Enzyme kinetics #3

Question 1

Kcat values can provides an ideas of…

Answer 1

the rate constant during the slowest step of catalysis

Question 2

Steady state kinetics

Answer 2

~ use low [E]

~ substrate consumption is slow = works in conventional practices

Question 3

Pre-steady state kinetics

Answer 3

a.k.a. burst

~ allows individual rate constants for reaction steps within an enzyme-catalysed reaction to be determined

Question 4

Name three methods used to measuring pre-steady state kinetics

Answer 4

~ temperature jump
~ stop-flow
~ quench-flow

Question 5

For enzyme-catalysed reactions with multiple reaction intermediates, the rate constants for the formation of those intermediates can be followed in real time IF:

Answer 5

1) catalytic cycle is slow enough

2) reaction intermediates have different spectroscopic properties

Question 6

Basic formulas for reaction with multiple intermediates

Answer 6

E + S ES EI1 –> EI2 –> P + E

EI here means intermediate

Question 7

Temperature jump

Answer 7

after a temp. jump, a new stay state will be reaches and the reaction is followed until a new equilibrium is reached

Question 8

Isoenzyme

Answer 8

~ forms of an enzyme that arise from genetically determined differences in a.a. sequences
~ categorised due to their migration patterns during electrophoresis

Question 9

Lactate dehydrogenase

Answer 9

pyruvate + NADH L-lactate + NAD+
~ isoenzyme
~ homolactic fermentation (gluconeogenesis)

Question 10

Types of lactate dehydrogenase

Answer 10

~ 5 types (LD1-5)
~ all have different combinations of M (muscle) and H (heart) subunits
~ expression lvls vary in diff tissues

Question 11

LD1

Answer 11

~ low Km (high affinity) for substrate
~ pyruvate = allosteric inhibitor
~ 4 H subunits

Question 12

LD5

Answer 12

~ high Km (low affinity) for substrate
~ pyruvate = not an allosteric inhibitor
~ 4 M subunits

Question 13

Lactate dehydrogenase + heart (5)

Answer 13

LD1
~ heart = aerobic = constant energy demand
~ high affinity = lactate efficiently scavenged + converted
~ allosteric inhibition = efficient regulation
~ use of lactose spare blood-sugar lvls for other organs

Question 14

Lactate dehydrogenase + skeletal muscles

Answer 14

LD5
~ lactate produced during muscle contraction - needed to re-generate NAD+
~ high [pyruvate] = no need for LD5 to have high affinity
~ allosteric inhibition would be BAD as need to generate NAD+

Question 15

Lactate dehydrogenase + liver

Answer 15

LD5
~ lactate required for gluconeogenesis
~ pyruvate abundant + allosteric inhibition not useful

Question 16

Glucokinase (5)

Answer 16

~ isoenzyme of hexokinase
~ found in liver
~ high Km for glucose (low affinity)
~ inulin = postive regulation
~ works mainly on glucose

Question 17

Hexokinase (5)

Answer 17

~ isoenzyme of glucokinase
~ found in muscles 
~ low Km (high affinity)
~ inhibition by G6P
~ works on all hexoses

Question 18

Significance of glucokinase

Answer 18

~ glucose transported in liver after meals
~ to facilitate steady and efficient metabolism, enzyme needed that gets more active as conc increases
~ therefore not inhibited by product (G6P) formation