Mathematics and statistics for self-pollinating plant breeding

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Doffer
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Mathematics and statistics for self-pollinating plant breeding

#1

Post: # 38585Unread post Doffer
Sun Jan 17, 2021 8:32 am

Every self pollinating generation is giving more homozygous genes. To understand this I was doing some math.

Parent one P1 is homozygous for the dominant AA gene.
Parent two P2 is homozygous for the recessive aa gene.

Crossbreeding P1xP2 is giving:
F1.jpg
We see all F1 plants are uniform and have the genes Aa (uniform genotype in then F1 generation).
To get the next generation one F1 is self-pollinated. The number of plants to grow to get all the segregation is 4^genelocation (in this case 4^1= 4 plants):
F2.jpg
Now there is segregation in the F2 generation. 25% is homozygous for the dominant gene, 25% is homozygous for the recessive gene and 50% is heterozygous. The phenotype (what the fruit looks like) is similar for all the plants that contain the dominant gene A. Hence the change to select the dominant homozygous plant is 1/3 is 33%.

We notice the amount of homozygous gene combinations is 2^genelocation = 2^1= 2 combinations (AA or aa)

If all four F2 plants are selfpollinated, there is a F3 generation of 16 plants:
F3.jpg
From the 16 plants there are 6 homozygous AA (dominant) plants (37,5%), 4 heterozygous plants (25%) and 6 homozygous aa (recessive) plants (37,5%).

Conclusion at this point:
From F1 with 0% homozygosity there will be 50% homozygosity in the F2 generation and in the F3 generation there is already 75% homozygosity. Hence every generation there is change for more homozygosity.

Lets calculate what will happen when this is going on till the F10 generation:
F10.jpg
Here we see that in generation F8 the change is bigger as 99% to get a homozygous plant. Only 50% of these plants are homozygous for aa and the other are homozygous for AA.
If u want a plant that have the dominant AA gene homozygous, selection is the only option. When every generation get selection and the homozygous aa plants are removed from the breeding program we see that at generation F8 there is 98,4% change to select a homozygous AA plant.
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Re: Mathematics and statistics for self-pollinating plant breeding

#2

Post: # 38586Unread post Doffer
Sun Jan 17, 2021 8:35 am

What will happen if u want to select for two genes:
Parent one P1 is homozygous for the AAbb genes
Parent two P2 is homozygous for the aaBB genes

Crossbreeding P1xP2 is giving:
F1 2.jpg
We see all F1 plants are uniform and have the genes AaBb (uniform genotype in then F1 generation).
To get the next generation one F1 is self-pollinated. The number of plants to grow to get all the segregation is 4^genelocation (in this case 4^2= 16 plants):
F2 2.jpg
Now there is segregation in the F2 generation. 25% is homozygous for both genes. One plant is homozygous for both AA and BB genes (6,3%). 9 plants contain minimal one dominant gene of both genes. Hence with selection there is 11,1% change to select the homozygous plant.

We notice the amount of homozygous gene combinations is 2^genelocation = 2^2= 4 combinations (AABB, AAbb, aaBB, aabb)


Continue this till the F10 generation:
F10 2.jpg
Now we see there is more as 99% homozygosity at generation F9. 24,8% of the plants is homozygous for both dominant genes AABB.
If u want a plant that have the dominant AABB genes homozygous, selection is the only option. When every generation get selection and the homozygous aa and bb plants are removed from the breeding program we see that at generation F9 there is 98,4% change to select a homozygous AABB plant.

Conclusion: How more genes we want homozygous in a plant how more generations (F9 for 2 genes and F8 for one gene) we need to get 99% change for homozygosity.
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Re: Mathematics and statistics for self-pollinating plant breeding

#3

Post: # 38588Unread post Doffer
Sun Jan 17, 2021 8:37 am

Let’s check this for 3 genes:

Parent one P1 is homozygous for the AABBcc genes.
Parent two P2 is homozygous for the aabbCC genes.

Crossbreeding P1xP2 is giving a uniform F1 generation AaBbCc.

For the F2 generation we need 64 plants to get all the segregation options. (4^genelocation, in this case 4^3)
F2 3.jpg
Now there is segregation in the F2 generation. 12,5% is homozygous for all 3 genes. One plant is homozygous for both AA, BB and CC genes (1,6%). 27 plants contain minimal one dominant gene of the three genes. Hence with selection there is 3.7% change to select the homozygous plant.

We notice the amount of homozygous gene combinations is 2^genelocation = 2^3= 8 combinations (AABBCC, AABBcc, AAbbCC, AAbbcc, aaBBCC, aaBBcc, aabbCC, aabbcc)

Continue the same mathematics till the F10 generation we get:
F10 3a.jpg
F10 3b.jpg
Now we see there is more as 99% homozygosity at generation F10. =12,4% of the plants is homozygous for all three dominant genes AABBCC.
If u want a plant that have the dominant AABBCC genes homozygous, selection is the only option. When every generation get selection and the homozygous aa or bb or cc plants are removed from the breeding program we see that at generation F10 there is 98,8% change to select a homozygous AABBCC plant.

Mathematics conclusion:
From all there mathematics I conclude the following rules for self pollination plants:
- To get all the segregation in the F2 u need to grow number of plants : 4^genelocations
- Final homozygous combinations : 2^genelocations
- Number of F generation to get 99% change for homogosity :
Fxx.jpg
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Re: Mathematics and statistics for self-pollinating plant breeding

#4

Post: # 38589Unread post Doffer
Sun Jan 17, 2021 8:43 am

Number of plants to get all the segregation in the F2:
1 gene 4 plants in F2 for segregation
2 genes 16 plants in F2 for segregation
3 genes 64 plants in F2 for segregation
4 genes 256 plants in F2 for segregation
5 genes 1.024 plants in F2 for segregation
6 genes 4.096 plants in F2 for segregation
7 genes 16.384 plants in F2 for segregation
8 genes 65.536 plants in F2 for segregation
9 genes 262.144 plants in F2 for segregation
10 genes 1.048.576 plants in F2 for segregation

Number of final Homozygous combinations:
1 gene 2 homozygous combinations
2 genes 4 homozygous combinations
3 genes 8 homozygous combinations
4 genes 16 homozygous combinations
5 genes 32 homozygous combinations
6 genes 64 homozygous combinations
7 genes 128 homozygous combinations
8 genes 256 homozygous combinations
9 genes 512 homozygous combinations
10 genes 1024 homozygous combinations

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Re: Mathematics and statistics for self-pollinating plant breeding

#5

Post: # 38639Unread post bower
Sun Jan 17, 2021 2:05 pm

[mention]Doffer[/mention] that is a really useful bit of math. Thanks for sharing it. I had no idea that the addition of a trait would prolong the generations needed to be completely stable for both traits.
Breeding projects are often considered or expected to be stable enough after 7 generations. But I guess we shouldn't be surprised if there are off types.
The tomato is said to have 35,000 protein encoding genes.
https://www.uniprot.org/proteomes/UP000004994
Seems like variation might be the rule.
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Re: Mathematics and statistics for self-pollinating plant breeding

#6

Post: # 38731Unread post MissS
Mon Jan 18, 2021 2:20 pm

Thank you Doffer. This was very well presented and easy to understand.
~ Patti ~

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Re: Mathematics and statistics for self-pollinating plant breeding

#7

Post: # 42041Unread post bower
Sat Feb 27, 2021 6:00 pm

[mention]Doffer[/mention] I have a cross which I was growing out, and unexpectedly in the F3 generation there was a cross between two F3 siblings. The F4 was obviously crossed because the number of plants with recessive traits was far more than expected. I've been growing two lines selected from the F4 for three more generations, and they appeared in the F7 last year to be identical to the F6 parents in all traits that mattered.
My question is, mathematically how does a sibling cross in the F3 affect the number of generations to become stable. Should it take longer than growing a plant from an F3 that didn't cross with its sibling?
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Re: Mathematics and statistics for self-pollinating plant breeding

#8

Post: # 42056Unread post Doffer
Sun Feb 28, 2021 7:03 am

If you cross an F3 with another varaty you start again at the front. Your F3 = P1 and the other parent is P2.
So this results in an F1 and not an F4!

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Re: Mathematics and statistics for self-pollinating plant breeding

#9

Post: # 42059Unread post bower
Sun Feb 28, 2021 7:31 am

Yes Doffer but the two F3's that crossed were siblings. So they had a certain percentage of genes in common, maybe still not many homozygous but a degree of relatedness all the same. The question I wondered, how different is it from selfing, when a same-generation cross occurs between sibling plants.
The cross produced a lot of determinate plants, and more pink-black vs brown-black fruit than expected. It was a win for me, to get a diverse group of determinates in one generation to select fruit traits from.
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Re: Mathematics and statistics for self-pollinating plant breeding

#10

Post: # 42082Unread post Rockoe10
Sun Feb 28, 2021 12:44 pm

Without going through the impressive detail that Doffer committed to, I'll offer what i can deduce.

When a plant 'Self Pollinates' it is pollinating with what could be considered a sibling. A sibling that is identical to itself.

If it pollinates with a sibling that is not itself, then there would be more randomness in its genotype/phenotype. Almost immeasurable, without gene sequencing.

My thoughts on it.
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Re: Mathematics and statistics for self-pollinating plant breeding

#11

Post: # 42136Unread post bower
Sun Feb 28, 2021 9:56 pm

I'm thinking something like this: Just looking at the chart at the bottom of the page, similar to the more complicated version which Doffer provided to consider multiple genes being looked for in selection process:
http://kdcomm.net/~tomato/gene/genes2.html
I see that in the F2, 50% of genes are expected to be homozygous and 50 % heterozygous.
In the F3, an additional 25% of genes should be newly homozygous and 25% heterozygous.
Now if you have two F3 plants from the same parent, they should be 50% similar to each other in the genes which are already homozygous in the parent. So the F3 sibling cross cannot be less homozygous than the F2 parent.

Working through the most extreme scenarios, the worst case for homozygosity would be if the 25% newly homozygous genes were the same ones but opposite ie XX vs xx. In this case all the 25% gained in F3 would revert to heterozygous condition. The remaining 25% heterozygous genes would be the same in both plants and segregate normally to produce a gain of 12.5% homozygous genes.
So the worst case for an F3 sibling cross is to result in 50% + 12.5% or 62.5% homozygosity, which is between F2 and F3 normal ratios.
In the opposite scenario where new homozygous genes in one sib are paired with heterozygous genes in the other and vice versa, the segregation produces a total of 25% homozygous genes gained. The total of 50% (parental) + 25% = 75%. This would be the normal level of homozygosity for an F3.
Realistically, the opposite scenario is more likely to produce the same 75%, if half of the homozygous genes were mismatched XX/xx in the sib, instead of the less likely case that all would be opposites. So the math is telling me, this sib cross of F3's produces an F3 level of stability at 75%, compared with the F4 produced by selfing which has 82.5% homozygosity.
So my conclusion is that only one generation (or in the worst case, less than two generations) of stabilization can be lost in a sib cross. :)
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Re: Mathematics and statistics for self-pollinating plant breeding

#12

Post: # 42143Unread post Doffer
Mon Mar 01, 2021 5:55 am

Bower wrote: Sun Feb 28, 2021 9:56 pm I'm thinking something like this: Just looking at the chart at the bottom of the page, similar to the more complicated version which Doffer provided to consider multiple genes being looked for in selection process:
http://kdcomm.net/~tomato/gene/genes2.html
I see that in the F2, 50% of genes are expected to be homozygous and 50 % heterozygous.
In the F3, an additional 25% of genes should be newly homozygous and 25% heterozygous.
Now if you have two F3 plants from the same parent, they should be 50% similar to each other in the genes which are already homozygous in the parent. So the F3 sibling cross cannot be less homozygous than the F2 parent.
Your interpretation of that picture is not entirely correct. You can see that in the F2 generation for 1 gene is homozygous in 50% of the plants. In the F3 generation you see that the same one gene is homozygous in 75% of the plants.
It is not that 50% of all genes in the F2 generation are already homozygous.

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Re: Mathematics and statistics for self-pollinating plant breeding

#13

Post: # 42144Unread post bower
Mon Mar 01, 2021 6:10 am

Yes, the picture refers to a population, however I believe the math of segregation is the same if you consider the genes of a single plant.
Normal ratios of segregation will produce 1/4 XX, 1/4 xx, and 2/4 Xx on average, genome wide.
The huge differences between one F2 and another F2 seed is because those random segregations affected different genes differently. They do not have the same 50% of genes homozygous.
Overall however, assuming mendelian inheritance, you would expect half of the genes to be homozygous in each F2 plant.
Since the two F3 siblings came from the same F2 parent, they would have the same parental homozygous set in common.
They would differ in the segregation of the remaining 50% of heterozygous genes, in which half again by mendelian ratios will become homozygous.
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Re: Mathematics and statistics for self-pollinating plant breeding

#14

Post: # 42172Unread post Pippin
Mon Mar 01, 2021 3:24 pm

I think you have answered all of your interesting questions very successfully, [mention]Bower[/mention]. I enjoyed reading the hypothetical problem and your line of thinking. Food for thoughts, I would say. :D
The cross produced a lot of determinate plants, and more pink-black vs brown-black fruit than expected.
The sibling cross pollination theory works for me but it could still be a matter of chance (and result of selfing). People win lotteries even when the odds are against it.

sp seems to be hiding where it should not be, and recessive color genes are popping-up more often than they should. One problem with small sample sizes is that 1:3 for trivial recessive traits does not always apply in practise. Last summer, I got 2 reds and 4 tangerine in F2 when growing only 6 plants, i.e. 67% of recessive phenotypes. :D
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Re: Mathematics and statistics for self-pollinating plant breeding

#15

Post: # 42175Unread post bower
Mon Mar 01, 2021 4:15 pm

Pippin, I do agree about lotteries. I've often had both type of results, either better or worse than expected.
I'm pretty certain that the F3 in question was a cross though. When two recessive traits turn up at 50% or more all of a sudden, and when the sibs were outdoors and heavily worked by bumblebees, it seems the most logical answer. The "F4" growout was a lot more like an F2 or F3 as well, with regards the diversity of fruit shape, size and other traits. However the selected lines do seem to be pretty stable now ("F7" last year if it was a selfing), so I'm hoping to see the same this summer and be able to share them more widely as stable varieties.
Meanwhile on the subject of maths, our colleague Crmauch at TV provided some excellent maths which incorporate probability in addition to the 'chance' that is the expected ratios of gene segregation. I found this really useful from a practical standpoint. The probability of finding your recessive with 6 plants is much better than 4. But I admit, I have sometimes not found it with 6 either! And another time, got it from two... Such is luck. ;) (and don't forget linkage, oh my that can really mess with the ratios. But there's no fun math to deal with that one.)
http://tomatoville.com/showthread.php?p=506648
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Re: Mathematics and statistics for self-pollinating plant breeding

#16

Post: # 42176Unread post Pippin
Mon Mar 01, 2021 4:32 pm

Overall however, assuming mendelian inheritance, you would expect half of the genes to be homozygous in each F2 plant.
One can probably think this as a classical coin flipping experiment (e.g. in https://en.wikipedia.org/wiki/Central_limit_theorem): heterozygote gene in the parent is a flip of coin having 50-50 chance of being homozygote in the seed. The flip of the coin is repeated as many times there are heterozygote genes (10 flips/genes in the figure). Each of the seed represents a repeat of the same (10 flip) experiment. The average of fixed homozygote genes is 50% in each seed, however, the seeds are different in practise and distributes to a normalized curv format. The more seeds one sows (repeats), the more extreme samples one can get (like few plants having up to 80% or as few as 20% of the genes fixed to homozygote).

https://www.vskills.in/certification/t ... t-theorem/
B63BA355-47B0-4AA3-AA7E-2D6E4581FF8E.jpeg
(Edit: moved the more commercial link next to the figure and added a link to Wikipedia as the first link)
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Re: Mathematics and statistics for self-pollinating plant breeding

#17

Post: # 42249Unread post Doffer
Tue Mar 02, 2021 12:12 pm

Bower wrote: Sat Feb 27, 2021 6:00 pm The F4 was obviously crossed because the number of plants with recessive traits was far more than expected.
Why do you think that only a cross between two siblings can explain that you suddenly got a lot of recessive traits?
Did the F3 sibling already have much more recessive traits? If not, you may not have had any cross at all in the F3.

P.s. if the desired traits are recessive traits, then you can get homozygous plants in many earlier generations because you can select directly for these traits as soon as they become visible.

Now I am also curious which breeds you crossed and what the generations looked like?

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Re: Mathematics and statistics for self-pollinating plant breeding

#18

Post: # 42257Unread post bower
Tue Mar 02, 2021 1:59 pm

[mention]Doffer[/mention] I'll make a separate thread about them when I get growing this season. I have photos of all generations but still getting those things settled into a new computer.
I did get a few determinates in F2 and F3 but they didn't have other desired qualities. Either fruit were not tasty or the plants were not robust, so I continued growing the indeterminates that were great tomatoes and disease resistant plants.
Either way if the F4 seeds were crossed or not, at least I have some idea now, how far the lines from that generation would be from stable. :) This year would be F8 if there was no cross.
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