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pony-tail

Anybody here have good knowlege of genetics ? As concerns line breeding .

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I have a Female Rhadinocentrus rainbow that I was breeding with ( not at present though )

This female throws about 5 in 100 red tailed females ( very unusual ) she also throws 1/2 of the females ( almost exactly ) with a congenital defect in the caudal fin but otherwise healthy fish ( feeders , I am sad to say )

All of her male Fry are normal ( 50% half Red 50% red tail ) The male fry from her are the best coloured ones I have .

Is there anyway of knowing if the male offspring from her will carry the congenital defect as a recessive gene .

Or is the defect solely in the X Chromosome .

I only went to year 12 ( no uni ) so a little over my head .

I would like to breed the Half red males back with the red tailed females to produce a population with half red males and red tail females but without congenital defects .

Just a note - I am also breeding a colony that is just a random selection of multiple colour morphs ( pretty much just like in the swamp )

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So what you're dealing with is a mutation which is causing the congenital defect.

This mutation by the looks of things would have to be recessively inherited (as most are) because you say that only 50% of the females she throws are affected.

What this means is that the mutation is recessively inherited on the X chromosome, and therefore can only affect (create the congenital defect phenotype) rainbows with two copies of the mutation (one on each of its X chromosomes).

Because males only carry one X chromosome, the congenital defect mutation can never be present twice in any male individual, preventing it from ever manifesting in any male.

So to answer your question, if:

- the original female HAS the congenital defect herself, then both her X chromosomes are mutated and all of her male offspring will be carriers for the trait.

- the original females is unaffected by the mutation herself, she is a carrier and therefore 50% of her male offspring will receive her healthy X chromosome and not carry the mutation while the other 50% receive her mutated X chromosome and exist as carriers for it.

From your post I'm assuming your original female does not have the congenital defect, which means we can assume the second scenario to be true and move forward.

Now that we know the original female is a carrier for the mutation, we can deduce that 50% of her female offspring will inherit her mutated X chromosome, while the other 50% inherit her healthy X chromosome (assuming that she is being crossed to a non-carrier male).

This means that you are most likely (ignoring the possibility that the red-tailed gene is somehow linked to the mutation) going to have half of those red tailed females carrying the congenital mutation and half of them 'healthy'.

So in theory what you are wanting to do is simple. Call your new population with mutation free half red males and red tail females population R. You need to prevent the congenital mutation from entering population R by breeding two 'healthy' (ie non-carrier) rainbows. This will create a population identical to the one you have now but minus the mutation.

How do you do that? Well the problem is we cannot tell if any fish is a carrier or 'healthy' just by looking at them. So selecting a non-carrier half red male to breed with a non-carrier red tail female just by staring into your tank and studying them is not possible. What you can do is determine wether a fish is a carrier or not by crossing them to a fish that you know is a carrier.. I.e your original female. The results of that cross will tell you wether or not your 'test fish' carries the mutation.

But this is going to be really time consuming for you. So you are probably best off working with odds. Here's what I would do.

Set up 4 new lines, say A, B, C and D. Each line is a different half red male red tailed female pair. From each line you then study what you see in the offspring. If line A for example results in any fish that have the congenital mutation, then you know straight away know that the mutation was present in one of the two parents that created line A, so scrap them - they are 'unclean'. Repeat until you find a pair that does not give birth to any mutated offspring at all, and once you succeed that will be your population R.. A half red male bred to a red tailed female where the congenital mutation is absent.

Normally you'd need to work through these sorts of genetic problems with these little diagrams geneticists love called punnet squares, so if you didn't follow any of that let me know and I'll draw things out in a much more logical manner, it should make the genetics of it a lot easier to picture in your head. Otherwise happy experimenting! :)

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[MENTION=10706]Robbo89[/MENTION]

FIRST OFF !

Thank you !

First up the original female has no sign of defect .

I did follow what you were saying .

It does sound like it could take a couple of years with growout times and such but does sound very interesting .

Is there a possibility that the red tail and the deformed tail are linked somehow ?

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Set up 4 new lines, say A, B, C and D. Each line is a different half red male red tailed female pair. From each line you then study what you see in the offspring. If line A for example results in any fish that have the congenital mutation, then you know straight away know that the mutation was present in one of the two parents that created line A, so scrap them - they are 'unclean'. Repeat until you find a pair that does not give birth to any mutated offspring at all, and once you succeed that will be your population R.. A half red male bred to a red tailed female where the congenital mutation is absent.

Go easy on me if I'm confusing the situation @Robbo89, as it's been nearly 20 years since I played with punnet squares in 1st year vertebrate biology...

The mutation is recessive on the X chromosome, hence it's not readily visible in males (XY) that may or may not be a carrier, & is visible only in females that have been bred from a carrier mother (XX) or mutated mother (XX), & carrier father (XY). When line breeding in the above example (A, B, C, D), would I be correct in assuming that a pairing of female carrier (XX) with clean male (XY) would still produce an ongoing set of offspring that would be visibly clean, but still contain carrier females (XX) & males (XY)?

So...the process could take more than one generation down any A/B/C/D line, & cross-breeding between visibly "clean" lines, to determine when a legitimately "clean" line had been achieved?

Edited by satchiel

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It'll take a bit of breeding to work out which fish carry the defective gene or not nut you should be able to.find it eventually. I'd suggest drawing up some punnet squares as they are easy to work with and will give you a visualisation of what's going on.

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Go easy on me if I'm confusing the situation @Robbo89, as it's been nearly 20 years since I played with punnet squares in 1st year vertebrate biology...

The mutation is recessive on the X chromosome, hence it's not readily visible in males (XY) that may or may not be a carrier, & is visible only in females that have been bred from a carrier mother (XX) or mutated mother (XX), & carrier father (XY). When line breeding in the above example (A, B, C, D), would I be correct in assuming that a pairing of female carrier (XX) with clean male (XY) would still produce an ongoing set of offspring that would be visibly clean, but still contain carrier females (XX) & males (XY)?

So...the process could take more than one generation down any A/B/C/D line, & cross-breeding between visibly "clean" lines, to determine when a legitimately "clean" line had been achieved?

You are definitely right. Had overlooked that completely. It would still be possible to achieve what is trying to be done but yep it would definitely take more than just a few simple pairings. Would pretty much need to perform a test cross, which is really easily explained in a few different YouTube videos. Basically it's a method used to determine the genotype of an individual, so in this case could be used to identify two 'clean' specimens to be used as population R founders

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You are crossing the line from Rainbows to ornamental fish, in simple terms line breeding reduces your Within-Population Variability. Of course you will keep these to youself

They are kept in Aquaria they can not be returned to the wild - aren’t they already ornamental fish ?

Are they not still rainbows ?

They are not hybrids .

Snapper ck has at least 8 morphs of Rhads , varying from plain silver with black fins . to nearly whole body red , with reds that vary from rusty brown to blood red , then there are the blue morphs , of which there are none in this population . All of these fish are from a single location . ( from a single catch site ) .It is just trying to isolate out a single morph . Why is this a problem ? most of the rainbows that get sold go into display tanks , some just thrown into community tanks . I would not sell them to anybody without informing them as to what they are - and how they came about ,

As soon as you remove them from the wild - possession limit 20 - you have reduced the gene pool and created a genetic bottle neck .

They are a single morph that exists in the wild population - just being isolated out .

I have a breeding population of the wild phenotype - I may be better off waiting for some more of the red taled female morph to grow out see if I can find one that does not throw mutant young .

Unfortunately I can not get back up there to get fresh breeders or new blood ( due to health ) or I would go and get a half dozen red tail females fresh from the wild that hopefully do not have the mutated tail and breed from them instead .

Edit - There is also the issue that most people buying for display tanks only want the pretty red ones - which means they only buy the pretty red ones , but they are not breeding them .

Edited by pony-tail

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It looks like it is going to take more time and Tanks than I have .

I might see if I can get up to Tin Can and get a half dozen or so of the red tail female morph and see If I can start with a mutant free population .

I had not considered the carrier male in my calculations and what it would take to eliminate the hidden faulty x gene from the male breeders .

Thanks all for your assistance

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Quick question did any of the red tail females have the deformity.

I only ask as it might help eliminate the linking of the red tail colour to the deformity ( ie they are on the same gene)

Edited by aussieorchid
Typo

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In that case I'd say that they aren't linked, because with such a low frequency rate you're looking at the two genes being pretty far apart (ie on different chromosomes) and therefore very frequently being recombined to eliminate gametes where the two traits exist together = statistically working against creating red tailed female gametes and reducing the number of such fish (only 1)

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It looks like it is going to take more time and Tanks than I have .

I might see if I can get up to Tin Can and get a half dozen or so of the red tail female morph and see If I can start with a mutant free population .

I had not considered the carrier male in my calculations and what it would take to eliminate the hidden faulty x gene from the male breeders .

Thanks all for your assistance

Would definitely not be an easy endeavour, but to say it would be interesting would be a massive understatement. Something worth trying for sure

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Would definitely not be an easy endeavour, but to say it would be interesting would be a massive understatement. Something worth trying for sure

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The red tailed females occur at a very low percentage in the wild population I only wound up wit one in my original breeders .

My thought was to go back and collect 6 or so of just the red tailed morph to see if I can get one with the red tail without the crunched tail mutation ( which is also in the wild population in small numbers )

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Yeah, that's one way that could work without a doubt. Will be interested to see how it goes so you should keep us updated!

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You could try and get a female without the deformity and red tail. If you can not get another female with the red tail trait another option is to try another male. If the male does not carry the deformity gene only approx 25 % of the babies would have the gene but none of the females would show signs of the deformity. If you were successful in obtaining the red tail trait you would still then need to do a few more crosses to weed out the deformity gene. Final test would be to breed a male who carries the deformity gene with your potential clean female. If none of the offspring show the deformity trait you should have been successful but this batch would then have carriers of the gene so do not breed from them or you risk contaminating the gene pool again.

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