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Genetic Punnett Squares: Predicting Offspring and the Future of your Program

Genetics sometimes seem quite convoluted and confusing but as a breeder it is essential to understand the ins and out of genetics and how to generate predictions amongst your program to determine potential offspring and plan the future of your program. Note this article is a complementary second part to our Elysian Bengals Color Genetics blog so this article will not cover specific color genes but should be referenced.

Let us first cover some basic terminology

Allele- versions of the same gene. Eg. Agouti gene can include the following alleles A, a, APb, A2, H2

Dominant - a gene that will express over another gene often masking the effect of the recessive gene. Denoted by an uppercase letter such as A, B, C, D, E, I

Genotype- what GENETIC genes the organism contains

Heterozygous- two alleles of a gene that are different eg. A/a C/cs ETC.

Homozygous- two alleles of a gene that are exactly the same eg. A/A, a/a ETC.

Phenotype- what the organism PHYSICALLY looks like

Probability - the chance that an offspring will have, express or carry specific genes from the parents

Punnett Square- a tool used to predict the genotypical and thus phenotypical outcome of the offspring of two parents.

Recessive - a gene that is often masked by a dominant gene and not expressed. Must contain two copies, one from each parent to express. Denoted by a lowercase letter such as a, b, cs, cb, e, d, i

Sex Linked- a gene that is located on a sex chromosome and can often only be expressed in a specific sex (male or female)

Now that we have a grasp on common terminology, let’s get in the thick of working with a Punnett Square. Let’s start simple with one trait.

First collect the Genotype of each of your parents. The best way to know these traits is through the knowledge of parentage where you can deduce the genotype of the offspring and/or genetic testing . An example of genotype deduction via parentage would be Parent 1 has the A/A agouti gene alleles. Parent B has the A/A agouti gene alleles. We know that each parent will contribute a gene to the offspring. Parent 1 can only donate an “A” and Parent B can only donate an “A” so no matter which genes come together the offspring will 100% of the time be A/A.

What if the parents have two different genes? Parent 1 has the A/A agouti gene alleles. Parent B has the a/a non agouti gene alleles. We know that each parent will contribute a gene to the offspring. Parent 1 can only donate an “A” and Parent B can only donate a “a” so the offspring can only be A/a 100% of the time. In most cases (as some genes haven’t been translated into a genetic test), genetic testing is the sure way to know what your cat’s genotype is. Some of the popular genetic testing companies include Wisdom Panel, Base Paws, UC Davis and Langsford. Below is an example of the Coat Color portion of a Wisdom Panel report

This cat is A/A (on this report that is the default so not listed) which is perfect for our first set up. Next we need to set up a Punnett Square this is the tool we are going to use to predict what genotypes the offspring will have from two parents both with the agouti gene alleles A/A

The Punnett Square acts almost like an addition problem. Above each box you will put the alleles of the parents.

The next step is to add the genes together by taking mom’s first allele and adding it to dad’s first, then mom’s first allele and add to dad’s second allele. Follow with mom’s second allele and dad’s first and finally mom’s second allele and dad’s second allele.

We can see 4/4 of the squares have an AA in them; this means according to these two parents 100% of the offspring will be A/A as well. Let's try an example where the parents have the same genotype but two different alleles. Parent A and Parent B both are A/a. Repeat the same steps as before.

As you can see just one change in an allele has changed our outcome! Instead of 100% of the kittens being A/A now only 25% can be A/A, 50% can be A/a and 25% can be a/a.

We should take a moment to note that these probabilities apply to EACH possible kitten NOT the whole litter. Over time yes you may find the same pairing leads to these percentages but each kitten is a new roll of the dice NEVER influenced by the previous kitten or previous litters.

Our third example includes two parents with two different genotypes each containing two different allele

Each box has something different in it now. 25% of the kittens can have the genotype A2/a, 25% can be A/a, 25% can be APb/A2 and 25% can be APb/A. Ok now determining one genotype like the agouti gene is great but cats don’t just have one gene and the set of alleles they have are a whole genotype code primarily consisting of A, B, C, D, E, I and sometimes O alleles. Let's start off with a dihybrid cross, this type of cross will work with two genes instead of one. Instead of up to four possible offspring genotypes, we are now working with 16. The process is primarily the same. 1. Determine the allele pairings of dad and mom

2. Add the alleles together 3. Determine probabilities of potential offspring

In step one we now have two traits, not one so we must combine these alleles in a different way. Let’s start simple with heterozygous traits, alleles that are not the same so we can visually see the combinations. Lets work with the Agouti gene and the Inhibitor gene. Both Parent A and Parent B will be A/a I/i. We must figure out the various gene combinations that each parent can produce. Each allele from each gene will need to pair.

Here you want to make four possible combinations for each parent where:

A and I are paired

A and i are paired

a and I are paired

a and i are paired

Our pairings for both parents will be AI, Ai, aI and ai.

Just like our previous set up, our allele pairs will be placed above each of the four boxes

The next step is to add similar alleles together all agouti alleles and all inhibitor alleles need to combined into the box.

Now we have 16 boxes so any genotype represents 1/16 or 6.25% however any same genotypes will increase that percent. A Punnett Square for a tetrahybrid cross or four traits contains 256 boxes. Given this complexity, Punnett Squares are not the best method for calculating genotype and phenotype ratios for crosses involving more than a few traits (although not impossible). Now that we understand what a punnett square is, how to set it up and how to fill it out with single and multiple gene traits, we need to address sex linked genes such as the Orange gene that creates the red or orange coloring. Most sex linked genes are found on the X chromosome so they can be called X-linked as well. This gene is sex linked meaning they are located on a sex chromosome. In this case, they are located on the X chromosome. XY denotes a male and XX denotes females

Male XY XOY- means the gene is located on the X sex chromosome resulting in an variation of orange color XoY- means there is no Red gene and the cat is not expressing any orange color

Female XX XO/XO- means there is a copy of O on each sex chromosomes and expresses orange Xo/Xo-means there are no copies of the dominant O so no orange is expressed XO/Xo- means there is a copy of O on one of the X sex chromosomes but expresses as a tortoiseshell (a coat menagerie of black, red, orange, yellow, or cream)

The final step in using your Punnett Square is understanding what the genotype combinations mean and how they express phenotypically.You may want to refer to our Elysian Bengals Color Genetics Article. Let’s use the following for an example: Parent A A/a C/C D/d E/E I/i and Parent B is the same A/a C/C D/d E/E I/i. This will generate the following possible offspring genotypes with the following ratios and percentages A/A C/C D/D E/E I/I - 1:64 (1.5625%) A/A C/C D/D E/E I/i - 1:32 (3.125%) A/A C/C D/d E/E I/I - 1:32 (3.125%) A/A C/C D/d E/E I/i - 1:16 (6.25%) A/a C/C D/D E/E I/I - 1:32 (3.125%) A/a C/C D/D E/E I/i - 1:16 (6.25%) A/a C/C D/d E/E I/I - 1:16 (6.25%) A/a C/C D/d E/E I/i - 1:8 (12.5%) A/A C/C D/D E/E i/i - 1:64 (1.5625%) A/A C/C D/d E/E i/i - 1:32 (3.125%) A/a C/C D/D E/E i/i - 1:32 (3.125%) A/a C/C D/d E/E i/i - 1:16 (6.25%) A/A C/C d/d E/E I/I - 1:64 (1.5625%) A/A C/C d/d E/E I/i - 1:32 (3.125%) A/a C/C d/d E/E I/I - 1:32 (3.125%) A/a C/C d/d E/E I/i - 1:16 (6.25%) A/A C/C d/d E/E i/i - 1:64 (1.5625%) A/a C/C d/d E/E i/i - 1:32 (3.125%) a/a C/C D/D E/E I/I - 1:64 (1.5625%) a/a C/C D/D E/E I/i - 1:32 (3.125%) a/a C/C D/d E/E I/I - 1:32 (3.125%) a/a C/C D/d E/E I/i - 1:16 (6.25%) a/a C/C D/D E/E i/i - 1:64 (1.5625%) a/a C/C D/d E/E i/i - 1:32 (3.125%) a/a C/C d/d E/E I/I - 1:64 (1.5625%) a/a C/C d/d E/E I/i - 1:32 (3.125%) a/a C/C d/d E/E i/i - 1:64 (1.5625%)

When we have a case like this we can reduce our load by eliminating genes that are homozygous and the same in both parents. By default the offspring will have the same gene sets. So we can eliminate C/C and E/E. We now have: A/A D/D I/I - 1:64 (1.5625%) A/A D/D I/i - 1:32 (3.125%) A/A D/d I/I - 1:32 (3.125%) A/A D/d I/i - 1:16 (6.25%) A/a D/D I/I - 1:32 (3.125%) A/a D/D I/i - 1:16 (6.25%) A/a D/d I/I - 1:16 (6.25%) A/a D/d I/i - 1:8 (12.5%) A/A D/D i/i - 1:64 (1.5625%) A/A D/d i/i - 1:32 (3.125%) A/a D/D i/i - 1:32 (3.125%) A/a D/d i/i - 1:16 (6.25%) A/A d/d I/I - 1:64 (1.5625%) A/A d/d I/i - 1:32 (3.125%) A/a d/d I/I - 1:32 (3.125%) A/a d/d I/i - 1:16 (6.25%) A/A d/d i/i - 1:64 (1.5625%) A/a d/d i/i - 1:32 (3.125%) a/a D/D I/I - 1:64 (1.5625%) a/a D/D I/i - 1:32 (3.125%) a/a D/d I/I - 1:32 (3.125%) a/a D/d I/i - 1:16 (6.25%) a/a D/D i/i - 1:64 (1.5625%) a/a D/d i/i - 1:32 (3.125%) a/a d/d I/I - 1:64 (1.5625%) a/a d/d I/i - 1:32 (3.125%) a/a d/d i/i - 1:64 (1.5625%)

Our Punnett Square will look like this (where the same color represents the same exact genotype)

Finally we can take our list of genotypes and translate them to their phenotype and figure out the percent of kittens that will express that phenotype.

A/A D/D I/I - 1:64 (1.5625%)- Silver A/A D/D I/i - 1:32 (3.125%)- Silver A/A D/d I/I - 1:32 (3.125%)- Silver A/A D/d I/i - 1:16 (6.25%)- Silver A/a D/D I/I - 1:32 (3.125%)- Silver A/a D/D I/i - 1:16 (6.25%) - Silver A/a D/d I/I - 1:16 (6.25%) - Silver A/a D/d I/i - 1:8 (12.5%) - Silver A/A D/D i/i - 1:64 (1.5625%)- Brown A/A D/d i/i - 1:32 (3.125%)- Brown A/a D/D i/i - 1:32 (3.125%)- Brown A/a D/d i/i - 1:16 (6.25%) - Brown A/A d/d I/I - 1:64 (1.5625%) - Blue Silver A/A d/d I/i - 1:32 (3.125%) - Blue Silver A/a d/d I/I - 1:32 (3.125%) - Blue Silver A/a d/d I/i - 1:16 (6.25%) - Blue Silver A/A d/d i/i - 1:64 (1.5625%) - Blue A/a d/d i/i - 1:32 (3.125%) - Blue a/a D/D I/I - 1:64 (1.5625%) - Smoke a/a D/D I/i - 1:32 (3.125%) - Smoke a/a D/d I/I - 1:32 (3.125%) - Smoke a/a D/d I/i - 1:16 (6.25%) - Smoke a/a D/D i/i - 1:64 (1.5625%) - Black a/a D/d i/i - 1:32 (3.125%) - Black a/a d/d I/I - 1:64 (1.5625%) - Blue Smoke a/a d/d I/i - 1:32 (3.125%) - Blue Smoke a/a d/d i/i - 1:64 (1.5625%)- Blue There are 19 genotypes: 4/19 or ~21% will be Brown 4/19 or ~21% will be Blue Silver 3/19 or ~15.8% will be Blue 4/19 or ~21% will be Smoke 2/19 or ~10.5% will be Black 2/19 or ~10.5% will be Blue Smoke

While Punnett Squares aren’t always ideal for a whole genotype due to the sheer size (it's possible of course) we can use Punnett Squares for any known trait! The process is fairly simple and straightforward once you understand it and practice a few times. Learning to use this tool can help you not only predict potential offspring as well as aid in planning the future of your program.

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