Question

In the summer squash, Cucurbita pepo, the fruit is normally disk-shaped. Two recessive mutations, g (globe) and r, (round), are known in different geographical races of the species, both of which cause fruit to be spherical. In crosses between homozygous g/g and r/r stocks, the F1 is always disk-shaped. However, when these F1 hybrids were interbred, it was discovered that squash plant homozygous recessive at both g and r make fruit that is elongated. What are the genotypes and phenotypes of the F2 from the cross of the F1 hybrids in Question 1?

Answer 1

While test crosses in pea plants can typically reveal the genotype of an unknown parent, crossing with a homozygous recessive cannot differentiate between homozygous dominant and heterozygous without a larger sample size. Epistasis in summer squash results in a 12 white:3 yellow:1 green phenotypic ratio when WwYy heterozygotes are crossed. The shepherd's-purse plant exhibits a 15 triangular:1 ovoid phenotypic ratio when AaBb plants are crossed.

Step-by-step explanation:

In pea plants, the allele for round peas (R) is dominant to the allele for wrinkled peas (r). When performing a test cross between a homozygous recessive plant with wrinkled peas (rr) and a plant of unknown genotype displaying round peas, and if all resulting offspring have round peas, we cannot conclusively determine if the round pea plant is homozygous dominant (RR) or heterozygous (Rr). This is because both genotypes can produce round pea offspring when crossed with a homozygous recessive (rr) plant. However, if the unknown parent were heterozygous (Rr), the probability of all three offspring displaying round peas is 1/23 because each offspring has a 1/2 chance of inheriting the dominant allele for round peas.

In the case of summer squash exhibiting epistasis, a cross between white heterozygotes (WwYy) for both the W and Y genes produces a phenotypic ratio of 12 white:3 yellow:1 green. Here, the presence of at least one dominant W allele masks the expression of the Y alleles, leading to white fruit regardless of the genotype of the Y gene.

Considering shepherd's-purse plant (Capsella bursa-pastoris), where seed shape is controlled by two genes, A and B, and triangular seeds occur if any dominant allele is present from either gene, the expected phenotypic ratio for a cross between plants that are heterozygous at both loci (AaBb x AaBb) is 15 triangular : 1 ovoid, as there are 15 genotypic combinations that contain at least one dominant allele, and only one genotypic combination that is homozygous recessive at both loci (aabb).

Answer 2

Let the interbred alleles be represente dy the letters A and B for two genes. Gene A will exist in two alleles, that will icnlude A and g while Gene B exists in two alleles, B and r.

The uppercase alleles are dominant to the lowercase alleles. The interbred long-shaped squash is AABB (aabb) and the true-breeding disk-shaped is ggrr AABB.The F1 offspring are AaBb.

The outcome of interbred of the F1 plants and genotypes and phenotype sof F2 for:

1) the disk-shaped phenotype, an offspring must inherit at least one dominant allele from both genes.

1 ggrr+ 2 Agrr+ 2 ggrB+ 4 AgBr = 9 disk-shaped offspring

2) To get the round phenotype, an offspring must inherit at least one dominant allele for one of the two genes but must be homozygous recessive for only one of the two genes.

1 ggrr+ 1 ggBB+ 2 AArB+ 2 gABB = 6 round-shaped offspring

3) To get the long phenotype, an offspring must inherit all recessive alleles:

1 AABB= 1 long-shaped offspring

Hence, the phenotype and genotype of F2 is is 9 disk-shaped offspring and 1 ggrr+ 2 Agrr+ 2 ggrB+ 4 AgBr; 6 round-shaped offspring and 1 ggrr+ 1 ggBB+ 2 AArB+ 2 gABB; and 1 long-shaped offspring and 1 ggrr+ 1 ggBB+ 2 AArB+ 2 gABB respectively.