Question

If glyceraldehyde-3-phosphate dehydrogenase in red blood cells is completely inhibited, which intermediates in glycolysis accumulate most? Please use the following equation and the provided information to explain your reasoning.

ΔG=ΔG°+RTln[C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ​
Fructose 1,6-bisphosphate → Glyceraldehyde 3-phosphate + Dihydroxyacetone phosphate The standard free energy for the above reaction is: ΔG°′=+23.9KJ/mol T°=298
R=8.314 J mol−1 K−1​

Answer 1

Inhibition of glyceraldehyde-3-phosphate dehydrogenase leads to the accumulation of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate in red blood cells. This results in the disruption of glycolysis, as the conversion to 1,3-bisphosphoglycerate is prevented.

Step-by-step explanation:

If glyceraldehyde-3-phosphate dehydrogenase is completely inhibited in red blood cells, the intermediates in glycolysis that would accumulate the most are glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. This inhibition would halt the conversion of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate, thereby causing an upstream accumulation of its precursors. According to the question's equation (ΔG=ΔG°'+RTln[CD]/[AB]), the positive value of ΔG°' indicates that the reaction is not favored in the forward direction under standard conditions.

However, the buildup of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate will alter the concentration ratio, leading to an increase in the reverse reaction's favorability until equilibrium is re-established. Given the enzyme's pivotal role in glycolysis, its inhibition also disrupts the production of ATP and NADH in erythrocytes, evidencing the importance of this enzyme in maintaining cellular energy balance.