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taurus [48]
3 years ago
6

This is a reaction going on in your muscle cells right this very minute:The enzyme triose phosphate isomerase catalyzes this rea

ction 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?
Chemistry
1 answer:
Arisa [49]3 years ago
5 0

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.

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