Question

Evolution

1. Assume this population is in Hardy-Weinberg equilibrium. In a population of 120 cats, 35 are black. Black cats have the bb genotype. Find the allelic frequency of the dominant and recessive allele. 2. Assume this population is in Hardy-Weinberg equilibrium. A litter of 10 puppies has both tipped ears (a) and floppy ears (A). If 4 puppies have tipped ears in the litter. What is the frequency of the recessive allele? 3. Assume this population is in Hardy-Weinberg equilibrium. The frequency of the homozygous dominant genotype in the tree shrew is 0.49. What's the frequency of the heterozygote genotype? 4 Assume this population is in Hardy-Weinberg equilibrium. The heterozygote frequency of a pulation is 0.32. The frequency of the dominant allele is 0.8. What is the frequency of the recessive phenotype?

Answer 1

Aus 1) bb is homozygous recessive. So; q = 35/120 = 0.29 So; 10. 29 = 0.54 ; from Handy- Weinburg equillibrium we kuow that Peq=1 So; p= 0.46 Hence, frequency of dominant allele in population = p = 0.46. The frequency of excessive allele in Population = q = 0.54

2) aa is homozygous recessive So; q 4/10 - 014 So, 0.4 0.63 frequency of recessive allelet in - Population & q = 0.63 3) The frequency of homorygous dominant is 0.49. So; p² * 0.49 P=10.49 = dit from Hardy-Weinburg Equillibruum we know that P+q=1 So; q = 013 Hence frequency of heterozygous genotype in the population will be = 2pq = I 2X0.7 x0.3 = 0.42. 47 frequency of dominant alleld - P=0.8 Genotypic frequency of heterozygous = 0.32 2pq = 0.32 frequency of recessive allele a 0.32 2P So, 0.2

secessar allele Now; frequency of recessive phenotype - q = 0.2² = 0.04 Please Rate my Auswen.