Co-Dominant?

[JimGERcream]

what's up with the piedsided discussion?

I remember someone wonderd if the very low expressed pieds were "just" het. pied and the mid to extreme expressed pieds are the hom. form.

I know this is not proven and I don't own one but I think it might be useful to keep that in mind?? :shrugs:

 

[Ssthisto]

Never heard the "at least 17 offspring" thing....What's the significance of that number?

It's something like this:

One offspring that's Tessera only proves that the Tessera parent is at least het Tessera.

But the more offspring that are Tessera, the less likely it is that the Tessera parent is heterozygous Normal - and 17 offspring is the point at which one can say with reasonable certainty that the animal cannot be heterozygous normal.

PaulH understands it much better than I do - and I cannot for the life of me find the post (which might not be on this forum!) where he explained the numeric reason why it's 17 and not some other number.

 

[paulh]

Here's the significance of the number 17.

You breed a snake that is heterozygous albino (or heterozygous for another recessive gene) to another snake that may or may not be heterozygous for the same recessive mutant gene. If both parents are heterozygous, from the Punnett square, each egg has a 3/4 (0.75 = 75%) probability of looking normal and a 1/4 (0.25 = 25%) probability of being albino. On the other hand, if one parent is normal rather than heterozygous albino, then none of the babies will be albino; they will all look normal. So at least one albino baby means that the possible het albino snake is a het albino snake.

If you have two eggs, then the first egg has a 3/4 probability of looking normal, and the second egg also has a 3/4 probability of looking normal. If either one hatches an albino, then the test is over. To find the probability that both eggs produce normal-looking babies, we multiply 3/4 * 3/4 = 9/16.

The probability of both eggs producing albinos is 1/4 * 1/4 = 1/16.

The probability of the first egg producing a normal and the second producing an albino is 3/4 * 1/4 = 3/16.

The probability of the first egg producing an albino and the second producing a normal is 1/4 * 3/4 = 3/16.

There are no other possibilities with two eggs, and 9/16 + 3/16 + 3/16 + 1/16 = 1 (100%). And the probability of getting at least one albino is 1 - 9/16 = 7/16.

If you want, you can repeat this with three eggs, four eggs, and so on. If you do, you will find that the probability of getting all normal-looking babies from a het recessive X het recessive mating is 3/4 * 3/4 * ... * 3/4 where each 3/4 represents one egg. If you have 17 eggs hatch, and all look normal, the odds of that happening are just under 0.01 (1%). The odds of getting at least one albino baby are just over 0.99 (99%). As no albinos appeared among the 17 babies, the odds are 0.99 (99%) that the possible het is actually a normal. That's a standard cutoff in statistics, so we stop the test there.

Personally, I would mate a tessera to a normal. If the tessera has a tessera mutant gene paired with a normal gene, then the odds of getting normal babies are 1/2 (0.5 = 50%) per egg. If the tessera has two tessera mutant genes, then the odds of a normal are zero per egg. 1/2 * 1/2 * ... * 1/2 where each 1/2 represents an egg gets to the 0.01 cutoff point after seven eggs. But this is only for one snake. This test should be repeated with a number of tessera snakes.

Hope this helps.

 

[Ssthisto]

Thanks, Paul - I couldn't for the life of me work out why dividing 100 by 2 (which sorta works out to the same thing) made "less than 1 percent" at seven instead of seventeen

 

[dendro19]

There are no "proven" codominant traits in corns. Tessera could be, but that is a whole different story. In fact, the ultra and amel genes are not codiminant, they are intermediate dominant. Such as a red flower crossed to a white flower makes a pink flower. So there is an even, but incomplete amount of phenotypic penetrance from each gene. It is the same with ultramels, given that there will be some small amount of phenotypic variation from snake to snake. A good example of true codominance is bloodtype. Both A and B blootype are both dominant over O bloodtype. So AO is expressed as A, and BO is expressed as B. But in the event that you have AB bloodtype, both A and B are fully expressed. So it shows a complete copy of A expressed with a complete copy of B at the same time. Very different from the ultramels that show an intermediate amount of expression from both genes.

Note: These concepts are commonly misused. I myself made the same mistake until i took a genetics course. Also, amel and ultra are found on the same locus on the same chromsome. And incomplete dominance is not the same as intermediate dominance. Pasodoma did an excellent job with explaining intermediate dominance. As for the motley/stripe question: Motley and stripe are neither codominant nor intermediate dominant. Motley is dominant over stripe, but stripe has an incomplete penetrance upon the phenotype. In simpler terms, motley is expressed over stripe, but stripe has some expression and the amount of expression depends upon the amount of pentrance the gene has in each individual snake.

 

[slangenbroed]

There is one other gene that is dominant

There is one other, i bred with this gene several years.And it made bigg discussions and therefore i,ve been at the background at this moment.
But i will come back.
For me dominants meen.
You have one animal that looks different call it A, when i bred this animal to what so ever ( normal-amel-anery-caramel-charcoal-hypo -lavender ect ect ) and in the F1, the A is showing up, then that's dominant.

 

[bitsy]

You have one animal that looks different call it A, when i bred this animal to what so ever ( normal-amel-anery-caramel-charcoal-hypo -lavender ect ect ) and in the F1, the A is showing up, then that's dominant.

Would another possibility be that the "what so ever" snake is het for A, and therefore A is co-dominant?

 

[slangenbroed]

Would another possibility be that the "what so ever" snake is het for A, and therefore A is co-dominant?

if its het for A then it is reccesive ore genderlinked, a dominant mutant has no hets you are ore you are not A

 

[Ssthisto]

For me dominants meen.
You have one animal that looks different call it A, when i bred this animal to what so ever ( normal-amel-anery-caramel-charcoal-hypo -lavender ect ect ) and in the F1, the A is showing up, then that's dominant.

Either dominant or codominant, depending on whether the A looks exactly like the "het A" offspring, and when you breed it out homozygous A looks exactly like heterozygous A.

In royal pythons, for example, Pinstripe is dominant; a homozygous pinstripe is indistinguishable from a heterozygous pinstripe. But Mojave is incompletely dominant - breed a brown, patterned Mojave to a normal, and you'll get some Mojaves and some normals, but a homozygous Mojave is a mostly white snake with a grey head, visually distinctive from heterozygous Mojave.

Until and unless there's a proven homozygous Tessera, we won't know whether Tessera is dominant or incomplete/codominant.

I'd love to see a gene that ACTED like a codominant, though - in corns or royals - patches of two different colours like a tortoiseshell cat

 

[dendro19]

There are no "proven" codominant traits in corns. Tessera could be, but that is a whole different story. In fact, the ultra and amel genes are not codiminant, they are intermediate dominant. Such as a red flower crossed to a white flower makes a pink flower. So there is an even, but incomplete amount of phenotypic penetrance from each gene. It is the same with ultramels, given that there will be some small amount of phenotypic variation from snake to snake. A good example of true codominance is bloodtype. Both A and B blootype are both dominant over O bloodtype. So AO is expressed as A, and BO is expressed as B. But in the event that you have AB bloodtype, both A and B are fully expressed. So it shows a complete copy of A expressed with a complete copy of B at the same time. Very different from the ultramels that show an intermediate amount of expression from both genes.

Note: These concepts are commonly misused. I myself made the same mistake until i took a genetics course. Also, amel and ultra are found on the same locus on the same chromsome. And incomplete dominance is not the same as intermediate dominance. Pasodoma did an excellent job with explaining intermediate dominance. As for the motley/stripe question: Motley and stripe are neither codominant nor intermediate dominant. Motley is dominant over stripe, but stripe has an incomplete penetrance upon the phenotype. In simpler terms, motley is expressed over stripe, but stripe has some expression and the amount of expression depends upon the amount of pentrance the gene has in each individual snake.

I meant to say tessera could be incomplete dominance. Sorry for any confusion.

 

[paulh]

In fact, the ultra and amel genes are not codiminant, they are intermediate dominant.

There are at least a dozen terms for conditions in which creatures have three phenotypes--one for two identical mutant genes, one for two identical normal genes, and one for the mutant gene paired with the normal gene. FOR SIMPLICITY, codominant is used for all such mutant genes because it is the shortest to type.

 

[paulh]

if its het for A then it is reccesive ore genderlinked, a dominant mutant has no hets you are ore you are not A

"Heterozygous" applies to any pair of genes where the two genes are not the same. A snake with a dominant mutant gene paired with a normal gene fits the definition. Check a good dictionary.

Ssthisto, tortoiseshell in cats is not an example of codominance. It's an example of the Lyon hypothesis. One X chromosome is inactivated in female mammals. It just isn't the same X chromosome in all cell lines.

 

[sscsgem]

There are at least a dozen terms for conditions in which creatures have three phenotypes--one for two identical mutant genes, one for two identical normal genes, and one for the mutant gene paired with the normal gene. FOR SIMPLICITY, codominant is used for all such mutant genes because it is the shortest to type.

Yes there are several terms for the relationships between genes,and yes some are complicated, but they are used to make it SIMPLE to explain the differences between certain gene interactions. What good would it do to call a particular gene interaction codominant and then have to explain ten minutes worth of information on how it is not the same as another interaction deemed codominant. If you read the rest of my post, then you know that ultramel does not in anyway fit a codominant interaction. Just as motley and stripe are not codominant. These two are great examples of how gene interactions can appear similar, but are in fact very different. That is why i bothered to type my previous post in the first place.

Note: I am also aware that the term intermediate dominance may be confusing because ultra and amel are recessive genes. The term intermediate dominance is still applicable because even though they are recessive to a normal, they are still intermediate dominant with each other. And because it would be even more confusing to start throwing out terms like intermediate recessive, which to my knowledge is not used at all.

 

[dendro19]

Yes there are several terms for the relationships between genes,and yes some are complicated, but they are used to make it SIMPLE to explain the differences between certain gene interactions. What good would it do to call a particular gene interaction codominant and then have to explain ten minutes worth of information on how it is not the same as another interaction deemed codominant. If you read the rest of my post, then you know that ultramel does not in anyway fit a codominant interaction. Just as motley and stripe are not codominant. These two are great examples of how gene interactions can appear similar, but are in fact very different. That is why i bothered to type my previous post in the first place.

Note: I am also aware that the term intermediate dominance may be confusing because ultra and amel are recessive genes. The term intermediate dominance is still applicable because even though they are recessive to a normal, they are still intermediate dominant with each other. And because it would be even more confusing to start throwing out terms like intermediate recessive, which to my knowledge is not used at all.

Sorry, my girlfriends computer automatically logged me on under her name. That was actually my post

 

[dendro19]

"Heterozygous" applies to any pair of genes where the two genes are not the same. A snake with a dominant mutant gene paired with a normal gene fits the definition. Check a good dictionary.

Ssthisto, tortoiseshell in cats is not an example of codominance. It's an example of the Lyon hypothesis. One X chromosome is inactivated in female mammals. It just isn't the same X chromosome in all cell lines.

Exactly, but there is more to the explanation than that. The black and orange coloring are both carried on the x chromosome, which makes it a sex linkage issue. In order to have a bicolor cat, it must be a female with one x-chromosome carrying the black gene and the other x-chromosome carrying the orange gene. This is why male cats cannot be orange and black bicolor, because they have only one x chromosome and can therefore not carry both colors. Anyway, so how does this become expressed the way it does? Paul, hit the nail on the head with his explanation. Because each cell does not need two x-chromosomes to function, so one is shut down within each different cell creating a denser chromosomal mass known as a bar-body. Depending upon which chromosme is shut down, you will either show black, or orange. In a bicolor cat this is evident in their patchy coloring. Sections of orange here and sections of black there. Representing the differences in which x-chromosome is dormant in each cell.


note: Yes a cat can be visually bicolor if it has the piebald trait, making it white and another color. I am not sure if this is considered a bicolor or not. Also, there have been instances where a male bicolor cat has been made. This is stricty a chance occurrence. Genetically the male would have to be XXY, meaning that in the meiotic division of his mothers sex cell, there was a non-disjunction event where two x-chromosmes did not differentiate into seperate gametes and instead both x-chromosomes went to one gamete. So this XX gamete came into contact with a Y gamete, creating an XXY individual. Again, a very rare occurence.

 

[paulh]

The black and orange coloring are both carried on the x chromosome, which makes it a sex linkage issue. In order to have a bicolor cat, it must be a female with one x-chromosome carrying the black gene and the other x-chromosome carrying the orange gene.

This part is straight out of many textbooks, which have not caught up with the cat genetics literature. It needs correction; the rest of the post is right. This part would be right, too, if the word "black" was changed to "gray tabby (AKA wild type)".

A black and orange female tortoiseshell cat has two pairs of mutant genes, O//o a//a where O is orange, o is normal, and a is nonagouti (black). O and o are on the X chromosomes, so that part is a sexlinkage issue. The a genes are not on the X chromosomes; they are in a pair of autosomes. The O gene masks the effect of the a genes in the cell lines where the O gene's X chromosome is active. The effect of the a genes is not masked in the cell lines where the o gene's X chromosome is active.

 

[paulh]

Yes there are several terms for the relationships between genes,and yes some are complicated, but they are used to make it SIMPLE to explain the differences between certain gene interactions. What good would it do to call a particular gene interaction codominant and then have to explain ten minutes worth of information on how it is not the same as another interaction deemed codominant.

The genes A and a can make three gene pairs--AA, Aa and aa.

A is dominant to a, and a is recessive to A:
AA makes phenotype #1
Aa makes phenotype #1
aa makes phenotype #2
Two phenotypes from three genotypes. Striped and motley fit this model, though the Aa phenotype is more variable than the AA phenotype.

A is codominant to a, and a is codominant to A:
AA makes phenotype #1
Aa makes phenotype #2
aa makes phenotype #3
Three phenotypes from three genotypes. Ultra, amelanistic, and ultramel fit this model, as do the A, B, and AB blood types.

IMO, this is a lot simpler to explain to a newbie than trying to explain the difference between incomplete dominance, intermediate dominance, transdominance, semidominance, partial dominance, overdominance, less than fully dominant, etc.

 

[dendro19]

This part is straight out of many textbooks, which have not caught up with the cat genetics literature. It needs correction; the rest of the post is right. This part would be right, too, if the word "black" was changed to "gray tabby (AKA wild type)".

A black and orange female tortoiseshell cat has two pairs of mutant genes, O//o a//a where O is orange, o is normal, and a is nonagouti (black). O and o are on the X chromosomes, so that part is a sexlinkage issue. The a genes are not on the X chromosomes; they are in a pair of autosomes. The O gene masks the effect of the a genes in the cell lines where the O gene's X chromosome is active. The effect of the a genes is not masked in the cell lines where the o gene's X chromosome is active.

Nice post, i was unaware of this relationship. As you said, the textbooks are not caught up to the cat lit.

 

[dendro19]

The genes A and a can make three gene pairs--AA, Aa and aa.

A is dominant to a, and a is recessive to A:
AA makes phenotype #1
Aa makes phenotype #1
aa makes phenotype #2
Two phenotypes from three genotypes. Striped and motley fit this model, though the Aa phenotype is more variable than the AA phenotype.

A is codominant to a, and a is codominant to A:
AA makes phenotype #1
Aa makes phenotype #2
aa makes phenotype #3
Three phenotypes from three genotypes. Ultra, amelanistic, and ultramel fit this model, as do the A, B, and AB blood types.

IMO, this is a lot simpler to explain to a newbie than trying to explain the difference between incomplete dominance, intermediate dominance, transdominance, semidominance, partial dominance, overdominance, less than fully dominant, etc.

I can understand it might be hard for someone not interested in genetics to understand, but I still think it is important to give a detailed explanation. That is the purpose of a forum, to exchange information and educate. I try to keep it as true to the book as i can, so people who do not know can learn it correctly the first time instead of being told they are wrong later. It is up to them whether they read into it and actually understand it or not. Your models do fit , but it is very simplistic and does not truly cover it. There is more to a model description than just a table of genotypes.

 

[paulh]

The idea is to start simple. As knowledge increases, the student can cope with the complications and exceptions.