Question

.) (3pts) If radioactive glucose labeled at carbon #6 with 14C, were incubated in a cell-free liver homogenate under anaerobic conditions, which carbon in the lactate produced would be labeled with 1℃ (refer to the carboxyl carbon of lactate as C1, the central carbon as C2, and the methyl carbon as C3)? What if theglucose were labeled at carbon #3 instead? In both cases, explain your answer.

Answer 1

Answer:

• Radioactive glucose, named at carbon #6 with 14C will deliver lactate with named 14C at third Carbon (C3, methyl carbon).

• This could be effectively comprehended by observing the system of glycolysis (anaerobic condition).

• C6 particle (out of the heterocyclic ring) is 14C molecule.

• Amid different advances glucose changes over into Fructose-6-phosphate and still this C6-iotas stays at a similar position in 5-membered ring.

• At that point this ring breaks into Glyceraldehyde 3-phosphate and Dihydroxy CH3)2CO phosphate. here 14C is available in Glyceraldehyde 3-phosphate at Carbon no. 3.

• This Glyceraldehyde 3-phosphate convers into Pyruvate through different advances yet at the same time the C-iota is available at a similar third position in methyl gathering.

• This pyruvate changes over into lactate by decrease and the C-iota (14C) is still here at third position.

• On the off chance that the glucose were marked at carbon 3, at that point it will be at CH2OH a mass in Dihydroxy CH3)2CO phosphate and this compound additionally changes over into Glyceraldehyde 3-phosphate however the 14C will be available in the aldehyde gathering.

• so finally the C1 carbon particle of lactate will be 14C.