Question

if 10 individuals out of a population of 1000 are suffering from phenylketonuria, find out the number of carriers?​

Answer 1

If the incidence of an autosomal recessive disorder is known, then it is possible to calculate the carrier frequency using some relatively simple algebra. If, for example, the disease incidence equals 1 in 10000, then q2 = 1/10000 and q = 1/100 . As p + q = 1, therefore p = 99/100 .

Answer 2

Out of 1000 individuals in which 10 are suffering from phenylketonuria, the carriers will be 2 in number.

Explanation:

Phenylketonuria is an autosomal recessive condition in an  organism. To express the phenotype its both alleles should be mutant. Wheh both the parents have one copy of gene for phenylketourania it gets passed to the progeny as homozygous trait and gets expressed.

from the question

10 out of 1000 individuals have phenolketonuria

number of cariiers=?

frequency of less frequent allele is given by q

$q^{2}$  (frequency of homozygous dominant trait)= $\frac{10}{1000}$

q   = $\frac{1}{100}$  or 0.01

from the Hardy Weinbergh equation:

p+q =1

putting the values of q in the equation: p(wild type allele)

p= 1-q

  =1 - $\frac{1}{100}$

  = $\frac{99}{100}$

The carrier frequency( 2pq) = 2 ($\frac{1}{100}$x $\frac{99}{100}$)

                                      = 0.0198

0.0198 is the frequency of carriers i.e nearly 2 people in 1000 are carriers.

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