Undergrad Biochem (Cont)

Front 1

What is the effect of decreasing the concentration of 2,3-BPG on the affinity of hemoglobin for O2?
A. Decreased affinity
B. No effect
C. Cannot determine from information provided
D. Increased affinity
E. Release of superoxide ion

Back 1

D. Increased affinity

Hint 1

2,3-BPG stabilizes the T state, which is the state without oxygen off the heme group. Reducing 2,3-BPG concentration therefore makes the T to R state transition easier, so that there will be more molecules in the R state. The R state has oxygen bound. So the net effect of reducing 2,3-BPG is to INCREASE affinity of Hb for O<sub>2</SUB>.

Front 2

Why does increasing the CO2 concentration of a solution cause reduced pH?
A. CO2 is an acid.
B. CO2 absorbs protons from the solution.
C. CO2 combines with H2O to form carbonic acid, which then dissociates to release a proton.
D. CO2 stabilizes the T state of hemoglobin by forming a covalent bond at the amino terminus.
E. None of the above.

Back 2

CO2 combines with H2O to form carbonic acid, which then dissociates to release a proton.

Front 3

Why is affinity for O2 higher in fetal hemoglobin than it is in maternal hemoglobin?

Back 3

B. Amino acid differences in fetal hemoglobin cause it to bind less tightly to 2,3-BPG than is the case for maternal hemoglobin.

Front 4

Which of the following molecular events is the trigger that initiates the change in hemoglobin structure responsible for cooperative binding of O2?
A. Binding of the proximal histidine residue to the iron atom in heme.
B. Protonation of His146 when the pH is reduced.
C. Binding of heme to the hemoglobin apoprotein to form the holoprotein.
D. Reduction in the electron density of the iron atom in heme after O2 binding, which allows it to move into the plane of the porphyrin ring. 100%
E. Binding of 2,3-BPG into the central cavity of deoxyhemoglobin.

Back 4

D. Reduction in the electron density of the iron atom in heme after O2 binding, which allows it to move into the plane of the porphyrin ring.

Front 5

Which of the following statements about the comparison of hemoglobin and myglobin is TRUE?
A. Both hemoglobin and myoglobin are tetrameric proteins.
B. Hemoglobin has a higher degree of O2 saturation than myoglobin at the partial pressure of O2 that occurs in muscle tissue.
C. Hemoglobin exhibits cooperative binding of O2 whereas myoglobin does not.
D. Hemoglobin shows a hyperbolic O2 binding curve whereas myglobin shows a sigmoidal curve.
E. O2 binding to both hemoglobin and myoglobin is regulated by pH.

Back 5

C. Hemoglobin exhibits cooperative binding of O2 whereas myoglobin does not.

Front 6

Why does reducing the pH cause increased release of O2 from hemoglobin?
A. Lower pH favors dissociation of the quaternary structure of hemoglobin.
B. Lower pH causes stabilizes the interaction between O2 and heme.
C. Lower pH causes 2,3-BPG to bind more tightly to hemoglobin.
D. Lower pH changes the structure of the heme group.
E. Lower pH causes protonation of His146, which forms a salt bridge that stabilizes the T (deoxy) state.

Back 6

E. Lower pH causes protonation of His146, which forms a salt bridge that stabilizes the T (deoxy) state.

Hint 6

Figure 7.19.

Front 7

Which of the following statements about active sites of enzymes are TRUE?
A. Active sites are accessible to the solvent environment.
B. Active sites contain multiple amino acid side chains that are capable of forming weak bonds with the substrate.
C. Active sites bind the transition state of the catalyzed reaction more tightly than they bind the reactant or product molecules.
D. Acitve sites are formed into their final structure only when they interact with the substrate.
E. All of the above.

Back 7

E. All of the above.

Front 8

The following table lists the kcat/KM values for the activity of chymotrypsin towards substrates with different amino acid R groups. For which of these substrates does chymotrypsin act most efficiently?
Amino acid side chain kcat/KM (s-1 M-1)
Glycine 1.3 x 10-1
Valine 2.0 x 100
Leucine 3.2 x 102
Isoleucine 3.6 x 102
Phenylalanine 10 x 105
A. Glycine
B. Valine
C. Leucine
D. Isoleucine
E. Phenylalanine

Back 8

E. Phenylalanine

Front 9

Which of the following statements about Gibbs free energy values is FALSE?
A. In conditions when ΔG = 0 the reaction is at equilibrium.
B. The magnitude of the ΔG value for a reaction does not influence the rate at which that reaction proceeds.
C. If the ΔG value of a reaction is positive, then that reaction cannot proceed spontaneously.
D. The ΔG°' value applies to a reaction at a specified set of reference conditions, whereas the ΔG value applies to a reaction at the actual conditions of the moment.
E. If the ΔG°' value of a reaction is positive, then that reaction cannot proceed spontaneously.

Back 9

E. If the ΔG°' value of a reaction is positive, then that reaction cannot proceed spontaneously.

Front 10

Which type of reversible enzyme inhibitor binds only to the ES complex?
A. Noncompetitive inhibitor
B. Competitive inhibitor
C. Suicide inhibitor
D. None
E. Uncompetitive inhibitor

Back 10

E. Uncompetitive inhibitor

Front 11

The figure show a Lineweaver-Burk plot for an enzyme in the presence or absence of a noncompetitive inhibitor. Which parameter of the kinetics of this enzyme appears to change in the presence of the inhibitor?
A. kcat
B. Vmax
C. KM
D. kcat/KM
E. The concentration of enzyme.

Back 11

B. Vmax

Front 12

Which of the following statements about KM are TRUE?
A. KM is equal to the concentration of substrate at which a reaction is at 1/2 of the maximal velocity.
B. KM is equal to the dissociation constant for formation of the ES complex, if k2 is much smaller than k-1.
C. KM appears to change in the presence of a competitive inhibitor.
D. Answers a, b and c are all true.
E. None of the above.

Back 12

D. Answers a, b and c are all true.

Front 13

Which of the following would result from a mutation in subtilisin that changed Ser 221 to isoleucine?
A. A large change in KM.
B. A small change in KM.
C. A large change in kcat.
D. Both a and c.
E. Both b and c.

Back 13

C. A large change in kcat.
or
E. Both b and c.

Front 14

Which functional group indicated in the figure below acts as the catalytic acid in the aspartyl protease mechanism?
A. Group a
B. Group b
C. Group c
D. Group d
E. Group e

Back 14

E. Group e

Front 15

Which of the following is NOT a way in which an enzyme could reduce the free energy of a transition state?
A. Using acid/base catalysis to accommodate charged atoms.
B. Using covalent intermediates.
C. Using a different transition state than would occur in the uncatalyzed reaction.
D. Using energy that comes from binding interactions between the transition state and the induced fit of the enzyme.
E. Bringing the reactive groups into close contact with each other.

Back 15

C. Using a different transition state than would occur in the uncatalyzed reaction.

Front 16

Which functional group indicated in the figure below is the base of the catalytic triad in a serine protease?
A. Group a
B. Group b
C. Group c
D. Group d
E. Group e

Back 16

C. Group c

Front 17

You have synthesized two transition state analogs for a particular enzyme catalyzed reaction. One contains a positive charge at the reacting center, the other a negative charge. The one with the positive charge binds less tightly than the substrate, while that with the negative charge binds much more tightly. What do you infer regarding the true transition state for the catalyzed reaction?
A. It contains a positive charge at this position.
B. Nothing can be inferred about the transition state from these results.
C. It is not charged at this position.
D. There is a covalent enzyme-bound intermediate in the reaction.
E. It contains a negative charge at this position.

Back 17

E. It contains a negative charge at this position.

Front 18

Why is deformation of the DNA backbone by a Type II restriction endonuclease required for catalysis?
A. Backbone deformation is required in order for the enzyme to bind to the DNA.
B. Backbone deformation creates a binding site for a Mg2+ ion that is required for catalysis.
C. Backbone deformation makes the phosphoester bond to be cleaved more reactive.
D. Backbone deformation is required for methylation of an adenine residue within the target sequence.
E. All of the above.

Back 18

B. Backbone deformation creates a binding site for a Mg2+ ion that is required for catalysis.

Front 19

Which of the following statements about allosteric regulation is TRUE?
A. Allosteric effectors of an enzyme usually are the same as the substrates of that enzyme.
B. Allosteric regulation involves binding of an effector at a location on the enzyme other than the active site.
C. Allosteric regulation is a slow means of modifying enzyme activity compared to other means of regulation.
D. Allosteric regulation involves covalent modification of the enzyme.
E. Monomeric enzymes are frequently subject to allosteric regulation.

Back 19

B. Allosteric regulation involves binding of an effector at a location on the enzyme other than the active site.

Front 20

A hypothetical metabolic pathway is shown in the diagram below. Which statement is most likely true about this pathway?
Enzyme 1 > 2 >3 > 4
Metabolite A>B>C>D

Back 20

D. Metabolite E is likely to be an allosteric inhibitor of enzyme 1

Front 21

Which of the following general mechanisms of enzyme regulation is results in the most rapid change in activity?
A. Changes in gene regulation that alter enzyme concentration
B. Post-translational modification
C. Proteolytic cleavage
D. Allosteric regulation
E. Protein degradation

Back 21

D. Allosteric regulation

Front 22

Why does protein phosphorylation regulate enzyme activity?
A. Introduction of two negative charges to a previously uncharged hydroxyl group results in conformational changes in polypeptide folding.
B. The phosphate group is added in the active site and blocks substrate access.
C. The phosphate group delivers metal ions to the protein.
D. The phosphate group targets the protein for degradation.
E. The negative charges on the phosphate group are used in substrate binding.

Back 22

A. Introduction of two negative charges to a previously uncharged hydroxyl group results in conformational changes in polypeptide folding.

Front 23

Which of the following types of enzyme regulation is responsible for the activation of protein kinase A?
A. Acetylation of lysine residues
B. Altered gene regulation
C. Protein phosphorylation
D. Allosteric regulation
E. Activation by specific proteolytic cleavage

Back 23

D. Allosteric regulation

Front 24

Which of the following proteins is the common activator of the pancreatic zymogens?
A. Protein kinase A
B. Chymotrypsin 0%
C. Trypsin
D. Thrombin
E. Elastase

Back 24

C. Trypsin

Front 25

Which of the following structures is the anomer of ß-D-glucopyranose?

Back 25

C. Structure C

Front 26

Which of the following compounds is L-glyceraldehyde?

Back 26

C. Compound

Front 27

Human cell surface carbohydrates are involved in:
A. Cell-cell contact
B. Virus contact
C. Bacterial contact
D. Immune self/non-self recognition
E. All of the above
F. a, b, and c above

Back 27

E. All of the above

Front 28

What is the difference between amylose and cellulose?
A. They have different branch linkage frequencies.
B. Amylose has α-(1→4) linkages and cellulose has α-(1→6) linkages.
C. They have different stereochemistry at the anomeric carbon.
D. They have different numbers of non-reducing ends.
E. They have different numbers of reducing ends.

Back 28

C. They have different stereochemistry at the anomeric carbon.

Front 29

Which of the following polysaccharides serves as an energy storage molecule in mammalian cells?
A. Chondroitin sulfate
B. Amylopectin
C. Glycogen
D. Cellulose
E. N-linked oligosaccharides

Back 29

C. Glycogen

Front 30

Why is the anomeric carbon of a monosaccharide more reactive than the other carbon atoms in the molecule?
A. It is more exposed to the solute than the other carbons.
B. It is modified with functional groups that make it more reactive.
C. There are no differences in reactivity between the anomeric carbon and other carbons in a monosaccharide.
D. It has less steric hindrance than other carbons so is more flexible.
E. It is attached to two oxygen molecules, which withdraw electrons and make the carbon more susceptible to nucleophilic attack.

Back 30

E. It is attached to two oxygen molecules, which withdraw electrons and make the carbon more susceptible to nucleophilic attack.

Front 31

How do bacteria modify their membranes to cope with increasing temperature?
A. By adding ions to bind to charged groups in the polar head groups.
B. By replacing 18-C fatty acids in their lipids with 16-C fatty acids.
C. By replacing polar phospholipids with neutral phospholipids.
D. By replacing unsaturated fatty acids in their lipids with saturated fatty acids.
E. By incorporating more cholesterol into their membranes.

Back 31

D. By replacing unsaturated fatty acids in their lipids with saturated fatty acids.

Front 32

Which of the following is the reason that fatty acids form micelles whereas phospholipids form bilayers?
A. Phosphate groups on phospholipids would prevent micelle formation.
B. The two non-polar tails of the phospholipids are too bulky to fit into the spherical interior of a micelle.
C. The nonpolar tails of phospholipids form a more fluid hydrophobic core than those of fatty acids.
D. Phospholipids can form covalent bonds with each other whereas fatty acids cannot.
E. Phospholipids are more polar than fatty acids.

Back 32

B. The two non-polar tails of the phospholipids are too bulky to fit into the spherical interior of a micelle.

Front 33

Which of these statements regarding biological membranes is true?
A. None of these statements are true.
B. They are static.
C. They are passive barriers.
D. They are symmetrical.
E. They are dynamic barriers.

Back 33

E. They are dynamic barriers.

Front 34

Which of the following membranes would be the most fluid?
A. All of the above are equivalent in fluidity.
B. A bilayer made of lipids with polyunsaturated 18 carbon-fatty acids.
C. A bilayer made of lipids with saturated 18 carbon-fatty acids.
D. A bilayer made of lipids with saturated 16 carbon-fatty acids.
E. A bilayer made of lipids with polyunsaturated 16 carbon-fatty acids.

Back 34

E. A bilayer made of lipids with polyunsaturated 16 carbon-fatty acids.

Front 35

Which of the following statements about integral membrane proteins is FALSE?
A. They can be removed from the membrane by treatment with mild salt solutions.
B. They require strong detergents to be removed from the membrane.
C. The can be involved in transporting specific molecules across the membrane.
D. They can have a largely alpha-helical structure.
E. They can have a largely beta-sheet structure.

Back 35

A. They can be removed from the membrane by treatment with mild salt solutions.

Front 36

Explain why a mutation that changed Asn155 of subtilisin to an alanine would reduce the catalytic rate of the enzyme?

Back 36

Asparagine is a polar amino acid with a side chain that can participate in hydrogen bonds with negatively charged atoms. During the transition state of the subtilisin reaction a negative charge develops on an oxygen molecule, and that atom moves close to Asn155. The hydrogen bond between the two molecules reduces the free energy of the transition state by masking what would otherwise be an exposed charge and thus reduces the free energy change between the reactant and the transition state. If Asn155 were replaced by the non-polar alanine residue, the H bond could not occur and the free energy change between the reactant and transition state would be greater than in the normal enzyme. Because that free energy change is proportional to the reaction rate, the mutant enzyme with Ala155 would work slower than the normal enzyme with Asn155.

Front 37

Briefly explain why the cooperative binding of hemoglobin to oxygen, diagrammed in the figure, is necessary for sufficient transfer of oxygen to mammalian tissues.

Back 37

Answer The sigmoidal binding curve of oxygen affinity for hemoglobin allows a large percentage of the ligand that binds at the pO2 of the lungs to dissociate at the pO2 of the tissues. Regulation by CO2 concentration and pH exaggerates this effect. If Hb did not bind oxygen cooperatively, then only a small amount of the bound ligand would dissociate in the tissues, owing to the hyperbolic binding curve.