Answer:
2.Less than 73% of the populations would have only one allele present.
Explanation:
The two alleles chosen do not affect the fitness of flies in the lab environment, so Kerr and Wright could be confident that if changes in the frequency of normal and forked phenotypes occurred, they would not be due to natural selection.
Using a larger breeding population would not be expected to alter the outcome of the experiment.
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
The answer is dominant.
A monohybrid cross involves organisms that are heterozygous for only one character. In autosomal dominant traits, the phenotype is present if both copies of the dominant allele (A) are present (homozygous individuals AA) or only one copy of the dominant allele is present (heterozygous individuals Aa). <u>Thus, t</u><span><u>he characteristic that results from a monohybrid cross is the dominant trait.</u></span>
