Building a Punnett square

Definition

Punnett square

A Punnett square is a grid that combines the alleles carried by the gametes of two parents to show every possible genotype of their offspring.

Definition

Monohybrid cross

A monohybrid cross is a genetic cross that follows the inheritance of a single gene from two parents to their offspring.

A Punnett square is a grid that combines the gametes of two parents so that every possible offspring genotype appears in the table.
The grid is built in the same six steps every time.
1. Write the genotype of each parent, for example $B b$ for the father and $B b$ for the mother.
2. Work out the gametes each parent can make, writing single alleles only, so a $B b$ parent makes $B$ and $b$ .
3. Draw a $\times 2$ grid, place one parent's gametes above the two columns and the other parent's gametes beside the two rows.
4. Fill each box by combining the allele at the head of its column with the allele at the start of its row.
5. Write the dominant allele first inside each box, so a box holding $b$ and $B$ is written $B b$ .
6. Read off the phenotype of each box, remembering that one dominant allele is enough to give the dominant characteristic.
Two gamete types from each parent give $\times 2 = 4$ boxes, and each box is one equally likely outcome of fertilisation.
A parent that is homozygous makes only one kind of gamete, so its allele is simply repeated along its side of the grid.

Hint

Put single alleles on the edges of the grid and never whole genotypes, because a gamete carries one allele of the gene and not both.

Common Mistake

Do not read the four boxes as four children. Each box is one possible outcome, and the same four outcomes apply afresh to every offspring.

Reading ratios from the grid

The genotype ratio counts the boxes by genotype, so $Bb \times Bb$ gives $bb1\,BB : 2\,Bb : 1\,bb$ .
The phenotype ratio counts the boxes by appearance, so the same cross gives $3 : 1$ dominant to recessive.
The cross $Bb \times bb$ gives $B b$ , $B b$ , $bb$ and $bb$ , which is $1 : 1$ by genotype and $1 : 1$ by phenotype.
The cross $BB \times bb$ fills all four boxes with $B b$ , so every offspring shows the dominant characteristic.
The cross $BB \times Bb$ gives $B B$ , $B B$ , $B b$ and $B b$ , which is $1 : 1$ by genotype but gives no recessive offspring at all.
A ratio is converted to a fraction using $parts\text{fraction} = \dfrac{\text{parts of interest}}{\text{total parts}}$ .
A ratio of $3 : 1$ therefore means $3$ parts out of $4$ , which is $34\frac{3}{4}$ , and $34×100=75%\frac{3}{4} \times 100 = 75\%$ .
A ratio of $1 : 1$ means $1$ part out of $2$ , which is $12\frac{1}{2}$ and therefore $50%50\%$ .
A ratio is simplified by dividing both parts by their highest common factor, so $27 : 9$ becomes $3 : 1$ .

Example

A brown eyed father of genotype $B b$ and a brown eyed mother of genotype $B b$ have children, where $B$ codes for brown eyes and $b$ codes for blue eyes.
Each parent makes $B$ gametes and $b$ gametes in equal numbers.
Filling the grid gives the offspring genotypes $B B$ , $B b$ , $B b$ and $bb$ .
Three of the four boxes hold at least one $B$ , so the phenotype ratio is $3$ brown to $1$ blue.
Written the other two ways, that is $34\frac{3}{4}$ brown and $14\frac{1}{4}$ blue, or $75%75\%$ brown and $25%25\%$ blue.

Common Mistake

Do not give a genotype ratio when the question asks for a phenotype ratio. The cross $Bb \times Bb$ is $1 : 2 : 1$ by genotype and $3 : 1$ by appearance, and only one of those answers fits each question.

Probability for one offspring

Each box of a completed grid is one equally likely outcome, so a single box carries a probability of $14\frac{1}{4}$ .
The probability of any genotype is the number of boxes showing it divided by the total number of boxes.
From $Bb \times Bb$ the probability of $bb$ is $14\frac{1}{4}$ , of $B b$ is $24=12\frac{2}{4} = \frac{1}{2}$ and of $B B$ is $14\frac{1}{4}$ .
The probability of the dominant phenotype from that cross is $34\frac{3}{4}$ , which is $0.75$ as a decimal and $75%75\%$ as a percentage.
Every probability lies between $0$ and $1$ , and the probabilities of all possible outcomes of a cross add up to $1$ .
Each fertilisation is an independent event, so the outcome of one offspring has no effect on the next.
Parents who already have three children showing the dominant characteristic still face a probability of $14\frac{1}{4}$ that the fourth shows the recessive one.

Note

A probability can be written as a fraction, a decimal or a percentage, and a question will usually name the form the answer should take.

Expected and observed numbers

The expected number of offspring of one type is found from $offspring\text{expected number} = \text{probability} \times \text{total number of offspring}$ .
From $Bb \times Bb$ with $40$ offspring, the expected number showing the recessive characteristic is $14×40=10\frac{1}{4} \times 40 = 10$ .
The expected number showing the dominant characteristic is $34×40=30\frac{3}{4} \times 40 = 30$ .
Observed numbers rarely match expected numbers exactly, because which gamete fuses with which is a matter of chance.
The more offspring are counted, the closer the observed proportions usually come to the expected ones.
Observed results close to the expected ratio support the genotypes proposed for the parents.
Observed results a long way from the expected ratio suggest that one of the parent genotypes was wrong, or that too few offspring were counted.

Practical

Modelling a monohybrid cross with beads

Aim: to model the cross $Bb \times Bb$ using beads drawn at random, and to compare the offspring ratio actually obtained with the $3 : 1$ ratio predicted by a Punnett square.
Apparatus: two opaque containers such as beakers covered with card, $50$ red beads, $50$ white beads, a marker pen, a tally chart on paper, a pencil and a calculator.
Method:
- Label one container as the male parent and the other as the female parent, since each container will supply one gamete for every offspring.
- Place $25$ red beads, standing for the allele $B$ , and $25$ white beads, standing for the allele $b$ , into each container.
- Check that both containers hold equal numbers of the two colours, because a heterozygous parent makes its two gamete types in equal numbers.
- Shake each container for several seconds so the beads are thoroughly mixed and no colour is left sitting on top.
- Look away from the containers and draw one bead from each with the other hand, so that neither draw is influenced by what can be seen.
- Record the pair as a genotype with the dominant allele written first, so one red bead and one white bead is recorded as $B b$ .
- Return each bead to the container it came from and shake again before the next draw, because taking beads out would change the proportion of alleles left behind.
- Repeat the draw until $100$ offspring have been recorded, tallying in groups of five so the totals can be counted quickly.
- Add up the tallies to give the observed numbers of $B B$ , $B b$ and $bb$ , then group $B B$ and $B b$ together as the dominant phenotype.
Results: a typical run gives roughly $25$ $B B$ , $50$ $B b$ and $25$ $bb$ , which is about $75$ offspring with the dominant phenotype and $25$ with the recessive one.
Maths: the expected numbers are $34×100=75\frac{3}{4} \times 100 = 75$ dominant and $14×100=25\frac{1}{4} \times 100 = 25$ recessive, and dividing the observed $75 : 25$ by $25$ simplifies it to $3 : 1$ .
Watch out: failing to replace the beads changes the allele proportions and drags the ratio away from $3 : 1$ , glancing into the container makes the draw non-random, and stopping after only $20$ draws can give a ratio as far out as $2 : 1$ purely by chance.
Safety: the hazard is low, but keep spilled beads off the floor because they are a slipping hazard, and never put beads near the mouth.

Self review

What goes along the two edges of a Punnett square, and why?
Give the genotype ratio and the phenotype ratio produced by the cross $Bb \times Bb$ .
What is the probability that a single offspring of the cross $Bb \times bb$ shows the recessive characteristic?
Calculate the expected number of recessive offspring when $60$ offspring come from the cross $Bb \times Bb$ .
Explain why observed offspring numbers do not always match the expected ratio.

Building a Punnett square

Definition

Punnett square

A Punnett square is a grid that combines the alleles carried by the gametes of two parents to show every possible genotype of their offspring.

Definition

Monohybrid cross

A monohybrid cross is a genetic cross that follows the inheritance of a single gene from two parents to their offspring.

A Punnett square is a grid that combines the gametes of two parents so that every possible offspring genotype appears in the table.
The grid is built in the same six steps every time.
1. Write the genotype of each parent, for example $B b$ for the father and $B b$ for the mother.
2. Work out the gametes each parent can make, writing single alleles only, so a $B b$ parent makes $B$ and $b$ .
3. Draw a $\times 2$ grid, place one parent's gametes above the two columns and the other parent's gametes beside the two rows.
4. Fill each box by combining the allele at the head of its column with the allele at the start of its row.
5. Write the dominant allele first inside each box, so a box holding $b$ and $B$ is written $B b$ .
6. Read off the phenotype of each box, remembering that one dominant allele is enough to give the dominant characteristic.
Two gamete types from each parent give $\times 2 = 4$ boxes, and each box is one equally likely outcome of fertilisation.
A parent that is homozygous makes only one kind of gamete, so its allele is simply repeated along its side of the grid.

Hint

Put single alleles on the edges of the grid and never whole genotypes, because a gamete carries one allele of the gene and not both.

Common Mistake

Do not read the four boxes as four children. Each box is one possible outcome, and the same four outcomes apply afresh to every offspring.

Reading ratios from the grid

The genotype ratio counts the boxes by genotype, so $Bb \times Bb$ gives $bb1\,BB : 2\,Bb : 1\,bb$ .
The phenotype ratio counts the boxes by appearance, so the same cross gives $3 : 1$ dominant to recessive.
The cross $Bb \times bb$ gives $B b$ , $B b$ , $bb$ and $bb$ , which is $1 : 1$ by genotype and $1 : 1$ by phenotype.
The cross $BB \times bb$ fills all four boxes with $B b$ , so every offspring shows the dominant characteristic.
The cross $BB \times Bb$ gives $B B$ , $B B$ , $B b$ and $B b$ , which is $1 : 1$ by genotype but gives no recessive offspring at all.
A ratio is converted to a fraction using $parts\text{fraction} = \dfrac{\text{parts of interest}}{\text{total parts}}$ .
A ratio of $3 : 1$ therefore means $3$ parts out of $4$ , which is $34\frac{3}{4}$ , and $34×100=75%\frac{3}{4} \times 100 = 75\%$ .
A ratio of $1 : 1$ means $1$ part out of $2$ , which is $12\frac{1}{2}$ and therefore $50%50\%$ .
A ratio is simplified by dividing both parts by their highest common factor, so $27 : 9$ becomes $3 : 1$ .

Example

A brown eyed father of genotype $B b$ and a brown eyed mother of genotype $B b$ have children, where $B$ codes for brown eyes and $b$ codes for blue eyes.
Each parent makes $B$ gametes and $b$ gametes in equal numbers.
Filling the grid gives the offspring genotypes $B B$ , $B b$ , $B b$ and $bb$ .
Three of the four boxes hold at least one $B$ , so the phenotype ratio is $3$ brown to $1$ blue.
Written the other two ways, that is $34\frac{3}{4}$ brown and $14\frac{1}{4}$ blue, or $75%75\%$ brown and $25%25\%$ blue.

Common Mistake

Probability for one offspring

Each box of a completed grid is one equally likely outcome, so a single box carries a probability of $14\frac{1}{4}$ .
The probability of any genotype is the number of boxes showing it divided by the total number of boxes.
From $Bb \times Bb$ the probability of $bb$ is $14\frac{1}{4}$ , of $B b$ is $24=12\frac{2}{4} = \frac{1}{2}$ and of $B B$ is $14\frac{1}{4}$ .
The probability of the dominant phenotype from that cross is $34\frac{3}{4}$ , which is $0.75$ as a decimal and $75%75\%$ as a percentage.
Every probability lies between $0$ and $1$ , and the probabilities of all possible outcomes of a cross add up to $1$ .
Each fertilisation is an independent event, so the outcome of one offspring has no effect on the next.
Parents who already have three children showing the dominant characteristic still face a probability of $14\frac{1}{4}$ that the fourth shows the recessive one.

Note

A probability can be written as a fraction, a decimal or a percentage, and a question will usually name the form the answer should take.

Expected and observed numbers

The expected number of offspring of one type is found from $offspring\text{expected number} = \text{probability} \times \text{total number of offspring}$ .
From $Bb \times Bb$ with $40$ offspring, the expected number showing the recessive characteristic is $14×40=10\frac{1}{4} \times 40 = 10$ .
The expected number showing the dominant characteristic is $34×40=30\frac{3}{4} \times 40 = 30$ .
Observed numbers rarely match expected numbers exactly, because which gamete fuses with which is a matter of chance.
The more offspring are counted, the closer the observed proportions usually come to the expected ones.
Observed results close to the expected ratio support the genotypes proposed for the parents.
Observed results a long way from the expected ratio suggest that one of the parent genotypes was wrong, or that too few offspring were counted.

Practical

Modelling a monohybrid cross with beads

Aim: to model the cross $Bb \times Bb$ using beads drawn at random, and to compare the offspring ratio actually obtained with the $3 : 1$ ratio predicted by a Punnett square.
Apparatus: two opaque containers such as beakers covered with card, $50$ red beads, $50$ white beads, a marker pen, a tally chart on paper, a pencil and a calculator.
Method:
- Label one container as the male parent and the other as the female parent, since each container will supply one gamete for every offspring.
- Place $25$ red beads, standing for the allele $B$ , and $25$ white beads, standing for the allele $b$ , into each container.
- Check that both containers hold equal numbers of the two colours, because a heterozygous parent makes its two gamete types in equal numbers.
- Shake each container for several seconds so the beads are thoroughly mixed and no colour is left sitting on top.
- Look away from the containers and draw one bead from each with the other hand, so that neither draw is influenced by what can be seen.
- Record the pair as a genotype with the dominant allele written first, so one red bead and one white bead is recorded as $B b$ .
- Return each bead to the container it came from and shake again before the next draw, because taking beads out would change the proportion of alleles left behind.
- Repeat the draw until $100$ offspring have been recorded, tallying in groups of five so the totals can be counted quickly.
- Add up the tallies to give the observed numbers of $B B$ , $B b$ and $bb$ , then group $B B$ and $B b$ together as the dominant phenotype.
Results: a typical run gives roughly $25$ $B B$ , $50$ $B b$ and $25$ $bb$ , which is about $75$ offspring with the dominant phenotype and $25$ with the recessive one.
Maths: the expected numbers are $34×100=75\frac{3}{4} \times 100 = 75$ dominant and $14×100=25\frac{1}{4} \times 100 = 25$ recessive, and dividing the observed $75 : 25$ by $25$ simplifies it to $3 : 1$ .
Watch out: failing to replace the beads changes the allele proportions and drags the ratio away from $3 : 1$ , glancing into the container makes the draw non-random, and stopping after only $20$ draws can give a ratio as far out as $2 : 1$ purely by chance.
Safety: the hazard is low, but keep spilled beads off the floor because they are a slipping hazard, and never put beads near the mouth.

Self review

What goes along the two edges of a Punnett square, and why?
Give the genotype ratio and the phenotype ratio produced by the cross $Bb \times Bb$ .
What is the probability that a single offspring of the cross $Bb \times bb$ shows the recessive characteristic?
Calculate the expected number of recessive offspring when $60$ offspring come from the cross $Bb \times Bb$ .
Explain why observed offspring numbers do not always match the expected ratio.

5.1.9 Predicting single gene crosses

Building a Punnett square

Reading ratios from the grid

Probability for one offspring

Expected and observed numbers

Modelling a monohybrid cross with beads

Review 5.1.9 Predicting single gene crosses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.9 Predicting single gene crosses

Building a Punnett square

Reading ratios from the grid

Probability for one offspring

Expected and observed numbers

Modelling a monohybrid cross with beads

Review 5.1.9 Predicting single gene crosses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.9 Predicting single gene crosses

Building a Punnett square

Reading ratios from the grid

Probability for one offspring

Expected and observed numbers

Modelling a monohybrid cross with beads

Review 5.1.9 Predicting single gene crosses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.9 Predicting single gene crosses Revision Guide

5.1.9 Predicting single gene crosses

Building a Punnett square

Reading ratios from the grid

Probability for one offspring

Expected and observed numbers

Modelling a monohybrid cross with beads

Review 5.1.9 Predicting single gene crosses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.9 Predicting single gene crosses Revision Guide