Question

In a population where hair color gene can be determined by 4 alleles B(brown)=0.210654 , b(blonde) =0.319212, g(black) =0.206791, e(red) =0.216791. The gene for hair texture can be determined by 3 alleles C(curly)=0.225879 c=(not curly)=0.412122 w(wavy)=0.345621. The gene for hair porosity can be determined by 2 alleles T(thin)=0.456123 and u(coarse)= 0.515123. Assume Hardy-Weinberg equlibrium, what is the frequency of getting Black, wavy, thin hair? (p² + 2pq + q² + 2pr + 2qr + r² = 1). ?

Answer 1

According to hardy - weinberg equilibrium

As allele for black and wavy hair is reccesive therefore only possible genotype for wavy and thin hair are gg, ww respectively

frequency for black hair = f(gg) =g^2=(0.206791​)^2=0.0427625177

frequency for​ wavy hair = f(ww)= w^2=(0.345621)^2=0.1194538756

But allel for thin hair is dominant therefore possible genotype for thin hair are TT and Tu

which implies total frequency for​​ thin hair = f(TT)+ f(Tu)= p^2+2pq = 0.208048191​+0.469918896​ = 0.677967087

now in probablity we have learnt that in case of more than one different events probabilty for them to occur together is calculated by adding indiviual probabilty.

hence frequency for black wavy thin hair is f(gg) ​+ f(ww) + f(TT)+ f(Tu)

i.e, 0.0427625177+0.1194538756+ 0.677967087=0.8401834803