Problem 15 - Biochemistry: Concepts and Connections

Question

The concentration of glucose in your circulatory system is maintained near \(5.0 \mathrm{mM}\) by the actions of the pancreatic hormones glucagon and insulin. Glucose is imported into cells by protein transporters that are highly specific for binding glucose. Inside the liver cells the imported glucose is rapidly phosphorylated to give glucose-6-phosphate (G-6-P). This is an ATP- dependent process that consumes \(1 \mathrm{~mol}\) ATP per mol of glucose. (a) The process of phosphorylating the glucose after it has been transported into the cell is considered a form of active transport-called "transport by modification"-even though ATP is not bound by the transporter protein, nor is ATP hydrolysis directly involved in the movement of glucose across the membrane. Explain the thermodynamic basis for this form of active transport. (Hint: Consider Le Chatelier's principle.) (b) Given ATP \(=4.7 \mathrm{mM} ; \mathrm{ADP}=0.15 \mathrm{mM} ; \mathrm{P}_{\mathrm{i}}=6.1 \mathrm{mM}\), calculate the theoretical maximum concentration of \(\mathrm{G}-6-\mathrm{P}\) inside a liver cell at \(37{ }^{\circ} \mathrm{C}, \mathrm{pH}=7.2\) when the glucose concentration outside the cell (i.e., [ glucose ] outside) is \(5.0 \mathrm{mM}\) : \(\mathrm{ATP}+\) glucose \(_{\text {inside }} \rightarrow \mathrm{ADP}+\) glucose-6-phosphate \(+\mathrm{H}^{+}\) For ATP \(+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}}+\mathrm{H}^{+} \Delta G^{\circ 1}=-32.2 \mathrm{~kJ} / \mathrm{mol}\) and for $$ \text { G-6-P }+\mathrm{H}_{2} \mathrm{O} \rightarrow \text { Glucose }+\mathrm{P}_{\mathrm{i}} \Delta G^{\circ \prime}=-13.8 \mathrm{~kJ} / \mathrm{mol} $$

Step-by-Step Solution

Answer

The theoretical maximum concentration of G-6-P is approximately 1.97 M.

Step 1: Understand Thermodynamic Basis of 'Transport by Modification'

The process described involves the phosphorylation of glucose after it's imported into the cell. This phosphorylation lowers the concentration of free glucose inside the cell, maintaining a concentration gradient favoring glucose import from outside. According to Le Chatelier's principle, removing a product of a reaction (glucose in free form) encourages more of the reactants to come together, i.e., more glucose to be imported, thus supporting active transport.

Step 2: Calculate Reaction Gibbs Free Energy Change (ΔG') Inside Cell

To calculate ΔG' for the phosphorylation reaction inside the cell, use the equation:\[ΔG' = ΔG^{0'} + RT \, \ln \frac{[ADP][Glucose-6-P]}{[ATP][Glucose_{inside}]}\]Given:\(R = 8.314 \, \text{J/mol·K}, \, T = 310.15 \, \text{K} (37^{\circ} \text{C}), \, ΔG^{0'}\) provided for breaking down reactions.Add the ΔG', therefore:\[ΔG' = (-32.2 + 13.8) \, \text{kJ/mol} = -18.4 \, \text{kJ/mol}\].

Step 3: Set Up Equation for Equilibrium Concentrations

At equilibrium, ΔG' becomes zero, thus:\[0 = -18.4 \, \text{kJ/mol} + 8.314 \, \text{J/mol⋅K} \cdot 310.15 \, \text{K} \, \ln \left(\frac{0.15[Glucose-6-P]}{4.7 \times 5.0}\right)\]Rearrange to solve for \([Glucose-6-P]\) using given concentrations for ATP, ADP, and \([Glucose_{inside}]=5.0 \, \text{mM}\).

Step 4: Solve for Theoretical Maximum Concentration of G-6-P

Convert -18.4 kJ/mol to J/mol (\(-18400 \, \text{J/mol}\)) and rearrange:\[\ln \left(\frac{0.15[Glucose-6-P]}{23.5}\right) = \frac{18400}{8.314 \times 310.15}\]Solving gives:\[\ln \left(\frac{0.15[Glucose-6-P]}{23.5}\right) = 7.144\]\[\frac{0.15[Glucose-6-P]}{23.5} = \mathrm{e}^{7.144}\approx 1267.2\]Finally solve for \([Glucose-6-P]\):\[[Glucose-6-P] \approx \frac{1267.2 \times 23.5}{0.15} \approx 1.97 \, \text{M}\].

Key Concepts

Glucose Phosphorylation

When glucose enters the cell, it is quickly phosphorylated, which means a phosphate group is added to it, forming glucose-6-phosphate (G-6-P). This process requires energy from the molecule ATP, where one phosphate group is transferred to glucose. This phosphorylation is crucial because it

• prevents glucose from diffusing back out of the cell;
• helps maintain a concentration gradient that facilitates further glucose import;
• initiates glucose's entry into several metabolic pathways.

Glucose phosphorylation is driven by an enzyme known as hexokinase or glucokinase, depending on the tissue type. These enzymes not only speed up the reaction but also ensure specificity, meaning they will only add phosphate to glucose and no other molecules. It's important to understand that the phosphorylation of glucose is not only an essential metabolic step but also acts strategically to sequester glucose within the cell, allowing it to be stored or used without exiting back into circulation.

Active Transport

Active transport involves moving molecules across a cell membrane against their concentration gradient, which typically requires energy. However, in the case of glucose phosphorylation, this form of active transport is slightly different as it doesn't require the direct use of ATP to transport glucose across the membrane. Instead, it's called "transport by modification."
As glucose enters the cell, phosphorylation converts it into glucose-6-phosphate. This change means glucose, in its original form, is no longer present in the cell. This transformation

• ensures that the concentration of free glucose remains low inside the cell;
• maintains a gradient that continues to pull more glucose into the cell from outside.

This is a clever biochemical strategy ensuring a continuous inward flow of glucose without needing a transporter that actively pumps it inside the cell. The energy used for phosphorylation indirectly supports the active transport by maintaining the driving force for glucose entry into the cell.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a change in conditions affects a chemical equilibrium. In simpler terms, it states that if a system at equilibrium is disturbed, the system will adjust to counteract that disturbance and restore a new equilibrium.
Applying this principle to glucose phosphorylation, we see how the continual removal of free glucose (by converting it to glucose-6-phosphate) shifts the reaction toward continuously bringing more glucose into the cell.
The constant removal of substrate ensures that the gradient remains favorable for glucose import. This dynamic feed supports the cell's metabolic needs. This principle is crucial for understanding how cellular processes remain adaptive and efficient, ensuring that cells can adequately respond to internal and external changes in available resources or cellular demand.