Question

In an anaerobic muscle preparation, lactate formed from 4-14C-glucose would yield:

Answer: CH3-CH(OH)-14COO-

But why is this the answer?

Answer 1

The carbon C4 is radio labeled (C14).

C4 is present in the following conversion,

Glucose - Glucose 6 Phosphate - Fructose 6 phosphate - Fructose 1,6 bisphosphate

When fructose 1, 6 bisphosphate is split into dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3 phosphate (G3P), 1st three carbon is formed as DHAP and the carbon 4, 5 and 6 is formed as G3P.

Therefore, that 4th carbon becomes 1st carbon of G3P and which is further converted to intermediates and finally to form lactate. The carboxy carbon atom holds the radio labeled C14.