Deducing a bird’s genotype – a case study
I’d like to talk a bit more about deductive genotyping – working out the genes that a bird carries based on its appearance and progeny. As previously commented, birds very inconsiderately don’t come with a little label detailing their genes, and we are forced to do our own investigative work. Where the bird comes from a line that we have bred for several generations we often have a very good idea of their genes; where the bird is a recent acquisition things can be much less clear.
There was a recent, very interesting conversation regarding a bird acquired for colour breeding, which I’d like to use as a case study; an example of how we can go about doing some deductive genotyping. I’ve probably simplified the case a bit, but hopefully I have the salient points correct.
A breeder and BYP member had recently acquired what they believed to be a Splash hen, in the hopes of introducing Blue into their current black line. The hen had the appropriate phenotype, white with black spots. She was put to the breeder’s black rooster.
As we know, Splash is homozygous Blue. Here is the cross in Punnet Square form:
As you can see, the expected outcome is 100% Bl/bl+, so all blue offspring.
However, the breeder in question got quite a surprise when her initial small hatch of eggs all from this cross resulted in:
* two black birds
* two white birds with black spots
* one blue bird.
They were understandably confused.
If we look at this as a logic problem, we can say that, based on the Punnet Square above, “If this bird is Splash, then when mated with a Black rooster, all the offspring will be blue.” The cross having been performed, we then look at the resulting offspring. Based on the variable colouring of the chicks we reach the following conclusion: “Not all of the offspring are blue, ergo this bird is not
So if this hen is not Splash, what is she?
Here’s where we talk again about genotype and phenotype. The genotype is the genes that the bird carries, the phenotype is what they look like. Don’t be confused into thinking ‘but the bird has a Splash phenotype, so it must be Splash!’. The bird’s phenotype is ‘white with black spots’. Now we know of two genes that will produce this phenotype; the Blue gene when present in two doses, and the Dominant White gene. There are in fact three genotypes that will produce this phenotype. They are:
Bl/Bl Homozygous Blue (Splash)
I/I Homozygous Dominant White
I/i+ Heterozygous Dominant White.
Lets entertain the possibility that the hen’s phenotype is produced instead by Dominant White. If so she could have either one copy (heterozygous) or two copies (homozygous). Lets have a look at the relevant Punnet Squares.
100% I/i+ = 100% Dominant White.
So, “If this bird is Homozygous Dominant White, when crossed to a black bird, all the offspring will be white with black spots”.
50% I/i+ = 50% Dominant White
50% i+/i+ = 50% Black
Or “If this bird is Heterozygous Dominant White, when crossed to a black bird, half the offspring will be black and half will be white with black spots”.
Now lets look at the results of the cross again. The breeder reported that there were two black chicks, one blue chick and two white spotty chicks. For the bird to be Homozygous Dominant White all the chicks produced must be white, therefore the bird is not Homozygous Dominant White. However the Heterozygous cross produces half white and half black chicks, and looks like a very real possibility. So lets tentatively say that the bird in question is I/i+.
But what about the blue chick? One would assume that the breeder would have bought the bird in question from a line that contained blue. We also know that white can hide other colours underneath. However we know from the production of the black chicks and our first punnet square that the hen is definitely not Splash. But perhaps they could have one copy of Blue?
50% Bl/bl+ = 50% Blue
50% bl+/bl+ = 50% Black
As we can see here, a blue bird crossed to a black bird will give half and half blue and black chicks. So it’s not unreasonable to assume that the hen in question is hiding one dose of blue under the single dose of Dominant White.
Now, because colour genes interact, we really want to look at the probabilities of two genes together. This is where we need to go the next level up with the Punnet Square.
First we array the mum’s Dominant White genes, then we do her blue genes. This way we can predict each possible combination that she can pass on. So she could pass on an I or an i+, and she could pass on a Bl or a bl+. This means that she has four possible combinations that she can pass on, because these genes 'segregate' (sort into the egg cell) randomly. These combinations are: I and Bl, I and bl+, i+ and BL, i+ and bl+.
Then we add Dad and do the same. He can only pass on i+ and bl+.
Scarey huh? In health care this is usually where people wimp out and pay people like me to do this for them; I am grateful as it gives me a job. However in this case it’s actually not that hard. The Black rooster’s genotype is i+/i+, bl+/bl+ so he only has one combination of genes to pass on. Thus we can simplify the Punnet square like so:
Filled out it looks like this:
This gives us the following ratios:
25% I/i+, Bl/bl+ = 25% Dominant White (carrying blue)
25% I/i+, bl+/bl+ = 25 Dominant White
25% i+/i+, Bl/bl+ = 25% Blue
25% i+/i+, bl+/bl+ = 25% Black
So phenotypically we would see:
50% White with black spots
And this is pretty close to what the breeder got, bearing in mind that 5 chicks is a very small sample size. A couple more clutches should fill out the ratios more evenly. Assuming that the breeder continues to get chicks in approximately these ratios, the conundrum is solved; the hen has one dose of Dominant White and one dose of Blue, and her genotype is I/i+, Bl/bl+.
It still took several of us nutting this through over a few days to work it out.