Biology · Introductory biology · Concept

Punnett squares and inheritance probability

A Punnett square predicts the chance of each offspring genotype from the alleles its parents carry. Each parent passes one of its two alleles, each with probability $\frac{1}{2}$ , so Aa × Aa gives $\frac{1}{4}$ AA, $\frac{1}{2}$ Aa and $\frac{1}{4}$ aa. Under complete dominance that is $3$ dominant to $1$ recessive in phenotype. The same rules, multiplied gene by gene, predict two-gene crosses and the chances in a family pedigree.

A concrete case

Mendel’s peas

Gregor Mendel crossed pea plants with round seeds and plants with wrinkled seeds. Every seed of the first generation was round. He grew those plants and let them fertilize themselves, and their $7, 324$ seeds came out $5, 474$ round and $1, 850$ wrinkled.

That is $2.96$ round seeds for every wrinkled one, close to $3$ to $1$ . The wrinkled form had not been lost in the first generation. It was hidden, and it came back in about a quarter of the seeds, $1, 850$ against the $1, 831$ that a quarter would be.

A ratio this close to $3$ to $1$ is what happens if each plant carries two copies of a factor for seed shape and passes one of them at random to each seed. A seed is round whenever it has at least one round copy. Those factors are now called alleles, and the square that follows them is a Punnett square.

\begin{array}{l} 5474 + 1850 = 7324 \\ \frac{5474}{1850} \approx 2.96 \\ \frac{7324}{4} = 1831 \end{array}

The rule

The model and its assumptions

A Punnett square follows allele copies from parents to offspring. It models one autosomal gene, on a chromosome other than X or Y, in a diploid organism. Each parent’s two alleles are equally likely to enter a gamete, and the two parents contribute independently. Not every trait follows this model.

Distinguish an allele, a genotype and a phenotype

A and a are two alleles of the same gene. AA, Aa and aa are genotypes, the allele pairs an offspring carries. AA and aa are homozygous, and Aa is heterozygous. A phenotype is the trait category the offspring shows, and the rule that turns genotype into phenotype must be stated separately.

What does dominant actually mean?

With complete dominance, a heterozygote has the same phenotype category as the homozygote carrying the dominant allele. Dominant does not mean more common, stronger, healthier or more likely to be inherited. Each allele from an Aa parent still has probability $\frac{1}{2}$ .

List the gametes before filling the square

An Aa parent contributes A or a, each with probability $\frac{1}{2}$ . Write one parent’s gametes along the top and the other’s down the side. Each cell combines one contribution from each parent.

P (A) = \frac{1}{2}, P (a) = \frac{1}{2}

Combine one gamete from each parent
Parent 1 ↓ / Parent 2 →	A $(\frac{1}{2})$	a $(\frac{1}{2})$
A $(\frac{1}{2})$	AA $(\frac{1}{4})$	Aa $(\frac{1}{4})$
a $(\frac{1}{2})$	Aa $(\frac{1}{4})$	aa $(\frac{1}{4})$

Multiply within a cell, then add equivalent outcomes

A particular pair of independent gametes has probability $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . Two cells give the heterozygous genotype, because A from the first parent with a from the second gives the same genotype as the other way around.

P (A a) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Apply the phenotype rule after finding genotypes

Under complete dominance, AA and Aa share the dominant phenotype, so their probabilities add to $\frac{3}{4}$ . The aa genotype has probability $\frac{1}{4}$ and shows the recessive phenotype. Incomplete dominance or codominance changes this grouping, not the genotype probabilities of the same cross.

P (dominant phenotype) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}

Two genes at once

Genes on different chromosomes assort independently, so a cross involving two genes can be solved one gene at a time and the probabilities multiplied. Two double heterozygotes, RrYy × RrYy, give their four phenotypes in a $9 : 3 : 3 : 1$ ratio.

\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}

Reading a pedigree

A pedigree charts a trait through a family. Two unaffected parents with an affected child point to a recessive allele. An affected daughter of an unaffected father rules out X-linked recessive inheritance. A probability for an unaffected relative counts only the outcomes that are still possible.

Why it works

Why the square works

The square is meiosis drawn as a table. In meiosis I the two homologs of each pair go to different cells, so a gamete gets exactly one of a parent’s two alleles. For an Aa parent, half the gametes carry A and half carry a. This is Mendel’s law of segregation.

Fertilization pairs one sperm with one egg, and which allele the sperm carries does not depend on which allele the egg carries. Independent events multiply, so each cell of the square has probability $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . Different cells are outcomes that cannot happen together, so their probabilities add.

Two genes on different chromosomes sit on different homologous pairs, and each pair lines up in meiosis I independently of the others. The allele a gamete gets for one gene says nothing about the other, which is why two-gene probabilities multiply.

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

The method

Solve a cross

Work the genotypes first, then the phenotypes, then any counts. The table runs the steps for Aa × Aa, as the first worked example does.

Solving Aa × Aa
Step	What to do	For Aa × Aa
1	Write each parent’s genotype and the inheritance rule.	Aa × Aa, complete dominance of A.
2	List each parent’s gametes with their probabilities.	A and a, $\frac{1}{2}$ each, from both parents.
3	Fill the square, multiplying the probabilities.	AA, Aa, aA and aa, $\frac{1}{4}$ each.
4	Add the cells that give the same genotype.	$\frac{1}{4}$ AA, $\frac{1}{2}$ Aa, $\frac{1}{4}$ aa.
5	Group the genotypes by phenotype.	$\frac{3}{4}$ dominant, $\frac{1}{4}$ recessive.
6	Multiply by the number of offspring for expected counts.	Among $100$ , about $75$ and $25$ .

Two genes, or a family

For two genes on different chromosomes, solve each gene as its own cross and multiply, as the second worked example does, rather than filling all $16$ cells. For a pedigree, decide dominant or recessive first, then autosomal or X-linked, then assign every genotype the family forces, as the third worked example does.

Where it breaks

The square assumes one gene with two alleles, complete dominance, and genes that assort independently. Each assumption fails for some traits.

Incomplete dominance is the case students most often misread. Red and white snapdragons give pink offspring, and pink crossed with pink gives red, pink and white in a $1 : 2 : 1$ ratio. The genotype probabilities are those of Aa × Aa, but every genotype now has its own phenotype. In codominance, as in the human AB blood type, both alleles show at once.

Genes close together on one chromosome are linked and are inherited together more often than the multiplication rule predicts. Genes on the X chromosome follow different rules in males, who have one X. Most traits, such as height, depend on many genes and on the environment, and no single square predicts them.

A probability is not a count. A $\frac{1}{4}$ chance of aa does not promise one aa child in every four. Each birth is a new draw, whatever earlier children were.

Check your answer

A genetics answer can be tested before you look anything up.

The probabilities add to $1$ . The genotype fractions of any cross must total $1$ , and so must the phenotype fractions. Aa × Aa gives $\frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1$ .

The offspring match the parents. A homozygous parent can pass only one allele, so a cross with aa can never give AA offspring. An answer with a genotype no parent could supply has a slip.

The counts fit the fractions. Expected counts are the fractions times the number of offspring. Real counts scatter around them, as Mendel’s $1, 850$ wrinkled seeds scatter around the expected $1, 831$ .

Each product and each sum of probabilities can go in a Math box as fractions, one per row, such as $3 / 4 \cdot 3 / 4 = 9 / 16$ or $320 \cdot 9 / 16 = 180$ . Chalk marks a row ✓ when the fractions agree and ✗ when they do not. Chalk does not check which genotypes belong in the square. Punnett square, in Biology, fills the square from the parents you enter.

How close is close to

3

1

If each of Mendel’s $7, 324$ seeds was wrinkled with probability $\frac{1}{4}$ , the number of wrinkled seeds would vary from one harvest to the next. Its expected value is $1, 831$ , and its standard deviation is the square root of $7, 324 \times 0.25 \times 0.75$ , about $37.06$ . The $1, 850$ he counted is $19$ above, about half a standard deviation, an ordinary scatter.

Counting also tells linked genes from independent ones. Two genes close together on a chromosome give far more of the parents’ combinations than $9 : 3 : 3 : 1$ predicts, and the fraction of offspring with new combinations measures how far apart the genes are. Genes $1$ centimorgan apart recombine in $1 %$ of gametes, and the first genetic maps were drawn this way, in fruit flies.

\begin{array}{l} \sqrt{7324 (0.25) (0.75)} \approx 37.06 \\ \frac{1850 - 1831}{37.06} \approx 0.51 \end{array}

Common mistakes

Treating dominant as more common or more beneficial. Dominance describes only how a heterozygote looks, and a dominant allele can be rare or harmful, as in Huntington’s disease.
Counting Aa and aA as different genotypes. They are two cells of the square with the same genotype, so their probabilities add to $\frac{1}{2}$ .
Applying a single-gene autosomal model to every trait. Linked genes, X-linked genes and traits shaped by many genes break the simple ratios, so state the model first.
Expecting exactly one aa child in every four. A probability of $\frac{1}{4}$ describes many births, and each birth is a new, independent draw.
Giving an unaffected sibling of an affected child a $\frac{1}{2}$ chance of being a carrier. Being unaffected rules out aa, which leaves three equally likely cells, two of them carriers, so the chance is $\frac{2}{3}$ .
Reading the $3 : 1$ phenotype ratio as a genotype ratio. Aa × Aa gives genotypes in a $1 : 2 : 1$ ratio, and only grouping AA with Aa makes $3 : 1$ .

Key terms

Allele: One version of a gene. A diploid organism usually carries two copies of each gene, one from each parent, and the two can be different alleles without looking different.
Genotype: The alleles an organism carries for the gene or genes being studied, such as Aa. It differs from the phenotype, the trait you can observe.
Phenotype: A trait you can observe or measure, such as flower color or blood type. It comes from genotype and environment, and one phenotype can come from different genotypes, such as AA and Aa.
Allele segregation: The separation of a gene’s two alleles into different gametes during meiosis. A heterozygote (Aa) passes A to half its gametes and a to the other half.
Independent assortment: Alleles of different genes go into gametes independently of each other, because each chromosome pair lines up on its own in meiosis. Genes close together on one chromosome usually don’t assort independently.
Dihybrid cross: A cross between two parents that are both heterozygous for two genes, such as RrYy × RrYy. With independent assortment and complete dominance, the offspring phenotypes come in a $9 : 3 : 3 : 1$ ratio.
Test cross: Crossing an individual with the dominant phenotype to a homozygous recessive one. The recessive parent passes on only recessive alleles, so the offspring’s phenotypes reveal the tested parent’s genotype.
Pedigree: A family-tree diagram showing who has a trait across generations. It narrows down how the trait is inherited, but it can’t reveal unknown genotypes without assumptions.
Genetic carrier: Someone with one copy of the allele for a recessive condition who does not have the condition, such as Aa. A carrier can pass the allele to their children.

Work through an example

Under complete dominance, what genotype and phenotype probabilities follow from two heterozygous parents?

Punnett square for an Aa × Aa cross →

Solve a dihybrid cross: $9 : 3 : 3 : 1$ phenotypes →

Read a pedigree to find the inheritance pattern →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 9 key terms Try the Punnett square Check the probabilities in Math Open worked example on a board Genetics reference

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