enzyme kinetics

Question 1

S:

Answer 1

solute concentration

Question 1

V0:

Answer 1

initial rate (velocity) of reaction

Question 2

VMAX:

Answer 2

the maximum rate of reaction when all enzyme active sites are saturated with substrate velocity

Question 3

Km(Michaelis constant):

Answer 3

the substrate concentration that gives half maximal velocity

Question 4

Km = [S] at

Answer 4

0.5(VMAX)

Question 5

Low Km =

Answer 5

enzyme only needs a small amount of substrate to work at half max velocity (+ reverse)

Question 6

Km =

Answer 6

(k-1+k2)k1

Question 7

k1 =

Answer 7

forward rate for enzyme association with substrate

Question 8

k-1 =

Answer 8

backwards rate for enzyme dissociation with substrate

Question 9

k2 =

Answer 9

forward rate of enzyme conversion from energy to product

Question 10

VMAX =

Answer 10

intersection of line with Y axis

Question 11

KM =

Answer 11

intersection of line with X axis

Question 12

competitive reversible inhibition

Answer 12

• Binds to active site
• VMAX stays the same
• Km varies

Question 13

non-competitive reversible inhibition

Answer 13

• Binds to allosteric site
• Km stays the same
• VMAX varies

Question 14

Irreversible inhibition

Answer 14

Non-competitive - usually involves formation/breakage of covalent bonds in the enzyme complex

Question 15

Enzyme regulation

Answer 15

Allosteric control - inhibition of rate limiting enzymes by end products
- Sigmoidal curve

Question 16

what is an example of enzyme regulation

Answer 16

binding of oxygen to haemoglobin