Exam III

Question 1

Which of the following statements is false for an enzyme that follows Michaelis-Menten kinetics? Option A: The relationship between substrate concentration and reaction rate is sigmoidal.

Option B: The initial velocity of the reaction is dependent on substrate concentration.

Option C: The Michaelis-Menten equation assumes that ES maintains a steady state.

Option D: Maximal velocity occurs when the enzyme is entirely in the ES form.

Answer 1

Option A: The relationship between substrate concentration and reaction rate is sigmoidal.

Question 2

In uncompetitive inhibition, the inhibitor binds only to the ______.

Option A: substrate

Option B: enzyme

Option C: active site

Option D: ES complex

Answer 2

Option D: ES complex

Question 3

For a reaction A + B → C, if the concentration of B is much larger than A so that [B] remains constant during the reaction while [A] is varied, the kinetics will be Option A: sigmoidal.

Option B: pseudo-first-order.

Option C: unimolecular.

Option D: zero-order.

Option E: hyperbolic.

Answer 3

Option B: pseudo-first-order.

Question 4

An enzyme is near maximum efficiency when

Option A: its turnover number is near Vmax.

Option B: k_cat/K_M is near 10⁸ M-1s-1.

Option C: k1 << k-1.

Option D: k_cat/K_M is equal to kcat.

Option E: K_M is large when k2 exceeds k1.

Answer 4

Option B: k_cat/K_M is near 10⁸ M-1s-1.

But It can be no greater than k1; that is, the decomposition of ES to E + P can occur no more frequently than E and S come together to form ES.

Question 5

Pseudo-first-order reaction kinetics would be observed for the reaction A + B → C

Option A: if [A] or [B] > [C].

Option B: if [C]>[A] and [C]>[B].

Option C: if [A] or [B] = 0.

Option D: if [C] = 0.

Option E: None of the above.

Answer 5

Option E: None of the above. answer is [B]>>[A]

Question 6

Different enzymes that catalyze the same reaction, although may be found in different tissues, are known as ______.

Answer 6

isozymes

Question 7

A two-substrate enzymatic reaction in which one product is produced before the second substrate binds to the enzyme has a ______ mechanism.

Answer 7

ping-pong

Question 8

A compound that distorts the active site, rendering the enzyme catalytically inactive is called

Option A: a uncompetitive inhibitor

Option B: an allosteric effector

Option C: an inactivator

Option D: a competitive inhibitor

Option E: none of the above

Answer 8

Option A: a uncompetitive inhibitor

Question 9

Allosteric activators

Option A: bind via covalent attachment.

Option B: stabilize conformations with higher Ks.

Option C: stabilize conformations with higher substrate affinity.

Option D: All of the above.

Option E: None of the above.

Answer 9

Option C: stabilize conformations with higher substrate affinity.

Question 10

Activator ATP preferentially binds to ATCase’s _____ state

Answer 10

R

Question 11

Inhibitor CTP preferentially binds to ATCase’s _____ state

Answer 11

T

Question 12

How do enzymes differ from ordinary chemical catalysts?

Answer 12

1. Higher reaction rates
2. Milder reaction conditions <100°C and neutral pH
3. Greater reaction specificity for substrates and products; rarely have side products.
4. Capacity for regulation: capacity for activities varies in response to concentration of other substances such as those that cause inhibitor, covalent modification of enzyme, and amounts of enzymes synthesized

Question 13

Oxidoreductases

Answer 13

Enzyme that catalyzes oxidation reduction reactions

Question 14

Tranferases

Answer 14

Enzyme that catalyzes the transfer of functional groups

Question 15

Hydrolases

Answer 15

Enzyme that catalyzes the hydrolysis of reactions

Question 16

Lyases

Answer 16

Enzyme that catalyzes group elimination to form double bonds

Question 17

Isomerases

Answer 17

Enzymes that catalyze isomerization

Question 18

Ligases

Answer 18

Enzyme that catalyzes bond formation coupled with ATP hydrolysis

Question 19

what are the 2 forms of complementarity that enzymes have for substrates?

Answer 19

1. geometric: enzyme active site contains indentation that is complementary in shape to substrate

2. electronic: AA residues that form the binding site are arranged to specifically attract the substrate

^both non-covalent

^^this complementarity is the basis for the “Lock and Key” model

Question 20

the substrate binding site is mostly pre-formed, but does undergo some conformational change upon binding the substrate. What term describes this phenomenon

Answer 20

induced fit

Question 21

Enzymes are stereospecific because?

Answer 21

They are themselves chiral and form asymmetric sites.

Question 22

geometric specificity refers to ?

Answer 22

the specificity for substrate based on the identities of the chemical groups on their substrates

Question 23

T/F: geometric specificity is a more stringent requirement than is stereospecificity

Answer 23

True

partly d/t the fact that steric hindrance may block entry in the first place

Question 24

T/F: Most enzymes catalyze the reactions of a small range of related compounds although with different efficiencies.

Answer 24

true

the difference is determiend by K value: Ethanol has a lower K value

For example: alcohol dehydrogenase catalyzes the oxidation of ethanol (CH3CH2OH) to acetaldehyde (CH3CHO) faster than it oxidizes methanol (CH3OH) to formaldehyde (H2CO) or isopropanol [(CH3)2CHOH] to acetone [(CH3)2CO], even though methanol and isopropanol differ from ethanol by only the deletion or addition of a CH2 group.

Other Variations in Geometric Specificity:

1. A few enzymes are absolutely specific for only one compound.
2. Some enzymes, particularly digestive enzymes, are so permissive in their ranges of acceptable substrates that their geometric specificities are more accurately described as preferences.

E.g chymotrypsin, in addition to its ability to mediate peptide bond hydrolysis, also catalyzes ester bond hydrolysis.

Question 25

__lower/higher__ the K value, the more the substrate binds to the enzyme

Answer 25

lower

Question 26

How would you describe the geometric specificity of chymotrypsin

Answer 26

Permissive (ie not very geometrically stringent)

it can hydrolyze both peptides and esters, 2 very different groups

Question 27

Preferred substrates for chymotrypsin are _____, __ terminal amino acids

Answer 27

bulky, C terminal

^{<em>Recall: the AA to the right of C terminal end can’t be proline.</em>}

If a lot of peptides have C terminal ends of bulky AAs, then we can assume that chymotrypsin is involved.

Question 28

metal ions that are not natural cofactors can do what type of damage?

Answer 28

they can replace Zn²⁺ and render the active sites of certain enzymes inactive

Question 29

which amino acid is essential in the functionality/folding of Zinc finger?

Answer 29

His

(Also Cys) - these bind with Zn to form active site

Question 30

what is the K value for the reaction in which ethanol is turned into acetaldehyde?

Answer 30

K= 10^-6

^{the K values for the other alcohols that have less binding affinity would be larger}

Question 31

NAD+ is an _____________agent in the alcohol dehydrogenase (ADH) reaction

Answer 31

obligatory oxidizing

Question 32

T/F: in order to complete the catalytic cycle, the coenzyme must return to its original state

Answer 32

True

Question 33

What stereospecific enzymes are used for

Answer 33

Binding of chiral molecules
Reactions involving prochiral molecules

Question 34

2 types of processes that enzymes are unable to catalyze alone

Answer 34

Oxidation-reduction reactions
Group-transfer reactions

Question 35

Cofactors that only transiently associate with a given enzyme molecule

Answer 35

Cosubstrates

Question 36

4 examples of cosubstrates

Answer 36

NAD
NADH
FAD
FADH2

Question 37

Enzymatically inactive protein resulting from the removal of a holoenzyme’s cofactor

Answer 37

Apoenzyme

Question 38

Ratio of the rate of the catalyzed to uncatalyzed reaction

Answer 38

rate enhancement

Question 39

T/F: catalyst lowers the free energy barrier by the same amount for both the forward and reverse reactions

Answer 39

true

Question 40

How enzymes achieve their enormous rate accelerations (2)

Answer 40

1. Reducing the free energy of the transition state 2.Stabilizing the transition state of the catalyzed reaction

Question 41

Steps in RNase A mechanism (6)

Answer 41

1. His 12 extracts a proton from RNA 2’OH group
2. Nucleophilic attack on adjacent phosphorus atom
3. His 119 protonates the leaving group
4. Water enters active site
5. His 12 (now an acid) and His 119 (now a base) facilitate hydrolysis of 2’3’-cyclic intermediate
6. enzyme returns to original state

Question 42

T/F: pH effects on an enzymatic rate may reflect denaturation of the enzyme rather than protonation or deprotonation of specific catalytic residues.

Answer 42

true

Question 43

transformation of an amine and carboxylic acid into an imine is an example of what kind of catalysis?

Answer 43

covalent

Question 44

T/F: The nucleophilicity of a substance is closely related to its basicity.

Answer 44

True

Question 45

possible electrophiles for catalytic mechanisms include: ?

Answer 45

H⁺

Metal Ions (Mⁿ⁺)

Carbonyl Carbon

Cationic imines

Question 46

An important aspect of covalent catalysis is that the more/less stable the covalent bond formed, the more/less easily it can decompose in the final steps of a reaction.

Answer 46

more

less

Question 47

what are the residue groups that can function as a nucleophile in covalent catalysis?

Answer 47

1. Asp
2. Cys
3. His (unprotonated)
4. Lys (unprotonated)
5. Ser

mnemonic: ACHeLS

“K” sound for Covalent Catalysis

Question 48

The active residues in lysozyme’s active site are ?

and which acts first

Answer 48

Asp and Glu

Glu35 acts first by giving off a hydrogen to O1, which stays attached to NAG

Question 49

lysozyme cleaves the bond between carbon 1 of _____ and carbon 4 of _____

by ____________catalysis

Answer 49

carbon 1 of NAM and carbon 4 of NAG

general acid

Question 50

how does a lysozyme degrade bacterial cell wall

Answer 50

by hydrolyzing the β(1→4) glycosidic linkages from N-acetylmuramic acid (NAM or MurNAc) to N-acetylglucosamine (NAG or GlcNAc) in cell wall peptidoglycans

Question 51

what segment of the peptidoglycan is released first from a lysozyme

Answer 51

the segment with the NAG end

Question 52

An enzyme that catalyzes a reaction by the preferential binding of its transition state has a greater binding affinity for an inhibitor that…

Answer 52

inhibitor that has the transition state geometry (a transition state analog) than it does for its substrate.

*What is mimicked is shape and size that’s occupied, not the exact structure

Question 53

Chymotrypsin, trypsin, and elastase are strikingly similar: The primary structures of these ~__#__-residue enzymes are ~_#__% identical

Answer 53

240

40%

Question 54

the residues that make up the catalytic triad are:

what is their spatial relationship?

Answer 54

Ser, Asp, His

Asp His and Ser are close in tertiary structure to eachother, not close in primary structure.

Question 55

what happens if you replace a Trypsin Asp to Ser by site directed mutagenesis?

Answer 55

you get a poor non-specific protease, contrary to what you might think, because the enzyme requires a precise arrangement to position the scissile bond correctly

Question 56

_______in Trypsine is equivalent to ________ in Chymotrypsin

Answer 56

Asp 189

Ser 189

Question 57

chymotrypsin and trypsin hydrolyze small neutral residued peptide bonds extremely slowly. why?

Answer 57

these small substrates cannot be sufficiently immobilized on the enzyme surface for efficient catalysis to occur.

Question 58

How do the “shapes” of each digestive enzyme active site affect which groups they can accommodate?

Answer 58

Gly is present on chymotrypsin because it’s small, which allows bulkier AAs to enter

Thr and val are bulkier, so it attracts smaller AAs

Question 59

What differentiates the 2 tetrahedral intermediates in serine protiease reactions?

Answer 59

Amide bond in first intermediate, but the second intermediate does not have the amide bond

*in between these states there’s also an acyl enzyme intermediate

Question 60

a powerful inhibitor of serine protease would most likely resemble

Answer 60

the transition state (in which the tetrahedral conformation has an oxygen pointing toward the oxyanion hole)

Question 61

DIPF is an effective inhibitor of serine proteases because ?

Answer 61

its tetrahedral phosphate group makes this compound a transition state analog.

Question 62

T/F: Mutating any or all of the residues in chymotrypsin’s catalytic triad yields enzymes that still enhance proteolysis by ~5 × 10⁴-fold over the uncatalyzed reaction (versus a rate enhancement of ~10¹⁰ for the native enzyme).

Answer 62

True - because of strong preferential binding of transition state (which relies more on hydrogen bonding than specificty of residue side chains)

Question 63

initial velocity is taken to mean

Answer 63

operationally taken as the velocity measured before more than ~10% of the substrate has been converted to product

Question 64

Which is the better substrate?

Answer 64

The lower the number the higher the affinity. so fumarate is preferred substrate to malate

Question 65

Vmax \_increases/decreases__ as [E]_T increases.

Answer 65

increases

Question 66

The best estimates of kinetic parameters are obtained by collecting data over a range of [S] from ___ to ___

Answer 66

~0.5 K_M to ~5 K_M.

Question 67

what is the competitive inhibitor for succinate?

Answer 67

malonate (because it is structurally similar but cannot be dehydrogenated)

Question 68

product inhibition is most closely linked to which type of inhibitor

Answer 68

competitive

product inhibition = product may accumulate and block the active site from which it came, thereby competing with substrate

Question 69

transition state analogs most closely resemble which type of inhibitor?

Answer 69

competitive because of their involvement in the active site

Question 70

what is the reaction scheme for competitive inhibitor?

Answer 70

Question 71

Km increases with competitive inhibitor. why? (mathematical explanation)

Answer 71

Km will increase with a competitive inhibitor because Km is unbound/bound and competitive inhibitor will give you more unbound

Question 72

The Degree of Competitive Inhibition Varies with ?

Answer 72

the Fraction of Enzyme That Has Bound Inhibitor.

Question 73

in competitive inhibition, K_I represents dissociation constant for inhibitor-enzyme complex… does K_I increase or decrease the more the competitive inhibitor is bound.

Answer 73

decrease

from K_I = [E][I] / [EI]

therefore, more inhibition → alpha increases

Question 74

in uncompetitive inhibition, alpha is expressed in lineweaver burk plot as what?

Answer 74

-alpha/km

and -alpha/Vmax

^reason for parallel lines

as opposed to competitive which only expresses -alpha/Km

Question 75

The activities of many enzymes are also (in addition to allosteric effectors) controlled by covalent modification, which are usually

Answer 75

phosphorylation and dephosphorylation of specific Ser, Thr, or Tyr residues.

…catalyzed by enzymes known as protein kinases and protein phosphatases

Question 76

The Feedback Inhibition of ATCase Regulates ________Synthesis.

how?

Answer 76

Pyrimidine

• Aspartate TransCarbamoylase (ATCase) from E. coli catalyzes the formation of N-carbamoyl aspartate from carbamoyl phosphate and aspartate:
• formation of N-carbamoyl aspartate is the first step unique to the biosynthesis of pyrimidines.

Question 77

ATCase is allosterically inhibited by ____, a pyrimidine nucleotide, and is allosterically activated by _____, a purine nucleotide.

Answer 77

inhibited by cytidine triphosphate (CTP)

activated by adenosine triphosphate (ATP)

Question 78

A random displacement mechanism would result from substrates with what pka relationship?

Answer 78

similar pkas with moderate binding affinity. Similar so that they will bind at equal rates.

Question 79

T/F The v_o versus [S] curve for ATCase is sigmoidal, rather than hyperbolic

Answer 79

True, because it undergoes cooperative binding

Question 80

ATP binding to transcarbomylase corresponds to a shift to the _\_right/left___

Answer 80

ATP - shift to left because it is an effector increasing affinity, CTP - shift to the right because it is an inhibitor, decreasing affinity

Question 81

How does feedback inhibition of ATCase work?

Answer 81

• CTP, which is a product of the pyrimidine biosynthetic pathway, is an example of a feedback inhibitor, since it inhibits an earlier step in its own biosynthesis.
• When CTP levels are high, CTP binds to ATCase, thereby reducing the rate of CTP synthesis.
• When cellular [CTP] decreases, CTP dissociates from ATCase and CTP synthesis accelerates.

Question 82

•The phosphorylation of ____promotes phosphorylase’s T (inactive) → R (active) conformational change.

Answer 82

Ser 14

•The Ser 14–phosphate group functions as a sort of internal allosteric effector that shifts the enzyme’s T<=> R equilibrium in favor of the R state

Question 83

• ATP and glucose-6-phosphate [G6P; to which the G1P product of the glycogen phosphorylase reaction is converted] preferentially bind to the ______\_ state of phosphorylase b
• whereas AMP preferentially binds to the ________ state of phosphorylase b

Answer 83

T state of phosphorylase b and, in doing so, inactivate the enzyme

R state of phosphorylase b and activates it

Question 84

•Phosphorylase a’s only allosteric effector is _______, which binds to the enzyme’s T state and inactivates the enzyme.

Answer 84

glucose

Question 85

Which of the following is TRUE regarding cofactors?

Option A: Prosthetic groups can dissociate readily and be regenerated for use in another enzyme.

Option B: Metal ions must be covalently attached to function as a cofactor.

Option C: An apoenzyme implies that a cofactor is present.

Option D: Cofactor is a broad term used for all enzyme “helpers”.

Option E: Coenzymes are often separate from the enzyme and do not need recharged.

Answer 85

Option D: Cofactor is a broad term used for all enzyme “helpers”.

Question 86

If you add enzyme to a solution containing only the product(s) of a reaction, would you expect any substrate to form?

Option A: All of the above may determine if product forms.

Option B: It depends on the concentration of products added.

Option C: It depends on the time interval and temperature of reaction.

Option D: None of the above determines if product forms.

Option E: It depends on the energy difference between E + P and the transition state.

Answer 86

Option E: It depends on the energy difference between E + P and the transition state.

Question 87

Which of these amino acid groups would not make a good nucleophilic catalyst?

Option A: methyl

Option B: imidazole

Option C: sulfhydryl

Option D: hydroxyl

Option E: amino

Answer 87

Option A: methyl

Question 88

T/F: If an enzyme-catalyzed reaction requires a group with a low pK to be deprotonated and a group with a higher pK to be protonated, the pH vs. rate curve will have a peak in the middle of the two pK values.

Answer 88

True

Question 89

T/F: Only subsite D of lysozyme binds the saccharide unit NAM with a positive binding free energy.

Answer 89

True

ie half chair

Question 90

Glu 35 of lysozyme is found in a nonpolar environment. Which of the following is true?

Option A: Its pK is lower than the usual value in this environment.

Option B: Its pK is higher than the usual value in this environment.

Option C: None of the above is correct.

Option D: Its pK would depend on the sample buffer.

Option E: Its pK would not change in this environment.

Answer 90

Option B: Its pK is higher than the usual value in this environment.

ie it’s less acidic

Question 91

A new serine protease was discovered that preferentially cleaves a peptide bond adjoining a negatively charged side chain. Which of the following is true?

Option A: The specificity pocket would mimic that of trypsin.

Option B: It likely reacts much slower than chymotrypsin.

Option C: The specificity pocket would mimic that of chymotrypsin.

Option D: The specificity pocket is likely lined with amino acids such as Arg and Lys.

Option E: It likely reaction much faster than chymotrypsin.

Answer 91

Option D: The specificity pocket is likely lined with amino acids such as Arg and Lys.

Question 92

Why are chymotrypsin and subtilisin considered examples of convergent evolution?

Option A: Because their polypeptide chains have different folds, but their active sites have identical residues.

Option B: Because their polypeptide chains have different folds, and their active sites have different residues.

Option C: Because their polypeptide chains have the same fold, and their active sites have identical residues.

Option D: Because their polypeptide chains have the same fold, but their active sites have different residues.

Answer 92

Option A: Because their polypeptide chains have different folds, but their active sites have identical residues.

Question 93

t/f: The Enzyme Commission supplies two names and a classification number for each enzyme to unequivocally establish its identity.

Answer 93

True

Question 94

The enzymatically inactive protein resulting from the removal of a protein’s cofactor is referred to as what?

Option A: Proenzyme

Option B: Apoenzyme

Option C: Holoenzyme

Option D: Zymogen

Answer 94

Option B: Apoenzyme

Question 95

Why is decomposition of the transition state to products usually considered the rate-determining process of the overall reaction?

Option A: The transition state is a highly-populated state.

Option B: The transition state is a long-lived state.

Option C: Passage through the transition state is very rapid, so the concentration of the transition state is small.

Option D: Passage through the transition state is very rapid, so the concentration of the transition state is large.

Answer 95

Option C: Passage through the transition state is very rapid, so the concentration of the transition state is small.

Question 96

T/F: In enzyme reaction mechanisms the normal rules of chemical kinetics do not always apply.

Answer 96

False

Question 97

Which of the following is NOT normally one of the three stages of covalent catalysis?

Option A: The nucleophilic reaction between the catalyst and the substrate to form a covalent bond

Option B: The elimination of the catalyst

Option C: The withdrawal of electrons from the reaction center by the electrophilic catalyst

Option D: Proton transfer by the enzyme

Answer 97

Option D: Proton transfer by the enzyme

Question 98

A lead compound would be most promising if it had:
A) K I = 4.7 × 10 5 M.
B) K I = 1.5 × 10 8 M.
C) K I = 1.5 × 10 -8 M.
D) K I = 4.7 × 10 -5 M.
E) K M = 4.7 × 10 5 M.

Answer 98

C) K I = 1.5 × 10 -8 M.

Question 99

What is the velocity of a first-order reaction at 37 o C when the reactant concentration is 6 × 10^-2 M and the rate constant is 8 × 10 ³ sec -1 ?
A) 1.33 × 10 ⁵ M -1 •sec -1
B) 1.33 × 10 ⁵ M•sec
C) 7.5 × 10 ^-2 M•sec
D) 4.8 × 10 ² M•sec -1
E) Not enough data are given to make this calculation

Answer 99

v= k[A]

D) 4.8 × 10 ² M•sec -1

Question 100

15. K M is
    A) a measure of the catalytic efficiency of the enzyme.
    B) equal to half of V max
    C) the rate constant for the reaction ES  E + P.
    D) the [S] that half-saturates the enzyme.
    E) a ratio of substrate concentration relative to catalytic power.

Answer 100

Ans: D

D) the [S] that half-saturates the enzyme.

B is a trap!!

Question 101

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 of its
maximum velocity,
A) [S] would need to be equal to K M
B) [S] would need to be ½ K M
C) [S] would need to be 3K M
D) [S] would need to be ¾ K M
E) not enough information is given to make this calculation

Answer 101

Ans: C

Question 102

The K M can be considered to be the same as the dissociation constant K S for E + S binding if
A. the concentration of [ES] is unchanged.
B. ES E + P is fast compared to ES E + S.
C. k1 >> k2.
D. k2 << k-1.
E this statement cannot be completed because KM can never approximate KS.

Answer 102

Ans: D

Question 103

Find k cat for a reaction in which V max is 4 × 10 -4 mol•min -1 and the reaction mixture contains
one microgram of enzyme (the molecular weight of the enzyme is 200,000 D).
A) 2 × 10 ^-11 min -1
B) 8 × 10 ⁷ min -1
C) 8 × 10 ⁹ min -1
D) 2 × 10 ^-14 min -1
E) 4 × 10 ⁸ min -1

Answer 103

B) 8 × 10 ⁷ min -1

Question 104

[S] = K M for a simple enzymatic reaction. When [S] is doubled the initial velocity is
A) 2 V max
B) equal to V max
C) (1/3) V max
D) 0.5 V max
E) 2 K M /[S]

Answer 104

Ans: C

trick question

Question 105

Which of the following is (are) true?
A) The [ES] will remain constant if k 2>k 1 and k −1<k .></k>B) The reaction is zero order with respect to [S] if [S]>>[E]
C) It describes a double displacement reaction
D) All of the above are true.
E) None of the above is true.

Answer 105

Ans: B

Question 106

The Michaelis constant K M is defined as
I. (k _–1 + k ₂ )/k ₁
II. ½ V max
III. [S] = [ES]
IV. [ES]/2

A) I
B) I, II
C) II
D) I, IV
E) II, IV
Ans: A

Answer 106

A) I

Question 107

Which of the following is correct in regards to the diagram above?

A) X=A, Y=B, Z=P
B) X=B, Y=A, Z=Q
C) X=E, Y=A, Z=E
D) X=E, Y=B, Z=Q
E) X=E, Y=B, Z=P

Answer 107

Ans: B

Question 108

Enzyme activity in cells is controlled by which of the following?
I. covalent modifications
II. modulation of expression levels
III. feedback inhibition
IV. allosteric effectors
A) I
B) II
C) III
D) III, IV
E) I, II, III, IV

Answer 108

Ans: E

Question 109

Protein kinases are involved in

Answer 109

the phosphorylation of a wide variety of proteins.

Question 110

I propose to design a new drug which will act as an inhibitor for an enzyme. If I have used all current information about the mechanism of this enzyme to design this inhibitor and I
carefully engineer it with similar chemical properties of the transition state, what type of inhibitor am I attempting to engineer and how will I know if I have succeeded?

A) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and
watch for a change in V _max .
B) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and
watch for a change in K _M .
C) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor
and watch for a change in K _M .
D) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor
and watch for a change in V max .
E) None of the above.

Answer 110

Ans: A

Question 111

42. An extremely efficient enzyme called “efficase” catalyzes the conversion of “A” to “B.” A researcher decides to mutate the enzyme in order to try to improve its performance. Following active site mutations, a significant reduction in the value of K M and V max was observed. Which of the following may have occurred?

A) The affinity of the enzyme for the substrate was increased to a point which did not favor
propagation (continuation) of the reaction.
B) The decrease in V max was not related to the decrease in K M .
C) If the reaction was first-order, the change in K M cannot have affected V max .
D) The stability of E+S (E+A as written above) was increased, thereby increasing the K M .
E) The reverse reaction (breakdown of EA to E+A) was favored, slowing the V max .

Answer 111

Ans: E

Question 112

Answer 112

answer is B

because Km for substrate B (NAD+) is decreasing, which means the affinity for substrate B is higher which only happens in random

Question 113

At substrate concentrations much lower than the enzyme concentration,
A) the rate of reaction is expected to be inversely proportional to substrate concentration.
B) the rate of reaction is expected to be directly proportional to substrate concentration.
C) first order enzyme kinetics are not observed.
D) the K M is lower.
E) the rate of reaction is independent of substrate concentration.

Answer 113

Ans: B

Question 114

Bisubstrate reactions in which all substrates must combine with the enzyme before a reaction can occur and products be released are known as __________

Answer 114

single displacement /aka sequential

In such reactions, the group being transferred, X, is directly passed from A (=P—X) to B, yielding P and Q (=B—X).

Question 115

in group transfer reactions, A= ? and Q = ?

Answer 115

A = P–X

Q = B–X

Question 116

Many NAD+- and NADP+-requiring dehydrogenases follow an Ordered bisubstrate mechanism in which the coenzyme is the _____ substrate

Answer 116

leading

Question 117

Some dehydrogenases and kinases operate through__________ mechanisms (kinases are enzymes that transfer phosphoryl groups from ATP to other compounds)

Answer 117

Random bisubstrate

Question 118

Many enzymes, including trypsin (in which F is the acyl–enzyme intermediate; Section 11-5), transaminases, and some flavoenzymes, react with_________`

Answer 118

Ping Pong mechanisms.

Question 119

T/F:

Answer 119

zymogens have distorted active sites

Question 120

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 of its maximum velocity,

Option A: [S] would need to be ½ KM

Option B: [S] would need to be equal to KM

Option C: [S] would need to be ¾ KM

Option D: not enough information is given to make this calculation

Option E: [S] would need to be 3KM

Answer 120

Option E: [S] would need to be 3KM

Question 121

Parallel lines on a Lineweaver-Burk plot indicate

Pick more than one

Option A :decrease in Vmax.

Option B :an increase in KM.

Option C :decrease in KM.

Option D :uncompetitive inhibition.

Answer 121

A, C and D

Question 122

t/f: The effects of uncompetitive inhibition on Vmax are not reversed by increasing substrate concentration.

Answer 122

True

Question 123

T/F: All enzymes can be analyzed such that their reaction rates and their overall efficiencies can be quantitated.

Answer 123

false

Question 124

Which expression containing the free energy of activation (∆G‡) is proportional to the rate of a reaction?

Option A: e(-∆G‡ /RT)

Option B: -∆G‡ /RT

Option C: +∆G‡ /RT

Option D: ln(∆G‡ /RT)

Answer 124

Option A: e(-∆G‡ /RT)

Question 125

T/F: The double-reciprocal plots for irreversible inhibition resemble those for competitive inhibition.

Answer 125

False

Question 126

If ATP is a substrate for an enzyme, and ADP was found to be a more potent competitive inhibitor than adenosine for the reaction, which of the following would be a plausible explanation?

Option A: The phosphate groups of ADP must be important for binding.

Option B: Adenosine is too large to bind at the active site.

Option C: Adenosine must be binding outside of the active site.

Option D: The ribose ring must be important for binding.

Answer 126

Option A: The phosphate groups of ADP must be important for binding.

Question 127

Pyrimidine synthesis is regulated by feedback inhibition of ______by ______

Answer 127

of aspartate transcarbamoylase by CTP