Double Recessive Morphs
by Steve Sykes
Combining two or more recessive traits can create new leopard gecko morphs. Examples of these morphs are patternless albinos (expressing both patternless and albino traits) and blazing blizzards (expressing both albino and blizzard traits).
I have designed this page to show you how to breed for double recessive morphs and how to create Punnett Squares to determine the expected ratios of offspring from a particular breeding. I am using patternless albinos for my example, but you could substitute any two recessive traits for patternless and albino (such as blizzard and albino to create a blazing blizzard).
There are many ways to get patternless albino babies, but basically both parents need to be carrying at least one copy of the patternless and albino genes. This could be achieved by breeding two double hets (het for both patternless and albino) together. Only 1 out of 16 from this cross would be a patternless albino, with albinos, patternless, and “normal” phenotype (spotted) animals also produced. But, the problem with this cross is that not all of the albino babies would be carrying the patternless gene, not all patternless babies would be carrying the albino gene, and not all “normal” looking (spotted) babies would be carrying the patternless and albino genes. This is a problem because you will not be able to tell the genotype of each baby from its phenotype (example: you will not be able to tell which of the albino babies are het for patternless). All the babies would be “possible hets” (i.e. the albino babies would be possible hets for the patternless trait, patternless babies would be possible hets for albino, and “normal” babies would be possible hets for both traits).
When working with Punnett Squares you denote the traits with letters, typically using an uppercase letter for the dominant trait (black spotting) and a lowercase letter for the recessive trait. Here are the possibilities we will be using:
= “normal” gene
Here are some examples of the genotypes of specific geckos that you may be using to create patternless albinos:
Het = AaPp
As you can see the gecko needs to have two copies of the recessive trait for it to be visible in the phenotype. For example, an albino het for patternless has two copies of the albino trait, (aa) and only one copy of the patternless trait (Pp) so the albino trait is visible but the patternless trait is not.
The genes for albino are located at a specific place on the gecko’s chromosomes, and the genes for patternless are located at a different place. At the place where the albino gene would be located, the gecko can have either albino genes (denoted as lowercase a, from above), or non-albino (denoted as uppercase A). It is important to note that all geckos have genes at these specific places, but they may or may not be of the recessive trait. For example, a blizzard (that is not carrying any other recessive genes) has the AA genes (from above) at the place where the albino genes would be present.
Every organism has two copies of each gene, one copy inherited from each of its parents. Sperm and eggs only carry one copy of the parent’s chromosomes, and when the sperm and egg meet the resulting embryo has two copies of all the chromosomes.
The purpose of a Punnett Square is to determine the ratios of specific offspring that could be expected from a particular breeding. So when setting up a Punnett Square you need to determine the possible genes that could occur in each of the sperm or egg cells when the parental genes segregate, then you combine the genes in the box (on the Punnett Square table) where the two meet. The possible genes carried by the sperm are listed on the top of the table, and the possible genes carried in the eggs are listed on the left side of the table. You will see exactly how to set this up in the examples below.
Here are the possible combinations of genes that could be carried by each sperm or egg from a double het parent (genotype = AaPp):
AP, Ap, aP, ap
When setting this up I usually start with the first gene in the genotype (in this case A) and combine it with the first copy of the other trait (in this case P), then combine the first gene (A) with the second copy of the other trait (in this case p). This is continued for the second copy of the first gene (in this case a), and it is combined with both the first and second copies of the second gene.
Here are the possible combinations of genes that could be carried by each sperm or egg from an albino, het for patternless (genotype aaPp).
aP, ap, aP, ap
you have the possible combinations you can construct your Punnett Square.
Put the possible combinations for the father on the top of the Punnett
Square, and the possible combinations for the mother on the left side.
Then combine the genes where the two boxes meet, signifying the combination
of the sperm and the egg.
All the following genotypes result from the above breeding:
= “Normal” phenotype, not carrying patternless or albino
The above genotypes would be observed in the following ratios (grouped by phenotype).
As mentioned above, the phenotype will not indicate the genotype of the offspring from this cross. Out of the patternless offspring, 2 out of 3 will be het for albino, but 1 out of 3 will not be carrying the albino gene. Hence, the patternless babies will be 66% possible het for patternless as only 2/3 of them will be carrying the patternless gene. Similarly, the albino babies from this cross would be 66% possible hets for patternless as only 2/3 of them will be carrying the patternless gene. The normal phenotype babies from this cross would be 44% possible double hets as only 4 out of 9 would be carrying both patternless and albino genes.
Breeding two albinos het for patternless or two patternless het for albino can also create patternless albinos, but possible hets will also be created in the process.
It is important to note that the probability of getting a specific genotype is independent of the genotypes that preceded it, and this confuses many people. For example, from the above double het x double het breeding we would expect 1 out of 16 babies to be a patternless albino. If the first 15 babies are not patternless albinos, what is the probability that the 16th baby will be a patternless albino? It is still a 1 out of 16 chance! Think of the 16 possible genotypes from the Punnett Square above as 16 slips of paper with genotypes on them in a hat. Each time an egg is fertilized, pull a genotype out of the hat. That genotype is not removed from the possible genotypes pool for the following embryos; all the same 16 slips are still in the hat each time an egg is fertilized.
It is also important to note that the ratios from the Punnett Square only indicate the expected ratios. In practice the observed ratios will likely deviate from these expected proportions, but the more eggs are hatched the closer you should get to the expected ratios. So if you only hatch 16 eggs from a double het x double het breeding it is possible you may not get any patternless albinos, or you may hatch more than one if luck is on your side. However, if you hatched 1,000 eggs from the same breeding your observed ratio of patternless albinos would likely be close to the expected 1 out of 16 ratio.
A "CLEAN", FUN WAY TO PRODUCE PATTERNLESS ALBINOS
A “clean” way to produce patternless albinos is to breed a patternless het albino with an albino het patternless. This is a “clean” way because you will be able to tell the genotype of the babies from their phenotype, i.e. all normal phenotype babies will be double hets, all patternless babies will be het albino, and all albinos will be het patternless. The Punnett Square below shows the details of this cross:
het patternless parent = aaPp (top of square)
= Double het = 4 out of 16
This is fun breeding because you will be hatching four different phenotypes of leopards (in equal ratios) from two different looking parents. I really enjoy breedings like this because each baby is a surprise! This is also a very interesting cross to use to teach kids about genetics.
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Owners: Steve and Debra Sykes (916) 62-GECKO (624-3256), 10AM-10PM PST