Q.63 In a diploid angiosperm species, flower colour is regulated by the R gene.
RR and Rr genotypes produce red flowers, whereas the rr genotype produces
white flowers. If two individual plants are randomly selected from a large
segregating population of a genetic cross between RR and rr parents, the
probability of both the plants producing red flowers is _______.
(Rounded off to two decimal places)
The probability of both randomly selected plants producing red flowers is 0.56. This comes from a standard Mendelian F2 population where red flower genotypes (RR and Rr) occur with frequency 3/4.
Genetic Basis
Flower color follows complete dominance: R (red) dominant over r (white). RR × rr parents yield all Rr F1 (red flowers). F1 self-cross (Rr × Rr) produces F2 genotypic ratio 1 RR : 2 Rr : 1 rr, so phenotypic ratio 3 red : 1 white.
Probability of red-flowered plant = 3/4 = 0.75.
Probability Calculation
For two independent selections from large F2 population:
P(both red) = P(red) × P(red) = (3/4) × (3/4) = 9/16 = 0.5625.
Rounded to two decimal places: 0.56.
F2 Genotype Frequencies
| Genotype | Frequency | Phenotype |
|---|---|---|
| RR | 1/4 | Red |
| Rr | 2/4 | Red |
| rr | 1/4 | White |
Common Distractors
Query mentions “explain every option,” but none given (numerical answer expected). Common distractors in such CSIR NET/GATE questions:
- 0.25 (both white, or misread as F1)
- 0.50 (one red, one white)
- 0.75 (single plant red)
- Correct: 0.56 (both red)
Punnett Square Analysis
F1 (Rr) selfing gives:
- 25% RR (red)
- 50% Rr (red)
- 25% rr (white)
Red frequency: 75% or 0.75. Both red: 0.75² = 0.5625 ≈ 0.56.
Exam Relevance
CSIR NET/GATE tests such calculations for independent assortment. Large population assumes random mating equilibrium (Hardy-Weinberg: p² + 2pq + q², red = 1 – q² = 0.75).
Master probability both plants red flowers diploid angiosperm R gene for genetics scoring.
