IChO 1985 · Theory problem 8

Glycolysis equilibria of glucose-1-phosphate, glucose-6-phosphate and fructose-6-phosphate

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The following reaction scheme represents part of anaerobic degradation of saccharides, i.e. the glycolysis, involving equilibrium constants \(K_1\) and \(K_2\):

Figure 1

Problems:

8.1 Give the structural formulae for all the three reactants (compounds) that are mutually interconverted, i.e. α-D-glucose-1-phosphate, α-D-glucose-6-phosphate and α-D-fructose-6-phosphate.

8.2 In the beginning of the reaction the reaction mixture contained 1 mmol of glucose-6-phosphate. Calculate the amounts of glucose-6-phosphate, glucose-1-phosphate and fructose-6-phosphate in the mixture at equilibrium. (As the reaction take place in a constant volume, the ratio of the amounts of substances equals that of their concentrations.)

Official English text of the problem.

Answers

  • 8.1 The structures, written with fully protonated phosphate groups, are:
  • α-D-glucose-1-phosphate, in its glucopyranose form:
    OC[C@H]1O[C@H](OP(=O)(O)O)[C@H](O)[C@@H](O)[C@@H]1O
  • α-D-glucose-6-phosphate, in its glucopyranose form:
    O=P(O)(O)OC[C@H]1O[C@H](O)[C@H](O)[C@@H](O)[C@@H]1O

  • α-D-fructose-6-phosphate, in its fructofuranose form:
    O=P(O)(O)OC[C@H]1O[C@@](O)(CO)[C@@H](O)[C@@H]1O

  • 8.2 The equilibrium amounts, rounded to two significant figures, are:

  • Glucose-6-phosphate: \(0.64\ \mathrm{mmol}\).
  • Glucose-1-phosphate: \(0.034\ \mathrm{mmol}\).
  • Fructose-6-phosphate: \(0.32\ \mathrm{mmol}\).

Solution

8.1 — Structures and stereochemistry

The phosphate ester group is represented as \(\ce{O-P(=O)(OH)2}\). This specifies the acid form; the problem does not prescribe a pH or phosphate protonation state.

For the two glucose derivatives, draw a six-membered pyranose ring. Place the ring oxygen, O5, at the upper right, followed clockwise by C1, C2, C3, C4 and C5.

For the fructose derivative, draw a five-membered furanose ring. Place O5 at the top, followed clockwise by C2, C3, C4 and C5. Here C2 is the anomeric carbon. The ring results from addition of the C5 hydroxyl group to the C2 carbonyl group; C6 carries the phosphate ester.

The complete Haworth substituent orientations are:

Compound Substituents below the ring Substituents above the ring
α-D-glucose-1-phosphate \(\ce{O-P(=O)(OH)2}\) at C1; hydroxyl groups at C2 and C4 Hydroxyl group at C3; \(\ce{CH2OH}\) at C5
α-D-glucose-6-phosphate Hydroxyl groups at C1, C2 and C4 Hydroxyl group at C3; \(\ce{CH2-O-P(=O)(OH)2}\) at C5
α-D-fructose-6-phosphate Hydroxyl groups at C2 and C4 \(\ce{CH2OH}\) at C2; hydroxyl group at C3; \(\ce{CH2-O-P(=O)(OH)2}\) at C5

In each structure, the side chain containing C6 points above the ring. The α configuration places the anomeric oxygen substituent on the opposite side: the phosphate ester at C1 for glucose-1-phosphate, the hydroxyl group at C1 for glucose-6-phosphate, and the hydroxyl group at C2 for fructose-6-phosphate.

Hydrogens on the remaining ring stereocentres point opposite the listed substituents. These Haworth descriptions specify configuration; they do not imply that the actual rings are planar.

8.2 — Equilibrium composition

Let the equilibrium amounts of glucose-1-phosphate, glucose-6-phosphate and fructose-6-phosphate be \(a\), \(b\) and \(c\), respectively.

The equilibrium constants refer to the forward directions shown in the scheme. Since the volume is constant,

\[ K_1=\frac{b}{a}=19, \qquad K_2=\frac{c}{b}=0.50. \]

Thus,

\[ a=\frac{b}{19}, \qquad c=0.50b. \]

Each interconversion replaces one molecule with one molecule, so the total amount of the three compounds is conserved:

\[ a+b+c=1\ \mathrm{mmol}. \]

Substituting gives

\[ \frac{b}{19}+b+0.50b=1\ \mathrm{mmol}, \]

and hence

\[ b=\frac{1}{1+\frac{1}{19}+0.50}\ \mathrm{mmol} =\frac{38}{59}\ \mathrm{mmol}. \]

The individual amounts are therefore

\[ \begin{aligned} n(\text{glucose-6-phosphate}) &=\frac{38}{59}\ \mathrm{mmol} =0.6440678\ldots\ \mathrm{mmol} \approx \boxed{0.64\ \mathrm{mmol}},\\[2mm] n(\text{glucose-1-phosphate}) &=\frac{2}{59}\ \mathrm{mmol} =0.0338983\ldots\ \mathrm{mmol} \approx \boxed{0.034\ \mathrm{mmol}},\\[2mm] n(\text{fructose-6-phosphate}) &=\frac{19}{59}\ \mathrm{mmol} =0.3220339\ldots\ \mathrm{mmol} \approx \boxed{0.32\ \mathrm{mmol}}. \end{aligned} \]

Before rounding, these amounts sum to \(1\ \mathrm{mmol}\) and reproduce both equilibrium constants exactly.

The solution was checked against the official solution.