Question

6. A farmer breeds cattle. In his current herd of males and females, in which the bulls and cows are allowed to mate at random, 96% are hornless (polled) and 4% are horned. Polled is dominant over horned. Then, the farmer purchased some new cattle (coming from randomly breeding herd in the ranch) which has 75% polled and 25% horned. The total number of new cattle is equal to 25% the number of the original herd, thus 20% of the combined herd is made up of new cattle). If the farmer allows the combined herd to undergo random mating, what (a) allelic frequencies, and (b) genotypic frequencies are expected after the first generation?

Answer 1

Here, it is given that there are 4% horned cattle in the original herd.

Since the allele for the horned trait is recessive, the bulls showing horned phenotype is recessive homozygote.

Since according to Hardy Weinberg principles

p+q=1 and

p2+q2+2pq=1

where p and q are the allelic frequencies for dominant and recessive alleles

p2 is the genotype frequency for dominant homozygote, q2 for recessive, and 2pq for heterozygotes.

now, for the original herd, q12= 0.04, so, q1=0.2 or 20%, so p1 = 1-q=0.8

For the new herd, q22= 0.25, so, q2=0.5 or 50%, so p2 = 1-q=0.5

For the combined herd

pc=0.2*p2+0.8*p1 ( since 80% of new population is original herd. 20% is new herd)

=0.74

qc=0.2*q2+0.8*q1 = 0.2*0.5 +0.8*0.2 =0.1+0.16 = 0.26

Check, pc+ qc = 0.74+0.26 =1

So, the allelic frequency in the combined herd is 0.74 for polled allele and 0.26 for horned allele.

The genotype frequency would be 0.74*0.74 =0.5476 =54.76% dominant homozygote (polled),

2*0.74*0.26 =0.3848 = 38.48% heterozygotes (polled) and

0.26*0.26 = 0.0676 = 6.76% horned cattle ( recessive homozygote)

Assuming Hardy weinberg equilibrium, since the population undergoes random mating, the genes are not linked and there are no other selection pressures, the allele and genotype frequencies would remain the same for F1 as well.