl methyl ether layers.

Solution: The correct answer is B. A. According to the passage, the MCS precursor is lipophilic, so it is not water soluble. B. The passage described the MCS precursor as being lipophilic, which means that it would not dissolve as readily in the aqueous layer. Therefore, it should be found in the tert-butyl methyl ether layer, which is hydrophobic. C. According to the passage, the MCS precursor is lipophilic, so it prefers to dissolve in organic solvents. D. The MCS precursor should dissolve in the organic layer.

Which reactant and product (one equivalent each) are necessary to balance Reaction 1? A. Reactant = H2O, product = NO3_ B. Reactant = H2O, product = NH4+ C. Reactant = O2, product = NO3_ D. Reactant = O2, product = NH4+

Solution: The correct answer is B. A. Adding one water molecule to the reactant side and one nitrate to the product side leaves the reaction unbalanced. There are more H atoms on the reactant side and too many O atoms on the product side. B. The reactant loses an ammonium ion and gains a water molecule. This balances both charge and mass on both sides of the chemical equation. C. If oxygen is added as a reactant, and nitrate as a product, the reaction is still unbalanced. There are too many O atoms on the product side and two H atoms too many on the reactant side. D. If oxygen is added as a reactant and ammonium is added as a product, the reaction remains unbalanced. There are two H atoms too many on the product side.

What is the identity of an atom that contains six protons and eight neutrons? A. Nitrogen B. Carbon C. Oxygen D. Silicon

Solution: The correct answer is B. A. The atom cannot be nitrogen because nitrogen contains seven protons. B. Carbon contains six protons because it also contains six electrons as a neutral atom. C. Oxygen contains eight protons because as a neutral atom it contains eight electrons. D. The silicon atom contains 14 protons.

If the wavelength of a light beam were doubled, its frequency would be: A. quartered. B. halved. C. doubled. D. quadrupled.

Solution: The correct answer is B. A. The frequency being quartered corresponds to the wavelength being quadrupled, not doubled. B. The product between wavelength and frequency is constant when light travels through a certain medium, thus doubling the wavelength results in halving the frequency. C. The frequency being doubled corresponds to the wavelength being halved, not doubled. D. The frequency being quadrupled corresponds to the wavelength being quartered, not doubled.

A 2 kg mass and a 5 kg mass are connected by a massless cord suspended over a massless and frictionless pulley. If the acceleration due to gravity is g, what will be the acceleration of the masses after they are released from rest? A. 2g/7 B. 3g/7 C. 5g/7 D. g

Solution: The correct answer is B. A. This is smaller than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.5. It implies that only the 2-kg mass is moving after the release. B. According to Newton's second law, the net force acting on the 5-kg mass is given by the expression Fnet= 5 kg × a1 = 5 kg × g - T, where a1 is the acceleration after the release and T is the tension in the cord. The net force acting on the 2-kg mass is given by the expression Fnet= 2 kg × a2 = 2 kg × g - T. Because the two masses move simultaneously but in opposite directions after they are released, a1 = -a2 = a. Substituting the expression T = 5 kg × (g - a) into the equation of motion of the 2-kg mass yields -2 kg × a = 2 kg × g - 5 kg × (g - a) = -3 kg × g + 5 kg × a. Then 7 kg × a = 3 kg × g, hence a = 3g/7. C. This is larger than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.7. It implies that only the 5-kg mass is moving after the release. D. This implies the two masses fall freely and separately without being connected by the cord.

Based on Figure 2, what is the approximate Ki of the MCS oligomers? inflection point at 0.03uM A. 12 nM B. 30 nM C. 170 nM D. 800 nM

Solution: The correct answer is B. A. This is the MCS oligomer concentration at almost the maximum specific activity. B. In the semi-log plot shown in Figure 2, the Ki is the inflection point of the sigmoidal curve because, by definition, the Ki is the concentration of the inhibitor at which the reaction rate is half of the maximum reaction rate. The inflection point is at approximately 0.030 mM, or 30 nM. C. This is the MCS oligomer concentration at about 0.05 × specific activity. D. This is the MCS oligomer concentration at complete enzyme inhibition.

What is the molecular formula of the heterocyclic aromatic compound pyrrole? A. C2H3N B. C4H5N C. C6H7N D. C8H9N

Solution: The correct answer is B. A. This is the molecular formula of the non-aromatic heterocycle 2H-azirine. B. This is the molecular formula of pyrrole, a five-membered aromatic heterocycle containing one nitrogen atom. C. This is the molecular formula of various forms of azepine, a seven-membered heterocycle, none of which are aromatic. D. There is a cyclohexene-fused pyrrole with this molecular formula, but not pyrrole itself.

What is the chemical structure of a component found in four of the five cofactors (flavins) used by Na+-NQR?

Solution: The correct answer is B. A. This is the structure of adenine. It is only found in FAD. B. This is the structure of flavin, found in four of the five cofactors used by Na+-NQR. C. This is the structure of ubiquinone. It is a substrate, but not a cofactor. D. This is the structure of histidine. It is an amino acid, not a cofactor.

When the covalent attachment to alliinase is broken, PLP is still held rigidly in the active site by a salt bridge and a π-stacking interaction. These interactions are most likely provided by the side chains of which amino acids? (Note: The salt bridging amino acid is listed first.) A. Asp and Tyr B. Glu and Ser C. Arg and Tyr D. Lys and Ser

Solution: The correct answer is C. A. Although Tyr can form a π-stacking interaction, Asp has a negatively charged side chain that will repel, not attract, the phosphate in PLP. B. Glu has a negatively charged side chain, which will repel, not attract, the negatively charged phosphate in PLP. Lacking an aromatic ring, Ser is not capable of a π-stacking interaction. C. PLP has a negatively charged phosphate that can make a salt bridge with the positively charged side chain of Arg, and the aromatic ring in the side chain of Tyr can form a π-stacking interaction. D. Although the positively charged side chain of Lys can form a salt bridge with the phosphate of PLP, Ser does not contain an aromatic ring for a π-stacking interaction.

If the energy of a photon is doubled, which of the following properties of the photon will also double? A. Amplitude B. Wavelength C. Frequency D. Intensity

Solution: The correct answer is C. A. Amplitude is a characteristic of waves, whereas photons constitute the particle description of light. Therefore, photons cannot be characterized by the property of amplitude. B. The energy of a photon is given by the relationship E = hc/l. If E is doubled, then wavelength l is halved not doubled. C. The energy of a photon is given by the relationship E = hf. If E is doubled, then frequency f is doubled, too, as Planck's constant h does not change. D. Intensity is a characteristic of waves, whereas photons constitute the particle description of light. Therefore, photons cannot be characterized by the property of intensity unless as the number of photons of a certain energy that pass through a surface per unit time.

Based on the passage, what most likely causes W96 to be accessible to CHCl3 at 75°C? A. Peptide bonds are broken, releasing W96. B. Reduction of disulfide bonds occurs. C. The protein unfolds and exposes W96 to the buffer. D. CHCl3 extracts W96 from the protein interior.

Solution: The correct answer is C. A. As stated in the passage, there is no change in primary structure when carbonic anhydrase, is modified respectively by CHCl3 at 75°C, so peptide bonds are not broken. B. There is no reducing agent present. C. As stated in the passage, at 75ºC, carbonic anhydrase fully denatures, which means that the protein is unfolded and exposes W96 to the buffer. D. At 10 mM, the reactant CHCl3 is not acting as a solvent. The buffer is the solvent, which is mostly water.

What structural feature(s) is(are) most important to the functioning of this compound as described in the passage? A. Specific configuration of numerous chirality centers B. Multiple hydrolysable linkages C. Combination of large hydrophobic and hydrophilic regions D. Presence of a reducing sugar

Solution: The correct answer is C. A. Chirality is important for specific binding to other compounds. This is not a requirement of a detergent. B. Multiple hydrolysable linkages would facilitate metabolism. Compound 2 was a detergent used to help isolate membrane proteins. C. Compound 2 was used as a detergent. It liberated a protein from a membrane so that it might be isolated. The combination of large hydrophilic and hydrophobic regions allows Compound 2 to function in this capacity. D. The presence of a reducing sugar makes the compound susceptible to oxidation and imparts aqueous compatibility. It does not provide all the necessary components of a functioning detergent.

Enantiomers can exhibit a difference in which chemical or physical property? A. Density B. Boiling point C. Smell D. IR spectrum

Solution: The correct answer is C. A. Enantiomers will not display different densities. B. The boiling points of enantiomers are identical, and they cannot be separated by distillation. C. Enantiomers have the same physical and chemical properties. They differ only in their three dimensional arrangement of atoms and their interactions with other chiral molecules. They can differ in their smell due to interacting differently with chiral odorant receptors. D. The IR spectra of enantiomers are identical when a normal light source is used. Circularly polarized light will potentially illustrate differences.

If only [I] is increased, then [ESI] or [EI] increases. This is an example of: A. the Bose-Einstein Principle. B. the Heisenberg Uncertainty Principle. C. the Le Châtelier's Principle. D. the Pauli Exclusion Principle.

Solution: The correct answer is C. A. The Bose-Einstein Principle states that a collection of atoms cooled close to absolute zero will coalesce into a single quantum state. B. The Heisenberg Uncertainty Principle states that one cannot know both the momentum and position of an object with absolute certainty. C. Le Châtelier's Principle states that in a reversible process, the application of stress to the system will cause the system to respond in a way that will relieve this stress. In this case, the reversible process is E + I forming EI or ES + I forming ESI. In either case, increasing [I] induces stress of the system, and the system relieves that stress by converting I to either more EI or ESI. D. The Pauli Exclusion Principle states that two or more identical fermions, e.g. electrons, cannot occupy the same quantum state.

Which experimental evidence suggests that the purified hMPRα obtained by the researchers was in its native state? The hMPRα that was obtained: A. retained both Compound 1 and (His)6 tags. B. was purified by two separate chromatography steps. C. exhibited a binding affinity for progesterone that was similar to that exhibited by native hMPRα. D. had a nearly identical molecular weight to hMPRα obtained elsewhere.

Solution: The correct answer is C. A. The retention of these tags does not confirm conformational stability. B. A protein can denature during any chromatography step. The fact that hMPRα required two steps for purification means that it was more likely to have denatured. C. The only way to tell for certain that a protein is still in its native state is to compare its functioning to the levels observed for protein believed to be in the native state. The researchers compared the binding affinity of progesterone of hMPRα both before (when it is assumed to be in its native state) and after releasing it from the membranes (which could conceivably cause denaturation). The fact that the affinities were the same suggests that the isolated hMPRα is indeed in the native state. D. Molecular weight is a measure of the stability of the primary structure. Activity requires full conformational stability.

What is the value of Kb for the conjugate base of a weak organic acid that has a pKa of 5? A. 10−​2.5 B. 10−​5 C. 10−​9 D. 10−​10

Solution: The correct answer is C. A. This is the Kb for the conjugate base of an acid with a pKa of 11.5. B. This is the Kb for the conjugate base of an acid with a pKa of 9. C. Because pKb = 14 - pKa, this is the Kb for the conjugate base of an acid with a pKa of 5. D. This is the Kb for the conjugate base of an acid with a pKa of 4.

Blood flows with a speed of 30 cm/s along a horizontal tube with a cross-section diameter of 1.6 cm. What is the blood flow speed in the part of the same tube that has a diameter of 0.8 cm? A. 7.5 cm/s B. 15 cm/s C. 60 cm/s D. 120 cm/s

Solution: The correct answer is D. A. A flow speed of 7.5 cm/s corresponds to a diameter of 3.2 cm, not 0.8 cm. B. A flow speed of 15 cm/s corresponds to a diameter of 2.25 cm, not 0.8 cm. C. A flow speed of 60 cm/s corresponds to a diameter of 1.125 cm, not 0.8 cm. D. The flow is characterized by the continuity equation because no amount of blood is lost between the two locations. The continuity equation is 30 cm/s × π × ((1.6 cm)/2)2 = v × π × ((0.8 cm)/2)2. Solving for v yields v = 4 × 30 cm/s = 120 cm/s.

If the 25 μF capacitor in the defibrillator in Figure 1 is replaced with a 30 μF capacitor, what new power supply setting would produce the same amount of charge? A. 3600 V B. 3500 V C. 3000 V D. 2500 V

Solution: The correct answer is D. A. A power supply setting of 3600 V would produce a charge of 108 mC, not 75 mC. B. A power supply setting of 3500 V would produce a charge of 105 mC, not 75 mC. C. A power supply setting of 3000 V would produce a charge of 90 mC, not 75 mC. D. On the 25 μF capacitor, the power supply stores a charge of 25 μF × 3000 V = 75 mC. On the 30 μF capacitor, the same charge is stored by a power supply that has a voltage of 75 mC/30 μF = 2500 V.

Based on the passage, which type of inhibitor will provide a flavorful onion that does not cause tearing? An inhibitor of: A. Compound 1 synthesis B. alliinase C. Reaction 2 D. LFS

Solution: The correct answer is D. A. Although inhibiting Compound 1 synthesis will stop the onion from causing tearing, because Compound 1 is the source of flavor components in onions, the onion will not be flavorful. B. Although inhibiting allinase will stop the onion from causing tearing, because allinase catalyzes the formation of flavor component precursors in onions, the onion will not be flavorful. C. Inhibiting Reaction 2 will stop the formation of Compound 3, which is one of the main flavor components of onions. D. Inhibiting LFS will stop the formation of Compound 4, a potent tearing agent.

What is the name of the ionic compound used to make Buffer B? (NH4CH3CO2) A. Ammonium formate B. Ammonium carbonate C. Ammonium bicarbonate D. Ammonium acetate

Solution: The correct answer is D. A. Ammonium formate is NH4HCO2. B. Ammonium carbonate is (NH4)2CO3. C. Ammonium bicarbonate is NH4HCO3, a component of Buffer A. D. Ammonium acetate is NH4CH3CO2, a component of Buffer B.

What is the name of the functional group containing the external oxygen on each ring of the MCS structures shown in Figure 1? A. Aldehyde B. Carboxylic acid C. Ester D. Ketone

Solution: The correct answer is D. A. An aldehyde has the structure R(C=O)H, where R = H or a carbon group. B. A carboxylic acid has the structure RCO2H, where R = a carbon group. C. An ester has the structure RCO2R', where R = H or a carbon group and R' = a carbon group. D. The structure R(C=O)R', where R and R' are carbon groups, is a ketone.

The ligand of hMPRα (progesterone) is derived from which compound? A. Glucose B. Phenylalanine C. Glycerol D. Cholesterol

Solution: The correct answer is D. A. Glucose is a simple sugar, which are not precursors to steroids. B. Phenylalanine is an amino acid, which are the components of proteins. C. Glycerol is a the trihydroxyl backbone for triacyl glycerols, which are fats. D. The ligand for hMPRα is the steroid progesterone. Steroids are a class of lipids that are derived from cholesterol.

An ice cube at 0°C and 1 atm is heated to form steam at 100°C and 1 atm. Ignoring heat loss to the surroundings, what part of the process uses the most heat? (Note: Specific heat of water = 1 cal/g°C. Heat of fusion = 80 cal/g. Heat of vaporization = 540 cal/g.) A. Melting the ice cube B. Heating all the water from 0°C to 50°C C. Heating all the water from 50°C to 100°C D. Vaporizing all the water

Solution: The correct answer is D. A. Melting the ice cube requires only 80 cal/g, which is less than the heat of vaporization. B. Heating all the water from 0°C to 50°C requires 50 cal/g, which is less than the heat of fusion. C. Heating all the water from 50°C to 100°C requires 50 cal/g, which is less than the heat of fusion and the heat of vaporization. D. Vaporizing all the water requires 540 cal/g, which is a greater heat requirement than specific heat or heat of fusion.

Which physical property does NOT change with the amino acid substitution made in TPMT*5? (L49S) A. Molecular weight B. Hydrophobicity C. Hydrogen bonding capability D. Net charge

Solution: The correct answer is D. A. The side chain of the L49S variant (TPMT*5) at position 49 has a different molar mass than the side chain of Leu. B. The L49S variant (TPMT*5) has a hydrophilic residue at position 49, but wild-type TPMT has a hydrophobic residue at that position. C. The L49S variant (TPMT*5) can hydrogen bond, but wild-type TPMT cannot. D. The L49S variant (TPMT*5) has the same charge as wild-type TPMT because the amino acid residues do not have charge.

A major obstacle to obtaining useful energy from a nuclear fusion reactor is containment of the fuel at the very high temperatures required for fusion. The reason such high temperatures are required is to: A. eliminate the strong nuclear force. B. remove electrical charge from reactants. C. decrease the density of the fuel. D. enable reactants to approach within range of the strong nuclear force.

Solution: The correct answer is D. A. The strong nuclear force cannot be eliminated by increasing the temperature because this force manifests whenever nucleons are present and within a range of picometers. B. The electrical charge is an intrinsic property of matter that is independent of temperature. C. Decreasing the density of the fuel is detrimental because the probability of fusion increases with the increase in fuel density because the attractive component of the strong nuclear force acts at short distances indicative of high density for the fuel material. D. The probability of fusion increases with the decrease in the average distance between fuel particles that enables attractive nuclear forces to overcome the repelling nuclear forces acting at medium and long distances. An increase in the temperature is equivalent to an increase in the root-mean-square speed of the fuel particles that will travel the average distance between fuel reactants in smaller times. The associated increase in the kinetic energy of the particles relative to the center of mass of the nuclear fuel system essentially correlates with a decrease in the electrostatic potential barrier that repels particles of the same electric charge. In turn, this increases the probability of particles to undergo the tunnel effect by penetrating the electrostatic barrier. These combined effects enable reactants to approach within range of the strong nuclear force.

Based on Reaction 1, when 1.0 atm of CO(g) completely reacts to form carbon suboxide at 550°C in a sealed container, what is the final pressure in the container? 4Co2--> C3O2 +CO2 A. 0.00 atm B. 0.10 atm C. 0.25 atm D. 0.50 atm

Solution: The correct answer is D. A. There is still gas present in the container. B. Since pressure is directly proportional to moles of gas at constant V and T, this is a smaller pressure than the balanced equation allows. C. This result ignores the pressure contribution of one of the gases. D. Based on Reaction 1, 4 mol CO(g) forms 2 mol of gases. Because of the direct relationship between P and n at constant V and T, that means 1.0 atm CO(g) makes 0.50 atm of gases.

Based on the passage, which of the following is closest to the pressure exerted on the chest by a 10 × 5 cm rectangular paddle during defibrillation? (Note: 1 Pa = 1 N/m2.) (F=100N) A. 5 kPa B. 10 kPa C. 15 kPa D. 20 kPa

Solution: The correct answer is D. A. This is 1/4 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2)=20 kPa. B. This is 1/2 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa. C. This is 3/4 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa. D. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa.

The reaction between NADH and ubiquinone is exergonic, but the reaction, when catalyzed by Na+-NQR, does not generate much heat in vivo. What factor accounts for this difference? The reaction catalyzed by Na+-NQR in vivo: A. is more exothermic as a result of the lower activation energy. B. occurs sequentially in several small steps. C. maintains a large separation between the reacting centers. D. is coupled to the movement of a charged particle against a concentration gradient.

Solution: The correct answer is D. A. This is impossible. Even if it were true, this would make the heat generation larger, not smaller, for the catalyzed reaction. Catalysis does not change thermodynamics. B. By Hess's Law, the heat of reaction will sum and be the same. The fact that the reaction can be broken down into steps will not change the overall thermodynamics. C. This is also impossible. The reactants ultimately must be close together to react. D. The movement of a charged particle against its concentration gradient is energetically costly. Coupling the two processes: the redox reaction between NADH and ubiquinone and the movement of Na+ up a concentration gradient makes the overall process less exothermic.

Which energy features characterize the energy profile of Reaction 2? A. High Ea and ΔG > 0 B. High Ea and ΔG < 0 C. Low Ea and ΔG > 0 D. Low Ea and ΔG < 0

Solution: The correct answer is D. A. This set of values describe a slow and non-spontaneous reaction. B. This set of values describe a slow but spontaneous reaction. C. This set of values describe a rapid but non-spontaneous reaction. D. Reaction 2 was described as occurring spontaneously and rapidly. This means that ΔG < 0 (spontaneous) and with a low Ea (rapid).

If the combined mass of the TPMT (transferase) substrate and cofactor was determined before the enzymatically catalyzed reaction and then compared to the combined mass of the product and the cofactor after the reaction, the net change in molecular weight will be: A. +15 g/mol. B. 0 g/mol. C. -15 g/mol. D. -16 g/mol.

0 g/mol bc TPTP is a transferase that transfers methyl group from cofactor to substrate

What is the number of neutrons in the nucleus of the atom used to produce laser radiations? A. 48 B. 49 C. 50 D. 51

86Kr+ 36 protons, so 86-36=50 (C)

What quantity of Compound 1 (483.5g) must be provided to prepare 100.00 mL of solution with a concentration equal to Ki (60.3uM)?

M=mol/v 1 umol=0.000001 moles (1x10^-6) 2.92mg