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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

## Answers

**Answer:**

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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**Answer:**

**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 .