Question

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 \(\mathrm{F}_{1}\) 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 \(4-\mathrm{cm}\) intervals from 6 to \(30 \mathrm{cm}(6,10,14,18,22,26, \text { and } 30) .\) Most had \(18-\mathrm{cm}\) tails, while \(1 / 64\) had \(6-\mathrm{cm}\) tails and \(1 / 64\) had \(30-\mathrm{cm}\) 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 \(18-\mathrm{cm} \mathrm{F}_{1}\) pigs is mated with one of the \(6-\mathrm{cm}\) \(\mathrm{F}_{2}\) pigs, what phenotypic ratio will be predicted if many offspring resulted? Diagram the cross.

Short answer

Answer: The phenotypic ratio of the offspring's tail lengths is 50% with 6-cm tails and 50% with 18-cm tails.

Step-by-step solution

Understanding the Problem and Terminology

To solve this exercise, it's important to understand several genetic concepts: - F1 generation: This refers to the first generation of offspring produced by crossing two parental individuals. - F2 generation: This refers to the generation produced by inbreeding the F1 generation. - Phenotype: The observable characteristics of an organism, such as tail length. - Genotype: The specific genetic makeup of an organism that determines its phenotype. - Mode of inheritance: The pattern in which a particular genetic trait is passed on from one generation to another.

Determine the Mode of Inheritance and Gene Pairs Involved

Given that F1 offspring had tail lengths intermediate between those of their parents (18 cm) and F2 piglets on inbreeding F1, show a range of tail lengths at regular intervals, this suggests that the inheritance falls under quantitative or polygenic inheritance. In quantitative inheritance, multiple gene pairs, also known as polygenes, act together to determine a trait with continuous variation. Each gene pair contributes a small additive amount to the trait. In this case, we are seeing variation in tail lengths in 4-cm intervals. There is a 24 cm difference between the shortest and the longest tails (30 cm - 6 cm), so 24 / 4 = 6 intervals. This suggests that 3 gene pairs (A, B, and C) are involved, as each can contribute two possible 4-cm intervals (A: 0 or 4 cm, B: 0 or 4 cm, and C: 0 or 4 cm).

Designate the Genotypes of Harvey, Erma and F1 Offspring

Now that we know three gene pairs are determining tail length, we can assign genotypes for Harvey, Erma and the F1 offspring. Harvey's tail length is 6 cm, which means he has the shortest tail possible (none of the gene pairs contribute to tail length). His genotype can be represented as AABBCC (where A, B, and C each represent 0 cm contribution to tail length). Erma has the longest tail length of 30 cm. All three gene pairs contribute 4 cm each to her tail length. Her genotype can be represented as aaBBcc (where a, b, and c each represent 4 cm contribution to tail length). Their F1 offspring have tails that are 18 cm long. Each parent contributes one allele of each gene pair to the offspring. So, the F1 offspring genotype would be AaBbCc (one allele from Harvey: ABC and one allele from Erma: abc).

Predict the Phenotypic Ratio for a Specific Cross

Now we have to find the phenotypic ratio when an F1 pig with an 18-cm tail (AaBbCc) mates with an F2 pig with a 6-cm tail (AABBCC). F1 pig (AaBbCc) x F2 pig (AABBCC) The possible gametes from the F1 pig would be ABC and abc. From the F2 pig, there is only one possible gamete, which is ABC (no variation since all the alleles are the same). So, the offspring genotypes would be AABBCC (50% of the offspring) and AaBbCc (50% of the offspring). This results in a phenotypic ratio of: - 50% 6-cm tails (AABBCC genotype) - 50% 18-cm tails (AaBbCc genotype)