Question

  Suppose that a particular gene that controls coat color in dogs is controlled by four alleles in the population. These alleles are called c^ch (chinchilla dog), c^d (white dog with dark eyes), c^b (pale gray dog) and c (albino dog). The alleles are given from the most dominant to the most recessive. Let’s assume that a particular population of these dogs is in Hardy-Weinberg equilibrium. The allele frequencies for our four alleles are:

c^ch = 0.37

c^d = 0.23

c^b = 0.25

c = 0.15

a.    What is the frequency of the c^chc^ch, c^bc, and cc genotypes in this population?

b.    Among a population of 2000 dogs, how many would you expect to have the pale gray phenotype, yet be heterozygous?

c.     Among a population of 5000 dogs, how many would you expect to have the white coat with dark eyes phenotype, yet be heterozygous?

d. Among a population of 15,000 dogs, how many would you expect to have c^chc^bor cc genotypes?

Answer 1

Answer:

Based on the given information, the population is in Hardy-Weinberg equilibrium. Gene that controls coat color in dogs is controlled by four alleles in the population. In order from most dominant to most recessive these alleles are:

• Allele frequency for (chinchilla dog), cch (p) = 0.37
• Allele frequency for (white dog with dark eyes) cd (q) = 0.23
• Allele frequency for (pale gray dog) cb (r) = 0.25
• Allele frequency for (albino dog) c (s) = 0.15

Allele Frequencies Equation: p + q + r + s = 1

Genotype Frequencies Equation: p² + 2pq + 2pr + 2ps + q² + 2qr + 2qs + r² + 2rs + s² = 1

Total number of genotypes with 4 alleles = 10

a.    What is the frequency of the cchcch, cbc, and cc genotypes in this population?

• Genotype frequency of cchcch = (0.37)² = 0.1369
• Genotype frequency of cbc = 2*0.25*0.15 = 0.075
• Genotype frequency of cc = 0.15*0.15 = 0.0225

b.    Among a population of 2000 dogs, how many would you expect to have the pale gray phenotype, yet be heterozygous?

• These would correspond to genotypes cbc (heterozygous) as c allele is recessive to cb allele.
• Expected number = 2000*genotype frequency of cbc = 2000*0.075 = 150

c.     Among a population of 5000 dogs, how many would you expect to have the white coat with dark eyes phenotype, yet be heterozygous?

• This would correspond to cdcb (heterozygous) and cdc (heterozygous) genotypes as cb and c are recessive to cd.
• Genotype frequency of cdcb = 2*0.23*0.25 = 0.115
• Gentotype frequency of cdc = 2*0.23*0.15 = 0.069
• Expected number = 5000*genotype frequency of cdcb + 5000*genotype frequency of cdc = 5000*0.115 + 5000*0.069 = 575 + 345 = 920

d. Among a population of 15,000 dogs, how many would you expect to have cchcb or cc genotypes?

• Genotype frequency of cchcb = 2*0.37*0.25 = 0.185
• Genotype frequency of cc = 0.15*0.15 = 0.0225
• Expected number of cchcb or cc genotypes = 15000*0.185 + 15000*0.0225 = 2775 + 338 = 3113