Genetics 102 - Using Simple Autosomal Punnett Squares
This page explains how to use a Punnett square to determine the offspring of any parents with autosomal mutations. For information about tracking sex-linked mutations, click here. For information about tracking multiple mutations, click here.
What are Punnett squares?
Punnett squares are notation devices used to show the possible distribution of parent genes to their offspring. They give a visual way to track how genes transfer from the parents to the offspring, to show the possible sets of genes the offspring will have.
How do genes get assigned in Punnett squares?
Each parent bird in an equation has two complete sets of chromosomes, which house all of the genes that make up the bird's genome. These genes each sit on their own locus within their set of chromosomes. Each parent donates only one complete set of chromosomes to their offspring, so we separate the pair of genes from each set into separate boxes to predict possible outcomes. One gene per box.
How do I know which locus a gene is on?
No one has completely mapped the peafowl genome for all the color and pattern mutations, so we don't know exactly which chromosomes they are on, or which loci they are on those chromosomes. This also means we often don't know which genes are alleles of other genes and which are not.
What DO we know?
We know that some genes (the sex-linked ones) sit on the Z chromosome. We also know that the genes for the white/pied patterns are alleles and that bronze and opal are NOT alleles. Lastly, we know that although Peach is purple + cameo, the genes cannot be separated in this one instance the way you could separate a het purple/het cameo's genes out to be just cameo or just purple. We'll worry about these later.
For peafowl, the wild type (blue) is haplosufficient, meaning will always show unless there is a dominant mutation or a pair of matching recessive mutations to replace it. The autosomal and sex-linked colors are all recessive mutations. White and white eye are incomplete dominant mutations (expresses in the phenotype in het and homo forms, but differently), and pied (the gene) is a dominant mutation (expresses in the phenotype in het and homo forms the same).
What are Punnett squares?
Punnett squares are notation devices used to show the possible distribution of parent genes to their offspring. They give a visual way to track how genes transfer from the parents to the offspring, to show the possible sets of genes the offspring will have.
How do genes get assigned in Punnett squares?
Each parent bird in an equation has two complete sets of chromosomes, which house all of the genes that make up the bird's genome. These genes each sit on their own locus within their set of chromosomes. Each parent donates only one complete set of chromosomes to their offspring, so we separate the pair of genes from each set into separate boxes to predict possible outcomes. One gene per box.
How do I know which locus a gene is on?
No one has completely mapped the peafowl genome for all the color and pattern mutations, so we don't know exactly which chromosomes they are on, or which loci they are on those chromosomes. This also means we often don't know which genes are alleles of other genes and which are not.
What DO we know?
We know that some genes (the sex-linked ones) sit on the Z chromosome. We also know that the genes for the white/pied patterns are alleles and that bronze and opal are NOT alleles. Lastly, we know that although Peach is purple + cameo, the genes cannot be separated in this one instance the way you could separate a het purple/het cameo's genes out to be just cameo or just purple. We'll worry about these later.
For peafowl, the wild type (blue) is haplosufficient, meaning will always show unless there is a dominant mutation or a pair of matching recessive mutations to replace it. The autosomal and sex-linked colors are all recessive mutations. White and white eye are incomplete dominant mutations (expresses in the phenotype in het and homo forms, but differently), and pied (the gene) is a dominant mutation (expresses in the phenotype in het and homo forms the same).
So let's have a look at what a basic, single-gene Punnett square looks like, with labels.
Let's start with the basic, wild type (WT) notation. The "WT" is the text used to represent one wild type gene on one chromosome, and since chromosome comes in pairs, we have to note both genes (WT/WT). Let's look at a bird with no mutations (WT/WT), bred to a bird with no mutations (WT/WT).
Since both sets of parent chromosomes are wild type (WT), and they can only donate one at a time, we will put "WT" in each of the parent cells along the top and bottom (bolded). In other words, if the parent genotype is WT/WT, we'd split it into WT and WT and put each in its own cell to represent how they can only donate one gene or the other, but not both.
For now, I've color coded the parent genotypes to demonstrate how splitting them up works.
- In genetic notation a forward slash "/" is used to separate genes on the same loci, but different chromosomes, but it's not necessary to show. I've left it out of most graphics, to save a little space for mobile viewing.
Since both sets of parent chromosomes are wild type (WT), and they can only donate one at a time, we will put "WT" in each of the parent cells along the top and bottom (bolded). In other words, if the parent genotype is WT/WT, we'd split it into WT and WT and put each in its own cell to represent how they can only donate one gene or the other, but not both.
For now, I've color coded the parent genotypes to demonstrate how splitting them up works.
To move these genes to their offspring, you take the gene notation from the cock row and add it to the notation from the hen column where they meet. I've added colors to the offspring cells to indicate their phenotype (what shows visually). The letters in the cells indicate the genotype (what genes they have). In this instance, the genotype (wild type blue) matches the phenotype (wild type blue).
Easy!
So now you have four results. Since all of them match, you know that 100% of the offspring will be the same, and since the genotype is "WTWT' (two copies of wild type genes), you know they will all be wild type birds.
But what happens when you introduce a color to this equation? Let's say that the hen is bronze (brbr) and the cock is wild type (WTWT) still. The cock has 2 sets of WT chromosomes, and the hen has 2 sets of chromosomes with the bronze mutation instead. Mutations are always alleles of the wild type genes, so they will always replace the wild type gene in that set of chromosomes.
Split the genes into their separate boxes as before.
So now you have four results. Since all of them match, you know that 100% of the offspring will be the same, and since the genotype is "WTWT' (two copies of wild type genes), you know they will all be wild type birds.
But what happens when you introduce a color to this equation? Let's say that the hen is bronze (brbr) and the cock is wild type (WTWT) still. The cock has 2 sets of WT chromosomes, and the hen has 2 sets of chromosomes with the bronze mutation instead. Mutations are always alleles of the wild type genes, so they will always replace the wild type gene in that set of chromosomes.
Split the genes into their separate boxes as before.
When you combine these genes for the offspring, you will again note the gene in the cock's row (in this case, WT) and the gene from the hen's column (in this case br) where they meet.
In this case the offspring phenotype (wild type blue) will not match the genotype (heterozygous bronze).
In this case the offspring phenotype (wild type blue) will not match the genotype (heterozygous bronze).
Now you have four results again. Since all of them match, you know that 100% of the offspring will be the same, and since the genotype is "WT/br" (one copy of the wild type gene, one copy of the bronze mutation), and bronze is a recessive autosomal mutation, you know the birds will have the wild type blue phenotype, but will be heterozygous bronze.
A lot of us deal with wild type parent birds that are only heterozygous for a gene. This means the parent bird can donate either a wild type gene or a mutated gene, but not both to one offspring. To show this, we only put the mutation in one of the boxes. Let's say the hen is heterozygous bronze (WT/br) and the cock is wild type (WT/WT).
Here is how the genes split into the Punnett square.
Here is how the genes split into the Punnett square.
To fill in the offspring sections, it's the same as before, taking the gene from the cock's row and combining it with the gene from the hen's column where they meet. Only one (the hen's second gene slot) will have the br gene notation, as only one half of the hen's chromosome pair has the gene.
Here is the filled out Punnett square.
Here is the filled out Punnett square.
You have four results again, but this time, half of them are different. Half of the offspring will be "WT/WT" which means full wild type birds. The other half will be wild type phenotype but will be heterozygous bronze. In this instance even though all birds look full wild type blue (WT/WT), only half the genotypes match the phenotypes. All WT/br will look like blues, because the wild type is haplosufficient, and the br is an autosomal recessive gene. There is no way to tell which babies are which genotype, because visually they all look the same.
In more colloquial terms, if someone asked "What do you get when you breed a blue to a split bronze" you would see the response "50% blue, 50% split bronze," the above image is visual representation of why and how they know that.
In more colloquial terms, if someone asked "What do you get when you breed a blue to a split bronze" you would see the response "50% blue, 50% split bronze," the above image is visual representation of why and how they know that.
On the next Punnett square, I've made both parents heterozygous bronze. This means both parents (as always) have two sets of chromosomes but only one set per parent has the mutation, so only one box from each of their boxes will contain the br notation.
To populate the offspring cells, we still take the gene from the cock's row and combine it with the gene from the hen's column where they meet.
Once again, you have 4 results, and this time there are 3 possibilities- wild type blue, heterozygous bronze (both highlighted in blue), and homozygous bronze (highlighted in bronze). The homozygous bronze will appear visually bronze, since there's no wild type gene left.
Since there are 4 results, each result represents 25% of the possible outcomes. So, 25% of the offspring will be wild type because wild type only has one box. 50% of the offspring will be heterozygous bronze, because het bronze has 2 boxes. 25% of the offspring will be bronze, because bronze only has one box.
Since there are 4 results, each result represents 25% of the possible outcomes. So, 25% of the offspring will be wild type because wild type only has one box. 50% of the offspring will be heterozygous bronze, because het bronze has 2 boxes. 25% of the offspring will be bronze, because bronze only has one box.
This same method will work no matter what single autosomal gene you want to track, be it color or pattern.
If you have questions about how to use simple, one-gene autosomal Punnett squares, please feel free to find me on the Peafowl Discord Server! This is a live chat app, and the server is like a live forum where you can post and ask questions and get answers.
Sex-linked and multi-gene Punnett squares are very similar, but slightly more complicated.
If you're ready to move on to learning about sex linked Punnett squares, click here!
If you have questions about how to use simple, one-gene autosomal Punnett squares, please feel free to find me on the Peafowl Discord Server! This is a live chat app, and the server is like a live forum where you can post and ask questions and get answers.
Sex-linked and multi-gene Punnett squares are very similar, but slightly more complicated.
If you're ready to move on to learning about sex linked Punnett squares, click here!