Respuesta :
Explanation:
The given reaction is as follows.
[tex]E + S \rightleftharpoons ES \xrightarrow[]{k_{2}} E + P[/tex]
Here, [E] = triose phosphate isomerase = 0.1 [tex] nm = 0.1 \times 10^{-9}m[/tex]
[S] = Dihydroxy acetone phosphate = 5 [tex]\mu m = 5 \times 10^{-6}m[/tex]
[P] = Glyceraldehyde-3-phosphate = 2 [tex]\mu m = 2 \times 10^{-6}m[/tex]
Therefore, velocity of the reaction will be as follows.
v = [tex]\frac{d[P]}{dt}[/tex] = [tex]\frac{K_{2}[E][S]}{K_{M} + [S]}[/tex]
where, [tex]K_{M}[/tex] = Michaelic menten constant = [tex]\frac{K_{1} + K_{2}}{K_{1}}[/tex]
v = [tex]\frac{900 \times 0.1 \times 10^{-9}m \times 5 \times 10^{-6}m}{10^{-5} + 5 \times 10^{-6}}[/tex]
= [tex]30 \times 10^{-9} m[/tex]
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.
