I was going through my own examples in some previous posts above and noticed I made a major mistake. Specifically I switched dominant/recessive and donor/parent in both the breeding in the dominant and recessive trait examples. Rendering the rest of it invalid. The basic principles above still apply but the examples were rather long so I asked Seamaiden to remove them (thanks!).
I will keep it shorter this time:
Cubing is not a myth, some supposed outcomes of backcrossing are.
In addition to using it for specific purposes depending on what you got to work with (the three above), you can't get achieve those three goals 'combined'. For example, you can't backcross a trait into a variety AND make it homozygous for that trait by backcrossing along. You can't save the genes of an elite cut from a hybrid AND at the same time make it homozygous by backcrossing alone.
The goal of the Grimm bros with c99 was not to backcross a specific trait into another variety. They sort of went for the first example in my post above, and ended up getting results more like the third example.
I made the same mistake as in the post quoted by the op of this thread, in my examples that is. Backcrossing a trait into another variety is not to be mixed with the goal of cubing a la c99.
Whether the donor trait is recessive or dominant it requires both selection and additonal steps to get at 100% uniformity for that trait, and even more if it needs to be bred true.
http://passel.unl.edu/pages/animation.php?a=BXbreed.swf&b=990818773
That's irrelevant for cubing as in backcrossing a population to make it as homozygous as the recurrent parent.
Cubing example: (H just being an example, applies to all others too).
Step1– Recurrent Parent [HH] × Donor Parent [hh] = [HHx hh] = F1 Hybrid generation [all Hh]
Step 2 – Cross the best of those F1 (Hh) plants to the recurrent parent [HH]. The generation produced is denoted BC1. [50%HH, 50%Hh, all same phenotype.]
Step 3 – Select plant
s from BC1 and cross with the recurrent parent; the resulting generation is denoted BC2. [take for example 8 males from BC1. Throw all the BC2 seeds you get from pollinating the recurrent parent together on a pile. Half of it will be HHxHH, half will be HHxHh, which means of 75% of the pile is HH and 25% is Hh. ]
Step 4 – Select plant
s from BC2 and hybridize with the recurrent parent; the resulting generation is denoted BC3. [select any 8 males from the mixed BC2 pile. 75% of the BC3 will be HHxHH and 25% will be HHxHh, so you get 87.5% HH in the mixed BC3 pile]
Step 5 – Grab 8 males again from BC3 and hybridize with the recurrent parent; the resulting generation is denoted BC4. As you probably guessed by now, the resulting pile of seeds will have 94% HH.
Those percentage are similar to P.50, P.75, P.88 (as in the Cindy example) but represent something different. Those P.xx refer simply to how much of Princess' genes are theoretically in the back cross, like so:
Princess Clone
x (
SSx
JH/?
) =
P.50 (50% Princess)
Princess Clone
x
P.50
=
P.75 (75% Princess)
Princess Clone
x
P.75
=
P.88 (87.5% Princess)
Princess Clone
x
P.88
=
P.99 AKA
Cinderella 99 (94% Princess)
Chimera refers to those percentages in the same context, i.e. how much of the genes of the recurrent parent are in the backcross, which indeed doesn't mean they actually express those genetics in the same way as the recurrent parent as that depends on the genotypes. But once you start using a lot of males (or females) to cross back to the recurrent parent the percentage of homozygous/truebred genotypes/traits increases in the same way in the resulting "combined" offspring. Both a homozygous (especially for the traits that matter) recurrent parent and using multiple plants from the backcross generations are required.