Tall and Short Pea Plants in the School Garden
In the school garden, students cross a pure tall pea plant with a pure short pea plant. All the F1 plants are tall. On self-pollination, the F2 generation has 787 tall and 263 short plants. In a second experiment, a student crosses one of the F1 tall plants with a short plant and obtains 51 tall and 49 short plants. In a third experiment, a tall plant from the garden is self-pollinated and produces 100 tall plants and no short plant.
The students argue: 'In the first experiment, shortness vanished in F1, so the short trait must have been destroyed. And a 3:1 ratio means the tall allele is stronger.' The teacher asks them to explain using alleles, gametes and fertilisation, and to say how they could find out whether the tall plant in the third experiment is TT or Tt without waiting for self-pollination.
1. Which trait is dominant? Write the genotypes of the parents, the F1 and the F2 in the first experiment. Explain how the 787:263 ratio supports your answer.
Read the worked answer
Tall is dominant. Let T = tall allele and t = short allele.
Parents: pure tall TT × pure short tt. The gametes are T and t, so all F1 plants are Tt and are tall, because T masks t in the heterozygote.
F1 selfing: Tt × Tt gives gametes T and t from each plant. The combinations are TT : Tt : Tt : tt = 1 TT : 2 Tt : 1 tt. TT and Tt are tall and tt is short, so the expected ratio is 3 tall : 1 short.
Observed: 787 : 263 = 2.99 : 1, very close to 3 : 1. This is exactly what segregation of a single pair of alleles predicts, so the data support tall being dominant and short recessive. Shortness did not disappear: the t allele was hidden in the F1 and reappeared when two t gametes met in the F2.
2. Explain the 51:49 result in the second experiment. What is this cross called, why is it useful, and what result would you expect if the F1 tall plant had been TT?
Read the worked answer
The F1 tall plant is Tt. Crossed with a short plant (tt), it produces T and t gametes in equal numbers, while the short parent produces only t. The offspring are Tt (tall) and tt (short) in a 1 : 1 ratio.
From 100 seeds we expect about 50 tall and 50 short. The observed 51 : 49 is a very close fit; the small difference is chance variation in a finite sample.
This is a test cross - crossing an individual showing the dominant phenotype with a homozygous recessive. It is useful precisely because a tall plant can be TT or Tt and looks the same in both cases; the test cross reveals the hidden genotype. TT × tt gives all Tt (all tall), while Tt × tt gives 1 tall : 1 short. If the F1 plant had been TT, all 100 offspring would have been tall, with no short plants.
3. The third experiment gives 100 tall and 0 short. Can the students conclude the plant is TT? Explain using probability and sample size, describe a quicker test, and correct the statement that 'the tall allele is stronger'.
Read the worked answer
The result makes TT very likely but does not prove it with certainty. A Tt plant selfed gives tall and short in a 3 : 1 ratio, so each seed has a 3/4 chance of being tall. The chance that 100 seeds in a row are all tall by luck is (3/4)^100 ≈ 3 × 10^-13 - so tiny that the plant is almost certainly TT. But 'almost certainly' is not proof; a larger sample or a direct test is better.
Quicker test: cross the tall plant with a short (tt) plant. If any short offspring appear, the tall plant is Tt; if all offspring are tall, it is probably TT, and more offspring give stronger evidence.
The statement 'the tall allele is stronger' is incorrect. Dominance means that in a heterozygote Tt the T trait is expressed and t is masked; it does not mean T is stronger, more numerous or better. The t allele remains unchanged and can reappear whenever two t gametes combine, as the 263 short F2 plants show.