Question

In the red blood cell, glucose is transported into the cell against its concentration gradient. The energy for this transport is supplied by the hydrolysis of ATP: ATP+H2O→ADP+P (Δ rG=−31.0kJ/mol) Assume that the overall transport reaction is 100% efficient and given by ATP+H2O+2glucose( out )→2glucose( in )+ADP+P iy (a) At 25∘C, under conditions where [ATP], [ADP], and [ Pi] are held constant at 0.01M by cell metabolism, find the maximum value of [glucose(in)] ÷ [glucose(out)]. Assume all activity coefficients are equal to 1 . (b) If the stoichiometry of transport were 1 mol of glucose transported per mole of ATP hydrolyzed, what would be the maximum concentration gradient of glucose under the same conditions as part (a)?

Answer 1

Explanation:(a) To find the maximum value of [glucose(in)] ÷ [glucose(out)] under the given conditions, we can use the concept of equilibrium and the principles of thermodynamics.

Since the transport reaction is 100% efficient, it can be assumed that the transport process is reversible and at equilibrium. At equilibrium, the free energy change (ΔrG) for the overall reaction is zero.

The overall reaction can be represented as:

ATP + H2O + 2glucose(out) ⇌ 2glucose(in) + ADP + Piy

Given that ΔrG = -31.0 kJ/mol for ATP hydrolysis, we can use the equation:

ΔrG = ΣνΔG∘ (products) - ΣνΔG∘ (reactants)

where ν represents the stoichiometric coefficient of each species and ΔG∘ represents the standard free energy change.

Considering the stoichiometry of the reaction, the equation becomes:

0 = 2ΔG∘ (glucose(in)) + ΔG∘ (ADP) + ΔG∘ (Piy) - ΔG∘ (ATP) - ΔG∘ (H2O) - 2ΔG∘ (glucose(out))

Given that ΔG∘ (ADP) = ΔG∘ (Piy) = ΔG∘ (ATP) = 0 (since they are held constant), and ΔG∘ (H2O) = 0 (pure water), we have:

0 = 2ΔG∘ (glucose(in)) - 2ΔG∘ (glucose(out))

Substituting the value ΔG∘ = -31.0 kJ/mol, we get:

0 = 2(-31.0 kJ/mol) - 2(-31.0 kJ/mol)

Simplifying the equation gives:

0 = -62.0 kJ/mol + 62.0 kJ/mol

Therefore, the equation holds true at equilibrium, and there is no net change in free energy.

This implies that the maximum value of [glucose(in)] ÷ [glucose(out)] is 1:1, meaning the concentrations of glucose inside and outside the cell would be equal.

(b) If the stoichiometry of transport were 1 mol of glucose transported per mole of ATP hydrolyzed, the maximum concentration gradient of glucose can be calculated under the same conditions.

Using the same equation as in part (a), but considering the stoichiometry of 1 mol of glucose transported, we have:

0 = ΔG∘ (glucose(in)) - ΔG∘ (glucose(out))

Substituting the value ΔG∘ = -31.0 kJ/mol, we get:

0 = -31.0 kJ/mol - ΔG∘ (glucose(out))

Rearranging the equation, we find:

ΔG∘ (glucose(out)) = -31.0 kJ/mol

Therefore, the maximum concentration gradient of glucose under these conditions would be 31.0 kJ/mol.