Question

58. A pure breeding pea plant with round
yellow seeds was crossed with pea
plant having wrinkled green seeds. On
selfing of F, hybrid of his cross, 64
progenies were obtained in F2
generation. Find out the number of F2
progenies showing non-parental
characters.​

Answer 1

The number of F₂ progenies showing non-parental characters is 24.

EXPLAINATION

GIVEN

• Number of progeny of F₂ generation is 64

PHENOTYPIC RATIO

The phenotypic ratio of a dihybrid cross is

(Round & Yellow): (Round & Green) :(Wrinkle & Yellow) :(Wrinkle & Green) = 9:3:3:1

CALCULATION

Using the phenotypic ratio, let

• 9x = No. of progeny with Round & Yellow seed
• 3x = No. of progeny with Round & Green seed
• 3x = No. of progeny with Wrinkled & Yellow seed
• 1x = No. of progeny with Wrinkled & Green seed

As total number of progeny = 64

⇒ 9x + 3x + 3x + 1x = 64

⇒ 16x = 64

⇒ x = 4

Number of progeny with non-parental characters

= (No. of progeny with Round & Green seed) + (No. of progeny with Wrinkled & Yellow seed)

= 3x + 3x

= 6x

= 6*4

= 24