Question

Glucose is actively transported into red blood cells by coupling the transport with the hydrolysis of ATP. The overall reaction can be written as $$ \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O}+n \text { Glucose(outside) } \rightleftharpoons n \text { Glucose(inside) }+\mathrm{ADP}+\mathrm{P}_{\mathrm{i}} $$ If the ratio of \([\mathrm{ATPJ} / \mathrm{ADP}]=1\) and \(\left[\mathrm{P}_{\mathrm{j}}\right]=10 \mathrm{mM}\), what is the concentration gradient, [glucose(inside)]/[glucose(outside)]. that is established at \(298 \mathrm{~K} ?\) Calculate this ratio for \(n=1, n=2\), and \(n=n\). Does this suggest a method for determining the value of \(n\) ? Use the value of \(\Delta G^{\circ}\) for the hydrolysis of ATP in Appendix 2 for these calculations.

Short answer

Calculating the glucose gradient gives approximately 4.43×10⁻⁵ for n=1 and 0.0066 for n=2. Measuring the gradient experimentally can help determine the correct value of n.

Step-by-step solution

Understanding the Reaction

The reaction described is:\[ ATP + H_2O + n \text{ Glucose(outside)} \rightleftharpoons n \text{ Glucose(inside)} + ADP + P_i \]This is an energy-coupled transport process where ATP hydrolysis facilitates the movement of glucose molecules into cells.

Identify Given Values

We are given:- Ratio \([\text{ATP}] / [\text{ADP}] = 1\)- \([P_i] = 10\, \text{mM}\)- Temperature \( T = 298\, \text{K} \)- Standard free energy change for ATP hydrolysis, \( \Delta G^\circ = -30.5\, \text{kJ/mol} \) in aqueous solution.

Calculating Free Energy Change for ATP Hydrolysis

The free energy change \( \Delta G \) for the reaction involving ATP hydrolysis is given by:\[ \Delta G = \Delta G^\circ + RT \ln\left(\frac{[ADP][P_i]}{[ATP]}\right) \]Since \([\text{ATP}] / [\text{ADP}] = 1\), the equation simplifies to:\[ \Delta G = \Delta G^\circ + RT \ln(10) \]Using \( R = 8.314\, \text{J/mol} \cdot \text{K} \) and converting \( \Delta G^\circ \) to \( \text{J/mol} \), \( \Delta G^\circ = -30500\, \text{J/mol} \).

Substitute Known Values

Substituting into the equation:\[ \Delta G = -30500 + 8.314 \times 298 \times \ln(10) \]Calculating this gives:\[ \Delta G = -30500 + 8.314 \times 298 \times 2.302 \approx -30500 + 5710.56 \approx -24789.44 \text{ J/mol} \]

Calculating the Concentration Gradient for Glucose

For the concentration gradient:\[ \Delta G = nRT \ln\left(\frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}}\right) \]Thus,\[ \ln\left(\frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}}\right) = \frac{\Delta G}{nRT} \]Substituting in \( \Delta G = -24789.44 \text{ J/mol} \) for each \( n \):

Compute for n=1

For \( n=1 \):\[ \ln\left(\frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}}\right) = \frac{-24789.44}{1 \times 8.314 \times 298} \approx -10.03 \]Thus,\[ \frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}} \approx e^{-10.03} \approx 4.43 \times 10^{-5} \]

Compute for n=2

For \( n=2 \):\[ \ln\left(\frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}}\right) = \frac{-24789.44}{2 \times 8.314 \times 298} \approx -5.015 \]Thus,\[ \frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}} \approx e^{-5.015} \approx 0.0066 \]

Analyze General Case for n=n

For a general \( n \):\[ \ln\left(\frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}}\right) = \frac{-24789.44}{n \times 8.314 \times 298} \]This implies that the concentration gradient is a function of \( n \), and can be calculated for specific values of \( n \).

Determine a Method to Find n

The evaluation of different concentration gradients as a function of \( n \) suggests that measuring \( \frac{\text{[Glucose(inside)]}}{\text{[Glucose(outside)]}} \) under controlled conditions and matching it to calculated values can reveal \( n \). Experimentally determining the concentration gradient and comparing it to these calculations can help identify the value of \( n \).

Key concepts

Understanding Glucose Transport

Glucose transport is a fundamental process for energy management within cells, especially in red blood cells. This process involves the movement of glucose from outside to inside of the cells, often requiring energy due to the cell's preference to maintain a particular internal environment. The transport can be passive or active; in this exercise, it is facilitated actively by something known as **ATP hydrolysis**.

• Active transport requires energy input, typically from ATP.
• Glucose, being a crucial energy source, needs regulated transport irrespective of internal concentrations.

By coupling glucose transport with ATP hydrolysis, the cell can effectively transport glucose even against a concentration gradient, ensuring it maintains energy supply for metabolic processes.

The Role of ATP Hydrolysis

ATP hydrolysis provides the necessary energy by breaking down ATP into ADP and inorganic phosphate. This process releases free energy, which drives endergonic reactions like glucose transport into cells.

• ATP (adenosine triphosphate) is the cell's "energy currency".
• Hydrolysis involves the cleavage of phosphate bonds, releasing energy.

This energy coupling is crucial because it allows glucose to move into cells regardless of their external concentration. This ensures that cells always have access to this primary energy source when needed.

Concentration Gradient and Its Significance

A concentration gradient refers to the difference in concentration of a substance between two regions. It forms the basis for passive transport, where substances move from high to low concentration until equilibrium is reached. However, glucose transport through active means involves moving against the gradient.

• Glucose typically moves from an area of high concentration (outside) to low concentration (inside).
• Active transport allows cells to maintain glucose levels in varying external conditions.

For glucose, maintaining a concentration gradient means the cell can regulate its energy intake and utilize it efficiently, ensuring metabolic flexibility and stability.

Energetics of Glucose Movement

The energetics of glucose transport involves energy derivation from ATP hydrolysis to facilitate movement against a concentration gradient. Understanding how much energy is required is key to estimating how feasible the transport process is.

• Energetics involves calculating energy changes or transfers during a biochemical reaction.
• Glucose transport demands an input of energy proportional to the concentration gradient and environmental conditions.

This energetic evaluation is essential for understanding the efficiency of cellular processes and the impact of different variables like temperature or ATP availability.

Thermodynamics and Biological Systems

Thermodynamics in biochemistry examines how energy transformations occur in biological processes. The movement of glucose via active transport must comply with thermodynamic principles, where energy changes can be quantified and predicted.

• Key principles include the laws of energy conservation and entropy.
• The free energy change (\( \Delta G \)) is central to understanding whether a process can occur spontaneously.

For glucose transport, calculating \( \Delta G \) helps predict the direction and feasibility of the process within the cellular environment, linking molecular actions to broad thermodynamic laws.