Question

In a genetic cross between a true-breeding tall parent bearing red flowers and a true-breeding dwarf parent bearing white flowers, only tall plants with red flowers are obtained in the F$_1$ population. Considering these two traits segregate independently, if one tall individual is selected from the F$_2$ population, the probability that it would be genotypically homozygous for plant height and make red flowers is ........... {(Round off to two decimal places)}

Hint:

When dealing with probabilities involving conditions (like "given the individual is tall"), always use conditional probability: divide favorable outcomes by total of the given condition.

Answer 1

Let: 
    Tall = dominant (T),     Dwarf = recessive (t) 
    Red = dominant (R),     White = recessive (r) 
Given: 
P generation: TT RR (tall red) × tt rr (dwarf white) 
F₁ generation: All offspring will be heterozygous: Tt Rr (tall red) 
F₂ generation: Cross Tt Rr × Tt Rr 
Use independent segregation: 
    For height (Tt × Tt): 
        Genotypes = TT, Tt, tt → Probabilities = \(\frac{1}{4}\), \(\frac{1}{2}\), \(\frac{1}{4}\) 
    For flower color (Rr × Rr): 
        Genotypes = RR, Rr, rr → Probabilities = \(\frac{1}{4}\), \(\frac{1}{2}\), \(\frac{1}{4}\) 
Now, we select a tall individual, so tt (dwarf) is not considered. Total probability of tall = TT + Tt = \(\frac{3}{4}\) 
We are asked to find the probability that an individual is: tall (i.e. TT or Tt), but specifically TT and red (RR or Rr). 
Let’s compute the favorable cases: 
Favorable genotype: TT and (RR or Rr) 
    TT and RR = \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\) 
    TT and Rr = \(\frac{1}{4} \times \frac{1}{2} = \frac{2}{16}\) 
    So, total favorable = \(\frac{1}{16} + \frac{2}{16} = \frac{3}{16}\) 
We only consider tall individuals, so we normalize this over total tall probability: \(\frac{3}{4}\) 
\[ \text{Required probability} = \frac{\frac{3}{16}}{\frac{3}{4}} = \frac{3}{16} \times \frac{4}{3} = \frac{1}{4} = 0.25 \]