asked by surajpuja2041, 17 days ago

# 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

57

4860

Explanation:

ACCORDING TO HARDY WEINBERG EQUILIBRIUM

ressecive individuals - qq=60

i.e q^2 =60/6000=0.01

i.e q=0.1

1-0.1=0.9=p

p^2=0.81

total homozygous normal population =p^x6000

0.81x6000

=4860

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0

Total heterozygous individual is out of .

Explanation:

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 .

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