Continue reading »]]> Punnet Squares are a handy way to visually represent the possible results of a breeding. Figure A is a simple Punnet Square. In this case we are trying to determine the possible results of a breeding between animals that are both heterozygous for the recessive albino trait (a). There are two possibilities for each parent. Pass on their Albino gene or pass on there normal (+) gene. The two possible genes for one parent are written across the top with each possibility representing a column of the square. The two possible genes for the other parent are written down the side with each possibility representing a row of the square. The genes across the top are then written down each respective column (‘a’ goes in both squares in the 1st column and ‘+’ goes in each square in the second column). Do the same for the genes written down the side going across each row.

Heterozygous Parents

There you have it. Each square represents one possibility. The top left square with ‘aa’ shows the result of an albino animal. The squares on the top right and bottom left would be heterozygous for albino animals, just like the parents. The bottom right would be a normal animal. the other two square are ‘a+’ and ‘+a’. These result in the same type of animal (heterozygous for albino)

The Punnet square also helps with the calculation of odds of any specific result. One of the Four squares is ‘aa’ for albino so this result is likely one out of four times or 25 percent. The same is true for a Normal offspring (one of four squares or 25%).

One thing to keep in mind: Let’s say you breed to heterozygous albino ball pythons. Let’s also say your female lays 4 eggs and you hit the odds exactly as the punnet square says. What will hatch is one albino, 2 het albinos, and one normal. The problem you have is that there is no visible way to tell which of the 3 normal appearing animals is actually normal and which two are hets. All you know about those three is that each has a two out of three chance of being a het ( or one out of three chance of being normal ). The only way to know for sure is to grow them up and breed them and see what the results are. Until then each animal would be described as 66% heterozygous albino. The actual percentage is 66.666…etc or 67% but is typically described as 66%.

Double Heterozygous Parents

Here is another, more complex, punnet square. In this case we have 2 ball pythons that are heterozygous for 2 different (albino and piebald) recessive traits (Double Hets). Look below for the odds.

Look Here for a full description of the results of this punnet square.

Continue reading »]]> Double Heterozygous x Double Heterozygous

Phenotype: Wild Type

Genotype: Both animals are 100% Heterozygous for Albino and 100% Heterozygous for Piebald

Gene Pair Description: Both animals are a+p+

There are 9 possible outcomes (4 different Phenotypes):

• aapp – Homozygous albino and Homozygous piebald (shows the recessive albino trait and the recessive piebald trait). Jackpot! The Albino Piebald.
• a+p+ (+a+p, a++p, +ap+) – Heterozygous for albino and piebald (does not show either trait)
• aap+ (aa+p) – Homozygous for albino and Heterozygous for piebald (shows the albino trait only)
• a+pp (+app) – Homozygous for piebald and Heterozygous for albino (shows the piebald trait only)
• a+++ (+a++) – Heterozygous for albino (does not show the albino trait)
• ++p+ (+++p) – Heterozygous for piebald (does not show the piebald trait)
• aa++ – Homozygous for albino (shows the albino trait)
• ++pp – Homozygous for piebald (shows the piebald)
• ++++ – Wild Type (Big time strike out)

Percentages:

We’ll break this down based on the 4 Phenotypes possible:

Albino Piebald – Homozygous for albino and Homozygous for piebald. This is the goal of this breeding. There is a 1 in 16 chance of obtaining this result. In other words each offspring has a 6.25% chance of being an Albino Piebald.

Albino – Homozygous for albino. There two different versions of this possibility. 2 in 16 (12.5%) will be Homozygous albino and Heterozygous for piebald. 1 in 16 (6.25%) will be Homozygous for albino and will not have acquired the piebald gene from either parent. All of these animals would be considered 66% possible for piebald. In other words, each offspring has and 18.75 % chance of being an Albino, 66% Heterozygous piebald.

Piebald – Homozygous for piebald. There two different versions of this possibility. 2 in 16 (12.5%) will be Homozygous piebald and Heterozygous for albino. 1 in 16 (6.25%) will be Homozygous for piebald and will not have acquired the albino gene from either parent. All of these animals would be considered 66% possible for albino. In other words, each offspring has and 18.75 % chance of being a Piebald, 66% Heterozygous albino.

Wild Type – 9 of 16 (56.25%) will be normal looking snakes that do not show either the albino or piebald traits. 4 of 16 (25%) will be Heterozygous for albino and piebald (just like the parents). 2 of 16 (12.5%) will be Heterozygous for albino only. 2 of 16 (12.5%) will be Heterozygous for piebald only. 1 in 16 (6.25%) will not have acquired either albino or piebald genes for the parents. How could these be described? I don’t like this one but you could describe them as 44% (44.4) possible for Double Heterozygous for Albino and Piebald. Personally, I like the description of 66% Heterozygous for Albino, 66% Heterozygous for Piebald.

The End Result:

Each offspring for this pairing will be one of these 4 possibilities:

• Albino Piebald
• Albino (66% Heterozygous for Piebald)
• Piebald (66% Heterozygous for Albino)
• 66% Heterozygous for Albino and 66% Heterozygous for Piebald