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I have a homework question that I am not able to solve. No solution is available, so I was hoping someone here could help. The question is this:

Calculate the maximal concentration ratio that can be obtained for transport of Ca$^{2+}$ ions if two ions are transported per ATP hydrolyzed at 310 K, and they are transported to a potential that is 70 mV higher at the receiving side. Assume that the $\Delta G$ for hydrolysis of ATP = -45 > kJ\mol.

Relevant formulas:

\begin{align} \Delta G_{\text{electrostatic gradient}} &= ZF\Delta \Psi\\ \Delta G_{\text{chemical gradient}} &= RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}}\\ \Delta G_{\text{hydrolysis of ATP}} &= -45000~\mathrm{J/mol}\\ \end{align}

Moving two positive $\ce{Ca^2+}$ ions against both a concentration gradient and an electrostatic gradient implies that $G_{\text{chemical gradient}}$ and $\Delta G_{\text{electrostatic gradient}} > 1$. The negative value for $\Delta G_{\text{hydrolysis of ATP}}$ means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, $\Delta G_{\text{chemical gradient}}$ is multiplied by 2. Also (I presume) the concentration of $\ce{Ca^2+}$ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.

The combined expression for $\Delta G$ for the unassisted transport is

$$\Delta G = RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi$$

Rearranging this yields

$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$

I am not sure what to plug in for $\Delta G$, but it needs to account for the push from ATP. I believe this is done by summing the two free energies affecting the transport, ergo the sum of $\Delta G_{\text{electrostatic gradient}}$ and $\Delta G_{\text{hydrolysis of ATP}}$. However, this gives me a concentration ratio much smaller than 1.

Can anyone see where I am incorrect?

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At equilibrium, the entire process (transport of $2~\ce{Ca^2+}$ ions and hydrolysis of $1~\text{ATP}$) involves no change in G.

\begin{align} 0 &=\Delta{G_\mathrm{total}}\\[2ex] 0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+}}}\\[2ex] 0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+},\text{ concentration gradient}}} + 2\Delta{G_{\ce{Ca^2+},\text{ electrostatic}}}\\[2ex] 0 &=\Delta{G_\mathrm{ATP}} + 2RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi\\ \end{align}

So basically I'm saying you made two mistakes, one a sign error and one a factor of 2 error.

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  • $\begingroup$ Thank you! So the system is at equilibrium? That seems counter-intuitive because of the concentration gradient and the charge gradient (at equilibrium I would think that these gradients were zero?). $\endgroup$ – Yoda Oct 19 '15 at 12:07
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    $\begingroup$ @AndersMB The ATP hydrolysis and Ca2+ transport are coupled, and when the maximum concentration is reached, the system is at equilibrium. The concentrations stop changing with time at equilibrium. The equilibrium would be different if the ATP reaction were not be coupled to the transport. $\endgroup$ – DavePhD Oct 19 '15 at 12:23
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ΔGelectrostatic gradientΔGchemical gradientΔGhydrolysis of ATP=ZFΔΨ=RTlnCinCout=−45000 J/mol

Moving two positive Ca2+ ions against both a concentration gradient and an electrostatic gradient implies that Gchemical gradient and ΔGelectrostatic gradient>1. The negative value for ΔGhydrolysis of ATP means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, ΔGchemical gradient is multiplied by 2. Also (I presume) the concentration of Ca2+ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.

The combined expression for ΔG for the unassisted transport is Rearranging this yields

$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$

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    $\begingroup$ Welcome to Chemistry! Please use better formatting on your answer. As it stands this is almost unreadable. For information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance. $\endgroup$ – bon Oct 19 '15 at 17:24

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