Genetics ?? Hypo gene specifically

GENETICS 101

Nicely put. LOL
And well undrstood................Now
I read that pea theory thing long ago actually. A little differently put from page to page, this time around. It did help though.
Thanks for the lessons. You guys are great. LOL He has had enough to convince him now. HA HA

I'll probably be back for more, (I don't get enough), but for now, I got some studying to do.

Ciao,
Rick

BTW here's a link to my online genetics tutorial that works great for getting started, or as a refresher.

http://www.serpwidgets.com/Genetics/genetics.html

OK, I'm back.
Had to brush up on some math to take a test for the bosses "TSPS" to prove I know what I do as a surveyor. Go figure. I been proving it for 5 yrs. but that little card makes him feel better.

Back to genetics. Thanks Charles for the link. I will give it a look.
Where were we?..........Oh Yeah, (told ya i don't get enough) Co-Dominant.
You both express what leads me to my belief.

Originally Posted by M.Dwight

However, sometimes a locus will inherit two different or unlike (heterozygous) alleles and neither is recessive in relation to the other.

If neither is recessive then both must be dominant.

Remember dominant just means visable in the phenotype

Since both alleles are dominant and are working TOGETHER at the same locus we call this a CODOMINANT gene.

With a codominant gene Both allele will be visible in the phenotype.

I agree with this completely, and believe you can see the differennces in the three phenotypes.

Originally Posted by Serpwidgets

If the alleles are codominant to each other, the three possible genotypes create three resulting phenotypes. That is, AA, Aa, and aa can all be visually identified.

Absolutely. Let me explain........
Perhaps in one way I am wrong to call it co-dom, when actually I believe it CAN be co-dom.

When the Wt alelle acts as dominant you get WT. We agree on that?
When the Salmon alelle acts dominant to the WT, the WT goes recessive, and you end up with a salmon het WT. (first two pics are of 2 I am leaning towards to purchase, the third an example only)
This would express the Salmon phenotype in it's heterozygous state, or mathematically, Salmon > WT..........



And when both alelles are paired as salmons, you get salmon expressed in its Homozygous state, or Salmon equal to Salmon, or Supers..........



And for the third phenotype, where both the salmon and WT phenotypes are expressed in their heterozygous state would be a salmon with visible signs of WT, or Salmon = WT, or Co-Dominant..........


Now I know i don't usualle talk all that loci and alelle stuff, I prefer plain ol' english, but i have read a little bit about this stuff also. I really do understand what you both are trying to say. And textbook speaking, I would have to agree with you both. I cannot say you are wrong.
But you guys have made me think ALOT about what you have said, and who knows, I may still be wrong. I can't afford that kind of research. But I believe that in a lot of salmons, both alelles are expressing their phenotypes in a co-dominant relationship. The last pic is not a prime example, but all I could find.

What say Ye?

Ciao,
Rick

Originally Posted by crotalusadamanteus

But I believe that in a lot of salmons, both alelles are expressing their phenotypes in a co-dominant relationship. The last pic is not a prime example, but all I could find.

What say Ye?

Ciao,
Rick

Salmon is as variable as wild type or normal.
In a litter of normal baby boas you will get some with light pigment, some with dark pigment and some that are in-between. But there is only one wild type gene.

The same is true of salmon. Once you produce enough salmons and supers you will see there is no real consistant difference between the two. Some salmons will be very dark and some will be very light with both reduced saddles and black pigment. The same is true of supers. This is why these days they are sold as "possible super salmons." Because the salmon gene is so variable there is no way to be 100% sure it is a super unless both of its parents are supers. Or, it was proved out to be a super through breeding trials.

Paul Hollander said it best in this post....
http://forums.kingsnake.com/view.php?id=926912,927843

That's a good point also. Maybe that's the connection I am missing. I do tend to get a little literal with things sometimes.

You are right though, they do almost always say "possible Super" LOL

Ciao,
Rick

Back to your original question. Yes if you breed an albino to a het albino, you would STATISTICALLY expect 50% albino, and 50%.

The salmon gene is at a different locus (location on the chromasome), and does not enter into the "albino" equation. However, I don't know if they could both be expressed, producing something that wasn't exactly either one.

If I understand what I've read here, and assuming the salmon trait is dominant, here are the possibilities:

Super Salmon phenotype = S
Salmon phenotype = s
Wild type = w

S X S = 100% Super Salmon

S x s = 50% Super; 50% Salmon

S x w = 100% Salmon

s x w = 50% Salmon; 50% Wild type

Second line should read: would STATISTICALLY expect 50% albino, and 50% normal.

Originally Posted by jsrocket

The salmon gene is at a different locus (location on the chromasome), and does not enter into the "albino" equation. However, I don't know if they could both be expressed, producing something that wasn't exactly either one.

Yes, both salmon and albino can be expressed in the phenotype. They are called sunglows.

If a mutant gene mask the presence of another mutant gene on a different locus it is called a epistatic gene. And the gene that is being masked is called the hypostatic gene. Neither salmon or albino are epistatic/hypostatic to each other so both are expressed in the phenotype.

Originally Posted by jsrocket

If I understand what I've read here, and assuming the salmon trait is dominant, here are the possibilities:

Super Salmon phenotype = S
Salmon phenotype = s
Wild type = w

S X S = 100% Super Salmon

S x s = 50% Super; 50% Salmon

S x w = 100% Salmon

s x w = 50% Salmon; 50% Wild type

IMO the best way to work these is to use the actual genotypes, since the phenotypes are not what are being crossed. It's like trying to learn mutiplication by simply memorizing times tables and never learning counting or addition. (Then if the number is not on the memorized table, you're screwed. )

That is, use both genes present at the locus instead of one phenotype. The results are the same, but would make more sense to someone who is trying to understand what is going on underneath the phenotypes.

So instead of starting out by memorizing the results of "S X S" (and every other cross under the sun for every locus) you can use S for the dominant salmon mutant allele and + for the wild-type allele. There are then three possible genotypes at the salmon locus:
SS = "Super" Salmon
S+ = Salmon
++ = Wild-type

When you cross any two of these genotypes to each other, each baby receives one of the dad's genes, one of mom's genes. Using FOIL is an easy way to get the four possible outcomes. For example, if you are crossing 12 X 34 at a given locus you get:
First = 13
Outside = 14
Inside = 23
Last = 24

This is the same thing as doing a one-locus Punnett square:
<table border=1><tr><td></td><td>1</td><td>2</td></tr><tr><td>3</td><td>13</td><td>23</td></tr><tr><td>4</td><td>14</td><td>24</td></tr></table>

You can just substitute the 1 and 2 for the father's two genes, and 3 and 4 for the mother's two genes, and this will cover every cross at every (paired) locus, no matter what the parents' genotypes are.

-----

So, given that, the prediction for the original cross (Albino X hypo het albino) is:
At the Salmon Locus you are crossing ++ X S+
Results are: +S, ++, +S, ++ (half salmon, half normal)

At the Albino locus you are crossing aa x a+
Results are: aa, a+, aa, a+ (half albino, half normal)

You can then combine those two sets of results in a grid similar to a Punnett square:
<table border=1><tr><td></td><td><B>Salmon</B></td><td><B>Non-salmon</B></td></tr><tr><td><B>Albino</B></td><td>Albino, Salmon</td><td>Albino, Non-salmon</td></tr><tr><td><B>Non-albino</B></td><td>Non-albino, Salmon</td><td>Non-albino, Non-salmon</td></tr></table>

Absolutely. That is exactly how I did it, too. (except for your final step of combining the results). I didn't memorize that, I calculated it, but should have "showed the work" a little better. Your explanation is more understandable. Thanks.