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Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right)\). The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 M\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 M .\) The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\), normal body temperature? When \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

Short Answer

Expert verified
The work required to transport 1.0 mol of K+ ions from the blood to the inside of a muscle cell at 37°C is approximately 2.99 x 10^4 J/mol. The transportation of other ions is necessary during this process to maintain charge neutrality.

Step by step solution

01

Understand Gibbs free energy formula

The work done in transporting ions across a membrane can be calculated using the Gibbs free energy formula: \(\Delta G = \Delta G^0 + RT \ln\frac{Q}{K}\) where: - \(\Delta G\) is the Gibbs free energy change - \(\Delta G^0\) is the standard Gibbs free energy change (0 for this process) - R is the gas constant (8.314 J/mol K) - T is the temperature in Kelvin (in this case, 37°C = 310 K) - Q is the reaction quotient (calculated by dividing the concentration of K+ in muscle cells by the concentration in blood plasma) - K is the equilibrium constant (1 for this process since the concentrations remain constant)
02

Calculate the reaction quotient (Q)

We are given that the concentration of K+ in muscle cells is 0.15 M, and the concentration in blood plasma is 0.0050 M. We can calculate the reaction quotient Q as follows: \(Q = \frac{\text{concentration in muscle cells}}{\text{concentration in blood}} = \frac{0.15 M}{0.0050 M} = 30\)
03

Calculate the Gibbs free energy change (\(\Delta G\))

Using the given formula, we can now find the Gibbs free energy change: \(\Delta G = \Delta G^0 + RT \ln\frac{Q}{K} = 0 + (8.314 J/mol \cdot K)(310 K) \ln(30)\) Solve for \(\Delta G\): \(\Delta G = 8.314 \times 310 \times \ln(30) \approx 2.99 \times 10^4 J/mol\)
04

Determine the necessity of transporting other ions

The transportation of K+ from blood to cells results in a change in the overall charge balance across the cell membrane. As K+ ions are moved into the cell, more positive charges are transferred inside the cell. In order to maintain charge neutrality, an equivalent amount of anionic charge (negative ions) must be moved across the membrane or an equivalent amount of cationic charge (positive ions) must be moved out of the cell to maintain the overall charge balance. Therefore, the transportation of other ions is necessary to maintain charge neutrality during this process. #Result# The work required to transport 1.0 mol K+ from the blood to the inside of a muscle cell at 37°C is approximately 2.99 x 10^4 J/mol. Yes, the transportation of other ions is necessary to maintain charge neutrality during this process.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy is a crucial concept in understanding biochemical reactions. In simple terms, it represents the amount of energy available to do work during a chemical reaction. It's especially important when discussing what happens on a microscopic level inside our cells.
To calculate Gibbs free energy, scientists use a specific formula:
  • \(\Delta G = \Delta G^0 + RT \ln\frac{Q}{K}\)
Here,
  • \(\Delta G\) is the change in Gibbs free energy, which tells us how much work is done.
  • \(\Delta G^0\) is the standard Gibbs free energy change. For ion transport in cells, this is often considered zero.
  • \(R\) is the universal gas constant \(8.314 \text{ J/mol K}\).
  • \(T\) is the temperature in Kelvin.
  • \(Q\) is the reaction quotient, reflecting concentration differences.
  • \(K\), the equilibrium constant, often equals 1 in biological processes.
This formula helps determine the energy required to move ions like \(\text{K}^+\) into muscle cells against concentration gradients. By using concentrations in muscle cells and blood plasma, we can calculate the energy changes and analyze how cells perform essential tasks.
Ion Transport
Ion transport is a fascinating biological process involving the movement of ions across cell membranes. This movement is important for maintaining cellular functions and often requires energy, as ions move against their concentration gradients.
Specific proteins or pumps embedded in the cell membrane help facilitate this transport. For example, the sodium-potassium pump is one such protein that uses energy to exchange \(\text{K}^+\) and \(\text{Na}^+\) ions across the membrane.
In the exercise, transporting \(1.0\) mol of \(\text{K}^+\) from blood plasma into muscle cells isn't merely passive; it relies on biochemical energy calculated from the Gibbs free energy. Such transport mechanisms are critical in maintaining the ion balance essential for muscle contraction and nerve impulse transmission.
These mechanisms showcase how cells actively manage their internal environments, ensuring optimal performance and survival.
Charge Neutrality
Charge neutrality is vital in biological systems to maintain a stable environment within cells. Simply put, cells need to balance the number of positive and negative charges inside and outside their membranes.
During ion transport, exchanging \(\text{K}^+\) ions often impacts the overall charge balance. If positive ions are brought in, like in our exercise, cells need to adjust by moving negative ions or other positive ions out to keep balanced charges. This is crucial for preventing unwanted voltage imbalances across membranes, which could otherwise disrupt cellular functions.
The mechanisms used by cells to maintain charge neutrality are best demonstrated by the interplay of multiple ion channels and pumps. For instances like the constant pumping of \(\text{Na}^+\) and \(\text{K}^+\), often paired with counter-ion movements. This not only helps in maintaining electrical neutrality but also supports various cellular activities, like neuron communications and muscle contractions.
The careful balance of electrolyte movement epitomizes cellular intelligence, and maintaining charge neutrality is a fundamental aspect of this balance.

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Most popular questions from this chapter

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{~kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) and \(\mathrm{Br}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) were mixed in a \(1.00\) -L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ}\), and \(\Delta S^{\circ}\).

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of \(\mathrm{AB}\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {sumr }} ? \Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

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