Question

1) For an enzyme that displays Michaelis-Menten kinetics, what is the initial velocity as a function of Vmax when:

i) [S] = Km

ii) [S] = 0.1 Km

iii) [S] = 50Km

2) An enzyme (follows Michaelis-Mentin Kinetics) has Km = 0.5 M. The initial velocity is 0.2 Mmin-1 at substrate concentration of 50 M. What is the Vmax? What is the initial velocity when a) [S] = 2 M and b) [S] = 0.5 M?

3) What will be the initial velocity (vo) for an enzyme that has Km = 2.5 [S]? Your answer will be a fraction of Vmax.

Answer 1

V = VmaxKm/Km+ [S]

if [S] = Vmax , then V = Vmax/2

if [S] 0.1 Km, then V = Vmax * Km/Km(1+0.1). So, V = 0.91 Vmax.

if [S] = 50kM, then V = Vmax * Km /Km + 50Km . So, V = V = 0.02 Vmax.

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2(a)

V = Vmax * Km/ Km + [S]

V = 0.2 M /min, Km = 0.5 M, [S] = 50M

Vmax = (0.2 M /min(0.5 M + 50M ))/0.5M =20.2 M/ min

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3. V = Vmax Km/Km + [S]

= Vmax 2.5[S]/ (1+2.5)[S] = 0.71 Vmax