kinetics2

Question 1

Initial assumptions about michaelis-menton

Answer 1

• k-2 is irrelevant bc theres no product
• rate limiting step is k2 Vo = k2[ES]
• [S] is constant
• [ES] is constant steady state assumption

Question 2

Vo =

Answer 2

Vmax[S]/(Km + [S])

Question 3

1/Vo =

Answer 3

Km/Vmax[S] + 1/Vmax

Question 4

Km =

Answer 4

(k2 + k-1)/k1

Question 5

when k2

Answer 5

then Km = k-1/k1 which is the dissociation constant Kd so km would be a measure of the affinity of an enzyme for it’s substrate

Question 6

Kcat

Answer 6

turnover number, number of product molecules per second (gives us an idea of how quickly the enzyme is actually going)

Question 7

catalase

Answer 7

One of the fastest enzymes, very impressive turnover number, breaks down hydrogen peroxide(from oxygen), if you look at the reaction being catalyzed it gives you a clue as to why the kcat is so high

Question 8

when [S]

Answer 8

Vo = kcat/Km * [Et][S]

Question 9

kcat/Km is

Answer 9

a second order rate constant for the interaction of [Et] and [S]

Question 10

The upper limit for Kcat/km

Answer 10

imposed by how fast E and S can diffuse together
(10^8M-1S-1)
an enzyme close to this limit is diffusion controlled and has reached catalytic perfection
(catalytic efficiency can’t be greater than this or it would break the diffusion control limit

Question 11

arrhenius equation

Answer 11

Kcat = Aexp(-Ea/RT)

predicts that the enzyme rate doubles for every 10˚C increase in temperature until enzyme denatures

Question 12

ternary comples

Answer 12

when an enzyme binds two substrates