The answers B we're using them faster than they can form
Answer:
Option (D).
Explanation:
ATP (adenotriphosphate) molecules are considered as energy currency of the cells as molecules provide energy for various cellular functions.
The energy is stored in ATPs in two high-energy phosphate bonds, known as phosphoanhydride bonds. This stored energy is released in during hydrolysis of ATP, which involves removal of terminal phosphate groups bound the carbon backbone.
Thus, the correct answer is option (D).
Assuming dragon genetics follow the same rules as fruit flies, we would get the same possible genotype for all 16 offspring provided that the genes are not linked.
Considering dragon genetics, flame eyes (F) are dominant to blue eyes (f) and burbling (B) is dominant to whistling (b).
Now, a dihybrid cross between two homozygous blue-eyed, whistling dragons will yield 16 offspring all with the same possible genotype .i.e. homozygous blue-eyed, whistling type.
Morgan through experiments on fruit flies observed that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combination were much higher than the non-parental type.
He attributed this due to the physical association or linkage of the two genes and coined the term 'linkage' to describe the physical association of genes on a chromosome. The term 'recombination' is to describe the generation of non-parental gene combination.
To learn more about dihybrid cross here
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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
