Answer:
0.095
Explanation:
Phenylkentonuria is a disease caused by a recessive allele.
The frequency of the recessive allele + the frequency of the dominant allele equals 1.
The frequency of the recessive allele is q = 0.05
The frequency of the dominant allele then is p = 1 - q = 0.95
If people mate randomly, the frequency of the homozygous dominant genotype will be p², the frequency of the heterozygous genotype will be 2pq and the frequency of the homozygous recessive genotype will be q² .
2pq=2× 0.05 × 0.95
2pq=0.095
The heterozygote frequency in the population is 0.095
The correct answer is - Its electrons are used to form NADPH.
On receiving light energy, electrons are expelled from the reaction center of photosystem II. The expelled electrons finally reduces oxidized NADH⁺ to NADH. The oxidized reaction center of photosystem II split water into protons, electrons and oxygen. The electrons released from water reduces oxidized reaction center of photosystem II. Thus, oxidized reaction center of photosystem II gets back its expelled electrons. Therefore, electrons from water forms NADH.
<span>A seatbelt should be fastened so as to run diagonally across the chest and to fit low and tight across the waist. This ensures that, during the rapid deceleration experienced during a crash, the seatbelt will perform the intended function, that is, to keep the passenger in place within the vehicle.</span>
Answer:
C AND E?
Explanation:
If it's right let me know. If not, then you can tell me the answer so I can change it for everyone to get the answer
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.
