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.
(a) What will be the phenotypic ratio in the F2?
(b) If an F1 plant is backcrossed to the bitter, yellow-spotted parent, what phenotypes and
proportions are expected in the offspring?
(c) If an F1 plant is backcrossed to the sweet, non-spotted parent, what phenotypes and proportions
are expected in the offspring?
(d) If you testcross the F1 plant, what phenotypes and proportions are expected in the offspring?

(a) 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)

(b) 100%, bitter, yellow spotted.

(c) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.

(d) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.

Explanation:

(a) The parental cross is BBSS x bbss.

We know that the F1 will be all BbSs (see attached punnet square). The only possible gametes the parental generation can pass on to the F1 are BS and bs, respectively.

The F1 intercross is therefore BbSs x BbSs. the alleles that each can pass on are BS, Bs, bS, or bs. The punnett square attached shows the phenotype ratio is 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)

(b) An F1 plant is BbSs. If it is backcrossed with a bitter, yellow spotted parent (BBSS), the cross is BbSs x BBSS (see attached punnett square). The resulting genotypes all give a bitter, yellow spotted phenotypes, so the phenotype of the offspring is 100% bitter, yellow spotted

(c) An F1 plant is BbSs, if it is backcrossed with a sweet, non spotted parent (bbss), the cross is BbSs x bbss. (see attached punnett square). The resulting phenotypes are 1:1:1:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. I.e. 25% of each genotype

(d) A test cross is a cross between an individual with a known genotype and another plant. The plants showing the recessive trait have a known genotype (they must be homozygous). So in this case, the F1 plant would be crossed with the homozygous recessive plants (BbSs x bbss). Therefore, the resulting phenotypes would be as in option c, 1:1:1:1