Biochemistry, The Molecular Basis of Life, 6th Edition
Biochemistry, The Molecular Basis of Life, 6th Edition
6th Edition
ISBN: 9780190259204
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 10, Problem 5Q
Summary Introduction

To analyze:

The maximum number of adenosine triphosphate (ATP) molecules generated from a mole of sucrose.

Introduction:

Glycolysis is a process in which glucose is broken down and gets converted to pyruvate. This process leads to the generation of energy. Energy is released in the form of ATP. These ATP molecules help in carrying out other reactions in the body.

Expert Solution & Answer
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Explanation of Solution

Sucrose is a disaccharide made up of two monosaccharides, glucoseand fructose.

For thebreakdown of one mole of glucose, ATP can be calculated at three stages.

1. Glycolysis: 6.5 ATP (2 ATP are utilized in the preparatory phase and 4 ATParereleased by substrate-levelphosphorylation. 4.5 ATPareproduced through the conversion of 2 nicotinamide adenine dinucleotide (NADH) to 2 NAD+ through oxidative phosphorylation)

4ATP(Substratelevelphosphorylation)+4.5ATP(glycerolphosphateshuttle)-2ATP(preparatoryphase)=6.5ATP

2. Pyruvate to acetyl-CoA: 5 ATP (2 NADH are converted to 2 NAD+ and 5 ATP are released).

3. Citric acid cycle: 19.5 ATP (6 NADH produce 15 ATP, 2 flavin adenine dinucleotide (FADH2)produce 3 ATP through oxidative phosphorylation and 1.5 ATP molecules are produced through the formation of guanosine triphosphate (GTP) ).

15ATP(through6NADH)+3ATP(through2FADH2)+1.5ATP(through2GTP)=19.5ATP

Total ATP generated through glycolysis:

6.5ATP(throughglycolysis)+5ATP(throughlinkreaction)+19.5ATP(throughcitricacidcycle)=31ATP

The number of ATP molecules generated during the breakdown of one mole of fructosecan be calculated at three stages.

1. Fructose metabolism: 6.5 ATP (2 ATP are utilized in the preparatory phase and 4 ATP arereleased by substrate-level phosphorylation. 4.5 ATP areproduced through the conversion of 2 NADH to 2 NAD+ through oxidative phosphorylation)

4ATP (Substratelevelphosphorylation)+4.5ATP(glycerolphosphateshuttle)2ATP(preparatoryphase)=6.5ATP

After forming fructose 6-phosphate from fructose by Hexokinase IV, it enters in gluconeogenic pathway.

2. Pyruvate to acetyl-CoA: 5 ATP (2 NADH are converted to 2 NAD+ and 5 ATP are released).

3. Citric acid cycle: 19.5 ATP (6 NADH produce 15 ATP, 2 FADH2 produce 3 ATP through oxidative phosphorylation and 1.5 ATP molecules are produced through the formation of GTP).

15ATP(through6NADH)+3ATP(through2FADH2)+1.5ATP(through2GTP)= 19.5ATP

Total ATP generated through glycolysis:

6.5ATP(throughglycolysis)+5ATP(throughlinkreaction)+19.5ATP(throughcitricacidcycle)=31ATP

The breakdown of one mole of sucrose will generate 31 ATP from glucose and 31 ATP from fructose.

31ATP(fromglucose)+31ATP(fromsucrose)=62ATP(total)

Conclusion

Therefore, it can be concluded that the maximum number of ATP that can be generated from one mole of sucrose is 62.

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