Punnett squares, dihybrid crosses and Mendelian ratios, worked and checked

A Punnett square is a table that lists one parent's possible gametes along the top and the other's down the side, and fills each box with the offspring genotype those two gametes make. Every box is equally likely, so the ratio of boxes is the predicted ratio of offspring. For one gene with two heterozygous parents that gives 1:2:1 by genotype and 3:1 by phenotype; for two genes, 9:3:3:1. Below, the method, a dihybrid cross worked box by box and checked with a one-line probability, and the two ratios students mix up.

Mendelian genetics: the vocabulary in six lines

• Allele: one version of a gene. Written as a letter: A or a.
• Dominant and recessive: the dominant allele is the one whose trait shows when both are present; the recessive one shows only when it is the only kind there. Dominant describes what shows, not what is common.
• Homozygous and heterozygous: two of the same allele (AA or aa) or one of each (Aa).
• Genotype and phenotype: the alleles an individual carries, and what it looks like.
• P, F₁, F₂: the parent generation, their offspring, and the next.
• Mendel's two laws: segregation, each gamete gets one allele of each gene; and independent assortment, the alleles of two genes on different chromosomes sort into gametes independently of each other. The first is why a heterozygote makes two kinds of gamete; the second is what makes the 16-box square valid.

Both laws are facts about meiosis, the division that makes the gametes, which is why that page is the prerequisite for this one.

How to do a Punnett square: the one rule

Five steps, and one rule under them.

1. Write both parents' genotypes.
2. List each parent's possible gametes: one allele of each gene per gamete. A heterozygote Aa gives A and a; a homozygote AA gives only A.
3. Put one parent's gametes across the top and the other's down the side.
4. Fill each box by combining the row's gamete with the column's.
5. Count the boxes by genotype, then group them by phenotype.

The rule: every box is equally likely. Each gamete is made with equal probability, and which sperm meets which egg is random, so each box is one equally likely outcome. That is why counting boxes gives a probability, and it is also why a square can be checked by multiplying probabilities instead of drawing it, which is the trick in the dihybrid section.

Worked warm-up: the monohybrid cross Aa × Aa (genotype vs phenotype ratio)

Both parents heterozygous for one gene.

	A	a
A	AA	Aa
a	aA	aa

Four boxes. By genotype: 1 AA, 2 Aa (the box written aA is the same genotype as Aa; the order of the letters means nothing), 1 aa. The genotype ratio is 1 AA : 2 Aa : 1 aa.

By phenotype, with A dominant: AA and Aa both show the dominant trait, aa shows the recessive one. 3 dominant : 1 recessive.

Check by probability: P(aa) = P(a from one parent) × P(a from the other) = ½ × ½ = ¼, which is one box in four.

The sentence that stops the first swap: a genotype ratio counts the letter combinations; a phenotype ratio counts what shows; and a dominant phenotype hides two genotypes. Students who count aA and Aa as different write 1:1:1:1, which is the common wrong line here, and it is wrong because the two heterozygotes are one genotype.

Worked example: the dihybrid cross RrYy × RrYy (the 9:3:3:1 ratio)

Two genes, each parent heterozygous for both. Our letters: R is round seed and r is wrinkled; Y is yellow seed and y is green; round and yellow are dominant. These are the pea traits Mendel used, named as traits only.

Line 1, the gametes. One allele of each gene per gamete, found by pairing each R-allele with each Y-allele: RY, Ry, rY, ry. Four kinds, each with probability ¼. Both parents make the same four.

Line 2, the square. Four gametes across, four down, 16 boxes.

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Line 3, the genotypes, counted. 1 RRYY, 2 RRYy, 2 RrYY, 4 RrYy, 1 RRyy, 2 Rryy, 1 rrYY, 2 rrYy, 1 rryy. Add them: 1 + 2 + 2 + 4 + 1 + 2 + 1 + 2 + 1 = 16. Nine genotype classes, sixteen boxes.

Line 4, the phenotypes. Round yellow (R_ Y_, at least one R and one Y): 1 + 2 + 2 + 4 = 9. Round green (R_ yy): 1 + 2 = 3. Wrinkled yellow (rr Y_): 1 + 2 = 3. Wrinkled green (rryy): 1. Check: 9 + 3 + 3 + 1 = 16. So 9:3:3:1.

Line 5, the check without the square, by the product rule. The two genes are independent, so each phenotype's probability is the product of two one-gene probabilities. From the monohybrid warm-up, P(round) = ¾ and P(wrinkled) = ¼; the same for yellow and green.

• P(round and yellow) = ¾ × ¾ = 9/16.
• P(round and green) = ¾ × ¼ = 3/16.
• P(wrinkled and yellow) = ¼ × ¾ = 3/16.
• P(wrinkled and green) = ¼ × ¼ = 1/16.

The square and the product rule must agree, and they do. This is how you check a square on a test in ten seconds: compute one phenotype's fraction by multiplying and see that the boxes match. A miscounted box shows up at once.

Line 6, expected counts for a real-sized sample. If this cross gave 320 offspring, the predicted counts are 320 × 9/16 = 180 round yellow, 320 × 3/16 = 60 round green, 60 wrinkled yellow, and 320 × 1/16 = 20 wrinkled green. Check: 180 + 60 + 60 + 20 = 320. Real counts scatter around these; "predicted" is the word to use, and whether a real sample's scatter is more than chance is a question for probability and expected counts in another course's toolbox.

The test cross, and reading a square backwards

A test cross answers one question: is this dominant-phenotype individual AA or Aa? Cross it with a homozygous recessive, aa. If it is AA, every offspring is Aa and shows the dominant trait. If it is Aa, about half the offspring are aa and show the recessive trait. One recessive offspring proves the parent was a heterozygote.

The dihybrid version: RrYy × rryy. The heterozygote's gametes are RY, Ry, rY, ry; the homozygote's are all ry. Four boxes: RrYy, Rryy, rrYy, rryy. Four phenotypes, one each: 1:1:1:1.

Now read ratios backwards, which is what a test often asks:

• 3:1 means both parents were heterozygous for one gene.
• 1:1 means a heterozygote was crossed with a homozygous recessive.
• 9:3:3:1 means both parents were dihybrid heterozygotes.
• 1:1:1:1 means a dihybrid heterozygote was crossed with a double homozygous recessive.

The two things students swap

Genotype ratio and phenotype ratio. 1:2:1 against 3:1 for one gene; nine genotype classes against four phenotypes for two. Read the question for the word it uses, and when it says "ratio" without the word, give both and label them.

Dominant and common. A dominant allele shows in a heterozygote. That is all the word means. It says nothing about how frequent the allele is in a population; some dominant alleles are rare.

A third, if your course includes X-linked crosses: writing the alleles on the X and then forgetting that a male's single X is his whole genotype for that gene. He has no second copy to hide a recessive allele behind.

Beyond Mendel, in one paragraph each

Incomplete dominance. The heterozygote is in between the two homozygotes (a pink flower from red and white parents), so the phenotype ratio is 1:2:1, the same as the genotype ratio.

Codominance. Both alleles show at once in the heterozygote, each fully, rather than blending.

Multiple alleles. A gene may have more than two alleles in the population, though each individual still carries two. The ABO blood groups are the usual example, with codominance between two of the alleles and simple dominance over the third.

Sex linkage. A gene on the X chromosome. Females have two copies and males one, so a recessive X-linked trait shows in males far more often.

Each of these is the same square with a different reading rule. Knowing the names tells you which rule to apply; the method does not change.

Check: a cross to set up cold

With the page closed: set up RrYy × rryy and predict the ratio before you draw (then draw and count); then compute P(wrinkled green) from a RrYy × RrYy cross by the product rule and find the box in the 16-box square that matches it. Open the page and mark it. When you practise more of these, mix the kinds, a monohybrid, a test cross and a dihybrid in one sitting rather than ten of one kind: Rohrer and Taylor (2007) found that shuffling problem types produced better test performance than blocking them, because the student has to decide which method applies. Redo it cold, then a sibling is the routine, and how to study biology has the week's plan around it.

Answer. RrYy × rryy: 1:1:1:1 (RrYy, Rryy, rrYy, rryy), four boxes, four phenotypes. P(wrinkled green) = ¼ × ¼ = 1/16, and the one box is rryy, the bottom-right corner of the 16-box square where the ry gamete meets the ry gamete.

Where this sits in the course

This is idea 21 of the 27 in the SBI4U course map, Patterns of inheritance, the last idea of strand D. Before it comes idea 20, Biotechnology and genetic engineering, and before that the DNA ideas, where the gene behind the allele is traced from DNA to protein. After it, idea 22, Homeostasis and negative feedback, opens strand E. Meiosis, the division that makes one allele per gamete true, is the Grade 11 prerequisite the course assumes.

Punnett squares on one screen

• Five steps: parents' genotypes; each parent's gametes (one allele per gene); one set across, one down; fill each box; count by genotype, then group by phenotype.
• Monohybrid Aa × Aa: 1 AA : 2 Aa : 1 aa by genotype; 3 : 1 by phenotype; P(aa) = ½ × ½ = ¼.
• Gametes for a dihybrid: pair each allele of the first gene with each of the second: RY, Ry, rY, ry.
• Dihybrid RrYy × RrYy: 16 boxes, nine genotype classes, phenotypes 9:3:3:1; check by ¾ × ¾, ¾ × ¼, ¼ × ¾, ¼ × ¼.
• Backwards: 3:1 both heterozygous; 1:1 a test cross; 9:3:3:1 both dihybrid heterozygotes; 1:1:1:1 a dihybrid test cross.
• The two swaps: genotype ratio against phenotype ratio; dominant against common.