MCM 2-29 Enzymes as Biological Catalysts

Question 1

enzymes are biological catalysts that perform what 3 functions?

Answer 1

carry out almost all chemistry required by living systems

permit a wide range of chemical reactions in a narrow set of conditions

allow rapid, efficient adjustments to environmental conditions.

Question 2

enzymes work by..

Answer 2

binding a substrate or substrates through an induced fit; the active site isn’t perfectly formed to fit the substrate, but is close enough so that it can adjust through conformation changes. They lower the activation energy of a reaction, increasing the reaction rate. This will not affect reaction spontaneity (the change in Gibbs free energy, delta G).

Question 3

how do enzymes effect reaction spontaneity?

Answer 3

This will not affect reaction spontaneity (the change in Gibbs free energy, delta G).

Question 4

reactions with

Delta G O

Answer 4

if delta g is negative, can occur spontaneously
if delta g is 0, is in equilibrium between the products and the starting material
if delta g is positive, it will take energy to push the reaction forward

Question 5

how do enzymes effect delta G and Ea

Answer 5

enzymes do not change delta G, they lower the activation energy Ea to make reaction go faster

Question 6

6 common features of active sites

Answer 6

1. occupy small part of enzyme
2. 3d structure
3. bind through multiple weak non-covalent interactions (van der waals, electrostatic, hydrophobic, etc.)
4. water is excluded, active sites in clefts of proteins
5. highly specific binding of substrate
6. can include non-protein prosthetic groups and cofactors

Question 7

E+S equation

Answer 7

E+S —-k1—> ES —k3—> P+ S

Question 8

michaelis menton equation

Answer 8

v = Vmax * [S]
[S]+Km

Vmax = k3 * [E]t

Km = (k2+k3)/k1

Question 9

k3 =

Answer 9

kcat

Question 10

Km =

Answer 10

the substrate concentration [S] that gives 1/2 Vmax

Question 11

high Km vs Low Km

Answer 11

High Km = high dissociation constant = low affinity

Low Km = low dissociation constant = high affinity

Question 12

Km values of hexokinase and glucokinase

Answer 12

Hexokinase has a low Km = founds in all cells, high affinity, saturated quickly, wants to find glucose.

Glucokinase = at low glucose concentration does not grab much, does not want to steal from cells that need it. but after big meal and high level of glucose, rate increases to soak up excess.

Question 13

Vmax =

Answer 13

max velocity of a specific enzyme concentration [E]

Question 14

kcat =

Answer 14

constant that is independant of enzyme concentration
kcat = Vmax/Et

Et = total enzyme concentration

Question 15

what values change if you have a mutated allele for one enzyme?

Answer 15

Km will stay the same because it has to do with the enzyme itself. Kcat will stay the same

Vmax will change

Question 16

competitive inhibitors bind to?

overcome by?

what increases? what does not change?

graph looks like?

Answer 16

competitive - bind to same sites as substrate. can be overcome by increasing substrate. Km appears to increase, Vmax does not change

intersections at the Y-axis

Question 17

non-competitive inhibitors bind to?

overcome by?

what increases? what does not change?

graph looks like?

Answer 17

site on enzyme other than active site

cannot be overcome by increasing [S]

vmax appears to decrease, Km does not change

intersecting at X axis

Question 18

Ki?

Answer 18

dissociation constant at equilibrum

Question 19

Which of the following HIV-1 protease mutations is MOST LIKELY found in a strain of HIV-1 that has become resistant to indinavir:

A. A mutation changing the catalytic site aspartate to asparagine.

B. Three mutations changing amino acids in the active site cleft as follows: valine at position 32 to lysine, methionine at position 46 to asparagine, isoleucine at position 47 to histidine.

C. Three mutations changing amino acids in the active site cleft as follows: valine at position 32 to isoleucine, methionine at position 46 to leucine, isoleucine at position 47 to valine.

D. A single mutation that changes the substrate specificity of the protease.

Answer 19

C. Three mutations changing amino acids in the active site cleft as follows: valine at position 32 to isoleucine, methionine at position 46 to leucine, isoleucine at position 47 to valine.

Key - the changes are phobic to phobic, you arent changing the phobicity.

This was in the practice problems I provided in your packet. C describes a typical resistance mutant; there are multiple changes of hydrophobic to hydrophobic amino acids which collectively increase the Ki, but still allow substrate to bind to the enzyme and be converted to product with reasonable efficiency. A and D are incorrect because they would not allow the normal substrate to be recognized and processed. B is incorrect both because these changes would prevent recognition of the hydrophobic substrate sequences that must be processed and because they might disturb folding of the protease.

Question 20

Which of the following will change if the concentration of enzyme added to a reaction mixture is tripled? 
Km
kcat
Vmax
all
none

Answer 20

Vmax. Km and kcat are independent of the total enzyme concentration, but Vmax is not. Therefore, Vmax will triple if the [E]t is tripled.

Vmax = Kcat * Et

Question 21

Which of the following enzymatic properties is MOST LIKELY to be characteristic of the HIV protease in a resistant strain of HIV that arose in a patient after prolonged treatment with indinavir?

Answer 21

A Kcat/km comparable to the WT and a higher Ki for the drug than the WT

The mutant protease must be able to process its substrate polyprotein with efficiency comparable to wild-type (i.e. with comparable kcat/Km), but not bind the inhibitor (increase Ki).

Question 22

BEST describes the reason that drugs (enzyme inhibitors) are often designed to mimic the transition state of the enzymatic reaction?

Answer 22

Many enzymes bind the transition state more tightly than the substrate. Inhibitors that mimic the transition state may, therefore, have a lower Ki for the enzyme than inhibitors that mimic the substrate.

Because enzymes often show “preferential binding of the transition state” of a substrate, inhibitors that mimic the transition state often bind very tightly (i.e. have a lower Ki). The transition state of a substrate is a transient, not a stable state, so A and D are wrong. The comparison of Ki and Km (choice C) is not really relevant.

Question 23

It is possible for a mutant HIV protease with more than one amino acid change to cleave a substrate peptide more efficiently than a mutant HIV protease with only one amino acid change

Answer 23

true

Question 24

accepting a proton makes you a….

donating a proton makes you a…

Answer 24

general base

general acid

Question 25

There are five different isozymes of lactate dehydrogenase (LDH). Which of the following properties is LEAST LIKELY to differ between the isozymes?

Answer 25

The chemical reaction catalyzed by the different isozymes

By definition, isozymes catalyze the same reaction, but all of the other characteristics may vary.

Question 26

A reaction for which the ΔG is negative will occur more rapidly than a reaction for which the ΔG is positive. TF

Answer 26

False - ΔG determines spontaneity, but does not determine the rate (rate depends on activation energy).

Question 27

In a set of blood tests, you observe that a patient exhibits elevated levels of creatine kinase (CK) activity. Assuming (however unlikely) that it is difficult for you to get a good history or description of other symptoms from this person, what additional tests could help you to distinguish damage to heart muscle (myocardial infarction) from a disease causing breakdown of skeletal muscle (such as muscular dystrophy)?

Answers:

A. A determination of the level of lactate dehydrogenase (LDH) activity in blood.

B. A test to determine which CK isoform(s) are present in the blood.

C. A repeat determination of the level of creatine kinase activity in several days.

D. All of the above could be helpful.

E. None of the above would be helpful.

Answer 27

All of the tests could help distinguish MI from muscle disease.

Question 28

The Km of an enzymatic reaction depends on the concentration of enzyme present in the reaction mixture. TF

Answer 28

false - Vmax depends on enzyme concentration, but Km does not.

Question 29

Tissue-specific isozymes carry out the same enzymatic reaction, so they must be encoded by the same gene and have the same amino acid sequence. TF

Answer 29

False - Isozymes do carry out the same reaction, but they have different sequences and different properties.

Question 30

Which of the following is MOST likely to differ between different serine proteases?

Answer 30

The substrate specificity of the proteases.

Question 31

why HIV-positive patients who become resistant to one protease inhibitor after prolonged therapy are often resistant to other protease inhibitors?

Answer 31

Most of the protease inhibitors have somewhat similar structures because they were designed to fit the active site of the HIV protease. Therefore, mutations that affect binding to one inhibitor often affect binding of other inhibitors.

Question 32

The activity (reaction velocity (v)) for an enzyme is directly proportional to the total enzyme concentration under which of the following conditions? (Assume that the enzyme has a single substrate and product.)

Answer 32

A. The substrate concentration is saturating.

Vmax is proportional to total enzyme concentration, and v = Vmax when substrate concentration is saturating (A is correct). C and D both refer to the rapidly rising part of the v vs. [S] curve where velocity is determined by substrate, not enzyme concentration. kcat and Km aren’t really comparable so B is wrong.

Question 33

The Ki of an enzyme for an inhibitor will depend on the concentration of inhibitor present in the enzymatic reaction.

Answer 33

false - Ki is a constant that does not depend on inhibitor concentration

Question 34

A genetic disease that caused a person to express very low levels of tissue plasminogen activator would be MOST LIKELY to hahttps://www.brainscape.com/decks/5398468/cards/quick_new_cardve which of the following effects?

Answer 34

TPA is part of the zymogen activation cascade that reverses clot formation, so in its absence thrombosis would probably occur).

Question 35

You have an HIV-1 positive patient who has not responded to treatment with an HIV-1 protease inhibitor and continues to have high levels of virus in his blood. However, when the lab sequences viral DNA from his blood, they find no mutations in the HIV-1 protease that can account for the apparent resistance to the inhibitor. Which of the following is MOST LIKELY to be TRUE?

A. The lab must have made a mistake, because he must have a protease-resistant strain of HIV-1 for the treatment to fail.

B. He has a small number of mutant viruses that are resistant to the protease, but they didn’t show up in the lab isolation because they don’t replicate as efficiently as the wild-type HIV-1.

C. Treatment is failing for reasons other than the development of resistance mutations, such as too rapid clearance of the inhibitor from the body or poor compliance by the patient.

D. The patient was infected by someone who had already developed resistance to HIV-1 protease inhibitors.

Answer 35

C. Treatment is failing for reasons other than the development of resistance mutations, such as too rapid clearance of the inhibitor from the body or poor compliance by the patient.

See Group E problems from CP2. C is correct-this patient must be failing to respond for reasons other than development of resistance mutations. As discussed in CP2, treatment can fail for reasons other than resistance (A is wrong). B is wrong because it couldn’t explain treatment failure (the predominant wild-type form of the virus should still have been inhibited). D is inconsistent with the lack of mutations.

Question 36

what role does water play in the catalytic triad?

Answer 36

Water is acting as a nucleophile and attacking the carbonyl group of the acyl enzyme intermediate formed by serine

his - general base aacepting proton from water

Question 37

Assume that isolated wild-type HIV-1 protease exhibits a Km of 200 mM, a kcat of 20 sec-1, and a Ki for the inhibitor indinavir of 0.1 nM. Choose the set of catalytic constants below that is MOST LIKELY to correspond to those of a mutant HIV-1 protease isolated from a patient who has become resistant to indinavir.

A. Km = 100 mM, kcat = 10 sec-1, Ki = 10 nM

B. Km = 1000 mM, kcat = 10 sec-1, Ki = 10 nM

C. Km = 100 mM, kcat = 10 sec-1, Ki = 0.01 nM

D. Km = 100 mM, kcat = 1 sec-1, Ki = 0.01 nM

E. Km = 200 mM, kcat = 0.2 sec-1, Ki = 10 nM

Answer 37

A.

Resistant strain must have higher Ki than WT, and have efficiency (Kcat/Km) similar to WT

An HIV protease from a resistant strain MUST have Ki for the inhibitor that is greater than the wild-type (A, B, or E), and an efficiency (kcat/Km) comparable to wild-type. kcat/Km for wt is 20/200=0.1, so A, which would have a kcat/Km of 10/100 would be correct.