Biochemistry I MCAT Biology Diagnostic Exam 1A

Question 1

Which of the following situations would NOT stimulate gluconeogenesis?
  	A.  High glucagon levels
  	B.  High levels of AMP 
  	C.  High levels of citrate
  	D.  High levels of acetyl-CoA

Answer 1

B. Gluconeogenesis will be stimulated when energy charge is high and glycolytic and Krebs cycle intermediates are high. If energy is low, and/or Krebs cycle intermediates are low, then glycolysis will be stimulated. High levels of AMP indicate that energy charge is low and that ATP needs to be made. AMP inhibits the enzyme fructose-1,6-bisphosphatase, keeping levels of fructose-1,6-bisphosphate high and driving the cell towards glycolysis (choice B would not stimulate gluconeogenesis and is the correct answer choice). High levels of citrate and acetyl-CoA indicate that the Krebs cycle is active and does not need more precursors, so will shift the cell away from glycolysis and toward gluconeogenesis. Think of it as taking these biosynthetic precursors and stashing them for future use. They aren’t needed if the Krebs cycle is operating at capacity (choices C and D would stimulate gluconeogenesis and can be eliminated). Glucagon is released when blood glucose levels are low and drives pathways toward gluconeogenesis to keep blood glucose levels stable (choice A would stimulate gluconeogenesis and can be eliminated).

Concepts tested
Biochemistry: Carbohydrate Metabolism

Question 2

Proteins are macromolecules formed by joining together individual amino acids. Which of the following is true?

A. Protein synthesis proceeds in the NRight arrowC direction, joining the carboxyl group carbon of the first amino acid to the amino group nitrogen of the second amino acid.
B. Protein synthesis proceeds in the NRight arrowC direction, joining the amino group nitrogen of the first amino acid to the carboxyl group carbon of the second amino acid.
C. Protein synthesis proceeds in the CRight arrowN direction, joining the carboxyl group carbon of the first amino acid to the amino group nitrogen of the second amino acid.
D. Protein synthesis proceeds in the CRight arrowN direction, joining the amino group nitrogen of the first amino acid to the carboxyl group carbon of the second amino acid.

Answer 2

A. Protein synthesis proceeds in the NRight arrow.C direction; the next amino acid in the sequence is added to the free carboxy terminus of the growing peptide (choices C and D are wrong). The carbon of the carboxyl group on the first amino acid (or on the polypeptide) is joined to the nitrogen of the amino group on the incoming amino acid (choice A is correct and choice B is wrong).

Concepts tested
Biochemistry: Biomolecules: Amino Acids and Proteins Molecular Biology: Translation

Question 3

In prokaryotic cellular respiration, which of the following processes occur in the cytoplasm?

Glycolysis
Pyruvate decarboxylation (via the PDC)
Krebs cycle
  	A.  I only
  	B.  I and II only
  	C.  II and III only
  	D.  I, II, and III

Answer 3

D. Since prokaryotes (bacteria) lack cellular organelles, all of cellular respiration takes place in the cytoplasm. Electron transport occurs across the cell membrane. Statements I, II, and III are all true.

Concepts tested
Biochemistry: Glycolysis/Cellular Respiration

Question 4

In the translation of a fifty-amino acid peptide, which of the following steps requires the greatest overall amount of energy?

A. tRNA loading
B. Initiation
C. Translocation
D. Termination

Answer 4

A. The most energy-intensive process in translation involves the attachment of amino acids to their corresponding tRNAs, also known as tRNA loading (or tRNA charging, or amino acid activation). This requires two high-energy phosphate bonds per aa-tRNA pair (choice A is correct). Initiation requires only 1 GTP (choice B is wrong). Elongation includes A-site binding and translocation (for amino acid number two and onwards); one GTP is hydrolyzed to bring an aa-tRNA into the A site, and one GTP is required for translocation (choice C is wrong). Termination involves the binding of a release factor to a stop codon and does not directly require any energy (choice D is wrong). In this example, it would require 100 ATP to charge a sufficient number of tRNAs to generate the peptide, 1 GTP for initiation, 49 GTP for A-site binding, and 49 GTP for translocation, but note that the answer to this question does not depend upon the number of amino acids in the peptide.

Question 5

A non-competitive inhibitor:

binds to an allosteric site.
reduces the Vmax of a reaction.
can be overcome by adding more substrate.
  	A.  I only
  	B.  I and II only 
  	C.  II and III only
  	D.  I, II, and III

Answer 5

B. Item I is true: non-competitive inhibitors do not bind at the active site, they bind at allosteric sites and prevent the enzyme from catalyzing the reaction (choice C can be eliminated). Item II is true: by “turning off” the enzymes to which it is bound, the inhibitor lowers the effective enzyme concentration, and lowering the enzyme concentration lowers Vmax (choice A can be eliminated). Item III is false: non-competitive inhibition cannot be overcome by adding more substrate. This is only true of competitive inhibition, where the inhibitor and substrate both bind to the active site (choice D can be eliminated and choice B is correct).

Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition

Question 6

Which of the following alterations to the lac operon in E. coli would negatively impact the cell’s ability to utilize lactose as an energy source?

A. A single amino acid substitution in the repressor’s lactose binding site, preventing lactose binding
B. A single nucleotide mutation in the operator, blocking repressor association
C. A mutation in the DNA binding site of the repressor, markedly reducing its ability to bind to DNA
D. A mutation resulting in the constitutive expression of the repressor

Answer 6

A. Any alteration that would decrease the expression of the genes necessary for lactose transport and breakdown would negatively impact the cell’s ability to utilize lactose as an energy source. If the lactose binding site on the repressor could no longer bind lactose, the repressor would be able to bind to the operator region on DNA, but the presence of lactose would no longer cause dissociation of the repressor, and transcription of the necessary genes would not occur (choice A is correct). If a mutation occurred in the operator which prevented repressor binding, transcription of the lac operon genes would always occur (choice B is wrong). Similarly, a mutation in the DNA binding site of the repressor that prevented it from binding to DNA would lead to constitutive gene expression (choice C is wrong). Increased expression of the repressor itself would not necessarily decrease transcription of the lac operon genes, assuming that the regulation of the repressor by lactose is not affected (choice D is wrong).

Concepts tested
Molecular Biology: Mutations and DNA Repair Molecular Biology: Regulation of Gene Expression

Question 7

A reaction with a negative ΔG is:

A. spontaneous and has a rapid rate.
B. non-spontaneous and has a slow rate.
C. spontaneous and may or may not have a rapid rate.
D. non-spontaneous and may or may not have a rapid rate.

Answer 7

C. Reactions with a negative ΔG are spontaneous (choices B and D are wrong), but ΔG does not provide any information about the rate of the reaction. Spontaneous reactions can move forward slowly or rapidly (choice A is wrong and choice C is correct). The thermodynamics of a reaction says nothing about rates; to know something about rates, you must consider the kinetics of the reaction.

Concepts tested
Biochemistry: Thermodynamics/Kinetics

Question 8

Hexokinase catalyzes the first step in glycolysis, the phosphorylation of glucose to form glucose-6-P. High levels of glucose-6-P inhibit hexokinase. This is most accurately described as:

A. positive feedback.
B. negative feedback.
C. feedforward inhibition.
D. non-competitive inhibition.

Answer 8

B. When the product of a reaction inhibits the enzyme running the reaction, it is most accurately termed negative feedback. Positive feedback occurs when an end product stimulates the enzyme running the reaction (choice A is wrong). Feedforward implies that an end product of one reaction affects an enzyme further on in the series of reactions, and that is not the case here (choice C is wrong). Non-competitive inhibition occurs when a molecule distinct from the substrate binds to an allosteric site on the enzyme, turning it off. This may or may not be the case here, although it seems unlikely; if anything this might be competitive inhibition, since glucose and glucose-6-P are structurally similar (choice B is better than choice D).

Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition

Question 9

Which of the following is true about enzymes?

A. Enzymes stabilize the transition state of a reaction, thereby lowering the energy of activation for the reaction.
B. Enzymes convert non-spontaneous reactions into spontaneous ones.
C. By increasing the energy of activation for a reaction, enzymes make it easier to get to the transition state, thus increasing the rate of the reaction.
D. A single enzyme can catalyze a wide variety of reactions.

Answer 9

A. By stabilizing the transition state of a reaction, the energy required to produce the transition state (energy of activation) is lowered (choice C is false), and the reaction can proceed more quickly (choice A is correct). Enzymes do not affect the thermodynamics of a reaction; they cannot convert a non-spontaneous reaction into a spontaneous one, all they can do is take an already-spontaneous reaction and make it go faster (choice B is false). Enzymes are highly specific for a particular reaction (choice D is false).

Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition

Question 10

Which of the following is a difference observed between prokaryotic and eukaryotic replication?

A. Polyadenylation is observed in eukaryotes only.
B. Genes in prokaryotes can be polycistronic.
C. Multiple origins of replication are present in the eukaryotic genome.
D. Eukaryotic replication occurs around a single circular chromosome.

Answer 10

C. In order to ensure that replication is completed in a reasonable length of time, eukaryotes possess multiple origins of replication on each chromosome (choice C is correct). Eukaryotes utilize polyadenylation of RNA transcripts and prokaryotic genes are frequently polycistronic. However, neither of these statements (while true), answer the question because they are about transcription instead of DNA replication (choices A and B can be eliminated). Eukaryotic cells have multiple linear chromosomes while prokaryotic cells have a single circular chromosome (choice D is wrong).

Concepts tested
Molecular Biology: DNA Replication

Question 11

During the oxidative phase of the pentose phosphate pathway:

A. ribulose-5-phosphate is converted into glycolytic intermediates.
B. ribulose-5-phosphate is converted into ribose-5-phosphate for incorporation into nucleic acids.
C. glucose-6-phosphate is oxidized to ribulose-5-phosphate and 2 NADPH are generated.
D. glucose-6-phosphate is oxidized to ribulose-5-phosphate and 2 NADH are generated.

Answer 11

C. During the oxidative phase of the pentose phosphate pathway (PPP), glu-6-P is oxidized while NADP+ is reduced to NADPH (choice C is correct); NADPH is then used in other biosynthetic pathways as a reducing agent. NADH is generated during glycolysis and the Krebs cycle, not the PPP (choice D is wrong). The interconversion of ribulose-5-P to glycolytic intermediates or into ribose-5-P is not oxidative (choices A and B are wrong).

Concepts tested
Biochemistry: Carbohydrate Metabolism

Question 12

How many NADH molecules are produced from a single glucose molecule during cellular respiration?

A. 6
B. 8
C. 10
D. 12

Answer 12

C. A single glucose molecule generates a total of 10 NADH during cellular respiration. 2 NADH are made in glycolysis, 2 NADH at the pyruvate dehydrogenase complex, and 6 NADH in the Krebs cycle (choice C is correct and choices A, B, and D are wrong).

Concepts tested
Biochemistry: Glycolysis/Cellular Respiration

Question 13

Vmax is the maximum rate of product formation for a given enzyme. Which of the following would NOT affect Vmax?

Decreasing the substrate concentration
Adding a competitive inhibitor
Increasing the enzyme concentration
  	A.  I only
  	B.  I and II only 
  	C.  II and III only
  	D.  I, II, and III

Answer 13

B. Vmax depends only on the enzyme and the concentration of the enzyme. Item I would not affect Vmax: decreasing the substrate concentration would decrease V (the rate of product formation), but would not affect Vmax, which is a theoretical maximum rate of product formation, assuming you have enough substrate to get there (choice C can be eliminated). Item II would not affect Vmax: competitive inhibitors reduce V, but given enough substrate the reaction will ultimately reach the same Vmax as an uninhibited reaction (choice A can be eliminated). Item III would definitely affect Vmax (and can therefore be eliminated): more enzyme means that more product can be formed, and Vmax would increase (choice D is wrong and choice B is correct).

Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition

Question 14

Which of the following is true of the template strand in transcription?

A. The template strand possesses a sequence nearly identical to the RNA transcript.
B. Another name for the template strand is the coding strand.
C. The template strand possesses distinct codons necessary for transcription initiation.
D. RNA polymerase binds directly to the template strand.

Answer 14

D. The template strand is bound directly by the RNA polymerase and functions as a template for RNA transcript generation (choice D is correct). The template strand’s sequence is complementary to the transcript, not identical (choice A is wrong). The coding strand is the complementary strand of DNA (the strand opposite the template strand) and it is nearly identical in sequence to the RNA transcript (T in the coding strand and U in the RNA transcript, choice B is wrong). Transcription initiation is dictated largely by the promoter, which is found on the DNA, not codons which would be found on RNA and are necessary for translation (choice C is wrong).

Concepts tested
Molecular Biology: Transcription

Question 15

Both fats and carbohydrates can be used as energy storage molecules. Lipids are a more efficient means of energy storage because:

fats can be easily converted to glucose molecules during times of starvation.
fats ultimately generate more ATP per carbon than do carbohydrates.
fats are hydrophilic, thus bringing along a lot of water molecules.
A. I only
B. I and II only
C. II only
D. II and III only

Answer 15

C. Statement I is false: humans cannot convert fats to glucose because we lack the necessary enzymes to do so. Rather than convert the fat to sugar, we just metabolize the fat directly (choices A and B can be eliminated). Note that both remaining choices include Statement II, so Statement II must be true: fats are broken down into acetyl-CoA molecules through beta oxidation, producing NADH and FADH2. These ultimately go through the electron transport chain (ETC) to generate ATP. The acetyl-CoA molecules from beta oxidation then go through the Krebs cycle, generating even more NADH and FADH2, that generate even more ATP when they go through the ETC. Statement III is false: fats are hydrophobic. This is an advantage because more fat can be packed into a given space than glycogen, which, because it is polar, comes with a lot of water molecules (choice D is wrong and choice C is correct).

Concepts tested
Biochemistry: Protein/Fat Metabolism Biochemistry: Biomolecules: Lipids

Question 16

Which of the following is a true statement?

A. Glycogen synthase is activated by insulin and epinephrine in a similar manner.
B. Glucagon stimulates the activity of glycogen phosphorylase.
C. Fatty acid oxidation during starvation produces acetyl-CoA, which feeds into the gluconeogenesis pathway.
D. Insulin release during starvation activates glycogen synthase.

Answer 16

B. Glycogen phosphorylase catalyzes the first step in glycogen breakdown, and is stimulated in conditions of starvation. Glucagon is released when blood sugar is low; it stimulates glycogen phosphorylase to activate glycogen breakdown and increase blood glucose levels (choice B is correct). Glycogen synthase is the enzyme that produces glycogen. It would be stimulated by insulin but inactivated by epinephrine; epinephrine would stimulate glycogen breakdown, not synthesis, to increase blood sugar during fight-or-flight conditions (choice A is wrong). Further, insulin would not be released during starvation (choice D is wrong). Lastly, while fatty acid oxidation would be increased during starvation, acetyl-CoA does not feed into gluconeogenesis. It either enters the Krebs cycle or is converted to ketone bodies (choice D is wrong).

Concepts tested
Biochemistry: Protein/Fat Metabolism Biochemistry: Carbohydrate Metabolism

Question 17

Can prokaryotes participate in aerobic respiration?

A. Yes, they run all the reactions in the cytosol and run the electron transport chain across their cell membrane.
B. Yes, the majority of the reactions run in the mitochondrial matrix, and electron transport occurs across the inner mitochondrial membrane.
C. No, prokaryotes lack mitochondria and thus can only generate ATP through glycolysis.
D. No, however in symbiotic relationships, prokaryotes can use the ATP generated by their host organism.

Answer 17

A. Prokaryotes (bacteria) are capable of running aerobic respiration by using their cell membrane for the electron transport chain. Protons are pumped out of the cytosol into the space between the membrane and the cell wall, and ATP is generated as those protons flow back into the cytosol through an ATP synthase (choice A is correct, and choices C and D are wrong). Prokaryotes lack any kind of membrane-bound organelles, such as mitochondria (choice B is wrong).

Concepts tested
Biochemistry: Glycolysis/Cellular Respiration

Question 18

Which of the following is true of the mRNA splicing reaction observed in eukaryotic cells?

A. It results in changes to the genetic code.
B. It occurs in the cytoplasm.
C. It occurs in every transcript.
D. It results in the shortening of the mRNA transcript.

Answer 18

D. Splicing occurs in the nucleus following transcription (choice B is wrong) and results in the removal of introns, shortening the mRNA transcript (choice D is correct). Since it is the RNA transcript that is altered, no changes in the genetic code, or DNA, occur (choice A is wrong). While splicing is a common post-transcriptional process, it does not occur in every gene; some genes lack introns (choice C is wrong). Note that the word “every” is included in this answer choice. When the word every (or never) is used, be cautious, as exceptions to the rule are a possibility.

Concepts tested
Molecular Biology: Transcription

Question 19

Which of the following best characterizes a difference between eukaryotic DNA and mRNA?

A. The presence of thymine in mRNA
B. The absence of uracil in mRNA
C. The transient nature of mRNA compared to DNA
D. Polyadenylation of DNA in heavily transcribed regions

Answer 19

C. Numerous differences between DNA and RNA are present in the eukaryotic cell, including the presence of a 2’-OH group in RNA, uracil being substituted for thymine in RNA (choices A and B are wrong), and the short half-life of mRNA relative to a cell’s DNA (choice C is correct). While transcripts of RNA are polyadenylated at the 3’ end, DNA is not (choice D is wrong).

Concepts tested
Biochemistry: Biomolecules: Nucleic Acids

Question 20

β-oxidation is a means of creating acetyl-CoA from fatty acids. This acetyl-CoA can then enter the Krebs cycle. Each turn of the β-oxidation cycle produces one acetyl-CoA molecule and a fatty acid two carbons shorter than it was at the beginning of the cycle. Stearic acid is an 18-carbon saturated fatty acid. How many turns of the β-oxidation cycle would it take to completely break down stearic acid into acetyl-CoA groups?

A. 4
B. 5
C. 8
D. 9

Answer 20

C. Each turn of the β-oxidation cycle produces one acetyl-CoA and a fatty acid two carbons shorter than before. Stearic acid, with 18 carbons, would ultimately produce 9 acetyl-CoA. If stearic acid enters the β-oxidation cycle, then after one turn of the cycle we would have one acetyl-CoA and a 16C fatty acid. After two turns of the cycle we would have two acetyl-CoA and a 14C fatty acid. This would continue in this manner until after the 8th turn of the cycle we would produce our 8th acetyl-CoA molecule, and all that would be left over would be another two-carbon acetyl-CoA molecule (the 9th acetyl-CoA). This last 2-carbon molecule does not need to go through the β-oxidation cycle again, so only 8 turns of cycle are necessary.

Concepts tested
Biochemistry: Protein/Fat Metabolism

Question 21

In the absence of oxygen, fermentation occurs in order to allow glycolysis to continue. Pyruvate, an end-product of glycolysis, is converted to either lactic acid or ethanol. The conversion of pyruvate to lactic acid is a(n):

A. oxidation.
B. reduction.
C. decarboxylation.
D. isomerization.

Answer 21

B. In order to keep running glycolysis in the absence of oxygen, NAD+ must be regenerated from NADH. In lactic acid fermentation, the oxidation of NADH occurs via the reduction of pyruvate to lactic acid (choice B is correct and choice A is wrong). In alcoholic fermentation (the conversion of pyruvate to ethanol) pyruvate must first be decarboxylated, since ethanol is a two-carbon structure, whereas pyruvate is a three-carbon structure. However, lactic acid is also a three-carbon structure, so no decarboxylation is needed in lactic acid fermentation (choice C is wrong). Isomerization is simply a rearrangement of atoms, and this is not the case here (choice D is wrong).

Concepts tested
Biochemistry: Carbohydrate Metabolism

Question 22

Which of the following statements is NOT true?

A. Fructose-2,6-bisphosphate levels are elevated when glucose levels are high; this helps to drive glycolysis forward.
B. Phosphofructokinase and fructose-1,6-bisphosphatase are reciprocally regulated.
C. Phosphoenolpyruvate carboxykinase (PEP CK) and pyruvate carboxylase are inhibited by ADP.
D. High levels of ATP stimulate phosphofructokinase and pyruvate kinase to drive glycolysis forward.

Answer 22

D. Fructose-2,6-bisphosphate is a major regulator of glycolysis and gluconeogenesis; when blood glucose is high, fru-2,6-bisP levels are high and glycolysis is promoted (choice A is true and can be eliminated). Phosphofructokinase (PFK) and fructose-1,6-bisphosphatase are reciprocally regulated by fru-2,6-bisP and by citrate; fru-2,6-bisP stimulates PFK and promotes glycolysis and inhibits fructose-1,6-bisphophatase to inhibit gluconeogenesis. Citrate operates in the reverse manner, stimulating gluconeogenesis and inhibiting glycolysis (choice B is true and can be eliminated). PEP CK and pyruvate carboxylase are the first two enzymes in gluconeogenesis and they are inhibited by ADP; a high level of ADP indicates a need for ATP and thus promotes glycolysis, not gluconeogenesis (choice C is true and can be eliminated). However, high levels of ATP inhibit phosphofructokinase and pyruvate kinase to prevent glycolysis when energy levels (ATP) are high (choice D is not a true statement and the correct answer choice).

Concepts tested
Biochemistry: Glycolysis/Cellular Respiration

Question 23

A graduate student in a yeast lab that studies double-strand break (DSB) repair has a mutant strain that is unable to complete repair via homologous recombination. Which of the following is true about this strain?

A. It requires an active cell cycle to complete DSB repair.
B. It could have a mutation in a nuclease, a single-stranded DNA binding protein, or DNA ligase.
C. It can repair DNA using a crossing-over based mechanism.
D. It can repair DSB with great specificity.

Answer 23

B. The strain is unable to complete homologous recombination. This process uses many proteins and enzymes, including helicases, nucleases, single-stranded DNA binding proteins, DNA polymerase, and ligases, thus this strain could have a mutation in any one of these components (choice B is correct). Homologous recombination is completed using crossing over and formation of a joint molecule and requires that the cell be actively dividing. Since the strain cannot do this, choices A and C can be eliminated. This strain must be relying on nonhomologous end joining, which is not a very specific method of DNA repair (choice D can be eliminated), but is better than nothing. It does not require active cell growth, and involves identifying and stabilizing the ends of a DSB, then ligating two DSBs together.

Concepts tested
Molecular Biology: Mutations and DNA Repair

Question 24

Which of the following graphs represents cooperative binding?

A. A graph with V labeled on its vertical axis and Start Bracket: S: End Bracket labeled on its horizontal axis is shown. A straight line originating from the origin is shown which increases uniformly with the increase in the values of Start Bracket: S: End Bracket.
B. A graph with V labeled on its vertical axis and Start Bracket: S: End Bracket labeled on its horizontal axis is shown. A curve originating from the origin is shown which first increases uniformly with the increase in the values of Start Bracket: S: End Bracket and then gradually becomes parallel to the horizontal axis.
C. A graph with V labeled on its vertical axis and Start Bracket: S: End Bracket labeled on its horizontal axis is shown. A curve is nearly parallel to the vertical axis for the larger values of V and after a certain point it decreases gradually and hence becomes parallel to the horizontal axis.
D. A graph with V labeled on its vertical axis and Start Bracket: S: End Bracket labeled on its horizontal axis is shown. A curve originating from the origin is shown which increases nominally in the beginning and attains a rapid growth for some values of horizontal axis. Thereafter, retains its nominal growth.

Answer 24

D. Enzymes that exhibit cooperativity have sigmoidal curves; the binding of some substrate facilitates the binding of more substrate until saturation is reached (choice D is correct). Choice A shows a directly proportional linear relationship that does not saturate; this is not typical of any enzyme. Enzyme kinetics are such that the reaction will ultimately saturate and the curve will level off (choice A is wrong). Choice B is typical of a non-cooperative enzyme (choice B is wrong). Choice C is not typical of enzyme kinetics in general; if anything this might show some sort of inhibition as the rate of product formation (V) declines with increasing substrate concentration (choice C is wrong).

Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition

Question 25

Which of the following steps occurs earliest in the initiation of translation in eukaryotes?

A. Met-tRNAmet associates with AUG on mRNA.
B. Met-tRNAmet binds to the small ribosomal subunit.
C. The large ribosomal subunit binds to the mRNA.
D. The mRNA binds to the small ribosomal subunit.

Answer 25

B. The proper sequence for the initiation of translation in eukaryotes begins with the loaded tRNAmet (the initiator tRNA) binding to the small ribosomal subunit. This tRNA is the only one that can bind tightly to the small subunit in the absence of mRNA (choice B is correct). The next step is for the mRNA to bind to the small subunit; this occurs with the help of the 5’ guanine cap (choice D is wrong). The small subunit then scans along the mRNA until the initiator codon (AUG) is found; at that point the met-tRNAmet associates with the mRNA in the P site (choice A is wrong). The final step in initiation is the binding of the large ribosomal subunit to this complex (choice C is wrong).

Concepts tested
Molecular Biology: Translation

Question 26

A yeast culture is subjected to DNA-damaging UV light. A researcher sequences a select gene of interest and identifies a distinct nucleotide that has changed, but no difference in function can be detected in the translated protein. Which of the following mutation types best explains this observation?

A. Insertion of a DNA base
B. Silent mutation
C. Missense mutation
D. Nonsense mutation

Answer 26

B. Given that there is no change in protein function, the mutation must be minimally disruptive. Silent mutations result when a change in the DNA sequence produces a codon that specifies the same amino acid as the unmutated sequence (this is possible because of the degeneracy of the genetic code; amino acids are specified by more than one codon). Since there is no change in the amino acid sequence of the protein, there is no change in protein function (choice B is the best answer). The insertion of a DNA base in the sequence would alter the reading frame of the coding region, alter many amino acids, and would very likely impact function (choice A is wrong). A missense mutation (the substitution of one amino acid for another) may or may not allow for protein function, and if it does allow for function the function may or may not be identical to the unmutated protein. Without knowing more about the specific mutation, we cannot be certain if it is a missense or not, making silent mutation a better answer (choice B is better than choice C). A nonsense mutation results in a premature stop codon; this would likely affect the protein’s function (choice D is wrong).

Concepts tested
Molecular Biology: Mutations and DNA Repair

Question 27

During times of starvation:

ketone bodies are formed from acetyl-CoA.
fatty acid synthase is active in the cytoplasm.
β-oxidation is active in the cytoplasm.
A. I only
B. I and III only
C. II and III only
D. I, II, and III

Answer 27

A. Item I is true: when glucose levels are low and glycogen stores are depleted, acetyl-CoA can be converted to ketone bodies to be used as fuel by the nervous system (choice C can be eliminated). Item II is false: fatty acid synthase drives the production of fatty acids; during starvation they would be broken down, not synthesized (choice D can be eliminated). Item III is false: while β-oxidation is likely to occur during starvation, it occurs in the mitochondrial matrix, not the cytoplasm (choice B can be eliminated and choice A is correct).

Concepts tested
Biochemistry: Protein/Fat Metabolism

Question 28

Which of the following monosaccharides are linked together to form sucrose?

A. Glucose and maltose
B. Glucose and fructose
C. Fructose and galactose
D. Glucose and glucose

Answer 28

B. Sucrose is a disaccharide formed by joining a glucose molecule to a fructose molecule (choice B is correct and choice C is wrong). Maltose is already a disaccharide (choice A is wrong), formed by joining two glucose molecules together (choice D is wrong).

Concepts tested
Biochemistry: Biomolecules: Carbohydrates

Question 29

Which of the following codons does not code for an amino acid?

A. AUG
B. GUA
C. AAA
D. UAA

Answer 29

D. Three codons do not code for amino acids: UAA, UGA, UAG. These three codons are the “stop” codons; they specify that a release factor should bind in the P site instead of a loaded tRNA. This will end translation and result in the release of the newly-synthesized peptide and the mRNA (choice D is correct). The other three codons all specify amino acids. Note that the only other codon that should be memorized is AUG, which codes for methionine, and serves as the start codon in translation.

Concepts tested
Biochemistry: Biomolecules: Amino Acids and Proteins

Question 30

Which of the following is an accurate description regarding the activity of transposons?

A. If a single transposon inserts into the promoter or regulatory region of a protein coding gene, protein levels will decrease.
B. If a single transposon inserts into protein-coding DNA, the protein-coding region will be disrupted and proteins levels will likely decrease.
C. Two transposons in the same orientation can cause chromosomal inversion, where a section of a chromosome is flipped into the reverse orientation.
D. Two transposons in the opposite orientation can cause chromosomal deletions and translocations, which are both types of chromosomal rearrangements.

Answer 30

B. If a single transposon inserts into the promoter or regulatory region of a protein-coding gene, protein levels can either increase or decrease (eliminate choice A). If a single transposon inserts into protein-coding DNA, the protein coding region will be disrupted and proteins levels will likely decrease (choice B is correct). Two transposons in the opposite orientation cause chromosomal inversion, where a section of a chromosome is flipped to the reverse orientation; two transposons in the same orientation cause chromosomal deletions and translocations, which are both types of chromosomal rearrangements (eliminate choices C and D).

Concepts tested
Molecular Biology: Mutations and DNA Repair

Question 31

Which of the following initially binds to single-stranded DNA during replication?

A. Helicase
B. RNA polymerase
C. DNA polymerase III
D. DNA ligase

Answer 31

B. Primase, an RNA polymerase, binds to single-stranded DNA in its generation of an RNA primer during replication (choice B is correct). Helicase binds to double-stranded DNA and facilitates DNA unwinding (choice A is wrong). DNA polymerase III binds to an RNA/DNA duplex (i.e., parental DNA strand with an RNA primer attached) before beginning elongation (choice C is wrong) and DNA ligase binds to double-stranded DNA and anneals the phosophodiester backbone of DNA (choice D is wrong).

Concepts tested
Molecular Biology: DNA Replication