Question

In dogs, wire hair is due to a dominant gene (W), smooth hair id due to its recessive allele.a. If a homozygous wire-haired dog is mated with a smooth-haired dog, what type of offspring could be produced?b. What types of offspring could be produced in F2?c. Two haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two wire-haired dogs mate again, what are the chances that they will produce another smooth-haired pup? What are the chances that the pup will be wire-haired?d. A wire-haired male is mated with a smooth-haired female. The mother of wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce?

Answer 1

Answer:

a) All the offspring produced would be a heterozygous in the first generation (F1 generation) with genotype: Ww, Ww, Ww, and Ww. Since wire hair is dominant, all would be wired haired.

b) In the F2 (second generation), we breed the heterozygous offspring from the F1 generation together (Ww x Ww). This will produce 3:1 ratio of dominant:recessive phenotypes having 3/4 of the offspring wire-haired (1/4 WW homozygotes and 1/2 Ww heterozygotes) and 1/4 will be smooth-haired (ww). Also the genotype would have the ratio 1:2:1 (i.e 1 homozygote WW, 2 heterozygote Ww and 1 smooth hair)

c) the chances of producing a smooth-haired pup is 1/4, and the chances of producing a wire-haired pup are 3/4.

d) Therefore, 1/2 of the offspring will have the genotype Ww and be wire-haired, and 1/2 of the offspring will be ww and be smooth-haired.  Also the phenotype ratio is 1:1 (1/2 is heterozygote wired hair and 1/2 is smoth haired)

Explanation:

Since the wire hair is the dominant gene (W) and smooth hair (w) is the recessive allele

a)  If a homozygous wire-haired dog is mated with a smooth-haired dog, All the offspring produced would be a heterozygous in the first generation (F1 generation) with genotype: Ww, Ww, Ww, and Ww. Since wire hair is dominant, all would be wired haired.

b) In the F2 (second generation), we breed the heterozygous offspring from the F1 generation together (Ww x Ww). This will produce 3:1 ratio of dominant:recessive phenotypes having 3/4 of the offspring wire-haired (1/4 WW homozygotes and 1/2 Ww heterozygotes) and 1/4 will be smooth-haired (ww). Also the genotype would have the ratio 1:2:1 (i.e 1 homozygote WW, 2 heterozygote Ww and 1 smooth hair)

c) If two wire-haired dogs produce a smooth-haired pup, that means that both parents must be heterozygotes (Ww) having a pair of dominant W allele and recessive w allele to pass on to the offspring. Therefore, if these two dogs were to mate again (Ww x Ww), the chances of producing a smooth-haired pup is 1/4, and the chances of producing a wire-haired pup are 3/4.

d) If the mother of the wire-haired male was smooth-haired, that means that the recessive allele w had been passed on to the male making the male a   heterozygote (Ww). When this male mates with a smooth-haired female (ww), the cross is Ww x ww. Therefore, 1/2 of the offspring will have the genotype Ww and be wire-haired, and 1/2 of the offspring will be ww and be smooth-haired.  Also the phenotype ratio is 1:1 (1/2 is heterozygote wired hair and 1/2 is smoth haired)