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 -55.0 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 °C.
The maximum energy (or work) available from the transport of Na ions into the cell consists of two parts, that available from the transport down the chemical gradient, and that available from the transport down the electrical gradient, where R is the gas constant, T is the temperature in Kelvin, Z is the charge on the ion (+1), F is the Faraday constant (96.5 kJ/(mol·V), and Δψ is the membrane potential.
ΔG chem = RT ln[Na+ in / Na+ out] and ΔG elec = ZFΔψ
The minimum energy (or work) needed to pump glucose against a concentration gradient is given by:
ΔG = RT ln[glucose in/glucose out]
I figured out a ΔG chem of -4.181 and a ΔG elec of -5.3075. Do I add those two together to get my ΔG?
If so, ΔG is -9.7255 which I divided by .008314 KJ and divided that by 310.15 Kelvin.
That left -3.7716 = ln [glucose in/glucose out] which comes out to the WRONG answer.
Any help is greatly appreciated! Thank you!!
ΔG = ΔG chem + ΔG elec
Remember there are 2 Na+ transported:
2*ΔG = -RT*ln[glucose in/glucose out]
[glucose in/glucose out] = apx 2000
Well, looks like those numbers aren't playing nice with each other! Maybe they need a good therapist.
But fear not, I am here to attempt to untangle this mathematical mess for you! So, let's see if we can salvage this situation.
First, we need to determine the maximum ratio of [glucose]in to [glucose]out. This can be calculated using the formula:
ΔG = RT ln([glucose]in/[glucose]out)
Now, you already calculated the chemical gradient energy change (ΔG chem) and the electrical gradient energy change (ΔG elec). Great job on that! But here's the thing - you can't simply add those two together. They represent different aspects of the energy coupling process.
To calculate the overall ΔG, you need to consider both ΔG chem and ΔG elec, taking into account their respective contributions. The equation for the overall ΔG is:
ΔG = ΔG chem + ΔG elec
Now, here comes the kicker - you need to convert the membrane potential (Δψ) from millivolts to volts before plugging in the values. So, -55.0 mV becomes -0.055 V.
Once you have the overall ΔG, you can proceed with the calculation:
ΔG = -9.7255 kJ/mol
R = 8.314 J/(mol·K)
T = 37 °C + 273.15 K = 310.15 K
Now, divide ΔG by RT:
ΔG/RT = -9.7255 kJ/mol / (8.314 J/(mol·K) x 310.15 K)
After some number crunching, you should end up with a value for ln([glucose]in/[glucose]out).
But remember, we want the maximum ratio of [glucose]in to [glucose]out. So, you'll have to take the exponential of that value (e to the power of that value) to obtain the actual ratio.
I hope this helps you on your quest for the correct answer. Good luck, brave explorer of biochemical gradients!
To calculate the maximum ratio of [glucose]in to [glucose]out, we need to consider the maximum energy available from the transport of Na ions into the cell and the minimum energy required to pump glucose against its concentration gradient.
First, let's calculate the maximum energy available from the transport of Na ions into the cell:
ΔGchem = RT ln([Na+]in / [Na+]out)
Temperature (T) should be in Kelvin, so let's convert 37 °C to Kelvin:
T = 37 + 273.15 = 310.15 K
R is the gas constant (8.314 J/(mol·K)), but we need ΔGchem in kJ/mol, so let's convert it:
R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
Now let's substitute the given values and calculate ΔGchem:
ΔGchem = (0.008314 kJ/(mol·K)) * 310.15 K * ln(23.0 mM / 149 mM)
Using these values, you obtain a ΔGchem of -4.181 kJ/mol.
Next, let's calculate the energy available from the transport down the electrical gradient:
ΔGelec = ZFΔψ
Charge on the ion (Z) is +1 and the membrane potential (Δψ) is -55.0 mV (inside negative). However, ΔGelec should be in kJ/mol, so we need to convert the potential to volts:
Δψ = -55.0 mV = -55.0 * 10^-3 V
F is the Faraday constant (96.5 kJ/(mol·V)), so let's calculate ΔGelec:
ΔGelec = (1 * 96.5 kJ/(mol·V)) * (-55.0 * 10^-3 V)
Using these values, you obtain a ΔGelec of -5.3075 kJ/mol.
Now, to find the total ΔG, you need to subtract ΔGelec from ΔGchem:
ΔGtotal = ΔGchem + ΔGelec
ΔGtotal = -4.181 kJ/mol + (-5.3075 kJ/mol) = -9.4885 kJ/mol
Divide ΔGtotal by the gas constant (R) to get the natural logarithm:
ln([glucose]in / [glucose]out) = ΔGtotal / (0.008314 kJ/(mol·K))
Finally, rearrange the equation to find the ratio of [glucose]in to [glucose]out:
[glucose]in / [glucose]out = e^(ΔGtotal / (0.008314 kJ/(mol·K)))
Substitute the value of ΔGtotal and calculate the ratio.
I hope this helps! Let me know if you have any further questions.