You may remember from biology class (or Jurassic Park the movie <G>) that DNA is a double strand. When eggs and sperm are formed, the strand divides, and when an egg is fertilized the two single strands join to form a new double DNA strand - and a new individual. Thus, each individual (bird or whatever) receives one "set" of chromosomes from each parent. Now look at a particular *position* on the strand and what is there - let's say a gene for feather color (remember I'm being simplistic). If the adult (parent) has the "code" or gene for blue feathers on both strands of the DNA, its offspring will inevitably receive the gene for blue feathers from that parent. If it also receives the gene for blue feathers from the other parent, it will almost certainly have blue feathers and be "homozygous" (both the same) for that trait. If however, it receives a gene for green feathers from the other parent, which color will it be? "Dominance" of genes will determine which is expressed visually, and in this case the chick will be green as green is dominant over blue (usually). BUT it is heterozygous (different genes) for that trait, having the gene for blue on one DNA strand and the gene for green on the other. So, you can see that its offspring may receive a gene for blue feathers from it, and if it also gets the gene for blue from the other parent, the visually green bird's offspring may be blue! Thus two green birds could produce a blue chick if both are heterozygous (called "split to") for that trait, and this is why it is important to know the genetics of your birds.
Simple recessive (autosomal recessive) genes are carried on the DNA strand in a location not connected to the gene that determines gender. Thus each individual has two (simplistic, remember?) genes for that trait, one on each DNA strand, and either is equally likely to be passed on to offspring (50/50 chance). G = green (capitalized because it is dominant) and b = blue; it looks like this:
GG (homozygous) or Gb (heterozygous)
So, what is happening when we mate the two heterozygous green birds?
Gb x Gb = GG, Gb, Gb, bb
It is simple algebra, and there are always 4 possibilities. Now we can see that 25% (1 in 4) of these babies will be homozygous green, 50% will be heterozygous green (green split to blue), and 25% will be homozygous blue. Of course, a bird that visually expresses a recessive trait must be homozygous. Right? Because if it were heterozygous, the recessive trait would be hidden by the dominant.
What happens if we mate two blue birds? All blue babies, right?
bb x bb = bb, bb, bb, bb
Right. All blue.
Simple algebra can be used to determine offspring if we look at one trait at at a time. What happens if we mate a green split to blue (heterozygous green) to a blue bird?
Gb x bb = Gb, Gb, bb, bb
50% are green split to blue, 50% are blue. Simple, right? <G> This formula applies to all autosomal (simple recessive) traits, including (in most parrots) blue and pied.
Sex-linked recessive traits are carried on, or linked to, the gene for gender. The male gender chromosome is Y, and the male has two - YY. The female chromosome is Z, but the hen has one of each - YZ. Thus in parrots the hen "determines gender." The sex-linked trait is "carried" on the Y chromosome, so the hen has it on only one DNA strand. Thus a hen can only be or not be (visually) a sex-linked mutation - she can't be "split to" it. Cinnamon, opaline, lutino, and "pearl" (in cockatiels) are examples of sex-linked mutations.
The algebra works like this: N = normal (dominant), c
= cinnamon.
The female only has it on one side so we use an X to represent
the "missing" chromosome:
Cinnamon male (cc) x normal hen (Nx):
cc x Nx = cN, cN, cx, cx
Hmm. All the males (cN) are normal split to cinnamon, all the hens (cx) are visual cinnamons! Hey, we can sex these babies by color, and all our hens will be visual cinnamon even though only one parent has the cinnamon trait! (This is called color sexing, or a "color sexed" nest, and can be very useful in non-dimorphic species, or species like cockatiels that are not dimorphic when young.)
But what happens if we mate a normal male (NN) to a cinnamon hen (cx)?
NN x cx = Nc, Nc, Nx, Nx
Rats! No cinnamons. All the males are split to cinnamon (Nc), but the hens are normals that don't even get the cinnamon trait! Not a good pairing unless we are only trying to produce split to cinnamon males.
So, to be visually expressed in offspring, a sex linked mutation must be in the male parent. To get MALE offspring that visually express the trait, the hen must BE that color and the male must at least be split to the same color. In other words, to get male cinnamons we must have either:
a cinnamon male x a cinnamon hen
cc x cx = cc, cc, cx, cx
which results in all cinnamon babies
OR
a male split to cinnamon x a cinnamon hen
Nc x cx = Nc, cc, Nx, cx
which results in 50% cinnamons of both sexes.
Suppose we mate two apparently normal birds together and get a (surprise!) cinnamon chick. What happened? Obviously this:
Nc x nx = Nn, Nc, Nx, cx
25% of our chicks will be cinnamons (and necessarily hens), and we can then conclude that this male is split to cinnamon. Half of his male offspring will be split to cinnamon also, but we won't be able to tell which ones unless we "test breed" them.
MOST mutations work and are "transmitted" to offspring independently of each other, so if working with cinnamon sky blues or opaline violet pieds we can consider each trait separately and get some idea of the results. We must merely know whether each trait is autosomal recessive or sex-linked and whether each parent bird is homozygous or heterozygous for that trait.
For example, a male opaline sky blue is "oo bb" and if mated to a green split to blue hen (Nx Gb) gets us:
oo x Nx = oN, oN, ox, ox
AND
bb x Gb = bG, bG, bb, bb
We will get some opaline green hens, some opaline blue hens, and some green and blue males all split to opaline. We could work out the percentages if we really wanted to.
And there you have it - genetics for the "rest of us."
Hope it was helpful to someone.
-- Heike H. Ewing -- heike@ionet.net
--
-- Bear's
Den Aviary --
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Revised: October 20, 1999
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