1. Two years after the above frequencies were calculated, a biologist reexamined the cat population and counted 153 short-haired homozygotes, 568 shorthaired heterozygotes, and 397 long-haired cats. What are the frequencies of each allele now?

Respuesta :

Answer:

The new frequencies are 0.39 for short hairs and 0.61 for the long haired.

Explanation:

The total population is given as 153+568+397=1118

S is given as the short haired, s is given as the long haired so ss is the short haired homozygotes and Ss is the shorthaired heterozygotes.

  1. As there are 1118 cats, then the gene pool has 2 x 1118 alleles for hair length =2236 alleles
  2. f(S)=(2SS+ Ss ) / total alleles =[(2 x 153) + 568] / 2236 = 0.39
  3. f(s)=(2ss+Ss) / total alleles=[(2 x 397) + 568] / 2236 =  0.61