Q 5 Answer
Reaction measured : 2[ 3 H]glucose  Lactate + 3 H
2
O
Reaction Total Vol. = 1 ml
0.1ml PCA and 0.1 ml KOH added to stop reaction. Vol = 1.2 ml
0.5ml sampled and radioactivity counted for
– data in table)
3 H
2
O produced (say x dpm
Therefore total radioctivity in 3 H
2
O in the cell suspension after 12 min = x * 1.2/0.5
How much radioactivity has been added?
50
 l of a solution which contains 1.5

Ci/ml
Therefore 1.5 * 50/1000 = 0.075

Ci added per assay
This is equivalent to 0.075 *2.2*10 6 dpm
Or 165,000 dpm per assay
How much labelled glucose added?
0.075

Ci added per assay
Specific Radioactivity = 10 Ci/mmol or = 10

Ci/nmol nmol glucose added = 0.075

Ci / 10

Ci/nmol = 7.5*10 -3 nmol
(i.e. makes a negligible contribution to the total glucose see below)
How much unlabelled glucose added?
Final Glucose concentration :
(A) 50mM*50/1000 = 2.5 mM
(B) 400mM*50/1000 = 20 mM
(A) 2.5 mM or 2.5
 mol/ml or 2,500 nmol/assay
Therefore total glucose (Labelled and unlabelled) =
2500.0075 nmol
(B) 20 mM or 20
 mol/ml or 20,000 nmol/assay
Therefore total glucose (Labelled and unlabelled) =
20,000.0075 nmol
Therefore we can say that the labelled glucose makes virtually no difference to the total glucose in either case

So we can say that either (A)2.5 or (B) 20
 mol glucose : is equivalent to 165,000 dpm
Rate is calculated from the fraction of the original glucose transformed per min (12 min ) and per g cells (0.05g)
Rate A
2.5*1500*1.2/0.5* 1/165,000*1/12*1/0.05
= 0.091
 mol.min
-1 .g cells -1
Rate B
20*20045*1.2/0.5* 1/165,000*1/12*1/0.05
= 9.72
 mol.min
-1 .g cells -1
Rate A {plus glucagon}
2.5*1,495*1.2/0.5* 1/165,000*1/12*1/0.05
= 0.091
 mol.min
-1 .g cells -1
Rate B {plus glucagon}
20*3,120*1.2/0.5* 1/165,000*1/12*1/0. 05
= 4.34
 mol.min
-1 .g cells -1
Rate is unchanged by glucagon at 2.5 mM glucose but inhibited by glucagon by 55% at 20 mM glucose
Conclusion: The high rate of glycolysis at 20 mM glucose is related to the effect of glucose to increase the fructose 2,6-bisphosphate concentration at this concentration. The effect of glucagon is to lower fructose 2,6-bisphosphate levels in the liver (see mechanism in textbook) which reduces the rate of glycolysis
The highest % of glucose utilised is
Total dpm utilised=20,045 * 1.2*0.5 = 48,108 dpm
Dpm in glucose at start = 165,000
% incorporation = 48108*100/165000 = 29.16%