Question

Two pigs whose tails are exactly 25 cm in length are bred over 10 years and they produce 96 piglets with the following tail lengths: 6 piglets at 15 cm, 25 at 20 cm, 37 at 25 cm, 23 at 30 cm, and 5 at 35 cm

Answer 1

The additive effect of quantitative genes produce different gradual phenotypes. In the exposed example a) two diallelic genes are involved / b) 50% of the progeny will have 25 cm tails, 50% will have 20 cm tails.

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According to the given information, we have many phenotypes, which might be determined by the action of more than one gene. So this is probably an example of Quantitative heritability.

This term refers to the transmission of a trait expressed by the additive action of many quantitative genes.

Varying phenotypes result from the interaction of different quantitative genes. Many different combinations can cause genotypic graduation, expressing different phenotypes.  

Quantitative traits are those that can be measure, such as longitude, weight, eggs laid per female, among others.

These characters do not group individuals by any precise and clear categories. Instead, they group individuals in many different categories that depend on how the genes were intercrossed and distributed during meiosis.

The result depends on the magnitude in which each allele contributes to the final phenotype and genotype.

When they interact, they create a gradation in phenotypes, according to the level of contribution.  

Let us analyze the problem.

→ We know that two 25cm-tailed pigs mated and produced 96 piglets.

→ These piglets' tails ranged in 5-cm intervals from 15 to 35 cm

→ There are 5 different phenotypes: 15cm, 20cm, 25cm, 30 cm, 35 cm.

→ The frequency of each phenotype is

• 6/96 ⇒ 15 cm,
• 25/96  ⇒20 cm,
• 37/96 ⇒ 25 cm ⇒ The most frequent phenotype
• 23/96 ⇒ 30 cm,
• 5/96 ⇒ 35 cm

⇒ FREQUENCIES According to the frequencies, we can make the following analysis

• Individuals expressing the most frequent phenotype are heter0zyg0us for all the involved genes.

The majority of pigs had 25 cm tails, which means that these pigs are  heterozygote for all genes.

• The phenotypes expressing the shortest tails are the extreme phenotypes, that is the recessive h0m0zyg0te and the dominant h0m0zyg0te.

The least frequent phenotypes are 15 cm tails (6/96) and 35 cm tails (5/96).

• The rest of the phenotypes are possible combinations of genes and their alleles.

⇒ PHENOTYPES ,   COMBINATIONS and GENOTYPES According to the phenotypes and the possible combinations, we can get the possible genotypes

There are four possible combinations (apart from the 25 cm lenght form). This suggests that there are two diallelic genes involved in the expression of the trait.

• Gene 1: allele A and a
• Gene 2: allele B and b

Each allele contributes in 5 cm to each phenotype. So,

• when no dominant allele is present, the tail is 15 cm long.
• When only one dominant allele is present, tails are 20 cm long (adds 5 cm)
• When two dominant alleles are present, tails are 25 cm long (each adds 5 cm = 10 cm)
• When three dominant alleles are present, 15 cm are added to the phenotype
• When only dominant alleles are present, tails are 35 cm long.

According to this, the phenotypes and genotypes of the piglets are as follows

• 15 cm length = aabb (6/96) H0m0zyg0us recessive for both alleles.
• 20cm length = Aabb or aaBb(25/96) H0m0zyg0us recessive for one  allele.
• 25cm length = AaBb (37/96 ⇒ The majority) Heter0zyg0us for both alleles.
• 30cm length = AABb or AaBB (26/96) H0m0zyg0us dominant for one allele.
• 35cm length = AABB (5/96) H0m0zyg0us dominant for both alleles.

Now, we can answer the questions:

a. How many pairs of genes are regulating the tail length character?

Two diallelic genes are regulating this trait ⇒ A/a  and  B/b.

Each dominant allele contributes 5 cm to the tail's length.

The minimal length is 15 cm with no dominant allele (aabb).  

b. What offspring phenotype would you expect from a mating between a 15-cm and a 30-cm pig?

Cross: 15 cm   x   30 cm

Parentals) aabb   x   AaBB

Gametes) ab, ab, ab, ab

                AB, AB, aB, aB

Punnett square)    AB        AB        aB        aB

                   ab    AaBb    AaBb    aaBb    aaBb

                   ab    AaBb    AaBb    aaBb    aaBb

                   ab    AaBb    AaBb    aaBb    aaBb

                   ab    AaBb    AaBb    aaBb    aaBb

F1) 8/16 = 1/2 = 50% of the progeny is expected to have 25cm tails, being heter0zyg0us for both genes, AaBb.

   

     8/16 = 1/2 = 50% of the progeny is expected to have 20 cm tails, being heter0zyg0us for one gene, aaBb.

Half of the progeny is expected to have 25 cm tails, while the other half is expected to have 20 cm tails.

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