Explanation:
soil organisms which range in size from microscopic cells that digest decaying organic material to small mammals that live primarily on other soil organisms play an important role in maintaining fertility structure drainage and aeration of soil
The answer to number 12 is mate.
Answer:
1. G° = -RT ln (G1P/P)
3.1 = 8.314 × 310 × ln (G1P/P)
3.1 / 2577.34 = ln (G1P/P)
0.0012 = ln (G1P/P)
0.0012 = (log G1P/P)/log 2.71828
0.4342 × 0.0012 = log G1P/P
0.00052 = log G1P/P
G1P/P = 10^0.00052 = 1.0012
P/G1P = 1/1.0012 = 0.9988
2. The cleavage of glycogen phosphorolytically is beneficial for the cell to conduct the process as the discharged glucose is phosphorylated. A general hydrolytic cleavage would give rise to only a glucose, which has to be phosphorylated again with the help of ATP.
Another merit of phosphorylated glucose is that it comprises the negative charge and cannot diffuse out of the muscle cell. Thus, the reaction will not be at equilibrium under the physiological conditions and always encourages the generation of the products. The formation of products will amend the change in free energy in such a manner that the reaction will always carry in the forward direction.
3. Greater the ratio of [Pi]/[glucose-1-phosphate], higher will be the relative rate of glycogen phosphorylase in comparison to the phosphoglucomutase as the transformation of Glu-1-P becomes slow because of lesser accessibility of substrate.
