What will I get if I cross....

When setting up breeding groups, the breeder wants to know approximately what to expect in the hatchlings. There are a couple of ways to go about determining what the results will be. The first step is to become familiar with the basics of genetics and the characteristics of the morph (is it a recessive trait?) that you are interested in. Then you are ready to dive in and figure out what the results of your cross will be!

Punnett Squares

When dealing with one or two traits at a time, Punnett squares tell you at a glance what results to expect. The first step when using a Punnett square is to determine which alleles can be passed on by each parent.

One Trait - Monohybrid Cross

As an example, let's look at two geckos that are heterozygous (het) for blizzard. This is a monohybrid cross. We can write this as Bb x Bb. Both parents have an equal chance of passing on either a B (normal) or a b (blizzard) allele in their gametes. We will then put the possible alleles along the left side for one parent and along the top for the other parent. Then, we can fill in the squares. By convention, we put uppercase letters first for heterozygotes.

The results show that we will have a genotypic ratio of 1:2:1 homozygous dominant (BB): heterozygous (Bb): homozygous recessive (bb) and a phenotypic ratio of 3:1 normal: blizzard. When both parents are hets, you will always get these ratios. If the trait is codominant (Mack snow instead of blizzard), the phenotypic ratio will be the same as the genotypic ratio - 1:2:1 because the hets have a different appearance than homozygous dominant. Of course, you can make Punnett squares for crosses other than a monohybrid cross.
What results would you get for Bb x bb?

Two Traits - Dihybrid Cross

We can also use a Punnett Square when dealing with two traits. In a dihybrid cross, both parents are het for two traits. Let's look at blizzard and Tremper albino. So the parents would be AaBb x AaBb. The hardest part about using a Punnett square for two traits is to determine what alleles to put across the top and down the side. Remember that these letters represent which alleles are going into the gametes produced by that parent. That means that for each trait there will be one and only one allele. So, for our example, there will always be one "a" (A or a) and there will always be one "b" (B or b). You need to have all the possible combinations of "a" and "b" that each parent can produce. For two traits, there will be four possibilities. The easiest way to determine all the possible combinations without missing any is to use the FOIL method (you may remember this from math class). FOIL stands for: First Outer Inner Last.

For our dihybrid parents: AaBb we get
First: AB
Outer: Ab
Inner: aB
Last: ab

Notice that for the dihybrids all four combiniations are different. Which alleles would we use for Aabb x aaBb?

Here is the Punnett square when we add the alleles for each parent to the top and side:

Now we can fill in the squares. By convention, we put the alleles for each trait together: AABB not ABAB.

We can see from these results that we get a phenotypic ratio of 9 normal (A_B_): 3 albino (aaB_): 3 blizzard (A_bb): 1 albino (blazing) blizzard (aabb). This 9:3:3:1 ratio is always found in dihybrid crosses. What results would you get from the Aabb x aaBb cross mentioned above?

Product Rule

If you are interested in a cross that involves more than two traits, the Punnett square can become a bit unwieldy. We can instead use the product rule to determine the probability of getting the desired genotype. The disadvantage here is that you must calculate the probability of each genotype separately, while everything is visible at once in a Punnett square.

First, let's use the dihybrid cross for comparison purposes. If your goal is to produce a blazing blizzard (aabb) from two double hets, what is the chance of hatching one? Using the product rule, we multiply the individual probabilities for each trait together. Looking just at the albino trait (Aa x Aa), there is a 1 in 4 chance of getting an albino. If you need help figuring this part out, you can always use a basic, 1-trait Punnett square. The same is true for blizzard (Bb x Bb) - 1 in 4 chance of getting a blizzard from two hets. Multiplying 1/4 x 1/4 gives us 1/16 chance of hatching a blazing blizzard from double hets (just like we found in the Punnett square).

This method becomes much more valuable when dealing with more than 2 traits. For example, let's add in the eclipse trait to our blazing blizzard. What are the chances of hatching a ruby-eyed blazing blizzard from the following sets of parents?

Triple hets: AaBbEe x AaBbEe

We have a 1 in 4 chance of hatching out an albino, 1 in 4 for blizzard, and 1 in 4 for eclipse. Multiply each of these probabilities together to get the probability of hatching that ruby-eyed blazing blizzard:

1/4 x 1/4 x 1/4 = 1/64

Ok, let's try parents that are albino, het for blizzard and eclipse: aaBbEe x aaBbEe

All offspring will be albino so we have a value of 1 for that trait: 1 x 1/4 x 1/4 = 1/16

Finally let's try one where the parents are different. Red eyed albino het blizzard (aaBbee) x Eclipse het albino and blizzard (AaBbee)

We have a 50/50 chance (1 in 2) of getting an albino (Aa x aa).
1 in 4 chance of getting a blizzard (Bb x Bb).
All offspring will be eclipse (ee x ee).

1/2 x 1/4 x 1 = 1/8

One final note - remember that these are statistical probabilities. There is no guarantee that if you hatch exactly 8 offspring for that last example that one of them will be a ruby-eyed blazing blizzard. The more offspring you hatch, the closer you will come to the statistical averages.

Here are the answers to the questions posed above: