Mendel crossed true-breeding pea plants: the F1 all showed one trait, and the F2 showed both in about a 3:1 ratio. He explained it with paired factors, now called genes, that separate (segregate).
In Punnett Squares you used letters for alleles. Those letters began with one patient experimenter. In the 1850s and 1860s Gregor Mendel, a friar in Brno (now in the Czech Republic), grew thousands of pea plants and counted their traits. He reported his results in 1865 and published them in 1866.
Mendel started with true-breeding plants: plants that, pollinated by themselves, always gave offspring like themselves. He crossed two true-breeding parents that differed in one trait, such as round seeds and wrinkled seeds. That is the P generation. Their offspring are the F1, and the offspring of F1 plants pollinated by themselves are the F2.
🔑 Key idea
Every F1 plant showed only one form of the trait, the dominant one. In the F2, the hidden recessive form came back in about one plant in four: a 3:1 ratio. Each plant carries two copies of each "factor" (gene), and the two copies separate when gametes form (the law of segregation).
Two of Mendel's F2 counts. Neither is exactly 3:1, but both are close.
Trait (F2 seeds)
Dominant
Recessive
Ratio
Seed shape
5,474 round
1,850 wrinkled
2.96 : 1
Seed color
6,022 yellow
2,001 green
3.01 : 1
In each trait, the dominant form outnumbers the recessive one about three to one.
💡 Why it works
Why 3:1? Call the round allele R and the wrinkled allele r. The F1 are all Rr. Each F1 parent puts R into half of its gametes and r into the other half. Two gametes join at random, so the chances are RR ¼, Rr ½ and rr ¼. Only rr looks wrinkled, so ¾ of the F2 are round and ¼ wrinkled.
The F1 × F1 cross. One box in four is rr.
✏️ Worked example: Predicting the F2
True-breeding purple-flowered peas are crossed with true-breeding white-flowered peas. All F1 plants are purple. If the F1 pollinate themselves and give 400 F2 plants, about how many are white?
Purple shows in the F1, so purple is dominant: P. White is recessive: p.
The F1 are all Pp. A Pp × Pp cross gives ¼ pp.
¼ of 400 is 100.
Answer: About 100 white plants (and 300 purple).
✏️ Worked example: Reading real counts
Mendel's stem-length F2 had 787 tall and 277 short plants. Is that close to 3:1?
Divide: 787 ÷ 277 ≈ 2.84.
Chance makes real counts wobble around the expected ratio, the way 100 coin flips rarely give exactly 50 heads.
Answer: Yes, about 2.8 : 1, close to 3 : 1.
⚠️ Common mistake
People often think the recessive trait was destroyed in the F1, or that the parents' traits blend. If they blended, the F1 would be in between and the F2 could never show pure white again. The white allele was hidden, not lost: it reappears in about a quarter of the F2.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.