Are all dominant morphs 50% homozygous?

[JoshSloane]

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?

You are mistakenly using the term 'gene' when you should be using allele.

OP: I sent you a PM with my email. More than happy to teach you. I'm a scientist.