Expermant laveas test strach
1 answer:
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Answer:
Frequency of allele A1- 0.41
Explanation:
In Hardy weinberg equilibrium,
P refers to the dominant allele
q refers to the recessive allele
The allele frequency will be p+q=1
The genotypic frequency is- P²+q²+2pq=1
P²= genotype of dominant trait ( A1A1)- 77
2pq= genotype of heterozygotes (2pq)- 65
q²= genotype of recessive trait (A2A2)- 123
Total number of offsprings= 77+ 65+ 123
= 265
Now to calculate allele frequency of A1=
= 77/265 + 1/2( 65//265)
= 0.290+ 0.122
= 0.413
Thus, 0.41 is correct.