Recessive Mutations
The following mutations are all "autosomal recessive". That means that they aren't attached to an X chromosome, but they yield to the
dominant normal (which is the normal grey):
- Pied
- Whiteface
- Fallow
- Silver
- Pastelface
You can figure out recessive mutations very similarly to the way you figure out sex-linked mutations.
Here's an example of two normal grey birds. Both are split to whiteface. We write this as wN, the w=whiteface, the N=normal.
You can also think of N as meaning nothing. Since it is the dominant trait, if a bird carries it, it will override the recessive. Generally,
we use lowercase letters to represent recessive genes, and capital letters to represent dominant genes.
Now, we can cross the wN the same way as the sex-linked: using the FOIL method.
| X X wN |
X Y wN |
| ww |
wN |
| wN |
NN |
Notice how we calculated it *slightly* different. Autosomal recessives should be calculated by
themselves before you do the regular Punnett Square. Since they are not dependent upon the X chromosome, each
value we calculated above can be used for each mutation we use. That is:
| X X wN |
X Y wN |
| XX ww, wN, or NN |
XX ww, wN, or NN |
| XY ww, wN, or NN |
XY ww, wN, or NN |
See? So you can't predict if males and females will be any of the recessive mutations. Both females
and males have an equal chance of being a whiteface, split to whiteface, or normal. According to our earlier Punnet Square for
the whiteface mutation (in which both birds were split to whiteface), we had a 25% chance of getting a whiteface, 50% chance of
split to whiteface, and 25% of normal. You cannot tell if a bird is split to a mutation just by looking at him, with the exception
of pied. In the pied mutation, there will be a few yellow feathers at the back of the bird's neck (it is pretty obvious).
Let's do an example with the pied mutation. We'll use a hen that is a visual pied, and a cock that is split to pied
(although again, we could switch it around, and it would still turn out the same). I use the abbreviation pd for pied. The hen would be
pdpd and the male would be pdN.
| X X pdN |
X Y pdpd |
| pdpd |
pdN |
| pdpd |
pdN |
So, half of their offspring would be visual pieds, the other half would be split to pied. You can use this
method for any of the recessive mutations.
Now try it with two different recessive mutations. We'll use fallow (f) and whiteface (w). One bird will be
split to both whiteface and fallow, the other will be split to whiteface and a visual fallow. Got it? This gets a lot more complicated. :)
|
wN fN |
wN ff |
|
| ww fN |
wN fN |
wN fN |
NN fN |
| ww ff |
wN ff |
wN ff |
NN ff |
| ww ff |
wN ff |
wN ff |
NN ff |
| ww fN |
wN fN |
wN fN |
NN fN |
Wasn't that fun? Did you follow that? You do the FOIL theorem for the first set, which would be the
whiteface gene in this example. Then you do the FOIL theorem for the second set, and combine them! There are 16 combinations
if you do it right. Fiddle with it and see if you come up with the same thing!
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