Biology · Introductory biology · Worked example

Punnett square for an Aa × Aa cross

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

State the cross and the inheritance rule

Both parents are heterozygous, Aa. The model is one autosomal gene with complete dominance of A over a, ordinary segregation and independent gamete contributions.

A a \times A a

List each parent’s possible contributions

Each parent makes A and a gametes with equal probability. Each parent passes one allele to an offspring, not its whole Aa genotype.

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

Combine the gametes

Fill each cell by taking the allele from its row and its column. The four equally likely pairings are AA, Aa, aA and aa. A genotype summary writes both Aa and aA as Aa.

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 $(\frac{1}{4})$

Calculate the genotype probabilities

Each cell has probability $\frac{1}{4}$ . AA fills one cell, Aa fills two and aa fills one. As a check, the probabilities must total $1$ .

\begin{aligned} P (A A) & = \frac{1}{4} \\ P (A a) & = \frac{1}{2} \\ P (a a) & = \frac{1}{4} \\ \frac{1}{4} + \frac{1}{2} + \frac{1}{4} & = 1 \end{aligned}

Group genotypes by phenotype

Complete dominance groups AA and Aa together. The dominant phenotype therefore has probability $\frac{3}{4}$ , and the recessive phenotype $\frac{1}{4}$ . The familiar $3 : 1$ ratio refers to phenotypes, while the genotype ratio is $1 : 2 : 1$ .

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

Explain what the prediction does and does not say

For $100$ offspring, the expected counts are $25$ AA, $50$ Aa and $25$ aa. Those are expectations, not a guarantee. Punnett square, in Biology, keeps the parents’ alleles and the inheritance rule visible, so you can see which assumption a change affects.

E (count) = N P (outcome)

Result

Genotypes are $\frac{1}{4}$ AA, $\frac{1}{2}$ Aa and $\frac{1}{4}$ aa. Phenotypes are $\frac{3}{4}$ dominant and $\frac{1}{4}$ recessive.

Your turn

1. One like the example

Under the same complete-dominance model, cross Aa with aa. Find the genotype and phenotype probabilities.

Show the answer and explanation

The genotypes are $\frac{1}{2}$ Aa and $\frac{1}{2}$ aa. The phenotypes are $\frac{1}{2}$ dominant and $\frac{1}{2}$ recessive.

The Aa parent passes A or a with probability $\frac{1}{2}$ each, and the aa parent always passes a. $P (Aa) = \frac{1}{2} \cdot 1 = \frac{1}{2}$ , and the two genotypes add to $1$ . A cross with a homozygous recessive parent shows the other parent’s gametes directly.

\begin{array}{l} \frac{1}{2} \cdot 1 = \frac{1}{2} \\ \frac{1}{2} + \frac{1}{2} = 1 \end{array}

2. A variation

In snapdragons, flower color shows incomplete dominance. RR is red, Rr is pink and rr is white. Two pink plants are crossed. What fraction of the offspring is pink, and how many pink plants do you expect among $200$ ?

Show the answer and explanation

$\frac{1}{2}$ are pink, about $100$ of $200$ . The rest are $\frac{1}{4}$ red and $\frac{1}{4}$ white.

Rr × Rr gives the same genotypes as Aa × Aa, $\frac{1}{4}$ RR, $\frac{1}{2}$ Rr and $\frac{1}{4}$ rr. Under incomplete dominance each genotype has its own color, so the two Rr cells, $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ , are pink, and $200 \cdot \frac{1}{2} = 100$ .

\begin{array}{l} \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ 200 \cdot \frac{1}{2} = 100 \end{array}

3. Somewhere new

Mendel’s self-fertilized hybrids for seed color gave $8, 023$ seeds, $6, 022$ yellow and $2, 001$ green. Yellow is dominant. What is the observed ratio, and how many of each color does a $3 : 1$ ratio predict?

Show the answer and explanation

About $3.01$ to $1$ . A $3 : 1$ ratio predicts $6, 017.25$ yellow and $2, 005.75$ green.

$6022 / 2001 \approx 3.01$ . The expected counts are $8023 \cdot \frac{3}{4} = 6017.25$ and $8023 \cdot \frac{1}{4} = 2005.75$ . The observed counts differ from them by under $5$ seeds, well within ordinary scatter.

\begin{array}{l} \frac{6022}{2001} \approx 3.01 \\ 8023 \cdot \frac{3}{4} = 6017.25 \\ 8023 \cdot \frac{1}{4} = 2005.75 \end{array}

Keep exploring

Punnett square opens in Biology with two Aa parents and shows $\frac{1}{4}$ AA, $\frac{1}{2}$ Aa and $\frac{1}{4}$ aa. Change one parent to aa to see the practice cross.

Return to the concept →

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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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State the cross and the inheritance rule

List each parent’s possible contributions

Combine the gametes

Calculate the genotype probabilities

Group genotypes by phenotype

Explain what the prediction does and does not say

Result

Your turn

1. One like the example

The genotypes are 1221​ Aa and 1221​ aa. The phenotypes are 1221​ dominant and 1221​ recessive.

2. A variation

1221​ are pink, about 100100 of 200200. The rest are 1441​ red and 1441​ white.

3. Somewhere new

About 3.013.01 to 11. A 3:13:1 ratio predicts 6,017.256,017.25 yellow and 2,005.752,005.75 green.

Keep exploring

Study it, then try it

The genotypes are $\frac{1}{2}$ Aa and $\frac{1}{2}$ aa. The phenotypes are $\frac{1}{2}$ dominant and $\frac{1}{2}$ recessive.

$\frac{1}{2}$ are pink, about $100$ of $200$ . The rest are $\frac{1}{4}$ red and $\frac{1}{4}$ white.

About $3.01$ to $1$ . A $3 : 1$ ratio predicts $6, 017.25$ yellow and $2, 005.75$ green.