Punnett squares
Every cross on this course is the same four steps: work out each parent's gametes, fill the grid, count the genotypes, then translate to phenotypes. This page does that for monohybrid, dihybrid, incomplete dominance, blood types and sex-linked crosses — and shows the probability shortcut that skips the grid entirely.
- 1. The four steps
- 2. Monohybrid and test crosses
- 3. When the heterozygote shows
- 4. Blood types
- 5. Sex-linked crosses
- 6. Dihybrid, without a 16-box grid
- 7. Turning ratios into answers
The four steps
The idea: a Punnett square is just an organized way of pairing every gamete from one parent with every gamete from the other.
- Write both genotypes from the words in the question. "Heterozygous tall" is Tt; "homozygous recessive" is tt.
- Split each parent into gametes. Each gamete gets one allele of each gene — that is meiosis. Tt gives T and t.
- Fill the grid: gametes along the top and side, each box the combination of its row and column.
- Count genotypes, then convert to phenotypes using which allele is dominant.
Watch out: the top and side hold gametes, not parents. Writing "Tt" along the top is the most common wrecking error, because every box then has three or four alleles in it.
Monohybrid and test crosses
The idea: Tt × Tt is the cross to know cold. It gives 1 : 2 : 1 genotypes and 3 : 1 phenotypes, and explains why a trait can skip a generation.
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
- Genotypes: 1 TT : 2 Tt : 1 tt.
- Phenotypes: 3 tall : 1 short, because TT and Tt look the same.
- Backwards: two tall parents with a short child must both be Tt — the child's two t alleles had to come from somewhere.
The test cross
A tall plant is either TT or Tt, and you cannot tell by looking. Cross it with tt: every gamete from that parent carries t, so the offspring reveal the unknown.
- TT × tt → all Tt, all tall.
- Tt × tt → half Tt, half tt, so about half are short.
A single short offspring proves the parent was Tt. All-tall offspring only make TT likely, not certain — with few offspring you could be unlucky.
When the heterozygote shows
The idea: if the heterozygote has its own appearance, the phenotype ratio equals the genotype ratio — 1 : 2 : 1 instead of 3 : 1.
| Pattern | Heterozygote looks like | Example |
|---|---|---|
| Complete dominance | the dominant homozygote | Tt is tall, like TT |
| Incomplete dominance | a blend of the two | CRCW is pink, between red and white |
| Codominance | both alleles, fully and separately | IAIB is blood type AB, not "in between" |
Pink × pink (CRCW × CRCW) gives 1 red : 2 pink : 1 white. Roan cattle and the AB blood type are codominance: you can see both alleles at once, not a mixture.
Watch out: notation counts. Incomplete dominance and codominance use superscripts on a shared letter (CR, CW) because neither allele is "the recessive one".
Blood types
The idea: three alleles, two of them codominant and one recessive — the standard example of multiple alleles.
| Blood type | Possible genotypes |
|---|---|
| A | IAIA or IAi |
| B | IBIB or IBi |
| AB | IAIB |
| O | ii |
Worked: type O mother × type AB father
Mother is ii, so every egg carries i. Father is IAIB, so half his sperm carry IA and half carry IB.
Children are IAi (type A) or IBi (type B) — 50% A and 50% B. Neither AB nor O is possible, which is the point of these questions.
Type O is the only phenotype with a single genotype, which makes it the most informative: an O child tells you both parents carry i.
Sex-linked crosses
The idea: the alleles ride on the X chromosome, so the genotype must show the chromosome: XCXc, XcY. Males have only one X, so one recessive allele is enough.
| carrier mother × normal father | XC | Y |
|---|---|---|
| XC | XCXC — girl, normal | XCY — boy, normal |
| Xc | XCXc — girl, carrier | XcY — boy, affected |
- 25% of the children are affected — but 50% of the sons are. Read whether the question asks about children, sons or daughters.
- Daughters need two copies to be affected, so an affected daughter must have an affected father and a carrier or affected mother.
- A father never passes his X to a son — sons get his Y — so sex-linked traits pass father → daughter → grandson.
Watch out: write the Y. Leaving it out turns XcY into a female genotype and the whole ratio changes.
Dihybrid, without a 16-box grid
The idea: two genes that assort independently are two separate monohybrid crosses. Multiply the probabilities instead of drawing 16 boxes.
For RrYy × RrYy, each gene on its own is Rr × Rr, which is 3/4 dominant and 1/4 recessive.
| Phenotype | Probability | Out of 16 |
|---|---|---|
| Round, yellow | 3/4 × 3/4 = 9/16 | 9 |
| Round, green | 3/4 × 1/4 = 3/16 | 3 |
| Wrinkled, yellow | 1/4 × 3/4 = 3/16 | 3 |
| Wrinkled, green | 1/4 × 1/4 = 1/16 | 1 |
That is the 9 : 3 : 3 : 1 ratio, in four lines instead of a 4 × 4 grid. If you do need gametes for the grid, take one allele from each gene: RrYy gives RY, Ry, rY, ry.
Worked: of 160 offspring, how many are wrinkled and green?
1/16 × 160 = 10 offspring. Expected numbers are always the fraction times the total — and they are expectations, not guarantees.
Turning ratios into answers
The idea: two rules cover every probability question here. "And" multiplies; "or" adds.
- Product rule ("and"): the chance a child of Aa × Aa is affected and a girl is 1/4 × 1/2 = 1/8.
- Sum rule ("or"): the chance a child of Aa × Aa is AA or aa is 1/4 + 1/4 = 1/2.
- Each child is independent. Three unaffected children do not make the fourth more likely to be affected — it is 1/4 every time.
- Ratios are not counts. A 3 : 1 ratio in four children can easily come out 4 : 0.
The carrier question
Two carriers (Aa × Aa) have an unaffected child. What is the chance it is a carrier?
The four boxes are AA, Aa, Aa, aa — but "unaffected" rules out aa, leaving three equally likely boxes. Two of them are Aa, so the answer is 2/3, not 1/2. Extra information changes the pool you are choosing from.