"is that once you backcross to the parent that has the recessive trait you want, is that you kind of skew the whole cross towards the recessive trait plant, do you not? "
Yes but doesn’t matter cuz ur a breeder, so you grow seeds out and find what you want cuz you know what you want has just been created and is in one of those seeds. These are just guides or outlines. You need plant/structure/profile in hand first and a desired end result to know which methodology would be best to use…
Yeah punnet talk, but i love it
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Not to tut the selfing horn. But its one of the most effect ive way to deal with more delicate crosses like uber Recessive traits like duck foot. What you do is F1, F2 in F2 that trait will show, when you have selected the F2 parents you can self them. You will generally see a range from 25 - 75% of the s1 with uber traits. You keep selecting to S3 for those traits. This is best done with 2-4 planta from the F2 generation. Then you cross (F2aS3 x F2bS3) x (F2cS3 x F2dS3) this will give you vigor back and you will keep all you freaks intact. I hope i sparked some ideas for you 
Pz my man
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Haha holy smokes! Well first off I’m one of the ‘traditional pollination’ kind of guys. Something about selfing just doesn’t sit right with me. It’s one thing if the plant does it, but it’s another if I force it to.
One of the reasons that I want to combine them in this new way is that I have it on good authority that Freakshow and ABC do not blend the leaf types together. They tend to separate into their own types. As for other mutant crosses I have a blended Duckfoot and ABC plant. It’s in the F4 generation and has webbed ABC leaves. Really cool but they grow damn slow like ABC.
The traditional F1+F1=F2 and then pheno hunting F2 doesn’t work for the Freakshow+ABC cross. I thought by backcrossing to the Freakshow female (in my case BerryFreak) it may increase the likelihood that I can blend the two mutants.
If you can show me that selfing would get me to that ideal cross faster I’m all ears! You’ve sparked more questions than ideas my man 
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@crunkyeah Some mutations are in the same chromosome and they cant be blended. Dunno If freakshow and abc have this Recessive trait in the same chromosome but its a chance that is the case. I would be awesome If you could do a diary If your project. You can treat BXing as selfing aswell. Its not as efficient as selfing but it does the same thing… Isolating genes. So instead of selfing you could BX 4 F2s until they are stable enough and then recombine them 
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Will do! The F1s are going to be created whenever they begin flowering outside. Honestly doing that BX4 thing sounds way too intensive. The more I think about it the more I like Sebring’s idea. I will have to think on this, as this is only one of the crosses I want to work with. The one that I’m more excited for is a mutant x landrace possibility.
I’ll have to reread some of the materal I’ve read on selfing. Always something to think about when I read this thread… haha…
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No he said backcross four of your f2 plants. Bx1 4x plants. Not do bx4.
I find it interesting that chart said f1 x bx1 = f2. That’d be bx2 unless you’re using a different parent, then it’s an inline cross 
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Ahh ok, that makes sense then. BX4 seemed redundant that’s why I was a little confused. The reason for 4 specimens is to not bottleneck the strain?
The reason I did it that way was because of the original quote by Mithridate. It was outlined in that post and I copied it in my little drawing. I thought it was odd too. I didn’t understand how they considered a BX1 the same as an F1. 
Actually it would seem like a brand new F1 to me. It can’t be a BX2 because BX1 would have to be BX’d to the original P1. Or am I wrong?? BX1+F1= ??
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Oh you’re right! Duh 
so its an Inline Cross at best… or an F1 BX1. Instead of P1 BX1 
that’d be like me taking the ssdd f2 bx1 back to the Starlite DayDream/StarShine dad instead of the ssdd f2 mom. You’re backcrossing to the outcross… I don’t really see a reason for that unless you really want something from the outcross in there
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Here’s a link to Sebring’s post
Cannabis crosses are not f1s thread on og
Cannabis crosses are not f1s thread on the mag (different line of discussion)
The bc1 x f1 = f2, had me staring at the screen for a minute too. I know cannabis breeding nomenclature doesn’t always reflect the official botanical one. I brushed it off thinking my broscience had tainted my comprehension and moved on haha
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BX1+F1= ??
=F3 or higher
You expect a 9:3:3:1 in f2’s but that’s not what you get
Yeah BX1 x F1 is not F2. Because F2 is a generation term which specify where in the process you are and what you have done to get there. I think P1 BX1 x F1 would probably be it. But i dont really get why you would want to do a BX1 x F1? BX1 will have alot of Recessive traits present which will be present once again in the children of P1 BX1 x F1. I think going to F2 then split the line and bx to recombine would be better. That just my opinion 
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I find myself guilty of this same trait most of the time when it comes to plant genetics. I think the big thing to take away is, no matter how you describe what you’re doing with letters it’s about amplifying and expanding on recessive genetics. If two multi-polyhybrids don’t make an F1 it doesn’t matter to me as long as I can describe the process of combining two different mutant leaf types together. Sometimes the breeding lingo makes my head spin when I think about outcrossing, selfing, recessives, and so on. For most people, they’re looking at regular strains and improving things like terpenes, THC content, maybe trace CBD/CBG etc… it just depends.
In the case of bc1 x f1 = f2 I think it’s because Sebring was trying illustrate that the bc1 is similar to an f1 in terms of passing on desired recessive traits. I will have to take a look at those links when I have a moment. Thank you for sharing them! I appreciate it and I’m sure others will to.
With any F2 cross, I actually expect something like 20:20:20:10:10:5:5:4:3:2:1, regardless what punnett squares say. Plants are too complex to have 75/25 or 50/50 splits imo. The thing with a punnett square is they only show certain traits. The more traits you add to the square, the more complex and infinite the possibilities become. Let’s say you have a green bud, floppy stem, super resinous, low yielding landrace sativa, and you breed that to a firm stem indica with insane root growth, pink stigmas, and really dense hashy buds. If you breed these two plants together you’re going to find everything under the sun in just the F1 generation, as I see it. When you select F1 specimens to continue to F2, you’ll have an extremely large amount of unique specimens to search through with those since the recessive traits begin to show. This is where breeding is a numbers game.
I’ve thought about this over and over, it doesn’t make sense to me too. In theory you should be able to find both a male and female specimen in the BX1 to breed those together to get your theoretical F2.
In this example a purple male and female are selected from the BX1 generation to find 100% purple phenotype F2s. I’m not 100% on this either. I need to think some more. 
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Don’t mind me spouting nonsense, total layman. But I’m here to learn. As far as I understand, Bx x F1= F2 makes sense if you want to improve the line, not create a new one
Let’s say, I’m completely satisfied with the line, but I want it to be beautiful as well. To preserve its characteristics, but to make it purple.
a=purple (recessive)
A=green (dominant)
aa(P1) x AA(P2) = F1 => aA aA aA aA
aA(F1) x AA(P1)= Bx => aA aA AA AA
Now, one option would be Bx x Bx. It would better represent the chosen parent, but considering that the desired trait is recessive, it would be necessary to guess a lot with both parents. And in Bx x F1 you need to guess only one(Bx)
And when you find a Bx parent, the combinations for that trait are
aA(Bx) x aA(F1) => F2 aa aA Aa AA
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aA(Bx) x aA(F1) => F2 aa aA Aa AA
and what are the ratios you will see those offspring?
is it 1:1:1:1?
a Bx creates 1:1:1:1
an f2 is 9:3:3:1 (1:3:3:9 respective of his example)
Law of Segregation says if you cross and don’t get a 9:3:3:1 then you haven’t created an F2 have you.
Law of Segregation says if you backcross and get something different than 1:1:1:1 then you haven’t made a backcross. i didn’t write these rules guys, potential allele combinations dictate what they should be called.
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9:3:3:1 distribution is applicable when you consider two traits
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Very good…so after a Bx x f1…you probably dont have 2 traits anymore do you? …in fact in my experience a Bx x F1 produces a very uniform line of offspring.
I should reword that i think…regardless, the point is ur not making an f2 with a bx x f1…ur making something higher…idk exactly what. That’s why i said f3 or higher in my first response to the topic
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I am a little confused…
Why stick to 9:3:3:1 as a benchmark?
If P1 and P2 are two separate lines, they will surely differ in several traits.
The ratios for the three traits in F2 are 27:9:9:9:3:3:3:1
Why not stick to that ratio? Or for 4 traits…
You apply ratios for the characteristics you observe
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Why stick to 9:3:3:1 as a benchmark?
Cuz that’s what discerns punnet square from real world results…that’s the rule (law of segregation) output. Did i mess up somewhere? Does bx x f1=27:9:9:9:3:3:1?
“Yeah Mr White! Yeah Science!”
A lot of this is beyond me but quite interesting.
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I fkn hate it…lol
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