a) An F1 x F1 self-fertilization gives a 9:7 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?

b) An F1 x F1 self-fertilization gives a 9:3:4 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?


c) An F1 x F1 self-fertilization gives a 15:1 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?

Answers

Answer 1

According to the gene interaction that is operating in the trait determination, the phenotypic ratios of the progeny produced when test-crossing a dihybrid are: a) 4:12, b) 8:8, and c) 12:4

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To answer these questions, the first thing you should notice by looking at the phenotypic ratios provided, is that there are two genes involved in the trait, and there is interaction between them.

Let us assume diallelic genes. A and B are the dominant alleles, while a and b are the recessive alleles.

In the three cases (a, b, and c), F1 individuals are dihybrid. Their genotype is AaBb.

Observing the altered phenotype proportions of the offspring can reveal which interaction is operating in the trait determination.

We can use the phenotypic ratio to determine the interaction, so:

9:3:3:1   ⇒   No interaction9:7        ⇒    Double Recessive epistasis 9:3:4     ⇒    Recessive epistasis15:1        ⇒    Dominant duplicate genes.

When talking about epistasis, we are referring to an interruption.

The epistatic gene can alter, influence, or suppress the expression of a hypostatic gene.

When there is only one epistatic gene in h0m0zyg0us recessive state, the interaction is known as recessive epistasis. This gene supresses the expression of the other gene. Phenotypic ratio 9:3:4.

In double recessive epistasis, both genes are epistatic when they are in a h0m0zyg0us recessive state. This is, in the h0m0zyg0us recessive state, either of the genes will inhibit the expression of the other gene. And vice-versa. Phenotypic ratio 9:7.

In Duplicate gene action, there are two genes involved in the expression of a phenotype, and the presence of at least one dominant allele is enough to produce the dominant phenotype. Phenotypic ratio 15:1.

So now, let us make the test-cross.

Note: You will see that the cross is the same in the three cases, but the phenotypic ratio changes according to the interaction.

Cross) dihybrid individual with h0m0zyg0us recessive one

Parental)           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

F2) Phenotypic ratio

***Double recessive epistasis  ⇒  4:12

4/16 dihybrids12/16 heter0zyg0us for one gene and h0m0zyg0us recessive for the other gene.

Both genes in recessive state inhibit the expression of the trait.

***Recessive epistasis  ⇒  8:8

In h0m0zyg0us recessive state, Gene A inhibits the expression of B.

8/16 AaBb or Aabb → The trait coded by B or b is being expressed8/16 aaBb or aabb → The trait coded by B or b is inhibited by aa.

***Duplicate gene action   ⇒  12:4

Only one dominant allele, either A or B, is enough to express the trait.

12/16 AbBb, Aabb, or aaBb   4/16 aabb → No dominant allele, no trait expressed.

Phenotypic ratios:

a) 4:12 = 1:3

b) 8:8 = 1:1

c) 12:4 = 3:1

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