Exhibit 6A 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 glycolytic pathway. It follows simple Michaelis-Menten kinetics: Typical cellular concentrations: triose phosphate isomerase = 0.1 nM?dihydroxyacetone phosphate = 5 µM glyceraldehyde-3-phosphate = 2 µM

$k_{1}=10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}, k_{-1}=100 \mathrm{~s}^{-1}, k_{2}=900 \mathrm{~s}^{-1}$
-Refer to Exhibit 6A. What is the actual velocity of the forward reaction under physiologic conditions?

Question

Exhibit 6A 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 glycolytic pathway. It follows simple Michaelis-Menten kinetics: Typical cellular concentrations: triose phosphate isomerase = 0.1 nM?dihydroxyacetone phosphate = 5 µM glyceraldehyde-3-phosphate = 2 µM

$k_{1}=10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}, k_{-1}=100 \mathrm{~s}^{-1}, k_{2}=900 \mathrm{~s}^{-1}$
-Refer to Exhibit 6A. What is the actual velocity of the forward reaction under physiologic conditions?

Quizplus · Accepted Answer

Under physiologic conditions, the concentration of dihydroxyacetone phosphate is 5 µM and the concentration of glyceraldehyde-3-phosphate is 2 µM. Using Michaelis-Menten kinetics, the initial velocity (v0) for the forward reaction can be calculated using the following equation:

v0 = (Vmax[S])/(Km + [S])

where Vmax = k2[ET], [S] is the substrate concentration, and Km = (k-1 + k2)/k1.

Plugging in the given values, we get: 
Km = (100 + 900)/10^8 = 0.001
Vmax = k2[ET] = 900 x 0.0001 = 0.09
[S] = 2 µM

v0 = (0.09 x 2 x 10^-6)/(0.001 + 2 x 10^-6) = 30 nM/s

Therefore, the actual velocity of the forward reaction under physiologic conditions is 30 nM/s, which corresponds to choice D.

Exhibit 6A This Is a Reaction Going on in Your k1=108M−1 s−1,k−1=100 s−1,k2=900 s−1k_{1}=10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}, k_{-1}=100 \mathrm{~s}^{-1}, k_{2}=900 \mathrm{~s}^{-1}k1​=108M−1 s−1,k−1​=100 s−1,k2​=900 s−1

Exhibit 6A This Is a Reaction Going on in Your $k_{1}=10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}, k_{-1}=100 \mathrm{~s}^{-1}, k_{2}=900 \mathrm{~s}^{-1}$