Q.A true breeding homozygous pea plant with round seeds and yellow cotyledons is crossed with homozygous pea plant with wrinkled seeds and green cotyledons.
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Start your 14-day free trial to unlock the full solution →A dihybrid cross of RRYY x rryy gives an all-RrYy F1, and F2 shows the classic 9:3:3:1 ratio, proving that the two gene pairs assort independently of each other.
(a) The cross:
Round seed shape (R, dominant) vs wrinkled (r, recessive); Yellow cotyledon colour (Y, dominant) vs green (y, recessive).
Parents: RRYY (round, yellow) x rryy (wrinkled, green)
Gametes: RY (from RRYY) and ry (from rryy)
F1 generation: all offspring are RrYy -- Round, Yellow (since R and Y are both dominant, they mask r and y).
When F1 (RrYy) is self-pollinated, each parent produces 4 types of gametes in equal proportion: RY, Ry, rY, ry. A 4x4 Punnett square gives 16 combinations in F2:
- 9 parts R_Y_ -> Round, Yellow
- 3 parts R_yy -> Round, Green
- 3 parts rrY_ -> Wrinkled, Yellow
- 1 part rryy -> Wrinkled, Green
So the F2 phenotypic ratio is 9 : 3 : 3 : 1 (Round Yellow : Round Green : Wrinkled Yellow : Wrinkled Green), and the genotypic ratio is the standard 1:2:1:2:4:2:1:2:1 (9 genotype classes).
(b) Mendel's generalisation: …
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