Advertisements
Advertisements
Question
A homozygous plant having round (R) and yellow (Y) seed is crossed with a homozygous plant having wrinkled (r) and green (y) seeds.
- Draw a Punnett square to show F1 generation.
- Give the genotype of F1 generation
- Give the dihybrid phenotypic ratio and the phenotype of the offspring of the F2 generation when two plants of the F1 generation are crossed.
- Name and state the law which explains the dihybrid ratio.
- Mention the phenotype and genotype ratios of the offsprings in F2 generation.
Advertisements
Solution
Let R = round, r = wrinkled, Y = yellow, and y = green.
P generation:
RRYY × rryy
Gametes:
RY × ry
(i) Punnett square for F1 generation
| ry | |
| RY | RrYy |
Thus, all F1 plants are RrYy and have round yellow seeds.
(ii) Genotype of F1 generation
RrYy
(iii) F2 generation
F1 cross:
RrYy × RrYy
Gametes produced by each parent: RY, Ry, rY, ry
| RY | Ry | rY | ry | |
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
F2 phenotypic ratio:
9 Round yellow : 3 Round green : 3 Wrinkled yellow : 1 Wrinkled green
(iv) Law explaining the dihybrid ratio
Law of Independent Assortment:
During gamete formation, the alleles of one pair of characters assort independently of the alleles of another pair of characters.
(v) Phenotypic and genotypic ratios of F2
Phenotypic ratio:
9 : 3 : 3 : 1
Genotypic ratio:
1 RRYY : 2 RRYy : 1 RRyy : 2 RrYY : 4 RrYy : 2 Rryy : 1 rrYY : 2 rrYy : 1 rryy
