Here are the basic steps to using a Punnett Square when solving a genetics question. After learning this, you should be able to figure out any genetic question involving the breeding of two birds. 

Basics -

1. Determine the genotypes of the parent organisms. For one specific trait use two letters to represent the genotype. A capital letter represents the dominant form of a gene (allele), and a lowercase letter is the abbreviation for the recessive form of the gene (allele). Dominant genes are listed first.
2. Write down your "cross" (mating). 
3. Draw a p-square.
4. "Split" the letters of the genotype for each parent & put them "outside" the p-square. Take the genotype letters of one parent, split them and put them on the left, outside the rows of the p-square. Take the two letters of the second parent's genotype, split them up, and place them above each of the two columns of the p-square.
5. Determine the possible genotypes of the offspring by filling in the p-square. Do this by taking a letter from the left & matching it with a letter from the top.
6. Summarize the results (genotypes & phenotypes of offspring) by  reporting the findings. There will always be two letters in each of the four boxes.

Here is a basic example of a recessive trait Punnett Square using a blue Pacific (bb) paired with a green split for blue Pacific (gb):                                 

The results would be: 50% visual blues and 50% green split for blue.

Dominant mutations can only be carried in the visual form; there is no 'split' form as is in recessive mutations. Therefore you only need one bird to start producing dominant mutations. There are also both a single and a double-factor mutation. They are indistinguishable unless bred to see what offspring are produced.  

Here is a basic example of inherited dominance using a single-factor dominant pied Pacific (PP) paired with a normal Pacific (nn)

The results are 50% single factor dominant pied and 50% normal

Here is another example of a dominant trait Punnett Square using a double-factor dominant pied Pacific (PP) paired with a normal green Pacific:

The results would be 100% dominant single factor pieds. 

Finally, here is an example of inherited dominance using a double factor dominant blue pied (Pg) with a recessive blue split (gb):

The results would 50% single factor dominant green pied and 50% single factor blue dominant pied. 

In sex linked mutations, the mutation is carried on the chromosome that determines sex. Males have two identical chromosomes identified as "ZZ". Females have different sex chromosomes determined as "Zw" Only males can be split for a sex linked mutation, females cannot. In parrotlets, so far, the only sex-liked mutation is the sex-linked cinnamon or pallid. Here is a Punnett Square for determine sex linked mutations using a sex-linked male and normal female:

The expected results would be 50% sex-linked cinnamon split males and 50% sex-linked cinnamon visual females.

Here is another example using a sex-linked cinnamon female with a normal male:

The expected results would also be 50% sex-linked cinnamon split males and 50% sex-linked cinnamon visual females. 

In order to produce 100% sex-linked cinnamons, you would need both males and females to be sex-linked as follows:

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