Answer:
d.0.48
Explanation:
When a population is in Hardy Weinberg equilibrium the <u>genotypic </u>frequencies are:
freq (AA) = p²
freq (Aa) = 2pq
freq (aa) = q²
<em>p</em> is the frequency of the dominant <em>A</em> allele and <em>q</em> is the frequency of the recessive <em>a</em> allele.
In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.
Therefore:
q²=16/100
q² = 0.16
q=√0.16
q = 0.4
And p+q=1, so:
p = 1 - q
p = 1-0.4
p = 0.6
The frequency of heterozygotes is:
freq (Aa) = 2pq = 2 × 0.4 × 0.6
freq (Aa) = 0.48
Because people may make fun of you or call you some terrible things because stds are often linked by sleeping around (obvi not always true but that’s what people make it out to be)
and i’m good lol you?
Answer:
A was the answer to mine but u didn't post your diagram so I don't know
Explanation:
Answer:
25%
Explanation:
The child only comes out as homozygous dominant (YY) for the trait in 1/4 boxes in the Punnett Square. This translates to the child having a 25% chance of being homozygous dominant for the trait.