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.

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:

$\text{55,000}\times \text{0}\text{.01}\text{μM}=\text{0}\text{.55M}$

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.

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.

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.

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:${\text{C}}_{\text{A}}{\text{=2}}^{\text{t-1}}$ 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.

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 2^t chains. Thus, to calculate the number of chains in the outermost tier, the formula to be used is 2^t-1.

(e)

Summary Introduction

To show: - C_T, the total number of chains present in the particle is given by C_T=2^t-1. Thus G_T=g_e(C_T)= g_e (2^t-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.

Explanation of Solution

The total number of chains is:

$\left({\text{2}}^0\right)+\left({\text{2}}^1\right)+\dots \left({\text{2}}^{t-1}\right)={\text{2}}^{t-1}$.

g_e is the number if glucose molecules present in each chain. Thus, the total number of glucose molecules are $g_e\times \left({\text{2}}^{t-1}\right)$.

Conclusion

The total number of glucose residues in the particle is $\underset{\_}{g_e\times \left(2^{t-1}\right)}.$

(f)

Summary Introduction

To show: ${\text{G}}_{\text{PT}}{\text{= (g}}_{\text{e}}-{\text{4)(2}}^{\text{t-1}}\text{)}$ 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.

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 g_e-4 number of residues. There are 2^t-1chains, thus the enzyme can release G_PT number of glucose molecules which is given by (g_e-4)( 2^t-1).

Conclusion

There are 2^t-1 chains, thus $\underset{\_}{{\text{G}}_{\text{PT}}{\text{= (g}}_{\text{e}}-{\text{4)(2}}^{\text{t-1}}\text{)}}$ is the amount of glucose that is readily available to glycogen phosphorylase.

(g)

Summary Introduction

To show: That the volume of a particle $V_s$ is given by the equation $V_s=\frac{\text{4}}{\text{3}}\text{π}{t}^3{\left(0.12g_e+0.35\right)}^3{\text{nm}}^{\text{3}}$ 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.

Explanation of Solution

The volume of the glucose molecule with sphere is $V_s=\frac{\text{4}}{\text{3}}\text{π}{r}^3$. 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 $V_s=\frac{\text{4}}{\text{3}}\text{π}{t}^3{\left(0.12g_e+0.35\right)}^3{\text{nm}}^{\text{3}}$.

Conclusion

The volume of a particle V_s is $\underset{\_}{\frac{4}{3}\pi t^3{\left(0.12g_e+0.35\right)}^{3}n{m}^3}.$

(h)

Summary Introduction

To determine: The optimal value of g_e.

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.

Explanation of Solution

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

gc	CA	GT	GPT	VS	f
5	64	635	64	1,232	2,111
6	64	762	128	1,760	3,547
7	64	889	192	2,421	4,512
8	64	1,016	256	3,230	5,154
9	64	1,143	320	4,201	5,572
10	64	1,270	384	5,350	5,834
11	64	1,397	448	6,692	5,986
12	64	1,524	512	8,240	6,060
13	64	1,651	576	10,011	6,079
14	64	1,778	640	12,019	6,059
15	64	1,905	704	14,279	6,011
16	64	2,032	768	16,806	5,943

The optimal $g_e$ 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 $g_e$ that maximizes the value of “f” is independent of “t”. Thus, the optimal value of g_e is 13, even if the value of “t” is changed.

Conclusion

The optimal value of g_e is 13.