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

Correct Answer: 0.24

Let:
\quad Tall = dominant (T), \quad Dwarf = recessive (t)
\quad Red = dominant (R), \quad White = recessive (r)
Given:
P generation: TT RR (tall red) $\times$ tt rr (dwarf white)
F$_1$ generation: All offspring will be heterozygous: Tt Rr (tall red)
F$_2$ generation: Cross Tt Rr $\times$ Tt Rr
Use independent segregation:
\quad For height (Tt × Tt):
\quad\quad Genotypes = TT, Tt, tt → Probabilities = \(\frac{1}{4}\), \(\frac{1}{2}\), \(\frac{1}{4}\)
\quad For flower color (Rr × Rr):
\quad\quad 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)
\quad TT and RR = \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\)
\quad TT and Rr = \(\frac{1}{4} \times \frac{1}{2} = \frac{2}{16}\)
\quad 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: \(\frac{3}{4}\)
\[ {Required probability} = \frac{\frac{3}{16}}{\frac{3}{4}} = \frac{3}{16} \times \frac{4}{3} = \frac{1}{4} = 0.25 \]