Recent content by paradoxium

For (a), my friends and I came up with: Aa X aa --> 2 tailess (Aa) + 2 long-tailed (aa) Manx allele = A (dominant) Frequency = 2A/(6a + 2A) = 2/8 = 1/4 = 0.25 For (b): The number of individuals can be calculated by: 2Aa = 2 X 0.25 (1-0.25) = 0.375 For (c): Genetic drift...

I'm a first-year college student and I'm havnig a lot of trouble with a particular question. Here goes: Manx cats are tailess and when crossed with one another produceo n averaeg one long-tailed cat for every 2 Manx. The allele is lethal in homozygous condition due to problems arisiging in...