I’m new here. I accidently stumbled upon this interesting forum when searching for the true offspring of Schneezwerg. I saw some interesting discussions on types of repeat-flowering. (English isn’t my mothers tongue, so I appologise if I make some mistakes in my grammar.)
I.e. I’ve read that China repeat isn’t the same repeat as Rugosa repeat. I remember I’ve read similar things in the past. I’ve got several questions and suggestions that some of you could shine your light upon.
I’ve had the feeling before that there is a visible/physical difference in the repeat of a “modern repeat flowering rose” (china repeat) and Rugosa roses. Also between several Moschata hybrids. Something to do with repeating on old wood several times (creating new flowering stems on old wood) and creating new flowering stems upon new wood several times. Is this correct?
There are several species roses that are repeat flowering. Could it be that all of the repeat-flowering species in the section Cinnamomeae (R. fedtschenkoana, R. multibracteata, R. rugosa and R. webbiana) have the same type of repeat-flowering? Maybe R. beggeriana (of section Hymnocarpa) also shares this type of repeat-flowering? This rose also has an incompatible type of repeat flowering (I’ve read a book of Delbard, who did some experimentation with R. beggeriana and found him ending up with non-rebloomers after a cross with a good modern repeater.
Rosa moschata seems a different type of repeat. Maybe closer to the chinese type? Some insights I’d greatly appreciate.
I’ve found some clashing info on repeat-flowering in Rosa hugonis, Rosa roxburghii and Rosa californica. I believe them to be non-repeat flowering, but HMF claims otherwise. Maybe Rosa roxburghii plena has some old china bred into it and has some repeat-flowering, but the wild single form hasn’t? Rosa hugonis is maybe mixt up with very similar repeat-flowering hybrids? Is Rosa californica repeat-flowering?
I’ve a suggestion/idea on the incompatibility of repeat-flowering when crossing a Rugosa and a modern repeat-flowering rose. So both are on different genes, that I’ve learned. And both are recessive (not dominant). So when crossing i.e. a repeater with a non-repeater you get (a = recessive gene for repeat, A dominant gene for non-repeat):
aa x AA = 100% Aa
So all non-repeaters but with the recessive gene for repeat-flowering. If you hybridize these:
Aa x Aa = 25% aa, 25% AA and 50% Aa
That is known I saw several posts in this forum that refered to this. It’s basic genetics, I guess it may be a bit more complex in roses.
So with this simple model, if you hybridize a Rugosa and a modern repeat-flowering rose. You are mixing two different genes. So you would get (a = Rugosa repeat, b = China repeat):
aaBB x AAbb = 100% AaBb (non-repeates with recessive genes for both types of repeat)
To get a rose that has both types of repeat (if that is possible). You would hybridize the offspring:
AaBb x AaBb = 6.25% aabb, 6.25% AABB, 6.25% aaBB, 6.25% AAbb, 6.25% Aabb, 6,25% AaBB, 6.25% aaBb, 6.25% AABb, 50% AaBb
If the above is correctly executed (I’m not that good in simple maths) you would get 12.5% Rugosa type repeat-flowering offspring, 12.5% “modern type” repeat-flowering offspring and 6.25% offspring that has both types. Is this a correct assumption (possible math errors not taken into account)?