Question

In guinea pigs hair straightness or curliness is thought to be governed by a single pair of alleles showing incomplete dominance. Individuals with straight hair are homozygous for the S allele, those with curly hair are homozygous for the C allele, and individuals with wavy hair are heterozygous (SC). You collect data on 1,000 individuals from a population and discover that 244 have straight hair, 444 have curly hair, and 312 have wavy hair.a) Calculate the allele frequencies of the S and C alleles.b.) Is this population in Hardy-Weinberg equilibrium? Show work to support your answer.

Answer 1

Answer:

Frequency of allele S is p $= 0.4939$

Frequency of allele C is q $= 0.666$

The population is not in Hardy-Weinberg equilibrium

Explanation:

Given -

Number of individuals with straight hair $= 244$

Number of individuals with curly hair $= 444$

Number of individuals with Wavy hair $= 312$

Let "p" represents the frequency for allele for straight hair and "q" represents the frequency for allele for curly hair

$p^2$ represents the frequency of genotype "SS"

$p^{2} = \frac{244}{1000} \\= 0.244$

$q^2$ represents the frequency of genotype "CC"

$p^{2} = \frac{444}{1000} \\= 0.444$

$2pq$ represents the frequency of genotype "SC"

$2pq = \frac{312}{1000} \\= 0.312$

Frequency of allele S is p

$= \sqrt{0.244} \\= 0.4939$

Frequency of allele C is q

tex]= \sqrt{0.444} \\= 0.666[/tex]

For being in Hardy Weinberg's equation-

$p+q=1$

Substituting the values in above equation, we get -

$0.4939+0.666\neq 1$

hence, the population is not in Hardy-Weinberg equilibrium