In watermelons bitter fruit B is dominant over sweet fruit b

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. Remember to draw out the cross. a. What will be the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

Solution

A)

Let us first analyze the allele variant for each character –

Bitter fruit – B

Sweet fruit – b

Yellow Spots – S

No spots – s

B and S are dominant over b and s respectively.

Genotype for Bitter, Yellow spotted homozygous melon – BBSS

Genotype for Sweet, Not spotted homozygous melon – bbss

A cross between these two parents would give us the following progeny –

BBSS X bbss

Therefore all the F1 progeny would have the genotype as BbSs and the phenotype would be Bitter and Yellow Spotted.

Now, the question asks for the phenotypic ratios when F1 is crossed with another F1 generation-

We can put this as -

F1 X F1

which will be -

BbSS X BbSs

We can resort to a punnet square board for the following

Let us try to analyze the number of each phenotype now -

Bitter and yellow spots - 9
Bitter and No spots - 3
Sweet and Yellow spots - 3
Sweet and No spots - 1

Therefore, the phenotypic ratio is 9:3:3:1

B)

Genotype of F1 progeny - BbSs

Genotype of Bitter, Yellow spotted parent - BBSS

A cross between the two will be as follows -

BbSs X BBSS

we can again resort to a punner square board for the following -

All the progeny will be Bitter with Yellow Spots.

C)

Genotype of F1 progeny - BbSs

Genotype of sweet, No spotted parent - bbss

A cross between the two will be as follows -

BbSs X bbss

we can again resort to a punner square board for the following -

Now, let us try to analyze the different genotypes and corresponding phenotypes-

BbSs - Bitter, Yellow Spotted ( total 4)
Bbss - Bitter, No spots( total 4)
bbSs - Sweet , Yellow spots( total 4)
bbss - Sweet, No spots( total 4)

Therefore, the above phenotypes will be in a ratio of 1:1:1:1

	bs	bs
BS	BbSs	BbSs
BS	BbSs	BbSs