Example 2

Here I take the classical example used to explain genetics in budgerigars: the blue gene bl. There exist at least four different alleles (forms) of this gene. In this example we only use the two most frequent ones:

the not mutated wild allele: bl⁺ (the superscript '+' always indicates the wild allele)
the mutated 'blue' allele: bl

Since genes come in pairs there are 3 possible combinations of these two alleles:

bl⁺ ---: normal, lightgreen bl⁺

bl⁺ bl --- = ---: normal/blue (normal split blue), lightgreen/blue bl bl⁺

bl ---: blue, skyblue bl

A lightgreen/blue bird is in appearance (phenotypicaly) the same as a normal lightgreen. You cannot see that the bird has one bl gene.

Consider the following pairing:

Pairing 2.1: lightgreen/blue cock x lightgreen/blue hen

Genetic formula: bl⁺/bl x bl⁺/bl

Punnet square:

bl⁺ bl

bl⁺ bl⁺/bl⁺ bl/bl⁺

bl bl⁺/bl bl/bl

Solution:

25% lightgreen

50% lightgreen/blue

25% skyblue

As you can see a new colour (skyblue) emerges from this pairing. This is because the parents are split. In practice you often do not know all the recessive genes a bird carries hidden in its DNA. The lightgreen offspring from this pairing for instance may be split blue or might not be split blue, visually we cannot see the difference.

This sometimes leads to some confusion. when people ask me what to expect from a pair of normal lightgreens I anwser: normal lightgreens. After the breeding season they say I was wrong and all my rules are good for nothing because they also bred skyblues. Stricktly speaking my answer was correct for the posed question, but due to a lack of information (How was I to know his birds where split blue ?) the assumptions I made about the question where wrong. To avoid this, I now anwser: normal lightgreens, but any recessive gene may popup.

These splits can stay undetected in a population: if by change form this pairing no skyblues are born we may not know that the parents and some offspring are split. Another way these splits can stay in a population is by the following pairing:

Pairing 2.2: lightgreen cock x lightgreen/blue hen

Genetic formula: bl⁺/bl⁺ x bl⁺/bl

Punnet square:

bl⁺

bl⁺ bl⁺/bl⁺

bl bl⁺/bl

Solution:

50% lightgreen

50% lightgreen/blue

The pairing lightgreen/blue cock x lightgreen hen gives the same result
(you can try this yourself).

The previous two pairings didn't produce a lot of skyblues. If you want to breed skyblues the following pairings are more appropriate:

Pairing 2.3: lightgreen/blue cock x skyblue hen

Genetic formula: bl⁺/bl x bl/bl

Punnet square:

bl⁺ bl

bl bl⁺/bl bl/bl

Solution:

50% lightgreen/blue

50% skyblue

The pairing skyblue cock x lightgreen/blue hen gives the same result
(you can try this yourself).

Pairing 2.4: skyblue cock x skyblue hen

Genetic formula: bl/bl x bl/bl

Punnet square:

bl

bl bl/bl

Solution:

100% skyblue

We could have solved this last pairing very fast: since the bl allele is the only allele present in the parents, the offspring can only have this allele. Hence the offspring is bl/bl or skyblue.

There is but one pairing left:

Pairing 2.5: lightgreen cock x skyblue hen

Genetic formula: bl⁺/bl⁺ x bl/bl

Punnet square:

bl⁺

bl bl⁺/bl

Solution:

100% lightgreen/blue

The pairing skyblue cock x lightgreen hen gives the same result
(you can try this yourself).

Example 1 was about a dominant mutation (Sp), example 2 about a recessive mutation (bl). The method for calculating pairings is the same. First of all we need to know the genetic background of the characteristic:
Is it dominant with respect to the wild allele? (spangle, darkfactor, australian greyfactor, australian pied, dutch pied, ...)
Is it recessive with respect to the wild allele? (blue, danisch pied, greywing, ...)
Finding this genetic background is the hardest part. For most colour mutation we can look it up in books. But some characteristics (like size) are controled by many unknown genes and therefore we cannot simply calculate the offspring.