Are all dominant morphs 50% homozygous?

[Mysnakeislong]

Other than the spider? I'm just trying to double check that i understand all this stuff i'm learning.

[OhhWatALoser]

All dominant morphs are heterozygous (het), besides a select few that have been proven out to be homozygous.

[Mysnakeislong]

Oh. Do you know why? I've already read about the spider morph for example which dies if it is homozygous but i didn't know other dominant morphs had trouble becoming homozygous also.

[OhhWatALoser]

Confusion might come from the 10 different definitions of dominant within the hobby and obviously 9 are wrong.

Dominant means that the snake will look the same in both heterozygous and homozygous forms, there is no visual difference. The only difference will be that when the homozygous is bred to a normal, all babies will be that morph, where the heterozygous has a 50/50 chance. Most morphs called dominant in the hobby have not proven out the homozygous, so they really aren't dominant. I call them unproven dominant. However a few have been supposedly proven out like pin, leopard, Congo, and daddy gene.

They are not hard to make, they are just hard to prove. To make a super pin, you only have to breed a pin to a pin. However the super pins look the same as the normal pins, so you have to do breeding to find out if any throw all pins. With them not looking any different and taking time to prove out, it really doesn't make sense economically to be trying to produce them, so most people don't.

The daddy gene however was easy to prove out, since it is allelic with lesser. A super lesser is a white snake, a lesser daddy is a Hypo looking lesser. Rdr bred a lesser daddy to another lesser daddy, only 3 Snakes could come from this pairing, the super lesser , lesser daddy, and super daddy. He produced a nice normal looking animal from the pairing, which the daddy gene by itself looks nice normal. So it was proved dominant easily because it was allelic

The spider hasnt had a proven super, so it cannot be called dominant. If a lethal super is found, it would be considered incomplete dominant

[JoshSloane]

Other than the spider? I'm just trying to double check that i understand all this stuff i'm learning.

So I think you are getting the general info, but you are mixing up a few terms. When percentages are used, that is meant to denote the chance that the individual has inherited the recessive allele from the parent animal. For example, if you breed an albino x normal, all the babies will be 100% heterozygous. This means that each baby is guaranteed to retain one recessive allele. This 100% het then has a 50% chance of passing this recessive allele onto the next generation if that animal is bred.

Homozygous dominant refers to an individual inheriting two identical dominant alleles for a given gene.

Heterozygous, as mentioned earlier is when the individual inherits one recessive, and one dominant allele for a given gene.

In Co-dominance, the individual inherits two alleles which are both expressed independently of one another phenotypically.

[Mysnakeislong]

Sorry my title is kind of confusing I realize now. I understand all that so far, what I meant is if i have a dominant morph is there a 50% chance that it is homozygous?

[Mysnakeislong]

Tbh that just went way over my head.

[Mysnakeislong]

Nvm i'm actually starting to understand, just had to read it slower lol.

[JoshSloane]

I think you are getting confused about the difference between homozygous dominant and co-dominant.

When you have a co-dominant gene, both the alleles are expressed and the phenotype is different relative to the situation where one allele of that gene is present. Think BELs. Mojave gene is co-dominant. When the individual has one Mojave allele, they look phenotypically Mojave. When they have two Mojave alleles, they are BELs. Both Mojave alleles are being expressed to give a certain phenotype.

When considering a purely dominant gene, the phenotype of the homozygous dominant and the heterozygote are identical, even though the genotypes are different.

[paulh]

Sorry my title is kind of confusing I realize now. I understand all that so far, what I meant is if i have a dominant morph is there a 50% chance that it is homozygous?

Short answer: no. It depends on what genes the parents have.

Long answer: There are two genes, A and a. These symbols are algebraic -- they stand for whatever they are defined as. These two genes make 3 gene pairs, AA, Aa, and aa. These 3 gene pairs make 9 possible matings if we include the parents' sex.
1. AA x (= mated to) AA
2. AA x Aa
3. AA x aa
4. Aa x AA
5. Aa x Aa
6. Aa x aa
7. aa x AA
8. aa x Aa
9. aa x aa

If we disregard the parents' sexes, then mating 2 is the same as mating 4, mating 3 is the same as mating 7, and mating 6 is the same as mating 8. Deleting matings 4, 7, and 8 leaves 6 possible matings --
1. AA x AA --> (= produces) 1/1 AA
2. AA x Aa --> 1/2 AA, 1/2 Aa
3. AA x aa --> 1/1 Aa
4. Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
5. Aa x aa --> 1/2 Aa, 1/2 aa
6. aa x aa --> 1/1 aa
The fractions are the expected fraction of the total number of babies.

Recessive mutant gene:
A = the normal (dominant) gene
a = the mutant (recessive) gene (example = albino)
Only the aa gene pair produces the mutant morph. Only matings 4, 5, and 6 produce aa babies. In mating 4, a quarter of the babies are expected to be the mutant (albino) morph. The rest look normal. In mating 5, half of the babies are expected to be the mutant morph. And in mating 6, all the babies are the mutant morph.
1. AA (normal) x AA (normal) --> 1/1 AA (normal)
2. AA (normal) x Aa (normal looking, het albino) --> 1/2 AA (normal), 1/2 Aa (normal looking, het albino)
3. AA (normal) x aa (albino) --> 1/1 Aa (normal looking, het albino)
4. Aa (normal looking, het albino) x Aa (normal looking, het albino) --> 1/4 AA (normal), 2/4 Aa (normal looking, het albino), 1/4 aa (albino)
5. Aa (normal looking, het albino) x aa (albino) --> 1/2 Aa (normal looking, het albino), 1/2 aa (albino)
6. aa (albino) x aa (albino) --> 1/1 aa (albino)

Codominant mutant gene:
A = the normal (codominant) gene
a = the mutant (codominant) gene (example = pastel)
The AA gene pair produces the double dose mutant morph (super pastel). The Aa gene pair produces the single dose mutant morph (pastel). Only the aa gene pair produces normal snakes.
1. AA (super pastel) x AA (super pastel) --> 1/1 AA (super pastel)
2. AA (super pastel) x Aa (pastel) --> 1/2 AA (super pastel), 1/2 Aa (pastel)
3. AA (super pastel) x aa (normal) --> 1/1 Aa (pastel)
4. Aa (pastel) x Aa (pastel) --> 1/4 AA (super pastel), 2/4 Aa (pastel), 1/4 aa (normal)
5. Aa (pastel) x aa (normal) --> 1/2 Aa (pastel), 1/2 aa (normal)
6. aa (normal) x aa (normal) --> 1/1 aa (normal)

Dominant mutant gene:
A = the mutant (dominant) gene
a = the normal (recessive) gene (example = pinstripe)
Only the aa gene pair is found in normal babies only. The AA and Aa gene pairs produce the mutant morph. In other words, the super pinstripe has the AA gene pair and the pinstripe has the Aa gene pair. But the pinstripe and super pinstripe snakes look alike.
1. AA (super pinstripe) x AA (super pinstripe) --> 1/1 AA (super pinstripe)
2. AA (super pinstripe) x Aa (pinstripe) --> 1/2 AA (super pinstripe), 1/2 Aa (pinstripe)
3. AA (super pinstripe) x aa (normal) --> 1/1 Aa (pinstripe)
4. Aa (pinstripe) x Aa (pinstripe) --> 1/4 AA (super pinstripe), 2/4 Aa (pinstripe), 1/4 aa (normal)
5. Aa (pinstripe) x aa --> 1/2 Aa (pinstripe), 1/2 aa (normal)
6. aa x aa --> 1/1 aa (normal)

In mating 1, all the babies are super pinstripe. In mating 2, all the babies are 50% probrability super pinstripe. In mating 4, 3/4 of the babies are either super pinstripe of pinstripe, so those babies are 33% probability super pinstripe. In the other matings, there are no super pinstripe babies. So the probability of a (super pinstripe/pinstripe) baby actually being a super pinstripe depends on what genes the parents have.

Clear as mud?