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.
Count the ones expressing the trait in the F2 gen. Divide by how many animals there are in F2. Times by 100.
Answer:
Yes, Since the interphase cells in G2 would have twice the DNA as the interphase cells in G1 phase or start of S phase.
Explanation:
Interphase has three stages: G1, S and G2 phase. DNA replication occurs in the S phase and doubles the DNA content of the cell. So, the DNA content of the interphase cells of the same tissue depends on stages of interphase.
The interphase cells which are in the G1 phase has half the DNA content as it is present in the interphase that has entered the G2 phase after completion of DNA replication in S phase.
