Peafowl Genetics for Dummies (in other words us)

[Scott Sr]

I looked at your punnett square but did not see white-eyed on it.
Where can I get breeding info on the effects of white-eyed crosses?

 

[Scott Sr]

Conserning white eyed....
single factor w/e x i/b split to white = ?
single factor w/e x black shoulder = ?
single factor w/e x white =?
HELP

 

[Scott Sr]

What if the whie eyed was a double factor in the examples above??

 

[Rosa moschata]

Here's how to figure out the answer yourself:

First, animals have a pair of each chromosome (except for the sex chromosomes). Genes are located on the chromosomes, and since there are two of each (except for the sex chromosomes), that means there are potentially two of each gene. Mutations are changes in genes, and so if a gene has a mutation, then there are two (or more) different versions of that gene. If an animal has two copies of the same version of a given gene, it is said to be homozygous (homo = same) for that gene. If an animal has one copy each of two different versions of the same given gene, it is said to be heterozygous (hetero = different) for that gene.

OK, once you've gotten that down, here's the next bit:

When an animal reproduces sexually, it does so first by making cells which have just one of each chromosome. Which chromosome of each pair end up in the mix is random, and each of the pair has a 50% chance of being in a particular egg or sperm.

Now, getting back to genes and mutations, and homozygous and heterozygous...

If a bird is "split to" or "single factor" for a particular mutation, that means it has only one copy of that particular version of that gene, and the other chromosome in the pair has another version. So a "single factor White Eye" has one chromosome with that mutation, while its "pair-mate" has the "normal" non-mutated version of that gene.

What does that mean with regards to offspring? Well, remember that which chromosome of each pair end up in a particular sperm or egg is random, and each has an equal chance. So a peacock with one copy of WE will thus pass on the gene via 50% of his sperm. The other 50% will have the "normal" or non-WE version of the gene, and so the WE mutation will NOT be passed on through them.

However, when you have a bird which is homozygous for a particular mutation, that means that BOTH chromosomes in that pair have the mutation. Therefore, when producing egg or sperm cells, it's 100% that the mutation will be passed on.

OK, now here comes the math part...

Unless mutations are known to be "linked" (in other words, they are found on the same chromosome, and thus "travel together" most of the time, separating only during crossover), then we say they are inherited independently. Remember that the odds of a particular chromosome in a pair being passed on is 50%. So if you have a bird that is "split to" two different mutations, the odds of BOTH being passed on in the SAME egg or sperm cell would be 25%. How'd I get that? If you remember from Probability in Math, when trying to determine the odds of two things occurring together, you multiply the odds of each occurring independently. So 50% X 50% = 25%.


OK, getting back to your crosses question....

Here's the first one:
Single Factor WE X IB split to White

OK, so there are two mutations discussed, and each parent is split to just ONE of them. Remembering back to what I said about being split, this means that 50% of the offspring will get the mutation, and 50% won't. This means that for the Single Factor WE, 50% of his/her offspring will inherit one copy of WE, while the other 50% won't. Same with the IB split to White -- 50% of his/her offspring will inherit one copy of White, while the other 50% won't. So what are the odds of getting a Single Factor WE split to White? That's 50% X 50% = 25% Single Factor WE split to White.

You will have to remember that individuals get one of each chromosome from Mom, and the other from Dad (except for sex chromosomes). So for a pea to be homozygous (i.e. have two copies) of ANYTHING, that means that BOTH parents must EACH have at least one copy of the given mutation. And so you won't get ANY Double Factor WE from ANY cross where only one parent has WE. And the same goes for Blackshoulder -- for offspring to be anything other than splits, you need to have the mutation in both parents (either visual or split).

So, going back to your other crosses...

Single Factor WE X IB Blackshoulder = 50% Single Factor WE split to Blackshoulder, and 50% IB split to Blackshoulder

Single Factor WE X White = 50% Single Factor WE split to White, and 50% IB split to White



What if the parent was Double Factor WE instead of Single Factor? Well, you'd get the same results, except that ALL offspring will be Single Factor WE. Since a Double Factor WE has the mutation on both chromosomes of that pair, then ALL of its eggs/sperm will pass on the mutation -- there is no alternative "normal" version in it. So from a breeding perspective, you won't get Double Factor offspring from those crosses, but you will have ALL (rather than half) of the offspring being Single Factor WE.


Hope you were able to follow along.

 

[Scott Sr]

The fog is lifting. Thanks!