Answer:
[tex]0.565\\Or\\56.5[/tex]%
Explanation:
Given,
Number of plants with rough seed [tex]= 19[/tex]
Total number of plants [tex]= 100[/tex]
Genotype frequency for "plant with rough seed" i.e [tex]q^{2} \\[/tex] [tex]= \frac{19}{100} \\= 0.19[/tex]
So the allele frequency for "rough seed" [tex]= \sqrt{q^{2} } \\= \sqrt{0.19} \\= 0.435\\[/tex]
As per Hardy weinberg equation, we know that sum of allele frequencies around the locus is equal to one .
Thus,
[tex]p+q=1\\[/tex]
Substituting the value of q in above equation, we get -
[tex]p+0.435=1\\p = 1-0.435\\p =0.565 \\Or 56.5[/tex]%
p represent the frequency of smooth seed.
