HYB: punnett squares for TBs (Facebook discussion)
iris@hort.net
  • Subject: HYB: punnett squares for TBs (Facebook discussion)
  • From: L* M* <l*@lock-net.com>
  • Date: Fri, 07 Feb 2014 07:49:38 -0500
Kat West is trying to learn genetics, figured I'd store some of the discussion here for future reference. 


Kat West
Question: I bred a non-plicata/luminata to a Luminata and in the F1 generation got all solid maroon colors with hafts. I backcrossed the Luminata parent to one of the F1 seedlings and in the F2 generation got a lot more of the solid maroons, but also the one pictured Luminata. My simplified Punnet square calculation says I should get about 25% Luminata in the F2 generation. What I got was 12%. Did I get lucky with this one seedling or is my Punnet square calculation roughly right?
Tom L Waters The actual probability is 1/6, assuming random pairing, so your results are consistent with that 
    17 hours ago 7 Like 7 1
Kat West Yoohoo! Tom L Waters , any chance you can explain how you calculated the probability? I'm trying to teach myself genetics off of the hort.net list and really appreciate being able to check my assumptions here. 
    16 hours ago 7 Unlike 7 1
Tom L Waters Tetraploid probabilities take a little getting used to. I'm going to use A for the dominant allele (in this case self), a for the recessive (luminata). Your F1 seedling is AAaa. Since the probability of a single a is 1/2, you might think getting aa would be 1/2 x 1/2 = 1/4. But after you have the first a, what you have left to draw from is two As and only one remaining a. So the chance of getting that second a is 1/3. Hence aa = 1/2 x 1/3 = 1/6. This is also the probability of AA. The probability of Aa is the remaining 2/3. Since the luminata parent always gives aa, the seedlings should be 1/6 aaaa, 2/3 Aaaa, 1/6 AAaa. Make sense? 
    16 hours ago 7 Edited 7 Like 7 2
Linda Mann This was the part I had the <most> trouble getting to click - there are 36 possibilities when making a cross, not 16 as it first would seem. 
    15 hours ago 7 Like 7 2
Kat West Thanks Tom L Waters, I understand your calculation and it makes sense. So maybe you can help me understand where I'm going astray? When I plug the F1 generation (say for example, SSSS is for dominant solid and uuaa is for the luminata phenotype) into a Punnet square where the F1 could be SuSa or SuSa or SaSa and cross it with luminata which I show as uuaa, the calculation still shows 25% luminata in the F2 generation. Arggh, obviously one (or all) of my assumptions is wrong, know which one? 
    11 hours ago 7 Like
Kat West Linda Mann, are you saying my Punnet square should have 36 boxes, not 16? [yes, I was] 
    11 hours ago 7 Like 7 1
Tom L Waters Hi Kat - I wasn't considering the presence of pl-a, although we know it is there in luminatas. Can't be sure how many doses of it your original luminata parent has, but let's assume 2 as you have done. Then the F1 will be 1/6 SSuu, 2/3 SSua, 1/6 SSaa, all of which appear to be selfs. If the one you used was an SSua (most likely), then its gametes should be 1/6 SS, 1/6 ua, 1/3 Su, and 1/3 Sa. If you cross this with a luminata (again assuming uuaa), then 5/6 will have S and be selfs, and 1/6 will have only u and a, in some combination. All will be luminatas, because there is no way to get an aa gamete from the F1 parent unless it was SSaa, which was not what we assumed. If it were an SSaa, 1/6 of the seedlings would still be without S, but now 1/6 of those (1/36 of entire cross) would be aaaa glaciatas, the remainder (5/36 of entire cross) would be luminatas. The difference between an SSua F1 and an SSaa F1 would thus be very hard to detect - you'd either have to get lucky and bloom a glaciata from the cross, or raise a few hundred seedlings to be sure no chance of a glaciata was present. Either way, you'll be getting about 1/6 luminatas. 
10 hours ago 7 Like 7 2
    10 hours ago 7 Like 7 2
    Ann Borel I wish there were a way to save posts like this.
    10 hours ago 7 Like 7 1
Tom L Waters I don't know how you are getting 25%. Think of it this way: your F1 has to be SSxx (where x can be either u or a). Only 1/6 of its gametes will be xx; the other 5/6 will have at least one S and produce selfs. 
    10 hours ago 7 Like
Tom L Waters Regarding Linda's comment - yes, a Punnett square for a tetraploid is 6 x 6. 
    10 hours ago 7 Like 7 1
Tom L Waters Here's why. Four chromosomes, which we can call a, b, c, and d. How many unique pairs can we make from those four? Just count the combinations: ab. ac. ad. bc, bd, cd. That's six. 
    10 hours ago 7 Like

Linda Mann

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