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

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