To study the fate of carbon during glycolysis under aerobic conditions, cultured cells are fed with 14C-glucose labeled in various positions. You use a radiolabeled form of glucose where the radioactive 14C occupies position 1. Which of the following is correct?a. The carboxyl group of acetyl-CoA and CO2 become radiolabeledb. The methyl group of acetyl CoA and CO2 become radiolabeledc. The carbonyl group of acetyl-CoA becomes radiolabeledd. The methyl group of acetyl CoA becomes radiolabelede. CO2 becomes radiolabeledf. I don’t have enough information to conclude

D. The methyl group of acetyl CoA becomes radio-labeled

Explanation:

During the steps in glycolysis, the carbon at position 1, becomes C-1 in dihydroxyacetone phosphate during the cleavage of fructose-1,6-bisphosphate to dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. Subsequently on isomerization of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate, C-1 of dihydroxyacetone phosphate becomes C-3 of glyceraldehyde-3-phosphate.

Furthermore, in pyruvate, the end product of glycolysis, C-3 is converted to a methyl group which then becomes the methyl group in the acetyl-CoA molecule produced from the oxidative decarboxylation of pyruvate.

Since the radioactive 14-C of radio-labeled glucose occupies position 1, it will become the methyl group of acetyl-CoA.

i'm gonna put a couple so you can chose from

explanation:

if you know the fur color of the offspring mice, you can use the ratios to identify the genotypes of the parents.

if either parent mouse passes a dominant allele, the offspring will have black fur.

if the parental mice has two different fur color (i.e white and black) , the color of the offspring's fur would be either black, white, or the combination of both (grey)

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