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?

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 = (??)*(??)/(??)

Answer 2

a.) To find the equilibrium concentration of dihydroxyacetone phosphate (DHAP), we need to use the given equilibrium constant (Keq) and the initial concentration of fructose-1,6-biphosphate (FBP).

1. Start by writing the balanced equation for the aldolase reaction:

FBP ⇌ DHAP

2. Based on the equilibrium constant expression, we have:

Keq = [DHAP] / [FBP]

3. Substitute the given value of Keq into the equation:

6.43 * 10^-5 = [DHAP] / [FBP]

4. Note that the initial concentration of DHAP is zero since we haven't added any yet. Therefore, the concentration of DHAP at equilibrium will also be zero.

5. Rearrange the equation to solve for the equilibrium concentration of DHAP:

[DHAP] = Keq * [FBP]

Now, substitute the given initial concentration of fructose-1,6-biphosphate (FBP) to calculate the equilibrium concentration of DHAP.

b.) To find the percent conversion of fructose-1,6-biphosphate (FBP) to dihydroxyacetone phosphate (DHAP), we need to compare the equilibrium concentration of DHAP to the initial concentration of FBP.

1. Calculate the molar concentration of FBP that is converted to DHAP:

Conversion = [DHAP] / [FBP]

2. Convert the conversion to a percentage:

Percent Conversion = Conversion * 100

c.) Repeat the process for a 10 mM concentration of FBP.

1. Use the equation [DHAP] = Keq * [FBP], as explained in part a.

d.) Calculate the percent conversion for the 10 mM FBP experiment.

1. Follow the same procedure as in part b to find the percent conversion.

e.) To find the equilibrium concentration of FBP when a buffered solution containing triose phosphate isomerase is included:

1. Triose phosphate isomerase (TPI) catalyzes the reversible interconversion of dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP).

2. Balance the equation for this reaction:

DHAP ⇌ GAP

3. Use the equilibrium constant expression for this reaction (K_TPI):

K_TPI = [GAP] / [DHAP]

4. Substitute the given value of Keq and the equilibrium concentration of DHAP (which is zero, as discussed in part a):

6.43 * 10^-5 = [GAP] / 0

Since the equilibrium concentration of DHAP is zero, the equilibrium concentration of GAP will also be zero when triose phosphate isomerase is present in the buffered solution.