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Chapter 9: Problem 18

What is the \(\Delta G\) for transport of glutamate from the outside of the cell, where the concentration is \(0.1 \mathrm{mM}\), to the inside of the cell, where the concentration is \(10 \mathrm{mM}\) ? Assume the membrane potential is \(-70 \mathrm{mV}\).

Short Answer

Expert verified
The \(\Delta G\) for the transport is approximately \(18.15 \text{ kJ mol}^{-1}\).

Step by step solution

01

Identify Given Values

First, identify the concentration of glutamate outside \(C_{out}\) and inside \(C_{in}\) the cell, as well as the membrane potential \(\Delta \psi\). From the problem, \(C_{out} = 0.1 \text{ mM}\), \(C_{in} = 10 \text{ mM}\), and \(\Delta \psi = -70 \text{ mV}\).
02

Understand the 9;Delta G9; Equation

The equation for the change in Gibbs free energy (\(\Delta G\)) for the transport of ions across a membrane is given by:\[\Delta G = RT \ln\left(\frac{C_{in}}{C_{out}}\right) + zF\Delta \psi\]where:- \(R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}\) (universal gas constant),- \(T\) is the temperature in Kelvin (assume 298 K if not specified),- \(z\) is the valence of the ion (\(-1\) for glutamate), and- \(F = 96485 \text{ C mol}^{-1}\) (Faraday constant).
03

Calculate First Term

Calculate the first term of the equation:\[RT \ln\left(\frac{C_{in}}{C_{out}}\right)\]Assuming \(T = 298 \text{ K}\),\[RT = 8.314 \times 298 = 2477.572 \text{ J mol}^{-1}\]\[\ln\left(\frac{10}{0.1}\right) = \ln(100) = 4.605\]Hence, the first term:\[2477.572 \times 4.605 = 11398.78 \text{ J mol}^{-1}\]
04

Calculate Second Term

Calculate the second term \(zF\Delta \psi\):\[z = -1\] (for glutamate),\[zF = -1 \times 96485 = -96485 \text{ C mol}^{-1}\]\[\Delta \psi = -70 \text{ mV} = -0.070 \text{ V}\]\[-96485 \times -0.070 = 6753.95 \text{ J mol}^{-1}\]
05

Summarize the 9;Delta G9; Calculation

Add the two calculated terms to get \(\Delta G\):\[\Delta G = 11398.78 + 6753.95 = 18152.73 \text{ J mol}^{-1}\]Thus, the \(\Delta G\) for the transport of glutamate is \(18152.73 \text{ J mol}^{-1}\) or approximately \(18.15 \text{ kJ mol}^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Membrane Potential
Membrane potential refers to the difference in electric potential between the interior and exterior of a cell. This potential is mainly due to the difference in the concentration of ions across the cell membrane.
For cells, typically this value is negative, meaning the inside of the cell is more negatively charged compared to the outside.
This potential is vital for many cellular processes including the transport of ions like glutamate.
  • The membrane potential in the given problem is 70 mV.
  • This potential influences the movement of ions since it contributes to the energy landscape across the cell's membrane.
  • Negative membrane potentials generally help drive positive ions into the cell and repulse negative ones like glutamate.
This potential difference is used in the Gibbs free energy equation to determine the energetic favorability of ion transport.
Understanding membrane potential is key to comprehending why and how cells maintain their internal environment despite changes in the external surroundings.
Glutamate Transport
Glutamate, an amino acid neurotransmitter, is critical for brain function, as it is involved in synaptic transmission.
Transporting glutamate across the cell membrane is an essential step in regulating its concentration in the extracellular and intracellular spaces.
To cross the cell membrane, glutamate depends on transport proteins since it is negatively charged and cannot diffuse freely across the lipid barrier.
  • Active transport mechanisms help in moving glutamate against its concentration gradient.
  • Membrane proteins often use energy from ATP hydrolysis or the existing ion gradients to facilitate this process.
  • In this context, the transport of glutamate inside the cell from an area of lower to higher concentration would require energy, as calculated via the Gibbs free energy equation.
This transport is not only significant for signal transmission in neurons but also plays a role in metabolic processes and cellular health.
Ion Concentration Gradient
An ion concentration gradient occurs when there is a difference in the concentration of ions across a membrane.
This gradient is a primary source of potential energy for cells.
In the exercise, the concentration gradient for glutamate is from 0.1 mM outside the cell to 10 mM inside the cell.
  • This gradient is critical for determining the direction and energy requirement for ion movement.
  • When ions move down their concentration gradient, they release energy, which can be harnessed by the cell.
  • Conversely, moving ions against their gradient requires an input of energy.
The balance and maintenance of ion gradients are crucial for many cellular functions, including signal transduction, nutrient uptake, and waste removal.
For effective cellular functioning, cells invest energy to maintain these gradients to ensure survival and proper functioning.

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Most popular questions from this chapter

As a result of the activity of the acetylcholine receptor, muscle cells undergo depolarization, but the change in membrane potential is less dramatic and slower than in neurons. a. The acetylcholine receptor is also a gated ion channel. What triggers the gate to open? b. The acetylcholine receptor/ion channel is specific for \(\mathrm{Na}^{+}\)ions. Do \(\mathrm{Na}^{+}\)ions flow in or out? c. How does the \(\mathrm{Na}^{+}\)flow through the ion channel change the membrane potential?

Calculate the free energy change for the movement of \(\mathrm{K}^{+}\)into a cell when the \(\mathrm{K}^{+}\)concentration outside is \(15 \mathrm{mM}\) and the cytosolic \(\mathrm{K}^{+}\)concentration is \(50 \mathrm{mM}\). Assume that \(T=20^{\circ} \mathrm{C}\) and \(\Delta \psi=-50 \mathrm{mV}\) (inside negative). Is this process spontaneous?

A channel protein specific for calcium ions was found to have a pore lined with six glutamate residues that participate in coordinating the calcium ion. What would happen to the selectivity of the pore if the Glu residues were mutated to Asp residues?

Explain why carbon dioxide can cross a cell membrane without the aid of a transport protein.

The PEPT1 transporter aids in digestion by transporting di- and tripeptides into the cells lining the small intestine. There are three components of this system: (a) a symport transporter that ferries di- and tripeptides across the membrane, along with an \(\mathrm{H}^{+}\)ion, (b) \(\mathrm{a} \mathrm{Na}^{+}-\mathrm{H}^{+}\)antiporter \(\left(\mathrm{H}^{+} / \mathrm{Na}^{+}\right.\)exchanger), and (c) a Na,K-ATPase. Draw a diagram that illustrates how these three transporters work together to transport peptides into the cell.
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