Question

This is a reaction going on in your muscle cells right this very minute:The enzyme triose phosphate isomerase catalyzes this reaction in the forward direction as part of the glycolyticpathway. It follows simple Michaelis-Menten kinetics:Typical cellular concentrations: triose phosphate isomerase = 0.1 nMdihydroxyacetone phosphate = 5 μM glyceraldehyde-3-phosphate = 2 μM48. Refer to Exhibit A. What is the actual velocity of the forward reaction under physiologic conditions if KM = 10 μM?

Answer 1

Explanation:

The given reaction is as follows.

       $E + S \rightleftharpoons ES \xrightarrow[]{k_{2}} E + P$

Here, [E] = triose phosphate isomerase = 0.1 $nm = 0.1 \times 10^{-9}m$

         [S] = Dihydroxy acetone phosphate = 5 $\mu m = 5 \times 10^{-6}m$

         [P] = Glyceraldehyde-3-phosphate = 2 $\mu m = 2 \times 10^{-6}m$

Therefore, velocity of the reaction will be as follows.

          v = $\frac{d[P]}{dt}$ = $\frac{K_{2}[E][S]}{K_{M} + [S]}$

where, $K_{M}$ = Michaelic menten constant = $\frac{K_{1} + K_{2}}{K_{1}}$

            v = $\frac{900 \times 0.1 \times 10^{-9}m \times 5 \times 10^{-6}m}{10^{-5} + 5 \times 10^{-6}}$

               = $30 \times 10^{-9} m$

or,            = 30 nm/s

Hence, we can conclude that the actual velocity of the forward reaction under physiologic conditions if KM = 10 μM is 30 nm/s.