i want to know the result of the f1 if the parents are both Snow hognose. all the offspring will be Snow??

Yes, if both of the parents are homozygous (meaning visual) for a trait, then that trait will breed true. In other words, if you breed an Albino to an Albino, you will get all albinos just like you would for Snows. The only difference between a Snow and an Albino is that the Snow is the culmination of 2 recessive traits (Albino and Axanthic) instead of just 1 for the Albino.

I'll explain why...

Let us use the letters A & a to represent the genes for Albinism and X & x to represent the genes for Axanthism. The capital letter means that the gene is not present on it's DNA strand while the small letter means the gene is present on it's particular DNA strand. In DNA, there are two sets of genes, one from the mother and one from the father, therefore the letters will always be used in pairs such as AA, Aa (aA translates to the same thing) or aa.

An Albino with no known hets would have it's genes basically looking like this: aaXX (I'm only showing the X's to represent the Axanthic gene loci but which is not present in either pair of genes, I'll explain this in a little bit). The letters basically represent whether those specific genes for any specific trait are present and whether it is turned on or turned off. In order to be turned on, both genes for that loci (gene location in the DNA strand) have to be present (small caps) such as 'aa'. If only one recessive gene is present (for a recessive trait which is what we are dealing with), then the mutant gene (Albino) is turned off and the dominant trait is the only one seen such as 'Aa'. When the gene is not present at all, then we represent this as 'AA'. The only time these rules do not apply is when you are dealing with dominant or co-dominant traits (like the Anaconda morphs which is a co-dominant trait but that's another topic for another day).

So, if a pure Albino is aaXX, then a pure Axanthic would be AAxx. When you breed an Albino to an Axanthic you will get all normal looking hatchlings but they will be 100% het for both genes. Let us use a Punnett Square to demonstrate this by using the FOIL (First, Outside, Inside, Last) method. We need to determine every variation of each pair of genes for both parents.

Albino Parent
First: aaXX = aX
Outside: aaXX = aX
Inside: aaXX = aX
Last: aaXX = aX

Axanthic Parent
First: AAxx = Ax
Outside: AAxx = Ax
Inside: AAxx = Ax
Last: AAxx = Ax

Here's the Punnett Square combining both parents genes together. (Sorry if these don't align properly, just go by combining the top row with the left column)

___aX__aX__aX__aX
Ax AaXx AaXx AaXx AaXx
Ax AaXx AaXx AaXx AaXx
Ax AaXx AaXx AaXx AaXx
Ax AaXx AaXx AaXx AaXx

As you can see, every animal produced from this pairing will be AaXx which means that they carry the 'a' Albino gene and the 'x' Axanthic gene but they are both turned off by 'A' and 'X'. Again, in order for the animal to exhibit either of these traits (Albino or Axanthic), then both letters need to be small caps.

An 'AaXx' animal is considered a 100% het Snow. This does not mean that simply breeding these animals together will give you all Snows. Although the genes are present for both parents to be able to produce Snows, only half of the genes are present by both parents.

To produce a Snow, we need to breed 2 100% het Snow animals 'AaXx' (or any other animals that carry both genes). Again, let us break down the genes for 2 100% het Snow parents.

het Snow Parents
First: AaXx = AX
Outside: AaXx = Ax
Inside: AaXx = aX
Last: AaXx = ax

Since both parents have the same genes, we simply use this pairing for both parents.

___AX__Ax__aX__ax
AX AAXX AAxX aAXX aAxX
Ax AAXx AAxx aAXx aAxx
aX AaXX AaxX aaXX aaxX
ax AaXx Aaxx aaXx aaxx

Let us look at all the gene combinations and the different types of animals we'll produce. Since we're 2 pairs of genes for each parent, it gives us 16 variations of gene pairings. Remember that 'Aa' and 'aA' (or 'Xx' and 'xX') basically represent the same gene structure.

AAXX = Normal, no hets. 1 out of 16
AAXx = Normal het Axanthic. 2 out of 16
AaXX = Normal het Albino. 2 out of 16
AaXx = Normal het Albino & het Axanthic. 4 out of 16
AAxx = Axanthic. 1 out of 16
Aaxx = Axanthic het Albino. 2 out of 16
aaXX = Albino. 1 out of 16
aaXx = Albino het Axanthic. 2 out of 16
aaxx = Snow. 1 out of 16

This is where it gets really confusing. If you notice, there are 4 different types of Normals, 1 with no hets, 2 het Axanthic, 2 het Albino and 4 het Snow (het Albino & het Axanthic). How do you know which is which? You don't! This is where "possible" hets comes into play.

An easy example would be to use the visual Axanthics or Albinos. According to the ratio, we should get 3 Axanthics, two of which are het Albino. Since we cannot tell which is het and which isn't, we use simple division by dividing the number of hets (2) by the overall number of visual animals (3) and get 0.667 which is 66%. That is how you get a 66% het animal. Of course if you wanted a pure non-het Axanthic animal, then there's a 33% chance of getting one.

It's the same thing with the Normals. If you wanted to determine the chance of getting a double het Snow, then you figure that according to the Punnett Square there are going to be 4 Normals and only 1 of those is the double het Snow. Divide 1 by 4 and you will have a 25% chance of getting one. For the Albinos and Axanthics, it's 50% (2 out of 4 chance for each).

But it's all based on the Punnett Square ratio. You could end up with 5 Axanthics out of a 10 egg clutch from a double het Snow type of pairing. How can you determine which is het Albino and which isn't? Again, you can't! My understanding is that you should still use the same mathematics according to the Punnett Square ratios in labeling your animals.

Back to breeding morphs...

If you breed a Snow to an Albino het Axanthic (or an Axanthic het Albino), then your odds of producing a Snow should be reduced significantly by about 50% giving you half Snows and half Albinos het Axanthic.

Let us work it out in a Punnett Square.

Snow Parent
First: aaxx = ax
Outside: aaxx = ax
Inside: aaxx = ax
Last: aaxx = ax

Albino het Axanthic Parent
First: aaXx = aX
Outside: aaXx = ax
Inside: aaXx = aX
Last: aaXx = ax

___ax__ax__ax__ax
aX aaxX aaxX aaxX aaxX
ax aaxx aaxx aaxx aaxx
aX aaxX aaxX aaxX aaxX
ax aaxx aaxx aaxx aaxx

aaXx = Albino het Axanthic. 8 out of 16
aaxx = Snow. 8 out of 16

So, now to your original question, what will I get if I breed a Snow to a Snow. You know the drill, Punnett Square time!

Snow Parents
First: aaxx = ax
Outside: aaxx = ax
Inside: aaxx = ax
Last: aaxx = ax

___ax__ax__ax__ax
ax aaxx aaxx aaxx aaxx
ax aaxx aaxx aaxx aaxx
aX aaxx aaxx aaxx aaxx
ax aaxx aaxx aaxx aaxx

aaxx = Snow. 16 out of 16

I hope this helps you understand how the ratios of recessive genetics are worked out.

I apologize for any errors and/or confusion. I spent 2 hours writing this out off of the top of my head as well as proofreading it a few times to make sure what I wrote was correct but I'm sure I goofed somewhere along the way.
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Troy Rexroth
Rextiles

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