Punnett squares are a tool designed by British geneticist Reginald Punnett to help predict the likelihood of an offspring having a particular genotype.
They are a very useful tool to understand how to use. We will start with basic Punnet squares, and later once we understand more about genetics, we will test more difficult genetics.
Even though new evidence suggests that Punnett Squares are not very accurate, or as correct as they need to be, they are still a good way of understanding how genetics works.
It is a good idea to use this article in conjunction with the Hair Colour article.
Starter Square
The easiest way to understand Punnett squares is to draw a 2×2 grid. We label the horizontal axis buck and the vertical axis doe.
We then add the genes. Here we have a doe with the gene Aa (agouti-self) and a buck with the gene Aa (agouti-self). Remember that dominant alleles are written in capital letters and recessive ones are in lower case.
Next we match the pairs together in each of the grid squares. The dominant allele always is written first. If there are only recessive alleles, it doesn’t matter which order they are written in.
In this instance, we have:
- AA is a full agouti rabbit (AA are both dominant so it will take over any banding on the hair).
- Aa is an agouti carrying self colouring (so the hair will have some agouti markings)
- aa will be a self coloured bunny (solid colour, no banding).
Base Colour Square
Let’s look at another Punnett square, this time for brown/black, which uses the B allele.
When combining more genes together, the Punnett square gets larger. When this occurs, this is called epistasis.
A single Punnet square (2×2) does not suffice for predicting all of the colours you will get in the litter. A dihybrid (4×4) cross square can help you do that.
If we had a buck that was carrying the Agouti self gene (Aa), and also carrying the black carrying chocolate gene (Bb), we would separate his alleles, and because we are working with 4 genes, we need 4 squares…
Because we now have 4 squares, we double the A allele. So, we put the A allele above squares 1 and 2, and the a allele above squares 3 and 4 for the buck.
We repeat the process for the doe.
To add the B alleles, we grab the first one and stick it next to the first A. Then we take the second one and put it next to the second A. Then repeat that process for the set of a‘s.
That way each set of alleles above each square has each alternative.
We repeat the process for the doe, too.
Next we write all of the pairings in the squares.
Remember that the dominant alleles appear first in the set.
According to this Punnett square, we have the chance to get the following colours in the litter of rabbits:
- Agouti banding, black base
- Agouti banding, black base
- Agouti-self banding, black base
- Agouti-self banding, black base
- Agouti banding, black base
- Agouti banding, chocolate base
- Agouti-self banding, black base
- Agouti-self banding, chocolate base
- Agouti-self banding, black base
- Agouti-self banding, black base
- Self banding, black base
- Self banding, black base
- Agouti-self banding, black base
- Agouti-self banding, chocolate base
- Self banding, black base
- Self banding, chocolate base
Getting more detail
As an example, imagine our buck and doe have the following gene sets.
To fully show this example, we will need a huge Punnett square of 32×32.
That means there are 1024 squares in the centre of the Punnett square, and that means there can be a maximum of 1024 different variations. Even though this is highly unlikely when only using 5 alleles, the more you include, the more chances of having a different result in each square. But remember, the more alleles you include, the bigger the Punnett square gets.
With this particular gene set, the 32×32 Punnett square would look like this. Each of the coloured cells (yellow, pink, green and blue) show that this particular buck and doe pair will produce four different coat colours.
There is an easier way to figure out the gene pairings.
Here we separate the genes out. The buck will have 4 sets because of his Aa and Ccchd alleles.
We put each variation together in a line.
The doe will only have one set, because all of her allele pairings are identical, which means she has aa, bb, cchlcchl, dd and EsEs.
Kit 1
We grab the bucks first row of alleles and put them at the top. Beneath it, we put the does.
We then match them up, and this will become Kit 1‘s genes.
Kit 1, should really be named Coat 1 because we are only talking about the colour and hair banding, rather than the kit itself.
- Aa – Agouti with self
- Bb – Black base
- Ccchl – C is dominant and will take over the cchl. Chinchilla colouring with lighter nose, tummy and tail.
- Dd – Dense blue colour
- EsE – Es is dominant, and gives steel colouring
Kit 1’s coat will be a blue steel light chinchilla with a dark steel grey coloured undercoat. It’s hair banding will start steel grey, have a blue middle and a black band at the top. The kit will start black and then turn silver after 3 months of age. It’s hair will have patches of black flecks, and its body will be a darker steel colour than its tummy and nose.
Kit 2
We repeat the process to get Kit 2‘s genes.
- Aa – Agouti with self
- Bb – Black base
- cchdcchl – black will be turned into silver
- Dd – Dense blue colour
- EsE – steel colour
Kit 2’s coat will be a steel blue colour with a slate grey base. The banding on the hair will be variations of grey and silver. The rabbit’s coat will start black and turn silver after 3 months of age.
Kit 3
And again for Kit 3‘s genes.
- aa – Self/solid colour
- Bb – Black base
- Ccchl – C will cause full colour over the body
- Dd – Dense blue colour
- EsE – steel colour
Kit 3’s coat will be a solid blue colour and it won’t have any banding on its hairs.
Kit 4
Finally, Kit 4‘s genes.
- aa – Self/solid colour
- Bb – Black base
- cchdcchl – black will be turned into silver
- Dd – Dense blue colour
- EsE – steel colour
Kit 4’s coat won’t have banding on the hairs. It will be a steel grey blue with lighter tummy and nose.
An overboard example
Here we have a buck and a doe. It looks pretty simple, but is it?
This is what the buck may look like:
This is what the doe may look like:
There are many more combinations in the buck’s genetics (left)compared to the doe’s combinations (above).
The doe would have more combinations if her AA, dd and ee alleles weren’t identically paired. So if she had, say she had Aa, that would add more variations.
I tried to keep it reasonably simple, because the buck is pretty complex by himself.
By taking the first line of the buck’s gene combinations and working through the doe’s 4 lines of combinations, we get these coat colours.
Full agouti banding, black base, chinchilla, dense colour, non extension.
Result: A chestnut rabbit with agouti banding and chinchilla patterning. It’s undercoat will be a dark slate grey.
Full agouti banding, black base, chinchilla, dense colour, non extension.
Result: An orange rabbit with agouti banding and chinchilla patterning. It’s undercoat will be slate grey.
Full agouti banding, black base, dark chinchilla, dense colour, non extension.
Result: A frosty black coat with chinchilla patterning. It’s undercoat will be slate grey.
Full agouti banding, black base, dark chinchilla, dense colour, non extension.
Result: A chinchilla coloured rabbit with chinchilla patterning.
Full agouti banding, black base, chinchilla, dense colour, non extension.
Result: The same coat as 1.2
Full agouti banding, chocolate base, chinchilla, dense colour, non extension.
Result: Chocolate chinchilla with full agouti banding, and a slate grey undercoat.
Full agouti banding, black base, dark chinchilla, dense colour, non extension.
Result: A frosty black with hair that has a browny sheen.
Full agouti banding, chocolate base, dark chinchilla, dense colour, non extension.
Result: A frosty chocolate chinchilla, with full agouti banding and a slate grey undercoat.
I won’t go through all of the combinations here because it will be too daunting.
According to the calculations, there are 48 unique allele combinations just by using that particular buck and doe.
Thinking about the 1024 squares:
- there are 16 allele combinations that appear 32 times each
- there are 32 allele combinations that appear 16 times each
Once you’ve calculated how many times they appear, you can calculate the percentage rate:
- if the combination appears 32 times: 32 / 1024 * 100 = 3.1%
- if the combination appears 16 times: 16 / 1024 * 100 = 1.6% (rounded up)
That means you have a 3.1% chance of getting one of the 16 allele combinations, and a 1.6% chance of getting one of the remaining 32 allele combinations.
If we compare this to the first buck (Aa BB Ccchd DD EE) and doe (aa bb cchlcchl dd EsEs) we studied much earlier, they only produce 4 unique combinations:
- Aa Bb Ccchd Dd EsE
- Aa Bb cchdcchl Dd EsE
- aa Bb Ccchd Dd EsE
- aa Bb cchdcchl Dd EsE
This means there is a 25% chance of the a kitten having one of these colours. The odds are much better in knowing the colours you will get from the breeding.
A little bit of fun
Let’s figure out Maple and Lunar’s genetics.
Lunar
Aa (agouti self),
BB (black base),
cchd_ (chinchilla dark),
dd (blue),
E_,
EnEn (broken colour),
vv (Blue eyes – Vienna)
Maple
Orange eyes (dilute brown),
Aa (agouti self),
BB (black is needed for harlequin colour)
C_
DD
eje
EnEn – can see some spotting (white/lighter areas on face and side)
vv (Vienna – white tip on nose, pale eyes)
Sibling 1
Orange eyes (dilute brown),
Aa (agouti self),
BB (black is needed for harlequin colour)
Ccchd
DD
eje
EnEn (black removed and changed to white)
vv (Vienna – white front colour)
Sibling 2
Orange eyes (dilute brown),
Aa (agouti self),
BB (solid colour)
C_
DD
eje
enen (no spotting)
vv (Vienna – white splash on nose)
Sibling 3
Aa (agouti self),
BB (solid colour)
C_
DD
E_
EnEn (spotting)
vv (Vienna – blue eyes)
Mum
Agouti self (Aa),
Black base (Bb),
Chinchilla (C_),
Dilute black (Blue) dd,
Full extension of colour (E_),
Broken Colour (EnEn),
Blue eyes (vv)
I couldn’t get hold of a photo of their father; however, the breeder said that he was brown with black stripes – so I guess he was a harlequin.
Because we have the doe’s alleles (I think), we can then figure out which ones the father had.
Keep in mind this is using my powers of deduction. Feel free to disagree.
The buck may have had: AA BB cchd_ DD eje enen