in a randommating population,an autosomal recessive disorder is present in 60 individual out of 6000.what is the no of homozygous normal(AA)offspring in population
ACCORDING TO HARDY WEINBERG EQUILIBRIUM
ressecive individuals - qq=60
i.e q^2 =60/6000=0.01
total homozygous normal population =p^x6000
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Total heterozygous individual is out of .
The recessive individual (rr) is out of the total population .
Therefore, the frequency of recessive individual :
, where denotes the genotype of the recessive individual.
Therefore the recessive allele (r) would be
We know that, Hardy Weinberg Equilibrium :
where p denotes the frequency of dominant allele
q denotes the frequency of recessive allele
In this question, q = r = 0.1
Hence, frequency of dominant allele is
Heterozygous according to Hardy Weinberg Equilibrium
Substituting our values,
Hence, the frequency of heterozygous individual is
Now, given that the total number of individuals is
Therefore heterozygous individuals
So, total heterozygous individuals is out of .