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Reverse Punnett square: Working backwards to the parents

Published

Table of contents
  1. The four one-gene signatures
  2. Two genes: split the ratio, then recombine
  3. Worked example
  4. Common two-gene signatures
  5. Common mistakes
  6. Practice
  7. Related
  8. Sources

Level: High school · You’ll need: monohybrid crosses and ratios as fractions

A normal Punnett square goes forwards: you know the parents and predict the offspring. A reverse (or backwards) Punnett square goes the other way. You start with the offspring you observed and work out which parent genotypes could have produced them.

The trick is that each simple cross leaves a recognisable “signature” ratio. Learn the signatures, then match.

Assumptions: complete dominance, independent assortment for two genes, and enough offspring for the ratio to be clear. With small numbers, any conclusion is only the most likely answer.

The four one-gene signatures

Offspring phenotypesParent genotypesWhy
All recessiveaa × aaNeither parent has an A to pass on
About 1 dominant : 1 recessiveAa × aaHalf the gametes from Aa carry a, and aa can only give a
About 3 dominant : 1 recessiveAa × Aa¼ of offspring get a from both parents
All dominantAA × AA, AA × Aa or AA × aaAt least one parent is AA and passes A every time (with only a few offspring, Aa × Aa or Aa × aa can also give all dominant by chance)

Two quick rules fall out of this table:

  • Any recessive offspring means both parents carry at least one recessive allele. An aa child needs an a from each parent.
  • “All dominant” can’t tell you the other parent’s genotype. Three different crosses produce it. A test cross with an aa partner is how you’d find out.

If you know offspring genotypes, not just phenotypes, it gets easier. A 1 AA : 2 Aa : 1 aa result can only come from Aa × Aa. A 1 AA : 1 Aa result can only come from AA × Aa.

Two genes: split the ratio, then recombine

Don’t try to memorise every two-gene ratio. Treat each trait separately.

  1. Add up the offspring for trait 1 only (dominant vs recessive) and ignore trait 2.
  2. Match that ratio to the one-gene table.
  3. Repeat for trait 2.
  4. Combine the two answers into full genotypes.

Worked example

Pea plants: tall (T) is dominant to dwarf (t), and purple flowers (P) are dominant to white (p). A cross produces 120 offspring:

OffspringCount
Tall, purple47
Tall, white13
Dwarf, purple44
Dwarf, white16

Height only: tall = 47 + 13 = 60, dwarf = 44 + 16 = 60. That’s 1 : 1, so the parents are Tt × tt.

Colour only: purple = 47 + 44 = 91, white = 13 + 16 = 29. That’s about 3 : 1 (91 ÷ 29 ≈ 3.1), so the parents are Pp × Pp.

Combine: one parent is TtPp and the other is ttPp.

Check it forwards. Expected proportions are ½ × ¾ = 3/8 tall purple, ½ × ¼ = 1/8 tall white, 3/8 dwarf purple and 1/8 dwarf white. Out of 120 that’s 45 : 15 : 45 : 15, close to what was observed. Run TtPp × ttPp in the dihybrid calculator to see the full square.

Common two-gene signatures

Offspring phenotype ratioParent genotypes
9 : 3 : 3 : 1AaBb × AaBb
1 : 1 : 1 : 1AaBb × aabb or Aabb × aaBb
3 : 3 : 1 : 1AaBb × Aabb or AaBb × aaBb
3 : 1 (one trait varies, the other is uniform)One gene Aa × Aa; the other gene has at least one homozygous dominant parent, or both parents recessive

Notice the 1 : 1 : 1 : 1 row. Two different pairs of parents give exactly the same ratio, because each gene is a 1 : 1 cross either way. Splitting trait by trait tells you which genes are heterozygous, but not always which parent carries them.

Common mistakes

  1. Reading a rough ratio too literally. 91 : 29 isn’t exactly 3 : 1, and it doesn’t need to be. Real counts vary by chance.
  2. Claiming certainty from “all dominant”. Say “at least one parent is AA.”
  3. Matching the four-class ratio directly instead of splitting it by trait.
  4. Forgetting the model. Linked genes, incomplete dominance, epistasis (one gene masking another, giving ratios like 9:3:4 or 9:7) or lethal alleles change the signatures. A result far from every row above is a clue that something else is going on.

Practice

  1. Two black mice have 31 black and 10 white pups. Black is dominant. What are the parents’ genotypes?
  2. A round-seeded pea crossed with a wrinkled pea gives 52 round and 49 wrinkled offspring. What are the parents?
  3. Two plants give offspring in a ratio of 3 tall red : 1 tall white. Tall is dominant, and red is dominant to white. What can you say about each gene?
Answers
  1. About 3 : 1, so both parents are Bb (heterozygous).
  2. About 1 : 1, so Rr × rr. The round parent is heterozygous.
  3. Colour is 3 : 1, so both parents are Rr. Height is all tall, so at least one parent is TT. You can’t tell the other parent’s height genotype from this data alone.

Sources

Try it: check any cross instantly with the dihybrid cross calculator, or use the monohybrid and trihybrid versions.