EOC biochem

Question 4 Which of the following is a primary difference between microtubules and actin filaments? A. Only actin filaments are composed of monomeric subunits. B. Only microtubules are part of epithelial cells' cytoskeletons. C. Only actin filaments are assembled from nucleotide-bound subunits. D. Only microtubules have intrinsic structural polarity.

A is correct. Actin filaments are assembled from monomeric units of actin. However, microtubule lattices are assembled from dimeric tubulin-heterodimers.B: Both actin and microtubules are components in the cytoskeleton of cells, including epithelial cells.D: Both actin and microtubules have intrinsic structural polarity because they are assembled from asymmetric subunits. Microtubules are made up of heterodimers of 2 kinds of tubulin, while actin is comprised of a monomeric globular protein with an asymmetric shape.C: The synthesis of microtubules and actin require that the subunits being added have a bound nucleoside triphosphate. GTP is used for tubulin, while ATP is used for actin

Question 11 Which type of inhibitor binds only to the enzyme-substrate complex, and how does this form of inhibition affect the kinetics of the enzyme in question? A. Uncompetitive; both Km and Vmax are reduced. B. Competitive; Km is reduced, while Vmax is not changed. C. Noncompetitive; Vmax is reduced, while Km is not changed. D. Uncompetitive; Vmax is reduced, while Km is not changed.

A is correct. An uncompetitive inhibitor is an inhibitor of the enzyme-substrate (E-S) complex. This reduces the ability for the enzyme to catalyze the reaction, lowering Vmax. Simultaneously, however, it seemingly increases the enzyme's affinity for its substrate, as E-S complexes are subject to becoming enzyme-substrate-inhibitor complexes, effectively removing E-S complexes from solution and thereby driving the reaction equilibrium towards the formation of new E-S complexes. Thereby uncompetitive inhibition also reduces Km. Also note that answer choice B gives the wrong relationship between competitive inhibition and Km/Vmax. In competitive inhibition Km is increased.

Question 15 Grapefruit juice is often listed as a contraindication for individuals on statins. This is the result of multiple studies that have shown that components of grapefruit juice covalently bind one or more enzymes involved in statin metabolism. Which of the following best describes this supposed "grapefruit juice effect"? A. Components of grapefruit juice inhibit the enzyme(s), thus increasing the bioavailability of the statin. B. Components of grapefruit juice inhibit the enzyme(s), thus decreasing the bioavailability of the statin. C. Components of grapefruit juice activate the enzyme(s), thus increasing the bioavailability of the statin. D. Components of grapefruit juice cleave the enzyme(s) from their precursors, which activates them.

A is correct. Bergamottin, a component in grapefruit juice, covalently binds enzyme CYP3A4, which is responsible for metabolizing certain statins. This inhibits the enzyme and increases the systemic concentration of the statin, which can be dangerous. Choices B and C do not make sense, as the enzyme is described as playing a role in statin metabolism (breakdown). Thus, inhibiting the enzyme would increase statin bioavailability, while activating it would cause the statin to be broken down more rapidly and exhibit lower bioavailability. Nowhere is it indicated that the enzyme in question has a zymogen precursor, nor is it stated that the components of grapefruit juice lead to any cleavage.

Question 4 Based on the data in Figure 3, which of the following could be the residue found farthest from the cytoplasmic environment? A. C234 B. G387 C. M551 D. D703

A is correct. Farthest from the cytoplasmic environment means the amino acid which is the most buried beneath the surface of the protein. Examining Figure 3, we see that the residue with the greatest degree of burial is represented by the light-colored circles, and peaks around 80% burial for its carbonyl carbon. The figure shows that this amino acid has just one carbon in its side chain (Cß), followed by one more atom in the γ position (the α-C is adjacent to the carbonyl C, and is present on all amino acids). This means we need an amino acid with a 2-atom side chain. Only cysteine (C) fits this description. B: Glycine has no side chain attached to its α-carbon and has no signal beyond that position. C: Methionine has four linear atoms past its α-carbon, which makes it the longest of the depicted amino acids. D: Aspartic acid has a two-carbon carboxylic acid attached to its α-carbon, so it has up to a δ signal.

Histone deacetylase is an enzyme that removes acetyl groups from histone proteins. Which of the following is the most likely effect of the activation of histone deacetylase? A. The rate of transcription of the affected DNA will decrease. B. The affected histones will become less positively charged. C. The affected histones will become more negatively charged. D. DNA-histone attractions will become weaker.

A is correct. Histones are typically positively charged, which attracts them to the negatively-charged DNA backbone. The addition of acetyl groups, however, removes the positive charge on lysine residues. Here, the opposite is happening; acetyl groups are being removed by the deacetylase enzyme, causing the histone to become more positively charged (eliminate options B and C). This causes DNA to be more tightly attracted to the histone (eliminate choice D), limiting the ability of transcription factors to access the DNA. This will decrease the rate at which the DNA is transcribed (choice A is correct).

Question 5 Which of the following, if true, would most likely invalidate the results of the experiment? A. Participants' tissue samples were collected without fasting having taken place. B. Different cancer subtypes have different levels of glucose transporter expressivity. C. Different numbers of participants were enrolled with CRAs versus NETS. D. Previous studies have shown no significant correlation between proliferation index and GLUT1 activity.

A is correct. If participants did not fast before data collection occurred, each of their cells would have taken up different levels of glucose prior to measurement of GLUT and HK activity. If researchers were unable to differentiate between the extent to which GLUT and HK activity was correlated with participants' snacking versus the cancerous phenotype, the results of the experiment would certainly be called into question. B, C, D: None of these factors would invalidate the measures carried out in the experiment

Question 4 Experiments showed that protein unfolding in the presence of the denaturant proceeds in a unimolecular fashion, with rate constant, k. What are the units for k? A. s-1 B. M•s C. M-1•s-1 D. M•s-1

A is correct. If polypeptide unfolding proceeds via a unimolecular mechanism, the rate law for the reaction must be first order with respect to only the polypeptide and first order overall. The rate law for such a reaction is given by the expression: R = k [polypeptide] where R represents the rate of the reaction, given in units of M/s. Rearranging the expression to solve for k, you find k = R/[polypeptide]. B, C, D: In terms of unit, k = ([M]/[s])/[M] = 1/s (choice A). Choice B is incorrect because all rate constants include an inverse time term. Choice C gives the units of a second order rate constant. Choice D gives the units for a zeroth order rate constant, the units of an overall rate expression.

Question 9 In biological systems, glucose exists in equilibrium between two: A. anomers. B. structural isomers. C. enantiomers. D. polymers.

A is correct. In aqueous solution, glucose readily undergoes mutarotation between its α and β anomers. By definition, anomers are a specific category of epimers, which are stereoisomers that differ at only one chiral center. These are not structural isomers because they have the same formula and connectivity (eliminate B). Enantiomers would require the α and β anomers to be mirror images, which is not the case (eliminate choice C). Finally, they are not polymers because polymers are larger molecules consisting of repeated subunits (eliminate choice D).

Question 14 A cation-exchange chromatography apparatus: A. contains a negatively-charged stationary phase, causing anions to elute faster than cations. B. contains a negatively-charged stationary phase, causing cations to elute faster than anions. C. contains a positively-charged stationary phase, causing anions to elute faster than cations. D. contains a positively-charged stationary phase, causing cations to elute faster than anions.

A is correct. In cation-exchange chromatography, the stationary phase is designed to attract cations (positively-charged molecules). As such, the phase itself is anionic, or negatively- charged. Cations are attracted to this phase and are retained in the column, while anions (negatively-charged particles) elute through the column quickly.

Question 9 The first step of glycolysis involves the phosphorylation of glucose by the enzyme hexokinase. This step requires the presence of magnesium cations. In this context, magnesium acts as a: A. cofactor. B. coenzyme. C. prosthetic group. D. apoenzyme.

A is correct. Magnesium is a necessary cofactor for hexokinase. In fact, metal ions are very common examples of cofactors. The term "coenzyme" specifically refers to organic cofactors, and magnesium is inorganic (eliminate choice B). Prosthetic groups are cofactors that are tightly bonded to their enzymes, which is not the case here (eliminate choice C). Finally, "apoenzyme" refers to an enzyme without its required cofactor(s), making choice D incorrect as well.

Question 4 The terminal end of type M laminarans could be synthesized through which of the following mechanisms? A. Reduction of mannose B. Oxidation of mannose C. Oxidation of mannuronic acid D. Decarboxylation of mannuronic acid

A is correct. Mannitol is a sugar alcohol that can be formed by reducing the C=O group in mannose. B. Further oxidation of the terminal C=O of mannose would yield a carboxylic acid, not an alcohol. C: Carboxylic acids cannot be further oxidized. D: Decarboxylation of mannuronic acid involve the loss of a carbon and could not yield mannitol, as the information in Figures 1 and 2 confirm that both mannitol and mannuronic acid are 6-carbon compounds.

Kinesins are motor proteins that transport vesicles and organelles by traveling along: A. microtubules, and require the hydrolysis of ATP. B. microtubules without the need for ATP. C. microfilaments, and require the hydrolysis of ATP. D. microfilaments without the need for ATP.

A is correct. Motor proteins, which include dynein and kinesin, hydrolyze ATP to slide along microtubules with their bound cargo. Microfilaments form an intricate network that gives structure to the cell membrane and facilitates amoeboid movements, but they are not involved in vesicle traffic.

An enzyme catalyzing which of the following reactions would decrease the solubility of laminarans? A. Cleavage of β-1,6-glycosidic bonds B. Cleavage of β-1,3-glycosidic bonds C. Formation of β-1,3-glycosidic bonds D. Formation of β-1,6-glycosidic bonds

A is correct. Paragraph 2 of the passage states that the degree of branching is correlated with solubility and that branches in laminarans are formed by single β-D-glucose residues attached to C-6 of the β-D-glucose residues in the linear chain. In other words, β-1,6-glycosidic bonds form branches. Therefore, an enzyme that cleaves β-1,6-glycosidic bonds will remove these branches and decrease solubility. B, C: β-1,3-glycosidic bonds form the links in the linear chain of laminarans, not the branches. D: Forming more β-1,6-glycosidic bonds would promote solubility.

Question 13 The structure of glyceraldehyde, an important metabolic intermediate, is shown below: What is the absolute stereochemical configuration of the glyceraldehyde stereoisomer depicted above? A. (R) B. (S) C. (+) D. (-)

A is correct. The convention for drawing Fischer projections dictates that substituents branching horizontally from the vertical backbone of the molecule are directed out of the plane of the paper, toward the reader. Therefore, the hydrogen and oxygen atoms bonded to the chiral C-2 carbon atom are directed outward. The hydroxyl group has the highest priority, and the aldehyde carbon has the second highest priority. Moving around the molecule from highest priority downward, we travel counterclockwise, which typically denotes an S configuration. However, since our fourth-priority group (the hydrogen atom) points toward us, we must reverse this configuration. Our final answer should be R.

Question 9 By the end of the investment phase of glycolysis, each molecule of glucose has been converted to: A. two molecules of glyceraldehyde 3-phosphate, with a net loss of two ATP molecules. B. two molecules of glyceraldehyde 3-phosphate, with no net loss or gain of ATP. C. two molecules of pyruvate, with a net gain of two ATP molecules. D. two molecules of pyruvate, with a net gain of four ATP molecules.

A is correct. The investment phase of glycolysis refers to the first five steps. By the end of these steps, the original glucose molecule has been converted into two three-carbon glyceraldehyde 3-phosphate molecules. As this is the "investment" phase, we must have used some ATP (specifically, two molecules), making choice A correct. ATP is not produced until the payoff phase, which refers to the last five steps of the overall pathway.

Question 7 The kinetics of the ninhydrin reaction were shown to be first order with respect to 1,2,3-indantrione and α-alpha amino acid, and second order overall. The reaction rate was shown to vary with solution pH. The initial reaction rate was greatest when the product of which two concentrations was maximized? (Note: R = C8H4O2) A. [RC=O+H], [H2NCHRCOO-] B. [RC=O], [H3+NCHRCOO-] C. [RC=O+H], [H3+NCHRCOOH] D. [RC=O], [H2NCHRCOOH]

A is correct. The reaction between 1,2,3-indantrione and the α-alpha amino begins with attack by the nitrogen of the α-alpha amino acid's amino residue on a carbonyl carbon of 1,2,3-indantrione. The reaction requires an available lone pair on the amino nitrogen of the acid. No lone pair is available in the protonated, quaternary amino nitrogen shown in choices B and C. The reaction rate will be maximized when the carbonyl oxygen of the carbonyl attacked in the reaction is activated to attack by protonation (choice A). Because of a more favorable resonance contributor in the protonated versus the unprotonated carbonyl, the electrophilicity of the carbonyl carbon, and the reaction rate, are increased when compared to the carbonyl carbon electrophilicity and reaction rate of the unprotonated carbonyl-oxygen (choice D).

Question 6 Which of the following half-reactions are likely to be involved the reduction of ferrous ions by alginic acid? I. Fe2+ → Fe II. Fe2+ → Fe3+ III. -C-OH → -C=O IV. -COOH → -C-OH A. I and III only B. I and IV only C. II and III only D. II and IV only

A is correct. The reduction of ferrous ions is shown by the half-reaction Fe2+ → Fe (remember that in reduction reactions, an atom's oxidation state decreases, in this case from 2+ to 0). This reduction half-reaction needs to be paired with an oxidation reaction. As a shorthand, adding bonds to oxygen means that a carbon atom is oxidized, so -C-OH → -C=O is an oxidation half-reaction that could be paired with the reaction in Roman numeral I. B: This is a pairing of two reduction half-reactions. C: This is a pairing of two oxidation half-reactions. D: This pairing involves a reduction half-reaction and an oxidation half-reaction, but it is incorrect because it shows a ferrous ion being oxidized, not reduced.

Question 11 A dipeptide is formed through the reaction of one amino acid with another to form a peptide bond. During this reaction, the nucleophile is: A. the amino terminal nitrogen of one of the amino acids. B. the side chain of one of the amino acids. C. the carboxylic acid terminal oxygen of one of the amino acids. D. a carbonyl carbon

A is correct. This question overlaps the information in this chapter with organic chemistry, which is precisely what a real MCAT question might do. However, even if our organic chemistry is rusty, we can still reason through this question. Peptide bonds form between the amino terminal of one amino acid and the carboxylic acid terminal of another. They do not directly involve side chains (eliminate choice B). Additionally, in the formation of a peptide bond, the amino terminal nitrogen of one amino acid attacks the carbonyl carbon of another. The -COOH terminal oxygen is not a direct part of this attack and instead is lost from the molecule as water (eliminate choice C). Finally, note that carbonyl carbons are classic electrophiles due to the electron withdrawing effect of the double-bonded carbonyl oxygen (eliminate choice D). Our answer is thus option A, which makes logical sense as it is the amino terminal nitrogen that does the attacking in peptide bond formation.

Question 4 Additional research has shown vegetable juices in high quantities uncompetitively inhibit the catalytic activity of CP450 enzymes. Given this, an avid juice drinker would likely have what changes in CP450's Vmax and Km respectively? A. Vmax would decrease and Km would decrease. B. Vmax would be zero and Km would remain the same. C. Vmax would decrease and Km would increase. D. Vmax would decrease and Km would be zero.

A is correct. Uncompetitive inhibitors bind to the enzyme-substrate complex, essentially trapping the substrate and lowering Vmax. Km is defined as the substrate concentration at which the enzymatic reaction is proceeding at ½ the maximum velocity. Since the Vmax decreases as the enzyme is inhibited, less substrate is necessary to reach the lowered Vmax, thereby decreasing the Km. Note, it is important to keep in mind that while Km is lowered with uncompetitive inhibition, the affinity of the enzyme for a given substrate is not actually increased.

Question 11 Which of the following statements is true regarding the impact of surrounding salt concentrations on protein solubility? A. "Salting in" increases protein solubility and is most effective when salt concentrations are already very high. B. "Salting in" increases protein solubility and is most effective when salt concentrations are not yet very high. C. "Salting out" is accomplished by lowering the salt concentration in a protein-containing solution. D. Protein solubility increases linearly with salt concentration along the entire range of possible concentrations.

B is correct. "Salting in" refers to the addition of a salt to a solution that does not yet contain very high salt concentrations. The salt interacts with charged amino acid side chains, reducing the protein's ability to aggregate with other proteins. The net result is increased protein solubility, making choice B the correct answer. In contrast, when salt concentrations are already high, addition of more salt decreases protein solubility and is termed "salting out."

The metabolism of which biomolecule is LEAST likely to directly increase aerobic respiration in the cell? A. Pyruvate B. Cytosine C. A short-chain fatty acid D. Glycogen

B is correct. Aerobic respiration in the cell consists of glycolysis, the citric acid cycle, and the electron transport chain. The pyrimidines cytosine and uracil can be metabolized, but by doing so they are converted into beta-alanine and later to malonyl-CoA which is needed for fatty acid synthesis. A: Pyruvate is the end product of glycolysis, which can be converted into acetyl-CoA and fed into the citric acid cycle. C: Fatty acids such as 3-ketoacyl-CoA can be broken down via beta-oxidation to produce two carbon units which are then fed into the citric acid cycle via acetyl-CoA. D: Glycogen can be broken down into glucose, which can directly enter glycolysis.

Question 2 What is the absolute configuration of selenocysteine? A. R B. S C. D D. L

B is correct. Assigning priorities by Cahn-Ingold-Prelog rules, we can determine that the only chiral center present must be in an S configuration. Note that cysteine and selenocysteine are the only naturally occurring alpha-amino acids to have this configuration, due to higher priority of their substituents than the carbonyl group. D and L are relative configurations, based on molecular homology to glycerol. L-Selenocysteine is pictured.

Question 14 Compared to RNA, DNA exhibits much more physical flexibility in solution. This facilitates tight helical coiling and may partially explain the prevalence of DNA as genomic material. DNA is more flexible than RNA because: A. deoxyribose is a pentose, while ribose is a bulkier hexose. B. ribose has a hydroxyl substituent at C-2, which increases its steric bulk. C. ribose predominantly exists in the pyranose form when incorporated into a nucleotide. D. deoxyribose is a more energetically unstable enantiomer of ribose.

B is correct. Despite its relatively small size, the presence of the C-2 hydroxyl group does increase the steric bulk of ribose compared to deoxyribose. This reduces the flexibility of RNA compared to DNA. Both ribose and deoxyribose are pentoses (eliminate choice A), and they exist exclusively as furanoses when incorporated into nucleotides (eliminate choice C). Finally, if DNA was more energetically unstable, it would not help to explain why DNA is a prevalent genomic material, nor are ribose and deoxyribose isomers of any kind (eliminate D)

Question 12 All of the following enzymes are used in gluconeogenesis, but not in glycolysis, EXCEPT: A. pyruvate carboxylase. B. glucose 6-phosphate isomerase. C. fructose 1,6-bisphosphatase. D. glucose 6-phosphatase.

B is correct. Glucose 6-phosphate isomerase, also known as phosphohexose isomerase, is involved in both glycolysis and gluconeogenesis. Its name helps give this away; it interconverts glucose 6-phosphate and fructose 6-phosphate, which is required for both processes and is not highly exergonic in either direction. The remaining three enzymes are used specifically in gluconeogenesis to bypass steps that are highly energetically favorable in the glycolytic direction.

Question 14 A kinase is observed to act on a certain tetrapeptide. Of the following, which is most likely to represent the primary structure of this tetrapeptide? A. DEKH B. SYTF C. AGVL D. MCCI

B is correct. Kinases phosphorylate their substrates. The amino acids most likely to become phosphorylated are those which contain an -OH group on their side chains. Specifically, these are serine (S), threonine (T), and tyrosine (Y), all of which are listed in choice B. None of the other choices contain any of these amino acids.

Question 10 An enzyme is subjected to an inhibitor with an unknown reversible mechanism. Upon graphing the Lineweaver-Burk plots of the inhibited and uninhibited enzymes, researchers notice that the two plots share an x-intercept. The unknown inhibitor is most likely: A. competitive. B. noncompetitive. C. uncompetitive. D. either noncompetitive or uncompetitive.

B is correct. On a Lineweaver-Burk plot, the x-intercept represents -1/Km. If the inhibited and uninhibited plots have the same x-intercept, they must also have the same Km. Noncompetitive inhibitors do not alter Km, so the unknown inhibitor is most likely noncompetitive. Competitive inhibitors increase Km, while uncompetitive inhibitors decrease it.

Question 7 A student proposes converting a laminaran into a form suitable for energy extraction in humans by treating it with a β-1,3-glycosidase. Is this strategy, without any additional steps, likely to be successful? A. No, because a β-1,6-glycosidase would also be necessary B. Yes, because both α-glucose and β-glucose can enter glycolysis C. No, because an additional enzyme would be needed for anomerization to α-glucose D. Yes, because cleaving β-1,3-glycosidic bonds would completely convert the laminaran into glucose monomers

B is correct. Since laminarans are primarily linear (<10% branching) strings of β-glucose connected by β-1,3 glycosidic bonds, a β-1,3-glycosidase would break up those linear strings and result in many free β-glucose molecules, which can enter glycolysis. Note that α- and β-anomers of glucose interconvert under cellular conditions, so the precise anomer of a glucose monomer makes no significant difference. In fact, hexokinase can catalyze the first step of glycolysis using either anomer, although this fact goes beyond the knowledge required for the MCAT. A, D: A β-1,6-glycosidase would be needed to break up the branches of laminarans, but there are very few (<10%) such branches. Therefore, treatment with only a β-1,3-glycosidase would suffice to extract the vast majority of possible energy from laminarans. B: No specific enzyme is necessary for conversion to α-glucose; glucose interconverts between α- and β cyclic anomers and its linear form under cellular conditions. It is also not necessary for glucose to be in its α-anomer to enter glycolysis.

Question 2 Which of the following oligosaccharide sequences could NOT be isolated from alginic acid through enzymatic cleavage? A. B. C. D.

B is correct. The passage states that alginic acid can be composed of consecutive GulA residues [GulA-GulA-GulA-GulA]n, consecutive ManA residues [ManA-ManA-ManA-ManA]n, or alternating ManA and GulA residues [GulA-ManA-GulA-ManA]n. Alginic acid has been shown to be a promising antioxidant agent against oxidative stress induced. Based on a careful examination of the structures we see in Figure 2, we can identify that B contains an irregular pattern of alternations, ManA-ManA-GulA-ManA-GulA, that does not correspond to any of those options. A: This oligosaccharide contains consecutive GulA residues. C: This oligosaccharide contains consecutive ManA residues. D: This oligosaccharide contains regularly-alternating ManA and GulA residues.

Question 14 The rate-limiting step of glycolysis involves phosphorylation and the expenditure of ATP. Which enzyme catalyzes this irreversible step? A. Hexokinase B. Phosphofructokinase-1 C. Fructose-bisphosphate aldolase D. Pyruvate kinase

B is correct. The rate-limiting step of glycolysis is the phosphorylation of fructose 6-phosphate by the enzyme phosphofructokinase-1. This is also a key regulatory step because it is essentially irreversible, so it commits the glucose molecule to the glycolytic pathway. We can eliminate answer choice A because although it does add a phosphate at the expense of ATP, it is not the rate-limiting step in glycolysis. Answer choice C can be eliminated because fructose-bisphosphate aldolase does not utilize ATP in glycolysis. Answer choice D can be eliminated because pyruvate kinase creates ATP, not expends it.

Question 2 Analysis revealed that maximum Cenp-E activity was 104 s-1, according to the results in Figure 1, how much inhibitor is needed to change the activity to 5⨉103 s-1? A. 0.6 μM B. 7 μM C. 30 μM D. 110 μM

B is correct. This question asks for what concentration of the inhibitor, GSK92395, causes the Cenp-E activity to be reduced by half of its maximum value. This would be called the IC50 of GSK92395. According to the figure, the activity reaches 50% of its maximum activity around 7 μM GSK92395

Question 8 The structure of fructose is shown below: Which of the following is true of fructose? A. Fructofuranose is a ketose, while fructopyranose is an aldose. B. Elemental silver will be produced when fructose is present in a solution of Tollens' reagent. C. Fructofuranose is a pentose, while fructopyranose is a hexose. D. Fructose and glucose are stereoisomers.

B is correct. Tollens' reagent, like Benedict's reagent, is used to test for reducing sugars. Reducing sugars have free aldehyde groups that can be oxidized to carboxylic acids, with concomitant reduction of Tollens' reagent. Tollens' reagent contains an oxidized silver compound, which is reduced to elemental silver. Although fructose does not have an aldehyde group, in aqueous solution it can tautomerize to glucose and give a positive result. Therefore, fructose is considered a reducing sugar. Regarding choices A, C, and D, fructose, which includes fructose's isomers fructofuranose and fructopyranose, is a ketohexose, and fructose and glucose are not stereoisomers of each other.

In the π-helix shown below, how many complete turns will be found in the secondary structure? A. 0 B. 1 C. 2 D. 3

B is correct. We are told in the final paragraph that in a π-helix, there are 4.1 residues per turn. Thus with, 8 residues, we can fit 8/4.1. = 1.95, or 1 turn. Note that this is extremely close to 2, but the question specifically asks for complete turns.

The pKa of the protonated, positively charged form of the side chain of histidine is approximately 6. In a solution with a pH of 8, the side chain of histidine is most likely to be: A. twice protonated and charged. B. once protonated and uncharged. C. completely deprotonated and charged. D. once protonated and charged.

B is correct. When pH > pKa, the group in question (whether side chain, carboxylic acid terminal, or amino terminal) will be deprotonated - in this case, the conjugate acid (the charged form of the histidine residue), will lose a proton and become neutral. Since the side chain of histidine contains an N-H (which is specifically part of an imidazole ring), it will be positively charged when it is twice protonated and uncharged when it is singly protonated. In this case, since it is deprotonated, it will be uncharged. Note that the neutral form of an imidazole group has a pKa of ~14.5. Consequently, at very high pH (greater than 15), when both the N and C termini are already deprotonated, histidine can exhibit a net charge of -2 as the imidazole is further deprotonated.

Question 3 The cancerous phenotype described in this passage is most likely associated with making the production of which molecule more energetically favorable? A. Acetyl-CoA B. Fructose-6-Phosphate C. Glucose 6-Phopshate D. Malate

B is correct. With an increased accumulation of the enzyme hexokinase, the rate of production of G6P will increase. G6P is on one side of the chemical equilibria between G6P and F6P. Due to Le Chatelier's principle, we know that an increase in the concentration of reactants will push the reaction to form more product (F6P). This can be described by the equation ΔG = ΔG° + RT ln Q, where Q describes the reaction quotient ([product]/[reactants]) at a given moment. An increase in reactant concentration (G6P) therefore makes it more favorable for a system to produce more product (F6P).

Question 11 Aerobes are microorganisms capable of using oxygen as a final electron acceptor, while anaerobes cannot do so. A facultative aerobe must be capable of undergoing which metabolic process(es) in the presence of oxygen? A. Fermentation only B. Aerobic respiration only C. Both fermentation and aerobic respiration D. Neither fermentation nor aerobic respiration

C is correct. A facultative aerobe can use oxygen as an electron acceptor in the presence of oxygen, but it can also survive in an anaerobic environment by deriving energy solely from glycolysis and subsequent fermentation. In contrast, an obligate anaerobe would not be able to survive in the presence of oxygen, while an obligate aerobe would require oxygen to survive. A facultative anaerobe, on the other hand, can survive in the presence of oxygen, but does not use it as a final electron acceptor and is therefore not undergoing aerobic respiration.

Question 13 Cadherins are classified as which of the following types of protein? A. Globular proteins B. Enzymes C. Transmembrane proteins D. Monotopic proteins

C is correct. Cadherins are cell adhesion molecules that comprise adherens junctions, which serve to connect cells to each other. As such, it makes logical sense that a cadherin should extend through the membrane of a cell and out into the surrounding environment. Transmembrane proteins are those which extend through the entire membrane of a cell. Globular proteins are spherical and tend to float freely, which is not the case of cadherins (eliminating choice A). Cadherins are not monotopic proteins, which connect only to one side of the plasma membrane (eliminating choice D).

Which CP450 enzyme exhibited the highest catalytic efficiency? A. WT B. I135F C. Y273F D. None of the above, all CP450 enzymes exhibited the same catalytic efficiency

C is correct. Catalytic efficiency is expressed as kcat/Km. No kcat values were provided but Vmax = kcat x [Et]. Equal enzyme concentrations were used in the experiment, so Vmax is functionally equivalent (directly proportional) to kcat in this circumstance. No rigorous calculations are necessary to arrive at this answer if we realize Y273F has the largest Vmax and a Km identical to the other CP450 enzymes tested. A: The wild-type enzyme has the lowest catalytic efficiency of all enzymes used in the experiment. B: This variant has a greater catalytic efficiency than the wild type, but it is not as great as Y273F. D: This answer choice would be correct if catalytic efficiency depended on Km only.

Question 7 If researchers wished to determine if myosin is necessary for proper cell function, which motility function should be monitored in the experiment? A. Flagellar movement in spermatids B. Axoplasmic transport in sympathetic neurons C. Cleavage furrow formation in hepatocytes D. Ciliary movement in intestinal epithelia

C is correct. Cleavage furrow formation, which aids the process of cytokinesis (daughter cell separation at the end of cell division), is based on the contraction of an array of actin filaments that are anchored to the plasma membrane. The sliding of actin filaments is induced by the action of the motor protein myosin, which splits the parent cell into 2 daughter cells.A, D: Both ciliary motility in epithelia and movement of the spermatid tail arise from microtubules and microtubule-associated motor proteins. Sperm motility is due to the movement of the sperm tail (a modified flagellum).B: Axoplasmic transport is the movement of membrane-bound organelles, mitochondria, lipids, synaptic vesicles, proteins, and other cell parts to and from a neuronal cell body, through the cytoplasm of the axon. This is handled by the action of microtubule-associated motor proteins kinesin (anterograde transport) or dynein (retrograde transport).

Which of the following classifications does the heme group found within CP450 enzymes described in the passage belong to? I. Cofactor II. Coenzyme III. Prosthetic group IV. Apoenzyme A. I and II only B. II and III only C. I, II, and III only D. I, II, and IV only

C is correct. Cofactors are organic or inorganic molecules that assist in the function of an enzyme. Coenzymes are an organic subclass of a cofactors and prosthetic groups are a type of coenzyme that is tightly bound to the enzyme itself (I,II, III). A: This answer choice is incorrect because an apoenzyme is an enzyme that is missing necessary cofactors (IV). B: Heme can also be classified as a cofactor, making this answer choice incorrect. D: Heme can also be classified as a prosthetic group, making this answer choice incorrect.

Conversion between which forms of carrageenan would require both a dehydration reaction and a sulfation reaction? A. κ-carrageenan to λ-carrageenan B. ι-carrageenan to ξ-carrageenan C. ξ-carrageenan to ι-carrageenan D. λ-carrageenan to κ-carrageenan

C is correct. Figure 3 shows that ι- and λ-carrageenan are the forms of carrageenan that contain a sulfate group, so the answer choice must end with either ι- or λ-carrageenan in order to involve a sulfation reaction. Based on a comparison between Figure 2A and Figure 2B, we can see that a dehydration reaction, in which a molecule of H2O is lost as two molecules form a bond, must occur in the transition between Gal and AnGal (the term "anhydro" in 3,6-anhydrogalactose [AnGal]" is also a hint). Therefore, the product of this conversion must be on the left-hand side of the figure (Figure 2A)—in other words, either κ- or ι-carrageenan. Combining these two pieces of information, the conversion must yield ι-carrageenan. A: This conversion would involve a combination of sulfation and hydrolysis, not dehydration. B: This conversion would include hydrolysis, not dehydration, and the loss of a sulfate group, making it doubly incorrect. D: This conversion would involve dehydration and the loss of a sulfate group, not sulfation.

Question 15 In a G protein-coupled receptor system, the G protein: A. is activated by the phosphorylation of GDP into GTP, and is inactivated by the hydrolysis of GTP. B. is activated by the phosphorylation of GDP into GTP, and is inactivated when GTP dissociates and is replaced by GDP. C. is activated when a GDP molecule is exchanged for a GTP molecule, and is inactivated by the hydrolysis of GTP. D. is activated when a GDP molecule is exchanged for a GTP molecule, and is inactivated when GTP dissociates and is replaced by GDP.

C is correct. G proteins are activated when bound to GTP; this binding occurs when a GDP molecule is exchanged for GTP. Note that GDP is not simply phosphorylated; it is actually replaced by an entirely new GTP molecule (eliminate choices A and B). G proteins are rendered inactive by the hydrolysis of GTP (eliminate choice D), which is accomplished via an intrinsic GTPase activity.

Suppose it was found that most CP450 enzymes are excreted from a cell in an inactive form, only to become active once acted upon by a protease. This discovery most reasonably suggests that CP450 is: A. an apoenzyme. B. a holoenzyme. C. a zymogen. D. a prohormone.

C is correct. Given the new information in the question stem, it is reasonable to conclude that CP450 is a zymogen. This is because zymogens are inactive precursors of enzymes that require proteolytic cleavage prior to becoming active. A: While it is true that inactive CP450 is an apoenzyme, information presented in the question stem does not address that. B: Whether or not CP450 is an enzyme that requires additional cofactors in order to work properly is not suggested by new information in the question stem. D: Prohormones do require cleavage prior to becoming active, but information in the passage and question stem does not suggest that CP450 is a hormone.

Question 15 Glycolysis begins with the phosphorylation of D-glucose at its sixth position by the enzyme hexokinase. If a population of bacteria were raised in media containing L-glucose as the only carbon source, which of the following would be a likely result? A. Increased expression of downstream glycolytic enzymes B. Increased expression of enzymes that metabolize fructose C. Absence of any hexokinase activity D. Increased consumption of ATP by hexokinase

C is correct. L-glucose does not exist in nature, and virtually no naturally-occurring enzymes are able to metabolize it. Since enzymes have chiral active sites, the ability of hexokinase to utilize D-glucose as a substrate does not mean that it would also be able to accept L-glucose. Since hexokinase would not work under these conditions, this eliminates D because it would not increase ATP consumption if it is not being used, and would eliminate A because glycolysis would not be able to proceed past the initial step if hexokinase activity was absent. Finally, although the bacteria may compensate by looking for other energy sources (answer choice B), such as fructose, the absence of hexokinase activity (answer choice C) more directly answers this question.

Which of the following statements is NOT true of disulfide bonds? A. They can be broken through exposure to certain reagents, including 2-mercaptoethanol. B. Like peptide bonds, they constitute covalent interactions. C. They can form between any two sulfur-containing amino acid residues. D. Their formation involves oxidation of sulfur, while their breaking constitutes reduction of sulfur.

C is correct. Of the twenty standard amino acids, just two contain sulfur: cysteine and methionine. Disulfide bonds can form between two cysteine residues, but they cannot form between two methionine residues or between a cysteine and a methionine. The remaining statements are true. Formation of a disulfide bond occurs through oxidation, while breaking such a bond requires reduction; this can be accomplished using reducing agents like 2-mercaptoethanol. Both disulfide linkages and peptide bonds are covalent interactions.

Question 12 Immunoglobulin M (IgM) is most commonly found in the form of a pentamer. In total, how many heavy and light chains are present in one IgM pentamer? A. Five heavy chains and five light chains B. Five heavy chains and ten light chains C. Ten heavy chains and ten light chains D. Twenty heavy chains and twenty light chains

C is correct. One immunoglobulin (Ig) subunit contains two heavy chains and two light chains. A pentamer contains five Ig subunits, so in total, one pentamer must contain ten heavy chains and ten light chains.

Question 3 Variant enzymes showed a significant increase in catalytic activity compared to WT, but Km remained relatively unchanged. What is the most likely explanation for this data? A. The increased catalytic activity of the variants masked the increased Km. B. Amino acids changed in the active site of the proteins were not involved in binding the substrate. C. Substituted channel residues allowed for easier substrate entry into the active site. D. Increase in variant catalytic activity was due to cooperative binding made possible by the substituted amino acids.

C is correct. The passage states the variant enzymes had substitutions of amino acids within the channels leading from the surface of the protein to the active site. If the turnover rate increases but the binding affinity (Km) remains the same, then it's likely that the substitutions increased entry of substrate into the active site without altering the site itself. A: This answer choice is incorrect because an increased Km implies a decreased binding affinity, which would not result in an increase rate of catalysis. B: The passage explicitly states that the substituted amino acids were not in the active site. D: While cooperative binding could theoretically lead to increased catalysis, there is not enough information to conclude this.

Question 2 According to the data in Table 1, what is the turnover number for variant V245F? A. 1.5 x 104 (μM/min) B. 25 ± 0.010 μM C. 3 x 1011 min-1 D. 5 x 10-8 min-1

C is correct. The turnover number is also known as kcat, or the time it takes one enzyme to turnover a maximum amount of substrate molecules per unit time. No kcat values were given, but it can be obtained by using the equation kcat = Vmax / [Et]: kcat = (1.5 x 104 μM/min ) / (0.05 pM). Changing concentrations to molarity we obtain (1.5 x 10-2 M/min) / (5 x 10-14 M) which is equal to 3 x 1011 min-1. A: This is a very tempting answer choice, but the Vmax is not the turnover number, rather this term refers to kcat. B: This value corresponds to the Km of this variant. D: This answer results from a miscalculation.

Assuming that the pKa of the side chain of histidine is approximately 6, which of the following is closest to the isoelectric point of this amino acid? A. 4 B. 5.7 C. 7.5 D. 9

C is correct. To calculate isoelectric point, you must consider the pKa values of the amino acid in question. Histidine has a carboxylic acid group (pKa ~ 2), an amino group (pKa ~ 9), and its side chain (pKa ~ 6). Note that we should not average all three of these values; instead, since histidine is a basic amino acid, we should average the two most basic pKa values. (6 + 9) / 2 = 7.5.

Question 3 Starting from the center of Figure 1, which amino acids are bound to the second glutamic acid residue? A. Glutamine and asparagine B. Glutamine and leucine C. Asparagine and histidine D. Lysine and asparagine

C is correct. We are told in the passage that amino acids in an alpha helix hydrogen bond with every 4th amino acid. Looking at Figure 1, you can see that the second glutamic acid (E) is directly attached to a histidine (H) via a peptide bond. Counting four more residues gives us an asparagine (N), with which the glutamic acid would hydrogen bond.

A certain globular protein is typically found in an aqueous environment in the body. In this protein, which of the following amino acid residues would you expect to be facing outside, towards the environment? A. Valine B. Isoleucine C. Threonine D. Tryptophan

C is correct. When the external environment is aqueous (and thus polar), hydrophilic amino acids generally face towards the cytosol or external environment, while hydrophobic residues comprise the inner core of the protein. Here, threonine is the only polar amino acid listed, so it is the option most likely to face the exterior of the protein, which interfaces with the aqueous polar environment.

Question 8 Which of the following is most likely true of an enzyme with a Hill coefficient of 2.8? A. It exhibits negative cooperativity. B. It is monomeric. C. A graph of its saturation versus substrate concentration is linear. D. It binds its second substrate molecule significantly more readily than its first.

D is correct. A Hill coefficient greater than 1 reflects positive cooperativity (eliminating A), meaning that binding at one position or active site causes binding to take place more easily at the remaining active sites. The most famous example of positive cooperativity occurs in hemoglobin. Note that monomeric proteins contain only one subunit; since enzymes must contain multiple active sites or binding positions to display any form of cooperativity, it is more likely that the enzyme referenced in the question stem contains multiple subunits (eliminating B). Additionally, a graph of saturation versus substrate concentration for an enzyme that displays positive cooperativity would be sigmoidal, or S-shaped, not linear (eliminating C).

During the Gabriel synthesis of an amino acid, which of the steps listed below occurs? A. A malonic ester undergoes an SN1 reaction. B. An unprotected amine attacks the carbonyl carbon of an ester. C. The oxygen atom of an aldehyde is protonated. D. Heat is used to remove an extra -COOH group.

D is correct. A final step in the Gabriel synthesis of an amino acid is the administration of heat to decarboxylate the molecule, removing an extra -COOH. Choice A is incorrect because the Gabriel synthesis involves an SN2 mechanism, not SN1. Option B is wrong because the amine in a Gabriel synthesis is protected, and choice C is something that occurs during the Strecker synthesis.

Question 15 A tired research student is setting up an isoelectric focusing procedure. He accidentally adds a lysine-rich protein sample to the pH-gradient gel instead of his desired sample, which was largely composed of hydrophobic residues. How would this mistake impact his findings? A. The protein will be observed to move more rapidly toward the positively-charged pole. B. The protein will become stationary at a position farther from its isoelectric point. C. The protein will travel a greater distance toward the anode. D. The protein will become stationary at a position closer to the cathode.

D is correct. A protein rich in lysine (a basic amino acid) will have a higher isoelectric point (pI) than a protein largely formed from hydrophobic residues. A higher pI means the protein will tend to be more positively charged and will thus move closer to the negative end of the gel. In electrophoresis, the apparatus functions like an electrolytic cell, meaning that the anode is positive and the cathode is negative. The lysine-rich protein, which is positively charged at physiological pH, then, will move closer to the negatively-charged cathode. Note that in isoelectric focusing, a protein or amino acid always becomes stationary when it has reached the portion of the gel corresponding to its isoelectric point.

Question 2 Which of the following molecules is the product obtained at the end of the Edman degradation of the polypeptide H2N-Ala-Gly-Arg-Met-COOH? A. PTH-Ala B. PTH-Gly C. PTH-Arg D. PTH-Met

D is correct. According to the passage, the first residue labeled by reaction with phenylisothiocyanate is a terminal one. Figure 2 indicates that the N-terminal amino residue is specifically the first removed. The N-terminal residue of the polypeptide shown, alanine (Ala), will be isolated first (choice A). Choice B will be the second residue isolated. Choice C will be the third residue isolated. Choice D will be the final residue isolated.

Question 6 What conclusion is most strongly suggested based on the results of the experiment? A. Cells located in aerobic microenvironments are more likely to demonstrate increased cellular uptake of glucose. B. GLUT1, HK1, and HK2 subtypes are expressed more strongly in cancerous tissue derived from the colon, versus cancerous tissue derived from nervous system tissue. C. Increased expressivity of glycolytic genes is one cause of cancerous cell types. D. Glucose transporter activity is a predictor of neuroendocrine tumor proliferation.

D is correct. Figure 2 plots the correlation between GLUT1 gene expression and proliferation index in NET samples. The researchers calculated a statistically significant (p = 0.047, which is less than 0.05), moderate correlational coefficient (0.34) between GLUT1 expression and NET proliferative activity. Glucose transporter activity is therefore predictive of NET proliferation. C: If causality had been determined from this experiment (which it was not), this statement reverses the directionality of cause and effect. It is more likely that a cell becoming cancerous increases the expressivity of glycolytic genes. B: HK2 expression was not significantly different between the two types of cancer tissue studied. A: This is opposite of the information given in the passage.

Question 8 Glycolysis is a near-universal pathway of energy acquisition. This series of reactions can best be characterized as: A. the oxidation of glucose to carbon dioxide. B. the anabolism of glucose to pyruvate. C. the catabolism of glucose to acetyl-CoA. D. the oxidation of glucose to pyruvate.

D is correct. Glycolysis involves the breakdown (catabolism) of glucose to form two molecules of pyruvate, among other products. This process constitutes oxidation. (If you find it difficult to remember this, recall that the electron carrier NAD+ is reduced during glycolysis, so something must be oxidized in turn.) Acetyl-CoA is produced later, during pyruvate decarboxylation by the pyruvate dehydrogenase complex, and carbon dioxide is formed during the Krebs cycle.

Question 12 A biological molecule that catalyzes the transfer of a phosphate group onto its substrate is classified as which type of enzyme? A. Transferase B. Ligase C. Phosphatase D. Kinase

D is correct. Kinases are responsible for transferring a high-energy phosphate group from a donor molecule (typically ATP) to the substrate. This reaction produces ADP along with the phosphorylated substrate molecule. Phosphatases do the opposite of this; they remove phosphate groups from their substrates (eliminating C). Ligases are enzymes that are responsible for binding together two smaller components (eliminating B). Finally, transferases transfer functional groups from one molecule to another, but kinase is a more specific answer here (eliminating A).

All of the following are expected to be attracted to positively-charged species EXCEPT: A. ATP. B. ADP. C. the backbone of a nucleic acid. D. testosterone.

D is correct. Opposite charges attract; thus, negatively-charged molecules are attracted to positively-charged ones. Since this is a NOT question, we are looking for the lone option that does not carry a negative charge. ATP, ADP, and the backbone of DNA or RNA all contain phosphate groups, which are highly negatively charged. In contrast, testosterone is a steroid hormone; it is not charged at all, as evidenced by the fact that it can pass through the plasma membrane and is thus nonpolar.

Question 1 Based on information in the passage, what class of enzymes does CP450 most likely belong to? A. Hydrolases B. Transferase C. Isomerase D. Oxidoreductases

D is correct. Paragraph 1 describes the mechanism of CP450 enzymes, which includes the reduction of O2 to H2O. Redox reactions are catalyzed by oxidoreductases. A: This answer choice is incorrect because the described mechanism results in the formation of water, not the loss of water. B: Transferases move a functional group from one molecule to another without the use of ATP. The described mechanism is more consistent with an oxidoreductase enzyme. C: The proposed mechanism in the passage states that one oxygen from O2 is added to the substrate. Isomerases simply rearrange atoms within a molecule, making this answer choice incorrect.

Question 1 If native Hrt-34 protein is run on SDS-PAGE and then a spate sample of Hrt-34 is subjected to 4 rounds of Edman degradation and run on a similar gel, how will the movement through the gel of the respective products compare? A. The Edman degraded Hrt-34 will migrate four times as far as the native protein. B. The Edman degraded Hrt-34 will migrate one-fourth as far as the native protein. C. The Edman degraded Hrt-34 will migrate 5% further than the native protein. D. The Edman degraded Hrt-34 will migrate further than the native protein.

D is correct. SDS acts to disrupt the molecular forces which allow proteins to take their native conformation. The detergent also creates a uniform (-) charge across the protein. The protein then migrates a distance through the gel which is inversely proportional to its molecular weight. Edman degradation shrinks the size of a polypeptide, meaning the degraded protein will have a smaller molecular weight, and will migrate a greater distance. Because we do not know the exact size of Hrt-34, we cannot know the exact difference in migration distance

Question 10 D-mannose is an epimer of D-glucose. If a solution of D-glucose has a specific rotation of +52°, what is the specific rotation of a solution of D-mannose? A. +52° B. -52° C. 0° D. It cannot be determined from the given information.

D is correct. Specific rotation is an experimental value. Generally, the specific rotation of a particular compound does not relate in any predictable way to that of another compound. The exceptions to this rule are enantiomers, which have equal but opposite specific rotation values, and racemic solutions, which have a specific rotation of 0°

Question 14 A certain signaling molecule binds its receptor in the nucleus, causing transcription of a particular gene to increase. This signaling molecule most likely: A. exerts its physiological effects rapidly and for a short duration. B. promotes a second-messenger cascade within the target cell. C. is highly water-soluble. D. has a nonpolar fused-ring structure.

D is correct. Steroid hormones are signaling molecules that bind their receptors in the nucleus and impact transcription. These molecules are nonpolar and are derived from cholesterol, meaning they tend to share cholesterol's fused-ring hydrocarbon structure. Alternatively, we can use process of elimination. Peptide hormones are short-acting (eliminate choice A), promote second-messenger cascades (eliminate choice B), and are soluble in water (eliminate choice C). These traits are not true of steroid hormones.

Question 11 Which of the following is a disaccharide? A. Cellulose B. Galactose C. Ribose D. Sucrose

D is correct. Sucrose is a disaccharide of glucose and fructose. Cellulose is a large polymer that confers structural integrity to plant cell walls. Ribose and galactose are both monosaccharides.

Question 8 Non-enzymatic protein functions include all of the following EXCEPT: A. acting as part of the humoral immune system. B. adhering one cell to another. C. composing eukaryotic flagella. D. catalyzing biological reactions.

D is correct. The question stem asks about non-enzymatic protein function. It is enzymes which catalyze biological reactions; thus, choice D is our "odd man out" and is correct. Option A is a non-enzymatic function of proteins, specifically antibodies. Choice B is also accomplished by non-enzymatic proteins. Finally we can rule out option C because eukaryotic flagella are composed of microtubules, which are proteins that do not play an enzymatic role.

Question 6 Intermolecular forces between which of the following are most directly involved in the formation of the alpha helix backbone? A. The hydrophobic groups on the amino acid side chains B. The amine group on one amino acid and the amine group of another C. The carbonyl group on one amino acid and the carbonyl group of another D. The amine group on one amino acid and the carbonyl group of another

D is correct. This question asks us to consider what two species on an amino acid will hydrogen bond with one another. Since nitrogen is very electronegative, the N-H bond in the amine group results in a partial positive charge on the H, which can interact with the negative dipole on the O of the carbonyl group on another residue. Note that side group interactions contribute to tertiary structure, not alpha helices, so we can eliminate choice A. Choice B is incorrect because oxygen is more electronegative than nitrogen, and is thus a better hydrogen bond recipient. Finally, the oxygens in carbonyl groups are all partially negative, and therefore would be unlikely to be attracted to each other via hydrogen bonding. The carbonyl carbons are both electropositive, and would also repel. We can thus eliminate choice C.