Examine the reaction: B-Hydroxybutyrate + NAD* → Acetoacetate+ NADH + H* What is the AEº' of this redox reaction? The Relevant half reactions are: Acetoacetate + 2H* + 2e → B-hydroxybutyrate (E°' = -0.345V) NAD+ + H* + 2e → NADH (E°' = -0.320V) O +25mV +665mV -665mV -25mV

Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
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**Examine the Reaction:**

\[ \text{β-Hydroxybutyrate} + \text{NAD}^+ \rightarrow \text{Acetoacetate} + \text{NADH} + \text{H}^+ \]

**What is the \( \Delta E^\circ \) of this redox reaction?**

**The Relevant Half Reactions are:**

1. \[ \text{Acetoacetate} + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{β-hydroxybutyrate} \quad (E^\circ = -0.345\text{V}) \]

2. \[ \text{NAD}^+ + \text{H}^+ + 2\text{e}^- \rightarrow \text{NADH} \quad (E^\circ = -0.320\text{V}) \]

**Options:**

- ⃝ +25mV
- ⃝ +665mV
- ⃝ -665mV
- ⃝ -25mV

**Explanation:**

To find \( \Delta E^\circ \), use the formula:

\[ \Delta E^\circ = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

In this context:

- \( E^\circ \) for the reduction of NAD\(^+\) is \(-0.320 \text{V} \)
- \( E^\circ \) for the oxidation of β-hydroxybutyrate is \(-0.345 \text{V} \)

\[ \Delta E^\circ = (-0.320 \text{V}) - (-0.345 \text{V}) = +0.025 \text{V} \text{ or } +25 \text{mV} \]
Transcribed Image Text:**Examine the Reaction:** \[ \text{β-Hydroxybutyrate} + \text{NAD}^+ \rightarrow \text{Acetoacetate} + \text{NADH} + \text{H}^+ \] **What is the \( \Delta E^\circ \) of this redox reaction?** **The Relevant Half Reactions are:** 1. \[ \text{Acetoacetate} + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{β-hydroxybutyrate} \quad (E^\circ = -0.345\text{V}) \] 2. \[ \text{NAD}^+ + \text{H}^+ + 2\text{e}^- \rightarrow \text{NADH} \quad (E^\circ = -0.320\text{V}) \] **Options:** - ⃝ +25mV - ⃝ +665mV - ⃝ -665mV - ⃝ -25mV **Explanation:** To find \( \Delta E^\circ \), use the formula: \[ \Delta E^\circ = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \] In this context: - \( E^\circ \) for the reduction of NAD\(^+\) is \(-0.320 \text{V} \) - \( E^\circ \) for the oxidation of β-hydroxybutyrate is \(-0.345 \text{V} \) \[ \Delta E^\circ = (-0.320 \text{V}) - (-0.345 \text{V}) = +0.025 \text{V} \text{ or } +25 \text{mV} \]
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