The Na+ -glucose symport system of intestinal epithelial cells couples the \"dow

Question

The Na+ -glucose symport system of intestinal epithelial cells couples the "downhill" transport of two Na+ ions into the cell to the "uphill" transport of glucose, pumping glucose into the cell against its concentration gradient. If the Na+ concentration outside the cell ([Na+]out) is 149 mM and that inside the cell ([Na+]in) is 23.0 mM, and the cell potential is -51.0 mV (inside negative), calculate the maximum ratio of [glucose]in to [glucose]out that could theoretically be produced if the energy coupling were 100% efficient. Assume the temperature is 37 degree C.

Explanation / Answer

Gchem = RTln(Na+ in / Na+ out) = (0.008314 KJ/K * mol)(310 K)(23 mM / 149 mM) = -4.82 KJ/mol

Gelec = ZF(cell potential) = (+1)(96.5 KJ/V)(-0.051 V) = -4.9215 KJ/mol

G = 2 * (Gchem + Gelec) = 2 * (-4.82 KJ/mol + -4.9215 KJ/mol) = -19.483 KJ/mol

*Couple of things you should notice here - these formulas come from the hint box at the bottom of the question and secondly units are everything. You cannot add things together with different units.

**If you look closely at the calculations I have converted the cell potential from millivolts (mV) to volts (V) and that is because the Faraday constant is given to you in KJ/V. Of the same token I also used the gas constant (R) in KJ to also follow the units of the Faraday constant.

***Z is the charge of the ions we are dealing with - in this case Na has a +1 charge and we plug it into Z as such. In other scenarios, let's say Calcium, you would put a +2 in there.

****You might ask, "Why did you not convert millimolar to molar considering R and F use moles?" I did not convert though because 23 mM / 149 mM is a ratio and therefore if you convert it to 0.023 M / 0.149 M you will produce the same number.

*****Lastly, I multiplied the addition of Gchem and Gelec by 2 because the question makes note of 2 Na+ ions in the "downhill" transport.

Now on to the next part because the hint section also provides another equation:

G = RTln(Gluose In / Glucose Out)

The answer we are asked for is (Gluose In / Glucose Out); therefore we need to rearrange the problem and solve for (Gluose In / Glucose Out).

(Gluose In / Glucose Out) = e^(G/RT) = e^((-19.483 KJ/mol)/(0.008314 KJ/K * mol * 310 K)) = e^(7.5593) = 1918.6

1918.6 is the final answer and there are no units because they all cancel out. From the initial equation ln becomes e because it is the inverse.

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