Short answer. Every plant carries two copies (alleles) of each gene and passes one, chosen at random, to each seed. Cross two plants that each carry one hidden recessive allele and about a quarter of the offspring will show the recessive trait. Most traits growers care about (yield, potency, flavour) are not single genes, but Mendel's rules are still the foundation: they explain why F1s look uniform and F2s scatter.
What's happening
Mendel's two laws [1]:
- Segregation. The two alleles of a gene separate when pollen and ovules form, so each gamete carries one.
- Independent assortment. Alleles of different genes are inherited independently, unless the genes sit close together on the same chromosome (linkage).
Cannabis has 10 chromosome pairs, and recombination is concentrated near chromosome ends, so large central regions are inherited in blocks [5]. That matters: the THC and CBD synthase genes sit together in such a block [5].
Punnett square 1: a recessive trait (autoflowering)
Autoflowering behaves as one recessive gene in the mapped populations [3]. A = photoperiod, a = auto.
F1 (Aa) x F1 (Aa):
| A | a | |
|---|---|---|
| A | AA (photo) | Aa (photo) |
| a | Aa (photo) | aa (auto) |
Genotypes 1 AA : 2 Aa : 1 aa. Phenotypes 3 photo : 1 auto. Two of every three photoperiod F2s still carry the auto allele, and you cannot see which by eye.
Punnett square 2: a codominant trait (chemotype)
de Meijer et al. (2003) modelled chemotype with one locus, B, and alleles B_T (THC synthase) and B_D (CBD synthase), behaving codominantly [4]. Later work shows it is really two tightly linked genes inherited as a unit, with complications (see chemotype inheritance).
THC:CBD F1 (B_T/B_D) x THC:CBD F1 (B_T/B_D):
| B_T | B_D | |
|---|---|---|
| B_T | B_T/B_T (THC dominant, type I) | B_T/B_D (mixed, type II) |
| B_D | B_T/B_D (mixed, type II) | B_D/B_D (CBD dominant, type III) |
Expect 1 type I : 2 type II : 1 type III. Here you can see the heterozygote, because both alleles show.
Punnett square 3: two genes at once
Cross two plants that are heterozygous for both autoflowering (Aa) and chemotype (B_T/B_D). If the genes are unlinked, you get 16 combinations. The share that is auto and CBD dominant (aa and B_D/B_D) is ¼ x ¼ = 1/16. Each extra fixed requirement multiplies the odds down. That is the core reason pheno hunts need big numbers.
Numbers that matter
| Situation | Expected share | Plants to grow for 95% chance of at least one |
|---|---|---|
| Recessive trait in an F2 (¼) | 25% | 11 |
| Two unlinked recessives (1/16) | 6.25% | 47 |
| Three unlinked recessives (1/64) | 1.6% | 191 |
| Homozygous dominant you can't see (1/3 of dominant-looking F2s) | 33% of the dominant class | Needs a progeny test |
Formula: n = ln(0.05) / ln(1 - p). With p = 0.25, n = 10.4, so 11 plants [2].
Checking a ratio
Use a chi-square test. For 100 F2 plants with 70 photo and 30 auto against 75:25: χ² = (70-75)²/75 + (30-25)²/25 = 0.33 + 1.0 = 1.33, below 3.84 (1 degree of freedom, 5% level), so the data fit 3:1. Toth et al. used exactly this logic to call autoflowering recessive [3].
Why real ratios drift
- Small samples. Ten plants rarely show a perfect 3:1.
- Linkage. Genes near each other travel together.
- Selection before scoring. If weak seedlings die first, recessive classes can be under-counted.
- Many genes. Height, yield and terpene totals are quantitative; see quantitative traits.
- Dominance is not always complete. Autoflower heterozygotes flower earlier than pure photoperiods even though they are not autos [3].
Common mistakes
- Assuming "dominant" means "common" or "better". It only describes how alleles interact.
- Reading 3:1 into a 12-plant tent.
- Forgetting that a dominant-looking plant can be AA or Aa. Only a test cross (to aa) or a DNA marker tells you.
