Going From Pet Owner To Breeder


Here are the fundamental procedures for utilizing a Punnett Square to answer a genetics issue. After you’ve learned this, you should be able to solve any genetic problem involving the mating of two birds.

Basics –

  1. Determine the parent organisms’ genotypes. Use two letters to denote the genotype for one particular attribute. A capital letter denotes the dominant version of a gene (allele), whereas a lowercase letter denotes the recessive form of the gene (allele). Dominant genes appear first.
  2. Make a note of your “cross” (mating).
  3. Make a p-square.
  4. “Split” the genotype letters for each parent and place them “outside” the p-square. Split the genotype letters from one parent and place them on the left, outside the rows of the p-square. Split the two letters of the second parent’s genotype and insert them above each of the p-two square’s columns.
  5. Fill in the p-square to get the probable genotypes of the children. Match a letter from the left with a letter from the top to do this.
  6. Report the findings and summarize the outcomes (genotypes and phenotypes of offspring). Each of the four boxes will always contain two letters.

Here’s a simple example of a recessive trait Punnett Square with a blue Pacific (bb) and a green split for blue Pacific (gb):

Parentsgb
bgbbb
bgbbb

The end result would be a visible blue split of 50% and a green split of 50% for blue.

Dominant mutations can only be carried in the visible form; there is no’split’ form, as there is with recessive mutations. As a result, only one bird is required to begin creating dominant mutations. There are both single-factor and double-factor mutations. Unless they are bred to see what offspring they produce, they are indistinguishable.

Here’s a simple example of inherited dominance with a single-factor dominant pied Pacific (PP) and a normal Pacific (nn)

Parentsnn
PPnPn
ggngn

The final outcome is 50% single factor dominating pied and 50% normal.

Another example of a dominant trait Punnett Square using a double-factor dominant pied Pacific (PP) and a normal green Pacific:

Parentsnn
PPnPn
PPnPn

The outcome would be a single factor pied with a 100% dominance.

Finally, here’s an example of inherited dominance utilizing a dominant blue pied (Pg) and a recessive blue split (gb):

Parentsgb
PPgPb
PPgPb

The outcome would be 50/50 single factor dominant green pied and 50/50 single factor dominant blue pied.

The mutation in sex associated mutations is carried on the chromosome that determines sex. Males have two “ZZ” chromosomes that are identical. Females have distinct sex chromosomes designated as “Zw.” Males may be separated for a sex related mutation, but females cannot. So far, the only sex-liked mutation in parrotlets is the sex-linked cinnamon or pallid. Using a sex-linked male and a normal female, here is a Punnett Square to discover sex connected mutations:

ParentsZw
ZcZcZZcw
ZcZcZZcw

50% sex-linked cinnamon split men and 50% sex-linked cinnamon visual females are predicted.

Another case using a sex-linked cinnamon female and a regular male:

ParentsZcwc
ZZcZZwc
ZZcZZwc

50% sex-linked cinnamon split men and 50% sex-linked cinnamon visual females are also predicted.

To generate 100% sex-linked cinnamons, both males and females must be sex-linked as follows:

ParentsZcwc
ZcZcZcZwcZc
ZcZcZcZwcZc

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