A researcher labels C‑6 of glucose 6‑phosphate with C14 and adds it to a solution containing the enzymes and cofactors of the oxidative phase of the pentose phosphate pathway. What is the fate of the radioactive label?

A. C14 appears at C‑4 of erythrose 4‑phosphate
B. C14 appears at C‑7 of sedoheptulose 7‑phosphate
C. C14 appears at C‑5 of ribulose 5‑phosphate
D. C14 appears at C‑6 of fructose 6‑phosphate
E. C14 appears in the CO2 evolved by the oxidative phase.

D. C14 appears at C‑6 of fructose 6‑phosphate

Explanation:

Conversion of glucose B-phosphate via 6-phospho-

gluconate to pentose phosphate and CO2 followed by conversion

of the pentose phosphate to fructose 6-phosphate and glyceralde-hyde phosphate followed by conversion of the fructose 6-phos-

phate to glucose g-phosphate .

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b

idk what is it? is this a question that you know the answer or u dont