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Biochemistry exam 3 review sheet with questions and answers
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University of Houston BCHS 3304
Practice Exam III Instructor Dr. Liu, Name____________________,____________________
LAST, FIRST
Multiple choice, 2 pts each (100 pts) – Circle letter (i.e., a., b., c., d., or e.) of the most correct answer, then mark it on the Scantron sheet with a #2 pencil. Mark BOTH the exam AND the Scantron sheet. Show your work on the exam itself. A table of half reactions, the gas constant (R) and graph paper are at the end of this exam.
1). Zymogens are not enzymatically active because a. the active site amino acids have been mutated. b. they have not yet bound the proper cofactor. c. their environment has the wrong pH. d. they are not yet shaped such that essential proximity and orientation catalysis can occur. e. None of the above is correct.
2). Serine proteases use _________ to catalyze the cleavage of a peptide bond. I. covalent catalysis II. proximity and orientation effects III. general base catalysis IV. electrostatic catalysis
a. I, II b. II c. III d. I, II, III e. I, II, III, IV
3). In trypsin the specificity for one substrate over another comes from a. the negatively charged pocket. b. the positively charge pocket. c. the hydrophobic pocket. d. the amino acid serine. e. the amino acid histidine.
4). A recent discovery has suggested that a gene mutation results in several amino acid substitutions within an active site. The following substitutions have been identified and are each expected to properly fold and form the active site:
Normal protein amino acids Substitutions Valine Lysine Leucine Arginine Valine Arginine Serine Deleted and replaced with a coenzyme capable of covalent binding Histidine Histidine (note no mutation) Aspartate Glutamate
It was also noticed that a coenzyme molecule containing a hydroxyl (—OH) group binds to the mutated protein in approximately the same location as the original serine but does not alter the structure of the protein in any way. Based on your knowledge of amino acids, enzymes, and catalysis, which of the following is a REASONABLE conclusion? a. The mutant protein cleaves on the carboxy side of phenylalanine. b. The mutant protein cleaves peptide bonds on the carboxy side of glutamate or aspartate. c. The mutant protein cleaves peptide bonds on the carboxy side of lysine or arginine. d. The mutant protein functions identical to chymotrypsin.
5). Consider the nonenzymatic elementary reaction AB. When the concentration of A is 20 μM, the reaction velocity is measured as 200 nM B produced per minute. Calculate the rate constant for this reaction. a. 1 min - b. 0.1 min - c. 0.01 min - d. 2.5x10 -3^ min - e. 2.5x10 -4^ min -
6). If there is 11 μmol of the radioactive isotope 32P (half-life 11 days) at t=0, how much 32P will remain at 15 days? a. 0.5 μM b. 3.5 μM c. 4.3 μM d. 5 μM e. 7 μM
7). Calculate the half-life, in years, for the reaction 2X Y when the starting concentration of X is 6 μM and the rate constant is 3.6x10 -2^ M -1^ s -^. a. 0.147 yrs b. 1.47 yrs c. 4.73 yrs d. 14.7 yrs e. 27.8 yrs
8). Determine the type of inhibition of an enzymatic reaction from the following data collected in the presence and absence of the inhibitor. [S], mM v 0 , mM/s (no inhib) v 0 , mM/s (inhib) 0.20 0.036 0. 0.25 0.042 0. 0.36 0.048 0. 0.60 0.065 0. 2.00 0.075 0.
a. Competitive b. Uncompetitive c. Mixed
14). Compounds that function as “mixed inhibitors” I. interfere with substrate binding to the enzyme. II. bind to the enzyme reversibly. III. can bind to the enzyme/substrate complex.
a. I b. II c. III d. II, III e. I, II, III
15). Enzyme activity in cells is controlled by which of the following? I. covalent modifications II. modulation of expression levels III. feedback inhibition IV. allosteric effectors
a. I b. II c. III d. III, IV e. I, II, III, IV
16). Determine the K (^) M and V max from the following graph. (Note: On the x -axis the minor tick mark spacing is 0.005; on the y -axis the minor tick mark spacing is 0.002)
a. K (^) M = [0.006]; V max = 0.0075/s b. K (^) M = [0.196]; V max = 0.0075/s c. K (^) M = [165]; V max = 33/s d. K (^) M = [33]; V max = 167/s e. K (^) M = [270]; V max x = 68/s
-0.
0
-0.05 -0.025 0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0. 1/[S]
1/v
17). From the graph below, plotting data that were collected under steady state conditions, velocity on the y -axis in units of μM/s and substrate concentration of the x - axis in units of μM, what is the V max and K (^) M?
(V (^) max, K (^) M) a. 0.24 μM/s, 18 μM b. 18 μM, 0.24 μM/s c. 0.2 μM/s, 70 μM d. 0.24 μM, 18 μM/s e. 0.12 μM/s, 18 μM
18). Based on the figures below, which of the following expressions would be correct?
a. V max = 1/B b. C = 1/ V max c. D= V max d. D = 1/ V max e. A = 1/ V max
II. This reaction is exergonic under these conditions. III. This reaction is at equilibrium under these conditions. IV. This reaction is not favorable under standard conditions.
a. I, IV b. II only c. III only d. IV only e. None of the above
24). Which of the numbered arrows in the figure points toward a high-energy phosphoanhydride bond?
a. 1 b. 2 c. 3 d. 4 e. 5
25). Cells control or regulate the flux through metabolic pathways by means of I. allosteric control of enzymes. II. covalent modification of enzymes. III. genetic control of the concentrations of enzymes. IV. genetic expression of feedback inhibitors. a. I, II, III, IV b. II, III c. I, II, IV d. I, II, III e. I, IV
26). While ____ is always involved in reactions that require the transfer of 2 electrons, ____ can participate in reactions that transfer electrons 1 at a time. a. O 2 ; NAD + b. NAD +^ ; FAD c. NAD +^ ; O (^2) d. flavin; niacin e. FAD; NAD +
27). In eukaryotes, glycolysis typically occurs in the a. mitochondrion. b. cytosol.
c. lysosome. d. rough endoplasmic reticulum. e. smooth endoplasmic reticulum.
28). Acetyl-CoA contains a ____ bond which often provides the energy required for substrate-level phosphorylation. a. phosphoester b. phosphoanhydride c. phosphothioester d. ester e. thioester
29). I have discovered a new metabolic enzyme which utilizes NAD +^ as a coenzyme. Which of the following could be true based on this information? a. It is involved in a catabolic pathway. b. It is involved in an anabolic pathway. c. It performs acyl group transfer. d. It catalyzes a hydrolytic cleavage. e. It catalyzes the reduction of the enzyme's substrate.
30). Compound "K" is the ultimate product of a linear metabolic pathway consisting of ten enzymatically catalyzed reactions as shown below. 1 2 3 4 5 6 7 8 9 10 A → B → C → D → E → F → G → H → I → J → K Complete the following description.
If "K" inhibits activity of enzyme "5", then enzyme "5" is under ___ control. If enzyme "3" becomes activated in response to activity of protein kinase A only when compound "A" is present, then enzyme "3" is under ___ control. If enzyme "2" is only expressed in the presence of high concentrations of "A," then enzyme "2" is under ___ control. If all of these are true, high concentrations of "__" will decrease the flux through the pathway. a. covalent modification; feedback; genetic; A b. feedback; covalent modification; genetic; K c. genetic; feedback; genetic; K d. feedback; genetic; genetic; K e. feedback; genetic; genetic; A
31). How many stereoisomers are possible for ketohexose? a. 1 b. 2 c. 4 d. 8 e. 16
32). Deduce the structure of the disaccharide trehalose from the following information. Complete hydrolysis yields only D-glucose; it is hydrolyzed by α-glucosidase but not β-glucosidase; and it does not reduce Cu 2+^ to Cu +. a. β-D-glucose-(11)-α-D-glucose
c. C and D d. A and D e. None of the above
37). Which sugar shown in the figure above is the enantiomer of sugar A?
a. B b. C c. D d. B and D e. none of the above
38). Cellulose is a. a linear copolymer of glucose and galactose. b. a branched polymer of glucose. c. a linear polymer of glucose with β(1→4) linkages. d. a linear polymer of glucose with α(1→4) linkages. e. sometimes called starch.
- P i + Glucose ⇆ Glucose-6-P + H 2 O 13. Reactants ⇆ Products ∆G 0’(kJ/mol) - ATP + H 2 O ⇆ AMP + PP i -45. - Glucose-1-P + H 2 O ⇆ Glucose + P i -20. - Glucose + P i ⇆ Glucose-3-P + H 2 O 9.