4. The reaction shown below is part of the glycolytic pathway. The standard free energy change for this reaction is +0.4 kcal/mol. Glucose-6-phosphate (G6P) → Fructose-6-phosphate (F6P) C) Steady-state concentrations of these compounds in the cell (at 25°C) are 0.675 mM G6P and 0.0969 mM F6P. In which direction is this reaction allowed to proceed in the cell? (show calculations)

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### Glycolysis and Free Energy Change

#### 4. Reaction and Free Energy in Glycolysis

The glycolytic pathway involves several key reactions, one of which is indicated below. The reaction shows the equilibrium between glucose-6-phosphate (G6P) and fructose-6-phosphate (F6P), with a standard free energy change (ΔG°'):

\[ 
\text{Glucose-6-phosphate (G6P)} \leftrightarrow \text{Fructose-6-phosphate (F6P)}
\]

The standard free energy change for this reaction is +0.4 kcal/mol.

#### C) Steady-State Concentrations and Reaction Direction

Given the steady-state concentrations in the cell at 25°C:

- \[ \text{[G6P]} = 0.675 \, \text{mM} \]
- \[ \text{[F6P]} = 0.0969 \, \text{mM} \]

Determine the direction in which the reaction proceeds by calculating the actual free energy change under these conditions.

**Calculations:**

1. **Convert concentrations to Molarity:**
   
   \[
   \text{[G6P]} = 0.675 \, \text{mM} = 6.75 \times 10^{-4} \, \text{M}
   \]
   \[
   \text{[F6P]} = 0.0969 \, \text{mM} = 9.69 \times 10^{-5} \, \text{M}
   \]

2. **Use the Gibbs free energy equation:**
   
   The actual free energy change (ΔG) can be calculated using the formula:
   
   \[
   ΔG = ΔG°' + RT \ln \left(\frac{\text{[Products]}}{\text{[Reactants]}}\right)
   \]
   
   Where:
   - \( R \) is the universal gas constant \( (1.987 \, \text{cal/mol·K}) \)
   - \( T \) is the temperature in Kelvin (298 K, since \( 25°C = 298 K \))
   
   \[
   ΔG = 0.4 \, \text{kcal/mol} + 0.001987 \, \text{kcal
Transcribed Image Text:### Glycolysis and Free Energy Change #### 4. Reaction and Free Energy in Glycolysis The glycolytic pathway involves several key reactions, one of which is indicated below. The reaction shows the equilibrium between glucose-6-phosphate (G6P) and fructose-6-phosphate (F6P), with a standard free energy change (ΔG°'): \[ \text{Glucose-6-phosphate (G6P)} \leftrightarrow \text{Fructose-6-phosphate (F6P)} \] The standard free energy change for this reaction is +0.4 kcal/mol. #### C) Steady-State Concentrations and Reaction Direction Given the steady-state concentrations in the cell at 25°C: - \[ \text{[G6P]} = 0.675 \, \text{mM} \] - \[ \text{[F6P]} = 0.0969 \, \text{mM} \] Determine the direction in which the reaction proceeds by calculating the actual free energy change under these conditions. **Calculations:** 1. **Convert concentrations to Molarity:** \[ \text{[G6P]} = 0.675 \, \text{mM} = 6.75 \times 10^{-4} \, \text{M} \] \[ \text{[F6P]} = 0.0969 \, \text{mM} = 9.69 \times 10^{-5} \, \text{M} \] 2. **Use the Gibbs free energy equation:** The actual free energy change (ΔG) can be calculated using the formula: \[ ΔG = ΔG°' + RT \ln \left(\frac{\text{[Products]}}{\text{[Reactants]}}\right) \] Where: - \( R \) is the universal gas constant \( (1.987 \, \text{cal/mol·K}) \) - \( T \) is the temperature in Kelvin (298 K, since \( 25°C = 298 K \)) \[ ΔG = 0.4 \, \text{kcal/mol} + 0.001987 \, \text{kcal
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