PLZ HELP ME ANSWER THESE QUESTIONS

In document

Explanation:

Part 1

The freckles allele (F) is dominant to the no freckles allele (f). The parents are both heterozygotes, Ff, and both freckled. When they are crossed (Ff x Ff) in the resulting offspring:

There is 1 FF genotype, 2 Ff genotypes, and 1 ff genotype. FF and Ff give freckles, because F is dominant to F.

ff gives no freckles.

Therefore:

Genotypes: 1 FF: 2 Ff: 1 ff

Phenotypes: 3 freckles: 1 non-freckles

Part 2

Both parents are heterozygous at both traits, so their phenotypes are both PpRr.

Due to the law of independent assortment, the alleles in each trait are sorted into gametes independently. That means the gametes can have any combination of flower and seed alleles.  The possible gametes in both parents are:

Gamete 1: PR

Gamete 2: Pr

Gamete 3: pR

Gamete 4: pr

When they are crossed (PpRr x PpRr - punnett square shown). The possible genotypes for each phenotype are:

purple flowers, round seeds: PPRr, PpRr, PPRr, PPRR, PpRR, PpRr, PpRr, PpRR, PpRr

purple flowers, wrinkled seeds: PPrr, Pprr, Pprr  

white flowers, round seeds: ppRR, ppRr, ppRr

white flowers, wrinkled seeds: pprr

and the numbers of each phenotype are:

purple flowers, round seeds: 9

purple flowers, wrinkled seeds: 3

white flowers, round seeds: 3

white flowers, wrinkled seeds: 1

The classic ratio for a dihybrid cross between two heterozygote parents.

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