Enzyme Kinetics

Question 1

Vo

Answer 1

The number of moles of product formed per second when the reaction is just beginning

Question 2

Preconditions:

Answer 2

1. At Vo [P]=0

2. [S]»[E]

Question 3

Vo=

Answer 3

Vmax x [S] / [S] + KM

Question 4

The Michaelis constant

Answer 4

KM is a measure of the affinity of the enzyme for its substrate

Question 5

A low KM means

Answer 5

A high affinity of enzyme for substrate

Question 6

Vmax is not

Answer 6

A constant

Question 7

Catalytic constant =

Answer 7

kcat= Vmax/[Eo]

Question 8

Kcat is a

Answer 8

turnover number

Question 9

The specificity constant

Answer 9

kcat/KM -> describes catalytic efficiency of enzymes

Question 10

High specificity constant

Answer 10

Better substrate for an enzyme

Question 11

The Line-Weaver Burk Plot

Answer 11

1/Vo = KM/Vmax x 1/[S] + 1/Vmax

Question 12

Reversible inhibition can be

Answer 12

Competitive or non-competitive

Question 13

Irreversible inhibition can be

Answer 13

Non-competitive or uncompetitive

Question 14

In competitive inhibition

Answer 14

KM increased, Vmax unaffected

Question 15

In uncompetitive inhibition

Answer 15

KM reduced, Vmax reduced

Question 16

In non-competitive inhibition

Answer 16

KM unaffected, Vmax reduced

Question 17

Slope in Line-Weaver

Answer 17

KM/Vmax

Question 18

Line-Weaver Burk Plot crosses Y axis at

Answer 18

1/Vmax

Question 19

Line-Weaver Burk Plot crosses X axis at

Answer 19

-1/KM

Question 20

How does excess substrate affect competitive inhibition?

Answer 20

Inhibition relieved

Question 21

Differences between significance of competitive and non-competitive inhibition?

Answer 21

Competitive - drug action

Non-competitive - toxicological