9.1. Does trans-oleic acid have a higher or lower melting point than cis-oleic acid? Explain.
trans-Oleic acid has a higher melting point because, in the solid state, its hydrocarbon chains pack together more tightly than those of cis- oleic acid.
9.5. How many different types of triacylglycerols could incorporate
the fatty acids shown in Fig. 9-1?
Of the 4 x 4 = 16 pairs of fatty acid residues at C1 and C3, only 10 are unique because a molecule with different substituents at C1 and C3 is identical to the molecule with the reverse substitution order. However, C2 may have any of the four substituents
9.6. What products are obtained when 1-palmitoyl-2-oleoyl-3- phosphatidylserine is hydrolyzed by
(a) phospholipase A1;
(b) phospholipase A2;
(c) phospholipase C;
(d) phospholipase D?
(a) Palmitic acid and 2-oleoyl-3-phosphatidylserine;
(b) oleic acid and 1-palmitoyl-3-phosphatidylserine;
(c) phosphoserine and 1-palmitoyl-2-oleoyl-glycerol;
(d) serine and 1-palmitoyl-2-oleoyl-phosphatidic acid.
9.7. Which of the glycerophospholipid head groups listed in Table 9-2 can form hydrogen bonds?
All except choline can form hydrogen bonds.
9.11. In some autoimmune diseases, an individual develops antibodies that recognize cell constituents such as DNA and phospholipids. Some of the antibodies react with both DNA and phospholipids. What is the structural basis for this cross-reactivity?
Both DNA and phospholipids have exposed phosphate groups that are recognized by the antibodies.
9.12. Most hormones, such as peptide hormones, exert their effects by binding to cell-surface receptors. However, steroid hormones do so by binding to cytosolic receptors. How is this possible?
Steroid hormones, which are hydrophobic, can diffuse through the cell membrane to reach their receptors.
9.15. Why can't triacylglycerols be significant components of lipid bilayers?
Triacylglycerols lack polar head groups, so they do not orient them- selves in a bilayer with their acyl chains inward and their glycerol moiety toward the surface.
9.16. Why would a bilayer containing only gangliosides be unstable?
The large oligosaccharide head groups of gangliosides would prevent the necessary close packing of the lipids in a bilayer.
10.5. Indicate whether the following compounds are likely to cross a membrane by nonmediated or mediated transport: (a) ethanol, (b) glycine, (c) cholesterol, (d) ATP.
(a) Nonmediated; (b) mediated; (c) nonmediated; (d) mediated.
10.6. Rank the rate of transmembrane diffusion of the following compounds: A. Acetamide {CH3CONH2}, B. Butyramide {CH3(CH2)2CONH2}, C. Urea {NH2CONH2}
The less polar a substance, the faster it can diffuse through the lipid bilayer. From slowest to fastest: C, A, B.
10.7. What happens to K+ transport by valinomycin when the membrane is cooled below its transition temperature?
K+ transport ceases because the ionophore-K+ complex cannot diffuse through the membrane when the lipids are immobilized in a gel-like state.
10.10. Which amino acids would you expect to be particularly abun- dant at the entrance of a porin that is specific for phosphate ions?
Positively charged Lys and Arg would be relatively abundant at the entrance of the pore, so as to attract the negatively charged phosphate ions and repel cations.
10.15. The rate of movement (flux) of a substance X into cells was measured at different concentrations of X to construct the fol- lowing graph. (see pg. 313)
(a) Does this information suggest that the movement of X into the cells is mediated by a protein
(a) The data do not indicate the involvement of a transport protein, since the rate of transport does not approach a maximum as [X] increases. (b) To verify that a transport protein is involved, increase [X] to demonstrate saturation of the transporter at
10.16. Endothelial cells and pericytes in the retina of the eye have dif- ferent mechanisms for glucose uptake. The figure shows the rate of glucose uptake for each type of cell in the presence of increas- ing amounts of sodium. What do these results reve
The hyperbolic curve for glucose transport into pericytes indicates a protein-mediated sodium-dependent process. The transport protein has binding sites for sodium ions. At low [Na+], glucose transport is directly proportional to [Na+]. However, at high [
10.17. Explain why Na+ and K+ ions usually move more slowly through pumps than through channels.
A channel provides an open pore across the membrane, whereas a pump operates by changing its conformation in an ATP-dependent manner. The additional time required for ATP hydrolysis and protein conformation changes causes ion movement through a pump to be
10.18. Why would overexpression of an MDR transporter in a cancer cell make the cancer more difficult to treat?
Overexpression of an MDR transporter would increase the ability of the cancer cell to excrete anticancer drugs. Higher concentrations of the drugs or different drugs would then be required to kill the drug-resistant cells.
10.21. If the ATP supply in the cell shown in Fig. 10-21c suddenly vanished, would the intracellular glucose concentration increase, decrease, or remain the same?
In the absence of ATP, Na+ extrusion by the (Na+-K+)-ATPase would cease, so no glucose could enter the cell by the Na+ -glucose symport. The glucose in the cell would then exit via the passive-mediated glucose transporter, and the cellular [glucose] would
11.1. Choose the best description of an enzyme:
(a) It allows a chemical reaction to proceed extremely fast.
(b) It increases the rate at which a chemical reaction approaches
equilibrium relative to its uncatalyzed rate.
(c) It makes a reaction thermodyna
b
11.2. What is the relationship between the rate of an enzyme-catalyzed reaction and the rate of the corresponding uncatalyzed reaction? Do enzymes enhance the rates of slow uncatalyzed reactions as much as they enhance the rates of fast uncatalyzed reacti
As shown in Table 11-1, the only relationship between the rates of catalyzed and uncatalyzed reactions is that the catalyzed reaction is faster than the uncatalyzed reaction. The absolute rate of an uncatalyzed reaction does not correlate with the degree
11.5. On the free energy diagram shown (see pg. 353), label the intermediate(s) and
transition state(s). Is the reaction thermodynamically favorable?
There are three transition states (X�) and two intermediates (I). The reaction is not thermodynamically favorable because the free energy of the products is greater than that of the reactants. (see answer key)
11.6. Draw a transition state diagram of (a) a nonenzymatic reaction and the corresponding enzyme-catalyzed reaction in which (b) S binds loosely to the enzyme and (c) S binds very tightly to the enzyme. Compare ?G� for each case. Why is tight binding of
The tighter S binds to the enzyme, the greater the value of ?GE�. As the value of ?GE� for reaction c approaches that of ?G�N, the rate of the enzyme-catalyzed reaction approaches the rate of the nonenzymatic reaction. (see study guide answer key)
11.11. Explain why enzyme activity varies with temperature, as shown (see pg. 353).
As the temperature increases, thermal energy boosts the proportion of reactants that can achieve the transition state per unit time, so the rate increases. Above an optimal temperature, the enzyme becomes denatured and rapidly loses catalytic activity (re
11.12. The covalent catalytic mechanism of an enzyme depends on a single active site Cys whose pK is 8. A mutation in a nearby residue alters the microenvironment so that this pK increases to 10. Would the mutation cause the reaction rate to increase or d
The active form of the enzyme contains the thiolate ion. The increased pK would increase the nucleophilicity of the thiolate and thereby increase the rate of the reaction catalyzed by the active form of the enzyme. However, at physiological pH, there woul
11.13. Studies at different pH's show that an enzyme has two catalytically important residues whose pK 's are ~4 and ~10. Chemical modification experiments indicate that a Glu and a Lys residue are essential for activity. Match the residues to their pK 's
Glu has a pK of ~4 and, in its ionized form, acts as a base catalyst. Lys has a pK of ~10 and, in its protonated form, acts as an acid catalyst.
11.14. Explain why RNase A cannot catalyze the hydrolysis of DNA.
DNA lacks the 2'-OH group required for the formation of the 2',3'-cyclic reaction intermediate.
11.16. What feature of RNA would allow it to function as a ribozyme? Why are there no naturally occurring DNA enzymes?
The ability of RNA molecules to form complex tertiary structures allows them to bind substrates and catalyze reactions through proximity and orientation effects as well as by transition state stabilization, even though RNA lacks a wide variety of function
11.18. Suggest a transition state analog for proline racemase that differs from those discussed in the text. Justify your suggestion.
Two such analogs are Furan-2-carboxylate & Thiophene-2-carboxylate (see study guide answer key). Both of these molecules are planar, particularly at the C atom to which the carboxylate is bonded, as is true of the transition state for the proline racemase
11.19. Explain why lysozyme cleaves the artificial substrate (NAG)4 ~4000 times more slowly than it cleaves (NAG)6.
The lysozyme active site is arranged to cleave oligosaccharides between the fourth and fifth residues. Moreover, since the lysozyme active site can bind at least six monosaccharide units, (NAG)6 would be more tightly bound to the enzyme than (NAG)4, and t
11.20. Would you expect lysozyme to hydrolyze cellulose? Why or why not?
Although the polymeric chains of cellulose, which are �(1 to 4)-linked glucose residues, have the same overall configuration as the NAG�NAM repeating disaccharide of lysozyme's primary substrate peptidoglycan, cellulose chains typically occur in hydrogen-
11.28. Why is the broad substrate specificity of chymotrypsin advantageous in vivo? Why would this be a disadvantage for some other proteases?
As a digestive enzyme, chymotrypsin's function is to indiscriminately degrade a wide variety of ingested proteins, so that their component amino acids can be recovered. Broad substrate specificity would be dangerous for a protease that functions outside o
11.29. Tofu (bean curd), a high-protein soybean product, is prepared in such a way as to remove the trypsin inhibitor present in soybeans. Explain the reason(s) for this treatment.
If the soybean trypsin inhibitor were not removed from tofu, it would inhibit the trypsin in the intestine. At best, this would reduce the nutritional value of the meal by rendering its protein indigestible. It might very well also lead to intestinal upse
12.8. Explain why it is usually easier to calculate an enzyme's reaction velocity from the rate of appearance of product rather than the rate of disappearance of a substrate.
Enzyme activity is measured as an initial reaction velocity, the velocity before much substrate has been depleted and before much product has been generated. It is easier to measure the appearance of a small amount of product from a baseline of zero produ
12.9. At what concentration of S (expressed as a multiple of KM) will Vo = 0.95Vmax?
(see answer key)
12.11. Calculate KM and Vmax from the following data:
(see pg. 393)
(see answer key)
12.12. Explain why each of the following data sets from a Lineweaver-Burk plot are not individually ideal for determining KM for an enzyme-catalyzed reaction that follows Michaelis-Menten kinetics. (see pg. 393)
(see answer key)
12.13. Is it necessary for measurements of reaction velocity to be ex- pressed in units of concentration per time (M/s, for example) in order to calculate an enzyme's KM ?
Velocity measurements can be made using any convenient unit of change per unit of time. KM is, by definition, a substrate concentration (the concentration when Vo = Vmax/2), so its value does not reflect how the velocity is measured.
12.14. Is it necessary to know [E]T in order to determine (a) KM, (b) Vmax, or (c) kcat?
(a, b) It is not necessary to know [E]T. The only variables required to determine KM and Vmax (for example, by constructing a Lineweaver-Burk plot) are [S] and Vo. (c) The value of [E]T is required to calculate kcat since kcat = Vmax/[E]T.
12.16. You are attempting to determine KM by measuring the reaction velocity at different substrate concentrations, but you do not realize that the substrate tends to precipitate under the experimental conditions you have chosen. How would this affect you
The experimentally determined KM would be greater than the true KM because the actual substrate concentration is less than expected.
12.17. You are constructing a velocity versus [substrate] curve for an enzyme whose KM is believed to be about 2 �M. The enzyme concentration is 200 nM and the substrate concentrations range from 0.1 �M to 10 �M. What is wrong with this experimental setup
The enzyme concentration is comparable to the lowest substrate concentration and therefore does not meet the requirement that [E] << [S]. You could fix this problem by decreasing the amount of enzyme used for each measurement.
12.18. Enzyme X and enzyme Y catalyze the same reaction and exhibit the vo versus [S] curves shown below. Which enzyme is more efficient at low [S]? Which is more efficient at high [S]? (see pg. 393)
Enzyme Y is more efficient at low [S]; enzyme X is more efficient at high [S].
12.20. Enzyme A catalyzes the reaction S to P and has a KM of 50 �M and a Vmax of 100 nM/s. Enzyme B catalyzes the reaction S to Q and has a KM of 5 mM and a Vmax of 120 nM/s. When 100 �M of S is added to a mixture containing equivalent amounts of enzymes
Product P will be more abundant because enzyme A has a much lower KM for the substrate than enzyme B. Because Vmax is approximately the same for the two enzymes, the relative efficiency of the enzymes depends almost entirely on their KM values.
12.27. Determine the type of inhibition of an enzymatic reaction from the following data collected in the presence and absence of the inhibitor. (see pg. 394)
(see answer key)
12.28. Estimate KI for an apparent value of KM that a competitive inhibitor when [I] = 5 mM gives is three times the KM for the uninhibited reaction.
From Eq. 12-32, a is 3.
a = 3 = 1 +[I]/KI = 1 + 5 mM/KI
KI = 2.5 mM
12.29. For an enzyme-catalyzed reaction, the presence of 5 nM of a reversible inhibitor yields a Vmax value that is 80% of the value in the absence of the inhibitor. The KM value is unchanged. (a) What type of inhibition is likely occurring? (b) What prop
(a) Inhibition is most likely mixed (noncompetitive) with a=a' since it is reversible and only Vmax is affected.
(b) Since V app/max = 0.8 Vmax, 80% of the enzyme remains uninhibited. Therefore, 20% of the enzyme molecules have bound inhibitor.
(c) (see a
14.5. Rank the following compounds in order of increasing oxidation state.
A. CH3CH(OH)CH2OH
B. CH2(COO-)2
C. CH2(CH3)2
D. CH3CHCH2
E. CH3CH(O)COO-
C,D,A,E,B
14.7. Citrate synthase catalyzes the reaction
Oxaloacetate + acetyl-CoA -> citrate + HS-CoA
The standard free energy change for the reaction is -31.5 kJ/mol. (a) Calculate the equilibrium constant for this re- action at 37�C. (b) Would you expect this rea
(see answer key)
14.8. A certain metabolic reaction takes the form A -> B. Its standard free energy change is 7.5 kJ/mol. (a) Calculate the equilibrium constant for the reaction at 25�C. (b) Calculate ?G at 37�C when the concentration of A is 0.5 mM and the concentration
(see study guide answer key)
14.9. Choose the best definition for a near-equilibrium reaction:
(a) always operates with a favorable free energy change. (b) has a free energy change near zero.
(c) is usually a control point in a metabolic pathway.
(d) operates very slowly in vivo.
b
14.12. Assuming 100% efficiency of energy conservation, how many moles of ATP can be synthesized under standard conditions by the complete oxidation of 1 mol of palmitate?
The theoretical maximum yield of ATP is equivalent to (?G�' for fuel oxidation)/(?G�' for ATP synthesis)
= (-9781 kJ/mol)/(-30.5 kJ/mol) ~ 320 ATP
14.13. Does the magnitude of the free energy change for ATP hydrolysis increase or decrease as the pH increases from 5 to 6?
At pH 6, the phosphate groups are more ionized than they are at pH 5, which increases their electrostatic repulsion and therefore increases the magnitude of ?G for hydrolysis (makes it more negative).
14.15. The reaction for "activation" of a fatty acid (RCOO-),
ATP+CoA+RCOO- => RCO-CoA+AMP+PPi
has ?G�' = +4.6 kJ/mol. What is the thermodynamic driving force for this reaction?
The exergonic hydrolysis of PPi by pyrophosphatase (?G�' = -19.2 kJ/mol) drives fatty acid activation.
14.18. If intracellular [ATP] = 5 mM, [ADP] = 0.5 mM, and [Pi ] = 1.0 mM, calculate the concentration of AMP at pH 7 and 25�C under the condition that the adenylate kinase reaction is at equilibrium.
(see study guide answer key)
14.26. Under standard conditions, is the oxidation of free FADH2 by ubiquinone sufficiently exergonic to drive the synthesis of ATP?
(see study guide answer key)
14.27. A hypothetical three-step metabolic pathway consists of interme- diates W, X, Y, and Z and enzymes A, B, and C. Deduce the order of the enzymatic steps in the pathway from the following information:
1. Compound Q, a metabolic inhibitor of enzyme B,
Z --B--> W --C--> Y --A--> X
14.28. (see pg. 471)
(a) The step catalyzed by enzyme Y is likely to be the major flux- control point, since this step operates farthest from equilibrium (it is an irreversible step).
(b) Inhibition of enzyme Z would cause the concentration of D, the reaction's product, to de
15.1. Which of the 10 reactions of glycolysis are (a) phosphorylations, (b) isomerizations, (c) oxidation-reductions, (d) dehydrations, and (e) carbon-carbon bond cleavages?
(a) Reactions 1, 3, 7, and 10; (b) Reactions 2, 5, and 8; (c) Reaction 6; (d) Reaction 9; (e) Reaction 4.
15.4. Bacterial aldolase does not form a Schiff base with the substrate. Instead, it has a divalent Zn2+ ion in the active site. How does the ion facilitate the aldolase reaction?
The Zn2+ polarizes the carbonyl oxygen of the substrate to stabilize the enolate intermediate of the reaction.
(see study guide answer key for more detailed)
15.14. The catalytic behavior of liver and brain phosphofructokinase-1 (PFK-1) was observed in the presence of AMP, Pi, and fructose- 2,6-bisphosphate. The following table lists the concentrations of each effector required to achieve 50% of the maximal ve
The liver enzyme is far more sensitive than the brain enzyme to the three activators. It is possible that liver PFK-1 is subject to a greater degree of regulation than brain PFK-1. Fuel must be supplied to the brain continuously and thus glycolysis is alw
15.15. Although it is not the primary flux-control point for glycolysis, pyruvate kinase is subject to allosteric regulation. What is the metabolic importance of regulating flux through the pyruvate kinase reaction?
Pyruvate kinase regulation is important for controlling the flux of metabolites, such as fructose (in liver), which enter glycolysis after the PFK step.
15.17. Consider the pathway for catabolizing galactose. What are the potential control points for this pathway?
Galactokinase, which catalyzes a highly exergonic reaction, is a possible control point. Because galactose enters the glycolytic pathway as glucose-6-phosphate, the PFK reaction is also likely to be a major control point.
15.21. Compare the ATP yield of three glucose molecules that enter gly- colysis and are converted to pyruvate with that of three glucose molecules that proceed through the pentose phosphate pathway such that their carbon skeletons (as two F6P and one GAP)
The three glucose molecules that proceed through glycolysis yield 6 ATP. The bypass through the pentose phosphate pathway results in a yield of 5 ATP.
15.28. (see pg. 515)
(see study guide answer key)
16.2. Glucose binds to glycogen phosphorylase and competitively inhibits the enzyme. What is the physiological advantage of this?
This mechanism allows glycogen phosphorylase activity to be regulated by the concentration of glucose so that glycogen is not broken down when glucose is already plentiful.
16.3. Phosphoglucokinase catalyzes the phosphorylation of the C6-OH group of G1P. Why is this enzyme important for the normal function of phosphoglucomutase?
Phosphoglucokinase activity generates G1,6P, which is necessary to "prime" phosphoglucomutase that has become dephosphorylated and thereby inactivated through the loss of its G1,6P reaction intermediate.
16.4. Glucose-6-phosphatase is located inside the endoplasmic reticulum. Describe the probable symptoms of a defect in G6P transport across the endoplasmic reticulum membrane.
A defect in G6P transport would have the symptoms of glucose-6-phosphatase deficiency: accumulation of glycogen and hypoglycemia.
16.5. Individuals with McArdle's disease often experience a "second wind" resulting from cardiovascular adjustments that allow glucose mobilized from liver glycogen to fuel muscle contraction. Explain why the amount of ATP derived in the muscle from circu
The conversion of circulating glucose to lactate in the muscle generates 2 ATP. If muscle glycogen could be mobilized, the energy yield would be 3 ATP, since phosphorolysis of glycogen bypasses
the hexokinase-catalyzed step that consumes ATP in the first
16.6. A sample of glycogen from a patient with liver disease is incubated with Pi, normal glycogen phosphorylase, and normal debranching enzyme. The ratio of G1P to glucose formed in the reaction mixture is 100. What is the patient's most probable enzymat
The deficiency is in branching enzyme (Type IV glycogen storage disease). The high ratio of G1P to glucose indicates abnormally long chains of a(1->4)-linked residues with few a(1->6)-linked branch points (the normal ratio is ~10).
16.10. Subjects who consumed a meal containing a high amylose:amylopectin ratio showed a lower increase in blood glucose levels than subjects who consumed a meal with a low amylose:amylopectin ratio. Explain.
Amylose is a linear form of starch, so during digestion, glucose monomers can be released only one at a time from the end of the molecule. Amylopectin, a branched form of starch, can release glucose from each of its branches. Consequently, the rate of glu
16.11. Many diabetics do not respond to insulin because of a deficiency of insulin receptors on their cells. How does this affect (a) the levels of circulating glucose immediately after a meal and (b) the rate of glycogen synthesis in muscle?
(a) Circulating [glucose] is high because cells do not respond to the insulin signal to take up glucose.
(b) Insulin is unable to activate phosphoprotein phosphatase-1 in muscle, so glycogen synthesis is not stimulated. Moreover, glycogen synthesis is muc
16.12. Why does it make metabolic sense for the same hormone signal to stimulate glycogenolysis and inhibit glycolysis in the liver while stimulating both glycogenolysis and glycolysis in muscles?
The two tissues perform different physiological functions and therefore respond differently to the same hormone. Liver responds by promoting glycogenolysis and gluconeogenesis to produce glucose that can be released for use by other tissues. Muscles respo
16.19. Predict the effect of a fructose-1,6-bisphosphatase deficiency on blood glucose levels (a) before and (b) after a 24-hour fast.
(a) At the beginning of a fast, blood glucose levels are normal, because dietary sources or glycogenolysis can supply glucose. (b) After a fast, the blood glucose levels are very low, because dietary glucose and glycogen have been depleted and gluconeogen
16.21. Would you expect the allosteric regulator AMP to increase or decrease fructose-1,6-bisphosphatase activity?
A high level of AMP results from a high rate of ATP consumption in the cell, so it would act to promote flux through ATP-generating pathways such as glycolysis. Therefore, AMP would be expected to inhibit the activity of the gluconeogenic enzyme fructose-