Concept explainers
(a1)
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
(a1)
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:
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)
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)
Explanation of Solution
The concentration of free glucose would be 0.55 M, and then it would dominate or hamper the cell osmolarity.
(b)
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)
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)
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)
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)
To show:
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)
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)
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)
Explanation of Solution
The total number of chains is:
ge is the number if glucose molecules present in each chain. Thus, the total number of glucose molecules are
The total number of glucose residues in the particle is
(f)
To show:
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)
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).
There are 2t-1 chains, thus
(g)
To show: That the volume of a particle
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)
Explanation of Solution
The volume of the glucose molecule with sphere is
The volume of a particle Vs is
(h)
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)
Explanation of Solution
Tabular representation: This represents the value of ge 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
The optimal value of ge is 13.
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