Answer:
Option-A
Explanation:
Pentose phosphate pathway is the alternative pathway taking place in both eukaryotes and prokaryotes parallel to the glycolysis.
The pentose phosphate pathway forms the electron carriers called NADPH and the ribose-5 phosphate.
The NADPH is formed in large amounts through this pathway which contains a very high amount of energy. NADPH can donate its hydride ions, therefore, it can be involved in the reaction where it can donate energy in the form of hydrogen and electron.
The NADPH is therefore involved in the formation of reducing biosynthetic pathways like fatty acid synthesis, cholesterol synthesis, and bile acid synthesis.
Thus, Option-A is the correct answer.
A is for sure an answer , i’m pretty sure C -ISNT- an answer . for b and c, someone can comment on it.
Answer:
we will know that the allelic frequencies are for R 0.95 and r 0.05
Explanation:
We know that the population is in Hardy-Winberg equilibrium, we deduce the following formula:
p + q = 1
p2 + 2pq + q2 = 1
data
R: red flower allele
r: allele blor blanca
p would be equal to the allelic frequency R
q will be equal to the frequency allelic r
2p = RR
2q = rr
2pq = Rr
If there are 25 white flowers in 1000 plants, their frequency will be:
2pq frequency of the Rr genotype
white flower = 25/10000 = 0.0025 = rr = 2q = 0.0025
we deduce that q is equal to 0.05
we replace the data with the previous formula
p + q = 1
p = 1-0.05
we get as a result
p = 0.95
if p = 0.95 and q = 0.05
we will know that the allelic frequencies are for R 0.95 and r 0.05
