Practice Exam Chemistry and Physics

What is the work generated by a healthy adult who circulates 9 L of blood through the brachial artery in 10 min? A. 2 kJ B. 12 kJ C. 20 kJ D. 120 kJ

Solution: The correct answer is D. This is a Physics question that falls under the content category "Translational motion, forces, work, energy, and equilibrium in living systems." The answer to this question is D because a flow of 9 liters in 10 minutes means a flow rate of 900 mL/min, and according to the graph, it corresponds to a power of 200 W. The work is then 200 W x 600 s = 120 kJ. It is a Data-based and Statistical Reasoning question because you are asked to use, analyze, and interpret data in a graph.

What is the approximate concentration of reaction product in a solution that has an absorbance of 0.7 at pH 6.0? A. 0.50 μM B. 500 μM C. 5 mM D. 500 mM

The solution is B. A) This result is 1/1000 of the correct answer, most likely due to a decimal point placement error. B) Based on the equation A = εbc given in the passage, the concentration c is the absorbance A divided by the absorptivity ε in a 1 cm path length cell. 0.7 divided by approximately 1400 gives 500 μM. C) This result is 10 times the correct answer, most likely obtained by a mistake in converting M to mM. D) This result is 1000 times the correct answer, most likely due to confusing mM for μM.

What is the approximate energy of a photon in the absorbed radiation that yielded the data in Table 1? (Note: Use 1 eV = 1.6 × 10-19 J and hc = 19.8 × 10-26 J•m.) A. 1 eV B. 2 eV C. 3 eV D. 4 eV

The solution is B. A) This is the energy of a photon if radiation had a wavelength of 1250 nm. B) The photon energy is E = hc/λ = 19.8 × 10-26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 2 eV. C) This is the energy of a photon if radiation had a wavelength of 416 nm. D) This is the energy of a photon if radiation had a wavelength of 312 nm.

Which type of bond is formed by glycogen synthase upon release of UDP? A. α-1,4-Glycosidic bond B. α-1,6-Glycosidic bond C. β-1,4-Glycosidic bond D. β-1,6-Glycosidic bond

Solution: The correct answer is A. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is A because the bond that is formed by glycogen synthase is the main chain linkage of glycogen, which is an α-1,4-glycosidic bond. UDP release means that only glucose was added. This is a Knowledge of Scientific Concepts and Principles question because you must apply knowledge about the structure of glycogen to solve the problem.

Which amino acid was incorporated into Compound 1 as a future site of covalent attachment to HA prior to mineralization? A. Ser B. Ala C. Tyr D. Thr

Solution: The correct answer is A. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is A because serine is the residue that was phosphorylated, and phosphorylation enabled the protein to be subsequently attached to hydroxyapatite (HA). It is a Scientific Reasoning and Problem Solving question because you are asked to reason using scientific principles and models.

According to the developed chromatography plate shown below, what is the approximate Rf value of aspartic acid? A. 0.20 B. 0.50 C. 5 D. 10

Solution: The correct answer is A. This is an Organic Chemistry question that falls under the content category "Separation and purification methods." The answer to this question is A (0.20) because Rf is the ratio of the distance travelled by the analyte relative to the solvent front during a chromatographic separation. Aspartic acid travelled two units, while the solvent front travelled ten units, giving an Rf of 2/10 = 0.20 for aspartic acid. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in a figure to draw a conclusion.

In the chromatography of the reaction mixture, water absorbed on cellulose functioned as the stationary phase. What was the principal factor determining the migration of individual components in the sample? A. Hydrogen bonding B. Solute concentration C. Stationary phase concentration D. Thickness of paper

Solution: The correct answer is A. This is an Organic Chemistry question that falls under the content category "Separation and purification methods." The answer to this question is A because the relative amount of hydrogen bonding to the stationary phase will determine the relative rate of migration of the various components in the sample. It is a Scientific Reasoning and Problem Solving question because you are asked to evaluate an argument about causes and consequences (the primary reason for chemical separation during a chromatographic separation on cellulose).

If Reaction 2 (Figure 4) is repeated with HCl and the compound shown below, which of the following compounds is NOT a direct product (without rearrangement)?

Solution: The correct answer is A. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is A because the structure shown in this response requires an additional shift of hydride atoms prior to elimination and is therefore NOT a direct product. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and observations to draw a conclusion.

Compared to micellular Compound 1, Compound 2 is structurally more rigid as a result of what type of interaction? A. Intermolecular hydrogen bonding B. Intermolecular covalent bonding C. Intramolecular hydrogen bonding D. Intramolecular covalent bonding

Solution: The correct answer is B. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is B. It can be reasoned that the interaction described is intermolecular in nature, since multiple molecules of micellular Compound 1 come together to form Compound 2, which is a solid. Multiple pieces of information point to the fact that the interaction is disulfide bond formation, including the fact that an oxidant causes the formation of Compound 2, which can be reversed by the addition of a reducing agent. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and observations to draw a conclusion.

The graph below shows the relationship between the predominant form of iron as a function of solution pH and applied potential. Based on the graph, which of the following statements is true? A. At a potential of -0.4 V, as pH increases, Fe2+ is reduced and precipitates as Fe(OH)3. B. At a potential of -0.44 V, the equilibrium between Fe and Fe2+ is independent of solution pH below pH 6. C. At pH = 1, as the potential is changed from -0.2 to +0.8, Fe3+ is reduced to Fe2+. D. At pH = 8 and V = -0.1 V, Fe(OH)2 is the predominant form of iron.

Solution: The correct answer is B. This is a General Chemistry question that falls under the content category "Atoms, nuclear decay, electronic structure, and atomic chemical behavior." The answer to this question is B. The Pourbaix diagram for iron is given, but only one statement correctly conveys information presented by the diagram. At an applied potential of −0.44V and pH below 6, there is an equilibrium between Fe(s) and Fe2+(aq), as shown by the line of demarcation. The fact that this line moves horizontally with increasing pH up until pH = 6 means that this equilibrium is unaffected by changing the pH. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data presented in a figure to draw a conclusion.

What percentage of standard atmospheric pressure is the pulse pressure of a healthy adult? A. 10% B. 6% C. 2% D. 1%

Solution: The correct answer is B. This is a General Chemistry question that falls under the content category "Importance of fluids for the circulation of blood, gas movement, and gas exchange." The answer to this question is B because the pulse pressure in a healthy adult is (120 − 75) mmHg = 45 mmHg, and so the percentage is 45 mmHg/760 mmHg = 6%. It is a Knowledge of Scientific Concepts and Principles question because you are asked to use mathematical equations to solve problems.

What are the hybridization states of the carbon atoms involved in the conversion of trans to cis retinal? A. sp B. sp2 C. sp3 D. sp3d

Solution: The correct answer is B. This is a General Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is B because the reacting carbon atoms are both central to AX3 systems with 3 bonded atoms and no lone pairs. The preferred geometry for such a system is trigonal planar and the hybridization scheme that facilitates this geometry is sp2. It is a Knowledge of Scientific Concepts and Principles question because you are asked to recognize the relationship between closely related concepts (bonding geometry and hybridization.)

When the current in the micro-robot's circuit increases and the resistance of the circuit remains constant, the voltage of the circuit: A. decreases. B. increases. C. is constant. D. is zero.

Solution: The correct answer is B. This is a Physics question that falls under the content category "Electrochemistry and electrical circuits and their elements." The answer to this question is B because according to Ohm's law, current is equal to voltage divided by resistance. If current increases and resistance is constant, then voltage increases as well. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use scientific formulas to solve problems.

Knowing that the speed of light in the vitreous humor is 2.1 × 108 m/s, what is the index of refraction of the vitreous humor? (Note: The speed of light in a vacuum is 3.0 × 108 m/s.) A. 0.7 B. 1.4 C. 2.1 D. 3.0

Solution: The correct answer is B. This is a Physics question that falls under the content category "How light and sound interact with matter." The answer to this question is B because the index of refraction of a medium is equal to the ratio of the speed of light in vacuum to the speed of light in the medium, thus it is equal to (3.0 x 108 m/s)/(2.1 x 108m/s) = 1.4. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use scientific formulas to solve problems.

The production of a variety of opsins functions to: A. increase sensitivity to low light. B. enable the detection of different colors. C. ensure fast recovery of 11-cis-retinal after exposure. D. increase refractive index of the eye lens.

Solution: The correct answer is B. This is an Organic Chemistry question that falls under the content category "How light and sound interact with matter." The answer to this question is B. The wavelength of light absorbed by a molecule depends on its structure, and so the production of a variety of structurally related opsins functions to enable the detection of different colors. It is a Scientific Reasoning and Problem Solving question because you are asked to reason using scientific principles.

If a solution containing the compounds shown in Figure 4, is injected into a gas-liquid chromatograph, the first peak observed in the gc trace is attributable to which compound? A. 2-Methyl-2-butanol B. 2-Methyl-2-butene C. 2-Chloro-2-methylbutane D. 2-Bromo-2-methylbutane

Solution: The correct answer is B. This is an Organic Chemistry question that falls under the content category "Separation and purification methods." The answer to this question is B because 2-methyl-2-butene will exhibit the lowest molecular weight and also the weakest intermolecular forces of attraction. This substance will therefore migrate the fastest and be the first peak in the gas chromatograph (gc) trace. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and observations to draw a conclusion.

Addition of laforin to the reaction mixtures used to generate the data in Figure 2 will result in: A. a decrease in the intensity of the band representing WT without glucosidase when visualized with 14C decay. B. an increase in the intensity of the band representing WT without glucosidase when visualized with 14C decay. C. a decrease in the intensity of the band representing WT without glucosidase when visualized with 32P decay. D. an increase in the intensity of the band representing WT without glucosidase when visualized with 32P decay.

Solution: The correct answer is C. This Biochemistry question falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is C because laforin is an enzyme that removes phosphate from glycogen. Presumably, the removal of phosphate prior to gel electrophoresis will result in a smaller band corresponding to 32P incorporation. This is a Reasoning about the Design and Execution of Research question because you are asked to make a prediction based on the properties of the experimental design.

In addition to glucose, what other monosaccharide is part of the UDP-glucose structure? A. Xylose B. Fructose C. Ribose D. Arabinose

Solution: The correct answer is C. This Biochemistry question falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because UDP contains uridine, which is a nucleic acid used in RNA. The structure of RNA contains ribose. This is a Knowledge of Scientific Concepts and Principles question because you must apply knowledge of the structure of nucleic acids to solve to problem.

Which reaction is catalyzed by LipA? A. ATP hydrolysis B. Peptide bond cleavage C. Hydrolysis of triacylglycerides D. Transfer of carboxyl groups

Solution: The correct answer is C. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is C because of the fact that LipA is a lipase, which means it must hydrolyze fatty acids. This is a Knowledge of Scientific Concepts and Principles question because you are asked to identify the role of a lipase in biochemical reactions.

The source of the phosphate groups that are added to rhodopsin is: A. arrestin. B. rhodopsin kinase. C. ATP. D. all-trans-retinol.

Solution: The correct answer is C. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is C because the requirement of ATP for kinase activity implies the phosphate groups come from ATP. All the other responses do not have phosphate to transfer. It is a Knowledge of Scientific Concepts and Principles question because you are asked to recall and apply the concept of phosphate group transfer to a particular enzyme substrate relationship.

The data in Figure 1 show that: A. the unfolding of LipA proteins is a non-reversible process. B. exposure to high temperature eliminates any variant protein activity. C. variation in the flexible regions of LipA decreases the temperature sensitivity of activity. D. wild-type LipA loses all activity when first heated to 50°C.

Solution: The correct answer is C. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is C because the data show that variation of the flexible regions results in less temperature sensitivity of activity. Activity returns even when variants are exposed to high temperature. This is a Data-based and Statistical Reasoning question because you are asked to interpret trends in data to arrive at an answer.

Compound 1 was designed to exhibit pH-dependent self-assembly. What feature(s) of the molecule is(are) responsible for the pH dependence of aggregation (Equation 1)? A. Thiol side chains that can hydrogen bond B. Long alkyl tail that exhibits predominantly London forces C. Side chains whose net charge responds to pH D. Covalent linkages that reversibly hydrolyze

Solution: The correct answer is C. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is C because the molecule contains mainly acidic side chains that were deprotonated and negatively charged at high pH. This inhibited aggregation due to electrostatic principles. At low pH these groups are neutral and this allowed aggregation. It is a Reasoning about the Design and Execution of Research question because you are asked to identify the relationship among variables in a study (pH and solubility).

Why is the velocity of blood flow slower in capillaries than in arteries? A. Capillary walls are more elastic than arterial walls. B. Capillaries have less resistance to blood flow than arteries. C. The total cross-sectional area of capillaries exceeds that of arteries. D. Blood pressure is higher in the capillaries than in arteries.

Solution: The correct answer is C. This is a Biology question that falls under the content category "Importance of fluids for the circulation of blood, gas movement, and gas exchange." The answer to this question is C because the high number of capillaries in the body means that the total cross-sectional area of these vessels is larger than any other vessel type in the circulatory system. This causes the velocity of the blood to decrease. This is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of how the structure and arrangement of blood vessels influences blood flow.

The term "ideal gas" refers to a gas for which certain assumptions have been made. Which of the following is such an assumption? A. The law PV = nRT2 is strictly obeyed. B. Intermolecular molecular forces are infinitely large. C. Individual molecular volume and intermolecular forces are negligible. D. One gram-mole occupies a volume of 22.4 L at 25°C and one atmosphere pressure.

Solution: The correct answer is C. This is a General Chemistry question that falls under the content category "Importance of fluids for the circulation of blood, gas movement, and gas exchange." The answer to this question is C since a property of an "ideal" gas is that it is composed of particles that have negligible volume and do not exert intermolecular forces. It is a Knowledge of Scientific Concepts and Principles question since you are asked to recognize a component of the Ideal Gas law.

Which of the following will decrease the percentage ionization of 1.0 M acetic acid, CH3CO2H(aq)? A. Chlorinating the CH3 group B. Diluting the solution C. Adding concentrated HCl(aq) D. Adding a drop of basic indicator

Solution: The correct answer is C. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is C because HCl is a strong acid that will increase the amount of H+ in solution and thus decrease the percentage of CH3CO2H that ionizes. It is a Scientific Reasoning and Problem Solving question because you are asked to reason using a scientific principle (Le Châtelier's principle) to identify that adding a strong acid to a solution of weak acid will decrease the amount of ionization of the latter.

A person, whose eye has a lens-to-retina distance of 2.0 cm, can only clearly see objects that are closer than 1.0 m away. What is the strength S of the person's eye lens? (Note: Use the thin lens formula ​​.) A. -50 D B. -10 D C. 51 D D. 55 D

Solution: The correct answer is C. This is a Physics question that falls under the content category "How light and sound interact with matter." The answer to this question is C because the strength of the eye lens is equal to the inverse of the focal length of the eye lens. Its numerical value is given by 1^​​-1+(0.02 m)^-1=1 D+50 D=51 D. It is a Knowledge of Scientific Concepts and Principles question because you are asked to use mathematical equations to solve problems.

What is the relationship between Pp, PMPA, and Pd? A. Pp = PMPA + 3Pd B. Pp = 3PMPA + Pd C. Pp = 3PMPA − 3Pd D. Pp = PMPA − 3Pd

Solution: The correct answer is C. This is a Physics question that falls under the content category "Importance of fluids for the circulation of blood, gas movement, and gas exchange." The answer to this question is C because according to Equation 1, PPMA = Pd + (Ps - Pd)/3 = Pd +Pp/3; thus Pp = 3PMPA - 3Pd. It is a Knowledge of Scientific Concepts and Principles question because you are asked to use mathematical equations to solve problems.

What is the product of the reaction of Compound 1 (shown below) with HBr by the pathway shown in Figure 3? Compound 1 A. (R)-1-bromo-1-deuteriohexane B. (S)-1-bromo-1-deuteriohexane C. (S)-1-bromo-1-deuteriopentane D. (R)-1-bromo-1-deuteriopentane

Solution: The correct answer is C. This is an Organic Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is C because the incoming nucleophile displaces the leaving group form the opposite side of the reacting center during an SN2 reaction. It is a Scientific Reasoning and Problem Solving question because you are asked to bring together theory and observations to draw a conclusion.

What type of functional group is formed when aspartic acid reacts with another amino acid to form a peptide bond? A. An amine group B. An aldehyde group C. An amide group D. A carboxyl group

Solution: The correct answer is C. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is C because the functional group that forms during peptide bond formation is known as an amide group. It is a Knowledge of Scientific Concepts and Principles question since you are asked to recognize the structural relationship between free amino acids and peptides.

The intensity of the radiation emitted by the oxygen sensor is directly proportional to the: A. propagation speed of the radiation. B. wavelength of the radiation. C. polarization of photons emitted. D. number of photons emitted.

Solution: The correct answer is D. This is a Physics question that falls under the content category "How light and sound interact with matter." The answer to this question is D because the energy of electromagnetic radiation is directly proportional to the number of photons, and the intensity of electromagnetic radiation is defined as energy emitted per unit time. Thus, intensity is directly proportional to the number of photons emitted. It is a Knowledge of Scientific Concepts and Principles question because you are asked to identify a relationship between closely related concepts.

What assumption is being made if scientists conclude that aspartic acid was formed by the prebiological synthesis in the passage? A. Aspartic acid is unstable at temperatures below 150°C. B. All of the malic acid underwent the dehydration reaction to form fumaric/maleic acid. C. Compound A and cyanide were available on primitive Earth. D. The reaction between ammonia and fumaric acid was catalyzed by the presence of water.

Solution: The correct answer is C. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is C since, in order for the experimental reaction sequence to be relevant to the primordial formation of aspartic acid, the starting materials used (Compound A and cyanide) are assumed to have been available. It is a Reasoning About the Design and Execution of Research question because you are asked to reason about the appropriateness of particular research designs to evaluate a natural sciences question.

Which statement correctly describes how enzymes affect chemical reactions? Stabilization of: A. the substrate changes the free energy of the reaction. B. the transition state changes the free energy of the reaction. C. the substrate changes the activation energy of the reaction. D. the transition state changes the activation energy of the reaction.

Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Principles of chemical thermodynamics and kinetics." The answer is D because the stabilization of the transition state, not the substrate, provides binding energy that is used to lower the activation energy. It is a Knowledge of Scientific Concepts and Principles question because you are asked to recognize a fundamental principle of enzyme kinetics.

Compared to the wild-type LipA, what is the change in net charge in variant XI at pH 7? A. +4 B. +3 C. −3 D. −4

Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because the amino acid substitutions replace amino acids with charges of +1, +1, 0, 0, and 0 with ones that have charges of 0, −1, −1, 0, and 0 so the net charge goes from +2 to −2. This is a Scientific Reasoning and Problem Solving question because you are asked to determine the necessary calculation to solve for net charge.

Which structure represents a phosphorylated subunit of glycogen as described in the passage?

Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer is D because the phosphate residues are added to the C2 and C3 positions of the sugar. This is a Knowledge of Scientific Concepts and Principles question because you must recognize the numbering convention for carbohydrates.

Which amino acid residues were incorporated into Compound 1 to promote the adhesion of cells on the scaffold surfaces? A. Arg and Gly B. Cys and Gly C. Cys and Asp D. Asp and Arg

Solution: The correct answer is D. This is a Biochemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is D. The residues that were engineered into the peptide for cell adhesion are arginine and aspartate as can be reasoned based on the structure of the peptide provided, and the description of the roles of the various residues provided in the passage. It is a Scientific Reasoning and Problem Solving question because you are asked to reason using scientific principles and models.

Which of the following types of orbitals of the central atom are involved in bonding in octahedral compounds? A. sp B. sp3 C. p D. d2 sp3

Solution: The correct answer is D. This is a General Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is D because octahedral compounds have six σ bonds and no lone pairs. According to valence bond theory, the central atom requires the hybridization of six atomic orbitals, d2sp3. It is a Knowledge of Scientific Concepts and Principles question since you are asked to recognize the relationship between the closely related concepts of chemical bonding and hybridization.

The environment of the retinal binding site is most likely: A. hydrophilic. B. positively charged. C. negatively charged. D. hydrophobic.

Solution: The correct answer is D. This is a General Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is D because retinal is composed of mainly carbon and hydrogen, making it largely hydrophobic. It is a Scientific Reasoning and Problem Solving question because you are asked to reason using scientific principles.

By what factor is the proton concentration increased in the experiments shown in Figure 1B from those shown in Figure 1A? A. 2 B. 10 C. 200 D. 1000

Solution: The correct answer is D. This is a General Chemistry question that falls under the content category "Unique nature of water and its solutions." The answer to this question is D because the pH scale is logarithmic (pH = −log([H+]). A difference of 3 pH units corresponds to a 103 = 1000-fold difference in proton concentrations. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and then use a formula to solve a scientific problem.

What fraction of a 15O sample decays in 10 min? A. 1/8 B. 9/16 C. 3/4 D. 31/32

Solution: The correct answer is D. This is a Physics question that falls under the content category "Atoms, nuclear decay, electronic structure, and atomic chemical behavior." The answer to this question is D because the half-life of 15O is 2 minutes; thus, 10 minutes = 5 half-lives. Therefore, only (1/2)^5 = 1/32 of the sample will be left after 10 minutes, while 31/32 of the sample will decay. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use scientific formulas to solve problems.

What is the energy of the photons emitted by the LED at a frequency of 610 THz? (Note: h = 6.6 × 10-34J·s) A. 9.2 × 10-12 J B. 1.6 × 10-16 J C. 1.1 × 10-18 J D. 4.0 × 10-19 J

Solution: The correct answer is D. This is a Physics question that falls under the content category "How light and sound interact with matter." The answer to this question is D because the energy of a photon of frequency 610 THz is equal to 6.6 x 10^-34 J•s x 610 x 10^12 Hz = 4 x 10^-19 J. It is a Scientific Reasoning and Problem Solving question because you are asked to determine and use scientific formulas to solve problems.

Which of the following statements does NOT correctly describe the dehydration of malic acid to fumaric acid and maleic acid? A. The reaction occurs most readily with tertiary alcohols. B. The reaction involves the loss of a water molecule. C. The reaction has a carbocation intermediate. D. The reaction is stereospecific.

Solution: The correct answer is D. This is an Organic Chemistry question that falls under the content category "Nature of molecules and intermolecular interactions." The answer to this question is D because the fact that both fumaric and maleic acid are produced means that the dehydration of malic acid is NOT stereospecific. It is a Data-based and Statistical Reasoning question because you are asked to analyze and interpret data (the outcome of a particular organic chemical reaction) in order to draw a conclusion.

If 2-pentanol replaces 1-pentanol in the reaction shown in Figure 3, the rate of substitution is less because: A. the C-O bond in 2-pentanol is stronger than the C-O bond in 1-pentanol. B. there is a competing elimination reaction that slows the rate of substitution. C. there is more steric hindrance at the oxygen atom in 2-pentanol than in 1-pentanol, making protonation less likely. D. there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol.

Solution: The correct answer is D. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is D because the rate of substitution of protonated alcohols is subject to steric hindrance. This inhibits the ability of nucleophiles to collide with the reacting electrophilic center and slows the rate of reaction. It is a Scientific Reasoning and Problem Solving question because you are asked to evaluate an argument about causes and consequences (the reason a reaction proceeds more slowly when a different reactant is used).

In determining which reactant loses the -OH group, which of the following isotopic substitutions would be most useful? A. Replace the acidic H of acetic acid with D. B. Replace the alcoholic H of ethanol with D. C. Replace the carbonyl oxygen of acetic acid with O-18. D. Replace the hydroxyl oxygen of ethanol with O-18.

Solution: The correct answer is D. This is an Organic Chemistry question that falls under the content category "Structure, function, and reactivity of biologically-relevant molecules." The answer to this question is D because this experiment involves labeling a group which does not exchange with other groups present prior to reaction and will therefore give information about the true identity of the groups, which are exchanged during the reaction. It is a Reasoning about the Design and Execution of Research question because you are asked to reason about the appropriateness of particular research designs to evaluate a natural sciences question.

Which saturated fatty acid is the most soluble in water? A. CH3(CH2)10COOH B. CH3(CH2)12COOH C. CH3(CH2)14COOH D. CH3(CH2)16COOH

The solution is A. A) The fatty acid that is the most soluble in water will have the shortest alkyl chain. B) The fatty acid with the shortest alkyl chain will be the most soluble in water, and CH3(CH2)12COOH has 2 more carbons in its alkyl chain than CH3(CH2)10COOH . C) The fatty acid with the shortest alkyl chain will be the most soluble in water, and CH3(CH2)14COOH has 4 more carbons in its alkyl chain than CH3(CH2)10COOH. D) The fatty acid with the shortest alkyl chain will be the most soluble in water, and CH3(CH2)16COOH has 6 more carbons in its alkyl chain thanCH3(CH2)10COOH.

Suppose a defibrillator successfully returns a baby's heart to normal beating. Suppose further that 20 g of blood enters the heart at 25 cm/s and leaves 0.10 s later at 35 cm/s. What is the estimated average force on the 20 g of blood as it moves through the baby's heart? A. 0.020 N B. 0.20 N C. 20 N D. 2000 N

The solution is A. A) According to Newton's second law, the average force is equal to the mass of blood multiplied by the average acceleration of the blood. The average acceleration is (35 cm/s - 25 cm/s)/0.10 s = 100 cm/s2 = 1 m/s2. The average force is 20 g × 1 m/s2 = 0.020 kg × 1 m/s2 = 0.020 N. B) Either the mass is incorrectly used as 0.20 kg, or the average acceleration is incorrectly computed as 10 m/s2. C) Either the mass is incorrectly used as 2.0 kg, or the average acceleration is incorrectly computed as 100 m/s2. D) Either the mass is incorrectly used as 20 kg, or the average acceleration is incorrectly computed as 1000 m/s2.

Which forms of guanine and thymine are favored under physiological conditions? A. The keto form of guanine and the keto form of thymine B. The keto form of guanine and the enol form of thymine C. The enol form of guanine and the keto form of thymine D. The enol form of guanine and the enol form of thymine

The solution is A. A) Based on the pKa of the protons on the nitrogen atoms in the rings in Figure 2, the keto state is preferred for both guanine and thymine at physiological pH (7.2). The nitrogen atoms are protonated because pH 7.2 is approximately 2 pH units less than the pKa.​ B) Although it is true that the keto form of guanine is preferred because its pKa (9.2) is less than pH 7.2, the pKa of thymine is 9.7, which prevents the formation of the enol form. C) Although it is true that the keto form of thymine is preferred because its pKa (9.7) is less than pH 7.2, the pKa of guanine is 9.2, which prevents the formation of the enol form. D) At physiological pH (7.2), there would be no significant deprotonation of the nitrogen atoms in either the keto form of guanine or thymine, preventing the formation of the enol form.

According to the IUPAC, what is the systematic name for the hydrocarbon shown? A. Z-3-methylpent-2-ene B. E-3-methylpent-2-ene C. Z-3-ethylbut-2-ene D. E-3-ethylbut-2-ene

The solution is A. A) By IUPAC rules, first identify the longest unbroken chain of carbon atoms. Next, number the carbon atoms in this chain starting from the end that gives C=C the lowest numbers. The double bond is identified by the position of the carbon atom from the lowest numbered end (2), and then the methyl group is assigned at the 3-position. The stereochemical designator for the double bond is Z because the highest priority groups (methyl at C2 and ethyl at C3) occur on the same side of the double bond. The name is therefore Z-3-methylpent-2-ene. B) The stereochemical designator for the configuration of the double bond is Z, not E, since the highest priority groups occur on the same side of the double bond. C) The longest chain is five carbon atoms long, not four. D) The longest chain is five carbon atoms long, not four, and the stereochemical designator is Z, not E.

Compound 3 is prepared from Compound 2 (Figure 2) by: A. reduction of the ketone and lactonization of the gamma-hydroxyester. B. hydrolysis of one ester and formation of an acetal from the ketoacid. C. reduction of one ester and formation of an acetal from the gamma-hydroxyketone. D. reduction of one ester and the ketone followed by dehydration to a ketoether.

The solution is A. A) First, NaBH4 reduces the ketone to a secondary alcohol, the gamma-hydroxyester intermediate. Secondly, the alcohol group in this intermediate then reacts as a nucleophile with the carbonyl in the ethyl ester in the same molecule, forming a new ester by displacing C2H5OH as a leaving group. This cyclic ester is called a lactone, and the intramolecular transesterification yielding this lactone is called lactonization. B) NaBH4 is a reducing agent, not a hydrolyzing agent. Also, there is no acetal formed. Because an acetal has the structure R1O-CR2R3-OR4, where R1 and R4 = carbon groups and R2 and R3 = H and/or a carbon group, it may resemble a lactone (a cyclic ester), which also has two oxygen atoms attached to one carbon atom. C) Although NaBH4 is a reducing agent, it reduces aldehydes and ketones, not esters. Also, there is no acetal formed. Because an acetal has the structure R1O-CR2R3-OR4, where R1 and R4 = carbon groups and R2 and R3 = H and/or a carbon group, it may resemble a lactone (a cyclic ester), which also has two oxygen atoms attached to one carbon atom. D) Although NaBH4 is a reducing agent, it reduces aldehydes and ketones, not esters. Also, there is no acetal formed. Because an acetal has the structure R1O-CR2R3-OR4, where R1 and R4 = carbon groups and R2 and R3 = H and/or a carbon group, it may resemble a lactone (a cyclic ester), which also has two oxygen atoms attached to one carbon atom.

Based on the information in the passage, which description of an enzyme-substrate covalent intermediate is most likely correct? The substrate is covalently attached to: A. Asp14 through the phosphorus atom of the phosphate group. B. Asp14 through the oxygen atom of the phosphate group. C. Asp16 through the phosphorus atom of the phosphate group. D. Asp16 through the oxygen atom of the phosphate group.

The solution is A. A) The covalent intermediate will occur through the nucleophilic substitution by the side chain carboxyl of Asp14 at the electrophilic phosphorus atom in the substrate, displacing a leaving group. B) Oxygen in the substrate is not an electrophile because it has a partially negative charge and therefore cannot accept the nucleophilic carboxylate from Asp14. C) Incorrect. There is no P-O bond formation between Asp16 and phosphate. The side chain carboxylic acid of Asp16 acts as a general acid, not a nucleophile. D) There is no P-O bond formation between Asp16 and phosphate. The side chain carboxylic acid of Asp16acts as a general acid, not a nucleophile. Additionally, the oxygen in the substrate is not an electrophile because it has a partially negative charge.

Why was it important that the cuvettes containing the glucose oxidase and the blood sample were identical in terms of optical properties? A. To enable the comparison of the absorption spectra B. To reduce the absorption in the glass walls C. To decrease the uncertainty in the wavelength D. To increase the absorption in the solutions

The solution is A. A) The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to the presence of glucose in the blood and not due to the difference in the absorption features of the walls. B) Identical glass walls do not necessarily reduce absorption. In fact, it is possible that absorption is increased in some identical glass walls compared to other glass walls. C) While the glass walls have the same optical properties, this does not decrease the uncertainty in the wavelength. Wavelength uncertainty is related only to photon properties. D) Solution absorption is independent of the optical properties of the glass walls of the cuvettes.

Two vectors of magnitudes |A| = 8 units and |B| = 5 units make an angle that can vary from 0° to 180°. The magnitude of the resultant vector A + B CANNOT have the value of: A. 2 units. B. 5 units. C. 8 units. D. 12 units.

The solution is A. A) The magnitude of A + B is as small as 3 units (when A and B are anti-parallel and make an angle of 180°) and as large as 13 units (when A and B are parallel and make an angle of 0°). The magnitude of 2 units is smaller than the smallest possible magnitude of vector A + B. B) When A + B has 5 units, the angle between A and B is cos-1 (25 - 64 - 25)/(2 × 8 × 5) = 143°, thus 5 units is a possible magnitude. C) When A + B has 8 units, the angle between A and B is cos-1 (64 - 64 - 25)/(2 × 8 × 5) = 108°, thus 8 units is a possible magnitude. D) When A + B has 12 units, the angle between A and B is cos-1 (144 - 64 - 25)/(2 × 8 × 5) = 46°, thus 12 units is a possible magnitude.

What is the magnitude of the electric field in the electrical discharge produced in the excimer laser tube? A. 2.0 × 10^6 V/m B. 4.0 × 10^5 V/m C. 6.0 × 10^4 V/m D. 8.0 × 10^3 V/m

The solution is A. A) The magnitude of the electric field produced by 8.0 kV across 4.0 mm is 8000 V/0.004 m = 2.0 × 10^6 V/m. B) This magnitude implies that either the voltage is 1.6 kV or the distance is 20 mm. C) This magnitude implies the 8 kV voltage is applied over 15 cm. D) This magnitude implies the 8 kV voltage is applied over 1 m.

If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its approximate volume? A. 22.4 L B. 44.8 L C. 67.2 L D. 89.6 L

The solution is A. A) The quantity of Gas X was given as 1 mole. One mole of gas occupies 22.4 L at STP. B) Two moles of gas occupy 44.8 L at STP. C) Three moles of gas occupy 67.2 L at STP. D) Four moles of gas occupy 89.6 L at STP.

Compared to treatment with Compound 2 alone, which cell line shows the greatest enhancement of chemotherapeutic activity as a result of sensitization by Compound 1? A. Acute leukemia B. Fibrosarcoma C. Cervical carcinoma D. Noncancerous fibroblast

The solution is A. A) The ratio of the values of CC95 to CC95 Combination, gives the "fold enhancement" that is produced by using a combination of compounds 1 and 2 instead of just Compound 2 alone. Acute leukemia cells show a 74-fold enhancement in apoptotic activity by using the combination of compounds 1 and 2, while the other cancer cell lines show only a roughly 10-fold enhancement. B) Fibrosarcoma cells exhibit only a 10-fold enhancement in apoptotic activity using the combination of compounds 1 and 2 when compared to Compound 2 alone. C) Cervical carcinoma cells exhibit only a 10-fold enhancement in apoptosis using the combination of compounds 1 and 2 when compared to Compound 2 alone. D) Noncancerous fibroblast cells do not exhibit enhancement of apoptosis using the combination of compounds 1 and 2 when compared to Compound 2 alone.

Which conclusion can be drawn from the experimental results described in the passage? A. Compound 1 selectively sensitizes cancer cells to chemotherapeutic action by Compound 2. B. Compound 1 sensitizes all cells to chemotherapeutic action. C. Compound 2 induces apoptosis in cancer cells only in conjunction with treatment with Compound 1. D. Compound 2 inhibits the NF-κB signaling pathway, which leads to apoptosis.

The solution is A. A) The results shown in Figure 1 and Table 1 show that treatment of cancerous cell lines with Compound 1 sensitizes cancer cells to chemotherapeutic action as a result of exposure to Compound 2. B) Table 1 shows that noncancerous cells are not sensitized by Compound 1 to chemotherapeutic action of Compound 2. The CC95 value is 15 μM, but the CC95 Combination value is higher at 16 μM. C) Table 1 shows that Compound 2 effectively induces apoptosis, even in the absence of Compound 1. D) Compound 1 is the molecule that targets the NF-κB signaling pathway.

The UV-VIS spectrophotometer used by the researchers contained a detector that had low sensitivity and was unable to measure high absorbance samples. Which approach to the experiment makes the most sense with this limitation in mind? A. The experiments done at high pH should be diluted relative to those at low pH, or the path length of the high pH experiments should be decreased. B. The experiments done at low pH should be diluted relative to those at high pH, or the path length of the high pH experiments should be decreased. C. The experiments done at high pH should be diluted relative to those at low pH, or the path length of the low pH experiments should be decreased. D. The experiments done at low pH should be diluted relative to those at high pH, or the path length of the low pH experiments should be decreased.

The solution is A. A) To not have an absorbance reading that is below the limits of the detector, the experiments done at pH values that result in the highest absorptivity should be diluted. The alternative is to decrease the path length of the experiments that have the highest absorptivity. B) Although decreasing the path length of the experiments that have the highest absorptivity is reasonable, the experiments at low pH already have low absorbance relative to the high pH samples. Thus, there is no reason to dilute them. C) Although diluting the solutions in the experiments that have the highest absorptivity is reasonable, the experiments at low pH already have low absorbance relative to the high pH samples. Thus, there is no reason to decrease their path lengths. D) The experiments at low pH already have low absorbance relative to the high pH samples. Thus, there is no reason to dilute them or to decrease their path lengths.

What is the frequency of the pulses that deliver laser radiation to the cornea? A. 0.4 Hz B. 4.0 Hz C. 25 Hz D. 250 Hz

The solution is B. A) This frequency implies the pulse period is 2.5 s. B) The frequency is 1/(250 ms) = 4 Hz. C) This frequency implies the pulse period is 0.04 s. D) This frequency implies the pulse period is 4 ms.

Of the events listed, which occurs first during action potential generation? A. Voltage-gated sodium channels open at the axon hillock. B. Hyperpolarization stimulates the opening of ligand-gated potassium channels. C. Graded potentials propagate along the axon. D. Calcium influx stimulates vesicle fusion and release of neurotransmitter.

The solution is A. A) When threshold is met at the axon hillock, voltage-gated sodium channels open, generating an action potential. B) Voltage-gated (not ligand-gated) potassium channels open in response to depolarization, not hyperpolarization. C) Graded potentials occur in the cell body and dendrites, not the axon. D) Calcium influx and release of neurotransmitter does not occur until action potentials arrive at axon terminals.

A carbonyl group contains what type of bonding interaction(s) between the C and O atoms? A. One σ only B. One σ and one π only C. One π only D. One σ and two π only

The solution is B. A) A carbonyl group contains a C=O double bond, which involves two bonding interactions. B) A carbonyl group contains a C=O double bond. The first bonding interaction between atoms is always a σ bond. The second bond is formed from π symmetry orbitals. C) A carbonyl group contains a C=O double bond. The first bond is always a σ bond. D) A carbonyl group contains a C=O double bond. A combination of one σ bond and two π bonds gives a net triple bond, as is found in carbon monoxide or molecular nitrogen.

Which variant of DNA polymerase will most likely retain catalytic activity? A. D429A B. D429E C. D429K D. D429F

The solution is B. A) Alanine does not have a carboxylate side chain, essential to metal ion binding. B) Since the side chain carboxylate groups bind the metal ions, the only variant that would retain this function is D429E. C) Lysine does not have a carboxylate side chain, essential to metal ion binding. D) Phenylalanine does not have a carboxylate side chain, essential to metal ion binding.

Which of the following functional groups is present in pyrrolizidine (Figure 1)? A. An amide B. An amine C. An imine D. A carbamate

The solution is B. A) An amide has the structure R1(C=O)NR2R3, where R1 = H or a carbon group and R2 and R3 = H and/or a carbon group. Because pyrrolizidine contains a nitrogen atom, this answer may seem plausible, but there is no R1(C=O)NR2R3 present. B) Pyrrolizidine contains a nitrogen atom bonded to three noncarbonyl carbon atoms, which is an amine. An amine is defined as R3N, where R = H or a carbon group (but NOT C=O) and no more than two out of three R groups can be H. C) An imine has the structure R2C=NR, where R = a carbon group or H. Because pyrrolizidine contains a nitrogen atom, this answer may seem plausible, but there is no R2C=NR present. D) A carbamate has the structure R1O(C=O)NR2R3, where R1 = a carbon group and R2 and R3 = H and/or a carbon group. Because pyrrolizidine contains a nitrogen atom, this answer may seem plausible, but there is no R1O(C=O)NR2R3 present.

Roughly to what height would a 5 kg stone need to be raised in order to have the same stored energy as the energy stored in the defibrillator's capacitor? A. 4 m B. 8 m C. 16 m D. 32 m

The solution is B. A) At this height (4 m), the gravitational potential energy is 200 J, which is half the 400 J of energy stored in the defibrillator's capacitor. B) The gravitational potential energy is mass × gravitational acceleration × height. Equating 400 J = 5 kg × 9.8 m/s2 × height and solving for height leads to 400 J/(49 kg × m/s2) ≈ 8 m.​ C) At this height (16 m), the gravitational potential energy is 800 J, which is double the 400 J of energy stored in the defibrillator's capacitor. D) At this height (32 m), the gravitational potential energy is 1600 J, which is four times the 400 J of energy stored in the defibrillator's capacitor.

Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is: A. positive and less than TΔS. B. positive and greater than TΔS. C. negative and less than TΔS. D. negative and greater than TΔS.

The solution is B. A) Because ΔG = ΔH - TΔS > 0, the reaction does not occur. If 0 < ΔH < TΔS, then ΔH - TΔS < 0, the reaction would occur. The entropy term would be larger than the enthalpy term and favor the reaction. B) The reaction does not occur (is not spontaneous). This indicates that ΔG = ΔH - TΔS > 0. From inspection of the reaction, it can be concluded that ΔS > 0 (a gas evolves). Consequently, ΔH > TΔS explains why the reaction does not occur. C) Spontaneous reactions have ΔG = ΔH - TΔS < 0. From inspection of the reaction, it can be concluded that ΔS > 0 (a gas evolves). If ΔH < 0, then both the enthalpic and entropic terms would contribute to ΔG < 0. If ΔH is negative, then ΔG would also have to be negative, and the reaction would occur spontaneously. D) Spontaneous reactions have ΔG = ΔH - TΔS < 0. From inspection of the reaction, it can be concluded that ΔS > 0 (a gas evolves). If ΔH < 0, then both the enthalpic and entropic terms would contribute to ΔG < 0. If ΔH is negative, then ΔG would also have to be negative, and the reaction would occur spontaneously.

How many π-bonds do compounds 1 and 3 have? A. Both compounds 1 and 3 have 6 π-bonds. B. Compound 1 has 6 π-bonds, and Compound 3 has 7 π-bonds. C. Both compounds 1 and 3 have 7 π-bonds. D. Compound 1 has 5 π-bonds, and Compound 3 has 7 π-bonds.

The solution is B. A) Compound 3 has 7 π-bonds. B) Each double bond contains 1 π-bond and each triple bond contains 2 π-bonds. Therefore, Compound 1 has 6 π-bonds and Compound 3 has 7 π-bonds. C) Compound 1 has 6 π-bonds. D) Although Compound 3 has 7 π-bonds, Compound 1 has only 6 π-bonds.

The electric field inside each of the conductors that forms the capacitor in the defibrillator is zero. Which of the following reasons best explains why this is true? A. All of the electrons in the conductor are bound to atoms, and thus there is no way for an external electric field to penetrate atoms with no net charge. B. Free electrons in the conductor arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field. C. Free electrons in the conductor arrange themselves on the surface and throughout the interior so that the electric field they produce inside the conductor exactly cancels any external electric field. D. All electrons in the conductor, both free and bound, arrange themselves on the surface so that the electric field they produce inside the conductor exactly cancels any external electric field.

The solution is B. A) Conductors are characterized by the existence of free electrons that carry current. B) Conductors contain both atom-bound electrons and free electrons. Free electrons arrange themselves on the surface of conductors, and their collective electric field produced inside the conductor cancels any external electric field. The resulting electric field inside the conductor is zero. C) Free electrons arrange themselves only on the surface of the conductor. If they also arranged inside, the electric field inside the conductor would move the electrons even in the absence of a battery. D) Bound electrons cannot arrange themselves on the surface of the conductor due to the binding effects.

Which of the following atoms gains one electron most readily? A. Ar B. Br C. Co D. Na

The solution is B. A) Group 18 elements are relatively inert. They have a closed shell electronic configuration that inhibits them from gaining electrons easily. B) Atoms positioned to the right of the Periodic Table tend to gain electrons more readily than the metallic elements on the left, with the exception of Group 18. Bromine, Br, is a member of the Group 17 halogen family and gains an electron readily. C) Cobalt is a metallic transition metal and is more likely to lose an electron than it is to gain one. D) Sodium is a member of the Group 1 alkali metal family of elements and is much more likely to lose rather than gain an electron.

Assume that cellular uptake rates and drug delivery rates of compounds 1 and 2 are identical. If the proposed mechanism of sensitization by Compound 1 is correct, what cancer cell treatment protocol is most likely to produce the most apoptosis 20 h after treatment? A. Administration of both Compound 1 and Compound 2 simultaneously B. Administration of Compound 1 followed by Compound 2 after 0.5 h C. Administration of Compound 2 followed by Compound 1 after 0.5 h D. Administration of Compound 2 followed by Compound 1 after 1 d

The solution is B. A) If the chemotherapeutic agent is administered at the same time as the sensitization agent, there is good reason to believe that some cells will be exposed to Compound 2 prior to Compound 1 as a result of random probability. This will initiate the anti-apoptotic NF-κB signaling pathway in some cells and prevent cell death. B) The NF-κB signaling pathway is anti-apoptotic and is initiated by degradation and release of the inhibitor binding protein IκB. Once started, this process is not easily reversed. If the chemotherapeutic agent is administered at the same time as the sensitization agent, there is good reason to believe that some cells will be exposed to Compound 2 prior to Compound 1 as a result of random probability. The researchers found that administration of Compound 1 prior to Compound 2 by 0.5 h provided benefit. C) If the chemotherapeutic agent is administered before the sensitization agent, then some cells will be exposed to Compound 2 prior to Compound 1. This will initiate the anti-apoptotic NF-κB signaling pathway in some of those cells and prevent cell death. D) If the chemotherapeutic agent is administered well before the sensitization agent (by 1 whole day), then most cells will be exposed to Compound 2 prior to Compound 1. This will initiate the anti-apoptotic NF-κB signaling pathway in some of those cells and prevent cell death.

Based on the reaction scheme in Figure 1, what is the mechanism of substrate binding to RT? A. Random order B. Ordered C. Ping-pong D. Double-displacement

The solution is B. A) Random order would mean either substrate could bind first, and Figure 1 shows the TP substrate binds first and the dNTP substrate binds second. B) Figure 1 shows that the TP substrate binds first without any catalysis occurring and then the dNTP substrate binds. This is an ordered mechanism. C) In a ping-pong mechanism, no ternary complex is formed. However, the ternary complex RT/TP/dNTP does form. D) Double-displacement is another term for ping-pong mechanism, which means that no ternary complex is formed. However, the ternary complex RT/TP/dNTP does form.

Ca(OH)2 and CaCl2 in the reaction shown in Equation 1 are best described as compounds of: A. an alkali metal. B. an alkaline earth metal. C. a metalloid. D. a transition metal.

The solution is B. A) The alkali metal family of elements comprise Group 1 of the Periodic Table: Li, Na, K, Rb, and Cs. B) The alkaline earth metals comprise Group 2 of the Periodic Table. Calcium, Ca, is a member of this family found in period 4. C) Metalloid elements have properties in between metals and nonmetals. Examples of metalloid elements are Si, As, and Te. D) Transition metals comprise groups 3-12 of the Periodic Table.

Which of the following oxidative transformations is unlikely to occur? A. A primary alcohol to an aldehyde B. A tertiary alcohol to a ketone C. An aldehyde to a carboxylic acid D. A secondary alcohol to a ketone

The solution is B. A) The conversion of a primary alcohol to an aldehyde is readily accomplished. B) Oxidation of tertiary alcohols is difficult because it involves C-C bond breaking. C) The conversion of an aldehyde to a carboxylic acid is readily accomplished. D) The conversion of secondary alcohols to ketones is readily accomplished.

What is the effect produced by the PRK technique designed to correct nearsightedness? A. The density of the cornea is increased. B. The radius of curvature of the cornea is increased. C. The index of refraction of the cornea is increased. D. The thickness of the cornea at the apex is increased.

The solution is B. A) The density of the cornea is unchanged due to the PRK technique because it is an intrinsic property of the tissue. B) According to the passage, to correct nearsightedness, the laser beam is directed onto the central part of the cornea, resulting in a flattening of the cornea. This means that the radius of curvature of the cornea is increased. C) The refraction index of the cornea is unchanged due to the PRK technique because it is an intrinsic property of the tissue. D) The thickness of the cornea at the apex is decreased because the laser removes layers of tissue from the cornea at the apex.

Based on the information in the passage, compounds 1-3 are what type of inhibitor? A. Competitive B. Mixed C. Product D. Irreversible

The solution is B. A) The passage states that compounds 1-3 do not bind to the active site of RT. Therefore, they cannot be competitive inhibitors. B) The passage states that compounds 1-3 can bind to RT both before and after the substrate has bound. Also, the compounds do not exclude substrate binding. Therefore, mixed inhibition is most likely. C) Compounds 1-3 are not the products of the RT-catalyzed reaction. D) Based on equilibrium arrow notation, the binding of inhibitors shown in Figure 1 is reversible.

Which transition produces a photon with the longest wavelength? A. Transition 1 B. Transition 2 C. Transition 3 D. Transition 4

The solution is B. A) The wavelength that corresponds to Transition 1 is shorter than the wavelength that corresponds to Transition 2 because the energy level separation is larger than in Transition 2. B) The separation between the energy level equals the energy of the photon. The energy of the photon is inversely proportional to the wavelength. The longest wavelength corresponds to the smallest photon energy, which corresponds to the transition between the closest energy levels. C) The wavelength that corresponds to Transition 3 is shorter than the wavelength that corresponds to Transition 2 because the energy level separation is larger than in Transition 2. D) The wavelength that corresponds to Transition 4 is shorter than the wavelength that corresponds to Transition 2 because the energy level separation is larger than in Transition 2.

NH3 acts as a weak base in water with Kb = 1.76 × 10-5 at 25°C. The corresponding equilibrium is shown below. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) At 25°C, the equilibrium concentration of the NH4+ ion in a 10 M aqueous solution of NH3 would be closest to which of the following? A. 0.001 M B. 0.01 M C. 0.1 M D. 1 M

The solution is B. A) A concentration of 0.001 M is too low by a factor of 10. B) The concentration of NH4+ at equilibrium can be estimated using Kb and the equilibrium constant expression. Kb = 1.76 × 10-5 = [NH4+][OH-]/[NH3]. Plugging in 10 M for [NH3], which is a good approximation since very little NH3 ionizes, and noting that [NH4+] = [OH-], it is possible to solve for [NH4+]: 1.76 × 10-5 = [NH4+]2/10 → 1.76 × 10-4 = [NH4+]2 →1.32 × 10-2 = [NH4+]. C) A concentration of 0.1 M is too high by a factor of 10. D) A concentration of 1 M is too high by a factor of 100.

What is the concentration of the acute leukemia CC95 Combination value in micromolar (μM) units? A. 6 × 10^-10 μM B. 6 × 10^-7 μM C. 6 × 10^- 4 μM D. 6 × 10^-2 μM

The solution is C. A) 6 × 10^-10 μM is 6 × 10^-16 M or 6 × 10^-7 nM. B) 6 × 10^-7 μM is 6 × 10^-13 M or 6 × 10^-4 nM. C) The CC95 Combination value for acute leukemia cells was 0.6 nM, or 0.6 × 10^-9 M. This corresponds to 0.6 × 10^-9 M × (1 μM/1 × 10^-6 M) = 0.6 × 10^-3 μM = 6 × 10^-4 μM. D) 6 × 10^-2 μM is 6 × 10^-8 M or 60 nM.

The equation for calculating mean arterial pressure (MAP) is shown. MAP = (2 × diastolic) + systolic / 3 What is the MAP of a person whose blood pressure is 135/90 mmHg? A. 85 mmHg B. 90 mmHg C. 105 mmHg D. 110 mmHg

The solution is C. A) 85 mmHg is below both the diastolic and systolic pressures, and therefore cannot be the mean arterial pressure. B) 90 mmHg is only the diastolic pressure. C) The blood pressure reading shows that the diastolic pressure is 90 mmHg and the systolic pressure is 135 mmHg. The calculation is [(2 × 90) + 135]/3, which is 315/3 = 105 mmHg. D) 110 mmHg is higher than the mean arterial pressure.

What is the [H3O+] in the solution used in the experiment done at the lowest pH? A. 0.01 μM B. 0.1 μM C. 1.0 μM D. 100 μM

The solution is C. A) A [H3O+] of 0.01 μM is the same as 0.01 × 10-6 M or 1 × 10-8 M. The pH would therefore be -log(1 × 10-8) = 8. B) A [H3O+] of 0.1 μM is the same as 0.1 × 10-6 M or 1 × 10-7 M. The pH would therefore be -log(1 × 10-7) = 7. C) The lowest pH is 6. Because pH = -log([H3O+]), [H3O+] = 1.0 × 10-6 M or 1.0 μM. D) A [H3O+] of 100 μM is the same as 100 × 10-6 M or 1 × 10-4 M. The pH would therefore be -log(1 × 10-4) = 4.

The intermolecular forces that exist among the molecules of NH3 gas are: A. dipole-dipole forces only. B. London dispersion forces only. C. both dipole-dipole and London dispersion forces. D. neither dipole-dipole nor London dispersion forces

The solution is C. A) All "real" molecules and atoms will exhibit London dispersion forces. It is not possible to exhibit dipole-dipole forces only. B) Strictly nonpolar molecules will exhibit London dispersion forces. Since NH3 has a permanent dipole, it will also exhibit dipole-dipole forces of attraction. C) Since NH3 is a permanent dipole, it will exhibit dipole-dipole intermolecular forces in addition to the London dispersion forces exhibited by all molecules. D) All atoms and molecules exhibit London dispersion attractive forces. As a molecule with a permanent dipole, NH3 will also exhibit dipole-dipole forces.

How many hydrogen bonds are present in the two types of G•T mismatches discussed in the passage? A. 2 in the wobble pair, 2 in the enol T mismatch B. 3 in the wobble pair, 2 in the enol T mismatch C. 2 in the wobble pair, 3 in the enol T mismatch D. 3 in the wobble pair, 3 in the enol T mismatch

The solution is C. A) Although the wobble base pair would appear as an offset A•T pair and has two hydrogen bonds, the enol T mismatch mimics the G•C base pair and therefore has three hydrogen bonds. B) The wobble base pair would appear as an offset A•T pair and has two hydrogen bonds. The enol T mismatch mimics the G•C base pair and therefore has three hydrogen bonds. C) Since the wobble base pair appears as an offset A•T pair, it has two hydrogen bonds. The G•C mimic mismatch would have three hydrogen bonds. This can be directly inferred from it mimicking the G•C pair or by examination of the enol T structure. D) Although the enol T mismatch mimics the G•C base pair and therefore has three hydrogen bonds, the wobble base pair appears as an offset A•T pair and would have only two hydrogen bonds.

In which of the following steps is the carbamate amide group cleaved? A. Step 2 B. Step 3 C. Step 4 D. Step 6

The solution is C. A) Both the reactant Compound 1 and the product Compound 2 retain the carbamate R2NCO2C2H5. B) Both the reactant Compound 2 and the product Compound 3 retain the carbamate R2NCO2C2H5. C) The carbamate R2NCO2C2H5 in Compound 3 is cleaved by hydroxide to a secondary amine. D) There is no carbamate reactant in Step 6. However, the ester R2NCH2CO2C2H5 in Compound 5 resembles the carbamate R2NCO2C2H5 in Compound 3.

Given that Tris has a pKa of 8.07, for how many of the experiments would Tris have been an acceptable buffer? A. None of the experiments B. Only 1 of the experiments C. Only 2 of the experiments D. All three of the experiments

The solution is C. A) For this option to be true, all of the experiments would have to have pH values outside of the range of the Tris pKa ±1 pH unit. B) In addition to the experiment at pH 7.5, there is another experiment at a pH within the range of Tris pKa±1 pH unit. C) A buffer has a buffering capacity that is ±1 pH unit away from the pKa, which means that only two of the experiments would have used Tris. D) One of the experiments (at pH 6.0) falls outside of the pH range of the Tris pKa ±1 pH unit.

When limestone is heated during Step 1, an equilibrium is established. Which of the following expressions is the equilibrium constant for the decomposition of limestone? A. [CaO] B. [CaCO3] C. [CO2] D. [CaO] × [CaCO3]

The solution is C. A) From the Law of Mass Action, an equilibrium constant expression involves a ratio of products to reactants with exponents determined from the stoichiometry of the reaction. Solids are excluded from equilibrium constant expressions. Although CaO(s) is a product, it is a solid and will not appear in the expression for Keq. B) From the Law of Mass Action, an equilibrium constant expression involves a ratio of products to reactants with exponents determined from the stoichiometry of the reaction. Solids are excluded from equilibrium constant expressions. CaCO3(s) is a reactant and would therefore appear in the denominator of the equilibrium constant expression. However, as a solid, it is excluded from the expression entirely. C) From the Law of Mass Action, an equilibrium constant expression involves a ratio of products to reactants with exponents determined from the stoichiometry of the reaction. Furthermore, solids are excluded from equilibrium constant expressions. CO2(g), as the only non-solid material in the reaction, is the only substance that appears in the equilibrium constant expression. The exponent for [CO2] is 1 because the stoichiometric coefficient that appears in the presented balanced reaction is 1. D) From the Law of Mass Action, an equilibrium constant expression involves a ratio of products to reactants with exponents determined from the stoichiometry of the reaction. Solids are excluded from equilibrium constant expressions. Taking the product [CaO] × [CaCO3] is inappropriate for two reasons. Both substances are solids and will not appear in the expression for Keq. Furthermore, as a reactant, CaCO3should appear in the denominator of the expression (if it were a dissolved solute).

The Na2CO3 used in Step 5 of Figure 3 is necessary to: A. dissolve the bromoester. B. remove an α-hydrogen from the bromoester. C. convert the ammonium salt in Compound 4 to the amine. D. replace the chloride ion in Compound 4 with the carbonate ion.

The solution is C. A) Na2CO3 is a Brønsted base, not a solvent. B) In the SN2 reaction with the bromoester, only the bromide leaving group is displaced. There is no loss of an α-proton. C) Na2CO3 acts as a Brønsted base. Until a proton is removed from the alkylammonium salt in Compound 4, it does not have the lone pair needed to be a nucleophile. As a nucleophile, Compound 4 performs the SN2 reaction that furnishes Compound 5. D) Although there is probably exchange of Cl- with CO32-, Na2CO3 acts as a base in this step.

If a thin thread is placed between a screen and a bright source of light, a pattern of parallel dark and bright fringes appears on the screen. The phenomenon best explaining the formation of this pattern is: A. refraction. B. polarization. C. diffraction. D. reflection.

The solution is C. A) Refraction occurs when light passes through the interface between optical media with different indices of refraction. There is no change in optical media when light passes by the thread. B) Polarization is not caused by the thread because the thread has no optical properties that can preferentially select a certain oscillation direction of the electric field of light. C) The thin thread disrupts the propagation of light by impeding light to pass through it. Diffraction causes the initially plane-parallel wave-fronts of light to change direction and partially enter the shadow region behind the thread due to its narrowness. The overlapping of different wave-fronts on the screen causes the pattern of dark and bright fringes on the screen according to the phase difference between the wave-fronts that interfere there. D) Reflection occurs when light bounces back at the interface between optical media with different indices of refraction. There is no change in optical media when light passes by the thread.

What atom remains after the β- decay described in the passage? A. Si B. Al C. S D. Cl

The solution is C. A) Si can, in theory, be produced from P by a β+ decay. Si has one less proton in its nucleus than P. The passage indicates that the process was a β- decay. β- decay increases the number of protons in the nucleus. B) Al can be produced from phosphorous by α decay. Al has two fewer protons in the nucleus than P. C) 32P undergoes the β- decay according to the reaction 3215P → 3216S + e- + energy. D) Cl has two more protons in the nucleus than P. A Cl nuclide could be produced from P by two successive β- decays. Only one is indicated in the passage.

The specific heat of liquid water is 4.2 kJ/°C•kg. How much energy does it take to heat 2 kg of water from 20°C to 70°C? A. 220 kJ B. 320 kJ C. 420 kJ D. 520 kJ

The solution is C. A) Supplying 220 kJ of heat will raise the temperature of 2 kg by only 26°C to a final temperature of 46°C. B) Supplying 320 kJ of heat will raise the temperature of 2 kg by only 38°C to a final temperature of 58°C. C) The energy to raise the temperature of 2 kg of water 50°C (70°C - 20°C) can be calculated using dimensional analysis: 2 kg × 4.2 kJ/°C•kg × 50°C = 420 kJ. D) Supplying 520 kJ of heat will raise the temperature of 2 kg by 62°C to a final temperature of 88°C.

During Reaction 2, did the oxidation state of N change? A. Yes; it changed from -3 to -4. B. Yes; it changed from 0 to +1. C. No; it remained at -3. D. No; it remained at +1.

The solution is C. A) The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H+ → NH4+). Acid-base reactions do not involve oxidation state changes. B) The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H+ → NH4+). Acid-base reactions do not involve oxidation state changes. Furthermore, the oxidation state of N in NH3 is -3, not 0. C) The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H+ →NH4+). Acid-base reactions do not involve oxidation state changes. The oxidation state of N in NH3 is -3. Each H is +1 and is balanced by the -3 of N to make a neutral compound. The oxidation state of N does not change when the N is protonated. D) While acid-base reactions do not involve oxidation state changes, the starting oxidation state for N is -3, not +1.

One company sells a defibrillator for home use that uses a 9-volt DC battery. The battery is rated at 4.2 A•hr (amp•hour). Roughly how much charge can the battery deliver? A. 4.2 C B. 38 C C. 15,000 C D. 136,000 C

The solution is C. A) This is the charge delivered in one second based on the calculation 4.2 A × 1 s = 4.2 C. B) The magnitude is computed as an approximation of 4.2 × 9 = 37.8 ≈ 38. However, multiplying the corresponding units V × A•hr yields J, not C. C) The definition of current is flow of charge per unit time. Thus, charge equals current multiplied by time, hence 4.2 A × 1 hr = 4.2 A × 3600 s = 15,120 C ≈ 15,000 C. D) The magnitude is computed as an approximation of 4.2 × 9 × 3600 = 136,080 ≈ 136,000. However, multiplying the corresponding units V × A × s yields J, not C.

Based on the data in Table 1, what are the effects of increasing pH on catalytic efficiency and the maximum velocity of the reaction? A. Both the catalytic efficiency and the maximum velocity of the reaction increase. B. The catalytic efficiency increases with pH, but the maximum velocity of the reaction decreases. C. The catalytic efficiency decreases with pH, but the maximum velocity of the reaction increases. D. Both the catalytic efficiency and the maximum velocity of the reaction decrease.

The solution is D. A) As pH increases, both kcat and kcat/KM decrease. This is true because Vmax is directly proportional to kcat. Thus, when both Vmax and Vmax/KM are decreasing, correspondingly, kcat and kcat/KM are also decreasing. B) Although it is true that kcat decreases with increasing pH, so does kcat/KM. This is true because Vmax is directly proportional to kcat. Thus, when both Vmax and Vmax/KM are decreasing, correspondingly, kcat and kcat/KM are also decreasing. C)Although it is true that kcat/KM decreases with increasing pH, so does kcat. This is true because Vmax is directly proportional to kcat. Thus, when both Vmax and Vmax/KM are decreasing, correspondingly, kcat and kcat/KM are also decreasing. D) As pH increases, the kcat decreases because it is proportional to the Vmax. Also, the catalytic efficiency (kcat/KM) decreases because Vmax/KM is decreasing.​

Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-chlorobutane and 1-butanol by fractional distillation? A. Both 1-chlorobutane and 1-butanol are polar. B. Both 1-chlorobutane and 1-butanol are nonpolar. C. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol. D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol.

The solution is D. A) Both 1-chlorobutane and 1-butanol are polar molecules with dipole moments that are greater than 1.5 D. B) Both 1-chlorobutane and 1-butanol are polar, not nonpolar, molecules. C) The boiling point of 1-chlorobutane is 78°C. The boiling point of 1-butanol is 118°C. D) The fact that 1-chlorobutane will have a boiling point that is substantially lower than that of 1-butanol can be rationalized from chemical principles. The molecules have similar molecular weights, but 1-butanol has a hydroxyl functional group that can participate in hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.

If methane (CH4) were substituted for NH3 in the aluminum can, the crushing of the can would: A. occur because CH4, being polar, would dissolve in the water in the tray. B. occur because CH4, being nonpolar, would dissolve in the water in the tray. C. not occur because CH4, being polar, would not dissolve in the water in the tray. D. not occur because CH4, being nonpolar, would not dissolve in the water in the tray.

The solution is D. A) Can crushing will not occur. CH4 is a nonpolar molecule and will not dissolve in the water in the tray. B) Can crushing will not occur. While it is true that CH4 is a nonpolar molecule, this means that it will not dissolve in the water. C) CH4 is a nonpolar molecule. Polar molecules dissolve in water. D) Can crushing will not occur. CH4 is a nonpolar molecule and will not dissolve appreciably in water. Since the CH4 will remain in the gas phase, the can will not be crushed by the formation of a vacuum inside the can.​

Based on the passage, what is the primary type of interaction that RT makes with Compound 2? A. Covalent B. Hydrogen bonds C. Ionic D. Hydrophobic

The solution is D. A) Figure 1 shows that the interactions of the inhibitors with RT are reversible. Covalent interactions are usually irreversible. Also, the side chains of valine and phenylalanine cannot spontaneously form covalent bonds with the inhibitors. B) The passage states that the strongest RT interactions with the inhibitors come from the side chains of valine and phenylalanine. These side chains cannot form hydrogen bonds. C) The passage states that the strongest RT interactions with the inhibitors come from the side chains of valine and phenylalanine. These side chains cannot form ionic bonds. D) The passage states that the strongest RT interactions with the inhibitors come from the side chains of valine and phenylalanine. These side chains are hydrophobic, so they would make hydrophobic interactions with the inhibitor.

What allows small insects to walk on the free surface of water? A. Hydrostatic pressure B. Viscosity C. Buoyancy D. Surface tension

The solution is D. A) Hydrostatic pressure applies to objects that are at least partially submerged. B) Viscosity prevents the relative motion between water and objects in contact with water. C) Buoyancy applies to objects that are at least partially submerged. D) Small insects walk on the free surface of water because, at the air-water interface, all hydrogen bonds in water face downward, causing the molecules of the water surface to bond together. Due to cohesion, the polar water molecules do not cling to nonpolar molecules (such as oils) like those found in the insect feet. The insects do not sink through the free surface of water provided their weight is smaller than the surface tension.

Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the experiment. What modification will provide a more precise reading by data interpolation as opposed to extrapolation using the same standards? A. Increase the enzyme concentration. B. Increase the oxygen pressure. C. Decrease the content of oxygen acceptor. D. Dilute the sample with additional solvent.

The solution is D. A) Increasing the enzyme concentration will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation. B) Increasing the oxygen pressure will increase the quantity of absorber (chromogen) in solution, according to the protocol described in the passage. This will make the resulting absorbance value even further from the values in Table 1 and require additional extrapolation. C) The oxygen acceptor is glucose. Removing glucose from the solution is not feasible and defies the logic of the experiment, which involves quantifying the amount of glucose present. One would need to know exactly how much glucose was removed, and this would require a separate measurement. D) By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance will fall within the range of the standards. This is easily accomplished, and the resulting calculations that account for the dilution are not difficult.

Point A in this diagram is best described as which of the following? A. Boiling point B. Critical point C. Melting point D. Sublimation point

The solution is D. A) The boiling point is between Point B and Point C on the curve that separates the liquid state from the gas state. B) Point C is the critical point. C) The melting point is on the curve that separates the solid state from the liquid state. D) Point A is on the curve that separates the solid state from the gas state. Sublimation is the phase change from a solid to a gas.

Which of the following compounds is most likely the intermediate product formed in Step 4 of Figure 2 if four equivalents of base are used?

The solution is D. A) The lactone will also be hydrolyzed by -OH. In fact, it is more reactive than a carbamate and will hydrolyze first. B) The lactone is an ester and therefore undergoes saponification with -OH. Additionally, not only will -OH hydrolyze the alkyl carbamate to the carbamate salt, this salt is unstable and will decarboxylate in water to form CO2 and a secondary amine. C) This product would form with only one equivalent of hydroxide. The additional hydroxide would also hydrolyze the carbamate. D) When four equivalents of base (-OH) are used, one -OH will saponify the lactone to form the secondary alcohol and carboxylate, a second -OH will hydrolyze the carbamate, with the remaining -OH as excess.

An unknown amount of a radioactive isotope with a half-life of 2.0 h was observed for 6.0 h. If the amount of the isotope remaining after 6.0 h was 24 g, what would the original amount have been? A. 3 g B. 4 g C. 144 g D. 192 g

The solution is D. A) The original amount of radioactive isotope cannot be 3 g because this amount is less than the 24 g that is still radioactive after 6 h. B) The original amount of radioactive isotope cannot be 4 g because this amount is less than the 24 g that is still radioactive after 6 h. C) Because 24 g/144 g = 1/6 ≈ (1/2)2.6, the lapsed time would have been about 2.6 half-lives, meaning approximately 5.2 h, instead of the 6 h indicated. D) Given the half-life T = 2 h, the duration t = 6 h represents 3 half-lives because 2 h × 3 = 6 h. According to the radioactive decay law N(t) = N0(1/2)t/T, the amount of remaining isotope is (1/2)3 = 1/8 of the original amount. The original amount was then 8 × 24 g = 192 g.

The use of pulsed laser radiation in the PRK procedure, as opposed to continuous laser radiation, allows the cornea to: A. absorb more radiation. B. change its index of refraction. C. increase the area exposed to radiation. D. maintain a lower average temperature.

The solution is D. A) The pulsed laser radiation allows the cornea to absorb less radiation than from a continuous laser because the time allowed for the laser-cornea interaction is shorter than for a continuous laser. B) The index of refraction does not change due to either a pulsed laser or a continuous laser, because it is an intrinsic property of the tissue material. C) The area exposed to radiation is controlled with the opaque disk. D) The pulsed laser radiation interacts with the cornea for a shorter time than a continuous laser radiation, thus less heat is transferred to the cornea. This allows the cornea to maintain a lower average temperature by cooling off between pulses.

According to Table 1, what is the concentration of the glucose in the blood from which the diluted sample was taken? A. 60 mg/dL B. 90 mg/dL C. 120 mg/dL D. 150 mg/dL

The solution is D. A) This concentration corresponds to an absorbance of 0.08 in Table 1. B) This concentration corresponds to an absorbance of 0.12 in Table 1. C) This concentration corresponds to an absorbance of 0.16 in Table 1. D) From Table 1, the glucose concentration in the diluted sample is (0.20/0.24) × 6.0 mg/dL = 5.0 mg/dL. The blood then has a glucose concentration of 30 × 5.0 mg/dL = 150 mg/dL.

The ionic form of which metal atom is NOT likely to be found in the pocket of a catalytically active DNA polymerase? A. Zn B. Fe C. Mg D. Na

The solution is D. A) Zinc forms the divalent cation Zn2+. As shown in Figure 1, a divalent cation is essential for DNA polymerase catalytic activity. B) Iron forms the divalent cation Fe2+. As shown in Figure 1, a divalent cation is essential for DNA polymerase catalytic activity. C) Magnesium forms the divalent cation Mg2+. As shown in Figure 1, a divalent cation is essential for DNA polymerase catalytic activity. D) Sodium does not readily form a divalent cation, essential for DNA polymerase catalytic activity as shown in Figure 1.