You're examining two different mutant liver cell lines in which gpk is only activated 50% when the cells are treated with epinephrine. in cell line "a" gpk is rapidly inactivated when epinephrine is removed. in cell line "b" gpk activity persists longer before it is fully inactivated. which of the following accounts for the difference between the two cell lines?

Glycogen phosphorylase kinase (GPK) that plays an important role in glycogenolysis. The enzyme release the glucose 1- phosphate from the  alpha-1,4-glycosidic bond.

The alpha and beta subunit of cell line A get changed. GPK activates in case when the phosphorylation is removed or Ca2+-Calmodulin is eliminated. A cell line has reversed the Ca2+•Calmodulin activity dependent on GPK. This cell line only requires the calcium level back into cytosol. The cell line B PKA ( protein kinase A) activation that are dependent on GPK is slowoly reversed in the absence of the calcium ion beacuse protein dehydrogenase 1 (PDK) is not get activated. This causes the fall in cAMP level slowly and allow the longer persistence of the PKA activity & GPK phosphorylation.

Thus, the answer is In cell line "A" certain S or Ts on the α+ β subunits of GPK were changed to A. In cell line "B"; the ER Ca2+ channels could not bind IP3.

answer: 21

step-by-step explanation:

divide 630 by 30. 30 goes into 630 21 times

b - the respond is innate and varies over time.

good luck!