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.
TRUE.................................................
Desert plants will have very long roots to reach water. Aquatic plants may be able to float to collect water or have sturdy roots to hold them in place.
If the mutation occurs during meiosis, the mutation will be incorporated into a gamete. If that gamete is the one that eventually fuses with another gamete (i.e. if it's the sperm that fertilizes the egg), that mutation will be passed on to the offspring. As all the offspring's cells are the result of the first two gametes, all the organism's cells will have that mutation. Obviously this can have dire consequences for the offspring, if the mutation is harmful.
An organism that eats decomposed items. For example plant detritus.
