L144 X A. sp.(3)

[craigfries]

In my attempts to breed L144 I bought 7 young fish. Five turned out to be males. One was set up with the two females. The other four are very robust so I decided to investigate their relationship to the common albino Ancistrus by housing them with six gravid female A. sp.(3.) The first spawn occurred almost immediately and they just now hatched. So far the wigglers look like brown A. sp.(3.) While this is the result I was told to expect, I still do not understand where the gene for normal (brown) came from. What can be expected from the spawn of these fry with their sibs?

TIA,
Craig

[Borbi]

Hi Craig,

this is what I merged for myself from a lot of publications regarding this topic:

At first- what is sold as L144 is (at least here in Germany, don´t know if this holds true for your place) in fact not L144. The "true" L144 is nearly lost for the aquarist. What is sold as L144 is a xanthoristic (is this an english word..?) mutant of Ancistrus sp. "brown" or A. sp.3 as it is called here at PC.
The Albino Bristlenose is an albinotic mutant of Ancistrus sp. "brown" (sorry, I´m too used to that name).

I am by no means a molecular biologist. But as far as I have learned, the mutated information for xanthoristic/albinotic is located at different spots in the genetic information. Also, both mutations are recessive.
Therefore, by mixing any of these with the "natural" form should produce 100% naturally coloured fry, since 50% of the genetic code for every fry comes from the naturally coloured parent. Unless, of course, the natural coloured one has some suppressed albinotic/xanthoristic information in its genetic code.
All this is according to the rules of Mendel. By applying these rules, you can expect roughly 25% xanthoristic fry if you have your actual fry spawn for you. All the rest is supposed to be brown, again.
But you should not be able to generate albinotic mutants by simply crossing those two colour mutants, since the genetic information is located at a different place in the genetic code. Interestingly, according the things said above, the spawn of an Ancistrus sp. "Albino" with a "L144" (which is not L144, in fact) should produce 100% naturally coloured fry. And, as I have learned from a friend, it actually does.

Hope, that helps..?

Greetings, Sandor

[Shane]

Sandor is correct. To my knowledge, L 144s never made it to the US from Europe (someone correct me here if I am wrong) or if they did, it was in very, very small numbers. US hobbyists are starting to use L 144 for albino and/or amelanistic Ancistrus sp. 3 which is incorrect. You crossed albino or amelanistic A. sp. 3 with normal colored A. sp 3 and got more normal colored A. sp. 3. No surprise there.

OK, so why are so many aquarists not getting the mixes that Mendel promised? We all know about dominant and recessive genes and how things should work...

I offer the following as a an explanation. I freely admit that I have no scientific data to back it up, but believe it is both possible and plausible.

The basic theory is that the characteristics are not passed on because they originate from environmental factors and not line breeding to fix a genetic trait/mutation. Asian fish breeders have been known for years to expose fertilized fish eggs to radiation in order to increase the chances of mutations cropping up. I believe that many of the mutations associated with A. sp 3 may have been developed in this manner and thus do not breed true in the numbers we would expect.

Comments welcome.
-Shane

[MatsP]

I have had two crosses of "normal" bristlenose that breed to albinos. In each case, I got a handfull of albinos. Roughly, I'd say they are 25%.

But just like if you have four children, you statistically will get 2 girls and 2 boys, we all know some families with four girls or four boys, 1 boy, 3 girls, etc, etc.

Statistically you will only get 25% albino's if you have a VERY LARGE number of samples (i.e. lots of babies to count).

Same way if you toss a coin 4 times, you should get 2 heads and 2 tails, but I just did: 3 heads, 1 tail. Of course, if I continue to toss the coin, I'll eventually, end up with a near the same number.

Also, to Shane, I'm reading the original post as the females being ALBINO A. sp(3). I agree with Sandor that the albino gene and the amelanistic one isn't the same gene, so the brown ones will carry one or both the albino & amelanistic genes, but not two pairs of either.

So: a = albino gene (recessive), A = not albino (dominant)
b = amelanistic gene (recessive), B = not amelanistic (dominant)

Mother (albino): aaBB
Father (amelanistic): AAbb

Children: aAbB & AaBb - brown ones.

Cross them back, and you'll get a combination of several different permutaitions like aabb, AABB, aAbB, and so on. Some will be albino, some will be amelanistic, some will be amelanistic and albino and some will be brown. You'd need to draw up a chart of which ones get what from which side to know which combinations are occuring how many times - there will be 16 combinations possible, but from any pair of parents, I think you'll only get 4 different "actually different" combinations, but there will be a larger number of variants that will carry one gene or another.

--
Mats