Question

The enzyme monoamine oxidase catalyzes the oxidation of amines to aldehydes. For tryptamine, Km for the enzyme is 4.0 x 10 ^-4 M and ${\mathrm{V}}_{\max }={\mathrm{k}}_2\left[{\mathrm{E}}_0\right]=1.6\times {10}^{-3}\mathrm{\mu M}/\min$${\mathrm{V}}_{\max }={\mathrm{k}}_2\left[{\mathrm{E}}_0\right]=1.6\times {10}^{-3}\mathrm{\mu M}/\min$at pH 8. Find the concentration of a solution of tryptamine that reacts at a rate of 0.12 μm/min in the presence of monoamine oxidase under the

above conditions. Assume that [tryptamine]$<<{\mathrm{K}}_{\mathrm{M}}$

Short answer

The concentration of a solution of tryptamine is 0.03 M.

Step-by-step solution

Given Information

The concentration of a solution of tryptamine under given conditions should be determined.

Km for the enzyme = 4.0 x 10 ^-4 M and${\mathrm{V}}_{\max }={\mathrm{k}}_2\left[{\mathrm{E}}_0\right]=1.6\times {10}^{-3}\mathrm{\mu M}/\min$${\mathrm{v}}_{\max }={\mathrm{k}}_2\left[{\mathrm{E}}_0\right]=1.6\times {10}^{-3}\mathrm{\mu M}/\min$

Formula to be used

The rate of the reaction is given by,

$\frac{\mathrm{d}\left[\mathrm{P}\right]}{\mathrm{dt}}=\frac{{\mathrm{k}}_2{\left[\mathrm{E}\right]}_0\left[\mathrm{S}\right]}{{\mathrm{k}}_{\mathrm{m}}+\left[\mathrm{S}\right]}$

[P] – product concentration

k₂– dissociation constant for enzyme substrate complex into products and free enzyme.

[E]₀– initial enzyme concentration

[S] – substrate concentration

Km – Michaelis constant

Calculation

Rate= $\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{{\mathrm{k}}_2{\left[\mathrm{E}\right]}_0\left[\mathrm{S}\right]}{{\mathrm{k}}_{\mathrm{m}}+\left[\mathrm{S}\right]}$

When , the equation is reduced to,

$$\begin{gathered} Rate=\frac{{\mathrm{k}}_2{\left[\mathrm{E}\right]}_0\left[tryp\right]}{{\mathrm{k}}_{\mathrm{m}}} \\ {v}_{max}={\mathrm{k}}_2{\left[\mathrm{E}\right]}_0 \\ \mathrm{Rate}=\frac{{\mathrm{v}}_{\max }\left[\mathrm{tryp}\right]}{{\mathrm{k}}_{\mathrm{m}}} \\ 0.12\times {10}^{-6}\mathrm{M}/\min =\frac{1.6\times {10}^{-9}\mathrm{M}/\min \times \left[\mathrm{tryp}\right]}{4\times {10}^{-4}\mathrm{M}} \\ \left[\mathrm{tryp}\right]=\frac{0.12\times {10}^{-6}\mathrm{M}/\min \times 4\times {10}^{-4}\mathrm{M}}{1.6\times {10}^{-9}\mathrm{M}/\min } \\ \left[\mathrm{tryp}\right]=0.03\mathrm{M} \end{gathered}$$