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 + Pi at 298K

Assume the overall transport reaction is:

ATP + H2O + 2glucose(out) → ADP + Pi + 2glucose(in).

(a) At 37 °C, under conditions where [ATP], [ADP], and [Pi] are each held constant at 100 mM by cell metabolism, what is the maximum value of [glucose(in)]/[glucose(out)]? You can assume the activity coefficients of all species are 1.
(b) In an actual cell, the glucose inside the cell may exhibit non-ideal behavior. How would this affect the activity coefficient? Would this increase or decrease the maximum concentration gradient obtainable?

Question

10-03-2020
Chemistry

contestada

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 + Pi at 298K

Assume the overall transport reaction is:

ATP + H2O + 2glucose(out) → ADP + Pi + 2glucose(in).

(a) At 37 °C, under conditions where [ATP], [ADP], and [Pi] are each held constant at 100 mM by cell metabolism, what is the maximum value of [glucose(in)]/[glucose(out)]? You can assume the activity coefficients of all species are 1.
(b) In an actual cell, the glucose inside the cell may exhibit non-ideal behavior. How would this affect the activity coefficient? Would this increase or decrease the maximum concentration gradient obtainable?

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-03-21T12:45:00+01:00

Answer:

a) the maximum value of [tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] = 4,200

b) Due to non ideal behavior; In an actual cell, the glucose inside the cell may have an activity coefficient much less than 1 .

Since, [tex]\gamma <1[/tex] , we conclude that the [tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] ratio will be bigger which implies that the maximum concentration gradient obtainable will also increase.

Explanation:

The hydrolysis of ATP can be written as follows:

[tex]ATP + H_2O -----> ADP + P_i[/tex] [tex](\delta G = - 31.0 kJ)[/tex]

Free energy available from ATP hydrolysis can be expressed as:

[tex]\delta G = \delta G^0 +RTIn \frac{[ADP][P_i]}{[ATP]}[/tex]

Given that:

[ATP], [ADP], and [Pi] are each held constant at 100 mM = [tex]1.0 *10^{-2}M[/tex]

[tex]\delta G^0 = -31.0 kJ/mol[/tex]

R(rate constant) = 8.314 J/mol

Temperature (T) = 37 °C = (37+273)K = 310 K

replacing our value into the above equation; we have:

[tex]\delta G = -31.0kJ/mol(8.314 J/mol) (310K) In \frac{(1.0*10^{-2})(1.0*10^{-2})}{(1.0*10^{-2})}[/tex]

[tex]\delta G = -31.0kJ/mol+(2.58 kJ/mol) In{(1.0*10^{-2})}[/tex]

[tex]\delta G = -31.0kJ/mol-11.9 kJ/mol[/tex]

[tex]\delta G = -42.9 kJ/mol[/tex]

We therefore conclude that; [tex]-42.9 kJ/mol[/tex] is the energy available for the transport of glucose.

However, if the glucose transport energy is equal to free energy available out of ATP hydrolysis;

Then: [tex]\delta G_{ATP} = - \delta G_{glucose}[/tex]

[tex]\delta G_{ATP} = - \delta G_{glucose}[/tex]

As Such ; +42.9 kJ/mole = - [tex]\delta G_{glucose}[/tex]

From the question; we Assume that the activity coefficients of all species = 1

Thus K= 1

[tex]\delta G ^0 = 0[/tex] (since no energy input at equilibrium)

[tex]\delta G = \delta G^0 + RT In \frac{[glucose(in)]^2}{[glucose_{(out)}]^2}[/tex]

Making [tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] the subject of the formula and taking the square of the exponential; we have:

[tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] [tex]=[/tex] [tex]\sqrt{e^\frac{42,900j/mol}{(8.314J/mol)(310)} }[/tex]

[tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] [tex]=[/tex] [tex]\sqrt{e^{16.65}}[/tex]

[tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] [tex]=[/tex] 4125.74

[tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] = 4,200 (to the nearest thousand )

Thus, the maximum value of [tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] = 4,200

b) Due to non ideal behavior; In an actual cell, the glucose inside the cell may have an activity coefficient much less than 1 .

The expression for glucose transport reaction can be written as:

[tex]\delta G = + 42.9 kJ/mol[/tex]

= [tex]0+RT In \frac{[glucose (in)] \gamma^2 in}{[glucose{(out)}]^2}[/tex]

[tex]\frac{[glucose(in)] \gamma^2 in}{[glucose_{(out)}]}[/tex] = [tex]\sqrt{e^\frac{42,900j/mol}{(8.314J/mol)(310)} }[/tex]

[tex]\frac{[glucose(in)] \gamma^2 in}{[glucose_{(out)}]}[/tex] = 4200

[tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] = [tex]\frac{4,200}{\gamma in}[/tex]

From the foregoing; we can conclude that [tex]\gamma <1[/tex] , and as such; the [tex]\frac{[glucose(in)]}{[glucose_{(out)}]}[/tex] ratio will be bigger which implies that the maximum concentration gradient obtainable will also increase.