Two wavy-haired (WW) cattle one male (homozygous dominant) and one female (heterozygous dominant) Using the Punnett Square, determine the phenotype, the genotype, and the percentage distribution of their possible offspring.
Answers
1/10 = 0.1
In the larger population --
Frequency of the recessive phenotype = (q2)2 = 54/600
Frequency of the recessive allele = q2 = 1/10 = 0.3.
In the merged population --
Frequency of recessive allele q = ((400 x 0.1) + (600 x 0.3))/1000 = 0.22
Frequency of black cats in the next generation = q2 = (0.22)2 = 0.0484.
A potential source of error in this problem is to simply add the number of recessive individuals from the two populations and to derive q from that -- i.e., take the square root of (4 + 54). However, doing so would ignore the contribution of recessive alleles from the heterozygotes in each population.
3. (i) If only black cats are left standing after the virus goes through, then only the recessive (black) allele will be left in the population; the frequency of the black allele in the next generation will be 1.0 (= 100%).
(ii)
Before the virus comes through, the frequency of the three genotypes is:
Homozygous dominant = p2 = 0.25
Heterozygotes = 2pq = 0.5
Homozygous recessive = q2 = 0.25
After the viral epidemic, the only cats left are homozygous dominant and heterozygotes: