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 equilibrium constant for the enzyme-catalyzed reaction?

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 equilibrium constant for the enzyme-catalyzed reaction?

Quizplus · Accepted Answer

The equilibrium constant is the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this case, the equilibrium constant is equal to the ratio of the concentration of the product (P) to the concentration of the reactant (S). From Exhibit 6A, we can see that the concentration of P at equilibrium is 0.9 M and the concentration of S at equilibrium is 1.0 M. Therefore, the equilibrium constant is 0.9.

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 $k_{1}=10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}, k_{-1}=100 \mathrm{~s}^{-1}, k_{2}=900 \mathrm{~s}^{-1}$