Question

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

Answer 1

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 2

Answer:

Total heterozygous individual is $1080$ out of $6000$.

Explanation:

The recessive individual (rr) is $60$ out of the total population $6000$.

Therefore, the frequency of recessive individual : $\frac{60}{6000} =0.01$

$rr= 0.01$, where $rr$ denotes the genotype of the recessive individual.

Therefore the recessive allele (r) would be $0.1$

We know that, Hardy Weinberg Equilibrium :

$p + q = 1$

where p denotes the frequency of dominant allele

q denotes the frequency of recessive allele

In this question, q = r = 0.1

$p + 0.1 = 1$

$p = 0.9$

Hence, frequency of dominant allele is $0.9$

Heterozygous according to Hardy Weinberg Equilibrium $= 2pq$

Substituting our values,

$2pq = 2 * 0.9* 0.1 \\2pq= 0.18$

Hence, the frequency of heterozygous individual is $0.18$

Now, given that the total number of individuals is $6000$

Therefore heterozygous individuals $= 0.18 * 6000$

$= 1080$

So, total heterozygous individuals is $1080$ out of $6000$.