Final answer:
The allele frequencies in the cat population are 0.6 for the black allele (B) and 0.4 for the white allele (b), with genotype frequencies of BB = 0.36, Bb = 0.48, and bb = 0.16, resulting in 360 black homozygous, 480 black heterozygous, and 160 white cats respectively.
Step-by-step explanation:
When determining the allele frequency and the frequency of individuals per genotype in a population, we use the Hardy-Weinberg principle and the phenotype information provided. Given 840 black cats and 160 white cats in a population of 1000 cats, and knowing that black (B) is dominant over white (b), we can infer the following:
All 160 white cats are homozygous recessive (bb).
The allele frequency for b (q) can be found by taking the square root of the frequency of the homozygous recessive genotype (160/1000 = 0.16), which gives us q = 0.4.
The allele frequency for B (p) can be found using the equation p + q = 1, so p = 0.6.
The expected genotype frequencies are then p² (BB) = 0.36, 2pq (Bb) = 0.48, and q² (bb) = 0.16.
The number of individuals per genotype can be calculated by multiplying these frequencies by the total population: BB = 360, Bb = 480, bb = 160.
Thus, the genotype frequency and number of individuals per genotype in this cat population can be estimated using the allele frequencies derived from the observed phenotypes.