Basic Genetics Tutorial
“Normal” or “wildtype” is the referance used to describe the most common genes. Wild-type genes are typically dominant. In the case of simple recessive genetics, we will assume that all wild-type genes are dominant and all “mutations” are recessive. When one parent contributes a wild-type gene, a snake will display the normal coloration, because the wild-type allele is dominant to the recessive or mutated gene. Notice that the words “gene” and “allele” may sometimes be used interchangeably. If one parent contributes a wild-type allele, and one parent contributes a recessive allele, the result is called heterozygous, which will often be shortened to “het.” A snake that is “het. albino” will contain one normal gene and one albino gene at the Albino Locus. Heterozygous therefor means that the snake posesses two different genes at a given locus.
In order for a snake to display a recessive mutation, such as in the case of an amelanistic(albino) cornsnake, that snake MUST contain the amelanistic(amel) gene in a pair. This is known as being homozygous, often shortened to “homo”, therefor a homozygous snake has two copies of the same gene at that locus.
As a recap–An heterozygous snake contains two different genes at the locus for that mutation. This snake will most often appear normal. A “morph” will contain 2 copies of the recessive gene at that locus, known as being homozygous. Any snake displaying a recessive gene mutation(morph) visually is usually homozygous for that mutation.
Breeding for Morphs
Breeding for morphs is simply the act of pairing snakes in such a way that the offspring will be homozygous for the desired mutation or trait. Pairing a homozygous amel corn to a homozygous amel corn will produce offspring that are 100% amel. Why? Let’s look at it simply.
Since we know that each parent contributes one copy of each allele to every offspring, think of it in simple terms. An amel male has 2 copies of the amel gene. It can ONLY contribute the recessive amel gene to each of it’s offspring, because it does not contain a copy of the normal gene. The same is true for an amel female. So when an amel male is paired with an amel female, every offspring from this pairing will recieve the recessive amel gene from both parents, making every snake in the clutch homozygous for amelanism. Since neither parent posesses the wild type gene, neither parent can pass on the wild type gene…and no offspring can receive the wild type gene.
When one of the snakes is heterozygous, the equation changes:
A male which is het. amel and a female which is amel are paired. The male will give 50% of the offspring the amel gene, and 50% of the offspring the normal gene. The amel female will give 100% of the offspring the amel gene. All of the offspring which receive the amel gene from the father will be homozygous for amelanism, because they also receive the amel gene from the mother. All of the offspring that receive the normal gene from the father will be heterozygous for amelanism because they all recieve the amel gene from the mother. This gives you a clutch of 50% normal het. amel and 50% amel.
As simply as possible, this is the way in which ALL simple recessive mutations work in snakes. The more combinations of genes you are dealing with, the more intricate the mathematical equations become.
Let’s look at a double-recessive pairing:
A=normal gene at amel locus
B=normal gene at anerythristic locus
Male is a snow= aabb
female is anery het. amel= Aabb
Pairing these snakes gives you:
aabb X Aabb
Since both parents are homozygous for anery, all offspring will be anery. Since female is heterozygous for amel, 50% of the offspring will be homozygous amel, and 50% will be heterozygous amel. This clutch will be 50% anery het. amel and 50% snow. Snow is the combination of homo. anery and homo. amel.
Now…let’s bring in co-dominance. Co-dominance is a situation where different genes at a single locus can be visually expressed in a heterozygous state. This is the case with Motley and Stripe. At the motley locus, you have the potential for three different genes; normal, motley, and stripe. Normal is dominant to motley and stripe.
This means that in order for a snake to display a recessive pattern mutation, it can be homozygous motley, homozygous stripe, or heterozygous for motley AND stripe. All three combinations will display a mutated pattern. ONLY when a snake contains one normal gene will the pattern be normal in appearance.
M=normal gene at motley locus
m=motley gene at motley locus
s=stripe gene at motley locus
A motley male paired with a normal het. motley female:
mm X Mm
This pairing, like the pairings above, will produce 50% normal het. motley offspring, and 50% motley offspring, according to the rules established earlier.
A stripe male by a het. stripe female:
ss X Ms
This pairing produces 50% stripe offspring and 50% normal het. stripe offspring according to the rules established earlier.
A motley male paired with a stripe female:
mm X ss
This pairing will produce 100% heterozygous motley AND stripe offspring, but since motley and stripe are co-dominant, 100% of the offspring will display a mutated pattern, which is different than the rules established above. This is because motley and stripe are both recessive genes at the same locus. Being het. for both recessive genes at this locus produces a pattern that is somewhere in between a homo. motley and a homo. stripe snake.
A het. motley and stripe male by a het. motley and stripe female:
ms X ms
This pairing will produce 25% motley, 25% stripe, and 50% het. motley and stripe offspring. 25% will display a pure motley pattern, 25% display a pure stripe pattern, and 50% display the “somewhere in the middle” pattern of a het. motley and stripe snake.
Genetics might be overwhelming at first, but when you look at it in it’s most simplified terms, it can be very easy.
In order for a snake to display a recessive gene, it MUST contain two copies of that gene, known as being homozygous. If a snake contains only one copy of a recessive gene, it is known as being heterozygous.
Pairings consisting of both parents being homozygous for the same morph will produce 100% homozygous offspring.
Pairings where one parent is homozygous and one is heterozygous for the same morph will produce 50% homozygous and 50% heterozygous offspring.
Pairings where one parent is homozygous for one morph and one parent is homozygous for a different morph will produce offspring which are 100% normal and heterozygous for both morphs.
For help in calculating what your outcomes would be, there is a fantastic online Genetics Calculator found at:
http://kornnatterlexikon.de Follow the link, and click on “Calculator” in the left-hand menu.
Just input your homozygous and heterozygous genes for each parent, and it will automatically calculate the percentages of the offspring for you.
I certainly hope that this tutorial has helped you to understand a little bit about what exactly is going on in breeding for genetic morphs. Obviously…there is ALOT more information to be digested in this field. But I don’t want to overwhelm anyone, and as such, I will leave it at this. If you have specific questions, you can send me an e-mail, and I will try to clarify it for you.