Question

4. In summer squash, fruit color and shape are controlled by two independently assorting genes as follows:

Trait	Gene	Alleles	Phenotypes
Fruit color	W	W w	White fruit yellow fruit
Fruit shape	D	D d	Disk-shaped fruit sphere-shaped fruit

1. A homozygous White Disk plant has been crossed with a homozygous Yellow Sphere plant (this is the P₁ generation). List the expected genotype and phenotype of the F₁ plants?

P1   WWDD x wwdd

F1    WwDd

Genotype: WwDd

Phenotype: White, Disk

2. If the F₁ plants are self-fertilized, what proportion (expressed as a fraction) of the F₂ plants will be White Sphere?

3.     A testcross is made with a White Disc F₂ plant selected at random. The offspring of this testcross were ½ White Disk and ½ Yellow Disk. What was the genotype of the White Disk plant that was used in the testcross?

Answer 1

a) A homozygous White Disk plant has been crossed with a homozygous Yellow Sphere plant (this is the P1 generation). List the expected genotype and phenotype of the F1 plants?

P1   WWDD x wwdd

F1    WwDd

Genotype: WwDd

Phenotype: White, Disk

You are right: 100% heterozygous White, Disk

b) If the F1 plants are self-fertilized, what proportion (expressed as a fraction) of the F2 plants will be White Sphere?

In this case, without making a Punnett Square, we can predict the proportions of F2. For this, we need to use the Mendel’s second law of independent assortment where establish that the dihybrid cross between two heterozygous will result in the following proportions: 9:3:3:1

In this case we will obtain:

9/16 White disk

3/16 White sphere

3/16 Yellow disk

1/19 Yellow sphere

c) A testcross is made with a White Disc F2 plant selected at random. The offspring of this testcross were % White Disk and 22 Yellow Disk. What was the genotype of the White Disk plant that was used in the testcross? Since we need at least two recessive alleles in order to obtain a Yellow color. The white disk must be heterozygous. Remember that when we test cross an individual, we are crossing it with a homozygous recessive. Lets make the Punnett Square to test our prediction: WWDD x wwdd WD WD WD WD wd wwDd wwDd WwDd WwDd WwDd WwDd wd wwDd wd WwDd wwDd WwDd WwDd wwDd wwDd wwDd wa WwDd wwDd 8/16 = 22 White disk 8/16 = 72 Yellow disk So, the genotype of the white disk parent was WwDD