Two tall plants were crossed and all the F1 plants were tall. when the F1 plants were selfed, some of the F2…

Analyze the question

Two tall parents are crossed. All the F1 are tall. When the F1 are selfed, the F2 contain both tall and short plants. We need the genotypes of the original (parental) plants.

Concept

Tall (T) is dominant over short (t). A short plant must be tt, so it can only appear in the F2 if both F1 parents of that generation carried a t allele. That means the F1 plants must be heterozygous (Tt). Selfing Tt × Tt gives 3 tall : 1 short.

Working it out

1. The F2 contain short plants (tt), so the F1 plants must be Tt.
2. For the F1 to be Tt, each must have received one T and one t from the original parents. So at least one of the original parents must have carried a t allele.
3. The original parents were both tall, so neither can be tt. A tall parent carrying t must therefore be Tt.

Possible crosses of two tall parents:

• TT × TT gives all TT. Selfing gives only tall, so there is no short F2. ✗
• TT × Tt gives 1/2 TT and 1/2 Tt. All are tall, and the Tt ones give short plants when selfed. ✓
• Tt × Tt gives 3 tall : 1 short in F1. Short F1 would appear, but the question says all F1 were tall. This is not consistent with the question as stated, unless we were lucky with small numbers. ✗
• Option A (Tt × tt) has a short parent, which contradicts "two tall plants". ✗
• Option C (tt × tt) gives all short. ✗

Check of option D (TT × Tt)

F1: TT and Tt, all tall ✓. Selfing the Tt plants: Tt × Tt gives TT, Tt, Tt, tt, so some F2 are short ✓.

Answer: D (TT and Tt)

Tip: When a recessive trait reappears in a later generation, work backwards. The plants that produced it must have been carriers (heterozygous). If the parents show the dominant trait and all offspring do too, at least one parent is homozygous dominant (TT), and the other must carry the recessive allele to explain the reappearance.