Lecture 9: enzyme kinetics 1

Question 1

Kcat (2)

Answer 1

• conversion step of substrate to product

- rate limiting step = slower than k1/ k-1 because requires more energy to change structure of the substrate

Question 2

km

Answer 2

enzymes affinity/ dissociation of ES

Question 3

low km means what for efficiency? S binds (tightly/ loosely) to E?

Answer 3

low km = higher efficiency so S binds tightly to E [high affinity for substrate]

Question 4

high affinity for substrate means what for concentration?

Answer 4

higher affinity for substrate means smaller concentration of substrate needed to reach 1/2Vmax and vise versa

Question 5

kcat turnover #

Answer 5

of molecules of S turned over into product/ second/ molecule of enzyme [turnover #]

Question 6

rate order of kcat/km in the rxn of free E and free S

Answer 6

behaves as 2nd order rate in the rxn of free E and S

Question 7

higher kcat/ km means what for efficiency?

Answer 7

more efficient reaction

Question 8

kcat increases or km decreases means what for reaction?

Answer 8

reaction becomes faster

Question 9

adding more enzyme does what to Vmax?

Answer 9

gives a new Vmax

Question 10

kinetics of competitive inhibition

Answer 10

higher Km; constant Vmax = need more substrate to get to 1/2Vmax

Question 11

kinetics of noncompetitive inhibition

Answer 11

constant Km; Vmax permanently affected [no amount of substrate will outcompete]