There are two kinds of chromosomes, (and genes ..... and traits) autosomal and sex-linked. Each animal generally has one sex chromosome and the rest are autosomal or "normal" ones. Most birds have 5-9 autosomal and one sex chromosome. (see below) Since the sex chromosome is outnumbered by at least 5 to 1, the majority of genes are passed on by the "normal" or autosomal route. In this section, we'll focus on this route, and thus, "normal" inheritance.
Most autosomal inheritance is much like the information reviewed on the Genetics page and is ruled by simple dominant and recessive characteristics. The only Quaker color mutations that has been proven to exist as an autosomal traits are the Blue and Lutino mutations. If these behave as they have in other species, both will be completely "independent" of each other as far as inheritance goes and recessive to the natural green trait. For the purpose of this discussion, we'll make these assumptions and assign "B+" for green and "b" for blue at the blue site and "L+" for green and "l" for lutino at the lutino site. Remember, though, that a Quaker with the "b" or blue trait actually has that because it lacks the yellow psittacin pigment and the "l" means it lacks black (seen as blue in feathers) melanin pigment.
Single Gene Autosomal Inheritance:
Now that genes and genetic notation are understood, we can talk about inheritance. In this section, we'll limit the discussion to inheritance of a single trait, the blue trait. After the "independence" section, we'll tackle multiple gene inheritance.
We all know that each parent contributes equally to the genetic make-up of their offspring. Specifically, though, sperm from the cock and the hen's egg EACH contain a single copy of each chromosome rather than the usual pair of each chromosome. When they merge during fertilization, the chick ends up with the normal pairs of chromosomes and, hence, genes.
Each parent contributes this single chromosome (and gene) of their pair in a random fashion. If a parent has the same genetic information on both chromosomes, then they always contribute the same information to their offspring. In other words, a bird who is "B+B+" at the blue site will always contribute a "B+" to their young. Similarly, a bird who is "b b " at the blue site will always contribute a "b" to their young.
However, when a parent has two different messages at a given site (such as "B+b" cock) then it's random as to which one it passes on to it's chicks; 50% will get the "B+" gene and 50% will get the "b" gene. The complimentary gene for whichever of these two (B+ or b) the chick gets will come from the hen. If the hen is "B+B+", then she will always contribute a "B+" gene. The 50% of the chicks which get the cock's "B+" gene will end up "B+B+" while the 50% of the chicks that get the cock's "b" gene will end up "B+b".
This is depicted graphically below. The cock's possible contributions are tinted blue and the hen's possible contributions are tinted pink. By conformation, the cock's genes are written first and listed along the left side of the table while the hen's genes are written second and listed across the top of the table. So, the cross of a Green split Blue (Green/Blue) cock to a normal Green hen is . . .
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B+b x B+B+
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B+
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B+
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B+
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B+B+
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B+B+
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b
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B+b
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B+b
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This table, called a Punnett square, is the most traditional way to figure out what offspring will come from a given cross. Though in this example, one can "figure it out in your head", as we get to more complex crosses it'll become clear that this is the easiest way to keep everything straight.
As you can see, half of the offspring in this example will be normal Green or "B+B+" and half of them will be split Blue (Green/Blue) or "B+b". Also, you can see that since the hen always contributes the same "B+" gene, the two columns with pink at the top are redundant. For simplicity, redundant columns are dropped to leave this as the accepted way to represent a Green/Blue cross Green pairing.
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B+b x B+B+
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B+
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B+
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B+B+
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b
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B+b
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Simplifying the columns didn't affect the outcome of 50% "B+B+" offspring and 50% "B+b" offspring. Another example would be a Green cock (B+B+) cross a Blue hen. (b b)
Even though one of the parents is visual Blue, all of the offspring would be Green/Blue and appear Green. Although you wouldn't get any Blue chicks, you would know FOR SURE the exact genetic makeup of all the chicks. If both parents were split Blue, (B+b x B+b) the offspring would be .......
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B+b x B+b
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B+
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b
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B+
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B+B+
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B+b
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b
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B+b
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b b
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25 % of the offspring would be B+B+ or Green.
50 % of the offspring would be B+b or Green/Blue.
25 % of the offspring would be b b or Blue.
Since the "B+b" chicks would be visually Green, you couldn't tell the difference between them and the "B+B+" chicks. 3 out of 4 chicks would appear Green and 1 out of 4 would be Blue. There would be no way to tell which visual Green chicks are split Blue. Chicks like these are called "possibles" because it is possible that they are split for a certain trait; in this case blue. The only way to find out the genetic makeup of possibles is to test breed them in the following years. Obviously, it is better to set up pairings that produce offspring with known genetics.
A very good way to set up Blue pairs such that you know the genetics of all offspring is to cross a Green/Blue with a Blue.
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B+b x b b
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b
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b
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B+
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B+b
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B+b
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b
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b b
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b b
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50 % of the offspring would be B+b or Green/Blue.
50 % of the offspring would be b b or Blue.
So, all visual Green chicks are Green/Blue and all visual Blue chicks are, well, BLUE! The simplest of all blue trait pairings is Blue x Blue which gives all b b offspring.
All of the pairings and offspring for crosses involving only green and blue traits can be viewed at the Blue Tables page.
Independence:
Independence means that two given traits are governed by separate genes at separate sites and have no effect on the other's outcome. Since the blue and lutino traits are anticipated to be "independent", the fact that a given bird has 1 or 2 copies of one of these genes has no impact on whether it does or can have 1 or 2 copies of the other. To put it a different way, blue and lutino always "compete" with green but never with each other. To denote this in genetic notation, each gene site is described separately, one on top of the other. In the following example, the pair on the top line refers to the blue site and the pair on the bottom refers to the lutino site:
B+B+
L+L+
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Green
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B+b
L+L+
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Green/Blue
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b b
L+1
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Blue/Lutino
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b b
L+L+
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Blue
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B+B+
L+1
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Green/Lutino
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B+b
1 1
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Lutino/Blue
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B+B+
1 1
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Lutino
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B+b
L+1
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Green/Blue/Lutino
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b b
1 1
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Albino
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The slash or "/" is aviculture notation for "split". Thus, the first word is the visual color and all those following a "/" are split traits. As you can see, the blue and lutino traits are expressed as pairs of genes and independently from each other.
Multiple Gene Autosomal Inheritance:
Multiple gene inheritance is conceptually NO DIFFERENT than that of single gene inheritance. It's just a bit harder to keep track of all the genetic notation. The first step is to write down, in genetic notation, the genes each parent has. As in the above table, these are listed in pairs with each trait to be followed having a pair of its own. We will use the blue and lutino traits as examples with the top pair of genes being assigned to the blue site and the bottom pair assigned to the lutino site just as in the table above.
The following are some examples to make sure this notation is clear:
b b
L+L+
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x
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B+B+
l l
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= Blue cock cross Lutino hen
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B+b
L+ l
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x
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B+B+
L+L+
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= Green/Blue/Lutino cock cross Green hen
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B+b
L+L+
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x
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B+b
l l
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= Green/Blue cock cross Lutino/Blue hen
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B+b
l l
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x
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b b
L+ l
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= Lutino/Blue cock cross Blue/Lutino hen
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B+B+
L+ l
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x
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b b
l l
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= Green/Lutino cock cross Albino hen
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Once the genetic notation of the parents is written down, it is easy to determine the possible contributions each may make to their offspring. Consider this example.
b b
L+L+
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x
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B+B+
l l
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= Blue cock cross Lutino hen
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The cock can only contribute a "b" gene at the blue site and a "B+" gene at the lutino site. Therefore, the ONLY combination he can give is this.
Similarly, the hen can only give a "B+" gene at the blue site and a "l" gene at the lutino site. Her contribution will ALWAYS look like this.
Filling these into a Punnett square, their offspring can be predicted.
All of their offspring will be Green/Blue/Lutino. This is simple because each of the parents can only give one particular combination of genes from the blue and lutino sites. There is no randomness to this at all. However, if each parent was also split for the other trait, it would become more complicated.
b b
L+ l
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x
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B+b
l l
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= Blue/Lutino cock cross Lutino/Blue hen
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In this case, the cock could contribute one of two different combinations of genes from the blue and lutino sites.
Similarly, the hen could also contribute one of two combinations.
Filling these into a Punnett square, the result is a bit more complex.
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B+
l
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b
l
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b
L+
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B+b
L+ l
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b b
L+ l
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b
l
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B+b
l l
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b b
l l
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| 25 % of the offspring are Green/Blue/Lutino |
| 25 % of the offspring are Blue/Lutino |
| 25 % of the offspring are Lutino/Blue |
| 25 % of the offspring are Albino (lower right) |
A nice benefit of this pairing is that you KNOW the genetic makeup of each chick just by the visual color. One of the most complex examples would be a cross between two Green/Blue/Lutino parents.
B+b
L+ l
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x
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B+b
L+ l
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=
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Green/Blue/Lutino cock cross
Green/Blue/Lutino hen
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Since the genes from the blue site and the genes from the lutino site are contributed "independently", the cock could provide one of 4 combinations to their young.
B+
L+
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or
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B+
l
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or
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b
L+
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or
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b
l
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The hen could also contribute one of the same 4 combinations.
B+
L+
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or
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B+
l
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or
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b
L+
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or
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b
l
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This gives a rather intimidating Punnett square.
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B+
L+
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B+
l
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b
L+
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b
l
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B+
L
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B+B+
L+L+
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B+B+
L+ l
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B+b
L+L+
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B+b
L+ l
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B+
l
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B+B+
L+ l
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B+B+
l l
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B+b
L+l
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B+b
l l
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b
L
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B+b
L+L+
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B+b
L+ l
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b b
L+L+
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b b
L+ l
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b
l
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B+b
L+ l
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B+b
l l
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b b
L+ l
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b b
l l
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Each of the 16 possible offspring described in this table is 6.25% of the total. (100 divided by 16) The breakdown would be . . .
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6.25 % (1) of the offspring are Green |
| 12.5 % (2) of the offspring are Green/Blue |
| 12.5 % (2) of the offspring are Green/Lutino |
| 25.0 % (4) of the offspring are Green/Blue/Lutino |
| 6.25 % (1) of the offspring are Blue
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| 12.5 % (2) of the offspring are Blue/Lutino |
| 6.25 % (1) of the offspring are Lutino |
| 12.5 % (2) of the offspring are Lutino/Blue |
| 6.25 % (1) of the offspring are Albino (lower right) |
Take a moment to look at the table and see if you can determine how these percentages were arrived at. Not only is this a dreadful square to look at, but it produces 9 genetically different offspring and in only ONE of the 16 (6.25%) can you predict the genetic makeup with only the visual color. (the Albino chick) ALL of the other 15 outcomes (93.75%) are "possibles". One should only allow a cross like this if there is no other choice ........ unless you're REALLY into test breeding.
By now you should . . .
- understand what an autosomal chromosome, gene and trait is.
- be familiar with which Quaker mutations are inherited as autosomal.
- know whether autosomal Quaker traits are dominant or recessive.
- be familiar with how pairs of genes are passed from parent to chick.
- know what a Punnett square is and how it predicts offspring genetics.
- understand the term "independence".
- be comfortable with the use of "Green/Blue/Lutino" type notation.
- be familiar with the notation used to describe 2 genetic sites inthe same bird.
- be able to determine possible contributions a parent may make to his/her offspring just by seeing it's genetic notation such as this. (try it !)
- be able to plug this information into a Punnett square to determine offspring for a pairing such as this. (try it !)
If these concepts are not clear at this point, you may want to review this page again before going on.
Had enough? Go back to our
Home Page and surf some other stuff.
Still thirsty for more? Finish the tutorial at the Sex-Linked Traits page.
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