Enzyme Kinetics and Inhibition

Question 1

This parameter is the concentration of substrate at 1/2 the Vmax

Answer 1

Km

Question 2

The higher the Km, the ______ the affinity

Answer 2

lower

Question 3

True or False: You can change Vmax by changing substrate concentration

Answer 3

False

Question 4

What is the Michaelis-Menten equation?

Answer 4

V=Vmax [S] / Km + [S]

Question 5

What does the x intercept on a line-weaver burke plot tell you?

Answer 5

The X-intercept is = -1/Km

Question 6

What does the y intercept on a line-weaver burke plot tell you?

Answer 6

The Y-intercept is = 1/Vmax

Question 7

The following Lineweaver-Burk plot depicts an enzyme-catalyzed reaction (A) and the same reaction in the presence of an inhibitor of the enzyme (B). What kind of inhibition is it showing?

Answer 7

Uncompetetive. The inhibitor decreases Vmax and decreases Km.

Question 8

What happens to Km if a mixed inhibitor binds preferrably to the enzyme and not the enzyme-substrate complex?

Answer 8

When it binds preferrably to the enzyme, it increases Km.

Question 9

What kind of inhibition is shown in the following plot:

Answer 9

Competetive inhibition, Km is changed but Vmax is still the same.

Question 10

What kind of inhibition is shown in the following plot. Note that the red line is the unihibited plot:

Answer 10

Noncompetetive. The X intercept is the same but the y intercept has changed meaning Km is unchanged but Vmax is decreased.

Question 11

What kind of inhibition is shown in the following plot:

Answer 11

Competetive. Y intercept is unchanged meaning the vmax is unchanged. X intercept has increased in the presence of the inhibitor indicating a decrease in affinity for the substrate because it is competing.

Question 12

What type of inhibition is exhibited in the graph below if you start with blue?

Answer 12

Since the x intercept is brought in, that means that Km is increased. Vmax is decreased since the y intercept moved up on the graph. Think:

Competetive: Vmax unchanged, Km increased

NonComp: Vmax decreased, Km unchaged

Uncompetetive: Vmax decreased, Km decreased

Mixed inhibition is the only inhibition that could show this kind of plot

Question 13

Competetive inhibitors bind to the __________

Answer 13

active site

Question 14

Noncompetetive inhibitors bind to the _____________

Answer 14

allosteric site

Question 15

Uncompetetive inhibitors bind to the _____________

Answer 15

ES complex

Question 16

Mixed inhibitors bind to the ____________

Answer 16

E or ES complex with different affinities

Question 17

Mixed inhibitors always decrease _______ but they vary in how they change _______

Answer 17

always decrease Vmax, vary in how they change Km

Question 18

When will a mixed inhibitor increase Km?

Answer 18

When the inhibitor binds to the free enzyme (similar to competetive inhibitor)

Question 19

When will a mixed inhibitor decrease Km?

Answer 19

When it prefers to bind to the ES complex (so like an uncompetetive inhibitor)

Question 20

The rate of the carbonic anhydrase-catalyzed reaction is extremely fast (≈ 10^6 reactions per second) and can be performed in the forward or reverse directions. According to this info, which of the following properties of carbonic anhydrase are represented by the value 10^6?

I. Turnover number

II. [ET]

III. Vmax

IV. kcat

Answer 20

I and IV only

In enzyme kinetics, turnover number (aka kcat) is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration ([ET]). Turnover number can be calculated from the maximum reaction rate (Vmax) and enzyme concentration as follows (I and IV):

kcat = Vmax/[ET]

Question 21

What is Kcat?

Answer 21

Kcat is the turnover rate

kcat = Vmax/[ET]

Question 22

Which type of inhibitor does NOT alter the KM/Vmax ratio of an enzyme?

Answer 22

This Biochemistry question falls under the content category “Structure and function of proteins and their constituent amino acids.” The answer to this question is B because uncompetitive inhibitors do not alter the slope of the Lineweaver–Burk plot, which is equal KM/Vmax. It is a Scientific Reasoning and Problem Solving question because you must work with various models of enzyme inhibition to determine the correct answer.

Question 23

An enzyme is more effectively inhibited by uncompetitive inhibitors when:

the substrate concentration is decreased.

the substrate concentration is increased.

the inhibitor concentration is increased.

Answer 23

the substrate concentration is increased.

the inhibitor concentration is increased.

Question 24

What iscatalytic efficiency/

Answer 24

Kcat/Km

Question 25

Ternary Complex

Answer 25

A ternary complex can be a complex formed between two substrate molecules and an enzyme. This is seen in multi-substrate enzyme-catalyzed reactions where two substrates and two products can be formed. The ternary complex is an intermediate between the product formation in this type of enzyme-catalyzed reactions.

Question 26

Ping Pong Mechanism

Answer 26

Also known as a double displacement mechanism, this mechanism involves the binding of two substrates to an ezyme without forming a ternary complex (so no two substrates will bind at the same time)

Question 27

Ordered Mechanism

Answer 27

There is a specific order that the substrates bind

Question 28

Random Order Mechanism

Answer 28

Of two possible substrates, either can bind first