Biochemistry

Question 1

In a neutral solution, most amino acids exist as:
A. positively charged compounds.
B. zwitterions.
C. negatively charged compounds.
D. hydrophobic molecules.

Answer 1

B. Zwitterions
Most amino acids (except the acidic and basic amino acids) have two sites for protonation: the carboxylic acid and the amine. At neutral pH, the carboxylic acid will be deprotonated (-COO-) and the amine will remain protonated (-NHz+).
This dipolar ion is a zwitterion, so (B) is the correct answer.

Question 2

At pH 7, the charge on a glutamic acid molecule IS:
A. -2.
B. -1.
C. 0.
D. +1.

Answer 2

B. -1
Glutamic acid is an acidic amino acid because it has an extra carboxyl group. At neutral pH, both carboxyl groups are deprotonated and thus negatively charged. The amino group has a positive charge because it remains protonated at pH 7. Overall, therefore, glutamic acid has a net charge of -1, and (B) is correct. Notice that you do not even need to know the pI values to solve this question; as an acidic amino acid, glutamic acid must have a pI below 7.

Question 3

Which of the following statements is most likely to be true of nopolar R groups in aqueous solution?
A. They are hydrophilic and found buried within proteins.
B. They are hydrophilic and found on protein surfaces.
C. They are hydrophobic and found buried within proteins.
D. They are hydrophobic and found on protein surface

Answer 3

C. They are hydrophobic and found buried within proteins.
Nonpolar groups are not capable of forming dipoles or hydrogen bonds; this makes them hydrophobic. Burying hydrophobic R groups inside proteins means they don’t have to interact with water, which is polar. This makes (C) correct. (A) and (B) are incorrect because nopolar molecules are hydrophobic, not hydrophilic; (D) is incorrect because they are not generally found on protein surfaces.

Question 4

Scientists discover a DNA sequence for an uncharacterized protein. In their initial studies, they use a computer program designed to predict protein structure. Which of the following levels of protein structure can be most accurately predicted?
A. Primary structure
B. Secondary structure
C. Tertiary structure
D. Quaternary structure

Answer 4

A. Primary structure
The cDNA sequence is a DNA copy of the mRNA used to generate a protein. A computer program can quickly identify the amino acid that corresponds to each codon and generate a list of these amino acids. This amino acid sequence is the primary structure of the protein. These observations support
(A) as the correct answer. By contrast, the secondary, tertiary, and quaternary structures involve higher level interactions between the backbone and R groups and are increasingly difficult to predict.

Question 5

How many distinct tripeptides can be formed from one valine molecule, one alanine molecule, and one
leucine molecule?
A. 1
B. 3
C. 6
D. 27

Answer 5

C. 6
There are three choices for the first amino acid, leaving two choices for the second, and one choice for the third.
Multiplying those numbers gives us a total of 3 x 2 x 1 = 6
distinct tripeptides. (Using the one-letter codes for valine (V), alanine (A), and leucine (L), those six tripeptides are VAL, VLA, ALV, AVL, LVA, and LAV.)

Question 6

Which of the following best describes the change in entropy that occurs during protein folding?
A. Entropy of both the water and the protein increase.
B. Entropy of the water increases; entropy of the protein decreases.
C. Entropy of the water decreases; entropy of the protein increases.
D. Entropy of both the water and the protein decrease.

Answer 6

B
As the protein folds, it takes on an organized structure and thus its entropy decreases. However, the opposite trend is true for the water surrounding the protein. Prior to protein folding, hydrophobic amino acid residues are exposed and the water molecules must form structured hydration shells around these hydrophobic residues. During fold-ing, these hydrophobic residues are generally buried in the interior of the protein so that the surrounding water molecules gain more latitude in their interactions. Thus, the entropy of the surrounding water increases, making the correct answer (B).

Question 7

An α-helix is most likely to be held together by:
A. disulfide bonds.
B. hydrophobic effects.
C. hydrogen bonds.
D. ionic attractions between side chains.

Answer 7

C. Hydrogen bonds
The α-helix is held together primarily by hydrogen bonds between the carboxyl groups and amino groups of amino acids. Disulfide bridges, (A), and hydrophobic effects, (B), are primarily involved in tertiary structures, not secondary.
Even if they were charged, the side chains of amino acids are too far apart to participate in strong interactions in secondary structure.

Question 8

Which of the following is least likely to cause denaturation of proteins?
A. Heating the protein to 100°C
B. Adding 8 M urea
C. Moving it to a more hypotonic environment
D. Adding a detergent such as sodium dodecyl sulfate

Answer 8

C. Moving it to a more hypotonic environment
High salt concentrations and detergents can denature a protein, as can high temperatures. But moving a protein to a hypotonic environment-that is, a lower solute concentration-should not lead to denaturation.

Question 9

A particular α-helix is known to cross the cell membrane. Which of these amino acids is most likely to be found in the transmembrane portion of the helix?
A. Glutamate
B. Lysine
C. Phenylalanine
D. Aspartate

Answer 9

C. Phenylalanine
An amino acid likely to be found in a transmembrane portion of an α-helix will be exposed to a hydrophobic environment, so we need an amino acid with a hydrophobic side chain. The only choice that has a hydrophobic side chain is (C), phenylalanine. The other choices are all polar or charged.

Question 10

Which of these amino acids has a chiral carbon in
its side chain?
I. Serine
Il. Threonine
Ill. Isoleucine
A. I only
B. Il only
C. Il and Ill only
D. I, II, and III

Answer 10

C. II and III only
Every amino acid except glycine has a chiral a-carbon, but only two of the 20 amino acids— threonine and isoleucine— also have a chiral carbon in their side chains as well. Thus, the correct answer is (C). Just as only one configuration is normally seen at the a carbon, only one configuration is seen in the side chain chiral carbon.

Question 11

Following translation and folding, many receptor tyrosine kinases exist as monomers in their inactive state on the cell membrane. Upon the binding of a ligand, these proteins dimerize and initiate a signaling cascade. During this process, their highest element of protein structure changes from:
A. secondary to tertiary.
B. tertiary to quaternary.
C. primary to secondary.
D. secondary to quaternary.

Answer 11

B. tertiary to quaternary
In their inactive state, the receptor tyrosine kinases are fully folded single polypeptide chains and thus have tertiary struc-ture. When these monomers dimerize, they become a protein complex and thus have elements of quaternary structure. This change from tertiary to quaternary structure justifies (B).

Question 12

Which of these amino acids has a side chain that can become ionized in cells?
A. Histidine
B. Leucine
C. Proline
D. Threonine

Answer 12

A. Histidine
Histidine has an ionizable side chain: its imidazole ring has a nitrogen atom that can be protonated. None of the remaining answers have ionizable atoms in their side chains.

Question 13

In lysine, the pka of the side chain is about 10.5.
Assuming that the pK, of the carboxyl and amino groups are 2 and 9, respectively, the pl of lysine is closest to:
A. 5.5.
B. 6.2.
C. 7.4.
D. 9.8.

Answer 13

D. 9.8
Because lysine has a basic side chain, we ignore the pK, of the carboxyl group, and average the pK of the side chain and the amino group; the average of 9 and 10.5 is 9.75, which is closest to (D).

Question 14

Which of the following is a reason for conjugating proteins?
I. To direct their delivery to a particular organelle
Il. To direct their delivery to the cell membrane
III. To add a cofactor needed for their activity
A. I only
B. II only
C. Il and Ill only
D. I, Il, and III

Answer 14

D. I, II, III
Conjugated proteins can have lipid or carbohydrate “tags” added to them. These tags can indicate that these proteins should be directed to the cell membrane (especially lipid tags) or to specific organelles (such as the lysosome). They can also provide the activity of the protein; for example, the heme group in hemoglobin is needed for it to bind oxygen.
Thus, (D) is the correct answer.

Question 15

Collagen consists of three helices with carbon backbones that are tightly wrapped around one another in a “triple helix.” Which of these amino acids is most likely to be found in the highest concentration in collagen?
A. Proline
B. Glycine
C. Threonine
D. Cysteine

Answer 15

B. Glycine
Because collagen has a triple helix, the carbon backbones are very close together. Thus, steric hindrance is a potential problem. To reduce that hindrance, we need small side chains; glycine has the smallest side chain of all: a hydrogen atom.

Question 16

Consider a biochemical reaction A → B, which is catalyzed by A-B dehydrogenase. Which of the following statements is true?
A. The reaction will proceed until the enzyme concentration decreases.
B. The reaction will be most favorable at 0°C.
C. A component of the enzyme is transferred from A to B.
D. The free energy change (dG) of the catalyzed reaction is the same as for the uncatalyzed reaction.

Answer 16

D. The free energy change (dG) of the catalyzed reaction is the same as for the uncatalyzed reaction.
Enzymes catalyze reactions by lowering their activation energy, and are not changed or consumed during the course of the reaction. While the activation energy is lowered, the free energy of the reaction, dG, remains unchanged in the presence of an enzyme. A reaction will continue to occur in the presence or absence of an enzyme; it simply runs slower without the enzyme, eliminating (A). Most physiological reactions are optimized at body temperature, 37°C, eliminating (B). Finally, dehydrogenases catalyze oxidation-reduction reactions, not transfer reactions, eliminating (C).

Question 17

Which of the following statements about enzyme kinetics is FALSE?
A. An increase in the substrate concentration (at constant enzyme concentration) leads to proportional increases in the rate of the reaction.
B. Most enzymes operating in the human body work best at a temperature of 37°C.
C. An enzyme-substrate complex can either form a product or dissociate back into the enzyme and substrate.
D. Maximal activity of many human enzymes occurs around pH 7.4.

Answer 17

A. An increase in the substrate concentration (at constant enzyme concentration) leads to proportional increases in the rate of the reaction.
Most enzymes in the human body operate at maximal activity around a temperature of 37°C and a pH of 7.4, which is the pH of most body fluids. In addition, as characterized by the Michaelis-Menten equation, enzymes form an enzyme-substrate complex, which can either dissociate back into the enzyme and substrate or proceed to form a product. So far, we can eliminate (B), (C), and (D), so let’s check (A). An increase in the substrate concentration, while maintaining a constant enzyme concentration, leads to a proportional increase in the rate of the reaction only initially. However, once most of the active sites are occupied, the reaction rate levels off, regardless of further increases in substrate con-centration. At high concentrations of substrate, the reaction rate approaches its maximal velocity and is no longer changed by further increases in substrate concentration.

Question 18

Some enzymes require the presence of a nonprotein molecule to behave catalytically. An enzyme devoid of this molecule is called a(n):
A. holoenzyme.
B. apoenzyme.
C. coenzyme.
D. zymoenzyme.

Answer 18

B. Apoenzyme
An enzyme devoid of its necessary cofactor is called an apoenzyme and is catalytically inactive.

Question 19

Which of the following factors determine an enzyme’s specificity?
A. The three-dimensional shape of the active site
B. The Michaelis constant
C. The type of cofactor required for the enzyme to be active
D. The prosthetic group on the enzyme

Answer 19

A. The three-dimensional shape of the active site
An enzyme’s specificity is determined by the three-dimensional shape of its active site. Regardless of which explanation for enzyme specificity we are discussing (lock and key or induced fit), the active site determines which substrate the enzyme will react with.

Question 20

Human DNA polymerase is removed from the freezer and placed in a 60°C water bath. Which of the following best describes the change in enzyme activity
as the polymerase sample comes to
thermal equilibrium with the water bath?
A. Increases then decreases
B. Decreases then plateaus
C. Increases then plateaus
D. Decreases then increases

Answer 20

А. Increases then decreases
As the temperature of the DNA polymerase sample increases from 0°C to the usual physiological temperature, i.e. 37°C, the enzyme’s activity will increase. However, at temperatures above 37°C, the enzyme’s activity will rapidly decline due to denaturation.

Question 21

in the equation below, substrate C is an allosteric inhibitor to enzyme 1. Which of the following is another mechanism necessarily caused by substrate C?

A. Competitive inhibition
B. Irreversible inhibition
C. Feedback enhancement
D. Negative feedback

Answer 21

D. Negative feedback
By limiting the activity of enzyme 1, the rest of the pathway is slowed, which is the definition of negative feedback. (A) is incorrect because there is no competition for the active site with allosteric interactions. While many products do indeed competitively inhibit an enzyme in the pathway that creates them, this is an example of an allosterically inhibited enzyme. There is not enough information for (B) to be correct because we aren’t told whether the inhibition is reversible. In general, allosteric interactions are temporary.
(C) is incorrect because it is the opposite of what occurs when enzyme 1 activity is reduced.

Question 22

The activity of an enzyme is measured at several different substrate concentrations, and the data are shown in the table below.

Km for this enzyme is approximately:
A. 0.5
B. 1.0
C. 10.0
D. 50.0

Answer 22

A. 0.5
While the equations given in the text are useful, recognizing relationships is even more important. You can see that as substrate concentration increases significantly, there is only a small change in the rate. This occurs as we approach Vmax Because the Vmax is near 100mmol/min, Vmax/2 equals 50mmol/min. The substrate concentration giving this rate is 0.5mM and corresponds to K; therefore, (A) is correct.

Question 23

Consider a reaction catalyzed by enzyme A with a Km value of 5 x 10^-6 M and Vmax of 20 mmol
8. At a concentration of 5 x 10^-6 M substrate, the rate of the reaction will be:
A. 10 mmol/min
B. 20 mmol/min
C. 30 mmol/min
D. 40 mmol/min

Answer 23

А. 10mmol/min
As with the last question, relationships are important. At a concentration of 5 x 10^-6 M, enzyme A is working at one-half of its max because the concentration is equal to the Km of the enzyme. Therefore, one-half of 20 mmol/min is 10 mmol/min which corresponds to (A).

Question 24

Consider a reaction catalyzed by enzyme A with a Km value of 5 x 10^-6 M and Vmax of 20 mmol/min

At a concentration of 5 x 10^-4 M substrate, the rate of the reaction will be:
A. 10 mmol/min
B. 15 mmol/min
C. 20 mmol/min
D. 30 mmol/min

Answer 24

C. 20 mmol/min
At a concentration of 5 × 10^-4 M, there is 100 times more substrate than present at half maximal velocity. At high values (significantly larger than the value of Km), the enzyme is at or near its max, which is 20 mmol/min

Question 25

The graph below shows kinetic data obtained for flu virus enzyme activity as a function of substrate concentration in the presence and absence of two antiviral drugs. Based on the graph, which of the following statements is correct? A. Both drugs are noncompetitive inhibitors of the viral enzyme. B. Oseltamivir increases the Km value for the substrate compared to Relenza. C. Zanamivir increases the Vmax Value for the substrate compared to Tamiflu. D. Both drugs are competitive inhibitors of the viral enzyme.

Answer 25

B Based on the graph, when the substrate is present, oseltamivir results in the same Vmax and a higher Km, compared to when no inhibitor is added. These are hallmarks of competitive inhibitors. Noncompetitive inhibitors result in decreased Vmax and the same Km, as the uninhibited reaction, which is shown by the zanamivir line in the graph. Because the question is only comparing the values between the two inhibitors, and not the enzyme without inhibitor, the mechanism of inhibition is less important to determine than the values of Km, and Vmax. This is a great example of why previewing the answer choices works well in the sciences.

Question 26

The conversion of ATP to cyclic AMP and inorganic phosphate is most likely catalyzed by which class of enzyme? A. Ligase B. Hydrolase C. Lyase D. Transferase

Answer 26

C. Lyase Lyases are responsible for the breakdown of a single molecule into two molecules without the addition of water or the transfer of electrons. Lyases often form cyclic compounds or double bonds in the products to accommodate this. Water was not a reactant, and no cofactor was men-tioned; thus lyase, (C), remains the best answer choice.

Question 27

Which of the following is NOT a method by which enzymes decrease the activation energy for biological reactions? A. Modifying the local charge environment B. Forming transient covalent bonds C. Acting as electron donors or receptors D. Breaking bonds in the enzyme irreversibly to provide energy

Answer 27

D. Breaking bonds in the enzyme irreversibly to provide energy Enzymes are not altered by the process of catalysis. A molecule that breaks intramolecular bonds to provide activation energy would not be able to be reused.

Question 28

A certain enzyme that displays positive cooperativity has four subunits, two of which are bound to substrate. Which of the following statements must be true? A. The affinity of the enzyme for the substrate has just increased. B. The affinity of the enzyme for the substrate has just decreased. C. The affinity of the enzyme for the substrate is half of what it would be if four sites had substrate bound. D. The affinity of the enzyme for the substrate is greater than with one substrate bound.

Answer 28

D. The affinity of the enzyme for the substrate is greater than with one substrate bound. Cooperative enzymes demonstrate a change in affinity for the substrate depending on how many substrate molecules are bound and whether the last change was accomplished because a substrate molecule was bound or left the active site of the enzyme. Because we cannot determine whether the most recent reaction was binding or dissociation, (A) and (B) are eliminated. We can make absolute comparisons though. For enzymes expressing positive cooperativity, the unbound enzyme has the lowest affinity for substrate, and the enzyme with all but one subunit bound has the highest. The increase in affinity is not necessarily linear. Further-more, if all four sites have substrate bound, the enzyme cannot bind to any more substrate. Therefore, (C) is not true. An enzyme with two subunits occupied must have a higher affinity for the substrate than the same enzyme with only one subunit occupied; thus, (D) is correct.

Question 29

Which of the following is LEAST likely to be required for a series of metabolic reactions? A. Triacyglycerol acting as a coenzyme B. Oxidoreductase enzymes C. Magnesium acting as a cofactor D. Transferase enzymes

Answer 29

А. Triacyglycerol acting as a coenzyme Triglycerides are unlikely to act as coenzymes for a few reasons, including their large size, neutral charge, and ubig-uity in cells. Cofactors and coenzymes tend to be small in size, such as metal ions like (C) or small organic molecules. They can usually carry a charge by ionization, protonation, or deprotonation. Finally, they are usually in low, tightly regulated concentrations within cells. Metabolic pathways would be expected to include both oxidation-reduction reactions and movement of functional groups, thus eliminating (B) and (D).

Question 30

How does the ideal temperature for a reaction change with and without an enzyme catalyst? A. The ideal temperature is generally higher with a catalyst than without. B. The ideal temperature is generally lower with a catalyst than without. C. The ideal temperature Is characteristic of the reaction, not the enzyme. D. No conclusion can be made without knowing the enzyme type.

Answer 30

B. The ideal temperature is generally lower with a catalyst than without. The rate of reaction increases with temperature because of the increased kinetic energy of the reactants, but reaches a peak temperature because the enzyme denatures with the disruption of hydrogen bonds at excessively high temperatures. In the absence of enzyme, this peak temperature is generally much hotter. Heating a reaction provides molecules with an increased chance of achieving the activation energy, but the enzyme catalyst would typically reduce activation energy. Keep in mind that thermodynamics and kinetics are not interchange-able, so we are not considering the impact of heat on the equilibrium position.

Question 31

At what pH can protein A best be obtained through electrophoresis? (Note: MM = molar mass) A. 2.5 В. 3.5 С. 4.5 D. 5.5

Answer 31

Question 32

What is the function of sodium dodecyl sulfate (SDS) in SDS-PAGE? A. SDS stabilizes the gel matrix, improving resolution during electrophoresis. B. SDS solubilizes proteins to give them uniformly negative charges, so the separation is based purely on size. C. SDS raises the pH of the gel, separating multiunit proteins into individual subunits. D. SDS solubilizes proteins to give them uniformly positive charges, so separation is based purely on pH.

Answer 32

Question 33

Which of the following is NOT involved in cell migration? A. Dynein B. Flagella C. Actin D. Centrioles

Answer 33

Question 34

Which of the following proteins is most likely to be found extracellularly? A. Tubulin B. Myosin C. Collagen D. Actin

Answer 34

Question 35

Hormones are found in the body in very low concentrations, but tend to have a strong effect. What type of receptor are hormones most likely to act on? I. Ligand-gated ion channels II. Enzyme-linked receptors III. G protein-coupled receptors A. I only B. III only C. Il and III only D. I, II, and III

Answer 35

C. II and III only For a ligand present in low quantities to have a strong action, we expect it to initiate a second messenger cascade system. Second messenger systems amplify signals because enzymes can catalyze a reaction more than once while they are active, and often activate other enzymes. Both enzyme-linked receptors and G protein-coupled receptors use second messenger systems, while ion channels do not.

Question 36

Which of the following is most likely to be found bound to a protein in the body? A. Sodium B. Potassium C. Chloride D. Calcium

Answer 36

Question 37

Which of the following characteristics is NOT attributed to antibodies? A. Antibodies bind to more than one distinct antigen. B. Antibodies label antigens for targeting by other immune cells. C. Antibodies can cause agglutination by interaction with antigen. D. Antibodies have two heavy chains and two light chains.

Answer 37

Question 38

Which ion channels are responsible for maintaining the resting membrane potential? A. Ungated channels B. Voltage-gated channels C. Ligand-gated channels D. No ion channels are involved in maintenance of the resting membrane potential.

Answer 38

Question 39

Which of the following is NOT a component of all trimeric G proteins? A. Gα B. Gβ C. Gγ D. Gi

Answer 39

D. Gi All trimeric G proteins have a, B, and y subunits-(A), (B), and (C), respectively. Gs, Gi, and Gq are subtypes of the Gα subunit of the trimeric G protein and differ depending on the G protein-coupled receptor's function.

Question 40

Which of the following methods would be best to separate large quantities of the following proteins? (Note: MM = molar mass) A. Ion-exchange chromatography B. Size-exclusion chromatography C. Isoelectric focusing D. Native PAGE

Answer 40

B. Size-exclusion chromatography The proteins described in the question differ primarily in their molecular weights. Their pI values are very close, so ion-exchange chromatography, (A), is not a good choice. The question specifies a large quantity, which is better processed through chromatography than through electrophoresis- (C) and D) —because the gel can only handle a small volume of protein.

Question 41

Which amino acids contribute most significantly to the pl of a protein? I. Lysine Il. Glycine III. Arginine A. I only B. I and II only C. I and III only D. Il and III only

Answer 41

C. I and III only The overall pl of a protein is determined by the relative number of acidic and basic amino acids. The basic amino acids are arginine, lysine, and histidine, and the acidic amino acids are aspartic acid and glutamic acid. Glycine's side chain is a hydrogen atom, so it will have the least contribution of all the amino acids.

Question 42

How does the gel for isoelectric focusing differ from the gel for traditional electrophoresis? A. Isoelectric focusing uses a gel with much larger pore sizes to allow for complete migration. B. Isoelectric focusing uses a gel with SDS added to encourage a uniform negative charge. C. Isoelectric focusing uses a gel with a pH gradient that encourages a variable charge. D. The gel is unchanged in isoelectric focusing; the protein mixture is treated before loading.

Answer 42

C The gel in isoelectric focusing uses a pH gradient. When a protein is in a region with a pH above its pl, it is negatively charged and moves toward the anode. When it is in a pH region below its pI, it is positively charged and moves toward the cathode. When the pH equals the pI, the migration of the protein is halted.

Question 43

Which protein properties allow UV spectroscopy to be used as a method of determining concentration? A. Proteins have partially planar characteristics in peptide bonds. B. Globular proteins cause scattering of light. C. Proteins contain aromatic groups in certain amino acids. D. All organic macromolecules can be assessed with UV spectroscopy.

Answer 43

C. Proteins contain aromatic groups in certain amino acids. UV spectroscopy is best used with conjugated systems of double bonds. While the double bond in the peptide bond does display resonance, this is not adequate for UV absorp-tion. However, aromatic systems are conjugated, and phe-nylalanine, tyrosine, and tryptophan all contain aromatic ring structures.

Question 44

A protein collected through affinity chromatography displays no activity even though it is found to have a high concentration using the Bradford protein assay. What best explains these findings? A. The Bradford reagent was prepared incorrectly. B. The active site is occupied by free ligand. C. The protein is bound to the column. D. The protein does not catalyze the reaction of interest.

Answer 44

B. The active site is occupied by free ligand. Protein activity and concentration are generally correlated. Because we have a high concentration of protein, we expect a high activity unless the protein has been damaged or inactivated in some way. The protein could have been inactivated by experimental conditions like detergents, heat, or pH; however, these are not answer choices. Rather, we must consider how the experimental procedure works. Protein elutes off of an affinity column by binding free ligand. In this situation, the binding may not have been reversed and thus the free ligand competes for the active site of the enzyme, lowering its activity.

Question 45

What property of protein-digesting enzymes allows for a sequence to be determined without fully degrading the protein? A. Selectivity B. Sensitivity C. Turnover D. Inhibition

Answer 45

A. Selectivity The selective cleavage of proteins by digestive enzymes allows fragments of different lengths with known amino acid endpoints to be created. By cleaving the protein with several different enzymes, a basic outline of the amino acid sequence can be created.

Question 46

When glucose is In a straight-chain formation, it: A. is an aldoketose. B. is a pentose. C. has five chiral carbons. D. is one of a group of 16 stereoisomers.

Answer 46

D. is one of a group of 16 stereoisomers. Glucose is an aldohexose, meaning that it has one aldehyde group and six carbons. Given this information, (A) and (B) can be eliminated. In aldose sugars, each nonterminal carbon is chiral. Therefore, glucose has four chiral centers, not five, as mentioned in (C). The number of stereoisomers possible for a chiral molecule is 2", where n is the number of chiral carbons. Because glucose has four chiral centers, there are 24 = 16 possible stereoisomers.

Question 47

All of the following are true of epimers EXCEPT: A. they differ in configuration about only one carbon. B. they usually have slightly different chemical and physical properties. c. they are diastereomers (with the exception of glyceraldehyde). D. they have equal but opposite optical activities.

Answer 47

D. they have equal but opposite optical activities. Epimers are monosaccharide diastereomers that differ in their configuration about only one carbon. As with all diastereo-mers, epimers have different chemical and physical proper-ties, and their optical activities have no relation to each other. Enantiomers have equal but opposite optical activities. There-fore, (D) is the only statement that does not apply to epimers.

Question 48

Aldonic acids are compounds that: A. can be oxidized, and therefore act as reducing agents. B. can be reduced, and therefore act as reducing agents. C. have been oxidized, and have acted as reducing agents. D. have been oxidized, and have acted as oxidizing agents.

Answer 48

C. have been oxidized, and have acted as reducing agents. Aldonic acids form after the aldehyde group on a reducing sugar reduces another compound, becoming oxidized in the process.

Question 49

The formation of a-d-glucopyranose from B-D-glucopyranose is called: A. glycosidation. B. mutarotation. C. enantiomerization. D. racemization.

Answer 49

B. mutarotation Mutarotation is the interconversion between anomers of a compound. Enantiomerization and racemization, (C) and (D), are related: enantiomerization is the formation of a mirror-image or optically inverted form of a compound, whereas racemization is moving a solution toward an equal concentration of both enantiomers. Glycosidation, (A), is the addition of a sugar to another compound.

Question 50

Ketose sugars may have the ability to act as reducing sugars. Which process explains this? A. Ketose sugars undergo tautomerization. B. The ketone group is oxidized directly. C. Ketose sugars undergo anomerization. D. The ketone group is reduced directly.

Answer 50

A. Ketose sugars undergo tautomerization. Ketose sugars undergo tautomerization, a rearrangement of bonds, to undergo keto-enol shifts. This forms an aldose, which then allows them to act as reducing sugars. A ketone group alone cannot be oxidized. Anomerization, mentioned in (C), refers to ring closure of a monosaccharide, creating an anomeric carbon.

Question 51

What is the product of the following reaction?

Answer 51

B. When glucose reacts with methanol under acid catalysis, the hemiacetal is converted to an acetal via replacement of the anomeric hydroxyl group with an alkoxy group. The result is a type of acetal known as a glycoside. This corresponds with (B). The other choices all show alkoxy groups on the wrong carbon, or too many carbons.

Question 52

Which of the following enzymes cleaves polysaccharide chains and yields maltose exclusively? A. α-Amylase B. β-Amylase c. Debranching enzyme D. Glycogen phosphorylase

Answer 52

B. β-Amylase β-Amylase cleaves amylose at the nonreducing end of the polymer to yield maltose exclusively, while a-amylase, (A), cleaves amylose anywhere along the chain to yield short polysaccharides, maltose, and glucose. Debranching enzyme, (C), removes oligosaccharides from a branch in glycogen or starches, while glycogen phosphorylase, (D), yields glucose 1-phosphate.

Question 53

When the following straight-chain Fischer projection is converted to a chair or ring conformation, its structure will be:

Answer 53

C. Start by drawing out the Haworth projection. Recall that all the groups on the right in the Fischer projection will go on the bottom of the Haworth projection, and all the groups on the left will go on the top. Next, draw the chair structure, with the oxygen in the back right corner. Label the carbons in the ring 1 through 5, starting from the oxygen and moving clockwise around the ring. Now, draw in the lines for all the axial substituents, alternating above and below the ring. Remember to start on the anomeric C-1 carbon, where the axial substituent points down. Now start filling in the substituents. The substituent can be in either position on the anomeric carbon, so skip that one for now. The -OH groups on C-2 and C-4 should point downward while the -OH group on C-3 should point upward; (C), the B-anomer of D-glucose, is the only one that matches.

Question 54

Why is the α-anomer of D-glucose less likely to form than the β-anomer? A. The β-anomer is preferred for metabolism. B. The β-anomer undergoes less electron repulsion. C. The α-anomer is the more stable anomer. D. The α-anomer forms more in L-glucose.

Answer 54

B. The β-anomer undergoes less electron repulsion. The hydroxyl group on the anomeric carbon of the β-anomer is equatorial, thereby creating less nonbonded strain than the α-anomer, which has the hydroxyl group of the anomeric carbon in axial position.

Question 55

Which two polysaccharides share all of their glycosidic linkage types in common? A. Cellulose and amylopectin B. Amylose and glycogen C. Amylose and cellulose D. Glycogen and amylopectin

Answer 55

D. Glycogen and amylopectin Glycogen and amylopectin are the only polysaccharide forms that demonstrate branching structure, making them most similar in terms of linkage. Both glycogen and amylopectin use a-1,4 and a-1,6 linkages. Cellulose uses B-1,4 linkages and amylose does not contain a-1,6 linkages.

Question 56

Which of the following is digestible by humans and Is made up of only one type of monosaccharide? A. Lactose B. Sucrose C. Maltose D. Cellobiose

Answer 56

C. Maltose While maltose and cellobiose both have the same glucose subunits, only maltose is digestible by humans because the β-glycosidic linkages in cellobiose cannot be cleaved in the human body.

Question 57

The reaction below is an example of one step in: A. aldehyde formation. B. hemiketal formation. C. mutarotation. D. glycosidic bond cleavage.

Answer 57

C In solution, the hemiacetal ring of glucose will break open spontaneously and then re-form. When the ring is broken, bond rotation occurs between C-1 and C-2 to produce either the a- or the B-anomer. The reaction given in this question depicts the mutarotation of glucose. (A) is incorrect because the reactant is an aldehyde, not the product. (B) is incorrect because a hemiketal has an - OH group, an -OR group, and two - R groups. In addition, hemike-tals are formed from ketones, and our starting reactant is an aldehyde. Finally, (D) is incorrect because there is no glycosidic bond in the starting reactant.

Question 58

Galactose is the C-4 epimer of glucose, the structure of which is shown below. Which of the following structures is galactose?

Answer 58

A Galactose is a diastereomer of glucose, with the stereochemistry at C-4 (counting from the aldehyde) reversed. Being able to identify C-4 is enough to answer this ques-tion, even without looking at the glucose molecule. Because (B), (C), and (D) have identical stereochemistry at C-4, they are incorrect.

Question 59

Andersen's disease (glycogen storage disease type IV) is a condition characterized by a deficiency in glycogen branching enzyme. Absence of this enzyme would be likely to cause all of the following effects EXCEPT: A. decreased glycogen solubility in human cells. B. slower action of glycogen phosphorylase. C. less storage of glucose in the body. D. glycogen devoid of α-1,4 linkages.

Answer 59

D. glycogen devoid of α-1,4 linkages In Andersen's disease, glycogen is less branched than normal, thereby inducing lower solubility of glycogen. Branches reduce the interactions between adjacent chains of glycogen and encourage interactions with the aqueous environment. The smaller number of branches means that glycogen phosphorylase has fewer terminal glucose monomers on which to act, making enzyme activity slower than normal overall. Finally, without branches, the density of glucose monomers cannot be as high; therefore, the total glucose stored is lower than normal. Glycogen synthase is still functioning normally, so we would expect normal a-1,4 linkages in the glycogen of an individual with Andersen's disease but few if any) a-1,6 linkages.

Question 60

The cyclic forms of monosaccharides are: I. hemiacetals. Il. hemiketals. Il. acetals. A. I only B. III only C. I and II only D. I, II, and III

Answer 60

C. I and II only Monosaccharides can exist as hemiacetals or hemiketals, depending on whether they are aldoses or ketoses. When a monosaccharide is in its cyclic form, the anomeric carbon is attached to the oxygen in the ring and a hydroxyl group. Hence, it is only a hemiacetal or hemiketal because an acetal or ketal would require the -OH group to be converted to another - OR group.

Question 61

Which of the following is NOT a type of glycolipid? A. Cerebroside B. Globoside C. Ganglioside D. Sphingomyelin

Answer 61

D. Sphingomyelin Glycolipids contain sugar moieties connected to their back-bone. Sphingomyelin is not a glycolipid, but rather a phos-pholipid. This class can either have phosphatidylcholine or phosphatidylethanolamine as a head group and therefore contains a phosphodiester, not glycosidic, bond.

Question 62

During saponification: A. triacylglycerols undergo a condensation reaction. B. triacyglycerols undergo ester hydrolysis. C. fatty acid salts are produced using a strong acid D. fatty acid salts are bound to albumin.

Answer 62

B. Saponification is the ester hydrolysis of triacylglycerol using a strong base like sodium or potassium hydroxide to form glycerol and fatty acid salts. This is not a condensation reac-tion, as in (A), but a cleavage reaction. Fatty acids do travel in the body bonded to serum albumin, as in (D), but that is unrelated to the process of saponification.

Question 63

Which of the following best describes the structure of steroids? A. Three cyclopentane rings, one cyclohexane ring B. Three cyclohexane rings, one cyclopentane ring C. Four carbon rings, differing in structure for each steroid D. Three cyclic carbon rings and a functional group

Answer 63

В. Three cyclohexane rings, one cyclopentane ring The basic backbone of steroid structure contains three cyclohexane rings and one cyclopentane ring. Although the oxidation status of these rings varies for different steroids, the overall structure does not, as in (C).

Question 64

Soap bubbles form because fatty acid salts organize into: A. lysosomes. B. micelles. C. phospholipid bilayers. D. hydrogen bonds.

Answer 64

B. Micelles Fatty acid salt micelles are responsible for the formation of soap bubbles. While phospholipids can form bilayers, as in (C), the fatty acids in soap are free fatty acids, not phospholipids.

Question 65

Steroid hormones are steroids that: 1. have specific high-affinity receptors. Il. travel in the bloodstream from endocrine glands to distant sites. Ill. affect gene transcription by binding directly to DNA. A. I only B. Ill only C. I and II only D. I and III only

Answer 65

C. I and II only Steroid hormones are produced in endocrine glands and travel in the bloodstream to bind high-affinity receptors in the nucleus. The hormones receptor binds to DNA as part of the hormone-receptor complex, but the hormone itself does not.

Question 66

Why are tricylglycerols used in the human body for energy storage? A. They are highly hydrated and therefore can store lots of energy B. They always have short fatty acid chains for easy access by metabolic enzymes. C. The carbon atoms of the fatty acid chains are highly reduced and therefore yield more energy upon oxidation. D. Polysaccharides, which would actually be a better energy storage form, would dissolve in the body.

Answer 66

C. The carbon atoms of the fatty acid chains are highly reduced and therefore yield more energy upon oxidation. Triacyglycerols are highly hydrophobic and therefore not highly hydrated (which would add extra weight from the water of hydration, taking away from the energy density of these molecules), eliminating (A). The fatty acid chains produce twice as much energy as polysaccharides during oxidation because they are highly reduced. The fatty acid chains vary in length and saturation.

Question 67

Which of the following is correct about fat-soluble Vitamins? I. Vitamin E is important for calcium regulation. II. Vitamin D protects against cancer because it is a biological antioxidant. III. Vitamin K is necessary for the posttranslational introduction of calcium-binding sites. IV. Vitamin A is metabolized to retinal, which is important for sight. A. Ill only B. I and II only C. Ill and IV only D. II, II, and IV only

Answer 67

C. III and IV only Vitamin A is metabolized to retinal, which is important for sight. Vitamin D is metabolized to calcitriol, which is important for calcium regulation. Vitamin E is made up of tocopherols, which are biological antioxidants. Vitamin K is necessary for the introduction of calcium binding sites, such as during the posttranslational modification of pro-thrombin.

Question 68

Which of the following is true of amphipathic molecules? A. They form protective spheres in any solvent, with hydrophobic molecules interior and hydrophilic molecules exterior. B. They have two fatty acid chains and a polar head group. C. They are important to the formation of the phospholipid bilayer and soap bubbles. D. They have a glycerol base.

Answer 68

C. They are important to the formation of the phospholipid bilayer and soap bubbles. Phospholipids are amphipathic, as are fatty acid salts. Although amphipathic molecules take spherical forms with hydrophobic molecules interior in aqueous solu-tion, as in (A), the opposite would be true in a nonpolar solvent. (B) describes phospholipids and sphingolipids, and (D) describes triacylglycerols and phospholipids; both groups do not include fatty acid salts.

Question 69

Which of the following is/are true about sphingolipids? I. They are all phospholipids. II. They all contain a sphingosine backbone. Ill. They can have either phosphodiester or glycosidic linkages to their polar head groups. A. I only B. Ill only c. II and III only D. I, II, and III

Answer 69

B. III only Sphingolipids can either have a phosphodiester bond, and therefore be phospholipids, or have a glycosidic linkage and therefore be glycolipids. Not all sphingolipids have a sphingosine backbone, as in statement II; some have related (sphingoid) compounds as backbones instead.

Question 70

Which of the following statements about saturation is FALSE? A. It can describe the number of double or triple bonds in a fatty acid tail. B. It determines at least one of the properties of membranes. C. More saturated fatty acids make for a more fluid solution. D. Fully saturated fatty acids have only single bonds.

Answer 70

C. More saturated fatty acids make for a more fluid solution. More saturated fatty acids make for a less fluid solution. This is because they can pack more tightly and form more noncovalent bonds, resulting in more energy being needed to disrupt the overall structure.

Question 71

Which of the following is true about glycerophospholipids? A. Glycerophospholipids can sometimes be sphin-golipids, depending on the bonds in their head groups. B. Glycerophospholipids are merely a subset of phospholipids. C. Glycerophospholipids are used in the ABO blood typing system. D. Glycerophospholipids have one glycerol, one polar head group, and one fatty acid tail.

Answer 71

В. Glycerophospholipids are merely a subset of phospholipids. Glycerophospholipids are a subset of phospholipids, as are sphingomyelins. Glycerophospholipids are never sphingo-lipids because they contain a glycerol backbone (rather than sphingosine or a sphingoid backbone), eliminating (A). Sphingolipids are used in the ABO blood typing system, eliminating (C). Glycerophospholipids have a polar head group, glycerol, and two fatty acid tails, not one, as in (D).

Question 72

Which of the following statements about terpenes is FALSE? A. Terpenes are strongly scented molecules that sometimes serve protective functions. B. Terpenes are steroid precursors. C. A triterpene is made of three isoprene moieties, and therefore has 15 carbons. D. Terpenes are made by plants and insects.

Answer 72

C. A triterpene is made of three isoprene moieties, and therefore has 15 carbons. A triterpene is made of six isoprene moieties (remember, one terpene unit contains two isoprene units), and therefore has a 30-carbon backbone.

Question 73

Which of the following is true about cholesterol? A. Cholesterol always increases membrane fluidity in cells. B. Cholesterol is a steroid hormone precursor. C. Cholesterol is a precursor for vitamin A, which is produced in the skin. D. Cholesterol interacts only with the hydrophobic tails of phospholipids.

Answer 73

B. Cholesterol is a steroid hormone precursor. Cholesterol is a steroid hormone precursor that has variable effects on membrane fluidity depending on temperature, eliminating (A). It interacts with both the hydrophobic tails and the hydrophilic heads of membrane lipids, nullifying (D). It is also a precursor for vitamin D (not vitamin A), which can be produced in the skin in a UV-driven reaction, eliminating (C).

Question 74

Which of the following statements regarding prostaglandins is FALSE? A. Prostaglandins regulate the synthesis of CAMP. B. Prostaglandin synthesis is inhibited by NSAIDs. C. Prostaglandins affect pain, inflammation, and smooth muscle function. D. Prostaglandins are endocrine hormones, like steroid hormones.

Answer 74

D. Prostaglandins are endocrine hormones, like steroid hormones. Prostaglandins are paracrine or autocrine signaling mol-ecules, not endocrine-they affect regions close to where they are produced, rather than affecting the entire body. Think of the swelling that happens when you bash your knee into your desk: your knee will swell, become discol-ored, and possibly bruise. Luckily, however, your entire body won't swell as well.

Question 75

Which of the statements regarding waxes is FALSE? A. Waxes generally have melting points above room temperature. B. Waxes are produced only in plants and insects and therefore must be consumed by humans. C. Waxes protect against dehydration and parasites. D. Waxes are esters of long-chain fatty acids and long-chain alcohols.

Answer 75

B. Waxes are produced only in plants and insects and therefore must be consumed by humans. Waxes are also produced in animals for similar protective functions, Cerumen, or earwax, is a prime example in humans.

Question 76

In a single strand of a nucleic acid, nucleotides are linked by: A. Hydrogen bonds B. Phosphodiester bonds C. lonic bonds. D. van der Waals forces.

Answer 76

B. Phosphodiester bonds Nucleotides bond together to form polynucleotides. The 3' hydroxyl group of one nucleotides sugar joins the 5' hydroxyl group of the adjacent nucleotide's sugar by a phos-phodiester bond. Hydrogen bonding, (A), is important for holding complementary strands together, but does not play a role in the bonds formed between adjacent nucleotides on a single strand.

Question 77

Which of the following statements regarding differences between DNA and RNA is FALSE? A. In cells, DNA is double-stranded, whereas RNA is single-stranded. B. DNA uses the nitrogenous base thymine; RNA uses uracil. C. The sugar in DNA is deoxyribose; the sugar in RNA is ribose. D. DNA strands replicate in a 5' to 3' direction, whereas RNA is synthesized in a 3' to 5' direction.

Answer 77

D. DNA strands replicate in a 5' to 3' direction, whereas RNA is synthesized in a 3' to 5' direction. Because we are looking for the false statement, we have to read each choice to eliminate those that are true or find one that is overtly false. Let's quickly review the main differences between DNA and RNA. In cells, DNA is double-stranded, with a deoxyribose sugar and the nitrogenous bases A, T, C, and G. RNA, on the other hand, is usually single-stranded, with a ribose sugar and the bases A, U, C, and G. (D) is false because both DNA replication and RNA synthesis proceed in a 5' to 3' direction.

Question 78

Which of the following DNA sequences would have the highest melting temperature? A. CCAACCATCCG B. CGCAATAATACA C. CGTAATAATACA D. CATAACAAATCA

Answer 78

A. CCAACCATCCG The melting temperature of DNA is the temperature at which a DNA double helix separates into two single strands (denatures). To do this, the hydrogen bonds linking the base pairs must be broken. Cytosine binds to guanine with three hydrogen bonds, whereas adenine binds to thymine with two hydrogen bonds. The amount of heat needed to disrupt the bonding is proportional to the number of bonds. Thus, the higher the GC-content in a DNA segment, the higher the melting point.

Question 79

Which of the following biomolecules is LEAST likely to contain an aromatic ring? A. Proteins B. Purines C. Carbohydrates D. Pyrimidines

Answer 79

C. Carbohydrates Aromatic rings must contain conjugated i electrons, which require alternating single and multiple bonds, or lone pairs. In carbohydrate ring structures, only single bonds are present, thus preventing aromaticity. Nucleic acids contain aromatic heterocycles, while proteins will generally contain at least one aromatic amino acid (tryptophan, phenylala-nine, or tyrosine).

Question 80

For a compound to be aromatic, all of the following must be true EXCEPT: A. the molecule is cyclic. B. the molecule contains 4n + 2 m electrons. C. the molecule contains alternating single and double bonds. D. the molecule is planar.

Answer 80

C. the molecule contains alternating single and double bonds. For a compound to be aromatic, it must be cyclic, planar, conjugated, and contain 4n + 2 i electrons, where n is any integer. Conjugation requires that every atom in the ring have at least one unhybridized p-orbital. While most examples of aromatic compounds have alternating single and double bonds, compounds can be aromatic if they contain triple bonds as well; this would still permit at least one unhybridized p-orbital.

Question 81

Which of the following enzymes is NOT involved in DNA replication? A. Primase B. DNA ligase C. RNA polymerase I D. Telomerase

Answer 81

С. RNA polymerase I During DNA replication, the strands are separated by DNA helicase. At the replication fork, primase, (A), creates a primer for the initiation of replication, which is followed by DNA polymerase. On the lagging strand, Okazaki fragments form and are joined by DNA ligase, (B). After the chromosome has been processed, the ends, called telo-meres, are replicated with the assistance of the enzyme telo-merase, (D). RNA polymerase I is located in the nucleolus and synthesizes rRNA.

Question 82

How is DNA best characterized? A. CDNA results from a DNA transcript with noncod. ing regions removed. B. CDNA results from the reverse transcription of processed mRNA. C. CDNA is the abbreviation for deoxycytosine. D. CDNA is the circular DNA molecule that forms the bacterial genome.

Answer 82

B. CDNA results from the reverse transcription of processed mRNA. CDNA (complementary DNA) is formed from a processed mRNA strand by reverse transcription. cDNA is used in DNA libraries and contains only the exons of genes that are transcriptionally active in the sample tissue.

Question 83

Endonucleases are used for which of the following? 1. Gene therapy II. Southern blotting III. DNA repair A. I only B. Il only C. Il and III only D. I, II, and III

Answer 83

D. I, II, and III Endonucleases are enzymes that cut DNA. They are used by the cell for DNA repair. They are also used by scientists during DNA analysis, as restriction enzymes are endonu-cleases. Restriction enzymes are used to cleave DNA before electrophoresis and Southern blotting, and to introduce a gene of interest into a viral vector for gene therapy.

Question 84

Which of the following statements regarding polymerase chain reaction is FALSE? A. Human DNA polymerase is used because it is the most accurate. B. A primer must be prepared with a complementary sequence to part of the DNA of interest. C. Repeated heating and cooling cycles allow the enzymes to act specifically and replaces helicase. D. Each cycle of the polymerase chain reaction doubles the amount of DNA of interest.

Answer 84

A. Human DNA polymerase is used because it is the most accurate. The polymerase chain reaction is used to clone a sequence of DNA using a DNA sample, a primer, free nucleotides, and enzymes. The polymerase from Thermus aquaticus is used because the reaction is regulated by thermal cycling, which would denature human enzymes.

Question 85

How does prokaryotic DNA differ from eukaryotic DNA? I. Prokaryotic DNA lacks nucleosomes. II. Eukaryotic DNA has telomeres. IlI. Prokaryotic DNA is replicated by a different DNA polymerase. IV. Eukaryotic DNA is circular when not restricted by centromeres. A. I only B. IV only c. II and III only D. I, Il, and III only

Answer 85

D. I, Il, and III only Prokaryotic DNA is circular and lacks histone proteins, and thus does not form nudeosomes. Both prokaryotic and eukaryotic DNA are replicated by DNA polymerases, although these polymerases differ in identity. Eukaryotic DNA is organized into chromatin, which can condense to form linear chromosomes; only prokaryotes have circular chromosomes. Only eukaryotic DNA has telomeres.

Question 86

Why might uracil be excluded from DNA but NOT RNA? A. Uracil is much more difficult to synthesize than thymine. B. Uracil binds adenine too strongly for replication. C. Cytosine degradation results in uracil. D. Uracil is used as a DNA synthesis activator.

Answer 86

C. Cytosine degradation results in uracil. One common DNA mutation is the transition from cytosine to uracil in the presence of heat. DNA repair enzymes recognize uracil and correct this error by excising the base and inserting cytosine. RNA exists only transiently in the cell, such that cytosine degradation is insignificant. Were uracil to be used in DNA under normal circumstances, it would be impossible to tell if a base should be uracil or if it is a damaged cytosine nucleotide.

Question 87

Tumor suppressor genes are most likely to result in cancer through: A. loss of function mutations. B. gain of function mutations. C. overexpression. D. proto-oncogene formation.

Answer 87

A. Loss of function mutations Oncogenes are most likely to result in cancer through activation, (B), while tumor suppressor genes are most likely to result in cancer through inactivation.

Question 88

Which of the following is an ethical concern of gene sequencing? A. Gene sequencing is invasive, thus the potential health risks must be thoroughly explained. B. Gene sequencing impacts relatives, thus privacy concerns may be raised. C. Gene sequencing is very inaccurate, which increases anxiety related to findings, D. Gene sequencing can provide false-negative results, giving a false sense of security

Answer 88

B. Gene sequencing impacts relatives, thus privacy concerns may be raised. One of the primary ethical concerns related to gene sequencing is the issue of consent and privacy. Because genetic screening provides information on direct relatives, there are potential violations of privacy in communicating this information to family members who may be at risk. There are not significant physical risks, eliminating (A), and gene sequencing is fairly accurate, eliminating (C) and (D).

Question 89

Which of the following is NOT a difference between heterochromatin and euchromatin? A. Euchromatin has areas that can be transcribed, whereas heterochromatin is silent. B. Heterochromatin is tightly packed, whereas euchromatin is less dense. C. Heterochromatin stains darkly, whereas euchro-matin stains lightly. D. Heterochromatin is found in the nucleus, whereas euchromatin is in the cytoplasm.

Answer 89

D. Heterochromatin is found in the nucleus, whereas euchromatin is in the cytoplasm. Euchromatin has a classic "beads on a string" appearance that stains lightly, while heterochromatin is tightly packed and stains darkly. Heterochromatin is primarily composed of inactive genes or untranslated regions, while euchromatin is able to be expressed. All chromatin is found in the nucleus, not the cytoplasm.

Question 90

During which phase of the cell cycle are DNA repair mechanisms least active? A. G1 B. S C. G2 D. M

Answer 90

D. M Mismatch repair mechanisms are active during S phase (proofreading) and G2, phase (MSH2 and MLH1), eliminating (B) and (C). Nucleotide and base excision repair mechanisms are most active during the G1, and G2, phases, also eliminating (A). These mechanisms exist during interphase because they are aimed at preventing propagation of the error into daughter cells during M phase (mitosis).

Question 91

What role does peptidyl transferase play in protein synthesis? A. It transports the initiator aminoacyl-tRNA complex. B. It helps the ribosome to advance three nucleotides along the mRNA in the 5' to 3' direction. C. It holds the protein in its tertiary structure. D. It catalyzes the formation of a peptide bond

Answer 91

D. It catalyzes the formation of a peptide bond Peptidyl transferase is an enzyme that catalyzes the formation of a peptide bond between the incoming amino acid in the A site and the growing polypeptide chain in the P site. Initiation and elongation factors help transport charged tRNA molecules into the ribosome and advance the ribosome down the mRNA transcript, as in (A) and (B). Chaperones maintain a protein's three-dimensional shape as it is formed, as in (C).

Question 92

A mutation in which of the following components of the lac operon would lead to a significant reduction in the expression of lactase (one of the structural genes)? A. Operator B. Regulatory gene C. Promoter D. Structural genes

Answer 92

C. Promoter The promoter functions as a recruitment site for RNA polymerase and is required for gene expression. A mutation in this region would hinder RNA polymerase recruitment, resulting in a decrease in gene expression. These observations support (C) as the correct answer. By contrast, mutations in the operator and regulatory genes would render the repressor less able to block transcription, leading to the opposite effect, i.e. an increase in lactase expression. This result eliminates both (A) and (B). Finally, mutations in the structural genes could lead to the production of a nonfunctional enzyme, but are unlikely to affect gene expression, eliminating (D).

Question 93

Topoisomerases are enzymes involved in: A. DNA replication and transcription. B. posttranscriptional processing. C. RNA synthesis and translation. D. posttranslational processing.

Answer 93

A. DNA replication and transcription. Topoisomerases, such as prokaryotic DNA gyrase, are involved in DNA replication and mRNA synthesis (tran-scription). DNA gyrase is a type of topoisomerase that enhances the action of helicase enzymes by the introduc-ion of negative supercoils into the DNA molecule. These negative supercoils facilitate DNA replication by keeping the strands separated and untangled.

Question 94

Val-tRNAVal is the tRNA that carries valine to the ribosome during translation. Which of the following sequences gives an appropriate anticodon for this ERNA? (Note: Refer to Figure 7.5 for a genetic code table.) A. CAU B. AUC C. UAC D. GUG

Answer 94

C. UAC There are four different codons for valine: GUU, GUC, GUA, and GUG. Through base-pairing, we can determine that the proper anticodon must end with "AC. Remember that the codon and anticodon are antiparallel to each other, and that nucleic acids are always written 5' → 3' on the MCAT. Therefore, we are looking for an answer that ends with "AC" (rather than starting with "CA").

Question 95

Enhancers are transcriptional regulatory sequences that function by enhancing the activity of: A. RNA polymerase at a single promoter site. B. RNA polymerase at multiple promoter sites. C. spliceosomes and lariat formation in the ribosome. D. transcription factors that bind to the promoter but not to RNA polymerase.

Answer 95

A. RNA polymerase at a single promoter site Specific transcription factors bind to a specific DNA sequence, such as an enhancer, and to RNA polymerase at a single promoter sequence. They enable the RNA polymerase to transcribe the specific gene for that enhancer more efficiently.

Question 96

In the genetic code of human nuclear DNA, one of the codons specifying the amino acid tyrosine is UAC. If one nucleotide is changed and the codon is mutated to UAG, what type of mutation will occur A. Silent mutation B. Missense mutation C. Nonsense mutation D. Frameshift mutation

Answer 96

C. Nonsense mutation UAG is one of the three known stop codons, so changing tyrosine to a stop codon must be a nonsense (or truncation) mutation.

Question 97

Which of the following is NOT used by eukaryotes to increase the transcription of a gene? A. Gene duplication B. Histone acetylation C. DNA methylation D. Enhancers

Answer 97

C. DNA methylation In this question, the correct answer is the answer that is not associated with increased transcription. DNA methylation, (C), is associated with silencing of gene expression. Regions of DNA with poor gene expression, called heterochromatin, are heavily methylated. These modifications significantly decrease the ability of RNA polymerase to access DNA.

Question 98

When trypsin converts chymotrypsinogen to chymotrypsin, some molecules of chymotrypsin bind to a repressor, which in turn binds to an operator region and prevents further transcription of trypsin. This is most similar to which of the following operons? A. trp operon during lack of tryptophan B. trp operon during abundance of tryptophan C. lac operon during lack of lactose D. lac operon during abundance of lactose

Answer 98

B. trp operon during abundance of tryptophan The example given is a sample of repression due to the abundance of a corepressor. In other words, this is a repressible system that is currently blocking transcription. For the trp operon, an abundance of tryptophan in the environment allows for the repressor to bind tryptophan and then to the operator site. This blocks transcription of the genes required to synthesize tryptophan within the cell. The system described is a repressible system; the lac operon is an inducible system, in which an inducer binds to the repressor, thus permitting transcription.

Question 99

Which of the following RNA molecules or proteins is NOT found in the spliceosome during intron excision? A. snRNA B. hnRNA C. shRNA D. snRNPS

Answer 99

C. shRNA shRNA (short hairpin RNA) is a useful biotechnology tool used in RNA interference. It is not, however, produced in the nucleus for use in the spliceosome. It targets mRNA to be degraded in the cytoplasm; it is not utilized in splicing of the nRNA (heterogeneous nuclear RNA). sRNA (small nuclear RNA) and snRNPs (small nuclear ribonucleopro-teins), however, do bind to the nRNA to induce splicing.

Question 100

A 4-year-old toddler with cystic fibrosis (CF) is seen by his physician for an upper respiratory infection. Prior genetic testing has shown that there has been a deletion of three base pairs in exon 10 of the CFTR gene that affects codons 507 and 508. The nucleotide sequence in this region for normal and mutant alleles is shown below X denotes the missing nucleotide): What effect will this mutation have on the amino acid sequence of the protein encoded by the CFTR gene? (Note: Refer to Figure 7.5 for a genetic code table.) A. Deletion of a phenylalanine residue with no change in the C-terminus sequence B. Deletion of a leucine residue with no change in the C-terminus sequence C. Deletion of a phenylalanine residue with a change in the C-terminus sequence D. Deletion of a leucine residue with a change in the C-terminus sequence

Answer 100

A. In this table, we are given the sequence of the sense (coding) DNA strand. This will be identical to the mRNA transcript, except all thymine nucleotides will be replaced with uracil. With the deletion of these three bases, codon 507 changes from AUC to AUU in the transcript; these both code for isoleucine due to wobble. However, codon 508 (UUU in the transcript) has been lost. UUU codes for phenylalanine. The C-terminus sequence will remain unchanged because the deletion of three bases (exactly one codon) will not throw off the reading frame. For reference, the mutant reading frames would be: (Note: refer back to Figure 7.5 for a table of the genetic code)

Question 101

A gene encodes a protein with 150 amino acids. There is one intron of 1000 base pairs (bp), a 5'-untranslated region of 100 bp, and a 3'-untranslated region of 200 bp. In the final mRNA, about how many bases lie between the start AUG codon and final termination codon? A. 150 В. 450 С. 650 D. 1750

Answer 101

B. 450 The intron will not be a part of the final, processed mRNA, and the untranslated regions of the mRNA will not be turned into amino acids. Translation will begin with codon 1 (which would be AUG). Because there are 150 amino acids, we can surmise that there will be 151 codons. Each codon will use 3 nucleotides, so 150 x 3 = 450 because codon 151 will be the stop codon.

Question 102

Peptidyl transferase connects the carboxylate group of one amino acid to the amino group of an incoming amino acid. What type of linkage is created in this peptide bond? A. Ester B. Amide C. Anhydride D. Ether

Answer 102

B. Amide Peptidyl transferase connects the incoming amino terminal to the previous carboxyl terminal; the only functional group listed here with a carbonyl and amino group is the amide. Peptide bonds are thus amide linkages, and the correct answer is (B).

Question 103

A eukaryotic cell has been found to exhibit a truncation mutation that creates an inactive RNA polymerase I enzyme. Which type of RNA will be affected by this inactivation? A. rRNA B. tRNA C. snRNA D. hnRNA

Answer 103

A. rRNA RNA polymerase I in eukaryotes is found in the nucleolus and is in charge of transcribing most of the RNA for use during ribosomal creation. RNA polymerase Il is responsible for hRNA and sRNA. RNA polymerase III is responsible for tRNA and the 5S rRNA.

Question 104

You have just sequenced a piece of DNA that reads as follows: 5'-TCTTTGAGACATCC-3' What would the base sequence of the mRNA transcribed from this DNA be? A. 5'—AGAAACUCUGUAGG—3' B. 5'—GGAUGUCUCAAAGA-3' C. 5'—AGAAACTCTGTAGG—3' D. 5'-GGATCTCTCAAAGA-3'

Answer 104

B. 5'—GGAUGUCUCAAAGA-3' To answer this question correctly, we must remember that mRNA will be antiparallel to DNA. Our answer should be 5' to 3' mRNA, with the 5' end complementary to the 3' end of the DNA that is being transcribed. Thus, the mRNA transcribed from this strand will be 5' -GGAUGUCUCAAAGA-3'. mRNA contains uracil, rather than thymine.

Question 105

Double-stranded RNA cannot be translated by the ribosome and is marked for degradation in the cell. Which of the following strands of RNA would prevent mature mRNA in the cytoplasm from being translated? A. Identical mRNA to the one produced B. Antisense mRNA to the one produced C. mRNA with thymine substituted for uracil D. Sense mRNA to the one produced

Answer 105

B. Antisense mRNA to the one produced The mRNA produced has the same structure as the sense strand of DNA (with uracils instead of thymines). Because bonding of nucleic acids is always complementary but antiparallel, the antisense strand of mRNA would be the one that binds to the produced mRNA, creating double-stranded RNA that is then degraded once found in the cytoplasm.

Question 107

A student is trying to determine the type of membrane transport occurring in a cell. The student finds that the molecule to be transported is very large and polar, and when transported across the membrane, no energy is required. Which of the following is the most likely mechanism of transport? A. Active transport B. Simple diffusion C. Facilitated diffusion D. Exocytosis

Answer 107

C.

Question 108

A researcher treats a solution containing animal cells with ouabain, a poisonous substance that interferes with the sodium-potassium ATPase embedded in the cell membrane, and the cell lyses as a result. Which of the following statements best describes ouabain's effects? A. Treatment with ouabain results in high levels of extracellular calcium. B. Treatment with ouabain results in high levels of extracellular potassium and sodium. C. Treatment with ouabain increases intracellular concentrations of sodium. D. Treatment with ouabain decreases intracellular concentrations of sodium.

Answer 108

C.

Question 109

Resting membrane potential depends on: I. the differential distribution of ions across the membrane. Il. active transport processes. Ill. selective permeability of the phospholipid bilayer. A. I only B. I and III only C. Il and Ill only D. I, II, and III

Answer 109

D The polarization of the membrane at rest is the result of an uneven distribution of ions between the inside and outside of the cell. This difference is achieved through active pumping of ions (predominantly sodium and potassium) into and out of the cell and the selective permeability of the membrane, which allows only certain ions to cross.

Question 110

Which of the following is NOT a function of the celf membrane? A. Cytoskeletal attachment B. Protein synthesis C. Transport regulation D. Second messenger reservoir

Answer 110

B Ribosomes are the site of protein synthesis within a cell and are not coupled to the cell membrane. The cell membrane functions as a site for cytoskeletal attachment, (A), through proteins and lipid rafts. Transport regulation, (C), is accomplished through channels, transporters, and selective permeability, while the phospholipids act as a reagent for second messenger formation, (D).

Question 111

The dynamic properties of molecules in the cell membrane are most rapid in: A. phospholipids moving within the plane of the membrane. B. phospholipids moving between the layers of the membrane. C. proteins moving within the plane of the membrane. D. proteins exiting the cell through exocytosis.

Answer 111

A.

Question 112

Which lipid type is LEAST likely to contribute to membrane fluidity? A. Unsaturated glycerophospholipids B. trans glycerophospholipids C. Cholesterol D. Unsaturated sphingolipids

Answer 112

B.

Question 113

A membrane receptor is most likely to be a(n): A. embedded protein with catalytic activity. B. transmembrane protein with sequestration activity. C. membrane-associated protein with sequestration activity. D. transmembrane protein with catalytic activity.

Answer 113

D Membrane receptors must have both an extracellular and intracellular domain; therefore, they are considered transmembrane proteins. In order to initiate a second messenger cascade, they typically display enzymatic activity, although some may act strictly as channels.

Question 114

Which of the following is NOT a cell-cell junction animals? A. Desmosomes B. Gap junctions C. Plasmodesmata D. Tight junctions

Answer 114

C Plasmodesmata are cell-cell junctions that are found in plants, not animals. Gap junctions, tight junctions, des-mosomes, and hemidesmosomes are all found in animals, particularly in epithelia.

Question 115

Which of the following is true of diffusion and osmosis? A. Diffusion and osmosis rely on the concentration gradient of only the compound of interest. B. Diffusion and osmosis rely on the concentration gradient of all compounds in a cell. C. Diffusion and osmosis will proceed in the same direction if there is only one solute. D. Diffusion and osmosis cannot occur simultaneously.

Answer 115

A The movement of any molecule is dependent on multiple factors, including electric potential, membrane solubility, and concentration gradient. While the electric potential takes into account multiple molecules, the movement of any one solute or water by diffusion or osmosis is dependent only on the concentration gradient of that molecule.

Question 116

The bulk movement of liquid into a cell through vesicular infoldings is known as: A. phagocytosis. B. pinocytosis. c. exocytosis. D. drinking.

Answer 116

B The endocytosis (bulk uptake through vesicle formation) of fluid is known as pinocytosis. Phagocytosis, (A), is the endocytotic intake of solids, while exocytosis, (C), is a method of releasing vesicular contents. Drinking, (D), does not apply on a cellular level.

Question 117

Which of the following is LEAST likely to be the resting membrane potential of a cell? A. -70 mV B. -55 mV c. 0 mV D. +35 mV

Answer 117

C Cell membranes are most likely to have a resting membrane potential that is nonzero because the resting membrane potential creates a state that is capable of responding to stimuli. Signaling molecules and channels would not be as useful with a membrane potential of zero. The values given in the answer choices correspond to different stages of the action potential, but the key information is that a resting potential of 0 mV does not maintain gradients for later activity.

Question 118

How does the inner mitochondrial membrane differ from the outer mitochondrial membrane? A. The inner mitochondrial membrane is more permeable and lacks cholesterol. B. The inner mitochondrial membrane is less permeable and lacks cholesterol. C. The inner mitochondrial membrane is more permeable and has cholesterol. D. The inner mitochondrial membrane is less permeable and has cholesterol.

Answer 118

B The outer mitochondrial membrane is very permeable while the inner membrane is highly impermeable. The inner mitochondrial membrane is unique within the cell because it lacks cholesterol.

Question 119

For most cells, the extracellular calcium concentration is around 10,000 times higher than the intracellular calcium concentration. What is the membrane potential established by this electrochemical gradient? A. -123 mV B. -61.5 mV C. +61.5 mV D. +123 mV

Answer 119

D.

Question 120

Which of the following statements conflicts with the fluid mosaic model? A. The cell membrane is static in structure. B. Membrane components can be derived from multiple biomolecules. C. Hydrophobic interactions stabilize the lipid bilayer. D. Proteins are asymmetrically distributed within the cell membrane.

Answer 120

A The fluid mosaic model accounts for a dynamic membrane. In this model, membrane components contain both fatty and carbohydrate-derived components, eliminating (B). Further, the membrane is stabilized by the hydrophobic interactions of both fatty acid tails and membrane proteins, which may be found on the cytosolic or extracellular side of the membrane, or may run directly through the membrane; thus, (C) and (D) are also eliminated.

Question 121

Which of the following is a sphingolipid? A. Lecithin B. Phosphatidylinositol C. Cholesterol D. Ganglioside

Answer 121

D Gangliosides, along with ceramide, sphingomyelin, and cerebrosides, are sphingolipids.

Question 122

Which of the following transporters is used by cells in the liver to store excess glucose and by beta cells in the pancreas as a glucose sensor? A. GLUT 1 B. GLUT 2 C. GLUT 3 D. GLUT 4

Answer 122

B The GLUT 2 receptor, (B), has a high Km and is expressed in hepatocytes of the liver. When the concentration of glucose in the blood is high, glucose is transported into the liver with first order kinetics. Also, in the pancreas, GLUT 2 and glucokinase serve as a glucose sensor. These observations together support (B) as the right answer.

Question 123

Which of the following organs does NOT require a constant supply of glucose from the blood for energy during a fast? A. Red blood cells B. Brain C. Pancreas D. Liver

Answer 123

D The liver, like all cells, needs a constant supply of glucose; however, it is able to produce its own glucose through gluconeogenesis (cells in the kidney can also complete low levels of gluconeogenesis). The other cells listed here are absolutely dependent on a glucose source from the blood for energy, although they may also use other fuels in addition to glucose. For example, the brain can utilize ketone bodies during lengthy periods of starvation; however, it still requires at least some glucose for proper function.

Question 124

When insulin is released, it acts to increase the absorption of glucose into skeletal muscle predominantly through which of the following transporters? A. GLUT 1 B. GLUT 2 C. GLUT 3 D. GLUT 4

Answer 124

D GLUT is an abbreviation for glucose transporter and describes a family of sugar transporters with varying distributions and activities. GLUT 4 is found in adipose tissue and muscle, and mediates insulin-stimulated glucose uptake; in fact, it is the only insulin-responsive glucose transporter. Insulin acts via its receptor to translocate GLUT 4 to the plasma membrane. GLUT 4 in skeletal muscle is also stimulated by exercise through an insulin-independent pathway.

Question 125

After an overnight fast, which of the following enzymes would be expected to have little, if any, physiological activity? A. Malate dehydrogenase B. Glucokinase C. a-Ketoglutarate dehydrogenase D. Phosphofructokinase-1

Answer 125

B After an overnight fast, the liver is producing glucose and glucokinase activity would be insignificant. Glucokinase is used to trap extra glucose in liver cells as part of a storage mechanism; with low blood glucose, liver cells would be generating new glucose, not storing it. It is also in the pan-creas, where it serves as a glucose sensor; if glucose levels are low, it has little activity in this tissue as well. Malate dehy-drogenase, (A), and a-ketoglutarate dehydrogenase, (C), are citric acid cycle enzymes. Phosphofructokinase-1, (D), is a glycolytic enzyme. Other enzymes used in glycolysis, the citric acid cycle, or gluconeogenesis, such as phosphofruc-tokinase-1, would be expected to maintain normal activity after an overnight fast, using glucose derived from glycogen or gluconeogenesis, rather than orally ingested glucose.

Question 126

Which of the following enzymes is NOT used to trap sugar in the cell? A. Hexokinase B. Galactokinase C. Phosphofructokinase D. Fructokinase

Answer 126

C This question asks for an enzyme whose role is not to trap sugar in the cell. Phosphoructokinase (C) is the rate-limiting enzyme of glycolysis. Its role is to phosphorylate fructose-6-phosphate to fructose-1,6-bisphosphate during the third step of glycolysis. Importantly, the substrate of phosphoructokinase, i.e. fructose-6-phosphate, has already been once phosphorylated before reaching phosphof-ructokinase. Therefore, fructose-6-phosphate is already trapped in the cell before even reaching phosphoructoki-nase. For this reason, it cannot be said that phosphoruc-tokinase is used to trap sugar in the cell. By contrast, (A), (B), and (D), can be eliminated because hexokinase phos-phorylates glucose, galactokinase phosphorylates galactose, and fructokinase phosphorylates fructose, each reaction trapping the previously unphosphorylated sugar in the cell

Question 127

When fatty acid B-oxidation predominates in the liver, mitochondrial pyruvate is most likely to be: A. carboxylated to phosphoeno pyruvate for entry into gluconeogenesis. B. oxidatively decarboxylated to acetyl-CoA for oxidation in the citric acid cycle. C. carboxylated to oxaloacetate for entry into gluconeogenesis. D. reduced to lactate in the process of fermentation.

Answer 127

C Pyruvate is converted primarily into three main intermedi-ates: acetyl-CoA, (B), for the citric acid cycle (via pyruvate dehydrogenase complex); lactate, (D), during fermentation (via lactate dehydrogenase); or oxaloacetate, (C), for gluconeogenesis (via pyruvate carboxylase). High levels of acetyl-CoA, which is produced during B-oxidation, will inhibit pyruvate dehydrogenase and shift the citric acid cycle to run in the reverse direction, producing oxalo-acetate for gluconeogenesis. Acetyl-CoA also stimulates pyruvate carboxylase directly.

Question 128

A biopsy is done on a child with an enlarged liver and shows accumulation of glycogen granules with single glucose residues remaining at the branch points near the periphery of the granule. The most likely genetic defect is in the gene encoding: A. a-1,4 phosphorylase (glycogen phosphorylase). B. a-1,4:a-1,6 transferase (branching enzyme). C. a-1,4:a-1,4 transferase (part of debranching enzyme complex). D. a-1,6 glucosidase (part of debranching enzyme complex).

Answer 128

D The pattern described for this child's glycogen demonstrates appropriate production: there are long chains of glucose monomers, implying that glycogen synthase works. There are also branch points, implying that branching enzyme, (B) works. During glycogenolysis, it seems that the child is able to remove individual glucose monomers and process glycogen down to the branch point itself, which requires glycogen phosphorylase, (A), and a-1,4:a-1,4 transferase, (C). The metabolic problem here is removing the final glucose at the branch point, which is an a- 1,6 (not a-1,4) link. This requires (D), a-1,6 glucosidase.

Question 129

An investigator is measuring the activity of various enzymes involved in reactions of intermediary metabolism. One of the enzymes has greatly decreased activity compared to reference values. The buffer of the assay contains citrate. Which of the following enzymes will most likely be directly affected by the use of citrate? A. Fructose-2,6-bisphosphatase B. Isocitrate dehydrogenase C. Phosphofructokinase-1 D. Pyruvate carboxylase

Answer 129

C Citrate is produced by citrate synthase from acetyl-CoA and oxaloacetate. This reaction takes place in the mito-chondria. When the citric acid cycle slows down, citrate accumulates. In the cytosol, it acts as a negative allosteric regulator of phosphofructokinase-1, the enzyme that catalyzes the rate-limiting step of glycolysis.

Question 130

After a brief period of intense exercise, the activity of muscle pyruvate dehydrogenase is greatly increased. This increased activity is most likely due to: A. decreased ADP. B. increased acetyl-CoA. C. increased NADH/NAD+ ratio. D. increased pyruvate concentration.

Answer 130

D In most biochemical pathways, only a few enzymatic reactions are under regulatory control. These often occur either at the beginning of pathways or at pathway branch points. The pyruvate dehydrogenase (PDH) complex controls the link between glycolysis and the citric acid cycle, and decar-boxylates pyruvate (the end product of glycolysis) with production of NADH and acetyl-CoA (the substrate for the citric acid cycle). After intense exercise, one would expect PDH to be highly active to generate ATP. ADP levels, (A), should be high because ATP was just burned by the muscle. Acetyl-CoA, (B), is an inhibitor of PDH, causing a shift of pyruvate into the gluconeogenic pathway. A high NADH/ NAD + ratio, (C), would imply that the cell is already energetically satisfied and not in need of energy, which would not be expected in intensely exercising muscle.

Question 131

The metabolism of which of the following sugars would be LEAST affected by the inhibition of the enzyme phosphofructokinase-1? A. Lactose B. Fructose C. Glucose D. Galactose

Answer 131

B In this question, the right answer will be the molecule whose metabolism will be the least impacted by the inhibition of phosphofructokinase-1 (PFK1). The enzyme PFK1catalyzes the rate-limiting step of glycolysis. So sugars that enter the glycolytic pathway after this rate-limiting step, i.e. downstream of PFK1, would be least impacted by inhibition of PFK1. Given this criterion, recall the metabolic pathway of fructose, choice (B), which is phosphorylated and then cleaved into glyceraldehyde and DHAP, both of which enter glycolysis downstream of PFK1. Therefore the metabolism of fructose does not depend on PFK-1, and would therefore be least impacted by the inhibition of PFK1, making (B) the correct answer. By contrast, glucose, (C), is metabolised using the the usual glycolytic pathway, and would therefore be impacted by inhibition of PFK1. Similarly, galactose, (D), is phosphorylated and subsequently isomerized into glucose-6-phosphate, which then follows the usual path way and would therefore be impacted by inhibition of PFK1. And lactose, (A), is cleaved by lac. tase into glucose and galactose. For these reasons, (A), (B), and (D) can be eliminated

Question 132

A patient is given antibiotics to treat a urinary tract infection and develops an episode of red blood cell lysis. Further studies show weakness of the plasma membrane and Heinz bodies (collections of oxidized hemoglobin). Which of the following enzymes is most likely defective in this patient? A. Fructose-1,6-bisphosphatase B. Glucose-6-phosphate dehydrogenase C. Hexokinase D. Pyruvate kinase

Answer 132

B Based on the question stem, we can infer that the antibiotics must have been an oxidative stress on the patient (antibiotics, antimalarial medications, infections, certain foods like fava beans, and other common exposures can induce an oxidative stress). The pentose phosphate pathway is responsible for generating NADPH, which is used to reduce glutathione, one of the natural antioxidants present in the body. In individuals with glucose-6-phosphate dehydrogenase (G6PD) deficiency, NADPH cannot be produced at sufficient levels, and oxidative stresses lead to cell membrane and protein (hemoglobin) damage. Note that you do not need to actually know the disease to answer this question; merely knowing that the enzyme must be from the pentose phosphate pathway, which is involved in mitigating oxidative stress, is sufficient.

Question 133

The unique enzymes of gluconeogenesis are used to circumvent specific irreversible steps of glycolysis. Which of the following correctly pairs an enzyme from glycolysis with its corresponding enzyme(s) used in gluconeogenesis? A. Phosphofructokínase-1/fructose-1,6-bisphosphatase B. Pyruvate dehydrogenase/pyruvate carboxylase and phosphoenolpyruvate carboxykinase C. Hexokinase/ glucokinase D. Pyruvate kinase/glucose-6-phosphatase

Answer 133

A The irreversible enzymes in glycolysis are hexokinase (or glucokinase in liver and pancreatic B-cells), phosphofructokinase-1, and pyruvate kinase. Pyruvate dehydrogenase is not considered a glycolytic enzyme because it requires the mitochondria to function. The list below shows the correct pairing of glycolytic enzymes with gluconeogenic enzymes: • Hexokinase or glucokinase/glucose-6-phosphatase • Phosphofructokinase-1/fructose-1,6-bisphosphatase • Pyruvate kinase/pyruvate carboxylase and phosphoenolpyruvate carboxykinase (PEPCK)

Question 134

After an overnight fast, which of the following processes would be expected to occur at an elevated rate compared with the well-fed state? A. Glycolysis B. Glycogenolysis C. Glycogenesis D. Glycerol synthesis

Answer 134

B After a fast, the liver must contribute glucose into the bloodstream through two main processes: glycogenolysis (early to intermediate fasting) and gluconeogenesis (intermediate to late fasting). The other processes would continue at normal basal levels or have decreased activity after a fast.

Question 135

Andersen disease is caused by a mutation in branching enzyme and is associated with cirrhosis. Scientists theorize the cause is likely an immune response against abnormal glycogen. Which of the following best describes the ratio of glycogen a-1,4 to a-1,6 linkages in individuals with this disorder? A. Elevated B. Decreased C. No difference D. Cannot determine

Answer 135

A A deficiency in branching enzyme would result in chains of unbranched glycogen. Since glycogen synthase is not affected, the chain length would be normal, indicating a normal abundance of a -1,4 linkages. However, a loss of branching would correspond to a decrease in a-1,6, linkages. Thus, the ratio of a-1,4 to a-1,6 would increase as division by a smaller denominator yields a larger value, making (A) the correct answer.

Question 136

Each of the following catalyzes a rate-limiting step of a carbohydrate metabolism pathway EXCEPT: A. hexokinase. B. glycogen synthase. c. glucose-6-phosphate dehydrogenase. D. fructose-1,6-bísphosphatase.

Answer 136

A Hexokinase catalyzes an important irreversible step of glycolysis, but it is not the rate-limiting step Phosphofructokinase-1 catalyzes the rate-limiting step of glycolysis. Glycogen synthase, (B), catalyzes the rate-limiting step of glycogenesis; glucose-6-phosphate dehydro-genase, (C), catalyzes the rate-limiting step of the pentose phosphate pathway; and fructose- 1,6-bisphosphatase, > (D), catalyzes the rate-limiting step of gluconeogenesis.