Question

If 10 individuals out of a population of 1000 suffering from phenylketonuria find out the number of carriers

Answer 1

The calculator below uses the Hardy-Weinberg equation: p²+ 2pq+ q² = 1

to estimate the frequency of the carrier state (2pq) for an autosomal recessive trait .

SELECT % OR PROPORTION THEN ENTER VALUE

Percent of population that has recessive trait    

Example :        .04   %  

 %  

OR  

Proportion  of population that has recessive trait  

Example: 1 in   2500     1 in    

 

 p=  

Frequency of wild type allele

q=  

Frequency of mutant allele

p² 2pq q2 =1

 

 

 

=1

Homozygous  

wild type Carrier frequency  

 

Homozygous mutation  

Example : Cystic fibrosis (CF)

Where:

A is the wild type allele

a is the cystic fibrosis mutation.

 Maternal

A (p) a (q)

Paternal A (p) AA (p2) Aa (pq)

a (q) Aa(pq) aa (q2)

The Hardy–Weinberg principle states that the genotype frequencies A2, 2Aa, and a2 will not change if the allele frequencies remain constant from generation to generation (they are in equilibrium).

Expressed as:

    A2 + 2Aa+ a2 =1

Hardy-Weinberg equation for the general case:  

    p²+ 2pq+ q² = 1

The genotype frequencies and allele frequencies are in equilibrium if the following assumptions of the Hardy–Weinberg principle are met:

There is no natural selection

Matings are at random ,all individuals reproduce equally, all offspring survive

There is no appreciable rate of mutation.

There is no migration.

There has been no genetic drift.

For CF the frequency of aa in northern Europeans = 1 in 2500

    q2=1/2500

    q = 1/50

     

    Since the sum of the alleles   q + p = 1 , p = 1 - q

    p= 49/50

    Carrier frequency = 2pq= 2*(49/50)(1/50) = 98/2500 =.04