Question

Conduct a dihybrid cross of pea plants with the following combination of traits: Parent 1—Tall plant with white flowers (pure breeding); Parent 2—Dwarf plant with purple flowers (pure breeding). i) What are the genotypes of the parents? ii) What are the genotypes of their gametes? iii) Portray a cross and determine the F1 generation. Use a Punnett’s square. What are the genotypes and phenotypes of their offspring? iv) Conduct a F2 cross with the offspring. What are the genotypes and phenotypes of the offspring and their ratios.

Answer 1

i) Genotype of Parent 1= TTpp

Genotype of Parent 2= ttPP

ii) The genotypes of their gametes

Parent 1= Tp

Parent 2= tP

iii) F1 generation = All TtPp (tall and purple flower)

iv) F2 phenotype ratio= 9 tall, purple: 3 tall, white: 3 dwarf, purple : 1 dwarf, white.

F2 Genotype ratio= 1:2:1:2:4:2:1:2:1

Step-by-step explanation:

Let's assume that allele "T" gives tall plants while allele "t" imparts dwarfism. Likewise, the dominant allele "P" is responsible for purple flowers while recessive allele "p" gives white flowers.

i) Genotype of Parent 1—Tall plant with white flowers (pure breeding)= TTpp

Genotype of Parent 2—Dwarf plant with purple flowers (pure breeding)= ttPP

ii) The genotypes of their gametes

Parent 1= Tp

Parent 2= tP

iii) F1 generation = All TtPp (tall and purple flower)

iv) Each of the F1 plant with TtPp genotype will form four types of gametes= TP, Tp, tP and tP

A cross between TtPp x TtPp gives F2 progeny in following ratio=

F2 generation phenotype ratio= 9 tall, purple: 3 tall, white: 3 dwarf, purple : 1 dwarf, white.

Genotype ratio= 1:2:1:2:4:2:1:2:1