Reptile & Amphibian Forums

I am a hobbyist breeder and was told that the Kahl strain albinos hold their color better than the Sharp strain and that you can't breed a Kahl to a Sharp. Any truth to this?

Thanks,

Eric Jensen

These two strains of albinism are created because of mutations in two different areas of the DNA. In the first example, gene A(for tyrosinase) is defective but gene B (for carrier proteins) is functioning. In the second case, gene A is functioning and it is gene B that is defective. If these two albinos were crossed, the offspring would get "Defective A" and "Functioning B" from one parent and "Functioning A" and "Defective B" from the other. The result would be a clutch of normal looking snakes who all carried the genes: "Defective A" + "Functioning A" and "Defective B" + "Functioning B". The babies
wouls all be normal looking because the Functioning genes would take over (I'm assuming this is a dominant trait). All of these babies are now double heterozygous for the two albino genes. If you crossed the double hets, you'd only have a 1/16 chance of getting albinos, that is, animals with two copies of each defective gene. Since it's just two ways of making an albino the babies would just look albino. These two
lines are not compatible. They do not work together to influence the appearance of the snakes.
explained to me by Mike Lockwood from Too Scaley Reptiles
Bob

I agree with the explanation all the way to the end:

"If you crossed the double hets, you'd only have a 1/16 chance of getting albinos"

The 1/16 chance would be for an animal that was homozygous for BOTH forms of albinism. Math gets a little complicated but looking at each form of albinism seperately there would be a 25% chance being homozygous for that form of albinism. So each would have 4/16ths probability. The complication comes with the overlapping 1/4 probability of also being homozygous for the other form of albinism.

Bottom line, crossing DH's for both forms of albinism would result in:

1/16 homozygous for both
3/16 homozygous for "Kahl" possible het for "Sharp"
3/16 Homozygous for "Sharp" possible het for "Kahl"
(all of which would look more or less alike just like any litter including albinos or 7/16 probability of "abino" with unknown specific genetic makeup)

9/16 Normal possible het for "kahl" and/or possible het for "Sharp"
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Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC)
0.2 kids (CBB, selectively bred from good stock)

LOL, to many snakes to list, last count:
12.25 BRB
12.14 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

I am sorry about that Dave, I forgot to mention that we were discussing brb's but I knew it as soon as I hit the enter button
Bob

np Bob,

The original post was about Kahl vs Sharp.

I was wondering why Mike was talking albino but the concept was (IMO) correct when applied to Kahl/Sharp albino BCI. In either case, any visual recessive morph that has two possible genetic sources (as in your post) would have the 7/16 of the visual morph from the DH x DH pairing with a 1/16 probability of being homozygous for both.

>>I am sorry about that Dave, I forgot to mention that we were discussing brb's but I knew it as soon as I hit the enter button
>>Bob
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Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC)
0.2 kids (CBB, selectively bred from good stock)

LOL, to many snakes to list, last count:
12.25 BRB
12.14 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

I am trying to absorb all the knowledge I can so I can figure out how I am going to get my anery brb, but if I had some of the brains you had then wouldn't need to know anymore, haha.
Bob

It's the other way around. Sharps tend to hold and gain color as adults. Not to say some Kahls don't but Sharps are more known for this.