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An enzyme has a $V_{max}$ of 50 $mu$mol product formed $(min * ext{mg protein})^{-1}$ and a $K_m$ of 10 $mu$M for the substrate. When a reaction mixture contains the enzyme and a 5$mu$M substrate, which of the following percentages of $V_{max}$ will be closest to the initial reaction rate ($V_o$)

a) 5% b) 15% c) 33% d) 50% e) 66%

Can someone explain this to me; I know michaelis-menton but I don't understand the solution to this or how to find it.

I don't think you understand MM kinetics.

In your case the $V_{max}$ value depends on the amount of the protein. If we assume *1mg* protein, then $V_{max} = 50frac{mu{M}}{min}$.

According to MM kinetics:

$V = V_{max}.frac{S}{K_M+S}$

so

$V_0 = V_{max}.frac{S_0}{K_M+S_0} = V_{max}.frac{5mu{M}}{10mu{M}+5mu{M}} = V_{max}.frac{1}{3}$

or in other terms

$frac{V_0}{V_{max}} = frac{1}{3} = 33.3\%$