Question

4. Basic concepts of Michaelis-Menten kinetics. The Michaelis-Menten equation is expression of the relationship between the initial velocity, Vo, of an enzymatic reaction and substrate concentration, [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation: [S] <<Km; [S] = Km; [S] >> Km. Match each condition with the statement(s) that describe it. TV, Vmox[S] Vo =Vmax m . V Vo - Vmax [S] Km +[S] V. (um/min) max [S] (mm) (a) Doubling [S] will almost double the rate. (b) Half of the active sites are occupied by substrate. (c) All active sites are occupied by substrate. (d) Doubling (S] will have little effect on the rate. (e) Less than 10% of the active sites are occupied by substrate. (f) This condition will result in the highest rate. () [ES] is much lower than [Efree). (h) Rate is directly proportional to [S]. (1) [Efree) is equal to [ES].

Answer 1

a. [S] <<< Km
If the [S] is far less than Km, doubling the [S] would result in a similar increase in reaction rate.

b. [S] = Km
Km = The substrate concentration at which the reaction rate is Vmax/2
At Km, half od the enzymes' active sites are occupied.

c. [S] <<< Km
At elevated [S], all the active sites on the enzyme are occupied.

d. [S] <<< Km
At elevated [S], doubling the [S] will have the least effect on the rate

e. [S] >>> Km
At very low [S], very few (<10%) of the active sites are occupied.

f. [S] >>> Km
At elevated [S], all the active sites on the enzyme are occupied.
This condition results in the highest reaction rate.

g. [S] <<< Km
At very low [S], very few active sites on the enzyme are occupied.
i.e. The [ES] complex is much lower than the free [E]

h. [S] <<< Km
At low [S], the rate is directly proportional to [S]

i. [S] = Km
At Km, [E_Free] = [ES]

Please provide a POSITIVE RATING. Thank you