Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey's tail was only \(6 \mathrm{cm}\) long, while Erma's was \(30 \mathrm{cm}\) Their \(F_{1}\) piglet offspring all grew tails that were \(18 \mathrm{cm} .\) When inbred, an \(\mathrm{F}_{2}\) generation resulted in many piglets (Erma and Harvey's grandpigs), whose tails ranged in 4 -cm intervals from 6 to \(30 \mathrm{cm}(6,10,14,18,22,26, \text { and } 30) .\) Most had 18 -cm tails, while \(1 / 64\) had 6 -cm tails and \(1 / 64\) had 30 -cm tails. (a) Explain how these tail lengths were inherited by describing the mode of inheritance, indicating how many gene pairs were at work, and designating the genotypes of Harvey, Erma, and their 18 -cm-tail offspring. (b) If one of the 18 -cm \(\mathrm{F}_{1}\) pigs is mated with one of the 6 -cm \(\mathrm{F}_{2}\) pigs, what phenotypic ratio would be predicted if many offspring resulted? Diagram the cross.

1. Observe the given information: Harvey's tail is 6 cm, Erma's tail is 30 cm, their F1 offspring's tails are 18 cm, and F2 offspring's tails range from 6-30 cm in 4-cm intervals (6, 10, 14, 18, 22, 26, and 30). 2. Since the traits are quantitative and there are multiple F2 phenotypes, this suggests that the mode of inheritance is polygenic using additive alleles. 3. Notice that there are 7 phenotypic categories in F2, and the phenotypic difference between adjacent categories is 4 cm, suggesting that 3 gene pairs (A, B, and C) are involved. Each gene pair has two alleles: one additive allele (A, B, or C) that contributes 4 cm to the tail length, and the other non-additive allele (a, b, or c) that does not contribute. 4. Assign genotypes to Harvey, Erma, and their F1 offspring: - Harvey's tail length is 6cm, which is the minimum phenotypic value (6cm) that the offspring can inherit. Therefore, Harvey’s genotype must be aabbcc (three non-additive alleles with 0 contribution). - Erma's tail length is 30cm, which is the maximum phenotypic value (6cm + 3(4cm)) that the offspring can inherit. Therefore, Erma’s genotype must be AABBCC (three additive alleles contributing 12 cm in total). - The F1 offspring have 18-cm tails. They inherit one allele from each parent, resulting in a heterozygous genotype AaBbCc (each additive allele contributing 4 cm to tail length).

1. To predict the phenotypic ratio, we will cross an 18-cm F1 pig (AaBbCc) with a 6-cm F2 pig (aabbcc). 2. Set up a Punnett square to perform the cross: - The possible gametes of the F1 parent are ABC (contributing 12 cm), ABc (contributing 8 cm), AbC (contributing 8 cm), Abc (contributing 4 cm), aBC (contributing 8 cm), aBc (contributing 4 cm), abC (contributing 4 cm), and abc (contributing 0 cm). - The possible gametes of the F2 parent are all a, b, and c, which does not contribute to the tail length. 3. Perform the cross and calculate the phenotypic ratio among the offspring. Since all F2 parent's gametes do not contribute to the tail length, the phenotypic ratio can be obtained by simply observing the F1 parent's gametes contribution. - Offspring with 6cm tails - 1 combination (abc) - 1/64 probability - Offspring with 10cm tails - 3 combinations (AbC, aBC, ABc) - 3/64 probability - Offspring with 14cm tails - 3 combinations (Abc, aBc, abC) - 3/64 probability - Offspring with 18cm tails - 1 combination (ABC) - 1/64 probability 4. The predicted phenotypic ratio among the offspring when mating an 18-cm F1 pig (AaBbCc) with a 6-cm F2 pig (aabbcc) is 1:3:3:1 for tails of 6cm, 10cm, 14cm, and 18cm, respectively.