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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\%\$

## Watch the video: Kinetics: Initial Rates and Integrated Rate Laws (July 2022).

1. Daryll

I know exactly that this is the error.

2. Karayan

What great words