Respuesta :
Answer:
Option E, [tex]0.64[/tex]
Explanation:
The frequency of x-linked recessive disease is equal to
[tex]\frac{1}{25}\\ = 0.04[/tex]
As per Hardy-Weinberg equation, the frequency of recessive genotype is represented by "[tex]q^{2}[/tex]"
Thus, here [tex]q^{2}[/tex][tex]= 0.04\\[/tex]
Frequency of recessive allele [tex]= \sqrt{q^{2} }[/tex]
Substituting the given value, we get -
[tex]q = \sqrt{0.4} \\q=0.2[/tex]
As per hardy Weinberg equation-
[tex]p+q=1\\p+0.2+1\\p=0.8[/tex]
Now "p" signifies frequency of dominant allele. Thus, frequency of dominant genotype would be
[tex]p^{2} \\0.8^2\\0.64[/tex]
Hence, option E is correct
