Phrag. (Fritz Schomburg x Waunakee Sunset 4N) plus genetics quizz!

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besseae 67%
kovachii 16,70%
fischeri 16,70%

That'll be 100,40%. I know I'm a pea counter.

As of your own counting, besseae has 2n=24, fischeri 2n=30, kovachii 2n=32

The Barbara LeAnn would have a 2n of 27, right? 4n would be 54, paired with a 4n besseae with 4n=48. What is the actual count for the 4n Waunakee Sunset? The Fritz should have 2n=28, your hybrid should have something between 2n=38 and 2n=40.
What I wanted to say is: there are possibly less genes from besseae in the mix than the simple equation would suggest. On the other hand, you chose a Waunakee Sunset which favours besseae, so: who knows.

as for the flower: the cross makes sense to me, but I agree with you on this flower: colour is interesting. The Waunakee Sunset is a cracker, though.
 
and of course, the individual chromosomes also break and recombine with each other during each cycle of crossing so what the actual % of each species is in each individual hybrid plant is anyone's guess.
You'd need to cover each chromosome in genetic markers that distinguish between each species to work that out!
David
 
That'll be 100,40%. I know I'm a pea counter.

As of your own counting, besseae has 2n=24, fischeri 2n=30, kovachii 2n=32

besseae 66,666666666666666… % or 2/3
kovachi 16,66666666666666…. % or 1/6
fischeri 16,66666666666…..% or 1/6

Now it is = 100.000000.....%


Is it better for you?

(BTW I also believe the actual % of each species in each individual hybrid plant is anyone's guess. Answering to this quizz was only a game for me)
 
No insult intended, I too took it as a game. I was thinking about a "next level": considered that each chromosome pairing results in 50% of either parent, a cross of besseae with n=24 and fischeri n=30 would result in n=27, with 24 chromosomes having 50% of besseae and fischeri, plus 3 chromosomes with 100% fischeri. You'll get ca. 45% besseae and 55% fischeri in Barbara LeAnn. Theoretically.
 

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