# Equal dominance?



## gregjones (Sep 14, 2008)

I've been reading up on genetics and I know how to draw the punnet squares to determine how the offspring will look.

However, what happens when 2 genes have equal dominance? So let's say I have 2 different recessive morphs, breed them together...what will the offspring look like?

The reason I ask is because I read a thing suggesting the "Spider" gene is dominant. So I was wondering what would happen if this was bred to a normal?


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## Blackecho (Jun 30, 2008)

Different genes sit at different loci, so you can have more than one visible, you will effectively get a mix of them.


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## alan1 (Nov 11, 2008)

grantjames said:


> However, what happens when 2 genes have equal dominance? So let's say I have 2 different recessive morphs, breed them together...what will the offspring look like?


in this case, the "normal" gene is more dominant than the 2 recessive genes...
so, the offspring from an albino & pied pairing, would result in all of the clutch looking normal, but het for both albino & pied...

breed these offspring back to each other for a 1:16 chance (per egg) of the visual albinopied...


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## bladeblaster (Sep 30, 2008)

grantjames said:


> I've been reading up on genetics and I know how to draw the punnet squares to determine how the offspring will look.
> 
> However, what happens when 2 genes have equal dominance? So let's say I have 2 different recessive morphs, breed them together...what will the offspring look like?
> 
> The reason I ask is because I read a thing suggesting the "Spider" gene is dominant. So I was wondering what would happen if this was bred to a normal?


when genes are quoted as being dominant, this is usually in relation to the 'normal' gene. Spider is dominant to the normal gene so a spider to a normal will produce spiders and normals.

In recessive genes say albino is recessive so the normal gene is the dominant gene in this case.



Blackecho said:


> Different genes sit at different loci, so you can have more than one visible, you will effectively get a mix of them.


: victory:


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## eeji (Feb 22, 2006)

a het spider would give spider and normal when bred to a normal, and a **** spider to a normal would give all spiders


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## SNAKEWISPERA (Aug 27, 2007)

There is no het spider.


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## Blackecho (Jun 30, 2008)

SNAKEWISPERA said:


> There is no het spider.


Yes there is, there is no known homozygous spider though as far as I'm aware.

However, genetically speaking, Ian is correct.


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## SNAKEWISPERA (Aug 27, 2007)

Blackecho said:


> Yes there is, there is no known homozygous spider though as far as I'm aware.
> 
> However, genetically speaking, Ian is correct.



wait there.. so everything ive just been told has been thrown out the window... how is there a het spider?
spider is dominant?


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## Blackecho (Jun 30, 2008)

SNAKEWISPERA said:


> wait there.. so everything ive just been told has been thrown out the window... how is there a het spider?
> spider is dominant?


Heterozygous and Homozygous have nothing to do with Recessive, Co-Dominant and Dominant.

Genes come in pairs, is the pair are the same they are Homozygous, if they are different they are Heterozygous.

Recessive genes need to be homozygous to be visual as the 'normal' gene is more dominant.

Dominant genes such as Spider only need 1 gene to be visual as they are more dominant than 'normal'.

So basically every spider out there is Heterozygous as I don't think anyone has proven a Homozygous Spider.


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## eeji (Feb 22, 2006)

there is indeed a het spider and a **** spider - they just happen to look the same.

a het spider x het spider breeding will produce 1:4 **** spiders, 1:4 het spiders and 1:2 normals (statistically)


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## Blackecho (Jun 30, 2008)

eeji said:


> there is indeed a het spider and a **** spider - they just happen to look the same.
> 
> a het spider x het spider breeding will produce 1:4 **** spiders, 1:4 het spiders and 1:2 normals (statistically)


Ian, do you actually know of any **** Spiders though?


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## eeji (Feb 22, 2006)

i don't know of any spiders **** or het, but i'm sure they're out there. Don't forget they look the same so you won't know you've got one until its been bred


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## alan1 (Nov 11, 2008)

eeji said:


> a het spider x het spider breeding will produce 1:4 **** spiders, 1:4 het spiders and 1:2 normals (statistically)


nope...

(statistically) 3:4 will be spiders of some kind, and 1:4 will be normal...


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## Blackecho (Jun 30, 2008)

eeji said:


> i don't know of any spiders **** or het, but i'm sure they're out there. Don't forget they look the same so you won't know you've got one until its been bred


I have not heard of a homozygous Spider, I'm wondering if they were perhaps lethal.


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## eeji (Feb 22, 2006)

alan1 said:


> nope...
> 
> (statistically) 3:4 will be spiders of some kind, and 1:4 will be normal...


my bad you're correct. out of those 3:4 spiders, 2 would be het and 1 **** (i read my square wrongly :blush: )


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## Pauline (Aug 3, 2006)

The full breeding odds for spiders are :-

Het x Normal = 50% Het & 50% Normal
Het X Het = 50% Het, 25% **** & 25% Normal 
**** x Normal = 100% Het
**** x Het = 50% Het & 50% ****
**** x **** = 100% ****

The only way you could be sure if your spider is **** or het is to breed to a normal. If you get normals, it's het, but with royal odds being what they are you'd have to repeat breedings with a pos **** to prove it wasn't a fluke that there are no normals in a clutch.


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## paulh (Sep 19, 2007)

grantjames said:


> I've been reading up on genetics and I know how to draw the punnet squares to determine how the offspring will look.
> 
> However, what happens when 2 genes have equal dominance? So let's say I have 2 different recessive morphs, breed them together...what will the offspring look like?
> 
> The reason I ask is because I read a thing suggesting the "Spider" gene is dominant. So I was wondering what would happen if this was bred to a normal?


The spider bred to normal question has been answered. The breeding of two morphs produced by recessive genes still needs work.

The following is true for breeding any two different morphs together, not just morphs produced by recessive mutant genes. But I will use morphs produced by recessive mutant genes for convenience.

Every vertebrate has many thousands of gene pairs. Every gene pair has a locus, a specific location in a chromosome pair. The wild type or normal appearance or phenotype is the most common phenotype in the wild population. A wild type or normal gene is the most common gene at a given locus in the creatures in the wild population. A mutant gene is any gene that is not a wild type gene. Every gene pair contains either two normal genes, a normal gene and a mutant gene, two identical mutant genes or two non-identical mutant genes.

Many thousands of gene pairs are too many to handle in a breeding problem. So for simplicity, we ignore all the gene pairs that have two normal genes in both parents. An albino ball python has one gene pair containing mutant genes. A pied also has one gene pair containing mutant genes, but it is not the same gene pair as the gene pair with the albino mutant genes. So when breeding an albino to a pied, we ignore all but these two gene pairs in both parents.

An albino ball python has two albino genes at the albino locus and two normal genes at the pied locus. Each sperm (or egg) gets one gene from each gene pair. So our albino produces sperm (or eggs) with an albino gene at the albino locus and a normal gene at the pied locus.

A pied ball python has two pied genes at the pied locus and two normal genes at the albino locus. Each egg (or sperm) gets one gene from each gene pair. So our albino produces eggs (or sperm) with a normal gene at the albino locus and a pied gene at the pied locus.

All the babies (the F1 generation) from an albino x pied mating have a normal gene pired with an albino gene and a normal gene paired with a pied gene. It is impossible to pair a pied gene with an albino gene because they are at different locations in the chromosomes. The albino gene is recessive to the normal gene and the pied gene is recessive to the normal gene. And all the other gene pairs contain two normal genes. So all of the F1 babies look normal and are heterozygous albino and heterozygous pied.

Breeding two F1 heterozygous albino, heterozygous pied.snakes together produces the F2 generation. The genes in the normal//albino gene pair segregate into F2s that are expected to be
1/4 albino//albino (albino phenotype)
2/4 normal//albino (normal phenotype)
1/4 normal//normal (normal phenotype)

The genes in the F1's normal//pied gene pair segregate into F2s that are expected to be
1/4 pied//pied (pied phenotype)
2/4 normal//pied (normal phenotype)
1/4 normal//normal (normal phenotype)

The gene pairs assort independently. The contents of the gene pair at the albino locus does not determine the contents of the gene pair at the pied locus. So any one of the three possible gene pairs at the albino locus can be combined with any one of the three possible gene pairs at the pied locus.

A snake that has a pair of albino genes and a pair of pied genes is both albino and pied.

So when there are mutant genes at two loci (loci is the plural of locus), both the effect of the genes at one locus and the effect of the genes at the second locus are expected to appear.

Unfortunately, sometimes the effect of one pair of mutant genes masks the effect of a second pair of mutant genes. Albino in mammals is particularly notorious for this sort of thing. In mice, the effect of the albino mutant gene masks the effect of the pied gene, but this is because there is only one pigment system in mammals instead of the three pigment systems in reptiles.

Clear as mud?


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