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)
