General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN: 9781285853918
Author: H. Stephen Stoker
Publisher: Cengage Learning
bartleby

Concept explainers

Question
Book Icon
Chapter 24, Problem 24.10EP

(a)

Interpretation Introduction

Interpretation: To indicate whether the intermediate 2Phosphoglycerate in the glycolysis pathway is a C6 molecule or a C3 molecule.

Concept introduction: In the glycolysis metabolic pathway, a glucose molecule is converted into two pyruvate molecules. Two ATP molecules and NADH coenzymes are produced along with pyruvate.

The block diagram to represent an overview of glycolysis is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  1

From the above diagram, it is concluded that in the overall process of glycolysis, two stages are present.

a) Steps 1 to 3 represents a six-carbon stage (C6 stage).

b) Steps 4 to 10 represent a three-carbon stage (C3 stage).

In the C6 stage, the intermediates are derivatives of either glucose or fructose. Glucose and fructose contain six carbon atoms. Therefore, they are C6 molecules. The phosphate groups are present in the intermediates.

In the C3 stage, the intermediates are phosphorylated derivatives of glyceraldehyde, pyruvate, glycerate, or dihydroxyacetone, which are derivatives of either acetone or glycerol. Acetone and glycerol contain three carbon atoms. Therefore, they are C3 molecules.

(a)

Expert Solution
Check Mark

Answer to Problem 24.10EP

The intermediate 2Phosphoglycerate is a C3 molecule.

Explanation of Solution

The structure of 2Phosphoglycerate is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  2

Here,General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  3 denotes the PO32 unit.

The intermediate 2Phosphoglycerate is formed in the C3 stage and it is the phosphorylated derivative of glycerate. Therefore, it is a C3 molecule.

Conclusion

The intermediate 2Phosphoglycerate is formed in the C3 stage and it is the phosphorylated derivative of glycerate. 2Phosphoglycerate contains three carbon atoms. Therefore, it is a C3 molecule.

(b)

Interpretation Introduction

Interpretation: To indicate whether the intermediate 1,3Biphosphoglycerate in the glycolysis pathway is a C6 molecule or a C3 molecule.

Concept introduction: In the glycolysis metabolic pathway, a glucose molecule is converted into two pyruvate molecules. Two ATP molecules and NADH coenzymes are produced along with pyruvate.

The block diagram to represent an overview of glycolysis is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  4

From the above diagram, it is concluded that in the overall process of glycolysis, two stages are present.

a) Steps 1 to 3 represents a six-carbon stage (C6 stage).

b) Steps 4 to 10 represent a three-carbon stage (C3 stage).

In the C6 stage, the intermediates are derivatives of either glucose or fructose. Glucose and fructose contain six carbon atoms. Therefore, they are C6 molecules. The phosphate groups are present in the intermediates.

In the C3 stage, the intermediates are phosphorylated derivatives of glyceraldehyde, pyruvate, glycerate, or dihydroxyacetone, which are derivatives of either acetone or glycerol. Acetone and glycerol contain three carbon atoms. Therefore, they are C3 molecules.

(b)

Expert Solution
Check Mark

Answer to Problem 24.10EP

The intermediate 1,3Biphosphoglycerate is a C3 molecule.

Explanation of Solution

The structure of 1,3biphosphoglycerate is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  5

The intermediate 1,3biphosphoglycerate is formed in the C3 stage and it is the phosphorylated derivative of glycerate. Therefore, it is a C3 molecule.

Conclusion

The intermediate 1,3Biphosphoglycerate is formed in the C3 stage and it is the phosphorylated derivative of glycerate. Three carbon atoms are present in 1,3Biphosphoglycerate. Therefore, it is a C3 molecule.

(c)

Interpretation Introduction

Interpretation: To indicate whether the intermediate glucose 6phosphate in the glycolysis pathway is a C6 molecule or a C3 molecule.

Concept introduction: In the glycolysis metabolic pathway, a glucose molecule is converted into two pyruvate molecules. Two ATP molecules and NADH coenzymes are produced along with pyruvate.

The block diagram to represent an overview of glycolysis is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  6

From the above diagram, it is concluded that in the overall process of glycolysis, two stages are present.

a) Steps 1 to 3 represents a six-carbon stage (C6 stage).

b) Steps 4 to 10 represent a three-carbon stage (C3 stage).

In the C6 stage, the intermediates are derivatives of either glucose or fructose. Glucose and fructose contain six carbon atoms. Therefore, they are C6 molecules. The phosphate groups are present in the intermediates.

In the C3 stage, the intermediates are phosphorylated derivatives of glyceraldehyde, pyruvate, glycerate, or dihydroxyacetone, which are derivatives of either acetone or glycerol. Acetone and glycerol contain three carbon atoms. Therefore, they are C3 molecules.

(c)

Expert Solution
Check Mark

Answer to Problem 24.10EP

The intermediate glucose 6phosphate is a C6 molecule.

Explanation of Solution

The structure of glucose 6phosphate is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  7

The intermediate glucose 6phosphate is formed in the C6 stage and it is the phosphorylated derivative of glucose. Therefore, the intermediate glucose 6phosphate is a C6 molecule.

Conclusion

The intermediate glucose 6phosphate is formed in the C6 stage and it is the phosphorylated derivative of glucose. Six carbon atoms are present in glucose 6phosphate. Therefore, it is a C6 molecule.

(d)

Interpretation Introduction

Interpretation: To indicate whether the intermediate fructose 1,6biphosphate in the glycolysis pathway is a C6 molecule or a C3 molecule.

Concept introduction: In the glycolysis metabolic pathway, a glucose molecule is converted into two pyruvate molecules. Two ATP molecules and NADH coenzymes are produced along with pyruvate.

The block diagram to represent an overview of glycolysis is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  8

From the above diagram, it is concluded that in the overall process of glycolysis, two stages are present.

a) Steps 1 to 3 represents a six-carbon stage (C6 stage).

b) Steps 4 to 10 represent a three-carbon stage (C3 stage).

In the C6 stage, the intermediates are derivatives of either glucose or fructose. Glucose and fructose contain six carbon atoms. Therefore, they are C6 molecules. The phosphate groups are present in the intermediates.

In the C3 stage, the intermediates are phosphorylated derivatives of glyceraldehyde, pyruvate, glycerate, or dihydroxyacetone, which are derivatives of either acetone or glycerol. Acetone and glycerol contain three carbon atoms. Therefore, they are C3 molecules.

(d)

Expert Solution
Check Mark

Answer to Problem 24.10EP

The intermediate fructose 1,6biphosphate is a C6 molecule.

Explanation of Solution

The structure of fructose 1,6biphosphate is as follows:

General, Organic, and Biological Chemistry, Chapter 24, Problem 24.10EP , additional homework tip  9

The intermediate fructose 1,6biphosphate is formed in the C6 stage and it is the phosphorylated derivative of fructose. Therefore, the intermediate fructose 1,6biphosphate is a C6 molecule.

Conclusion

The intermediate fructose 1,6biphosphate is formed in the C6 stage and it is the phosphorylated derivative of fructose. Six carbon atoms are present in fructose 1,6biphosphate. Therefore, it is a C6 molecule.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Which enzyme is involved in the conversion of phosphoenolpyruvate to pyruvate?   a. enolase   b. ketolase   c. pyruvate kinase   d. pyruvate oxidase
In the electron transport chain, the hydrogen ions enter the inner compartment of mitochondria through special channels formed by A. ATP synthase. B. coenzyme A. C. acetyl CoA. D. oxygen.
110. The main purpose of glycolysis is to a. oxidize the acetyl group b. make NADH c. make pyruvate d. make protons e. make ATP

Chapter 24 Solutions

General, Organic, and Biological Chemistry

Ch. 24.2 - Prob. 6QQCh. 24.2 - Prob. 7QQCh. 24.3 - Prob. 1QQCh. 24.3 - Prob. 2QQCh. 24.3 - Prob. 3QQCh. 24.3 - Prob. 4QQCh. 24.3 - Accumulation of which of the following substances...Ch. 24.4 - Prob. 1QQCh. 24.4 - The net yield of ATP for the complete oxidation of...Ch. 24.4 - Prob. 3QQCh. 24.5 - Prob. 1QQCh. 24.5 - Prob. 2QQCh. 24.5 - Prob. 3QQCh. 24.6 - Prob. 1QQCh. 24.6 - Prob. 2QQCh. 24.6 - Prob. 3QQCh. 24.6 - Which of the following statements about ATP...Ch. 24.6 - Prob. 5QQCh. 24.7 - Prob. 1QQCh. 24.7 - Prob. 2QQCh. 24.8 - Prob. 1QQCh. 24.8 - Prob. 2QQCh. 24.8 - Prob. 3QQCh. 24.9 - Which of the following hormones promotes the...Ch. 24.9 - Which of the following pairs of hormones increases...Ch. 24.10 - Prob. 1QQCh. 24.10 - Prob. 2QQCh. 24.10 - Prob. 3QQCh. 24 - Where does carbohydrate digestion begin in the...Ch. 24 - Very little digestion of carbohydrates occurs in...Ch. 24 - Prob. 24.3EPCh. 24 - Prob. 24.4EPCh. 24 - Prob. 24.5EPCh. 24 - Prob. 24.6EPCh. 24 - Prob. 24.7EPCh. 24 - Prob. 24.8EPCh. 24 - Prob. 24.9EPCh. 24 - Prob. 24.10EPCh. 24 - Prob. 24.11EPCh. 24 - Prob. 24.12EPCh. 24 - Prob. 24.13EPCh. 24 - Prob. 24.14EPCh. 24 - Prob. 24.15EPCh. 24 - Prob. 24.16EPCh. 24 - Prob. 24.17EPCh. 24 - Prob. 24.18EPCh. 24 - Prob. 24.19EPCh. 24 - Prob. 24.20EPCh. 24 - Prob. 24.21EPCh. 24 - Prob. 24.22EPCh. 24 - Prob. 24.23EPCh. 24 - Prob. 24.24EPCh. 24 - Prob. 24.25EPCh. 24 - Prob. 24.26EPCh. 24 - Prob. 24.27EPCh. 24 - Prob. 24.28EPCh. 24 - Prob. 24.29EPCh. 24 - Prob. 24.30EPCh. 24 - Prob. 24.31EPCh. 24 - Prob. 24.32EPCh. 24 - Prob. 24.33EPCh. 24 - Prob. 24.34EPCh. 24 - Prob. 24.35EPCh. 24 - Prob. 24.36EPCh. 24 - Prob. 24.37EPCh. 24 - Prob. 24.38EPCh. 24 - Prob. 24.39EPCh. 24 - Prob. 24.40EPCh. 24 - Prob. 24.41EPCh. 24 - Prob. 24.42EPCh. 24 - Prob. 24.43EPCh. 24 - Prob. 24.44EPCh. 24 - Prob. 24.45EPCh. 24 - Prob. 24.46EPCh. 24 - Prob. 24.47EPCh. 24 - Prob. 24.48EPCh. 24 - Prob. 24.49EPCh. 24 - Prob. 24.50EPCh. 24 - Prob. 24.51EPCh. 24 - Prob. 24.52EPCh. 24 - Prob. 24.53EPCh. 24 - Prob. 24.54EPCh. 24 - Prob. 24.55EPCh. 24 - Prob. 24.56EPCh. 24 - Prob. 24.57EPCh. 24 - Prob. 24.58EPCh. 24 - Prob. 24.59EPCh. 24 - Prob. 24.60EPCh. 24 - Prob. 24.61EPCh. 24 - Prob. 24.62EPCh. 24 - Prob. 24.63EPCh. 24 - Prob. 24.64EPCh. 24 - Prob. 24.65EPCh. 24 - The liver, but not the brain or muscle cells, has...Ch. 24 - Prob. 24.67EPCh. 24 - Prob. 24.68EPCh. 24 - Prob. 24.69EPCh. 24 - Prob. 24.70EPCh. 24 - Prob. 24.71EPCh. 24 - Prob. 24.72EPCh. 24 - Prob. 24.73EPCh. 24 - Prob. 24.74EPCh. 24 - Prob. 24.75EPCh. 24 - Prob. 24.76EPCh. 24 - Prob. 24.77EPCh. 24 - Prob. 24.78EPCh. 24 - Prob. 24.79EPCh. 24 - Prob. 24.80EPCh. 24 - Prob. 24.81EPCh. 24 - Prob. 24.82EPCh. 24 - Prob. 24.83EPCh. 24 - Prob. 24.84EPCh. 24 - Prob. 24.85EPCh. 24 - Prob. 24.86EPCh. 24 - Prob. 24.87EPCh. 24 - Prob. 24.88EPCh. 24 - Prob. 24.89EPCh. 24 - Prob. 24.90EPCh. 24 - Prob. 24.91EPCh. 24 - Prob. 24.92EPCh. 24 - Prob. 24.93EPCh. 24 - Prob. 24.94EPCh. 24 - Prob. 24.95EPCh. 24 - Prob. 24.96EPCh. 24 - Prob. 24.97EPCh. 24 - Prob. 24.98EPCh. 24 - Prob. 24.99EPCh. 24 - Prob. 24.100EPCh. 24 - Prob. 24.101EPCh. 24 - Prob. 24.102EPCh. 24 - Prob. 24.103EPCh. 24 - Prob. 24.104EPCh. 24 - Prob. 24.105EPCh. 24 - Prob. 24.106EPCh. 24 - Prob. 24.107EPCh. 24 - Prob. 24.108EPCh. 24 - Prob. 24.109EPCh. 24 - Prob. 24.110EPCh. 24 - Prob. 24.111EPCh. 24 - Prob. 24.112EPCh. 24 - Prob. 24.113EPCh. 24 - Prob. 24.114EPCh. 24 - Prob. 24.115EPCh. 24 - Compare the biological functions of glucagon and...Ch. 24 - Prob. 24.117EPCh. 24 - Prob. 24.118EP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
Organic And Biological Chemistry
Chemistry
ISBN:9781305081079
Author:STOKER, H. Stephen (howard Stephen)
Publisher:Cengage Learning,
Text book image
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
Text book image
Chemistry In Focus
Chemistry
ISBN:9781305084476
Author:Tro, Nivaldo J., Neu, Don.
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9781305080485
Author:John E. McMurry
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning