This question asks us to calculate the quantity of Compound 1 needed to prepare a 100.0 mL solution with a concentration equal to Ki. This question requires us to refer to the passage to retrieve the concentration of Ki.

This answer likely results from miscalculation. Refer to option D for the correct calculation.

This answer likely results from improper unit conversion. Refer to option D for the correct conversion and calculation.

This answer likely results from using a Ki value of 120 μM instead of 60.3 μM. Refer to option D for the correct calculation.

The molarity of a solution is a measure of concentration equal to the number of moles of solute (in this case, Compound 1) per liter of solution. The passage states that Ki = 60.3 μM. Unit conversion will be important for this question.
Recall that 1 μM = 1 x 10-6 M, so 60.3 μM = 60.3 x 10-6 M or 6.03 x 10-5 M. Furthermore, 1000 mL = 1 L, so 100.00 mL = 0.10000 L
Molarity (M) = Moles solute / Liters solution
Moles solute = Molarity x Liters solution
Moles solute = (6.03 x 10­-5 M) x (0.10000 L) = 6.03 x 10-6 moles
Now, we must convert moles of Compound 1 to mass, using its molar mass (483.5 g/mol). To make the mental math easier, we can round 6.03 x 10-6 moles to 6 x 10-6 moles and 483.5 g/mol to 500 g/mol.
Number of moles= Mass / Molecular Weight
Remember that there are 1,000 milligrams in 1 gram.
6 x 10-6 moles = Mass (mg) / [(500 g / 1 mol) ∙ (1000 mg / 1 g)]
Mass = 6 x 10-6 mol ∙ (500 g / 1 mol) ∙ (1000 mg / 1 g)
This calculation can be performed more easily if 500 and 1000 are converted to scientific notation.
6 x 10-6 mol ∙ (5 x 102 g / 1 mol) ∙ (1 x 103 mg / 1 g) = 30 x 10-1 mg = 3 mg
Due to rounding, our answer differs slightly from the actual quantity of Compound 1, but it is nearest to option D.
Therefore, we can conclude that option D is correct.

This question asks us to identify the principle demonstrated by the concentration of ESI or EI increasing when only [I] is increased, prompting us to consider factors affecting reversible reactions. This question requires an understanding of the passage to answer.

The Bose-Einstein Principle states that when separate atoms or subatomic particles are cooled to absolute zero, they consolidate into a single quantum entity. This principle is not relevant to the question.

The Heisenberg Uncertainty Principle states that it is impossible to be absolutely certain of the position and momentum of an electron simultaneously. This principle is not relevant to the question.

The passage states that Compound 1 can bind and inhibit HIV-1 protease to form an EI (enzyme-inhibitor) or ESI (enzyme-substrate-inhibitor) complex. It also states Lineweaver-Burk pilots with or without Compound 1 generate parallel lines, which indicates an ‘uncompetitive inhibitor’. Therefore, the formation of each complex can be thought of as a reversible reaction:

E + I ↔ EI

ES + I ↔ ESI

The question states that if [I] is increased, then [ESI] or [EI] increases. Le Châtelier’s Principle states that changes to an equilibrium system will result in a predictable shift that counteracts the change. Accordingly, when [I] is increased, the equilibrium is disrupted, causing the forward reaction to be favored such that [EI] or [ESI] increases, counteracting the increase in [I].

Therefore, we can conclude that when [ESI] or [EI] increases as a result of [I] increasing, this is an example of Le Châtelier’s Principle.

The Pauli Exclusion Principle states that two electrons cannot share the same set of quantum numbers. In practical terms, this means that for an electron pair, one electron must have a +½ spin and the other -½. This principle is not relevant to the question.

This question asks us to identify which structural change would increase the water solubility of Compound 1, prompting us to consider polarity and hydrophobicity. To answer this question, we do not need to reference the passage.

Water is a polar solvent that is capable of dissolving polar molecules due to its dipole-dipole interactions and hydrogen bonding. Hydrogen bonds are an intermolecular force that results from the electrostatic attraction between hydrogen in one molecule and a highly electronegative atom in another. Hydrogen bonds are possible when a molecule contains an -FH, -OH, or -NH group. If a nitrogen atom was placed in the benzene ring of Compound 1, its lone pair could accept a hydrogen bond from water, making it more soluble.

Therefore, we can conclude that replacing benzene CH with N in the ring would make Compound 1 more water-soluble.

Other Ways to Solve: In essence, this question asks us to identify the answer option that increases the polarity of Compound 1. In options B, C, and D, the number of electronegative atoms decreases or remains the same as a result of the replacement. In option A, an electronegative atom is added, increasing the polarity of Compound 1 and making it more soluble in water. Therefore, we can quickly conclude that option A is correct.

Replacing C=O with C=CH2 would make the molecule less polar and thus less soluble in water.

Replacing N-N=N with CH-CH=CH would make the molecule less polar and thus less soluble in water.

Replacing NH with NCH3 removes a hydrogen bond donor, reducing the polar interactions between Compound 1 and water and decreasing its solubility.

This question asks us to calculate the values of kcat/KM and Ki, given the concentration of the inhibitor. This passage-based question requires us to refer to and interpret data from Table 1.

These values are one-fourth of the actual kcat/KM and Ki. It is impossible for Ki to increase as it is an equilibrium constant and is thus not affected by [I]. Moreover, the trend in Table 1 shows that kcat/KM remains unchanged as [I] increases.

These values are one-half of the actual kcat/KM and Ki. It is impossible for Ki to increase as it is an equilibrium constant and is thus not affected by [I]. Moreover, the trend in Table 1 shows that kcat/KM remains unchanged as [I] increases.

Catalytic efficiency is a measure of an enzyme’s efficiency in catalyzing a reaction equal to kcat/KM. KM is an important measurement of enzyme-substrate affinity, and kcat, also known as the turnover number, is the amount of product formed per second by an enzyme.

According to Table 1, as the concentration of Compound 1 ([I]) increases, there is no change in kcat/KM, as both are reduced at the same rate. Moreover, Ki (a constant) does not change as [I] increases, so it should remain at 60.3 μM when [I] = 180 μM.

Therefore, because kcat/KM remains at 150 μM/s and Ki remains at 60.3 μM, we can conclude that option C is correct.

These values are double the actual kcat/KM and Ki. It is impossible for Ki to increase as it is an equilibrium constant and is thus not affected by [I]. Moreover, the trend in Table 1 shows that kcat/KM remains unchanged as [I] increases.

This question asks us to identify the functional group transformation that occurs in the reaction catalyzed by Na+-NQR, prompting us to think about different types of enzymes. This question requires us to draw information from the passage and apply content knowledge.

The passage states that Na+-NQR “catalyzes the reaction between NADH and ubiquinone.” Ubiquinone, also known as Coenzyme Q, is an electron carrier that is reduced to ubiquinol, its alcohol form, by NADH in Complex III of the electron transport chain.

As indicated by the -one suffix, ubiquinone is in the ketone form, while ubiquinol is in the alcohol form, as indicated by the -ol suffix. Thus, we must identify the option that indicates a transformation from ketone to alcohol form.

Therefore, because option A shows a two-two-electron reduction of a ketone to an alcohol, we can conclude that it shows the functional group transformation catalyzed by Na+-NQR.

This reaction shows the removal of a phosphate group, which is catalyzed by a phosphatase. Na+-NQR is an oxidoreductase.

This reaction shows the transformation from an amide to a carboxylic acid, which is catalyzed by a protease or amidase. Na+-NQR is an oxidoreductase.

This reaction shows the cleavage of an ester, which is catalyzed by an esterase. Na+-NQR is an oxidoreductase.

This question asks us to identify the structure of a component in four of the five cofactors used by Na+-NQR, prompting us to recall significant organic structures. This passage-based question requires a reference to the second paragraph to answer.

This answer choice shows the structure of the nitrogenous base adenine. Of the five cofactors used by Na+-NQR, only FAD contains adenine.

Of the five cofactors used by Na+-NQR, four are flavins: FAD, FMNc, FMNb, and riboflavin. Flavins are organic compounds derived from riboflavin, often attached to nucleotides and used as cofactors. The structure of flavin is shown below:

Therefore, because the flavin structure is found in four of the five cofactors used by Na+-NQR, we can conclude that option B is correct.

This answer choice shows the structure of ubiquinone. It is a substrate of the reaction catalyzed by Na+-NQR, not a cofactor.

This answer choice shows the structure of histidine, an amino acid. Histidine is not a cofactor.

This question asks us to determine the ratio of the concentration of cations to the concentration of Na+-NQR in the experiments from the passage. This question requires reference to the third paragraph and the application of content knowledge.

This corresponds to a cation concentration of 0.375 mM. According to the passage, it is 0.150 M. Refer to option D for the correct calculation.

This corresponds to a cation concentration of 1.5 mM. According to the passage, it is 0.150 M. Refer to option D for the correct calculation.

This corresponds to a cation concentration of 15 mM. According to the passage, it is 0.150 M. Refer to option D for the correct calculation.

The passage indicates that Na+-NQR was diluted to a final concentration of 0.75 mM in a solution of 0.150 M of various cations. Thus, we must calculate the ratio of 0.75 mM to 0.150 M. Recall that 1 mM = 1 x 10-3 M, so 0.75 mM = 7.5 x 10-4 M.

[Cation] / [Na+-NQR] = (0.150 M) / (7.5 x 10-4 M) = (1.5 x 10-1 M) / (7.5 x 10-4 M)

[Cation] / [Na+-NQR] = (1.5 / 7.5) (10-1 / 10-4) = (⅕) x 103 = 0.2 x 103 = 2 x 102

[Cation] / [Na+-NQR] = 200

Therefore, we can conclude that the ratio of cation to the enzyme in the spectroelectrochemical experiments described in the passage is 200:1.

This question asks us to identify the likely reason why the reaction catalyzed by Na+-NQR in vivo does not generate much heat, despite the uncatalyzed reaction being exergonic. This question requires a careful understanding of the passage to answer.

The question stem states that the reaction generates less heat when catalyzed by Na+-NQR, meaning that it is less exothermic. On top of that, a catalyst directs reactants to their final destination, which is the product. Catalysts are more concerned with kinetics than with thermodynamics. Catalysts cannot generate kinetic energy.

Hess’s Law states that the total enthalpy change of a reaction is the sum of the changes of each step of the reaction. As a result, the heat of the reaction will sum and be the same. The fact that the reaction can be divided into steps does not affect the overall thermodynamics.

Enzymes work by bringing reacting centers close together in the proper orientation for the reaction to occur. For a reaction to occur, the reactants need to get close together.

An exergonic reaction is accompanied by the release of energy, corresponding to a negative ΔG value. ΔG is the change in Gibbs free energy, a measure of thermodynamic potential that determines whether a reaction will occur spontaneously. Exergonic reactions are often (but not always) exothermic (heat-releasing).

The passage states that Na+-NQR “catalyzes the reaction between NADH and ubiquinone coupled to the pumping of Na+ across the plasma membrane.” The movement of a charged particle against its concentration gradient is energetically costly. By coupling the exergonic (energy-releasing) reaction between NADH and ubiquinone with the energy-requiring movement of Na+ up a concentration gradient, the overall process becomes less exothermic, generating less heat.

Therefore, we can conclude that the reaction catalyzed by Na+-NQR in vivo is coupled to the movement of a charged particle against a concentration gradient.

This question asks how the pressures compare at the same depth beneath the same liquid in two flasks that are different shapes. We must use our content knowledge of the hydrostatic pressure formula to answer this question.

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. The hydrostatic pressure formula calculates the pressure beneath liquids. The formula is

p = ρgd,

where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured. In this question, the depth is 10 cm.

Because the densities and gravities are the same at the same depths in both flasks, the pressure experienced will be the same despite the different shapes of the containers, so option A is correct.

The pressure will be the same in both flasks according to the hydrostatic pressure formula.

The pressure will be the same in both flasks according to the hydrostatic pressure formula.

Although the pressure cannot be calculated with the information given, we know that the pressure will be the same in both flasks according to the hydrostatic pressure formula.

This question asks us to identity an atom that contains six protons and eight neutrons. We can use our content knowledge of atoms to answer this question.

Nitrogen always has seven protons.

The number of protons in an atom determines its identity, whereas the number of neutrons can determine which isotope it is. Isotopes are, by definition, atoms that are the same element, having the same number of protons and the same chemical properties, but have a different number of neutrons and are thus different atomic masses. The element that has six protons is carbon.

Because an atom that has six protons is always carbon, and the number of neutrons only determines the isotope, we can conclude that option B is correct.

Oxygen always has eight protons.

Silicon always has twelve protons.

This question asks us to identify which of the substances is polar. This prompts us to use our content knowledge of Lewis structures, molecular shapes, and polarity to answer this question.

Polarity is the distribution of electrical charge between atoms that are joined by a bond. A bond is polar if the distribution of charge is unequal, meaning part of the molecule will have a slightly positive charge and part of the molecule will have a slightly negative charge. This unequal polarity is often called a dipole. A bond is nonpolar if the distribution of charge is evenly balanced and there is no perceived charge. Polarity is generally determined by the electronegativity (a measure of an atom’s ability to attract shared electrons to itself and varies between elements) of the atoms in a molecule and the three-dimensional shape of the molecule.

Drawing Lewis structures of these molecules and determining their three-dimensional shape will help us determine which molecule is polar. If a Lewis structure is drawn and the molecule is in a shape where dipoles are evenly distributed, the overall molecule may be nonpolar although specific bonds are polar. Trifluoroamine is trigonal pyramidal due to the lone pair of electrons repelling the three bonds to nitrogen. The difference in electronegativity between nitrogen and fluorine will make each of these bonds polar, as determined by electronegativity trends on a periodic table.

Because of the polarity of each of the bonds, and the imperfect shape of the molecule, the dipoles do not cancel, and trifluoroamine is polar. Option A is correct.

Each bond in tetrachloride is polar due to differences in electronegativity, however, since the shape of the molecule is tetrahedral, the dipoles are evenly distributed making this molecule overall nonpolar.

Carbon dioxide is linear so the dipole moments in each bond cancel each other out because they are in opposite directions, making this molecule nonpolar.

Since this molecule is made of two of the same element, the electronegativities of each atom in the bond are the same and the bond is nonpolar.

This question asks us to identify which of the listed changes would NOT double the volume of a gas if all else is held constant. We can use our content knowledge of gas laws to answer this question.

The gas laws that help us solve this problem are:

Boyle’s Law: P1V1=P2V2

Charles’ Law: V1 / T1 = V2 / T2

and Avogadro’s Law: V1 / n1 = V2 / n2

In these equations, P is for pressure, V is for volume, T is for temperature, and n is for mols. By manipulating these equations according to the answer options’ directions, we can determine the correct option.

By using Boyle’s Law, we can conclude that pressure is inversely proportional to volume, and by doubling the pressure, we would decrease the volume rather than increase it. Thus, we can determine option A is correct.

By using Charles’ law we can conclude that temperature is directly proportional to volume and by doubling the temperature, we would double the volume as well.

By using Boyle’s law we can conclude that pressure is inversely proportional to volume and by halving the pressure, we would double the volume.

By using Avogadro’s law we can conclude that number of gas molecules is directly proportional to volume and by doubling the number of gas molecules, we would double the volume as well.

This question asks us to determine the number of neutrons in the nucleus of the atom used to produce laser radiations, 86Kr+. This passage-based question requires a reference to the third paragraph.

This answer corresponds to 84Kr, not 86Kr. Refer to option C for the correct calculation.

This answer corresponds to 85Kr, not 86Kr. Refer to option C for the correct calculation.

While the number of protons in an atom’s nucleus is equal to the element’s atomic number (Z), the number of neutrons can vary to produce isotopes, which are atoms of the same element that have different mass numbers.

The passage states that “the fluorescence spectrometer … employs a 86Kr+ laser” The 86 in 86Kr+ indicates that it is an isotope with an atomic mass of 86 amu. We can gather the number of protons from the periodic table (36 protons) and subsequently calculate the number of neutrons in the nucleus. Recall that the mass of each proton and neutron is equal to approximately 1 amu.

Atomic mass = (Number of protons) + (Number of neutrons)

Number of neutrons = (Atomic mass) – (Number of protons)

Number of neutrons = 86 – 36 = 50

Therefore, we can conclude that the nucleus of the atom used to produce laser radiations, 86Kr+, contains 50 neutrons.

This answer corresponds to 87Kr, not 86Kr. Refer to option C for the correct calculation.

This passage-based question asks us to determine the molecular formula of pyrrole, prompting us to recall and apply content knowledge. Although the structure of a tetrapyrrole is depicted in Figure 1, the passage is not required to answer this question.

This molecular formula describes non-aromatic 2H-azirine, which is a three-membered heterocycle consisting of 2 carbons, 3 hydrogens, and 1 nitrogen.

Pyrrole is a 5-member aromatic ring that contains 1 nitrogen atom. Aromatic compounds are organic molecules that display aromaticity, a pi-bond structure that grants increased stability due to resonance effects. Because it has two different atoms in its ring (4 carbons, 1 nitrogen), it is classified as a heterocyclic compound.

Counting the number of each atom, pyrrole contains 1 nitrogen, 4 carbons, and 5 hydrogens.

Therefore, we can conclude that the molecular formula of pyrrole is C4H5N.

This molecular formula describes azepines, which are a group of 7-membered heterocycles consisting of 6 carbons, 7 hydrogens, and 1 nitrogen.

This molecular formula describes a cyclohexene-fused pyrrole, but not pyrrole itself, which is a pyrrole ring fused on a cyclohexane ring consisting of 8 carbons, 9 hydrogens, and 1 nitrogen.

This question asks us to calculate the approximate number of moles of Kr+ contained in the laser tube at 0°C and 1 atm, prompting us to think about gas stoichiometry. This question requires us to refer to the third paragraph.

At STP, this number of moles of gas takes up 0.00672 mL. Refer to option D for the correct conversion and calculation.

At STP, this number of moles of gas takes up 0.0448 mL. Refer to option D for the correct conversion and calculation.

At STP, this number of moles of gas takes up 0.90 mL. Refer to option D for the correct conversion and calculation.

At standard temperature and pressure (273k or 0°C and 1 atm), 1 mole of any gas occupies a volume of 22.4 L.

The passage states that Kr+ occupies 11 cm3 in the laser tube. To calculate the number of moles in this volume, we must first convert it to liters. Then we can convert liters to moles by dividing it to 22.4 L per mol. Recall that 1 ml is equivalent to 1 cm3 and 1000 mL is equal to 1 L.

[ 11 cm3 Kr+ ∙ (1 mL / 1 cm3) ∙ (1 L / 1000 mL)] / (22.4 L / 1 mol)

To make the mental math easier, we can round 11 cm3 to 10 cm3 and 22.4 L to 20 L.

[ 10 cm3 Kr+ ∙ (1 mL / 1 cm3) ∙ (1 L / 1000 mL)] / (20 L / 1 mol) = 5 x 10^-4 mol Kr+

Therefore, we can conclude that approximately 5 x 10-4 moles of Kr+ are contained in the laser tube.

This passage-based question asks us to identify the reason why radiation of wavelength 605 nm CANNOT be used to produce fluorescence radiations depicted in Figure 3, prompting us to think about fluorescence.

Fluorescence is the emission of light by a substance that has absorbed energy in the form of electromagnetic radiation. In fluorescence, the emitted photon always has a smaller energy level than the absorbed photon. Radiation wavelength and energy are inversely proportional; higher wavelength corresponds to lower energy. Thus, although it could conceivably produce lower-energy fluorescence of 622 nm and 639 nm, radiation of 605 nm could not produce higher-energy fluorescence of wavelength 604 nm.

Therefore, we can conclude that the radiation of wavelength 605 nm cannot produce the fluorescence radiations depicted in Figure 3 because the energy of the absorbed radiation must be larger than the energy of the fluorescence radiation.

The energy of the absorbed radiation must be larger than the energy of the fluorescence radiation, per the principle of energy conservation.

Radiation of 605 nm has less energy than the radiation of 407 nm because the wavelength and energy of a photon are inversely proportional.

605 nm light falls within the visible light spectrum (400-700nm). This feature is not related to fluorescence characteristics.

This question asks us to identify the chiral atom that allows CD spectroscopy signals to be generated. This passage-based question requires reference to Figure 1 and the application of content knowledge.

The passage states that CD spectroscopy “is used to examine the secondary structure of proteins,” and the question stem indicates that signals arise from chiral atoms of the protein. Chiral atoms are bound to four unique groups, such that they have a non-superimposable mirror image.

In a protein, each amino acid residue except glycine has a chiral α carbon (the central carbon). In most amino acids, the β carbon is achiral, leaving the α carbon as the sole chiral center.

Therefore, we can conclude that the CD spectroscopy signal used to generate the data in Figure 1 arises from the chirality of the α carbon.

The amide nitrogen of an amino acid is achiral because it is not bound to four unique groups.

The carbonyl carbon of an amino acid is achiral because it is not bound to four unique groups.

The β carbon is achiral in each amino acid except isoleucine and threonine. Thus, it is more likely that CD spectra arise from the chirality of the α carbon.

This question asks us to identify which physical property does NOT change with the amino acid substitution present in the TPMT*5 variant of TPMT. This passage-based question requires reference to the second paragraph and the application of content knowledge on amino acids.

The side chains of serine and leucine have different molecular weights, so this substitution will change the molecules’ weight.

Wild-type TPMT has leucine, a hydrophobic residue, at position 49, but the L49S variant (TPMT*5) has serine, a polar residue, in this position. Therefore, this substitution will affect hydrophobicity.

The hydrocarbon side chain of leucine is not capable of hydrogen bonding. However, the hydroxymethyl side chain of serine is, so this substitution will alter hydrogen bonding capability.

The passage indicates that the TPMT*5 variant contains an L49S substitution, meaning that leucine at position 49 was swapped with serine.

Leucine is classified as an uncharged, nonpolar amino acid because of its hydrocarbon side chain, while serine is an uncharged, polar amino acid due to its hydroxymethyl side chain.

Therefore, because leucine and serine are uncharged at physiological pH, the amino acid substitution made in TPMT*5 will not change the net charge.

This question asks us to identify the expected appearance of protein bands that have undergone SDS-PAGE under reducing conditions, prompting us to think about electrophoresis. This pseudo-discrete question does not require details from the passage to answer.

This gel depicts the aggregation of protein subunits at the top of the gel over time. SDS-PAGE causes the separation of protein subunits as they move down the gel. Subunits of lower molecular weight move more quickly than larger ones, creating separation, not aggregation.

This gel depicts aggregation of protein subunits toward the top of the gel and does not change the amount of starting protein. SDS-PAGE causes the separation of protein subunits as they move down the gel. Subunits of lower molecular weight move more quickly than larger ones, creating separation, not aggregation.

Although this gel correctly depicts the separation of protein bands, the original protein band at the top of the gel should diminish over time. Refer to option D for the correct gel depiction.

Sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE) is a technique used to separate proteins according to their size. SDS is a surfactant that gives molecules a uniform negative charge, ensuring that separation in the polyacrylamide gel occurs solely based on molecular weight. Under reducing conditions, a reducing agent like β-mercaptoethanol is added to reduce disulfide bonds, breaking proteins into individual subunits (proteolysis).

In SDS-PAGE under reducing conditions, all proteins start at the top of the gel in a single, thick band. Over time, due to proteolysis, the separation of the band can be visualized as subunits of lower molecular weight move more quickly through the gel, collecting in bands towards the bottom.

Therefore, we can conclude that option D is correct because it depicts the original protein band (with the highest molecular weight) diminishing over time as the number of lower molecular weight bands increases.

This question asks us to determine the net difference in molecular weight of the molecules involved in the reaction catalyzed by TPMT, prompting us to think about different types of enzymes. This question requires an understanding of the passage to answer.

The substrate has a net change in molecular weight of +15 g/mol because it gains a methyl group, but there is no overall net change in molecular weight because the methyl is simply transferred from the cofactor to the substrate.

TPMT is a methyl transferase, meaning that it transfers a methyl group (weighing 15 g/mol) from the cofactor to the substrate. There is no net change in the molecular weight because it is simply transferred from the cofactor to the substrate.

Therefore, we can conclude that if the combined mass of the TPMT substrate and cofactor was compared to the combined mass of the product and the cofactor, the net change would be 0 g/mol.

The cofactor has a net change in molecular weight of -15 g/mol because it loses a methyl group, but there is no overall net change in molecular weight because the methyl is simply transferred to the substrate.

This is the molecular weight of methane (CH4), which is not involved in the reaction catalyzed by TPMT.

This passage-based question asks us to identify the compound from which the ligand of hMPRα is derived, prompting recall content knowledge and refer to the passage.

Glucose, a 6-carbon monosaccharide, is not a precursor to steroids.

Phenylalanine is an aromatic amino acid. It contributes to protein structure but is not a precursor to steroids.

Glycerol is a triol compound that forms the backbone of triglycerides. Glycerol is not a precursor to steroids.

The passage states that hMPRα is a membrane progestin receptor, so its ligand is progesterone, a steroid hormone. Like all steroid hormones, progesterone is derived from cholesterol.

Therefore, we may conclude that the ligand of hMPRα is derived from cholesterol.

This question asks us to identify which type of chromatographic separation is utilized in the second purification step. This question requires a careful understanding of the passage and application of content knowledge.

In affinity chromatography, proteins are separated from a mixture based on their affinity for a specific ligand. Beads in the chromatography column contain a ligand that the protein of interest binds with high specificity and affinity. Proteins with a lower affinity for the ligand will elute first, leading to the separation of the mixture.

The passage states that the researchers manipulated yeast cells to attach two different tags to hMPRα, each to be used for separate rounds of chromatography. The second tag “consisted of six consecutive histidine residues (His)6… [that] binds tightly to Ni2+ cations.” Then, in chromatography, (His)6 tags bind to the column, requiring an eluent to be displaced. This is a classic example of affinity chromatography.

Therefore, we can conclude that the second purification step utilizes affinity chromatography.

In size exclusion chromatography, molecules are separated based on their size. The chromatography column contains porous beads that trap small molecules to slow their elution while larger molecules travel down the column more quickly.

In cation exchange chromatography, the column is coated with negatively charged beads that interact with positively charged molecules to slow their elution while negatively charged/uncharged molecules elute quickly.

In anion exchange chromatography, the column is coated with positively charged beads that interact with negatively charged molecules to slow their elution, while positively charged/uncharged molecules elute quickly.

This question asks us to identify structural features of Compound 2 that are important to its function, as mentioned in the passage. To answer this question, we must refer to the final paragraph and apply content knowledge.

Chirality is crucial to allow specific binding to other molecules. Still, it does not contribute to the function of a detergent and thus does not permit Compound 2 to extract hMPRα from cell membranes.

The presence of multiple hydrolyzable linkages would facilitate metabolism. In the passage, Compound 2 was used as a detergent to extract hMPRα from cell membranes.

The passage states that hMPRα was “extracted from the membranes using n-decyl-β-D-maltopyranoside, Compound 2.” Extracting proteins from a membrane requires the membrane to be dissolved using a detergent. Detergents facilitate the mixture of hydrophobic and hydrophilic compounds, allowed by their large hydrophobic and hydrophilic regions.

Compound 2 contains large hydrophobic and hydrophilic regions, permitting it to act as a detergent for the purpose of extracting hMPRα from cell membranes.

Therefore, we can conclude that the combination of large hydrophobic and hydrophilic regions permit Compound 2 to extract hMPRα from cell membranes.

A reducing sugar permits oxidation but does not contribute to the function of a detergent and thus does not permit Compound 2 to extract hMPRα from cell membranes.

This question asks us to calculate the number of moles of NaCl present in 500 mL of the buffer solution used to elute hMPRα. This question requires us to refer to the final paragraph.

This is the quantity of NaCl in 7.5 mL of buffer. Refer to option D for the correct calculation.

This is the quantity of NaCl in 3.75 mL of buffer. Refer to option D for the correct calculation.

This is the quantity of NaCl in 37.5 mL of buffer. Refer to option D for the correct calculation.

The passage states that the buffer solution used to elute hMPRα contained 300 mM NaCl, among other compounds. Molarity (M) is a measure of the concentration of a solution, equal to the number of moles of solute per liter of solution.

M = Moles of solute / Liters of solution

Knowing the molarity of NaCl (300 mM) and volume of solution (500 mL), we can calculate the number of moles of NaCl present. Recall that 1000 mM = 1 M, so 300 mM = 0.300 M, and 1000 mL = 1 L, so 500 mL = 0.500 L.

0.300 M = Moles NaCl / 0.500 L

Moles NaCl = (0.300 M) x (0.500 L) = 0.150 moles

0.150 moles = 1.5 x 10^-1 moles

Therefore, we can conclude that 500 mL of the buffer solution used to elute hMPRα contained 1.5 x 10-1 mole of NaCl.

This question asks us to identify which piece of experimental evidence suggests that the purified hMPRα obtained by the researchers was in its native state, prompting us to consider the characteristics of native versus denatured proteins. This question requires an understanding of the passage and application of content knowledge.

Retention of Compound 1 and (His)6 tags suggests that the tag binding sites are intact, not that the protein is in its native state.

Any chromatographic procedure can cause a protein to denature. Because hMPRα required two stages for purification, it was more likely to be denatured.

In their native state, proteins are fully folded and interact with other molecules. When they are unfolded (denatured), the 3D structure of the protein is disrupted, changing the position of amino acid residues, which in turn impacts how the protein interacts with other molecules.

The passage states that researchers “found that Kd was similar” after purification. The dissociation constant is a term used to indicate the affinity of a ligand for its receptor. Kd can be used to calculate binding affinity (how strongly a ligand binds to a receptor). The smaller the value of Kd, the more tightly the ligand is attached and, thus, the higher the affinity between the ligand and the receptor. If the Kd is unchanged after purification, this would strongly suggest that the protein has not been denatured.

Therefore, because the obtained hMPRα exhibited a binding affinity for progesterone that was similar to that exhibited by native hMPRα, we can conclude that the purified hMPRα was in its native state.

Molecular weight measures the stability of a protein’s primary structure; folding status does not affect it. Thus, this would not suggest that purified hMPRα was in its native state.

This question asks us to identify why such high temperatures are required for fusion. Answering this question will require our content knowledge of atoms, fusion, and the strong nuclear force.

Because the strong nuclear force emerges whenever nucleons are present and within a range of picometers, increasing the temperature does not eliminate it.

The electrical charge is an intrinsic feature of matter which remains unaffected by temperature.

Fuel density decrease is undesirable because the probability of fusion increases as fuel density increases. This is because the attracting component of the strong nuclear force works at short distances, implying that the fuel material must be dense enough.

The possibility of fusion increases as the average distance between fuel particles decreases. This allows attractive nuclear forces to overcome repulsive nuclear forces at medium and long distances.

Since temperature is a measure of the average kinetic energy of the particles in a group, raising the temperature will increase the average kinetic energy of the particles. Kinetic energy and speed are connected by the equation,

KE = ½ x m x v^2

where KE is kinetic energy, m is mass, and v is velocity. This increased speed will allow particles to overcome the electrostatic potential barrier that causes these reactants to repel each other and increases the likelihood that they will undergo the tunnel effect (a phenomenon where a particle tunnels through a barrier that it normally could not overcome), allowing them to get within range of the strong nuclear force. The strong nuclear force is one of the four forces of nature and is the force that holds nuclei together. It is the strongest of the four, but its effects are only felt when nucleons (the subatomic particles of the nucleus) are very close together.

From this, we can conclude that increasing the temperature enables reactants to approach within range of the strong nuclear force and choose option D.

This question asks us to identify a chemical or physical property where enantiomers exhibit a difference. We must use our content knowledge of enantiomers to answer this question.

Enantiomers do not have different densities.

Enantiomers have identical boiling points and cannot be identified with distillation.

Enantiomers are molecules that are mirror images of each other, and thus have the same atom-to-atom connections, but have a different three-dimensional arrangement of atoms. Enantiomers have the same physical and chemical properties and are thus difficult to detect and separate. Enantiomers will rotate polarized light to the same degree but in opposite directions and interact with other chiral molecules differently. Odorant receptors can be chiral and thus enantiomers can be detected by their difference in smell.

Because chiral odorant receptors can detect a difference in enantiomers, we can choose option C.

Should a normal light source is used, the IR spectra of enantiomers would appear identical.

This question asks us to calculate the speed of blood when it switches from flowing 30 cm/s in a tube with a diameter of 1.6 cm to a tube that is 0.8 cm. Answering this question will require content knowledge of the continuity equation and the area of a circle.

A 7.5 cm/s flow rate corresponds to a 3.2 cm diameter.

A 15 cm/s flow rate corresponds to a diameter of 2.25 cm. This option may have been chosen because it is half of the original speed and the diameter was halved, however, the area of a circle is not directly related to the diameter or radius (it related via a square root).

A 60 cm/s flow rate corresponds to a diameter of 1.125 cm. This is half of the correct option.

The continuity equation can be applied to the flow since no blood disappears between the two settings.

By dividing both sides by A2, and entering the equation for the area in, we can get the equation we need in order to solve for the new velocity. Then we can use the values that were given to calculate the new velocity. Note that since the diameter is given, it is divided by two to get the radius, and the π cancel each other out.

V2 = V1A1/A2 = (V1 x (π x r12)) / (π x r22) = [30 cm/s x ((1.6 cm)/2)^2] / ((0.8 cm)/2)^2 = 120 cm/s.

We can then conclude that option D is correct.

This question asks us to identify the atom that has an electron configuration equivalent to that of a noble gas. We must use our content knowledge of atoms to answer this question.

Noble gases are unique because they have full electron shells. To find an atom that has an electron configuration equivalent to that of a noble gas, we must find an atom that has only full electron shells. One easy way to determine how many electrons noble gases have is to note that noble gases are located in the rightmost group of the periodic table. The number of protons, as noted just above each element, is how many electrons a noble gas, or an atom with an equivalent electron configuration has. When writing configurations, we count from left to right and fill each shell in order as shown by the periodic table below with the number on the left being the energy level, the letter being the subshell, and the number on the right being the number of electrons in the subshell.

A noble gas configuration simply would have the noble gas in brackets instead of the entire row for the configuration (e.g. for a regular calcium atom, it would be [Ar] 4s2 instead of 1s2 2s2 2p6 3s2 3p6 4s2). Since each electron has a -1 charge, removing electrons will give the atom a +1 charge for every electron removed. Since calcium only has 2 electrons more than a full shell normally and they have been removed as shown by the 2+ above this calcium ion, this calcium ion has an electron configuration equivalent to that of a noble gas, Ar.

Thus we can conclude that option A is correct.

Copper normally has 11 outer electrons as noted on the periodic table. The loss of 2 electrons leaves the cation with a [Ar] 3d10 configuration. Copper is actually a special exception for electron configurations, but the reason that it is not the correct option here is that it would still have 9 outer electrons, and thus have an incomplete subshell.

An oxygen atom has a 2p subshell that is partially filled.

A hydrogen atom has a 1s subshell that is half-filled.

This question asks us to identify the Trp residue of carbonic anhydrase that can be selectively modified with CHCl3 at 20 °C. We must use passage information to answer this question.

W4 is only modified when buffer A is used. Under these conditions, W190 and W243 are also modified, so W4 is not selectively modified.

From the table, W96 is never modified and thus couldn’t be selectively modified.

W190 is only modified when buffer A is used. Under these conditions, W4 and W243 are also modified, so W190 is not selectively modified.

The word selectively in the answer stem leads us to look for the residue that can be modified without other residues being modified along with it when using CHCl3. Looking at Table 1, we see that the first two rows are using CHCl3, and we are looking for a set of conditions that only leads to one residue being affected. The second row shows that only the last residue, W243, is affected when buffer B is used and thus is selectively modified. When buffer A is used, W4, W190, and W243 are all affected, thus, none of them are selectively modified when buffer A is used.

From this, we can conclude that option D, W243, is correct.

This question asks us to identify the classification of amino acids that applies to the Trp residues after photochemical modification by CCl3COH2. We must use passage information and our content knowledge of organic chemistry to answer this question.

In Figure 1, we can see that the group that was added to Tryptophan (Trp) was a carboxylic acid group. Carboxylic acid groups are acidic, polar (has a charge that is not canceled out), hydrophilic (attracted to water, typically due to being polar), and negative at physiological pH (7.4)

Thus, we can conclude that option A, acidic, is correct.

Carboxylic acid groups are acidic, not basic.

Carboxylic acid groups are hydrophilic, not hydrophobic. Hydrophobic groups are groups that are not attracted to water, typically because they are nonpolar.

Although carboxylic acid is polar, at physiological pH, which is around 7, carboxylic acid loses a hydrogen and develops a negative charge.

This question asks us to identify the type of heterocycle that is found on two amino acid residues blocking access to W15. We must use passage information and our content knowledge of amino acids and organic chemistry groups to answer this question.

The passage states that “access to W15 is blocked by two histidine (His) amino acids”. Histidine has an amino group, a carboxylic acid group, and an imidazole side chain, making it a positively charged amino acid at physiological pH.

Imidazole (C₃N₂H₄) is an aromatic heterocycle with non-adjacent nitrogen atoms in meta-substitution. A heterocyclic compound or ring structure is a cyclic compound that has atoms of at least two different elements as members of its ring.

Thus, we can conclude that option A, imidazole, is correct.

Tryptophan (trp) has an R group that is an indole heterocycle.

Pyrimidine is a heterocycle that is found on cytosine, uracil, and thymine in DNA and RNA.

Pyrrole is a five-membered heterocyclic aromatic organic molecule with the formula C4H4NH. The amino acids proline and hydroxyproline, heme (a component of hemoglobin), and bile pigments all contain the pyrrole ring structure.

This question asks us to identify what most likely causes W96 to be accessible to CHCl3 at 75°C. We must use passage information and our content knowledge of biochemistry to answer this question.

The passage states that the primary structure (the sequence of amino acids caused by amino acids being bound together via peptide bonds) remains intact. Because peptide bonds create the primary structure, we can assume that peptide bonds were not broken.

A reducing agent such as beta-mercaptoethanol is necessary to break disulfide bonds (a covalent bond between two sulfur atoms). The passage only states that the protein is denatured, so this is not the best option.

The passage states, “W15 and W96, which are in the interior of carbonic anhydrase, are 21% and 29% modified respectively by 10 mM CHCl3 in Buffer A at 75°C, a temperature which fully denatures carbonic anhydrase but leaves its primary structure intact.” In this case, a buffer was used to denature (unfold) the protein, but no covalent bonds were said to be broken, including disulfide bonds and peptide bonds (which create primary structure). Because the protein unfolds, W96 becomes exposed, and CHCl3 can react with it.

Thus, we can conclude that option C is correct.

There is not enough CHCl3 for it to act as the solvent that denatures the protein, and it is stated that the buffer denatures the protein, so it is the buffer that would extract W96, not CHCl3.

This question asks us to identify the name of the ionic compound used to make Buffer B. We must use our content knowledge of molecule names to answer this question.

The formula for ammonium formate is NH4HCO2. It is not mentioned in the passage.

The formula for ammonium carbonate is (NH4)2CO3. It is not mentioned in the passage.

The formula for ammonium bicarbonate is NH4HCO3. It is a component in buffer A, not buffer B.

A buffer is composed of a weak acid and its conjugate weak base. In Figure 1, beneath the horizontal arrow, we see that NH₄HCO₃ (ammonium + bicarbonate) is a component of buffer A and NH4CH3CO2 is a component of buffer B. NH4CH3CO2 is an ionic compound that dissociates into its constituent ions which are ammonium ion (NH4+) and acetate ion (C2H3O2-). Ammonium is a weak acid, and acetate is a weak base. This coincides with our knowledge of the definition of buffers.

From this, we can conclude that option D, ammonium acetate, is correct.

This question asks us to identify the chromatographic technique that would most likely separate a mixture of native carbonic anhydrase from carbonic anhydrase photochemically modified by CCl3CO2H. We must use our content knowledge of separation techniques to answer this question.

Chromatography is a separation technique that takes advantage of the different products’ solubilities and relative affinities for the stationary phase used. Chromatography separates a mixture using solids and liquids to separate its parts. All chromatography works by having a mobile phase (the part that moves), and a stationary phase (the part that stays still) which allow for the separation of the different fractions of the original mixture. Chromatography works because of the different affinities (strength of adhesion) of the various components of the mixture towards the stationary and mobile phases, which are dictated by two properties of the molecule: ‘adsorption’ and ‘solubility’. These different affinities allow for the separation of the components.

Anion-exchange chromatography and cation-exchange chromatography both separate molecules by their charge. In anion-exchange chromatography, beads in a tube are attached to positive molecules, which attract, and therefore slow down, anions or negatively charged molecules. The solution that is being separated enters the top, and different molecules come out of the bottom at different times. Those that are more negative are more attracted to the positive charges, and thus the more negatively charged molecules are eluted (come out of the bottom) later. Cation-exchange chromatography works similarly, except that the beads have negative molecules attached to them, which attract positive charges, and thus those that are more positive are eluted later.

Size-exclusion chromatography works similarly, but its beads have passageways in them. Those molecules that are larger do not fit in the passageways and pass right past the beads, whereas those molecules that are smaller take a longer route through the beads and elute later.

Gas chromatography involves samples being vaporized and passed through a liquid or solid stationary phase using a gaseous (nitrogen or argon) mobile phase. The molecules with the lowest boiling points come out of the column first. The molecules with the higher boiling points come out of the column last. At the end of the stationary phase, there is a detector that works out how many particles of each component are found at different times.

Because a carboxylic acid group was added, which makes the protein more negative, option A, anion-exclusion chromatography, would be correct.

Because both of the enzymes are negatively charged, and not positive, cation-exchange chromatography would not be effective since the proteins would elute together in the void volume (the volume that comes out at the very beginning that doesn’t have anything that the chromatography is sensitive to.

Proteins would degrade before they become vaporized, so gas-liquid chromatography would just destroy the proteins.

Although there is a difference in size between the proteins and their modified versions, the size difference is not big enough to separate them effectively.

This passage-based question asks us to identify the functional group containing the external oxygen on each ring of the MCS structures shown in Figure 1, prompting us to recall and apply content knowledge while referring to Figure 1.

An aldehyde group is a carbonyl group bonded to at least one hydrogen atom [R(C=O)H, where R = H or a carbon group]. The functional groups in question are bonded to two carbon groups.

A carboxylic acid is a carbonyl bonded to a hydroxyl (-OH) group and a carbon group (RCO2H).

An ester is a carbonyl bonded to a non-hydroxyl oxygen (RCO2R’, where R = H or a carbon group and R’ = a carbon group).

In organic chemistry, a functional group is a substituent that alters a molecule’s chemical properties. Functional groups can vary widely in their structures and properties. Carbonyls are a class of functional groups that consist of a carbon double-bonded to an oxygen atom. The carbon’s two remaining bonds vary based on the type of carbonyl.

A ketone group is a carbonyl group bonded to two carbon atoms. Each MCS structure shown in Figure 1 contains several ketone groups outside of the ring.

Therefore, we may conclude that option D, ketone, is correct.

This question asks us to determine the final pressure of the container after 1 mole of gaseous carbon monoxide reacts to form carbon suboxide. To answer this passage-based question, we must refer to Reaction 1 from the passage.

This would be the pressure of the container if all gas molecules were evacuated. However, there is gas present in the container after Reaction 1.

This would be the pressure of the container if the number of moles of gas was decreased by a factor of 10, not halved.

This would be the pressure of the container if the number of moles of gas was quartered, not halved.

In Reaction 1, which occurs at 550 °C, “CO(g) converts to carbon suboxide.” For every 4 moles of CO that react, 1 mole of suboxide (C3O2) and 1 mole of carbon dioxide (CO2) form. Thus, 4 moles of CO (g) form 2 moles of gases.

In a fixed container, volume is constant. If the temperature is also kept constant, then the moles of gas and the pressure of the container have a direct relationship, according to the ideal gas law.

PV = nRT
Thus, if the number of moles of gas is halved (from 4 mol to 2 mol), then the pressure must also be halved.

Therefore, we may conclude that the final pressure of the container is 0.50 atm.

This question asks us to identify which phase the MCS precursor will be predominantly found in after the extraction step, prompting us to think about the process of chemical extraction. Answering this passage-based question requires us to refer to the third paragraph.

According to the passage, the MCS precursor is lipophilic, so it is not particularly water-soluble. Thus, it will not be found in the aqueous layer.

The passage states that “This MCS precursor was isolated from plant roots through an extraction that involved mixing an aqueous emulsion with tert-butyl methyl ether.”

Extraction is a method used to isolate a substance from a mixture based on its relative solubility in two immiscible solvents. When two immiscible liquids are mixed (e.g., oil and water), they form two layers due to their differing densities. When the substance of interest is added to these liquids in a separatory funnel, it will dissolve in either the hydrophobic or hydrophilic (water) layer, depending on its hydrophobicity (or lack thereof).

In the extraction mentioned in the passage, an aqueous layer was combined with tert-butyl methyl ether, a nonpolar solvent. The MCS precursor is described as “lipophilic”, so it must be hydrophobic and thus insoluble in the aqueous layer. Thus, it will dissolve most readily in the tert-butyl methyl ether layer, which is hydrophobic.

Therefore, we may conclude that the MCS precursor will be found in the tert-butyl methyl ether layer.

According to the passage, the MCS precursor is lipophilic, so it is not particularly water-soluble. Thus, it will be largely found in the tert-butyl methyl ether layer, not the aqueous layer.

According to the passage, the MCS precursor is lipophilic, so it will likely dissolve in the organic (tert-butyl methyl ether) layer.

This passage-based question asks us to approximate the Ki of the MCS oligomers, prompting us to employ our data-based statistical reasoning skills while analyzing Figure 2.

At this MCS oligomer concentration, the enzyme would be at an almost maximal specific activity.

By definition, Ki is the concentration of the inhibitor that gives a reaction rate of half the maximum reaction rate. Graphically, this is the inflection point of Figure 2.

The inflection point occurs at about 0.030 mM, or 30 nM.

Therefore, we may conclude that Ki of the MCS oligomers is approximately 30 nM.

This is the MCS oligomer concentration at about one-twentieth of the maximum specific activity.

At this MCS oligomer concentration, the enzyme would be completely inhibited.

This question asks us to determine how the MCS oligomers affect inhibition, based on the reported Hill coefficient. This passage-based question requires us to refer to the final paragraph.

The passage states that “when those inhibition data were fitted with a Hill function, the resulting coefficient was 2.56.” The Hill coefficient is a measure of cooperativity in a binding process. A Hill coefficient of 1 indicates independent binding, while a value of greater than 1 indicates positive cooperative binding.

Cooperative binding refers to a phenomenon present in some inhibitors composed of multiple subunits, in which the binding of substrates by one subunit changes the affinity of other unoccupied subunits for the substrates. In positive cooperative binding, when one subunit binds a substrate, the rest become more likely to do the same. Thus, the MCS oligomers exhibit positive cooperative binding (Hill coefficient = 2.56; greater than 1).

Therefore, we may conclude that as one MCS oligomer binds to the ATPase, it makes it easier for the others to bind, leading to inhibition.

This describes negative cooperative binding, when the binding of one subunit makes it more difficult for the others to bind. Negative cooperative binding is associated with a fractional Hill coefficient.

This describes independent binding, which is associated with a Hill coefficient of 1.0.

The Hill coefficient indicates cooperativity; it does not indicate whether binding is random.

This question asks us to identify which inhibitor will prevent tearing without taking away the flavor of onion. This passage-based question requires reference to the third and fourth paragraphs.

An inhibitor of Compound 1 synthesis would stop the onion from causing tearing, but it would also remove the onion’s flavor because Compound 1 is the precursor to flavor components.

An inhibitor of alliinase would stop the onion from causing tearing by preventing Compound 4 synthesis, but it would also remove the onion’s flavor by preventing Compound 3 synthesis.

An inhibitor of reaction 2 would prevent the synthesis of Compound 3, “one of the main flavor components of onions.” It would also fail to prevent the synthesis of Compound 4, “a potent lachrymator (tearing agent).”

The passage states that Compound 3 “is one of the main flavor components of onions” and Compound 4 “is a potent lachrymator (tearing agent).” Compounds 3 and 4 are direct products of Compound 2, formed in Reaction 2 and Reaction 3, respectively. Thus, to retain onion flavor while preventing tearing, Reaction 3 must be inhibited. Reaction 3 is catalyzed by LFS when the onion is cut. Therefore, we may conclude that an inhibitor of LFS will provide a flavorful onion that does not cause tearing.

This question asks us to determine which amino acid side chains are most likely to provide a salt bridge and π-stacking interaction, prompting us to think about the categorization of amino acids. This question requires reference to the second paragraph.

Aspartate (Asp, D) is negatively charged at physiological pH and would thus be repulsed by the negative charge of PLP instead of forming a salt bridge.

Glutamate (Glu, E) is negatively charged at physiological pH and would thus be repulsed by the negative charge of PLP instead of forming a salt bridge. Furthermore, serine (Ser, S) lacks an aromatic ring and is thus not capable of participating in π-stacking interactions.

A salt bridge is an attractive force consisting of a combination of hydrogen bonding and ionic bonding. A positively charged amino acid side chain can form a salt bridge with PLP, which contains a negatively charged phosphate group. Arginine (Arg, R) is basic and is thus positively charged at physiological pH.

π-stacking refers to the attractive, noncovalent pi interactions that occur with the overlap of π bonds of aromatic rings. Tyrosine (Tyr, Y) is large, nonpolar, and aromatic. Thus, it can participate in π-stacking interactions with PLP.

Therefore, we can conclude that the side chains of Arg and Tyr most likely provide the salt bridge and π-stacking interaction with PLP.

Although lysine (Lys, K) is positively charged at physiological pH, serine (Ser, S) lacks an aromatic ring and is thus not capable of participating in π-stacking interactions.

This passage-based question asks us to identify which reactant and product will balance Reaction 1, requiring us to analyze Reaction 1.

In Reaction 1, Compound 1 loses ammonium, not nitrate. This would not balance the reaction in charge or mass.

In Reaction 1, Compound 1 loses an ammonium ion and gains a water molecule to form Compound 2 and pyruvate.

Adding water as a reactant and ammonium as a product balances both charge and mass on both sides of the chemical equation.

Therefore, we can conclude that to balance Reaction 1, water must be included as a reactant and ammonium as a product.

In Reaction 1, Compound 1 gains water and loses ammonium. This would not balance the reaction in charge or mass.

In Reaction 1, Compound 1 gains water, not molecular oxygen. This would not balance the reaction in mass.

This question asks us to identify the relative values of Ea and ΔG in Reaction 2, prompting us to think about the characteristics of spontaneous reactions. This question requires us to refer to the third paragraph.

If Reaction 2 had a high Ea, it would not occur rapidly. Furthermore, if it had ΔG > 0, it would not occur spontaneously.

If Reaction 2 had a high Ea, it would not occur rapidly.

If Reaction 2 had ΔG > 0, it would not occur spontaneously.

Change in Gibbs free energy (ΔG) is a thermodynamic measure that represents the ability of a system to do work. A reaction with a positive ΔG value will not occur spontaneously, and a reaction with a negative ΔG value will occur spontaneously. Activation energy (Ea) is the minimum amount of energy required for compounds to react. Even if a reaction is spontaneous, it will not occur rapidly unless activation energy is low.

The passage states that “Compound 2 spontaneously and rapidly condenses to form Compound 3” in Reaction 2.

Therefore, because Reaction 2 occurs spontaneously and rapidly, it must have low Ea and ΔG < 0.

This question asks us to identify which group will be deprotonated first as the pH of the solution is raised. We must use our content knowledge of organic chemistry groups and pKas to answer this question.

The three types of groups that are highlighted in this diagram are carboxylic acid groups (I, III), amines (IV), and hydroxyl groups (II) with pKas of about 5, 10, and 15, respectively. A pKa is a dissociation constant for a group that is a general measure of how weak or strong an acid is. Since strong acids (low pKas) are deprotonated first, we will be comparing the carboxylic acid groups. To compare the two, we use our knowledge of the inductive effect, or the local change in electron density (that affects the dissociation constants of the groups there) due to the electron donating or withdrawing effects of nearby groups. Since chlorine is an electronegative atom, it will cause a withdrawing effect on carboxylic acid group I, thus stabilizing the negative conjugate base (an anion) and dissipating the negative charge that will build up.

From this information, we can conclude that group I will be deprotonated first since carboxylic acids have the lowest pKas, and group I has a neighboring group that will lower the pKa even more.

Hydroxyl groups are not very acidic because their negative charge cannot be delocalized to nearby electronegative atoms.

This carboxylic acid group is not as acidic as group I because it does not have the electronegative groups nearby to stabilize the conjugate base.

Protonated amines are less acidic than carboxylic acid groups and thus require a higher pH to deprotonate them.

This question asks us to identify the expression which gives the activation energy for the reaction. We must use our content knowledge of thermodynamics to answer this question.

Option A represents the activation energy for the reverse reaction.

Option B represents the change in enthalpy for the reaction.

Activation energy is the energetic barrier that the reactants must overcome before forming product. The difference between the potential energy levels of the reactants and products is the thermodynamic energy of the equation or change in enthalpy.

In this graph, the activation energy is the difference between E3 and E2, which is option C.

Option C is the correct representation of the activation energy for the reaction.

This question asks us to calculate the acceleration of both the 2 kg and the 5 kg masses on the pulley once they have been released. We will need to use our content knowledge of Newton’s laws of motion to answer this question.

Option A suggests that just the 2-kg mass moves following the release.

According to Newton’s second law of motion,

F = ma,

Where F equals force, m equals mass, and a equals acceleration. To determine the acceleration on each of the masses (the acceleration will be the same but in the opposite direction for each one, thus a1 = -a2 = a), we must find the acceleration on one of the masses while considering the acceleration from the other one. Using an equation of motion, which uses forces to determine acceleration, is one way to calculate acceleration.

We know that the total force on the smaller object will be equal to the mass times the total acceleration which is equal to the force due to gravity (mass times gravity) minus the force due to tension.

Fnet = m1a = m1g – T

The tension in that case will be the mass of the heavier object times the net acceleration which is the acceleration of gravity (for the heavier object) minus the total acceleration.

T = m2(g-a)

We can then substitute and simplify. Note Fnet is negative because we know that the smaller weight will go up, in the same direction as the tension.

-2kg x a = 2(kg x g) – 5kg x (g – a) => -2(kg x a) = 2(kg x g) – 5(kg x g) + 5(kg x a) => -7 (kg x a) = -3(kg x g) => a = 3/7 g

Thus, option B is the correct option.

Other Ways to Solve: We can draw a free-body diagram with two boxes on opposite sides of a pulley with forces down on each of them (due to gravity), and forces up on each of them (due to tension). The acceleration of the masses can be determined by choosing one mass and finding the acceleration for that one mass because the masses are connected and will accelerate at the same acceleration, but in opposite directions. According to Newton’s second law of motion,

F = ma,

which rearranged to find the acceleration is,

a = F / m.

Since we want the net acceleration, we will be using the net forces and masses (acceleration will be the same and will be acting on both masses). The net force can be shown as

Fnet = F1 – F2 = m1g – m2g = g (m1 – m2)
and the total mass will be

m = m1 + m2.

When these are substituted into the equation, and we input our numbers, we can get the correct answer. We can choose either mass to be m1 but having m1 be the 5 kg mass will make the answer positive.

a = F / m = g (m1 – m2) / (m1 + m2) = g (5 – 2) / (5 + 2) = 3g / 7.

Option C suggests that just the 5-kg mass moves following the release.

Option D suggests that the two masses go down freely and independently, unconnected by the cable.

This question asks us to identify which part of converting an ice cube to steam uses the most energy when the specific heat of water is 1 cal/g°C, the heat of fusion is 80 cal/g, and the heat of vaporization is 540 cal/g. We will need to use our content knowledge of thermodynamics to answer this question.

Melting the ice cube will only take 80 cal/g, whereas vaporizing all the water will take 540 cal/g. See option D for the full explanation.

Heating all the water from 0°C to 50°C will only take 50 cal/g, whereas vaporizing all the water will take 540 cal/g. See option D for the full explanation.

Choice B Solution: Heating all the water from 50°C to 100°C will only take 50 cal/g, whereas vaporizing all the water will take 540 cal/g. See option D for the full explanation.

Since the heat of fusion (the amount of energy required to convert the solid ice into liquid water) is 80 cal/g, we know that melting the ice cube will require 80 cal/g. Heating all the water from 0°C to 50°C and heating all the water from 50°C to 100°C will take the same amount of energy since it is the same change in temperature (50°C). We can calculate this by multiplying the specific heat of water (the energy needed to raise the temperature of one gram of water by one degree) by the change in temperature giving us 50 cal/g.

Finally, since the heat of vaporization (the amount of energy required to convert the liquid water to steam) is 540 cal/g, we know that converting the water to steam will take 540 cal/g. The simple reason it takes more energy to go from liquid to gas than to go from solid to liquid is that, when going from solid to liquid, only the bonds that keep the molecules directly in place are broken, whereas when going from liquid to gas, all the bonds and intermolecular forces holding the molecules together must be overcome.

From our calculations, we can conclude that option D, vaporizing all the water, will require the most energy (540 cal/g).

This question asks us to calculate the pressure exerted on the chest by a 10 x 5 cm rectangular paddle, prompting us to think about the components of pressure. This passage-based question requires reference to the third paragraph.

This corresponds to 25% of the actual pressure. Refer to option D for the correct calculation.

This corresponds to 50% of the actual pressure. Refer to option D for the correct calculation.

This corresponds to 75% of the actual pressure. Refer to option D for the correct calculation.

Pressure is equal to the force applied perpendicular to a surface (F) divided by the area to which the force is applied (A).

The passage states that 100 N of force is applied to the chest during defibrillation. Assuming that this force is perpendicular to the chest, we can calculate the pressure after determining the surface area.

A = (10 cm) x (5 cm) = 50 cm2

Recall that 1 cm2 = 1 x 10^-4 m2, so 50 cm2 = 50 x 10^-4 m2 = 5 x 10^-3 m2

P = F / A = (100 N) / (5 x 10^-3 m2) = 0.2 x 10^5 N/m2 = 2 x 10^4 N/m2 = 2 x 10^4 Pa

2 x 10^4 Pa x (1 kPa / 10^3 Pa) = 20 kPa

Therefore, we can conclude that option D is correct

This question asks us to determine the electrical current that would flow through the paddles if the entire defibrillator’s charge were discharged through the patient in 10 ms. This question requires reference to Figure 1 to solve.

Electrical current (I) is a measure of the amount of charge passing through a space per unit of time, typically measured in amperes (A; A = C/s). The amount of charge passing through the defibrillator can be calculated using the voltage (V) of the power supply (3000 V) and the capacitance (C) of the capacitor (25 μF; F = coulombs/V).

Q (Coulombs) = Capacitance (Farads) ∙ Electrical Potential (Volts)

Charge = C ∙ V = 25 x 10^-6 F ∙ 3000 V = (25 x 10^-6 F) ∙ (3 x 10^3 V) = 75 x 10^-3 A = 7.5 x 10^-2 coulombs

Now, using the charge and time (10 ms), we can calculate the current. Recall that 1000 ms = 1 s, so 10 ms = 0.01 s.

I = charge / time = (7.5 x 10^-2 coulombs) / (0.01 s) = (7.5 x 10^-2 coulombs) / (10^-2 s)

I = 7.5 coulombs/s = 7.5 A

Therefore, we can conclude that option A is correct.

This corresponds to twice the actual current. Refer to option A for the correct calculation.

This corresponds to one and a half times the actual current. Refer to option A for the correct calculation.

This corresponds to four times the actual current. Refer to option A for the correct calculation.

This question asks us to determine which adjustment would increase the charge on the capacitor in Figure 1, prompting us to recall the factors that affect the charge on a capacitor. This question does not require specific information from the passage to answer.

The capacitance is directly proportional to the area of the parallel plates, so this will decrease the capacitance. The charge is directly proportional to the capacitance, so this will also correspond to a decrease in the charge.

Capacitance is a measure of an object’s ability to store electric charge. Parallel plate capacitors, such as the one seen in Figure 1, are composed of two conductive plates separated by an insulating material called a dielectric. The capacitance of the capacitor (C) depends on the surface area of the plates (A), the distance between the plates (d), and a constant associated with the dielectric (k).

Equation 1: C = kA/d

The amount of electric charge stored in each plate (Q) is directly proportional to the potential difference (voltage, V) between the two places:

Equation 2: Q = CV

By Equation 1, decreasing the separation between the two plates (d) will increase the capacitance (C). Then, by Equation 2, increasing the capacitance will result in an increase in the charge of the capacitor.

Therefore, we may conclude that decreasing the separation between the parallel plates will increase the charge on the capacitor.

Capacitance depends on the ability of the dielectric to store electrical potential, known as its permittivity. Removing the dielectric essentially decreases the permittivity to a level that does not permit capacitance. Thus, removing the dielectric from the capacitor will decrease the capacitance, which decreases the charge, per Equation 2, discussed in option B.

The charge on the capacitor is precisely proportional to capacitance multiplied by the power supply voltage. If the voltage is reduced, the charge decreases unless the capacitance is increased independently.

This question asks us to determine which power supply voltage would produce the same amount of charge as a 25 μF capacitor with a 3000 V battery, shown in Figure 1. This question does not require specific information from the passage to answer.

This corresponds to a charge of 108 mC. Refer to option D for the correct calculation.

This corresponds to a charge of 105 mC. Refer to option D for the correct calculation.

This corresponds to a charge of 90 mC. Refer to option D for the correct calculation.

The electric charge of a capacitor (Q) is equal to the product of the capacitance (C) of the capacitor and the voltage of the power supply:

Q = CV

Using this relationship, we can calculate the charge of the capacitor setup in Figure 1. Recall that 1 μF = 1 x 10^-6 F, so 25 μF = 25 x 10^-6 F.

Q = CV

Q = (25 x 10^-6 F) ∙ (3000 V) = (25 x 10^-6 F) ∙ (3 x 10^3 V)

Q = 75 x 10^-3 FV = 7.5 x 10^-2 coulombs

Now, we must calculate what voltage is necessary to produce a charge of 7.5 x 10^-2 coulombs if the capacitor has a capacitance of 30 μF.

Q = CV

V = Q/C

V = (7.5 x 10^-2 coulombs) / (30 x 10^-6 F) = (7.5 / 30) ∙ (10-2 / 10-6) V

V = (¼) ∙ (10^4) V = 0.25 x 10^4 V = 2.5 x 10^3 V = 2500 V

Therefore, we may conclude that option D is correct.

This question asks us to determine how the frequency of a light beam would change if the wavelength was doubled, prompting us to consider the relationship between frequency and wavelength. This pseudo-discrete question does not require the passage to answer.

Wavelength and frequency are inversely proportional, so doubling the wavelength will cause the frequency to be halved, not quartered.

The wavelength (λ) of light refers to the physical distance of one complete cycle. The frequency (f) is the number of complete cycles that pass a given point per second. The product of the wavelength and frequency is equal to the speed of light (c):

c = fλ

From this equation, it is apparent that frequency and wavelength are inversely proportional. When the speed of light is held constant, doubling the wavelength will cause the frequency to be halved.

Therefore, we may conclude that option B is correct.

Wavelength and frequency are inversely proportional, so doubling the wavelength will cause the frequency to be halved, not doubled.

Wavelength and frequency are inversely proportional, so doubling the wavelength will cause the frequency to be halved, not quadrupled.

This question asks us to identify which property of a photon will double if the photon’s energy is doubled, prompting us to think about the relationship between energy, amplitude, wavelength, frequency, and intensity of a photon. The passage contains key information that is necessary to answer this question.

Amplitude refers to the maximum displacement of a wave from its equilibrium position. Photons constitute the particle description of light. As a result, photons cannot be defined by an amplitude.

If the energy is doubled, the wavelength is halved, not doubled. Refer to option C for a detailed explanation.

The energy of a photon is related to the the speed of light and the wavelength by Planck’s constant, h = 4.1 x 10^-15 eV∙s:

E = hc/λ

If the energy of a photon is doubled, then by this equation, the wavelength will be halved if the speed is constant. The wavelength is inversely proportional to the frequency, per the following equation:

c = fλ

Because wavelength and frequency are inversely related, if the change in the photon’s energy causes the wavelength to be halved, then the frequency will be doubled.

Energy of a photon equation can be simplified to:

E = hf

Here it shows that photon energy is directly proportional to frequency since Planck’s constant can not change.

Therefore, we may conclude that option C is correct.

Waves have the property of intensity, whereas photons are the particle description of light. As a result, photons cannot be defined by a measure of intensity unless expressed as the number of photons of a specific energy that travel through a surface per unit time.

This question asks us to calculate the frequency of a red line in the Balmer series with a wavelength of 656 nm, prompting us to consider the relationship between frequency and wavelength. This question does not require details from the passage to answer.

The wavelength (λ) of light refers to the physical distance of one complete cycle. The frequency (f) is the number of complete cycles that pass a given point per second. The product of the wavelength and frequency is equal to the speed of light (c, c = 3 x 10^8 m/s):

c = fλ

The equation can be rearranged to solve for the frequency. Recall that 1 nm = 1 x 10-9 m, so 656 nm = 656 x 10-9 m.

f = c/λ = (3 x 10^8 m/s) / (656 x 10^-9 m) = (3 x 10^8 m/s) / (6.56 x 10^-7 m)

f ≈ (3 x 10^8 m/s) / (6 x 10^-7 m) = (3 / 6) ∙ (10^8 / 10^-7) s-1 = 0.5 x 10^15 s-1

f ≈ 5 x 10^15 s-1 = 5 x 10^15 Hz

Therefore, we may conclude that option A is correct.

This answer likely results from an error when summing the scientific notation exponents. Refer to option A for the correct calculation.

This answer likely results from dividing the frequency by the speed of light and from making an error when summing the scientific notation exponents. Refer to option A for the correct calculation.

This answer likely results from dividing the frequency by the speed of light, instead of multiplying. Refer to option A for the correct calculation.

This question asks us to calculate the wavelength of a photon whose energy is 2 eV, prompting us to consider the relationship between a photon’s wavelength and energy. Answering this question requires the equation mentioned in the second paragraph.

This corresponds to a photon with an energy of approximately 1.66 eV. Refer to option B for the correct calculation.

The passage indicates that the energy of a photon is related to the wavelength by the following equation:

λ = hc/E

In this equation, E is the energy of the photon, Planck’s constant (h) is 4.1 x 10^-15 eV∙s, and the speed of light (c) is 3 x 10^8 m/s.

Given the photon’s energy of 2 eV, we can solve for the wavelength.

λ = (4.1 x 10^-15 eV∙s) ∙ (3 x 10^8 m/s) / (2 eV) = (4.1 ∙ 3 / 2) ∙ (10^-15 ∙ 10^8)

λ ≈ (4 ∙ 3 / 2) ∙ (10^-15 ∙ 10^8)

λ ≈ 6 x 10^-7 m

6 x 10^-7 m ∙ (1 nm / 1 x 10^-9 m) = 6 x 10^2 nm ≈ 620 nm

Therefore, we may conclude that option B is nearest to the wavelength of a photon whose energy is 2 eV.

This corresponds to a photon with an energy of approximately 2.75 eV. Refer to option B for the correct calculation.

This corresponds to a photon with an energy of approximately 4 eV. Refer to option B for the correct calculation.

The question asks us to determine the value of Kb for the conjugate base of a weak organic acid that has a pKa of 5. To answer this question, we need to understand several key concepts related to acid-base equilibria.

This answer suggests a larger value for Kb, which would be inconsistent with a weak organic acid and its conjugate base.

This is the value of Ka for the weak acid. However, the question asks us for Kb, not Ka.

Firstly, we must understand what Ka, Kb, and Kw represent. Ka is the acid dissociation constant. It’s a measure of the strength of an acid in solution. The larger the Ka, the stronger the acid, and vice versa. The pKa is just the negative logarithm of the Ka and is used for ease of notation. The important trend for us to know is that a smaller pKa indicates a stronger acid.

Kb, on the other hand, is the base dissociation constant. It is a measure of the strength of a base in solution. The larger the Kb, the stronger the base, and vice versa. In a similar fashion to its acid counterpart, a smaller pKb indicates a stronger base.

Kw is the ion product of water, which at 25 degrees Celsius, is equal to 1.0 x 10^-14. This is a constant value that represents the propensity of water to autoionize (dissociate into ions in its pure state) into H+ and OH- ions. We know that Ka and Kb are related to each other through Kw in the equation Ka x Kb = Kw. We also know that pKa is the negative logarithm (base 10) of Ka. From this information, we can determine that the Ka for the weak acid is 10^-5.

From the equation Ka x Kb = Kw, we can solve for Kb by rearranging the equation to Kb = Kw / Ka. Plugging in the known values, we get Kb = (1.0 x 10^-14) / (10^-5), which simplifies to Kb = 10^-9.

We should always try to make physical sense of values by looking at the overall chemical picture. Here, we can see that this value makes logical sense in the context of the problem as the weak organic acid has a relatively small Ka value (10^-5), and so its conjugate base should also have a small Kb value.

This answer suggests a smaller value for Kb than is correct. The Kb value is a measure of the base’s strength, so a smaller Kb value would imply a weaker base than what is actually the case for the conjugate base of this weak acid.

The question asks for a modification that will NOT increase the final yield of an ester in a direct esterification reaction. Yield in this context refers to the quantity of ester product produced by the reaction. Esterification involves the combination of a carboxylic acid (RCOOH) and an alcohol (RCH2OH), catalyzed by an esterase enzyme, to form an ester (RCOOCH2R) and water.

Enzymes catalyze reactions by lowering the activation energy but do not alter the equilibrium position. They affect the rate at which equilibrium is reached, not the final quantities of reactants or products at equilibrium. Therefore, doubling the amount of enzyme will increase the rate of ester formation, but it won’t influence the final yield of the ester.

Therefore, we can conclude that option A, doubling the enzyme, is correct as it will not affect the reaction’s yield. In order to eliminate the remaining answer options, we must understand Le Chatelier’s principle, a fundamental concept in chemistry. The principle states that if a stress (such as a change in concentration, temperature, or pressure) is applied to a system at equilibrium, the system will adjust to counteract the change and re-establish equilibrium.

Increasing the concentration of RCOOH, a reactant, will shift the equilibrium to the right (toward the products) according to Le Chatelier’s principle. This will lead to a higher final yield of the ester.

Similarly to option B, increasing the concentration of RCH2OH, the other reactant, will also shift the equilibrium to the right and result in a higher final yield of the ester.

This action decreases the concentration of a product. According to Le Chatelier’s principle, the system will respond by shifting the equilibrium to the right, resulting in more ester being produced. Thus, this action will increase the final yield of the ester.

The question requires us to find the amount of work done by a person pushing a cart. This scenario is illustrated graphically with force on the y-axis and distance on the x-axis, whereby the force exerted by the person diminishes linearly with distance from a maximum of 100 N at 0 m to 0 N at 20 m.

This would imply an area under the graph of 500 N·m, which doesn’t match the given graph.

When a force is exerted on an object causing it to move over a certain distance, work is said to have been done on the object. It’s a measure of energy transfer and is calculated by multiplying the force exerted on the object by the distance the object moves as a result of this force (Work = Force x Distance). The unit of work is the joule (J), which is equivalent to a newton-meter (N·m). When force is not constant but varies with distance, the work done can be visualized as the area under the force-distance curve (the explanation behind this involves calculus, which is out of the scope of the MCAT).

In this case, because the force changes linearly with distance, the work done is equivalent to the area under the graph, which is a right triangle. The area of a triangle is given by 1/2 · base · height. Substituting the given values, the work done is 1/2 · 20 m · 100 N = 1000 J. Therefore, the person does 1000 J of work in pushing the cart, making option B correct.

This would suggest an area under the graph of 2000 N·m, which is not accurate based on the given scenario.

This answer would correspond to an area under the graph of 4000 N·m, but this is not consistent with the given graph.