Apologies that I gave you the incorrect solution in Study session 2
Here is the correct version:
Q4
Reaction : 2[3H]glucose  2[3H]glucose-6-phosphate
2[3H]glucose-6-phosphate  fructose-6-phosphate + 3H2O
Reaction Vol. = 1 ml
0.1ml PCA and KOH added to stop reaction. Vol = 1.1 ml
0.5ml sampled and radioactivity counted for 3H2O (say x dpm)
Therefore total radioctivity in 3H2O = x * 1.1/0.5
How much radioactivity has been added?
5 l of a solution which contains 25Ci/ml
Therefore 25 * 5/1000 = 0.125 Ci added per assay
This is equivalent to 0.125 *2.2*106 dpm
Or 275,000 dpm per assay
How much labelled glucose added?
0.125 Ci added per assay
Specific Radioactivity = 10 Ci/mmol or = 10 Ci/nmol
nmol added = 0.125 Ci / 10 Ci/nmol = 12.5*10-3 nmol
How much unlabelled glucose added?
4 mM or 4 mol/ml or 4,000 nmol/assay
Therefore total glucose (Labelled and unlabelled) =
4,000.0125
Therefore we can say that the labelled glucose makes virtually no
difference to the total glucose
So we can say that either 4 or 40 mol glucose :
is equivalent to 275,000 dpm
Rate is calculated from the fraction of the original glucose
transformed per min and per g cells
Rate 1
4*1500*1.1/0.5* 1/275,000*1/8*1/0.1
= 0.06 mol.min-1.g cells-1
Rate 2
40*3080*1.1/0.5* 1/275,000*1/8*1/0.1
= 1.232 mol.min-1.g cells-1
Rate 3 {plus fructose}
4*10,300*1.1/0.5* 1/275,000*1/8*1/0.1
= 0.412 mol.min-1.g cells-1
Rate 4 {plus fructose}
40*3,120*1.1/0.5* 1/275,000*1/8*1/0.1
= 1.248 mol.min-1.g cells-1
Rate increased 6.9 – fold at 4 mM glucose but not at 40 mM glucose
Conclusion: The effect is to change the Km of the enzyme for its
substrate at a physiologically relevant concentration of glucose