Answer
Data Analysis Problem
17. Optimal Glycogen Structure Muscle cells need rapid access to large amounts of glucose during heavy
exercise. This glucose is stored in liver and skeletal muscle in polymeric form as particles of glycogen.
The typical glycogen particle contains about 55,000 glucose residues (see Fig. 15–35b). Meléndez-Hevia,
Waddell, and Shelton (1993) explored some theoretical aspects of the structure of glycogen, as de-
scribed in this problem.
(a) The cellular concentration of glycogen in liver is about 0.01
M
. What cellular concentration of
free glucose would be required to store an equivalent amount of glucose? Why would this con-
centration of free glucose present a problem for the cell?
Glucose is released from glycogen by glycogen phosphorylase, an enzyme that can remove glucose
molecules, one at a time, from one end of a glycogen chain. Glycogen chains are branched (see
Figs 15–28 and 15–35b), and the degree of branching—the number of branches per chain—has a
powerful influence on the rate at which glycogen phosphorylase can release glucose.
(b) Why would a degree of branching that was too low (i.e., below an optimum level) reduce the rate
of glucose release? (Hint: Consider the extreme case of no branches in a chain of 55,000 glucose
residues.)
(c) Why would a degree of branching that was too high also reduce the rate of glucose release?
(Hint: Think of the physical constraints.)
Meléndez-Hevia and colleagues did a series of calculations and found that two branches per chain
(see Fig. 15–35b) was optimal for the constraints described above. This is what is found in glycogen
stored in muscle and liver.
To determine the optimum number of glucose residues per chain, Meléndez-Hevia and coauthors
considered two key parameters that define the structure of a glycogen particle: tthe number of tiers
of glucose chains in a particle (the molecule in Fig. 15–35b has five tiers); g
c
the number of glucose
residues in each chain. They set out to find the values of tand g
c
that would maximize three quantities:
(1) the amount of glucose stored in the particle (G
T
) per unit volume; (2) the number of unbranched
glucose chains (C
A
) per unit volume (i.e., number of chains in the outermost tier, readily accessible to
glycogen phosphorylase); and (3) the amount of glucose available to phosphorylase in these unbranched
chains (G
PT
).
(d) Show that C
A
2
t1
. This is the number of chains available to glycogen phosphorylase before
the action of the debranching enzyme.
(e) Show that C
T
, the total number of chains in the particle, is given by C
T
2
t
1. Thus G
T
g
c
(C
T
) g
c
(2
t
1), the total number of glucose residues in the particle.
(f) Glycogen phosphorylase cannot remove glucose from glycogen chains that are shorter than five
glucose residues. Show that G
PT
(g
c
4)(2
t1
). This is the amount of glucose readily avail-
able to glycogen phosphorylase.
(g) Based on the size of a glucose residue and the location of branches, the thickness of one tier of
glycogen is 0.12g
c
nm 0.35 nm. Show that the volume of a particle, V
s
, is given by the
equation V
s
4
3
t
3
(0.12g
c
0.35)
3
nm
3
.
Chapter 15 Principles of Metabolic Regulation S-181