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0.58
The individuals with extreme variations of a trait.
This is a Hardy Weindburg situation P represents the percentage of the population that has a dominant allele... now there can only be two alleles one is dominant and one is recessive... q is the recessive allele This means that p+q=1 and so q has to be equal to 0.32 If you do the square of p (p^2) then that gives you the number of people who are homozygous dominant If you do the square of q (q^2) then that gives you the number of people who are homozygous recessive If you do 2*(p*q) then that will give you the number of people who are heterozygous Hope this helps...
Ther is a 50% chance you will have two p alleles
50% as there are two alleles per gene and four gametes are made from every undeveloped sex cell.
0.78
0.58
0.11
0.65
Q is represented as follow: q = 1 - 0.68 = 0.32
0.93 0.58 0.32
q is 1-p = 0.81
It is 0.65
0.93
.93
0.58
.78 (Apex)