Lehninger Principles of Biochemistry
Lehninger Principles of Biochemistry
7th Edition
ISBN: 9781464126116
Author: David L. Nelson, Michael M. Cox
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 17DAP

(a1)

Summary Introduction

To calculate: The cellular concentration of free glucose that would be required to store an equivalent amount of glucose and also explain the reason that this concentration of free glucose would present a problem for the cell.

Introduction:

Glucose is stored in muscle and liver in the form of glycogen. Glycogen is a glucose polymer in which glucose molecules are bridged by glycosidic bonds. The concentration of glucose in the cell increases to a large extent in comparison to the glycogen.

(a1)

Expert Solution
Check Mark

Explanation of Solution

Each glycogen contains nearly 55,000 moieties of glucose. The concentration of glycogen in cell is 0.01 µM. Thus, the equivalent glucose concentration will be:

55,000×0.01μM=0.55M

Conclusion

The cellular concentration of free glucose would be 0.55 M and this concentration of free glucose would hamper the osmolarity of the cell.

(a2)

Summary Introduction

To explain: The reason that this concentration of free glucose would present a problem for the cell.

Introduction:

Glucose is storied in muscle and liver in the form of glycogen. Glycogen is a glucose polymer in which glucose molecules are bridged by glycosidic bonds. The concentration of glucose in the cell increases to a large extent in comparison to the glycogen.

(a2)

Expert Solution
Check Mark

Explanation of Solution

The concentration of free glucose would be 0.55 M, and then it would dominate or hamper the cell osmolarity.

(b)

Summary Introduction

To determine: The reason that low degree of branching reduces the rate of glucose release.

Introduction:

Muscle cells require rapid access to glucose during strenuous exercise. This glucose is stored in skeletal muscle and liver in the form of glycogen. The typical glycogen particle contains nearly 55,000 residues of glucose.

(b)

Expert Solution
Check Mark

Explanation of Solution

Glycogen phosphorylase converts glycogen into glucose by breaking α-glycosidic linkage. The activity exhibits only at non-reducing ends. The branching present in glucose increases non-reducing nature which increases the activity of enzyme leading to achievement of product in less time duration. Further, branching increases number of sites accessible to enzymes and also increases the solubility. On the other hand, chain of glucose residue is more insoluble and difficult to expose non-reducing end and then cleave.

(c)

Summary Introduction

To determine: The reason that high degree of branching also reduces the rate of glucose release.

Introduction:

Glucose is released from glycogen with the help of enzyme, glycogen phosphorylase that removes glucose molecule from one end of glycogen chain turn by turn. Glycogen chains are branched and number of branches per chain influences the rate of release of glucose by glycogen phosphorylase.

(c)

Expert Solution
Check Mark

Explanation of Solution

High degree of branching makes the structure more complicated and debranching enzyme has to act on all the alpha linkages at the branching points. This makes the reaction time consuming and complicated.

(d)

Summary Introduction

To show:CA=2t-1 that represents the number of chains available to glycogen phosphorylase before the action of the debranching enzyme.

Introduction:

Glucose is storied in muscle and liver in the form of glycogen. Glycogen is a glucose polymer in which glucose molecules are bridged by glycosidic bonds. The concentration of glucose in the cell increases to a large extent in comparison to the glycogen.

(d)

Expert Solution
Check Mark

Explanation of Solution

The number of chains doubles with succeeding tier. Tier 1 has single chain 20, tier 2 has 2 chains 21 and so on. Thus, the tier “t” should have 2t chains. Thus, to calculate the number of chains in the outermost tier, the formula to be used is 2t-1.

(e)

Summary Introduction

To show: - CT, the total number of chains present in the particle is given by CT=2t-1. Thus GT=ge(CT)= ge (2t-1)is the total number of glucose residues in the particle.

Introduction:

Glucose is released from glycogen with the help of enzyme, glycogen phosphorylase that removes glucose molecule from one end of glycogen chain turn by turn. Glycogen chains are branched and number of branches per chain influences the rate of release of glucose by glycogen phosphorylase.

(e)

Expert Solution
Check Mark

Explanation of Solution

The total number of chains is:

(20)+(21)+(2t1)=2t1.

ge is the number if glucose molecules present in each chain. Thus, the total number of glucose molecules are ge×(2t1).

Conclusion

The total number of glucose residues in the particle is ge×(2t-1)_.

(f)

Summary Introduction

To show: GPT= (ge4)(2t-1) is the amount of glucose that is readily available to glycogen phosphorylase.

Introduction:

Glucose is released from glycogen with the help of enzyme, glycogen phosphorylase that removes glucose molecule from one end of glycogen chain turn by turn. Glycogen chains are branched and number of branches per chain influences the rate of release of glucose by glycogen phosphorylase.

(f)

Expert Solution
Check Mark

Explanation of Solution

Glycogen phosphorylase can act upto four molecules of glucose residue far from the branching site. From each outer tier it can release ge-4 number of residues. There are 2t-1chains, thus the enzyme can release GPT number of glucose molecules which is given by (ge-4)( 2t-1).

Conclusion

There are 2t-1 chains, thus GPT= (ge4)(2t-1)_ is the amount of glucose that is readily available to glycogen phosphorylase.

(g)

Summary Introduction

To show: That the volume of a particle Vs is given by the equation Vs=43πt3(0.12ge+0.35)3nm3 based on size and location of branches of glucose.

Introduction:

Glucose is storied in muscle and liver in the form of glycogen. Glycogen is a glucose polymer in which glucose molecules are bridged by glycosidic bonds. The concentration of glucose in the cell increases to a large extent in comparison to the glycogen.

(g)

Expert Solution
Check Mark

Explanation of Solution

The volume of the glucose molecule with sphere is Vs=43πr3. In the case, where r is the thickness of each tire then the thickness of all the tiers will be (0.12g+0.35) nm. Thus, the volume of sphere is Vs=43πt3(0.12ge+0.35)3nm3.

Conclusion

The volume of a particle Vs is 43πt3(0.12ge+0.35)3nm3_.

(h)

Summary Introduction

To determine: The optimal value of ge.

Introduction:

Glucose is released from glycogen with the help of enzyme, glycogen phosphorylase that removes glucose molecule from one end of glycogen chain turn by turn. Glycogen chains are branched and number of branches per chain influences the rate of release of glucose by glycogen phosphorylase.

(h)

Expert Solution
Check Mark

Explanation of Solution

Tabular representation: This represents the value of ge that maximized f is independent of t.

gcCAGTGPTVSf
564635641,2322,111
6647621281,7603,547
7648891922,4214,512
8641,0162563,2305,154
9641,1433204,2015,572
10641,2703845,3505,834
11641,3974486,6925,986
12641,5245128,2406,060
13641,65157610,0116,079
14641,77864012,0196,059
15641,90570414,2796,011
16642,03276816,8065,943

The optimal ge value ranges from 12 to 14. However, the value of “t” may be any value due to no restriction over the range. Algebraically the value of ge that maximizes the value of “f” is independent of “t”. Thus, the optimal value of ge is 13, even if the value of “t” is changed.

Conclusion

The optimal value of ge is 13.

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!
Knowledge Booster
Background pattern image
Biochemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Biochemistry
Biochemistry
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:W. H. Freeman
Text book image
Lehninger Principles of Biochemistry
Biochemistry
ISBN:9781464126116
Author:David L. Nelson, Michael M. Cox
Publisher:W. H. Freeman
Text book image
Fundamentals of Biochemistry: Life at the Molecul...
Biochemistry
ISBN:9781118918401
Author:Donald Voet, Judith G. Voet, Charlotte W. Pratt
Publisher:WILEY
Text book image
Biochemistry
Biochemistry
ISBN:9781305961135
Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal
Publisher:Cengage Learning
Text book image
Biochemistry
Biochemistry
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Cengage Learning
Text book image
Fundamentals of General, Organic, and Biological ...
Biochemistry
ISBN:9780134015187
Author:John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Publisher:PEARSON
Biomolecules - Protein - Amino acids; Author: Tutorials Point (India) Ltd.;https://www.youtube.com/watch?v=ySNVPDHJ0ek;License: Standard YouTube License, CC-BY