Exam 2

Question 1

2006

Two protein sequences have been aligned and the amino acid differences between corresponding positions are examined. Which of the following substitutions would be the least conservative?

• A. D for E
• B. E for N
• C. K fo R
• D. I for L
• E. D for L

Answer 1

E. D for L

Question 2

2006

Two protein which have the same catalytic activity and very nearly the same residues in their active sites but very different sequences and tertiary structures would usually be considered an example of

• A. mistaken observation
• B. convergent evolution
• C. sequence homology
• D. divergent evolution
• E. none of these

Answer 2

B. convergent evolution

Question 3

2006

Two protein which have the have similar activity, similar amino acid sequences and similar tertiary structures would usually be considered an example of

• A. mistaken observation
• B. converent evolution
• C. sequence homology
• D. divergent evolution
• E. none of these

Answer 3

D. divergent evolution

Question 4

2006

In the α, β barrel motif often seen in proteins, the β-strands are

• A. hydrogen bonds between backbone groups on different amino acids
• B. hydrogen bonds between side chain groups on different amino acids
• C. hydrogen bonds between backbone and side chain groups on different amino acids
• D. hydrogen bonds between water molecules and side chain or backbone groups
• E. all of the above are found in protein interiors

Answer 4

E. all of the above are found in protein interiors

Question 5

2006

Which of the following is not found in the interior of globular proteins

• A. hydrogen bonds between backbone groups on different amino acids
• B. hydrogen bonds between side chain groups on different amino acids
• C. hydrogen bonds between backbone and side chain groups on different amino acids
• D. hydrogen bonds between water molecules and side chain or backbone groups
• E. all of the above are found in protein interiors

Answer 5

E. all of the above are found in protein interiors

Question 6

2006

In connecting which of the following will must the crossover have handedness, irrespective of whether the handedness is left or right?

• A. an a-helix and a B-strand
• B. two parallel B-strands
• C. two antiparallel B-strands
• D. more than one of the above
• E. none of the above

Answer 6

B. two parallel B-strands

Question 7

2006

If a protein - or any object for that matter - has a two-fold rotation axis of symmetry?

• A. rotation about that axis by 180º converts it into itself, ie. the result of the rotation cannot be distinguished from the original unrotated form
• B. rotation about that axis alone cannot convert it into itself; a second rotation about the axis of 180º is needed to do that
• C. the protein can only be converted into itself by a 180º rotation followed by a translation along the axis
• D. the protein can only be converted into itself by reflection in a mirror plane

Answer 7

A. rotation about that axis by 180º converts it into itself, ie. the result of the rotation cannot be distinguished from the original unrotated form

Question 8

2006

If the ΔS of the U to N transition of a protein in aqueous solution is negative, then

• A. entropy contributes favorably to the free energy of folding
• B. folding will not occur
• C. the enthalpy of folding will be positive
• D. the protein cannot contain disulfide bonds
• E. folding, if it occurs, i.e. if ΔG < 0, will be enthalpy dirven

Answer 8

E. folding, if it occurs, i.e. if ΔG < 0, will be enthalpy dirven

Question 9

2006

The prediction of secondary structures in proteins of unknown three dimensional structure is based on

• A. the frequency with which each type of amino acid is found in the various secondary structures in proteins of known three dimensional structure
• B. the number of amino acid residues which fall outside the allowed zones in a Ramachandran plot
• C. the difference between physiological pH and the protein’s PI
• D. tthe phase of the moon

Answer 9

A. the frequency with which each type of amino acid is found in the various secondary structures in proteins of known three dimensional structure

Question 10

2016

Which of the following is/are correct units for reaction velocity?

• M^-1
• moles / liter
• liters / mole
• moles / liter-minute
• mg / mL

Answer 10

moles / liter-minute

Question 11

2006

K_m for a Michaelis-Menten enzyme is a good estimate of the dissociation constant for the enzyme-substrate complex provided that

• A. conversion of ES to product is fast compared to substrate binding
• B. product dissociation is not the rate limiting step in the catalysis
• C. substrate binds to the enzyme and dissociates from it at the same rate
• D. substrate binding and release are fast compared to product formation on the enzyme and product release
• E. more that one of the above

Answer 11

D. substrate binding and release are fast compared to product formation on the enzyme and product release

Question 12

2006

After five half-lives of a first order reaction have passed, the percent of the initial material left is

• A. 3.1%
• B. 50%
• C. 20%
• D. 25%
• D. 5%

Answer 12

A. 3.1%

Question 13

2006

In the steady state assumption which underlies Michaelis-Menten kinetics

• S is converted to P at the same rate P is converted to S
• ES is formed at the same rate at which it is broken down
• The reaction rate does not change as substrate concentration rises
• More than one of the above
• None of the above

Answer 13

ES is formed at the same rate at which it is broken down

Question 14

2006

When the reaction velocity of a reaction catalyzed by a Michaelis-Menten enzyme is one half Vmax

• [S] << Km
• [S] = Km
• [S] >> Km
• [S] is unrelated to Km
• Km = Vmax

Answer 14

[S] = Km

Question 15

2006

An experimenter believes that an inhibitor of a Michaelis-Menten enzyme binds to the enzyme only if substrate is already bound. She would expect that

• Lineweaver-Burk plots of the data would have parallel lines
• Lineweaver-Burk plots of the data would meet at a point on the 1/V axis
• Lineweaver-Burk plots of the data would meet at a point on the 1/[S] axis
• more than one of the above
• none of the above

Answer 15

Lineweaver-Burk plots of the data would have parallel lines

Question 16

2006

In the Bohr effect of homoglobin one observes that the binding of oxygen causes dissociation of protons from the protein. This observation requires that

• the pH of the strongly buffered solution drop sharply
• the pH of the strongly buffered solution rise sharply
• oxygen binding causes the pKa one or more side chains of the hemoglobin to decrease
• oxygen binding causes the pKa one or more side chains of the hemoglobin to increase
• none of the above

Answer 16

oxygen binding causes the pKa one or more side chains of the hemoglobin to decrease

Question 17

2006

The enthalpy of folding (ΔH) of most small proteins is approximately - 200kJ/mol at 25°C under physiological conditions. Myoglobin, however, folds with an enthalpy of zero. Which of the following best explains why myoglobin folds with ΔH = 0

• the folding is entropy driven
• the entropy of folding is unfavorable
• the enthalpy and entropy of folding balance exactly thereby allowing folding to occur
• the enthalpy of folding is unfavorable
• none of the above is a possible explanation

Answer 17

the folding is entropy driven

Question 18

2006

The enthalpy of folding (ΔH) of most small proteins is approximately - 200kJ/mol at 25°C under physiological conditions. The free energy of folding under the same conditions is approximately - 50 kJ/mol. Which of the following best explains these results?

• the folding is entropy driven
• the entropy of folding is unfavorable
• the enthalpy and entropy of folding balance exactly thereby allowing folding to occur
• the enthalpy of folding is unfavorable
• none of the above is a possible explanation

Answer 18

the entropy of folding is unfavorable

Question 19

2006

Serine is found at positions 77 and 79 of a normal, wild type protein. In a series of site specific mutagenesis experiments, all nineteen of the other amino acids are substituted for ser 77, and in a second series of experiments, all nineteen amino acids are substituted for ser 79. The unfolding temperature of all thirty-eight mutant proteins is compared with that of the wild type. At position 77 only alanine had no effect on the unfolding temperature. All other amino acids lowered the Tm. At position 79, however, not only alanine but also many other amino acids had no effect on the stability. Which of the following best explains these results?

• A. Residue 79 is on the protein surface and interacts with few other residues, while residue 77 is an interior residue and normally interacts with many other residues in the three dimensional structure.
• B. Residue 77 is on the protein surface and interacts with few other residues, while residue 79 is an interior residue and normally interacts with many other residues in the three dimensional structure.
• C. Residue 79 is involved in an α-helix, while residue 77 is involved in a β-strand.
• D. Residue 77 is involved in an α-helix, while residue 79 is involved in a β-strand.
• E. None of the above can explain the results.

Answer 19

A. Residue 79 is on the protein surface and interacts with few other residues, while residue 77 is an interior residue and normally interacts with many other residues in the three dimensional structure.

Question 20

2006

It is observed that side chains such as valine, which are branched at the B-carbon, and very bulky side chains have a strong propensity to occur in B-sheets, whereas side chains which are unbranched at the B-carbon and are relatively long have a high probability of being found in a-helices. Which of the following might explain this observation?

• The peptide bonds of residues having branched or bulky side chains are too short to make helical hydrogen bonds
• The peptide bonds of residues having branched or bulky side chains are too long to make a-helical hydrogen bonds
• Side chain to backbone hydrogen bonds of branched or bulky residues stabilize the B-sheet but not the a-helix
• Steric interference betwee neighboring branched or bulky side chains prevents the formation of a-helices but does not occur in the B-sheet coformation. Unbranched and long side chains don’t have this problem and fit sterically in a-helices quite well
• more than one of the above

Answer 20

Steric interference betwee neighboring branched or bulky side chains prevents the formation of a-helices but does not occur in the B-sheet coformation. Unbranched and long side chains don’t have this problem and fit sterically in a-helices quite well

Question 21

2006

Anfinsen’s experiment with ribonuclease-A demonstrates that

• A. proteins have an absolute requirement for chaperonins and other cellular components in order to fold correctly.
• B. the amino acid sequence of a protein contains all the information necessary for the protein to fold to its native three dimensional conformation.
• C. you can’t unscramble an egg.
• D. more than one of the above E. none of the above

Answer 21

B. the amino acid sequence of a protein contains all the information necessary for the protein to fold to its native three dimensional conformation.

Question 22

2006

The carbohydrate unit(s) at the non-reducing end(s) of X is (are)

• Gal
• Gal, Glc
• Fru, Glc
• Man, Fru
• Gal, Fru

Answer 22

Gal, Fru

Question 23

2006

The carbohydrate unit at the reducing end is

• Gal
• Glc
• Fru
• Man
• None of the above

Answer 23

None of the above

Question 24

2006

The Hydrolysis of X in acid solution will give _____ monosaccharides in the proportion _____

• 2; 1:2
• 3; 1:1:1
• 1
• 3; 2:1
• X will not hydrolyse

Answer 24

3; 1:1:1

Question 25

2006

Which exoglycosidase(s) will act on X?

• B-glucosidase and a-fructosidase
• a-galactosidase and a-fructosidase
• B-galactosidase and B-fructosidase
• a-fructosidase
• B-fructosidase and B-mannosidase

Answer 25

B-galactosidase and B-fructosidase

Question 26

2006

Partial hydrolysis of X can give two disaccharides and they are

Answer 26

Question 27

2006

When two carbohydrates are epimers

• they rotate plane-polarized light in the same direction
• they differ in length by one carbon
• one is an aldose, the other a ketose
• one is a pyranose, the other a furanose
• they differ only in the configuration around one carbon atom

Answer 27

they differ only in the configuration around one carbon atom

Question 28

2006

Which of following is an anomeric pair?

• D-gluose and L-glucose
• D-gluose and D-fructose
• a-D-glucose and B-D-glucose
• a-D-glucose and B-L-glucose
• D-glucose and L-fructose

Answer 28

a-D-glucose and B-D-glucose

Question 29

2006

In glycoproteins, the carbohydrate moiety is always attached through the amino acid residues:

• tryptophan, aspartate, or cysteine
• asparagine, serine or threonine
• glycine, alanine, or aspartate
• aspartate or glutamate
• glutamine or arginine

Answer 29

asparagine, serine or threonine

Question 30

2006

Which of the following is a heteropolysaccharide?

• glycogen
• hyaluronate
• starch
• cellulose
• chitin

Answer 30

hyaluronate

Question 31

2009

Why do some amino acids in native proteins fall outside the allowed zones in Ramachandran plots?

• A. Glycine, lacking a side chain, has more conformational freedom than other amino acids and therefore does sometimes fall outside the allowed zones.
• B. Protein structures are based on experimental data, and experimental error leads some residues to have abnormal phi/psi angles.
• C. Although non-covalent interactions in a folded protein are individually weak, there are so many of them that their favorable folding energy can overcome some steric hindrance.
• D. The allowed zones are based on idealized bond lengths, bond angles and atomic sizes. For real amino acids all these can vary a little from the idealized values, leading some residues to fall outside the allowed zones without experiencing steric hindrance.
• E. all of the above.

Answer 31

E. all of the above.

Question 32

2009

Why, in enzyme assays to examine Michaelis-Menten kinetics, does one measures initial velocities (VO)?

• A. in order to avoid the effects on the reaction rate of back reaction of product to form substrate
• B. it is more convenient
• C. measurements at later times contain too much error
• D. because the reaction is fastest at time zero
• E. because one can only achieve a steady state at time zero

Answer 32

A. in order to avoid the effects on the reaction rate of back reaction of product to form substrate

Question 33

2009

The pitch of an alpha helix is 5.4 Å. One turn of a helix has 3.6 residues. What is the rise per residue along the helix axis, i.e. the distance traveled along the axis from a point opposite one alpha carbon to the point opposite the next alpha carbon in the helix?

• A. 5.4 Å
• B. 3.6 Å
• C. 1.5 Å
• D. zero
• E. none of these

Answer 33

C. 1.5 Å

Question 34

2009

The disappearance of the enzyme-substrate complex in Michaelis-Menten kinetics

• A. inactivates the enzyme
• B. is prevented by competitive inhibitors
• C. is a second order process
• D. is a first order process for both product formation and ES dissociation
• E. does not happen, as the complex does not disappear.

Answer 34

D. is a first order process for both product formation and ES dissociation

Question 35

2009

Which of the following would you be least likely to find in a protein interior?

• A. a hydrogen bond from a serine hydroxyl to a valine side chain
• B. a hydrogen bond from a serine hydroxyl to an amide carbonyl on the backbone
• C. interactions between isoleucine and leucine side chains
• D. a metal ion
• E. none of the above

Answer 35

A. a hydrogen bond from a serine hydroxyl to a valine side chain

Question 36

2009

In the steady state assumption in Michaelis-Menten kinetics

• A. the concentrations of substrate and product do not change with time.
• B. S is converted to P at the same rate as P is converted back to S.
• C. the concentration of enzyme-substrate complex remains constant, as it is made at the same rate it is broken down.
• D. product is made at a steady rate throughout the reaction.
• E. the affinity of the enzyme for substrate rises at higher substrate concentrations.

Answer 36

C. the concentration of enzyme-substrate complex remains constant, as it is made at the same rate it is broken down.

Question 37

2009

How many residues are found in beta (tight) turns?

• A. 2
• B. 4
• C. 6
• D. 8
• E. none of these

Answer 37

B. 4

Question 38

2009

The crossovers between parallel beta strands are usually

• A. right handed.
• C. hairpins.
• E. none of these.
• B. left handed.
• D. without hydrogen bonds within the crossover.

Answer 38

A. right handed.

Question 39

2009

Some of the toxic strains of E. coli make two toxic proteins which resembles very closely the toxic protein from Shigella dysenteriae. If these strains are eaten on contaminated food, severe illness and sometimes death results. One of these proteins, Stx2, buries 3212 Å2 of hydrophobic surface when it folds. The overall net free energy change on folding is -15.3 Kcal/mol protein. If the favorable contribution to the folding energy were to come solely from the hydrophobic effect and if the unfavorable contribution were to come only from conformational entropy, what is the free energy contribution of conformational entropy to the folding reaction? (Burial of 1 Å2 of hydrophobic surface yields 25 cal/mol protein.)

• A. 0.025 Kcal/mol
• B. -80.3 Kcal/mol
• C. +15.3 Kcal/mol
• D. +65.0 Kcal/mol
• E. -65.0 Kcal/mol

Answer 39

D. +65.0 Kcal/mol

Question 40

2009

Water molecules are sometimes found deeply buried in the interior of natively folded proteins. When this is so, their immediate environment is likely to involve

• A. side chains of ile, leu and val
• B. several backbone interactions with alpha carbons.
• C. backbone interactions with peptide bond atoms such as the carbonyl, the amide nitrogen and the amide hydrogen.
• D. side chains of ser, tyr and asn
• E. more than one of the above.

Answer 40

E. more than one of the above.

Question 41

2009

The observation that amino acid volumes calculated by the Voronoi method are essentially the same for buried residues in proteins and for crystalline amino acids implies that

• A. the protein interior is loosely packed as is the case for water.
• B. the protein interior is closely packed with little empty space within the folded molecule.
• C. the protein interior is crystalline in its structure.
• D. folded proteins are always more stable than unfolded proteins.
• E. proteins need assistance in order to fold correctly

Answer 41

B. the protein interior is closely packed with little empty space within the folded molecule.

Question 42

2009

In alpha/beta barrel proteins the eight beta strands are nearly always connected by

• A. eight other beta strands
• B. eight alpha helices
• C. non-periodic “worm-like” stretches of polypeptide
• D. four helix bundles
• E. other domains

Answer 42

B. eight alpha helices

Question 43

2009

What is the ionic strength of 0.1 mM trisodium-ATP (Na3ATP)?

• A. 0.1 mM
• B. 0.3 mM
• C. 0.09 mM
• D. 0.39 mM
• E. none of these

Answer 43

E. none of these

Question 44

2009

The number of loops and/or turns needed to connect the helices of a four helix bundle protein is

• A. one
• B. two
• C. three
• D. four
• E. more than four

Answer 44

C. three

Question 45

2009

The activation free energy of an enzyme catalyzed reaction running in the forward direction is

• A. > zero.
• B. zero.
• C. < zero.
• D. the same as G_products – G_reactants
• E. depends on the relative free energies of the substrate and product.

Answer 45

A. > zero.

Question 46

2009

The role of the zinc ion in the zinc finger motif is to

• A. prevent interaction with the DNA when it would not be appropriate
• B. interact with the edges of the DNA bases
• C. interact with the DNA phosphates
• D. stabilize the protein conformation
• E. none of the above

Answer 46

D. stabilize the protein conformation

Question 47

2009

In the lock-and-key model of enzyme catalysis, the active site

• A. is closed to substrate until another type of molecule binds and “unlocks” it.
• B. adapts its shape in order to accommodate the substrate.
• C. fits the substrate very well without significant conformational change.
• D. closes down upon the substrate as a clam closes its shell on being disturbed.
• E. none of the above

Answer 47

C. fits the substrate very well without significant conformational change.

Question 48

2009

Convergent evolution would be expected to yield

• A. proteins of similar three dimensional structure and the same or similar function.
• B. proteins of similar three dimensional structure and unrelated function.
• C. proteins of clearly different three dimensional structure and the same or similar function.
• D. proteins of clearly different three dimensional structure and unrelated function.
• E. none of the above

Answer 48

C. proteins of clearly different three dimensional structure and the same or similar function.

Question 49

2009

In the serine protease reaction mechanism the enzyme’s active site is complementary in shape to

• A. the protein substrate to be cleaved.
• B. the acyl-enzyme complex.
• C. the hydrolysis products. greatest for the transition state.
• D. the transition state.
• E. all of these, the complementarity of shape being

Answer 49

E. all of these, the complementarity of shape being

Question 50

2009

When a protein has multiple domains one can observe that

• A. the domains of some proteins are connected by a flexible linker region and can operate more or less independently.
• B. the domains of some proteins wrap around one another.
• C. the domains of some proteins are tightly associated with one another with extensive interfacial areas.
• D. the domains of some proteins have hinge regions between them which can close down upon ligands.
• E. all of the above

Answer 50

E. all of the above

Question 51

2009

The binding of oxygen to hemoglobin causes one or more groups on the protein to

• A. become less acidic
• B. become more acidic
• C. have a lower pKa
• D. have a higher pKa
• E. B and C

Answer 51

E. B and C

Question 52

What percent of a drug administered to a human being will remain unexcreted after five half-lives? (The answers below have been rounded off to one digit to the right of the decimal point.)

• A. 3.1%
• B. 6.3%
• C. 12.5%
• D. 25.0%
• E. 50%

Answer 52

3.1%

Question 53

2009

The binding of BPG to hemoglobin

• A. shifts the oxygen binding curve toward higher oxygen affinity.
• B. shifts the oxygen binding curve toward lower oxygen affinity.
• C. changes the Hill coefficient from cooperative binding (nH >1) to non-cooperative binding (nH = 1)
• D. lowers the maximum possible oxygen binding.
• E. more than one of the above

Answer 53

B. shifts the oxygen binding curve toward lower oxygen affinity.

Question 54

2009

The units of a first order rate constant are

• A. moles/L
• B. moles/(L-min)
• C. L/(mole-min)
• D. 1 / min
• E. none of these

Answer 54

D. 1 / min

Question 55

2009

When hemoglobin becomes carbamoylated,

• A. it helps the body get rid of carbon dioxide.
• B. becomes unable to bind oxygen.
• C. it unfolds.
• D. it changes the pH of the blood.
• E. none of these is correct

Answer 55

A. it helps the body get rid of carbon dioxide.

Question 56

2009

The half-time of a first order reaction is the time required for

• A. 0.693 (69.3%) of the reactant to be consumed.
• B. football teams to rest between halves of the game.
• C. half the amount of reactant present to be consumed.
• D. the time for the amount of product to decrease by half.
• E. more than one of the above.

Answer 56

C. half the amount of reactant present to be consumed.

Question 57

2009

The metabolic role of an allosteric stimulator is to

• A. prevent an enzyme from being active when the stimulator is bound to the protein.
• B. shift the substrate binding curve to the right on a Vo versus [S] plot.
• C. help ensure that the enzyme is active throughout the normal range of substrate concentration.
• D. prevent the cell from accessing abnormally high or low substrate concentrations.
• E. none of the above

Answer 57

C. help ensure that the enzyme is active throughout the normal range of substrate concentration

Question 58

2009

Why do relatively few amino acids in native proteins fall outside the allowed zones in Ramachandran plots?

• A. Falling outside the allowed zones would cause like charges to be close to one another.
• B. Residues outside the allowed zones cannot make hydrogen bonds with water.
• C. Residues outside the allowed zones usually experience steric hindrance with other residues.
• D. Too many residues outside the allowed zones would make the protein so stable it couldn’t vibrate enough to perform its function.
• E. none of the above

Answer 58

C. Residues outside the allowed zones usually experience steric hindrance with other residues

Question 59

2010

Sea water has approximately the same NaCl concentration as human blood, 0.15 M. If all other ions were to be ignored, the ionic strength of sea water would be

• A. 0.075 M
• B. 0.15 M
• C. 0.3 M
• D. 0.45 M
• E. None of these is correct.

Answer 59

B. 0.15 M

Question 60

2010

4. What would be the ionic strength of 0.15 M calcium chloride?

• A. 0.075 M
• B. 0.15 M
• C. 0.3 M
• D. 0.45 M
• E. None of these is correct.

Answer 60

D. 0.45 M

Question 61

2010

It is observed that approximately one healthy human being in 600 has a single amino acid mutation in the hemoglobin. There being approximately 3000 students at each of Douglass College and SEBS, around ten of these students are mutants. (You could be one of them!) This means that

• A. mutations are good for your health.
• B. the mutations probably occur in the protein interior.
• C. the mutations disrupt the folding of the protein.
• D. the mutations are probably on the protein surface and do not affect the hemoglobin’s function.
• E. none of the above

Answer 61

D. the mutations are probably on the protein surface and do not affect the hemoglobin’s function.

Question 62

2010

If disulfide bonds of a protein were to be reduced, for example, by mercaptoethanol, and the conformation were to remain native, the unfolding temperature of the protein would probably

• A. decrease
• B. increase
• C. remain unchanged
• D. increase or decrease depending on how many SS bonds were reduced
• E. none of the above

Answer 62

A. decrease

Question 63

2010

In an alpha-beta barrel structure the beta strands are

• A. parallel
• B. alternating antiparallel
• C. randomly mixed between parallel and antiparallel
• D. perpendicular to the barrel axis E. none of the above

Answer 63

A. parallel

Question 64

2010

The strands in an alpha-beta barrel protein such as triose phosphate isomerase are connected by

• A. other beta strands
• B. flexible hinge regions
• C. metal binding sites
• D. alpha helices with some loops
• E. none of the above

Answer 64

D. alpha helices with some loops

Question 65

2010

Which of the following would NOT be found in a folded native protein?

• A. a side chain to side chain hydrogen bond between a serine hydroxyl and an asparagine amide
• B. a backbone to side chain hydrogen bond between the amide part of a glutamine side chain and a peptide NH group
• C. a backbone to backbone hydrogen bond between a peptide NH and a different peptide CO
• D. a side chain to side chain bond between a serine hydroxyl and a valine methyl group
• E. all of the above would be found in folded native proteins.

Answer 65

D. a side chain to side chain bond between a serine hydroxyl and a valine methyl group

Question 66

2010

When a run of polypeptide enters a tight turn of either type I or type II and then leaves the turn, its average direction changes

• A. not at all.
• B. by approximately 90 degrees.
• C. by approximately 45 degrees.
• D. by approximately 180 degrees.
• E. by approximately 360 degrees.

Answer 66

D. by approximately 180 degrees.

Question 67

2010

Phosphoglycerate kinase is a key enzyme in glycolysis. When phosphoglycerate kinase binds its substrate, phophoglyceric acid, the enzyme’s two domains close down like a clam, completely shielding the substrate from contact with the solvent. The water which solvated the active site and the substrate is expelled from the biomolcular surface into the bulk solution. The free energy of contribution of this water to the binding is

• A. favorable.
• B. unfavorable.
• C. dependent on how many water molecules are involved.
• D. primarily a contribution to the binding enthalpy, ∆H.
• E. none of the above

Answer 67

A. favorable

Question 68

2010

The number of loops and/or turns needed to connect the helices of a four helix bundle protein is

• A. one
• B. two
• C. three
• D. four
• E. more than four

Answer 68

C. three

Question 69

2010

In the steady state assumption which underlies Michaelis-Menten kinetics

• A. ES is formed at the same rate at which it is broken down.
• B. the reaction rate does not change as substrate concentration rises.
• C. S is converted to P at the same rate as P is converted to S.
• D. more than one of the above
• E. none of the above

Answer 69

A. ES is formed at the same rate at which it is broken down

Question 70

2010

Which of the following best describes an equilateral triangle?

• A. It has a three-fold axis of rotational symmetry.
• B. It has a two-fold axis of rotational symmetry.
• C. It has three two-fold axes of rotational symmetry.
• D. It has three two-fold axes of rotational symmetry and one three-fold axis of rotational symmetry.
• E. cannot tell from data given

Answer 70

D. It has three two-fold axes of rotational symmetry and one three-fold axis of rotational symmetry.

Question 71

2010

In first order reactions

• A. the rate of the reaction depends on the square of the reactant concentration.
• B. the rate of the reaction is independent of the current reactant concentration.
• C. the rate of the reaction increases as the reactant concentration decreases.
• D. the rate of the reaction decreases as the reactant concentration decreases.
• E. none of the above

Answer 71

D. the rate of the reaction decreases as the reactant concentration decreases.

Question 72

2010

A chemical pollutant of the environment is observed to disappear such that the amount in the sediment of Newark Bay drops from 100 mg per gram of dried sediment to 12.5 mg per gram dry sediment in nine years. What is the half-time for its first order decay?

• A. three years
• B. six years
• C. nine years
• D. eighteen years
• E. none of these

Answer 72

A. three years

Question 73

2010

In catalysis by a normal Michaelis-Menten enzyme, when the substrate concentration is much greater than the enzyme concentration, doubling the substrate concentration will have what effect on the reaction rate?

• A. little or none.
• B. it will approximately double the rate
• C. it will slow the rate by a factor of approximately two
• D. it will inhibit the enzyme
• E. none of the above

Answer 73

A. little or none.

Question 74

2010

When human and sperm whale myoglobins are compared, it is found that of the 153 residues in each protein, 128 (84%) are identical. If the 16 conservative substitutions are included, the proteins are 94% homologous. These data suggest that

• A. humans are descended from whales.
• B. whales are descended from humans.
• C. whales and humans are descended from a common ancestor.
• D. the proteins are examples of convergent evolution.
• E. this is a whale of a problem….

Answer 74

C. whales and humans are descended from a common ancestor.

Question 75

2010

Rotations about the peptide C-N bond in proteins (ω rotations) are severely limited because

• A. steric hindrance would cause atoms to interpenetrate.
• B. the C-N σ (sigma) electrons are stiff.
• C. π (pi) electrons give the bond partial double bond character.
• D. hydrogen bonding restricts mobility about the bond.
• E. the definition of the zero point for rotation about the ω bond prevents much rotation.

Answer 75

C. π (pi) electrons give the bond partial double bond character.

Question 76

2010

Which of the following amino acid pairs could form a salt bridge such as might help hold protein subunits together in a multisubunit protein?

• A. asp and glu
• B. ala and phe
• C. gln and lys
• D. lys and asp
• E. more than one of these

Answer 76

D. lys and asp

Question 77

2010

Cations such as Ca++, Mn++, Zn++, and a number of others are frequently found in globular proteins as cofactors. In some cases they participate in catalysis, while in others they serve a structural role, helping to hold parts of the polypeptide chain in the native conformation. Which of the following pairs of side chains would be most likely to coordinate the cation?

• A. asp and glu
• B. ala and phe
• C. gln and lys
• D. lys and asp
• E. more than one of these

Answer 77

A. asp and glu

Question 78

2010

The steepness of the denaturation curves of native globular proteins indicates that

• A. their conformations are extremely stable.
• B. their conformations are extremely unstable.
• C. the interactions which maintain the native conformation give way nearly all at once as the denaturing stress increases.
• D. α-helices are more stable than β-sheets.
• E. more than one of the above.

Answer 78

C. the interactions which maintain the native conformation give way nearly all at once as the denaturing stress increases.

Question 79

2011

Which of the following best describes a square?

• A. It has a four-fold axis of rotational symmetry.
• B. It has four two-fold axes of rotational symmetry and one four-fold axis of rotational symmetry.
• C. It has four two-fold axes of rotational symmetry.
• D. It has a two-fold axis of rotational symmetry.
• E. cannot tell from data given (This question is adapted from last year’s exam.)

Answer 79

B. It has four two-fold axes of rotational symmetry and one four-fold axis of rotational symmetry.

Question 80

2011

In first order reactions

• A. the rate of the reaction decreases as the reactant concentration decreases.
• B. the rate of the reaction is independent of the current reactant concentration.
• C. the rate of the reaction increases as the reactant concentration decreases.
• D. the rate of the reaction depends on the square of the reactant concentration.
• E. none of the above (This question is from last year’s exam.)

Answer 80

A. the rate of the reaction decreases as the reactant concentration decreases.

Question 81

2011

In first order reactions

• A. the half time of the reaction decreases as the reactant concentration decreases.
• B. the half time of the reaction is independent of the current reactant concentration.
• C. the half time of the reaction increases as the reactant concentration decreases.
• D. the half time of the reaction depends on the square of the reactant concentration.
• E. none of the above

Answer 81

B. the half time of the reaction is independent of the current reactant concentration.

Question 82

2011

In the Michaelis-Menten formulation of enzyme kinetics, why do we neglect the back reaction of enzyme and product to form the enzyme-substrate complex?

• A. Back reactions do not occur in enzyme catalyzed reactions.
• B. After releasing the product, the enzyme becomes inactive.
• C. By using the velocity at the start of the reaction, the backward rate is small compared to the forward rate.
• D. By waiting until the reaction reaches equilibrium, the backward rate equals the forward rate, so we can use the forward rate instead.
• E. None of the above

Answer 82

C. By using the velocity at the start of the reaction, the backward rate is small compared to the forward rate.

Question 83

2011

Under steady state conditions in Michaelis-Menten kinetics

• A. the total enzyme concentration is maintained in a constant, steady ratio to the substrate concentration.
• B. substrate and product concentrations do not change with time.
• C. the enzyme-substrate complex breaks down as fast as it is formed.
• D. the system is at equilibrium
• E. more than one of the above.

Answer 83

C. the enzyme-substrate complex breaks down as fast as it is formed.

Question 84

2011

Which of the following is the correct unit for K_m?

• A. M^-1
• B. moles / (liter-minute)
• C. moles / liter
• D. mg / mL E. liters / mole

Answer 84

C. moles / liter

Question 85

2011

Which of the following is the correct unit for k_cat?

• A. moles enzyme / mol S
• B. L / (mol-deg)
• C. min^-1
• D. min / mol P
• E. none of these.

Answer 85

C. min^-1

Question 86

2011

At Vmax / 2,

• A. Km = kcat.
• B. kcat is at its maximum
• C. the enzyme is saturated
• D. the enzyme is inhibited
• E. none of these is correct

Answer 86

E. none of these is correct

Question 87

2011

The Michaelis constant

• A. can be estimated as the substrate concentration at which the reaction velocity is half the velocity at saturation.
• B. is an estimate of the dissociation constant of the enzyme-substrate complex.
• C. can be used to estimate the affinity of the enzyme for the substrate.
• D. is an estimate of the substrate concentration at which half the enzyme molecules have substrate bound and half do not.
• E. All of the above are true.

Answer 87

E. All of the above are true.

Question 89

2011

The specificity constants SCA and SCB of two proteins, A and B, are compared. The proteins come from different species and catalyze the same reaction. They are similar in sequence and structure. SCA > SCB. This implies that

• A. kcat A > kcat
• B . B. KmA > KmB .
• C. kcat A < kcat B .
• D. KmA < KmB .
• E. cannot tell from data given

Answer 89

E. cannot tell from data given

Question 90

2011

In the Lineweaver-Burk (double reciprocal) plot

• A. one can obtain Km from the 1/V axis intercept.
• B. one can obtain Km from the 1/[S] axis intercept
• C. one can obtain Vmax from the 1/V axis intercept
• D. one can obtain Vmax from the 1/[S] axis intercept
• E. more than one of the above

Answer 90

E. more than one of the above

Question 91

2011

If one has the values of Km and kcat for a Michaelis-Menten enzyme, one could calculate the reaction velocity at any substrate concentration provided that one also has

• A. the concentration of free enzyme.
• B. the concentration of enzyme with substrate in its active site.
• C. the free energy of the substrate to product reaction.
• D. the total concentration of enzyme.
• E. none of the above

Answer 91

D. the total concentration of enzyme.

Question 92

2011

In a study of an enzyme inhibitor the enzyme is assayed as a function of substrate concentration in the absence of inhibitor and then in the presence of several different inhibitor concentrations. It is found that the Lineweaver-Burk plots all intersect the 1/[S] axis at a single point. This implies that

• A. the inhibitor can only bind if substrate is already bound.
• B. the inhibitor and substrate can bind in either order (S first and then I or I first and then S).
• C. the substrate and inhibitor bind to the same site.
• D. more than one of the above
• E. cannot tell from data given

Answer 92

B. the inhibitor and substrate can bind in either order (S first and then I or I first and then S).

Question 93

2011

In competitive inhibition

• A. the apparent affinity of the enzyme for the substrate is increased.
• B. the maximal velocity seen in the absence of inhibitor can be attained at very high substrate concentrations even though the inhibitor is present.
• C. the maximal velocity seen in the absence of substrate can never be attained in the presence of the inhibitor even at very high substrate concentrations.
• D. Lineweaver-Burk plots show parallel lines.
• E. none of the above

Answer 93

B. the maximal velocity seen in the absence of inhibitor can be attained at very high substrate concentrations even though the inhibitor is present.

Question 94

2011

Why is rotation about the peptide bond in proteins restricted?

• A. The peptide unit would no longer be planar.
• B. Rotation would cause steric hindrance with other atoms of the protein.
• C. The partial double bond character of the bond raises the energy barrier to rotation.
• D. Rotations about the backbone single bonds are sufficient to specify the three dimensional conformation of the protein.
• E. none of the above.

Answer 94

C. The partial double bond character of the bond raises the energy barrier to rotation.

Question 95

2011

When horse liver alcohol dehydrogenase binds NAD+ or NADH, the enzyme’s two domains close down like a clam, completely shielding the substrate from contact with the solvent. The water which solvated the active site and the substrate is expelled from the biomolcular surface into the bulk solution. The free energy of contribution of this water to the binding is

• A. unfavorable.
• B. favorable.
• C. dependent on how many water molecules are involved.
• D. primarily a contribution to the binding enthalpy, ∆H.
• E. none of the above (Modified from last year’s exam)

Answer 95

B. favorable.

Question 96

2011

Heavy atoms in proteins are defined as all non-hydrogen atoms, i.e. all C, N, O, S and any other nonhydrogens covalently attached to the protein. Consider a pentapeptide. Would rotation about phi of residue 4 change the distance between any atoms of residue 2 and residue 3?

• A. Yes.
• B. No.
• C. It would depend on the values of other phi and psi.
• D. Yes provided that residue 4 is proline
• E. cannot tell from data given.

Answer 96

B. No.

Question 97

2011

Two protein sequences have been aligned and the amino acid differences between corresponding positions are examined. Which of the following substitutions would be the least conservative?

• A. D for E
• B. E for N
• C. K for Y
• D. I for L
• E. N for Q

Answer 97

C. K for Y

Question 98

2011

Which of the following represent right-handed crossovers of β-strands?

• A. (d) and (b)
• B. (a) and (e)
• C. (c) and (e)
• D. (c) and (b)
• E. more than one of these

Answer 98

C. (c) and (e)

Question 99

2011

The molar extinction coefficient of lysozyme at 280 nm is 39,800 M-1 cm-1 . What is the lysozyme concentration in a solution whose absorbance in a 1 cm cell at this wavelength is 0.796?

• A. 2 x 10^-5 M
• B. 2 mg/mL
• C. 5 x 104 M
• D. 1 mM
• E. none of the above.

Answer 99

A. 2 x 10^-5 M

Question 100

2011

Zinc fingers are stabilized primarily by

• A. helical hydrogen bonds.
• B. disulfides.
• C. a combination of side chain coordination of the zinc and hydrophobic interactions among a set of hydrophobic side chains
• D. interactions with the DNA itself
• E. none of these.

Answer 100

C. a combination of side chain coordination of the zinc and hydrophobic interactions among a set of hydrophobic side chains

Question 101

2011

The characteristic structure of an alpha/beta barrel protein such as triose phosphate isomerase has

• A. eight antiparallel beta strands connected by (usually) eight alpha helices.
• B. eight alpha helices surrounded and connected by parallel beta strands.
• C. eight parallel beta strands surrounded and connected by (usually) eight alpha helices.
• D. four alpha helices connected by loops.
• E. none of these.

Answer 101

C. eight parallel beta strands surrounded and connected by (usually) eight alpha helices.

Question 102

2011

When ribonuclease was unfolded and its four disulfide bonds broken in concentrated urea containing mercaptoethanol, the protein returned to a fully active conformation after the urea and mercaptoethanol were diluted out. This observation by Anfinsen’s group proved that

• A. proteins require auxiliary proteins called chaperones in order to fold to the native conformation.
• B. the protein sequence is determined by the DNA sequence.
• C. disulfide bonds control protein stability.
• D. the information needed to determine the three dimensional structure is contained in the amino acid sequence.
• E. none of the above.

Answer 102

D. the information needed to determine the three dimensional structure is contained in the amino acid sequence.

Question 103

2011

Which of the following would serve as a measure of the stability of a globular protein?

• A. the temperature at which half the molecules are in the native conformation and half are unfolded.
• B. the urea concentration at which the N and U forms are each 50% of the total protein present.
• C. the guanidine hydrochloride concentration at which the N and U forms are each 50% of the total protein present.
• D. the pH at which the N and U forms are each 50% of the total protein present.
• E. all these would be measures of the stability

Answer 103

E. all these would be measures of the stability

Question 104

2011

Amyloid is

• A. a highly unstable protein conformation which reverts readily to the native form.
• B. an insoluble aggregated form of proteins whose conformation involves many parallel beta sheets wrapped into large multimolecular helices.
• C. a protein form which is readily cut up by cellular “housekeeping” proteases to prevent cellular damage.
• D. a highly stable protein conformation which can arise when the native form is slightly destabilized by a conformational stress or when the native form partly unfolds spontaneously.
• E. more than one of the above.

Answer 104

E. more than one of the above.

Question 105

2013

An experimenter believes that an inhibitor of a Michaelis-Menten enzyme binds with equal affinity both to free enzyme and also to the ES complex. She would expect that

• A. Lineweaver-Burk plots of the data would meet at a point on the 1/[S] axis.
• B. Lineweaver-Burk plots of the data would have parallel lines.
• C. Lineweaver-Burk plots of the data would meet at a point on the 1/V axis.
• D. more than one of the above
• E. none of the above

Answer 105

A. Lineweaver-Burk plots of the data would meet at a point on the 1/[S] axis.

Question 106

2014

Which of the following represent left-handed crossovers of β-strands?

• A. (d) and (b)
• B. (a) and (b)
• C. (c) and (e)
• D. (c) and (b)
• E. none of these combinations

Answer 106

(a) and (b)

Question 107

2014

The molar extinction coefficient of lysozyme at 280 nm is 39,800 M^-1 cm^-1. What is the lysozyme concentration in a solution whose absorbance in a 1 cm cell at this wavelength is 0.199?

• A. 2 x 10^-5 M
• B. 2 mg/mL
• C. 5 x 10^-6 M
• D. 1 mM
• E. none of the above. (This question is adapted from the posted exam.)

Answer 107

C. 5 x 10^-6 M

Question 108

2014

Under steady state conditions in Michaelis-Menten kinetics

• A. the system is unstable
• B. the reaction is at equilibrium because the concentration of the enzyme-substrate complex does not change.
• C. the reaction is running at its maximum velocity,
• D. the reaction velocity does not depend on the substrate concentration.
• E. none of the above is correct.

Answer 108

E. none of the above is correct.

Question 109

2014

The protein coats of many viruses pack to give various forms of symmetry. One of these forms is a regular pentagon. Which of the following best describes the symmetry of these pentagons?

• A. It has five two-fold axes of rotational symmetry.
• B. It has a five-fold axis of rotational symmetry.
• C. It has five two-fold axes of rotational symmetry and one five-fold axis.
• D. It has five five-fold axes of rotational symmetry.
• E. cannot tell from data given

Answer 109

C. It has five two-fold axes of rotational symmetry and one five-fold axis.

Question 110

2014

Which of the following would most likely give the free energy of substrate binding to the active site of an enzyme?

• A. ∆G^o’ = -RT ln (Km)
• Β. ∆G^o’ = -RT ln (1/Km)
• C. ∆G^o’ = -RT ln (Vmax)
• D. ∆G^o’ = -RT ln (1/Vmax)
• E. none of the above

Answer 110

Β. ∆G^o’ = -RT ln (1/Km)

Question 111

2014

Which of the following best describes kcat?

• A. It is the equilibrium constant for an enzyme catalyzed reaction.
• B. It is the rate at which the ES complex breaks down to free enzyme and substrate.
• C. It catches a kmouse.
• D. It is the rate constant for the conversion of the ES complex to product and free enzyme.
• E. more than one of the above

Answer 111

D. It is the rate constant for the conversion of the ES complex to product and free enzyme.

Question 112

2014

In a study of an enzyme inhibitor the enzyme is assayed as a function of substrate concentration in the absence of inhibitor and then in the presence of several different inhibitor concentrations. It is found that the Lineweaver-Burk plots display parallel lines. This implies that

• A. the inhibitor can only bind if substrate is already bound.
• B. the inhibitor and substrate can bind in either order (S first and then I or I first and then S).
• C. the substrate and inhibitor bind to the same site.
• D. more than one of the above
• E. cannot tell from data given.

Answer 112

A. the inhibitor can only bind if substrate is already bound.

Question 113

2014

In uncompetitive inhibition

• A. the apparent affinity of the enzyme for the substrate is increased.
• B. the maximal velocity seen in the absence of inhibitor can be attained at very high substrate concentrations even though the inhibitor is present.
• C. the maximal velocity seen in the absence of inhibitor can never be attained in the presence of the inhibitor even at very high substrate concentrations.
• D. Lineweaver-Burk plots show parallel lines.
• E. more than one of the above

Answer 113

E. more than one of the above

Question 114

2014

Why do some amino acids in native proteins fall outside the allowed zones in Ramachandran plots?

• A. Glycine, lacking a side chain, has more conformational freedom than other amino acids and therefore does sometimes fall outside the allowed zones.
• B. Protein structures are based on experimental data, and experimental error leads some residues to have abnormal phi/psi angles.
• C. Although non-covalent interactions in a folded protein are individually weak, there are so many of them that their favorable folding energy can overcome some steric hindrance.
• D. The allowed zones are based on idealized bond lengths, bond angles and atomic sizes. For real amino acids all these can vary a little from the idealized values, leading some residues to fall outside the allowed zones without experiencing steric hindrance.
• E. all of the above.

Answer 114

E. all of the above.

Question 115

2014

In the zinc finger motif the zinc ion

• A. interacts with the DNA to stabilize binding of the protein to the DNA.
• B. stabilizes the motif’s conformation so that parts of the polypeptide are able to interact with the DNA.
• C. prevents binding of the protein to the DNA, so that binding occurs only when the zinc dissociates.
• D. is coordinated (bound) to four cysteine side chains.
• E. is coordinated (bound) to four histidine side chains.

Answer 115

B. stabilizes the motif’s conformation so that parts of the polypeptide are able to interact with the DNA.

Question 116

2014

Which of the following has the greatest conformational entropy?

• A. a native globular protein that has been crystallized
• B. a native globular protein in solution
• C. a globular protein which is unfolded in solution
• D. a globular protein which has been precipitated from solution as an amorphous solid
• E. all these have the same conformational entropy

Answer 116

C. a globular protein which is unfolded in solution

Question 117

2014

If the ∆S of the U to N transition of a protein in aqueous solution is negative, then

• A. entropy contributes favorably to the free energy of folding.
• B. folding will not occur.
• C. the enthalpy of folding will be positive.
• D. the protein cannot contain disulfide bonds.
• E. folding, if it occurs, i.e. if ∆G < 0, will be enthalpy driven.

Answer 117

E. folding, if it occurs, i.e. if ∆G < 0, will be enthalpy driven.

Question 118

2014

The left handed helices of collagen have 3.3 residues per turn and a rise per residue of 1.9 Å, while α helices have 3.6 residues per turn and a rise per residue of 1.5 Å. Which would be longer, a collagen helix having 100 residues or an α helix of 100 residues?

• A. the collagen
• B. the α helix
• C. they would have equal length
• D. cannot tell from data given.

Answer 118

A. the collagen

Question 119

2014

Voronoi polyhedra were used by Richards to calculate

• A. the solvent accessible surface area of proteins.
• B. the contact surface area between proteins and the solvent.
• C. the hydrophobicity of the protein interior.
• D. the packing volume of the protein interior.
• E. more than one of the above.

Answer 119

D. the packing volume of the protein interior.

Question 120

2014

When ribonuclease A is heated at neutral pH, the temperature at which it unfolds is approximately 75C. When it is heated at pH 2.1, it unfolds at approximately 30C. One can conclude that

• A. acidic pH stabilizes the protein.
• B. acidic pH destabilizes the protein.
• C. acid has nothing to do with the protein’s stability.
• D. basic pH stabilizes the protein.
• E. basic pH destabilizes the protein.

Answer 120

B. acidic pH destabilizes the protein.

Question 121

2014

The free energy of folding cytochrome c is –44 KJ/mol, while the enthalpy of folding it is –52 KJ/mol. The entropy of folding is thus

• A. favorable.
• B. unfavorable.
• C. approximately zero.
• D. cannot tell from the data given

Answer 121

B. unfavorable.

Question 122

2014

Mutants of a protein are found which differ from one another in that many different amino acids can occur at a particular position along the polypeptide chain without affecting the activity of the protein. It is most probable that

• A. this position begins an α helix.
• B. this position ends an α helix.
• C. this position is buried in the protein interior. the surface as in the interior.
• D. this position is on the protein surface.
• E. this position is as likely to be on

Answer 122

D. this position is on the protein surface.

Question 124

2014

Which of the following kinds of hydrogen bonds cannot occur in a protein?

• A. backbone to backbone
• B. backbone to side chain
• C. side chain to side chain
• D. amino group to carboxylate group
• E. peptide backbone carbonyl to proline nitrogen

Answer 124

E. peptide backbone carbonyl to proline nitrogen

Question 125

2014

The assumptions at the core of Michaelis-Menten kinetics are

• A. that enzymes have specific sites to which substrates bind and which can be saturated by a large excess of substrate.
• B. that the rate at which this “ES” complex is made is exactly balanced by the rate at which it is broken down once the reaction gets under way.
• C. that Vmax = kcat x [Et], where [Et] is the total enzyme concentration.
• D. that catalyzed reactions proceed faster than uncatalyzed reactions
• E. A and B

Answer 125

E. A and B

Question 126

2014

What protein folding motif was represented by the carved Halloween pumpkin shown in the class period before this examination?

• A. an α/β barrel
• B. a four helix bundle
• C. a β barrel
• D. an extended β sheet as is found in silk
• E. none of these

Answer 126

C. a β barrel

Question 127

2015

The resistance to stretching of silk is the result of

• A. stiffness in the coiled coil conformation of its poly-proline-like left handed helices whose stretching would bend or stretch covalent bonds.
• B. the difficulty of bending or stretching covalent bonds.
• C. the layered packing of β-sheets whose side chains move past one another easily as backbone bonds rotate.
• D. the difficulty of inserting water molecules into the hole in the center of its π helices.
• E. more than one of the above

Answer 127

B. the difficulty of bending or stretching covalent bonds.

Question 128

2015

The soft flexibility of silk is the result of

• A. stiffness in the coiled coil conformation of its poly-proline-like left handed helices whose stretching would bend or stretch covalent bonds.
• B. the difficulty of bending or stretching covalent bonds.
• C. the layered packing of β-sheets whose side chains move past one another easily as backbone bonds rotate.
• D. the difficulty of inserting water molecules into the hole in the center of its π helices.
• E. more than one of the above.

Answer 128

C. the layered packing of β-sheets whose side chains move past one another easily as backbone bonds rotate.

Question 129

2015

Deviations of the φ,ϕ angles determined experimentally by x-ray crystallography from the “allowed zones” in the Ramachandran plot may be due to

• A. small amounts of bond stretching and/or bending in real proteins as opposed to the models used to construct the plot.
• B. experimental error in the structure determination.
• C. intervention of the vital force.
• D. shrinkage of atoms in the protein when amino acids are joined to make it.
• E. more than one of the above.

Answer 129

E. more than one of the above.

Question 130

2015

In the α,β barrel motif often seen in proteins, the β-strands are

• A. all antiparallel
• B. all parallel
• C. alternating parallel and antiparallel
• D. required to be identical in length
• E. none of the above

Answer 130

B. all parallel

Question 131

2015

The large area of the Ramachandran plot which is “forbidden” exists because

• A. steric hindrance would cause atoms to interpenetrate.
• B. the C-N σ (sigma) electrons are stiff.
• C. π (pi) electrons give the bond partial double bond character.
• D. hydrogen bonding restricts mobility about the bond.
• E. the definition of the zero point for rotation about the Φ and ψ bonds prevents much rotation

Answer 131

A. steric hindrance would cause atoms to interpenetrate.

Question 132

2015

The atoms which are contained within the peptide plane are

• A. Cα, the peptide C=O, the backbone N, and the amide nitrogen’s proton.
• B. Cα, the hydrogen bonded to the Cα, the peptide C=O, and the backbone N.
• C. Cβ, the peptide C=O, the backbone N, and the amide nitrogen’s proton.
• D. the proline ring, the hydrogen bonded to the Cα, and the amide of asparagine side chains.
• E. all of the above.

Answer 132

A. Cα, the peptide C=O, the backbone N, and the amide nitrogen’s proton.

Question 133

2015

Aspartic carbamoyl transferase, an allosteric enzyme we will study next semester, contains six identical catalytic subunits arranged in a hexagon in the native form. This structure has

• A. six six-fold axes of rotation and one two-fold axis.
• B. several three-fold axes of rotation.
• C. six two-fold axes of rotation and one six-fold axis.
• D. one six-fold axis of rotation.
• E. cannot tell from data given.

Answer 133

C. six two-fold axes of rotation and one six-fold axis.

Question 134

2015

The steepness of the denaturation curves of native globular proteins indicates that

• A. their conformations are extremely stable.
• B. their conformations are extremely unstable.
• C. the interactions which maintain the native conformation give way nearly all at once as the denaturing stress increases.
• D. α-helices are more stable than β-sheets.
• E. more than one of the above.

Answer 134

C. the interactions which maintain the native conformation give way nearly all at once as the denaturing stress increases.

Question 135

2015

Avery, MacLeod and McCarthy incubated DNA from virulent pneumococci with live non-virulent pneumococci and injected the culture into mice, which died of pneumonia. Their conclusion that the DNA carried the genetic information for virulence was criticized because their DNA preparation might have contained trace amounts of protein, which could have carried the information. How did they show this was not the case? (from the posted exam)

• A. Their DNA showed only one band on a gel.
• B. Treatment of the preparation with proteases destroyed the transforming activity of the preparation but treatment with nucleases had no effect.
• C. Injection of pneumococcal proteins into the mice had no effect on them.
• D. Nuclease treatment of the preparation destroyed the transforming activity, but protease treatment did not destroy it.
• E. none of the above (from a posted exam)

Answer 135

D. Nuclease treatment of the preparation destroyed the transforming activity, but protease treatment did not destroy it.

Question 136

2015

The genome of adenovirus AD-2 has been sequenced and found to be a linear double stranded DNA containing 35,932 base pairs. In the B form, DNA has a rise per residue of 3.4 Å. How long would the viral DNA be if it were in the B form? 1 Å = 10-8 cm = 10-10 m. The Greek letter µ means “micro.” One micrometer = 10-6 m

• A. 35,932 Å
• B. 34 Å
• C. 108 µm
• D. 12.2 µm
• E. cannot tell from data given.

Answer 136

D. 12.2 µm

Question 137

2015

UV light is well known to produce mutations in bacteria and cancer in human beings. One of the ways it does these is to

• A. destroy nucleic acid bases by opening the aromatic rings.
• B. rupture the ribose-phosphate backbone.
• C. change one base into another.
• D. drive the formation of thymine dimers in places where two thymines are adjacent to one another.
• E. none of the above.

Answer 137

D. drive the formation of thymine dimers in places where two thymines are adjacent to one another.

Question 138

2015

When DNA is hydridized, the hybrid DNA must contain

• A. a strand from one species and a strand from another species.
• B. thymine dimers.
• C. a high GC content.
• D. a high AT content.
• E. none of the above.

Answer 138

A. a strand from one species and a strand from another species.

Question 139

2015

How many molecules of water would be removed to assemble a single molecule of ATP from adenine, ribose and phosphates?

• A. none
• B. two
• C. three
• D. four
• E. more than four

Answer 139

D. four

Question 140

2015

In the Bohr effect of hemoglobin one observes that the binding of oxygen to hemoglobin in the lungs causes the protein to give up protons to the surrounding solution. This observation requires that

• A. the pH of the strongly buffered solution will drop sharply.
• B. the pH of the strongly buffered solution will rise sharply.
• C. oxygen binding causes the pKa of one or more side chains of the hemoglobin to decrease.
• D. oxygen binding causes the pKa of one or more side chains of the hemoglobin to increase.
• E. none of the above

Answer 140

C. oxygen binding causes the pKa of one or more side chains of the hemoglobin to decrease.

Question 141

2015

Addition of a toxic compound to a solution containing an allosteric enzyme shifts the sigmoid curve of activity versus substrate concentration to the left on the substrate axis. In the absence of other information, it is plausible that the mechanism of toxicity involves

• A. a decrease in activity of the enzyme caused by binding of the compound.
• B. an increase in activity of the enzyme caused by binding of the compound.
• C. irreversible inhibition of the enzyme by the compound.
• D. competitive inhibition yielding linear Lineweaver-Burk plots and a Michaelis-Menten vo versus [S] curve.
• E. more than one of the above.

Answer 141

B. an increase in activity of the enzyme caused by binding of the compound.

Question 142

2015

What is the correct order of the following steps in the solid state synthesis of oligonucleotides?

1. reaction of a nucleotide which is activated at the 3’ position and blocked at 5’ and at the base
2. deblocking of the protected 5’ position
3. attachment of a nucleotide blocked at the 5’ position and at the base to the solid support
4. oxidation at the phosphorous atom by iodine
5. cleavage of the chain from the support
6. removal of blocking groups from the bases
7. removal of methyl groups from the phosphates `

• A. 3, (4, 1, 2) repeatedly 6, 7, 5
• B. 1, 2, (3, 4, 5) repeatedly 6, 7
• C. 1, 3, (2, 4, 7) repeatedly 6, 5
• D. 3, (2, 1, 4)repeatedly 6, 7, 5
• E. none of the above

Answer 142

D. 3, (2, 1, 4)repeatedly 6, 7, 5

Question 143

2015

The absence of which hydroxyl group(s) on the ribose in Sanger’s dideoxy nucleotide sequencing method causes termination of polymer chain elongation?

• A. 2’
• B. 3’
• C. 5’
• D. 2’ and 5’
• E. 3’ and 5’

Answer 143

B. 3’

Question 144

2015

Which of the following strands contains self-complementary regions which could therefore form double stranded hairpin loops? (The region which “turns the corner” need not be base-paired.)

• A. TTAGCACAAAAGTGCTAA
• B. TTAGCACCACGATT
• C. TTTTTTCCCCAAAAAA
• D. More than one of the above
• E. None of the above

Answer 144

D. More than one of the above

Question 145

2018

What structural form was represented on the pumpkin shown in the last class before this exam?

• A. a disordered structure
• B. an antiparallel beta pleated sheet
• C. a parallel beta pleated sheet
• D. a helix
• E. more than one of the above

Answer 145

B. an antiparallel beta pleated sheet

Question 146

2018

After four half-lives of a first order reaction have passed, the percent of the initial material left is

• A. 6.25%
• B. 50%
• C. 12.5%
• D. 25%
• E. None of these

Answer 146

A. 6.25%

Question 147

2018

Why is rotation about the peptide bond (the omega bond) more difficult than rotation about phi and psi?

• A. Steric hindrance interferes with omega rotation.
• B. π electrons facilitate phi and psi rotation.
• C. π electrons inhibit rotation about the peptide bond.
• D. A and C E. none of the above (from last year’s exam)

Answer 147

C. π electrons inhibit rotation about the peptide bond.

Question 148

2018

Which of the following cannot be a hydrogen bond in a protein?

• A. tyrosine hydroxyl to serine hydroxyl
• B. threonine hydroxyl to a backbone carbonyl
• C. a buried water molecule to a side chain and to a backbone atom
• D. an amino nitrogen of a peptide bond to the carbonyl of another peptide bond
• E. all of the above can be hydrogen bonds in proteins.

Answer 148

E. all of the above can be hydrogen bonds in proteins.

Question 149

2018

Injection with which of the following would lead mice used as test animals to die of pneumonia?

• A. virulent bacteria which had been killed by heat
• B. a mixture of live non-virulent bacteria and heat-killed virulent bacteria
• C. live non-virulant bacteria
• D. a mixture of DNA from heat-killed virulent bacteria and live non-virulent bacteria to which DNAase had been added prior to injection.
• E. more than one of the above

Answer 149

B. a mixture of live non-virulent bacteria and heat-killed virulent bacteria

Question 150

2018

The concentration of hemoglobin is determined in the laboratory by first oxidizing the iron atoms from the +2 to the +3 state by adding a few crystals of potassium ferricyanide to the solution. Hemoglobin in the +3 state is called methemoglobin. After a minute or two a few crystals of potassium cyanide are added. The cyanide ions bind to the iron, yielding cyanmethemoglobin, which has an absorption peak at 540 nm. The absorbance at this wavelength of a 1 mM solution of cyanmethemoglobin is 11.0. What is the concentration of a methemoglobin solution whose absorbance at 540 nm in a 1 mM cell is 0.785?

• A. 0.627 mM
• B. 0.714 M
• C. 0.0714 mM
• D. 0.714 mM
• E. none of the above (from the Beer-Lambert Law problem set)

Answer 150

D. 0.714 mM

Question 151

2018

A 1 µg/mL solution of double stranded DNA has an absorbance at 260 nm of 0.020. Two microliters of a stock DNA solution are diluted to 100 µL. The absorbance of the diluted solution at this wavelength is found to be 0.017. What is the concentration of the stock solution?

• A. 8.5 x 10-2 µM
• B. 42.5 µg/mL
• C. 42.5 µM
• D. 0.017 µg/mL
• E. none of these

Answer 151

B. 42.5 µg/mL

Question 152

2018

The spectra of oxyhemoglobin and deoxyhemoglobin differ at many wavelengths. When a nurse measures the percentage of your hemoglobin that is oxygenated, he or she clips a small instrument to the end of your finger. The instrument uses infrared light to make the measurement, as infrared light can penetrate the skin far enough for a reading. How many wavelengths must the instrument use to obtain the percentage of your hemoglobin that is in the oxy state?

• A. Three
• B. Four
• C. Two
• D. One
• E. none of these

Answer 152

C. Two

Question 153

2018

Which of the following processes will be first order?

• A. conversion of carbon 14 to nitrogen 14 by emission of a beta particle (a fast-moving electron)
• B. the decrease in the blood level of aspirin after taking two tablets to reduce the pain of a headache
• C. the disappearance from the soil of a toxic contaminant as a result of microbial action
• D. the reaction of ethanol with NAD+ to yield acetaldehyde and NADH, a reaction catalyzed by liver alcohol dehydrogenase
• E. A, B and C

Answer 153

E. A, B and C

Question 154

2018

A researcher puts a hot cup of coffee at 95°C into a thermos bottle, expecting to drink it during a break. She can drink it once the temperature falls to 65°C or below. If the half time for cooling is 75 minutes, approximately how long before she can take the break?

• A. 75 min
• B. 65 min
• C. 45 min
• D. 41 min
• E. cannot tell from data given

Answer 154

D. 41 min

Question 155

2018

In the Michaelis-Menten reaction sequence, which of the following is/are first order ?

• A. formation of the ES complex from E and S
• B. dissociation of the ES complex to E and S
• C. catalysis with release of the product
• D. back reaction in which E and P recombine
• E. B and C

Answer 155

E. B and C

Question 156

2018

In the steady state assumption underlying Michaelis-Menten kinetics

• A. substrate is consumed steadily until the reaction approaches equilibrium.
• B. product is formed steadily until the reaction approaches equilibrium.
• C. substrate is converted to product as fast as product is converted back to substrate.
• D. the enzyme-substrate complex breaks down as fast as it is formed.
• E. none of the above.

Answer 156

D. the enzyme-substrate complex breaks down as fast as it is formed.

Question 157

2018

At substrate concentrations much lower than the value of Km

• A. doubling the substrate concentration approximately doubles the reaction rate.
• B. doubling the substrate concentration has little or no effect on the reaction rate.
• C. the reaction rate is approximately one half Vmax.
• D. the reaction rate is twice the value of Vmax.
• E. none of the above

Answer 157

A. doubling the substrate concentration approximately doubles the reaction rate.

Question 158

2018

When the substrate concentration of a Michaelis-Menten enzyme equals the value of Km,

• A. one fifth of the enzyme molecules have substrate in the active site at any moment.
• B. all or nearly all of the enzyme molecules have substrate in the active site at any moment.
• C. half the enzyme molecules have substrate in the active site at any moment.
• D. the enzyme is inactive
• E. the free energy of substrate binding is zero.

Answer 158

C. half the enzyme molecules have substrate in the active site at any moment.

Question 159

2018

If kcat for a Michaelis-Menten enzyme were 500, how many enzyme molecules would be required to produce 35,000 molecules of product per minute at saturating substrate concentrations?

• A. 500
• B. 35
• C. 3.5 x 105
• D. 70
• E. cannot tell from data given

Answer 159

D. 70

Question 160

2018

Two proteins, X & Y, each bind a ligand, L. The binding of L changes the absorption spectrum of the proteins, so binding can be tracked by following the intensity of the spectrum. As the concentration of L is increased, X reaches half-saturation at [L[= 2.5 µM, while Y reaches half-saturation at [L] = 6 µM. Which protein has a greater affinity for L?

• A. X
• B. Y
• C. Since they both bind L, they have the same affinity.
• D. One would need the extinction coefficients of the proteins to determine the affinity.
• E. Even with the extinction coefficients there are not enough data to determine the affinity.

Answer 160

A. X

Question 161

2018

The partial pressure of oxygen at which myoglobin is half-saturated at approximately neutral pH is 0.26 kPa. The partial pressure of oxygen in the tissues is 4 kPa, while in the lungs it is 13 kPa. What percent of the myoglobin molecules would have oxygen bound at the pO2 in the lungs?

• A. zero
• B. 50%
• C. 94%
• D. 98%
• E. Cannot tell from data given

What percent of the myoglobin molecules would have oxygen bound at the pO2 in the tissues?

• A. zero
• B. 50%
• C. 94%
• D. 98%
• E. Cannot tell from data given

If myoglobin were the oxygen carrier from the lungs to the tissues, what percent of them would deposit their oxygen in the tissues?

• A. zero
• B. 44%
• C. 48%
• D. 4%
• E. none of these

Answer 161

D. 98%

C. 94%

D. 4%

Question 162

2018

What would be the effect of deamination of a cytosine base if there were no DNA repair mechanisms available?

• A. DNA replication would replace an AT base pair by a UG base pair.
• B. There would be a chain break upon replication leading to a nicked strand.
• C. DNA replication would replace a CG base pair by an AU base pair.
• D. DNA replication would stop.
• E. none of the above

Answer 162

C. DNA replication would replace a CG base pair by an AU base pair.

Question 163

2018

UV light exposure causes formation of cyclobutane structures from adjacent thymines. What effect would this have if there were no DNA repair mechanisms available?

• A. DNA replication would replace an AT base pair by a UG base pair.
• B. There would be a chain break upon replication leading to a nicked strand.
• C. DNA replication would replace a CG base pair by an AU base pair.
• D. DNA replication would stop.
• E. none of the above

Answer 163

D. DNA replication would stop.

Question 164

2018

28. In the polymerase chain reaction to produce DNA in the laboratory, the DNA to be replicated is heated, a high concentration of short primers is added, and the solution is cooled before DNA polymerase is added. Why is the solution cooled?

• A. to prevent premature activity of the polymerase
• B. to prevent denaturation of the polymerase
• C. to allow the primers to anneal to the appropriate regions of the two strands
• D. to prevent denaturation of the primers
• E. none of the above

After the solution has cooled and the polymerase added, the solution is heated and cooled repeatedly. Why?

• A. Heating separates the newly synthesized strands. Cooling then allows unused primers to anneal so that the polymerase can make new DNA
• B. Heating denatures and destroys mistakenly formed DNA so the final product will be pure.
• C. Heating unfolds the polymerase, releasing it from the DNA, while cooling allows it to refold.
• D. more than one of the above
• E. none of the above

Answer 164

C. to allow the primers to anneal to the appropriate regions of the two strands

A. Heating separates the newly synthesized strands. Cooling then allows unused primers to anneal so that the polymerase can make new DNA

Question 165

2019

What structural form was represented on the pumpkin shown in the last class before this exam?

• A. a disordered structure
• B. an antiparallel beta pleated sheet
• C. a parallel beta pleated sheet
• D. tyrosine
• E. none of the above

Answer 165

D. tyrosine

Question 166

2019

After five half-lives of a first order reaction have passed, the percent of the initial material left is

• A. 6.25%
• B. 50%
• C. 12.5%
• D. 25%
• E. None of these

Answer 166

E. None of these

Question 167

2019

Which of the following cannot be a hydrogen bond in a protein?

• A. tyrosine hydroxyl to serine hydroxyl
• B. threonine methyl to a backbone carbonyl
• C. a buried water molecule to a side chain and to a backbone atom
• D. an amino nitrogen of a peptide bond to the carbonyl of another peptide bond
• E. all of the above can be hydrogen bonds in proteins.

Answer 167

B. threonine methyl to a backbone carbonyl

Question 168

The free energy of folding a protein is -44 KJ/mol, while the enthalpy of folding it is -22 KJ/mol. The entropy of folding is thus

• A. favorable.
• B. unfavorable.
• C. approximately zero.
• D. cannot tell from the data given
• (adapted from last year’s exam)

Answer 168

A. favorable.

Question 169

2019

Bacterial rhodopsin, a single polypeptide chain, is used in the generation of cellular energy. It is present in the cell membrane and has seven helices that cross from one side of the membrane to the other. How many loops are needed to connect the helices?

• A. 3.6
• B. eight
• C. seven
• D. six
• E. cannot tell from data given

Answer 169

D. six

Question 170

2019

Divergent evolution would be expected to yield

• A. proteins of similar three dimensional structure and the same or similar function.
• B. proteins of similar three dimensional structure and unrelated function.
• C. proteins of clearly different three dimensional structure and the same or similar function.
• D. proteins of clearly different three dimensional structure and unrelated function.
• E. none of the above (adapted from last year’s exam)

Answer 170

A. proteins of similar three dimensional structure and the same or similar function.

Question 171

2019

Consider an oligopeptide of five amino acids. If rotation about the phi bond of residue two occurs and all other possible rotations are frozen, which of the following interatomic distances would not change?

• A. between Cα of residue two and Cα of residue five
• B. between Cα of residue two and Cα of residue four
• C. between Cα of residue two and Cα of residue three
• D. between Cα of residue two and the peptide nitrogen of that residue
• E. More than one of these distances would remain unchanged.

Answer 171

E. More than one of these distances would remain unchanged.

Question 172

2019

How many normally rotatable single bonds are there in the backbone of the peptide asn phe ala gly pro ser glu?

• A. none
• B. two
• C. seven
• D. thirteen
• E. none of these

Answer 172

D. thirteen

Question 173

2019

In a β-pleated sheet

• A. the side chains on adjacent residues in each strand point in the same direction.
• B. side chains on one strand that are adjacent to residues on the next strand point in the same direction.
• C. side chains in the valleys of the pleats point up out of the valley, while those at the peaks point down below the peak.
• D. side chains in adjacent strands point towards one another, forming an interlocking pattern.
• E. none of the above

Answer 173

B. side chains on one strand that are adjacent to residues on the next strand point in the same direction.

Question 174

2019

Approximately 80% of the peptides in unfolded proteins are trans, while the remaining peptide bonds are cis. Why are cis peptides somewhat less stable than trans?

• A. The π clouds of cis peptides are repelled by partial negative charges of nearby peptides.
• B. The π clouds of trans peptides are stabilized by partial positive charges of nearby peptides.
• C. Cis peptides cause some steric hindrance of atoms.
• D. Cis peptides are less rotatable about the peptide bond.
• E. none of the above

Why do the phi-psi pairs of much of the Ramachandran plot represent “forbidden” conformations?

• A. The π clouds of forbidden conformations are repelled by partial negative charges of nearby peptides.
• B. The π clouds of allowed conformations are stabilized by partial positive charges of nearby peptides.
• C. Unallowed conformations cause steric hindrance of atoms.
• D. Unallowed conformations are less rotatable about the peptide bond. E. none of the above

Answer 174

C. Cis peptides cause some steric hindrance of atoms.

C. Unallowed conformations cause steric hindrance of atoms.

Question 175

2019

How many symmetry axes are there in a hexagon whose sides are of the same length?

• A. three two-fold & one six-fold
• B. six two-fold
• C. six two-fold & one six-fold
• D. six three-fold & one six-fold
• E. None of these is correct

Answer 175

C. six two-fold & one six-fold

Question 176

How many of the images shown below are right handed?

• A. None
• B. one
• C. two
• D. three
• E. four

Answer 176

D. three

Question 177

2019

The image below applies to the next two questions. The Zika virus, of whose capsid you made a model in class, is an icosahedron, a polyhedron with twenty faces. Imagine that line 1 were to point straight up from the top of the image, that line 2 were to be perpendicular to the plane of the page, i.e. pointing toward you, and that line 3 were to be shifted slightly up so that it bisects the edge it touches and passes through the vertex that is close to it. Lines 1, 2 and 3 would then be five-fold, a three-fold and two-fold axes of symmetry respectively. 1

1. How many three-fold axes of symmetry does the virus have?

• A. three
• B. five
• C. ten
• D. twenty
• E. None of these

Considering only the parts of the icosahedron visible in the image, for how many five-fold axes of symmetry are all five faces visible?

• A. three
• B. five
• C. ten
• D. twenty
• E. None of these

Answer 177

D. twenty

A. three

Question 178

2019

The melting curves for ribonuclease and apomyoglobin are shown in the figure to the left. When 100% of the maximum signal is reached, all the molecules of the proteins are fully unfolded. The reaction may be represented as N (Native conformation) ⇄ U (Unfolded conformation). At 60% of the maximum signal, what is the equilibrium constant for the N ⇄ U reaction ?

• A. zero
• B. 1.0
• C. 0.6
• D. 1.5
• E. none of these

Answer 178

D. 1.5

Question 179

2019

The Anfinsen experiment on ribonuclease demonstrated that

• A. proteins cannot fold to their native conformations without the assistance of auxiliary proteins now called chaperones.
• B. all the information needed for a protein to fold to its native conformation is present in the primary sequence.
• C. protein form incorrect disulfide bonds before forming those found in the native conformation.
• D. protein denaturation is irreversible at high protein concentration.
• E. none of the above

Answer 179

B. all the information needed for a protein to fold to its native conformation is present in the primary sequence.

Question 180

2019

Protein folding proceeds via folding intermediates, most of which are stable only for very short times. Imagine that a mutation occurs which has no effect on the function of the native protein but which stabilizes a folding intermediate for a longer time than normal. Which of the following is a likely consequence?

• A. aggregation of protein molecules to form amyloid and leading to cell death
• B. no effect. The native conformation still has the lowest free energy of the available conformations.
• C. The cell will speed folding to prevent aggregation.
• D. Full unfolding will be favored by the mutation.
• E. none of the above

Answer 180

A. aggregation of protein molecules to form amyloid and leading to cell death

Question 181

2019

Chothia’s analysis of the free energy of transfer of amino acids from water to ethanol yielded a slope of 25 cal / (Å2 mol) of hydrophobic surface area transferred (cal, not kcal). The free energy of binding the ligand NAD+ to liver alcohol dehydrogenase is -4.6 kcal/mol. The enthalpy of binding is -1 kcal/mol. If all the entropy of binding is hydrophobic in origin, how many square Ångstroms of hydrophobic surface are withdrawn from contact with water?

• A. 3.6
• B. 3600
• C. 144
• D. 25 E. 12

Answer 181

C. 144

Question 182

2019

From Voronoi polyhedron volumes Richards found that the packing density in the protein interior is approximately 74%. Also using Voronoi polyhedra, Chothia discovered that amino acids which are buried in the protein interior have the same volumes as they do in crystals of the pure amino acid once the water of peptide bond formation is subtracted. The amino acid crystals are “dry,” having no water of hydration in them. Why, then, are water molecules found buried in the interior of many proteins?

• A. The 74% packing density is an average across the entire protein molecule. This allows for looser packing at some places, which can have water in them, and tighter packing at others.
• B. Polar groups forced into the interior during folding are more stable if they can make hydrogen bonds to buried water molecules.
• C. Active sites are often designed to have water in them, water which stabilizes the site or participates in catalysis or both.
• D. All the above
• E. None of the above

Answer 182

D. All the above

Question 183

2019

Rate constants measure the number of reaction events per unit time. For ligand binding the association rate constant is often called the “on rate”, while the dissociation rate constant is called the “off rate”. Kd, the equilibrium constant for ligand dissociation, is a ratio of these rate constants. A molecule that could become an inhibitor of an enzyme involved in disease is to be tested. Which of the following would bring a smile to the face of the researcher?

• A. kon = koff
• B. kon >> koff
• C. koff >> kon
• D. Kd << 1
• E. more than one of these

Answer 183

B. kon >> koff

Question 184

2019

Ligand binding equilibria can be described by an equation identical in form to the Henderson-Hasselbalch equation. The binding equation states that

• A. The concentration of unbound ligand depends on the total ligand concentration.
• B. The concentration of bound ligand depends on the total ligand concentration.
• C. The concentration of unbound ligand depends on the ratio of unliganded protein to liganded protein.
• D. The concentration of bound ligand depends on the total protein concentration.
• E. none of the above

Answer 184

C. The concentration of unbound ligand depends on the ratio of unliganded protein to liganded protein.

Question 185

2019

In studying oxygen binding to heme proteins, what does [O2]0.5 mean?

• A. the binding of a single atom of oxygen
• B. It means that half the available oxygen is bound.
• C. Half the protein molecules have oxygen bound.
• D. The oxygen concentration is half what is needed to bind any oxygen to protein.
• E. none of the above

Answer 185

C. Half the protein molecules have oxygen bound.

Question 186

2019

Ligand binding affinities shown in class for several binding complexes range from Kd ≈ 10-2 M to 10-15 M. Why are enzyme-substrate affinities at the low affinity (Kd ≈ 10-2 M) end of the range?

• A. Active sites fluctuate more than other parts of the protein.
• B. If substrate binding were very tight, catalysis and product release would be difficult.
• C. Low affinity allows for more rapid catalysis.
• D. All the above
• E. none of the above

Answer 186

D. All the above

Question 187

2019

The oxygen binding curve for hemoglobin is shown below. What percentage of the oxygen carried when the blood leaves the lungs is deposited in the tissues?

• A. Approximately 25%
• B. less than 5%
• C. Approximately 100%
• D. Approximately 40%
• E. cannot tell from data shown.

Answer 187

D. Approximately 40%

Question 188

2019

How do bacteria become resistant to penicillin?

• A. They develop “pumps” which cause the drug to be secreted from the cell as fast as it enters.
• B. They develop an enzyme which inactivates the drug.
• C. They add proteins to their plasma membranes which prevent the drug from entering the cell.
• D. They use clavulanic acid as an inactivating agent.
• E. none of the above

Answer 188

B. They develop an enzyme which inactivates the drug

Question 189

2019

It is estimated that one person in 600 carries a mutation in his or her hemoglobin molecule. Approximately 6000 students live on the combined Cook and Douglass campuses. That means that ten of them are mutants and suffer no adverse effects of the mutation. You could be one of them! It is highly likely that

• A. the mutation occurs deep within the folded polypeptide chains.
• B. the mutation occurs in the heme pocket.
• C. the mutation occurs at the interface between the alpha-1 and beta-1 chains
• D. The mutation occurs on the protein surface.
• E. none of these is likely

Answer 189

D. The mutation occurs on the protein surface.

Question 190

2019

29. In the serine protease catalytic mechanism, why does diisopropylfluorophosphate (DIPF) react only with serine 195 and not with other hydroxyl-containing side chains?

• A. DIPF can only reach serine 195; the other hydroxylated groups are not accessible to it.
• B. There are no other such side chains in these proteases.
• C. Only serine 195’s hydroxyl is strongly nucleophilic due to its interactions with specific other protein groups.
• D. DIPF would react with other groups after reacting with serine 195, but its reaction with this side chain changes the protein conformation.
• E. none of the above

Answer 190

C. Only serine 195’s hydroxyl is strongly nucleophilic due to its interactions with specific other protein groups.

Question 191

2020

Alcohol dehydrogenase (ADH) works by converting one ethanol molecule to acetaldehyde, reducing one NAD+to do so. Suppose you run an experiment comparing wild-type ADH with a mutant ADH. Both experiments have the same ADH concentration, ethanol concentration, NAD+ concentration, no NADH at the start, and are placed in the same buffered solution in two separate reaction tubes. After 90 seconds you find that [NADH] in the wild-type ADH experiment is 14 μM and [NADH] in the mutant ADH experiment is 1.9 mM. Which of the following could account for these results?

• We cannot conclude anything because we are not given substrate or product concentration before or after the reaction occurs
• Km for the mutant ADH is lower than for the wild-type
• Enzyme binding is looser for the mutant ADH because more product is formed.
• Vmax is higher for the wild-type ADH.
• kcat is lower for the mutant ADH.

Answer 191

Km for the mutant ADH is lower than for the wild-type

Question 192

2019

How do enzymes lower the activation energy of the reactions they catalyze?

• A. They bend and stretch bonds toward the shape of the transition state.
• B. The energy of binding reactants contributes to lowering the activation energy.
• C. They have side chains which take part in catalysis in just the right positions to act upon reactants.
• D. They may form covalent intermediates.
• E. all of the above

Answer 192

E. all of the above

Question 193

2020

Which of the following is the correct unit for the rate constant of a first order reaction?

• moles enzyme / mol S
• L / (mol-deg)
• min-1
• min / mol P
• none of these

Answer 193

min-1

Question 194

2020

How many hydrogen bonds would hold the oligonucleotide p-ATTGCGCT-p to its complementary sequence?

• none
• eight
• twenty
• sixteen
• none of these

Answer 194

twenty

Question 195

2020

Starting with the bases, ribose and phosphate, how many water molecules would have to be removed to make the oligonucleotide pAGTC-OH

• four
• three
• eleven
• seven
• none of these

Answer 195

eleven

Question 196

2020

In human DNA 15% of the bases are adenines. What percentage of the bases are guanines?

• 15%
• 30%
• 35%
• 70%
• cannot tell from data given

Answer 196

35%

Question 197

2020

What role in the serine protease reaction mechanism is played by the charge relay system?

• It is not involved in the catalytic mechanism at all.
• It destabilizes the transition state.
• It stabilizes the tetrahedral intermediates.
• It converts an otherwise chemically unreactive serine into a strong nucleophile.
• It helps maintain the active conformation of the protein.

Answer 197

It converts an otherwise chemically unreactive serine into a strong nucleophile.

Question 198

2020

Uncontrolled diabetes often leads to diabetic acidosis, in which the pH of the blood drops well below normal. What effect could this have on the ability of the hemoglobin to become oxygenated in the lungs?

• It would have no effect at all, as the system would readjust to compensate
• The hemoglobin would become oxygenated more easily.
• The oxygen saturation level on leaving the lungs would be reduced.
• The binding affinity for BPG would be reduced.
• The T to R conformational transition would be stimulated.

Answer 198

The oxygen saturation level on leaving the lungs would be reduced.

Question 199

2020

The release of oxygen from hemoglobin causes one or more groups on the protein to

• become less acidic
• become more acidic
• have a lower pKa
• have a higher pKa
• (A) and (D)

Answer 199

(A) and (D)

Question 200

2020

If the free energy of a reaction which converts a substrate, S to a product, P, is negative, it follows that

• the activation energy in the forward reaction exceeds that of the back reaction
• the entropy change of the S à P reaction is favorable.
• the transition state is less stable in the back reaction
• the free energy difference between the transition state and the substrate is less than the difference between the transition state and the product.
• none of these is correct

Answer 200

the free energy difference between the transition state and the substrate is less than the difference between the transition state and the product.

Question 201

What survival advantage do bacteria which make beta-lactamase have over those which do not make it?

• Their lactamase inactivates clavulanic acid.
• The forms that make it are not killed by penicillin.
• The forms that make it can kill the bacteria that don’t make it.
• The forms that make it are resistant to clavulanic acid.
• none of these

Answer 201

The forms that make it are not killed by penicillin.

Question 202

2020

In the serine protease reaction mechanism which of the following is/are a transition state?

• A- the ES complex
• B- the first tetrahedral intermediate
• C - the acyl enzyme complex
• D - the second tetrahedral intermediate
• (B) and (D)

Answer 202

(B) and (D)