Bio/Biochem Content Review
A researcher is running a PCR experiment to amplify her DNA of interest. Assuming she starts with a single DNA molecule, after 30 rounds of PCR, how many DNA molecules would she have? A. 2^30 B. 2 × 30 C. 1 + 30^2 D. 30^2
A. 2^30
During the metabolism of a fatty acid side chain containing 11 carbons, how many water molecules are used and how much FADH2 is produced? A. 4 molecules of water, 4 FADH2 B. 4 molecules of water, 8 FADH2 C. 5 molecules of water, 5 FADH2 D. 3 molecules of water, 6 FADH2
A. 4 molecules of water, 4 FADH2 Fatty acid metabolism occurs by removal of two carbons per round of β-oxidation until the final two or three carbons are reached, depending on whether the original fatty acid chain had an even or odd number of carbons in total. Since the side chain in the question has a total of 11 carbons, there will be four rounds of β-oxidation, leaving the final three carbons at the end of the process. Each β-oxidation uses one water molecule and produces one FADH2
How much net ATP would be produced if three molecules of glucose were used during anaerobic cellular respiration? A. 6 B. 24 C. 36 D. 12
A. 6 In an anaerobic environment, only glycolysis can occur. This process produces a net of 2 ATP per glucose, so three glucose molecules would produce 6 ATP total.
In terms of ATP, approximately how many glucose molecules would it take to translate a 60 amino acid polypeptide chain in a eukaryote undergoing aerobic respiration? A. 8 B. 10 C. 6 D. 12
A. 8 Aerobic respiration in a eukaryote results in 30 ATP per glucose. Translation requirements are as follows (note that these are commonly referred to as "ATPs" but many are actually ATP-equivalents, including ATP and other high energy bonds like GTP): 2 ATP per AA-tRNA loading; 1 ATP to form the initiation complex; and 2 ATP per AA joining the peptide chain. Thus, for a 60 AA peptide it would require 239 ATP and ~ 8 molecules of glucose.
A pregnancy test is an ELISA set up to detect human chorionic gonadotropin (hCG), a hormone specific to pregnancy. How would this ELISA be configured? A. A primary antibody specific for hCG, blood or urine sample, a secondary antibody specific for hCG B. Blood or urine sample, a primary antibody specific for hCG, a secondary antibody specific for the primary antibody's constant region C. Blood or urine sample, a primary antibody specific for hCG, a secondary antibody specific for the primary antibody's antigen binding site D. A primary antibody specific for hCG, blood or urine sample, a secondary antibody specific for the primary antibody
A. A primary antibody specific for hCG, blood or urine sample, a secondary antibody specific for hCG
Mutation of which of the following codons would be most disruptive to the initiation of translation? A. AUG B. UGA C. UUU D. UAA
A. AUG During translation initiation, the start codon (AUG) is recognized by the tRNA bound to methionine. By causing a mutation in this codon, the ribosome will either fail to initiate translation, or begin translation at an unintended site resulting in a protein which will likely be nonfunctional (choice A is correct). Also note that the codons UGA and UAA are stop codons (along with UAG).
How does ketogenesis trigger energy production? A. Acetyl-CoA is produced for use in the Krebs cycle B. Acetone is produced for use in the PDC C. β-hydroxybutyrate is produced for use in glycolysis D. Acetoacetate is produced for use in the electron transport chain
A. Acetyl-CoA is produced for use in the Krebs cycle
Fluoroquinolones are a class of antibiotics that inhibit DNA gyrase. Which of the following would be inhibited? A. Condensing of DNA B. Releasing of ribosomal subunits after transcription C. Connecting the Okazaki fragments during DNA replication D. unwinding of DNA at the origin of replication
A. Condensing of DNA
A codon is a segment of an mRNA molecule that codes for one amino acid in a polypeptide chain formed during protein synthesis. Which of the following correctly describes the chain of events that occurs in the synthesis of a polypeptide? A. DNA generates mRNA; mRNA moves to the ribosomes, where a tRNA anticodon binds to an mRNA codon, causing amino acids to join together in their appropriate order. B. DNA generates tRNA; the tRNA anticodon attaches to the mRNA codon in the cytoplasm; tRNA is carried by mRNA to the ribosomes, causing amino acids to join together in a specific order. C. Specific RNA codons cause amino acids to line up in a specific order; tRNA anticodons attach to mRNA codons; rRNA codons cause protein molecules to cleave into specific amino acids. D. DNA generates mRNA in the nucleus; mRNA moves to the cytoplasm and attaches to a tRNA anticodon; an operon regulates the sequence of events that causes amino acids to line up in their appropriate order.
A. DNA generates mRNA; mRNA moves to the ribosomes, where a tRNA anticodon binds to an mRNA codon, causing amino acids to join together in their appropriate order.
What type of reaction most commonly creates polymers from macromolecules? A. Dehydration synthesis B. Peptide bond formation C. Disulfide bridge formation D. Hydrolysis
A. Dehydration synthesis
During times of starvation: I. ketone bodies are formed from acetyl-CoA. II. fatty acid synthase is active in the cytoplasm. III. 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
A. I only 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
Compounds A and B react very slowly to form Compound C. Addition of a very small quantity of Enzyme X doubles the reaction rate. Addition of twice as much Enzyme X would most likely: A. Increase the reaction rate by the same amount as the first addition of enzyme. B. Have no effect, since reaction rate is independent of enzyme concentration. C. Have no effect since the enzyme will be saturated with substrate D. Increase the reaction rate by increasing the equilibrium constant.
A. Increase the reaction rate by the same amount as the first addition of enzyme.
A bacterial strain auxotrophic for lysine, isoleucine, and alanine production is mixed with an Hfr bacterial strain whose genome needs to be mapped. A sample taken after three minutes is able to grow on media supplemented with isoleucine and alanine. A sample taken after six minutes is able to grow on media supplemented with only isoleucine. A sample taken after nine minutes is able to grow on minimal media. What is the order of the amino acid synthesis genes on the Hfr strain? A. Lysine - Alanine - Isoleucine B. Lysine - Isoleucine - Alanine C. Isoleucine - Lysine - Alanine D. Alanine - Isoleucine - Lysine
A. Lysine - Alanine - Isoleucine It can be assumed that the bacteria are conjugating. After three minutes, the resultant new strain is able to grow in the absence of leucine, indicating that the leucine synthesis gene was transferred first. After six minutes, it was able to grow in the absence of leucine and alanine (the alanine synthesis gene was transferred second), and after nine minutes the new bacterial strain needed no supplementation, indicating that the isoleucine synthesis gene was the final gene transferred.
There are many structural similarities and differences between eukaryotic and prokaryotic cells. Which of the following cellular structures are the same whether isolated from a prokaryote or eukaryote? A. Plasma membrane B. Flagella C. Ribosomes D. Cell wall
A. Plasma membrane
In humans, which of the following enzymes is responsible for the transcription of most rRNA? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III D. RNA polymerase IV
A. RNA polymerase I
Which of the following lab techniques is described by the following steps? Step 1: Separate DNA fragments on a gel Step 2: Transfer fragments to a nitrocellulose filter Step 3: Probe the filter for the target DNA sequence with hybridized probes A. Southern blotting B. ELISA C. Conjugation D. Radioimmunoassay
A. Southern blotting
A yeast colony is subject to the mutagen EMS (ethyl methanesulfonate) and subsequently fails to divide. Further analysis reveals excessive supercoiling in the S-phase and failure to progress in the cell cycle. Which of the following was the most likely gene target of EMS? A. Topoisomerase B. Helicase C. DNA ligase D. Single-stranded binding protein
A. Topoisomerase Excessive supercoiling may result from unwinding DNA without the aid of topoisomerase which cleaves the DNA backbone to prevent such strain (choice A is correct). Helicase functions to unwind DNA; it produces supercoils in the DNA via unwinding, but these would not be excessive. The excessive nature of these supercoils is a failure in topoisomerase function (choice B is wrong)
All of the following are true statements concerning the Krebs cycle EXCEPT: A. oxygen is directly required for it to occur. B. two GTP are produced per glucose molecule. C. in eukaryotic cells it occurs in the matrix of the mitochondria. D. NAD+ and FAD get reduced during the cycle.
A. oxygen is directly required for it to occur.
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
A. tRNA loading 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).
An uncompetitive inhibitor binds to: A. the enzyme-substrate complex. B. the enzyme alone, at the active site. C. the enzyme alone, at an allosteric site. D. the substrate.
A. the enzyme-substrate complex.
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. Additionally, one NADH and one FADH2 are generated per turn. Lauric acid is an 12-carbon saturated fatty acid. Including those produced in the Krebs cycle, how many total NADH and FADH2 molecules would be generated from the complete β-oxidation of lauric acid and subsequent entry of the acetyl-CoA into the Krebs cycle? A. 24 NADH and 12 FADH2 B. 23 NADH and 11 FADH2 C. 12 NADH and 12 FADH2 D. 11 NADH and 11 FADH2
B. 23 NADH and 11 FADH2 5 NADH and 5 FADH2 would be generated during β-oxidation. When the 6 acetyl-CoA molecules go through the Krebs cycle, and additional 18 NADH would be generate (3 per turn of the Krebs cycle) and and additional 6 FADH2 would be generated (1 per turn), for a total of 23 NADH and 11 FADH2.
How many net ATPs are produced when a prokaryotic cell fully oxidizes one molecule of glucose? A. 36 ATP B. 32 ATP C. 38 ATP D. 30 ATP
B. 32 ATP
Some amino acids can be converted to pyruvate via several biochemical pathways. Pyruvate can then enter the cellular respiration pathways, either by decarboxylation to acetyl-CoA or by carboxylation to oxaloacetate. For a single pyruvate molecule, first converted to acetyl-CoA, then traveling through the Krebs cycle, how many NADH molecules are produced? A. 3 B. 4 C. 6 D. 8
B. 4 The decarboxylation of pyruvate to acetyl-CoA nets 1 NADH, and as that acetyl-CoA travels through the Krebs cycle, an additional 3 NADH are generated, resulting in a total of 4 NADH per pyruvate (choice B is correct and choices A, C, and D are wrong).
Epigenetics is the heritable increases or decreases in gene expression due to environmental (among other) causes. One of the mechanisms involved this differential expression is DNA methylation. What is the most likely mechanism for this process? A. Addition of a methyl group to ribose, decreasing the interaction with histones B. Addition of a methyl group to cytosine, resulting in methyl-cytosine base-pairing with guanine C. Addition of a methyl group to diphosphate, increasing the interaction with histones D. Addition of a methyl group to cytosine, resulting in methyl-cytosine mismatch base-pairing with thymine
B. Addition of a methyl group to cytosine, resulting in methyl-cytosine base-pairing with guanine
What substance might be expected to be in mosquito saliva that could inhibit the process of hemostasis? A. A vasoconstrictor B. An inhibitor of platelet aggregation C. A stimulator of coagulation D. A stimulator of T cells
B. An inhibitor of platelet aggregation Platelets are an essential part of the hemostatic process; they form large clumps (aggregation) as part of clot formation. Thus an inhibitor of platelet aggregation could inhibit hemostasis
How does standard DNA sequencing differ from running a polymerase chain reaction to synthesize DNA? A. PCR requires electrophoresis whereas DNA sequencing does not. B. DNA sequencing utilizes ddNTPs which lack the 3'OH group. This terminates polymerization, creating fragments. C. DNA sequencing is faster than PCR and yields higher copy numbers. D. PCR utilizes radiolabeling which would be deleterious in DNA sequencing.
B. DNA sequencing utilizes ddNTPs which lack the 3'OH group. This terminates polymerization, creating fragments.
Which of the following is NOT a function of cholesterol? A. Precursor for steroid hormones B. Energy storage C. Increase fluidity of plasma membranes D. Precursor for bile
B. Energy storage
Which of the following statements is true? A. When ADP levels are high, pyruvate kinase is inhibited. B. Fructose-2,6-bisphosphate levels are reduced when glucose levels are low; this helps to drive gluconeogenesis forward. C. The phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate is an important regulatory step in gluconeogenesis. D. High levels of citrate inhibit fructose-1,6-bisphosphatase to prevent gluconeogenesis when Krebs intermediates are plentiful.
B. Fructose-2,6-bisphosphate levels are reduced when glucose levels are low; this helps to drive gluconeogenesis forward.
A researcher hypothesizes that histone H2A.2 may have cytosolic functions. This hypothesis can best be tested by: A. His-tagging H2A.2 and performing affinity chromatography. B. GFP-tagging H2A.2 and performing fluorescent microscopy. C. knocking down H2A.2 expression using siRNAs. D. isolating H2A.2 and performing immunoprecipitation.
B. GFP-tagging H2A.2 and performing fluorescent microscopy. Adding a GFP (green fluorescent protein) tag to H2A.2 means this protein will be green under a fluorescent microscope. This will allow you to see where in the cell H2A.2 is expressed.
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.
B. Glucagon stimulates the activity of glycogen phosphorylase. 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;
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
B. Glucose and fructose
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
B. High levels of AMP 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).
A non-competitive inhibitor: I. binds to an allosteric site. II. reduces the Vmax of a reaction. III. 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
B. I and II only
Vmax is the maximum rate of product formation for a given enzyme. Which of the following would NOT affect Vmax? I. Decreasing the substrate concentration II. Adding a competitive inhibitor III. Increasing the enzyme concentration A. I only B. I and II only C. II and III only D. I, II, and III
B. I and II only
A graduate student attaches a fluorescent tag to β-galactosidase and performs site-directed mutagenesis to generate a mutation in the operator of the lac operon. The mutation completely prevents repressor binding. Which of the following is the most likely observation following mutagenesis of these cells? A. Increased fluorescence compared to control due to repressor inhibition, only in the presence of lactose B. Increased fluorescence compared to control due to failed repressor binding, in the presence or absence of lactose C. Decreased fluorescence compared to control, only in the presence of lactose D. Decreased fluorescence compared to control, in the presence or absence of lactose
B. Increased fluorescence compared to control due to failed repressor binding, in the presence or absence of lactose
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.
B. It could have a mutation in a nuclease, a single-stranded DNA binding protein, or DNA ligase. 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
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 nonhomologous end joining. Which of the following is true about this strain? A. It can only repair DSB with minimal specificity. B. It is able to form a joint molecule when repairing DSBs. C. It will accumulate many chromosomal aberrations over time. D. It can specifically repair DSB in a DNA polymerase-independent manner.
B. It is able to form a joint molecule when repairing DSBs.
Phosphatidylinositol-4,5-bisphosphate 3-kinase (PI3K) phosphorylates phosphatidylinositol (4,5)-bisphosphate (or PIP2) to generate the second messenger phosphatidylinositol (3,4,5)-trisphosphate (PIP3). The morpholine-containing chemical LY294002 is cell permeable and is a competitive inhibitor of ATP binding PI3K. Each of the following is true EXCEPT: A. LY294002 can simply diffuse into the cell across the plasma membrane, and increases the Km for the PI3K phosphorylation reaction. B. LY294002 likely undergoes simple diffusion to enter the cell and will decrease the Vmax but not change the Km for the PI3K phosphorylation reaction. C. LY294002 and adenosine triphosphate bind the same pocket of PI3K and must therefore have some structural similarities. D. LY294002 can be outcompeted for binding to PI3K by increasing the concentration of ATP in the reaction.
B. LY294002 likely undergoes simple diffusion to enter the cell and will decrease the Vmax but not change the Km for the PI3K phosphorylation reaction.
The bacterial flora of the gut is most likely characterized as which of the following? A. Mesophilic facultative aerobes B. Mesophilic obligate anaerobes C. Psychrophilic obligate aerobes D. Thermophilic tolerant anaerobes
B. Mesophilic obligate anaerobes
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.
B. Met-tRNAmet binds to the small ribosomal subunit. 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).
Which of the following initially binds to single-stranded DNA during replication? A. Helicase B. RNA polymerase C. DNA polymerase III D. DNA ligase
B. RNA polymerase 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).
During a polymerase chain reaction, high heat is used to denature and separate the DNA strands while cooling allows primers to anneal. Why does the extension step require additional heating? A. Nucleotides are synthesized only in high temperature conditions. B. The DNA polymerase used is heat sensitive and will not elongate in cooler conditions. C. Heat degrades the RNA primers so they do not become part of the permanent DNA sequence. D. Exonuclease activity requires elevated temperatures.
B. The DNA polymerase used is heat sensitive and will not elongate in cooler conditions.
If an electrophoretic gel is placed in reverse, but the DNA samples are still loaded into the established wells, what will happen to the sample? A. The samples will still run down the gel, but the separation will not be as good since they will be running toward the negative electrode. B. The samples will run out of the back of the gel towards the positive electrode rather than down the length of the gel toward the negative electrode. C. The samples will separate based on density rather than weight. D. The samples will not run at all.
B. The samples will run out of the back of the gel towards the positive electrode rather than down the length of the gel toward the negative electrode.
Which of the following would be true of the Lineweaver-Burk plot for a non-competitive inhibitor? A. The y-intercept of the graph would be bigger and the x-intercept would be smaller. B. The y-intercept of the graph would be bigger, and the x-intercept would not change. C. The y-intercept of the graph would not change and the x-intercept would be bigger. D. The y-intercept of the graph would be smaller and the x-intercept would not change.
B. The y-intercept of the graph would be bigger, and the x-intercept would not change.
During the absorption of dietary fat, the molecule is broken down before being reassembled following absorption. What is dietary fat broken down into before absorption? A. Three fatty acids and one monoglyceride B. Two fatty acids and one monoglyceride C. Triacylglycerol D. Triglycerides
B. Two fatty acids and one monoglyceride
When running an ELISA to test for the presence of anti-chickenpox antibody in a patient's blood, which of the following would be bound to the microtiter well plate? A. primary anti-chickenpox antibody B. chickenpox antigen C. secondary anti-chickenpox antibody D. conjugated enzyme
B. chickenpox antigen
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.
B. reduction. 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
Antibodies specific to the mitotic spindle apparatus would most likely recognize products of: A. transcription. B. translation. C. transformation. D. replication.
B. translation. Antibodies against the spindle would recognize proteins, which are produced during translation (choice B is correct).
How many NADH molecules are produced from a single glucose molecule during cellular respiration? A. 6 B. 8 C. 10 D. 12
C. 10
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
C. 8 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.
Which of the following situations would be most likely to cause glycogenolysis? A. Stimulation of the parasympathetic nervous system B. Stimulation of β-cells in the pancreas C. An overnight fast D. Consumption of a large meal
C. An overnight fast
Each of the following is true EXCEPT: A. PCR cycles through 3 temperatures: high temperature for template denaturing, low temperature for primer annealing and medium temperature for polymerization. B. During gel electrophoresis, negative DNA moves to the positive end of the gel in a size dependent manner. C. DNA is probed with DNA or RNA in a northern blot. D. Proteins are probed with antibodies in a western blot.
C. DNA is probed with DNA or RNA in a northern blot. A northern blot is when RNA is probed with either DNA or RNA ("DNA is probed with DNA or RNA in a northern blot" is a false statement and is the correct answer choice).
Which of the following enzymes is NOT required for eukaryotic DNA replication? A. RNA polymerase B. Topoisomerase C. Gyrase D. Ligase
C. Gyrase
Which of the following enzymes is NOT required for eukaryotic DNA replication? A. RNA polymerase B. Topoisomerase C. Gyrase D. Ligase
C. Gyrase
Antimycin is used as a piscicide (fish poison) because it inhibits Complex III of the electron transport chain. Blocking the flow of electrons through Complex III will produce which of the following effects? I.Complex I (NADH dehydrogenase) will persist in a reduced state. II.Complex IV (cytochrome C oxidase) will persist in a reduced state. III.Oxygen consumption will be decreased. A. I, II, and III B. II and III only C. I and III only D. III only
C. I and III only Because these earlier proteins, including Complex I, cannot pass their electrons off, they will persist in the reduced state ("Gain of Electrons is a Reduction"). Item II, however, is false: Proteins later in the chain can release their electrons, but cannot replenish them. These later proteins, including Complex IV, will persist in the oxidized state ("Loss of Electrons is an Oxidation"). Electrons meeting a roadblock at Complex III will never reach oxygen. Item III is true: Thus, oxygen consumption (reduction) will be decreased.
Which of the following produces a strand of DNA in the 5' to 3' direction? I.Eukaryotic DNA polymerase II.Prokaryotic DNA polymerase III III.Reverse transcriptase A. I and II only B. III only C. I, II, and III D. I only
C. I, II, and III
Both fats and carbohydrates can be used as energy storage molecules. Lipids are a more efficient means of energy storage because: I. fats can be easily converted to glucose molecules during times of starvation. II. fats ultimately generate more ATP per carbon than do carbohydrates. III. 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
C. II only
Which of the following is the most likely effect of a eukaryotic transcription factor? A. Provide the energy necessary for nascent RNA polymerization. B. Serve to inhibit translation to allow for completion of transcription. C. Increase the encounter rate of DNA with RNA polymerase. D. Bind at specific AUG sequences to start transcription.
C. Increase the encounter rate of DNA with RNA polymerase.
Which of the following best describes the cascade that leads to cellular apoptosis after exposure to intra- or extracellular death signals? A. Effector kinases cluster together and activate each other. The activation of effector kinases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. B. Effector caspases cluster together and activate each other. The activation of effector caspases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. C. Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. D. Effector kinases cluster together with initiator capsases, activating the initiator capsases. The activation of effector kinases leads to the deactivation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.
C. Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.
Which of the following is NOT a monosaccharide? A. Glucose B. Fructose C. Lactose D. Ribose
C. Lactose
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.
C. Multiple origins of replication are present in the eukaryotic genome.
Methoxy arachidonyl fluorophosphates (MAFPs) covalently link to the active sites of serine proteases. What best characterizes the activity of a serine protease following MAFP binding? A. Competition between the substrate and the MAFP which can be overcome with high substrate concentrations B. Increased protease activity C. Negligible protease activity D. Activation of the protease activity of the serine protease to cleave the MAFP
C. Negligible protease activity Here, a serine protease is treated with an MAFP that covalently binds to the active site. This would result in an irreversible loss of protease activity and no amount of substrate could displace the covalently linked MAFP (competition is not possible when the inhibitor binds covalently).
In a culture of mammalian skeletal muscle cells, the consumption of oxygen and glucose is measured. Which of the following would occur in response to inhibition of electron transport? A. Oxygen consumption will increase, and glucose consumption will increase. B. Oxygen consumption will decrease, and glucose consumption will decrease. C. Oxygen consumption will decrease, and glucose consumption will increase. D. Oxygen consumption will increase, and glucose consumption will decrease.
C. Oxygen consumption will decrease, and glucose consumption will increase.
Following the binding of a loaded tRNA to its codon during translation, which of the following steps occurs next? A. Translocation of the ribosome along the mRNA transcript B. Dissociation of the tRNA present at the P site C. Peptide bond formation D. A release factor binds
C. Peptide bond formation
Which of the following would NOT be required to run a polymerase chain reaction (PCR)? A. Primers B. Taq DNA polymerase C. RNA polymerase D. Template DNA
C. RNA polymerase
How is a standard curve formulated for a radioimmunoassay? A. Unlabeled antigen is mixed with antibody followed by radiolabeled antigen and the increases in radioactivity are measured until they asymptotically level. B. Radiolabeled antigen and unlabeled antigen are simultaneously mixed with antibody and competitive binding determines the shape of the curve. C. Radiolabeled antigen is mixed with antibody followed by antigen without the radiolabel and the decrease in radioactivity is measured as the amount of unlabeled antigen increases. D. Radiolabeled antigen is mixed with antibody followed by antigen without the radiolabel and the transfer of radioactivity between them is measured.
C. Radiolabeled antigen is mixed with antibody followed by antigen without the radiolabel and the decrease in radioactivity is measured as the amount of unlabeled antigen increases.
Alternative splicing permits somatic cells to contain the same genome while maintaining the capability to express widely differing proteins, based on the tissue in which the cell is located. Which experimental technique would be LEAST useful in detecting the differing cellular products created by alternative splicing? A. ELISA (enzyme-linked immunosorbent assay) B. Western blotting C. Southern blotting D. Northern blotting
C. Southern blotting Alternative splicing creates different mRNA sequences leading to different proteins, thus any technique that detects changes in mRNA transcripts or final protein products could be useful. Northern blotting is used to detect RNA and both Western blotting and ELISAs can be used to detect proteins. However, Southern blotting is used to detect DNA, and a point made by the question is that the genomes are the same (therefore the least useful technique).
A patient being treated with antibiotics for a severe bacterial infection appears to worsen when treatment begins, experiencing decreases in blood pressure. As treatment progresses, the infection resolves and the patient's blood pressure returns to normal. Why did the patient's condition degrade when starting treatment? A. The infection was likely cause by a strain of C. difficile that created a secondary opportunistic infection when the flora of the gut was eliminated by the antibiotic. B. The infection was likely caused by a strain of E. coli that was removing 5' caps from host cell transcripts for use on its own. C. The infection was likely caused by a Gram-negative bacterium producing endotoxins. D. The infection was likely caused by a Gram-positive bacterium producing exotoxins.
C. The infection was likely caused by a Gram-negative bacterium producing endotoxins. Gram-negative bacteria produce endotoxins as part of their outer lipid bilayer and these are released upon cell death.
Which of the following is a difference between eukaryotic and prokaryotic translation? A. The first translated codon B. The function of the codon UAA C. The mechanism by which ribosomes recognize the 5' end of mRNA D. The number of times a single mRNA transcript can be translated
C. The mechanism by which ribosomes recognize the 5' end of mRNA
A researcher is working with a certain strain of bacteria that will grow at 37°C, but only on plates that contain agar, glucose, arginine, and leucine. This organism can be classified as a(n): A. eukaryote. B. psychrophile. C. auxotroph. D. thermophile.
C. auxotroph.
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.
C. glucose-6-phosphate is oxidized to ribulose-5-phosphate and 2 NADPH are generated.
The function of galactoside permease is to transport lactose across the bacterial cell membrane. Galactoside permease is necessary because: A. lactose is hydrophobic and cannot cross the lipid bilayer. B. lactose needs a protein carrier to move it against its concentration gradient. C. lactose is hydrophilic and cannot cross the lipid bilayer. D. lactose needs to be converted to allolactose to enter the cell.
C. lactose is hydrophilic and cannot cross the lipid bilayer. Lactose, like other carbohydrates, is very hydrophilic, not hydrophobic (choice A is wrong). Hydrophilic molecules are unable to diffuse across a membrane into the cell
A researcher is making a transgenic mouse using gene targeted embryonic stem cells from a male black mouse, and a donor wild type morula from a brown mouse. This means that germline transmission of the targeted gene will most likely come from a: A. male chimeric mouse with mostly brown fur. B. female chimeric mouse with mostly black fur. C. male chimeric mouse with mostly black fur. D. female chimeric mouse with mostly brown fur.
C. male chimeric mouse with mostly black fur.
Both microarrays and flow cytometry involve labeling biological agents with fluorescent markers. These two techniques are different in that: A. microarrays give information on which proteins are expressed by a cell, while flow cytometry gives information on where fluorescent molecules are expressed in a tissue. B. microarrays give information on transcript levels, while flow cytometry gives information on where fluorescent molecules are expressed in a tissue. C. microarrays give information on transcript levels, while flow cytometry gives information on which fluorescent molecules are bound to individual cells. D. microarrays give information on which proteins are expressed by a cell, while flow cytometry gives information on which fluorescent molecules are bound to individual cells.
C. microarrays give information on transcript levels, while flow cytometry gives information on which fluorescent molecules are bound to individual cells. Microarrays are not used to determine directly which proteins are expressed by a cell. Flow cytometry looks at individual cells, as they are passed in a liquid stream through a laser. It does not look at cells in the context of a tissue.
During the blastula stage, neighboring cells are in close contact with each other. The formation of the blastula is contemporaneous with the formation of tight junctions. The function of tight junctions is to: A. provide a means through which induction occurs. B. act as pores connecting the cytoplasm of adjacent cells. C. seal off the blastocoel and unite connected cells into a tissue. D. transmit action potentials between cells.
C. seal off the blastocoel and unite connected cells into a tissue. Tight junctions hold cells together and can form a seal so that fluid does not leak between cells
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.
C. spontaneous and may or may not have a rapid rate.
In a male individual with Down's Syndrome (trisomy 21), how many chromosomes would be visible at metaphase I of spermatogenesis? A. 23 B. 24 C. 46 D. 47
D. 47 During metaphase I, the developing gametes are still diploid (separation of homologues has not yet occurred), so this individual would have the normal 46 chromosomes plus the extra copy of chromosome 21, for a total of 47 chromosomes
Which of the following is NOT necessary for prokaryotic translation? A. fMet B. GTP C. Shine-Dalgarno sequence D. 80S ribosome
D. 80S ribosome
Which of the following would most affect the Km of an enzyme? A. A mutation in a highly variable region of the protein coding sequence B. Addition of a noncompetitive inhibitor C. Increasing substrate concentration D. A mutation in the active site
D. A mutation in the active site
Northern blotting could be used to explore all of the following EXCEPT: A. Retroviral genome B. Viroid C. Transcripts D. Bacterial genome
D. Bacterial genome Northern blotting could be used to explore all of the following EXCEPT bacterial genome. Northern blotting is used to probe RNA sequences. Retroviruses are a type of (+)sense single-stranded RNA viruses, viroids are sequences of RNA, and transcripts are composed of RNA. However, bacterial genomes are composed of DNA; Southern blotting would be a better choice.
Which of the following would lengthen Okazaki fragments? A. Separating stop transcription sequences to a greater degree during replication B. Increasing the number of origins on the DNA strand C. Increasing the rate of all aspects of replication D. Decreasing the number of primers generated on the lagging strand during replication
D. Decreasing the number of primers generated on the lagging strand during replication
Which of the following is the most accurate, from least organized to most organized? A. Deoxyribose, nucleoside, nucleotide, DNA helix, chromatin, nucleosome B. Deoxyribose, nucleotide, nucleoside, DNA helix, nucleosome, chromatin C. Deoxyribose, nucleotide, nucleoside, nucleosome, DNA helix, chromatin D. Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin
D. Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin
In the creation of an expression vector, what step is necessary to prepare both a gene of interest and the plasmid that will carry it for ligation? A. Ensure both genetic pieces are single stranded. B. Treat both with proteases to avoid contaminating interactions from proteins. C. Allow exonucleases to proofread both sequences. D. Digest both genetic sequences with the same restriction endonuclease.
D. Digest both genetic sequences with the same restriction endonuclease.
Reaction coupling allows for: A. Formation of multimeric enzymatic complexes to decrease the overall energy of activation required. B. Addition of the equilibrium constants in a series of sequential steps to give a more spontaneous equilibrium constant. C. Increased reaction rates due to coupling of a fast reaction to a slow reaction. D. Generation of products that would not normally be formed spontaneously.
D. Generation of products that would not normally be formed spontaneously.
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.
D. High levels of ATP stimulate phosphofructokinase and pyruvate kinase to drive glycolysis forward.
Respiratory rate is regulated by many factors. Which of the following conditions is most likely to cause a decrease in breathing rate? A. Metabolic acidosis B. Low O2 concentration in the blood C. High CO2 levels in the blood D. High plasma pH
D. High plasma pH
A researcher cuts a thin slice of a fixed tissue, mounts it on a microscope slide and permeabilizes the cells. She can then: 1. analyze the tissue by in situ hybridization, to determine where a transcript is found in the cell. 2. analyze the tissue by immunohistochemistry to determine where a protein is found in the cell. 3. perform Bradford quantification, using a UV-Vis spectrophotometer, to determine how much protein the cell contains. A. I only B. I, II, and III C. II only D. I and II only
D. I and II only Both in situ hybridization and immunohistochemistry are performed on sections of tissue that have been mounted on a microscope slide. They give information on transcripts and proteins respectively. However, Bradford quantification is performed using a Bradford reagent and a UV-Vis spectrophotometer; it allows researchers to quantify the concentration of protein in a lysate mixture. In other words, it is performed on a solution of proteins, not on a fixed tissue.
In prokaryotic cellular respiration, which of the following processes occur in the cytoplasm? I. Glycolysis II. Pyruvate decarboxylation (via the PDC) III. Krebs cycle A. I only B. I and II only C. II and III only D. I, II, and III
D. I, II, and III Since prokaryotes (bacteria) lack cellular organelles, all of cellular respiration takes place in the cytoplasm
Enzymes are proteins that catalyze specific chemical reactions within a cell. Which of the following statements about enzymes is/are NOT true? I. A common means of regulating enzyme activity is through phosphorylation of the enzyme. II. Enzymes increase the energy of activation for a reaction, thereby making it go faster. III. Enzymes shift the equilibrium of a reaction towards products. A. I only B. II only C. I and II only D. II and III only
D. II and III only
Which of the following is an example of reciprocal regulation of glycogen metabolism? A. High levels of citrate stimulate phosphofructokinase and inhibit fructose-1,6-bisphosphatase. B. High levels of AMP inhibit glycolysis and stimulate gluconeogenesis. C. Glucagon inhibits glycogen phosphorylase and stimulates glycogen synthase. D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase.
D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase.
Short tandem repeat analysis for DNA fingerprinting utilizes patterns of repetitive DNA within what part of the genome to identify individuals? A. Exons B. Sex chromosomes C. Transposons D. Introns
D. Introns
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.
D. It results in the shortening of the mRNA transcript.
Which of the following western blotting techniques correctly pairs the material used as a probe with the material being detected? A. Labeled nucleic acids are used to detect certain proteins. B. Labeled nucleic acids are used to detect certain nucleic acid sequences. C. Labeled antibodies are used to detect certain nucleic acid sequences. D. Labeled antibodies are used to detect certain proteins.
D. Labeled antibodies are used to detect certain proteins. Western blotting uses antibodies to detect proteins. Proteins are separated via gel electrophoresis, transferred to a membrane, and then probed with a primary antibody. Northern and Southern blotting use the same general idea, but with nucleic acids; nucleic acids are first separated on a gel, then transferred to a membrane, then probed with nucleic acids. Nucleic acids are not used to detect proteins and antibodies are not used to detect nucleic acids.
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.
D. RNA polymerase binds directly to the template strand. 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).
Which of the following is LEAST likely to occur in an individual with uncontrolled diabetes? A. There is an increase in the amount of acetyl-coA produced. B. There is a decrease in the amount of NADPH produced in the pentose phosphate pathway. C. There is an increase in the amount of fatty acid oxidation. D. There is an increase in the amount of glucose-6-phosphate, which is shunted toward the pentose phosphate pathway.
D. There is an increase in the amount of glucose-6-phosphate, which is shunted toward the pentose phosphate pathway.
Which of the following codons does not code for an amino acid? A. AUG B. GUA C. AAA D. UAA
D. UAA Three codons do not code for amino acids: UAA, UGA, UAG. -- stop codons
Which of the following is NOT a known mechanism of direct enzymatic regulation? A. Interaction of the enzyme with downstream products B. Peptide hydrolysis C. Removal of a phosphate with the use of a phosphatase D. Varied activity of transcription factors affecting enzymatic expression
D. Varied activity of transcription factors affecting enzymatic expression
Within animal cells, the transport of Na+ out of the cell by the Na+/K+ ATPase involves: A. a symport. B. an antiport. C. facilitated diffusion. D. active transport.
D. active transport. An ATPase uses the energy of ATP hydrolysis to drive the transport of Na+ out of the cell against its electrochemical gradient, thus it is an example of active transport
High levels of ATP would: A. stimulate phosphofructokinase and inhibit pyruvate kinase, thus stimulating glycolysis. B. stimulate phosphofructokinase and stimulate pyruvate kinase, thus stimulating glycolysis. C. inhibit phosphofructokinase and stimulate pyruvate kinase, thus inhibiting glycolysis. D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.
D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.
The conformational change of a regulatory protein after the binding of a repressor most likely represents an alteration of the protein's: A. amino acid composition. B. primary structure. C. secondary structure. D. tertiary structure.
D. tertiary structure.
