Cubing.......a myth.
Here's breeder chimera's take on the subject:
"you’ve just discovered the biggest myth (IMNSHO) of marijuana breeding- it is a mistake that almost EVERYONE makes (including many of the most respected breeders!).
Backcrossing will not stabilize a strain at all- it is a technique that SHOULD be used to reinforce or stabilize a particular trait, but not all of them.
For e.g.- G13 is a clone, which I would bet my life on is not true breeding for every, or even most traits- this means that it is heterozygous for these traits- it has two alleles (different versions of a gene). No matter how many times you backcross to it, it will always donate either of the two alleles to the offspring. This problem can be compounded by the fact that the original male used in the cross (in this case hashplant) may have donated a third allele to the pool- kinda makes things even more difficult!
So what does backcrossing do?
It creates a population that has a great deal of the same genes as the mother clone. From this population, if enough plants are grown, individuals can be chosen that have all the same traits as the mother, for use in creating offspring that are similar (the same maybe) as the original clone.
Another problem that can arise is this- there are three possibilities for the expression of a monogenic (controlled by one gene pair) trait.
We have dominant, recessive, and co-dominant conditions.
In the dominant condition, genotypically AA or Aa, the plants of these genotypes will look the same (will have the same phenotype, for that trait).
Recessive- aa will have a phenotype
Co-dominant- Aa- these plants will look different from the AA and the aa.
A perfect example of this is the AB blood types in humans:
Type A blood is either AA or AO
Type B blood is either BB or BO
Type AB blood is ONLY AB
Type O blood is OO.
In this case there are three alleles (notated A, B, and O respectively).
If the clone has a trait controlled by a co-dominant relationship- i.e. the clone is Aa (AB in the blood example) we will never have ALL plants showing the trait- here is why:
Suppose the clone mother is Aa- the simplest possibility is that the dad used contributes one of his alleles,
let us say A. That mean the boy being use for the first backcross is either AA or Aa. We therefore have two possibilities:
1) If he is AA- we have AA X Aa- 50% of the offspring are AA, 50% are Aa. (you can do the punnett square to prove this to yourself).
In this case only 50% of the offspring show the desired phenotype (Aa genotype)!
2) If the boy being used is Aa- we have Aa X Aa (again do the punnett square) this gives a typical F2 type segregation- 25% AA, 50% Aa, and 25% aa.
This shows that a co-dominant trait can ONLY have 50% of the offspring showing the desired trait (Aa genotype) in a backcross.
If the phenotype is controlled by a dominant condition- see example #1- all 100% show the desired phenotype, but only 50% will breed true for it.
If the phenotype is controlled by a recessive condition- see example #2- only 25% will show the desired phenotype, however if used for breeding these will all breed true if mated to another aa individual.
Now- if the original dad (hashplant) donates an 'a' allele, we only have the possibilities that the offspring, from which the backcross boy will be chosen, will be either Aa or aa.
For the Aa boy, see #2.
For the aa boy (an example of a test cross, aa X Aa) we will have:
50% aa offspring (desired phenotype), and 50% Aa offspring.
Do you see what is happening here? Using this method of crossing to an Aa clone mother, we can NEVER have ALL the offspring showing the desired phenotype! Never! Never ever ever! Never!! LOL
The ONLY WAY to have all the offspring show a Aa phenotype is to cross an AA individual with an aa individual- all of the offspring from this union will be the desired phenotype, with an Aa genotype.
Now, all of that was for a Aa genotype for the desired phenotype. It isn't this complicated if the trait is AA or aa. I hope this causes every one to re-evaluate the importance of multiple backcrosses- it just doesn't work to stabilize the trait!
Also- that was all for a monogenic trait! What if the trait is controlled by a polygenic interaction or an epistatic interaction- it gets EVEN MORE complicated? AARRGH!!!!
Really, there is no need to do more than 1 backcross. From this one single backcross, as long as we know what we are doing, and grow out enough plants to find the right genotypes, we can succeed at the goal of eventually stabilizing most, if not all of the desired traits.
The confusion arises because we don't think about the underlying biological causes of these situations- to really understand this; we all need to understand meiosis.
We think of math-e.g. 50% G13, 50% hashplant
Next generation 50% G13 x 50% g13hp or (25% G13, 25%HP)
We interpret this as an additive property:
50% G13 + 25% G13 +25% HP = 75% G13 and 25% hashplant
This is unfortunately completely false- the same theory will apply for the so called 87.%% G13 12.5% HP next generation, and the following 93.25% G13, 6.25% HP generation; we'd like it to be true as it would make stabilizing traits fairly simple, but it JUST DOESN'T work that way. The above is based on a mathematical model, which seems to make sense- but it doesn't- we ignore the biological foundation that is really at play.
I hope this was clear, I know it can get confusing, and I may not have explained it well enough- sorry if that is the case, I'll try to clear up any questions or mistakes I may have made.
Have fun everyone while making your truebreeding varieties, but just remember that cubing (successive backcrosses) is not the way to do it!
-Chimera"
