Biochemistry I
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. Your Answer C. Have no effect since the enzyme will be saturated with substrate D. Increase the reaction rate by increasing the equilibrium constant. Concepts Tested: Biochemistry: Thermodynamics/Kinetics Biochemistry: Enzymes and Enzyme Inhibition
A. 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 increase the reaction rate by the same amount as the first addition of enzyme. Doubling the amount of enzyme doubles the number of available active sites, and so doubles the rate of the reaction. The addition of more active sites (without the addition of more substrate) will not lead to saturation, in fact it is the opposite. Reaction rate is clearly dependent on enzyme concentration; initially, at an enzyme concentration of zero, the reaction rate was stated to be "very slow". When enzyme was added, the reaction rate doubled. Enzymes have no effect on the equilibrium constant. They do not change the equilibrium point of a reaction, they only speed up the rate at which the reaction reaches equilibrium.
What best explains the reason for the inability of the human intestinal tract to digest cellulose? A. β(1,4) linkages B. (1,6) linkages at branch points C. 5'-3' phosphodiester bonds D. φ-ψ peptidyl linkages Concepts Tested: Biochemistry: Biomolecules: Carbohydrates
A. Humans are unable to digest cellulose due to the large numbers of β(1,4) glycosidic linkages. These cannot be cleaved by human amylases and pass though the GI tract undigested. (1,6) linkages are present in starch (not cellulose) at branch points and are digestible. Phosphodiester bonds link nucleotides in DNA and RNA while peptidyl linkages are what link amino acids in protein. Note that φ and ψ correspond to bond angles present around the alpha carbon in a protein.
Which of the following statements regarding enzymes is NOT true? A. Enzymes increase the kinetic barrier. B. Enzymes speed up reactions by lowering the energy of activation. C. Enzymes are not used up during reactions. D. Enzymes do not affect the thermodynamics of a reaction. Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
A. Regarding enzymes, the statement "Enzymes increase the kinetic barrier" is NOT true. Enzymes decrease the kinetic barrier, not increase it (the statement "Enzymes increase the kinetic barrier" is a false statement and is the correct answer choice here). Enzymes lower the activation energy (the statement "Enzymes speed up reactions by lowering the energy of activation" is true and can be eliminated), do not get used up in reactions (the statement "Enzymes are not used up during reactions" is true and can be eliminated), and do not affect the thermodynamics of a reaction (the statement "Enzymes do not affect the thermodynamics of a reaction" is true and can be eliminated).
What bonds/interactions are responsible for maintaining quaternary structure of a protein? A. The same bonds/interactions responsible for the tertiary structure B. Polar covalent bonds between alpha carbons C. The same bonds/interactions responsible for the primary structure D. The same bonds/interactions responsible for the secondary structure Concepts Tested: Biochemistry: Biomolecules: Amino Acids and Proteins
A. The bonds/interactions responsible for maintaining quaternary structure of a protein are the same bonds/interactions responsible for the tertiary structure. Quaternary structure, the interaction of protein subunits, is maintained by both intermolecular forces and disulfide bridges, which also are responsible for maintaining tertiary structure, the folding based on R groups. Primary structure is the sequence of amino acids linked by peptide bonds and secondary structure is maintained primarily by hydrogen bonds.
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. Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition Cell Biology: The Cell Membrane
B. All of the statements are true EXCEPT LY294002 likely undergoes simple diffusion to enter the cell and will decrease the Vmax but not change the Km for the PI3K phosphorylation reaction. LY294002 is described as a competitive inhibitor of ATP binding. Competitive inhibitors increase Km, but do not change Vmax because continuing to increase the concentration of substrate allows it to outcompete the inhibitor and still achieve saturation at the active site. The chemical is also described as cell-permeable, which means it likely undergoes simple diffusion to enter the cell. Further, competitive inhibitors bind to the same sites (the active site) as the substrate(s) for the reaction. It is therefore likely that they share structural similarities.
Carbon dioxide is a small, hydrophobic molecule that diffuses through cell membranes. Its movement can be described as: A. thermodynamically favorable because it decreases entropy. B. thermodynamically favorable because it increases entropy. C. thermodynamically unfavorable because it increases entropy. D. thermodynamically unfavorable because it decreases entropy. Concepts Tested: Biochemistry: Thermodynamics/Kinetics
B. Carbon dioxide is a small, hydrophobic molecule that diffuses through cell membranes. Its movement can be described as: thermodynamically favorable because it increases entropy. This is a two-by-two question, where two decisions are required to find the correct answer. Diffusion is a passive process and since this is how carbon dioxide moves, this is a thermodynamically favorable process. Thermodynamically favorable processes generally increase disorder (or entropy).
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. Concepts Tested: Biochemistry: Biomolecules: Nucleic Acids
B. DNA methylation occurs via addition of a methyl group to cytosine, resulting in methyl-cytosine base-pairing with guanine. Given that epigenetics involves change in gene expression, and is not the expression of mutant products, it is unlikely that cytosine will mismatch with thymine after methylation. In DNA, the nucleosides are separated by individual phosphates (which together are called nucleotides) and the sugar moiety is deoxyribose. Therefore the methyl group could not be added to diphosphate or ribose. This leaves methylation of cytosine and its association with guanine as the only viable answer choice.
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 Concepts tested: Biochemistry: Biomolecules: Lipids
B. Dietary fat is broken into two fatty acids and one monoglyceride prior to absorption. Due to the activity of lipase, triglycerides are broken down into two fatty acids and a monoglyceride before absorption and ultimate reassembly into a triglyceride in the enterocyte. Note that triglycerides are dietary fat, which are also known as triacylglycerol, and neither are therefore breakdown products.
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 Concepts Tested: Biochemistry: Biomolecules: Lipids
B. Energy storage is NOT a function of cholesterol, rather it is a function of triglycerides. Cholesterol is an important component of plasma membranes, increasing fluidity of the membrane. It is also the precursor to steroid hormones and bile.
Which of the following will change the ∆G° of an enzyme catalyzed reaction? A. Decreasing product concentration B. Increasing temperature C. ∆G is a constant and cannot be readily changed D. Increasing enzyme concentration Concepts Tested: Biochemistry: Thermodynamics/Kinetics
B. Increasing temperature will change the ∆G of an enzyme-catalyzed reaction. The ∆G of an enzyme-catalyzed reaction can be changed by influencing temperature but not by altering reactant or product concentrations (these do not affect Keq) as given by the following equation:
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. Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
B. On the Lineweaver-Burk plot for a non-competitive inhibitor, the y-intercept of the graph would be bigger and the x-intercept would not change. The y-intercept represents Vmax; in non-competitive inhibition, Vmax is reduced, so on an inverse plot this would appear as a bigger y-intercept. However, Km in non-competitive inhibition stays the same; Km is represented by the x-intercept, so the x-intercept would not change. Note that an unchanging y-intercept and a bigger x-intercept represents competitive inhibition (where Vmax stays the same and Km increases), and a bigger y-intercept with a smaller x-intercept represents uncompetitive inhibition (lowered Vmax and smaller Km). There are no inhibitors that make the y-intercept of a Lineweaver-Burk plot smaller, as this would represent an increase in Vmax, which is not characteristic of enzyme inhibition.
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 Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
C. Binding of an MAFP to the serine protease active site results in 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). Irreversible inhibitors are those that act on a specific enzyme and prevent its further activity, often by covalently binding to it. As MAFP is not a peptide, it is very unlikely that a serine protease would cause its cleavage
Which of the following is an example of feedback inhibition? A. Decreased activity of hexokinase due to a decreased concentration of its substrate glucose B. Decreased activity of hexokinase due to elevated levels of a competitive inhibitor, 2-deoxyglucose C. Decreased activity of PFK due to elevated levels of phosphoenolpyruvate, a glycolytic intermediate D. Increased activity of PFK due to high levels of AMP Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition Biochemistry: Carbohydrate Metabolism
C. Decreased activity of PFK due to elevated levels of phosphoenolpyruvate is an example of feedback inhibition. Feedback inhibition involves the decrease in activity of an enzyme by binding to a downstream product. PFK, the rate-limiting enzyme in glycolysis, is inhibited by phosphoenolpyruvate (PEP) which is an intermediate in the glycolytic pathway. Hexokinase is the first enzyme in the pathway which phosphorylates glucose. Decreasing concentrations of glucose may limit the activity of hexokinase, but as it is a substrate, it would not be characterized as feedback inhibition. 2-deoxyglucose does decrease hexokinase activity, but is an inhibitor of that enzyme, not a downstream product.
Which of the following is NOT a monosaccharide? A. Glucose B. Fructose C. Lactose D. Ribose Concepts Tested: Biochemistry: Biomolecules: Carbohydrates
C. Lactose is not a monosaccharide, it is a disaccharide composed of glucose and galactose. All other sugars listed are monosaccharides. Fructose and glucose are six-carbon sugars, while ribose is a five-carbon sugar.
A biochemist assesses the changes in reaction rate of an enzyme-catalyzed reaction in a solution containing saturating quantities of a known inhibitor. When substrate is added, the reaction rate increases, but fails to reach the published Vmax for this enzyme concentration, even at concentrations of substrate well above that of the inhibitor. Interestingly, the researcher notes that the concentration of substrate required to reach half of the observed maximal rate matches the published value. What is true of this enzyme and its inhibitor? A. The inhibitor binds to active site of the enzyme. B. Inhibitor concentration has no impact on reaction rate. C. Addition of inhibitor does not impact enzyme affinity for the substrate. D. The Vmax is independent of enzyme concentration. Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
C. The addition of inhibitor does not impact enzyme affinity for the substrate in the described reaction. The stem of the question describes a situation where an enzyme is in the presence of a saturating quantity of inhibitor and substrate is added. Given we have a decreased Vmax with an unchanged Km, this is likely a noncompetitive inhibitor. Km, or the concentration of substrate needed to reach half Vmax, is also a measure of enzyme affinity for the substrate and has not changed here. Vmax is impacted by enzyme concentration; more enzyme means an increased rate of product formation (although this is not being tested here) and inhibitor concentration does impact reaction rate (hence why they are inhibitors). For noncompetitive inhibition, the inhibitor binds to an allosteric site, not the active site. Binding to the active site indicates competitive inhibition, and this would be indicated by the reaction rate reaching Vmax at higher substrate concentrations (which does not happen in this case).
In which of the following situations would there be the greatest yield of products at equilibrium following the addition of an enzyme? A. A reaction favoring reactants and large activation energy. B. A reaction with more stable reactants and small activation energy. C. A reaction with less stable reactants and large activation energy. D. A reaction favoring reactants and small activation energy. Concepts Tested: Biochemistry: Thermodynamics/Kinetics Biochemistry: Enzymes and Enzyme Inhibition
C. The greatest yield of products at equilibrium following the addition of an enzyme would occur for a reaction with less stable reactants and a large activation energy. Enzymes serve to increase the rate of a reaction by decreasing activation energy but have no impact upon the reaction equilibrium. Therefore the activation energy is irrelevant and we must focus on the Keq or relative stabilities of the reactants and products. The greater the stability of the reactants, the more they are favored at equilibrium and the greater the stability of products, the more they are favored at equilibrium. Therefore the reaction with less stable reactants would favor products at equilibrium and generate a greater yield.
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 Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
D. A mutation in the active site of an enzyme would affect its Km.The Michaelis constant (Km) is the concentration of substrate necessary for a reaction to achieve ½ Vmax. It is also inversely proportional to affinity of the enzyme for its substrate. Therefore a mutation in the active site of the enzyme is likely to affect the Km for a given reaction. A mutation in a highly variable (less conserved) region of the protein coding sequence is less likely to impact the Km. Substrate concentration would have no impact on affinity and addition of an inhibitor that binds to an allosteric site (a noncompetitive inhibitor) would not impact the Km.
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. Concepts Tested: Biochemistry: Thermodynamics/Kinetics Biochemistry: Enzymes and Enzyme Inhibition
D. Reaction coupling allows for generation of products that would not normally be formed spontaneously. Reaction coupling allows for a spontaneous process to drive a nonspontaneous process to make the overall reaction spontaneous. As this is affecting a thermodynamic property, it is unlikely to have an impact on kinetics (reaction rates would not increase, but would be limited by the slow reaction). Reaction coupling may involve the use of an enzymatic complex but reaction coupling does not alter the rate of each step (note that this is also a kinetics answer). When reactions are coupled, their ∆Gs are additive but their equilibrium constants are not. To find the new equilibrium constant, you must multiply the Keq of each step.
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 Concepts Tested: Biochemistry: Enzymes and Enzyme Inhibition
D. Varied activity of transcription factors affecting enzymatic expression is not a mechanism of direct enzymatic regulation. Direct enzymatic regulation takes place via several possible mechanisms including (but not limited to) protein hydrolysis, phosphorylation/dephosphorylation (covalent modification), allosteric regulation (including interaction with a downstream product), or association with another protein. While varying activity of transcription factors will affect enzymatic expression, it is an indirect method of regulation.
Proinsulin, the precursor of insulin, is one peptide chain that contains three disulfide bonds. Protease cleavage releases a middle segment of the peptide chain, which leaves two peptide chains held together via two disulfide bonds. Which one of the following is most accurate? Concepts Tested: Biochemistry: Biomolecules: Amino Acids and Proteins
Figure A
