Erma and Harvey were a compatible barnyard pair, but a curious sight. Harveys tail was only 6 cm while Ermas was 30 cm. Their F1 piglet offspring all grew tails that were 18 cm. When inbred (F1X F1), an F2 generation resulted in many piglets (Erma and Harveys grandpigs) whose tails ranged in 4 cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). Most had 18 cm tails while 1/64 had 6 cm and 1/64 had 30 cm tails.

a) Explain how tail length is inherited by describing the mode of inheritance, indicating how many gene pairs are at work, and designating the genotypes of Harvey, Erma, and their 18 cm offspring.

b) If one of the 18 cm F1 pigs were mated with the 6 cm F2 pigs, what phenotypic ratio would be predicted if many offspring resulted?

(a). The mode of inheritance is quantitative inheritance.

There are 3 gene pairs at work.

Genotypes designating Emma 30cm, Harvey 6cm and their 18cm offspring are Emma as TT, Harvey as Tt and 18cm offspring as Tt since the phenotype is about height T.

(b) The phenotype ratio of 18cm of F1 and 6cm of F2 is 1:1.

Explanation:

(a) Quantitative inheritance is the mode of inheritance that deals with additive action of many genes resulting into a traits such as height, showing contiguous variation.

The frequency of either extreme phenotype give us n, the number of gene pairs.

Frequency of one extreme phenotype = (1/4)n = 1/64.

Number of gene pairs =

log 64/log 4 = 3.

Therefore 3 gene pairs were at work.

The phenotype here is about height either tall, T or short t. There genotype designation will be given as 30cm of Emma as TT homozygous tall, 6cm of Harvey as tt homozygous short, and 18cm of offspring as Tt, heterozygous tall.

(b) The phenotype of 18cm of F1 and 6cm of F2 offspring will come the crossing of 18cm, Tt of F1 and 6cm, tt of F2 to give 4 offspring as Tt, Tt, tt, tt.

Therefore, comparing the phenotype ratio (tall or short) we have 2:2 which gives 1:1.

eukarya because is is like alche and b is bacteria and d is animals

explanation: