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Is this something that has been observed? I have been raising Short Fin Albino bushynose for year now. After raising a large group from a club member in a 125 gallong tank for 1 year, I decided to take them out of the commnity tank to breed. When I was draining the tank I found 1 brown BN fry that was 2 inches long? Confused since all I have are albinos, didn't think that dual recessive can produce a heterozygous or dominant gene fry? I ask another fellow member and he said that common?... but now I have one isolated pair in a 15 gallon tank with some anubias, 10 albino Lamprologous birchardi fry and 8 Julidochromis marlieri (3/4" long) They were powerfed for 2 weeks and I added a terracotta pot for the male. After 1 week I did a waterchange and the next morning they was a clump of eggs like I expected. now the fry are 8 days post spawn and have absorbed most of their yold sac, they are obviously not albino, I was a bit concerned when their eyes were developing since they looked black, but now i am sure since their bodies are grey? Has this happened to anyone before, doesn't seen possible with parents that are both dual recessive? Any help would be awesome!!!

Hello Mouthbrooders, welcome to planetcatfish!

What you've seen, two albino parents producing all brown fry, is unusual, but i'm aware of it happening to several other fishkeepers.
Personally I'd guess this is due to the the albino allele originating at two different loci in the two albino parents, since these loci do not match up during mating (and they are homozygous wild type brown at the alternative loci), you end up with a batch of fry wth a genotype that is heterozygous for the albino allele at both loci, and an entirely brown brood due to the dominance of the brown allele over the albino allele at both sites. I hope you understand what I mean, It easier to explain with a diagram and a punnet square really.

Crossing the young should result in a number of albino fry, but the ratios are hard to guess due to the possibility of "crossing-over" (recombination). Also they way the two loci interacact would have to be taken into consideration, but assuming (!)brown is epistaic over the albino homozygous at the other locus, and no recombination takes place, mating two of the young should result in one 16th of the fry being albino and pure breeding with albino siblings for further generations.

Bear in mind however, this two loci theory is entirely out of my head and I have no way of proving it, it just appears to me to be the most obvious answer to what is going on, but there may be something entirely different at work!

Cheers, Conrad

Also, have a look at this thread http://www.planetcatfish.com/forum/view ... e+genetics

Actually, I think Loracidlover is a bit too pessimistic

If we follow Mendelian rules (and you were right observing a few problems with that) we only need to recombine one of the albino genes. I maen, say the father has his albinism on gene A, and the mother on gene B, we would expect 1/4th of the fishes to be A albinos, and 1/4 th of the fishes B albinos (but one would not be able to see, unles using expensive genetics) which of the fishes is an A of B albino

Further, one would assume 1/16th of all fry would be AB albinios - having only albino genes on both loci

Thus part of the A albinos are also B albinos, and thus one can not say 1/4 + 1/4 = 1/2. It would be 1/4 + 1/4 - 1/16 = 7/16 th. almost half

However, Mendel has pend a lot of time before finding a system which gave numbers he could use. In other words, quite often the Mendelian rules don't apply (better put: they apply only after a lot of other considerations)

Bas may well be right, but I'm sure he would admit, as I do, that predicting how the two loci will interact is something of a guessing game, my *guess* would still be that a brown allele at one loci would be expressed in the phenotype, and homozygous recessive at both loci would be needed to cause a albino phenotype, why no do the crosses and see!

Cheers, Conrad

It's hard to guess, that's right, but we know what homozygote alnino A and homozygotous brwon B does look like, as that is one of the parents (and the other is jhust the same)

So, within the Mendel rules (to keep matters simple) as soon as eithe A of B is homozygotous for albinism, the fish will be an albino

(therefore, if someone would be able to produce AaBbCcDd (Big A means bronw on allel A, small a means albina on A, and so on, thus having 4 traits of albinism in its blood) brown Ancistrus, crossing them within themselves would produce around half of the fry albinos, according to the Mendel Rules)

Now I better answer the question why an AABBCCdd fish must be an albino. That's not that hard: A, B, C and D allelles are supposted to set 1 step in the production of the brown color (making it, dispersing it, whatever) the a, b, c and d alleles don't do this, they are either defective or missing, and a gene can be seen as a blueprint for a feature. Well if an essential blue print is missing, the house will not be build (or not properly)

The same goes for the natural brown color

How do I know the A, B, C and D are essential? Because a fish homozygotous for aa, bb, cc or dd is an albino (after all, that is how I definied them)

I had a similar problem- a pair of long fin albinos that threw only 100% browns. I can say for sure that the offspring threw a mix of browns and albinos, but almost no long fins.

Thanks for all the info you guys, I am gonna split the pair and breed them back to another Albino and let's see who the culprit is? If all my Albinos I am growing are gonna do this I will be out 1.5 years of grow out, damn!

It should not take anywhere near 1.5 years. I had the last of the remaining brown offspring from the alfs parked in a 10 gal. They were between 6 and 9 months old when I got my first surprise spawn in the sponge filter. I had another batch parked in a 15 and had a similar result from about 9 month old offspring.

This was the second in sponge spawn- notice the albino fry from the first spawn below the sponge skeleton.