Type O is the remainder after subtracting types A, B and AB.
so percentage of type O blood supply
= 100%-(28+15+10)%
= 100% - 53%
= 47%
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.
B I’m pretty sure if not it’s d But I always say go with ur gut but B
I believe the correct answer is B. False
The seven SI units are: kilogram, meter, second, kelvin,mole, ampere and candela.
Out of these seven, the one that will probably be part of the definition of a lumen is candela.
The luminous intensity of a light source is the the luminous flux emitted by that light source in a unit solid angle. The SI unit of luminous flux is lumen and the SI unit of luminous intensity is lumen divided by steradian [lumen/steradian] which is called Candela.<span />
