Re: HYB: punnett squares for TBs (Facebook discussion)
iris@hort.net
  • Subject: Re: HYB: punnett squares for TBs (Facebook discussion)
  • From: C* C* <i*@aim.com>
  • Date: Sat, 8 Feb 2014 08:32:20 -0500 (EST)
One of the special features of tetraploid genetics is the possibility of doubling up on a gene that is present only once. something that is not shown or predicted in a punnett square.
This happens as chromosomes are doubled during meiosis, so you are choosing from 8 choices, not four. So you can, for example, get a plicata from a cross where there is only one plicata gene in each parent. Kidd calculated this to be 1/576, if plicata genes is at a distance from kinitochore of the gene it is on. Lower if it is close to kinitichore. Kidd used some data from Keith Keppel to show that it does happen and is not just a theory. 

Chuck Chapman



-----Original Message-----
From: Linda Mann <lmann@lock-net.com>
To: iris <iris@hort.net>
Sent: Fri, Feb 7, 2014 5:21 pm
Subject: [iris] HYB: punnett squares for TBs (Facebook discussion)

  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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