Question

Assuming that the value of the equilibrium constant for the aldolase reaction Keq=6.43*10^-5 at pH=7 a.) What will be the equilibrium concentration of dihydroxyacetone phosphate (DHAP) if 1 mM of fructose-1,6-biphosphate is added to a buffered solution (pH=7) of aldolase? b.) What percent of the original FBP concentration will be converted to DHAP? c.) If this experiment is repeated but with 10 mM of FBP what will be the equilibrium concentration of DHAP? d.) What % conversion does this represent? e.) if in part a the buffered solution also contains triose phosphate isomerase, what will be the equilibrium concentration of FBP?

Answer 1

I must tell you that I'm not a biochemist and I don't know the equilibrium reactions you need; however, two equations will solve your problem. The first one is the Henderson-Hasslebalch equation of
pH = pKa + log [(base)/(acid)]
and second the equilibrium constant expression of Ka = (??)*(??)/(??)