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
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
Section: Chapter Questions
Problem 1P
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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 \(-0.345 \text{V} \)
\[ \Delta E^\circ = (-0.320 \text{V}) - (-0.345 \text{V}) = +0.025 \text{V} \text{ or } +25 \text{mV} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff5c7587d-60b6-48ba-8c1a-ab4a0a8fbdbd%2F59d12156-8c74-47f5-88f5-2959e1422695%2F5nm2ben_processed.jpeg&w=3840&q=75)
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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