let's hear your thoughts...

[trichrider]

hey.

i want to produce some seeds with as much genetic diversity as possible, with as few plants as possible.
this is my theory.

i started about 40 plants from seed (regular seed).
out of those i chose the most vigorous, maybe 20.

instead of one spectacular specimen of male plant i've kept at least 3 (so far showing) and multiples of that for females.

this should avoid bottle-necking correct?

i'm not out to resurrect the strain, just to keep viable diversity of seed-stock on hand.

opinions are ok too. thnx.

 

[Water-]

You might find more help posting this in the Breeder's Laboratory, posts stick around a a lot longer there.

the more males the better, i think
you are going to have a shit load of seeds.

good luck

 

[Koondense]

For genetic diversity you need little selection, so more parents is better and keep also the not so vigorous individuals.
But you should start with all different lineages for example three different indica males with five different sativa females will give you many hybrids with huge variations. Not sure what to do with all the seeds, may be hard to search for a killer plant to keep, might be cool to throw them around in spring.
All depends on your goal.

I like crossing two different F1's and see what happens, similar approach to yours but with only two plants.

 

[Betterhaff]

For some reason the number 30 comes to mind. Obviously the more plants you have the more diversity there will be. But I think 30 is a number that will give you the most diversity with the fewest plants out of a given population.

I think Tom Hill made some comments about this but not sure, most of his posts are gone.

 

[zif]

This also depends heavily on where you're starting.

A true F1 requires just one selfed plant to recover all of the genes. Same for an idealized inbred line (cannabis requires too much heterozygosity to achieve this goal, but in principle). Both of these cases are true because the parents have essentially no genetic diversity.

The outcomes would be very different, though: uniformity in an inbred line, extreme diversity in the F2 generation.

We're stuck with some variation of the latter. SamS has posted a link to studies of just what number of plants are required in each stage. It's actually more like 1000s to preserve a complete population! I'm short of time otherwise I'd point you to it.

 

[Limeygreen]

Remove any off types, things that aren't what you care for. Examples may include, early flowering males, very vigorous and tall growing males, hermaphrodites, structure of the plant etc. Supposedly some would say hollow stems are good indicators of higher thc, but I have only heard this and not sure about it myself. If you compile a list of the things you require for the characteristics of the variety, you should be able to rouge out what you don't want and keep as many plants as possible for what you do want.

 

[MJPassion]

hey.

i want to produce some seeds with as much genetic diversity as possible, with as few plants as possible.
this is my theory.

i started about 40 plants from seed (regular seed).
out of those i chose the most vigorous, maybe 20.

instead of one spectacular specimen of male plant i've kept at least 3 (so far showing) and multiples of that for females.

this should avoid bottle-necking correct?

i'm not out to resurrect the strain, just to keep viable diversity of seed-stock on hand.

opinions are ok too. thnx.

If your desire is to keep some diversity within a given line, what you describe will work.
But... As suggested above, more parents will give more diversity.

I think I remember reading that it takes a field of 20,000 males & 20,000 females in order to not lose diversity. I doubt anybody has that kind of space to breed with.
Maybe Sam.

 

[zif]

FYI:

And a classic that explains the problems maintaining Cannabis landraces in a closet.
1,000 males and 1,000 females are required to avoid serious gene loss, because Cannabis is an dioecious obligate outcrosser.
-SamS

Theor Appl Genet (1993) 86:673-678
Statistical genetic considerations for maintaining germ plasm collections
J. Crossa , C. M. Hernandez , P. Bretting , S. A. Eberhart , S. Taba

 

[Yard dog]

FYI:

That was based on the Wang/crossa work that Rob Clarke looked at, the number was to conserve all the alleles, this is a post from the thread it was discussed in;

[FONT=Arial, Helvetica, sans-serif]

Sam, I should have metioned that my argument was based on my recollections of constructing those statistical tableaus with Ne on the left hand side and very small numbers on the top of the matix.So your argument about Ne at 2000 would be correct insofar as preserving almost all of the alleles. Anyways doing too much math makes my head hurt. The little ^ symbols I used are exponentials.
There are methods for estimating the probability of genetic drift and underscoring its devastating impact on unsystematic shifts in allele frequencies where Ne is small.
The formula is: Prob. = (1 - frequency) ^ 2 ^ Ne
For example, lets say there is a less common allele (w) in a white widow clone that occurs say 5% of the time (.05) and further for arguments sake Ne = 2.
So. (1.- .05) = .95
(.95) ^ (2)(2) = 81.4% probability of elimination
Make Ne = 50
(.95) ^ (2)(50) = .00592 probability of elimination.
As is the case with any algorithm, the answers will change depending upon what numbers one intends to plug into the formula.
The higher the frequency of an allele in a given population the greater its likelihood of surviving a genetic bottleneck with the opposite being true of lower frequency alleles surviving when Ne is very small.
To expand on the basic concept of Ne one must consider how Ne effects the F statistic (Coefficient of Inbreeding).
So let's assume an NE of say 4. The decline in heterozygosity or rate of inbreeding at from the first generation would be: F1 = (1) / (2 * Ne) or 1 / (2 * 4) = .125.
With each subsequent generation the decline in heterozygosity is cumulative. Take generation F6
F6 = 1 - (1 - F1) ^ 6 = 1 - (1 - .125) ^ 6 = .551
By the 6th generation 55% of the genetic diversity will have been lost in the line. By the 12th generation 1 - (.87.5) ^ 12 = 80 % of the genetic diversity will have been eliminated.
The larger the Ne the lower the rate is the actual loss of diversity through inbreeding.
Let's take a hypothetical example:
Breeder X grows out 20 plants and gets 6 males and 14 females.
Effective breeding population will be:
(4)*(6)M*(14)F / 20 Plants Total =16.8

Change in inbreeding per generation:
1 / (2)*(16.8)= 2.9%

The "best" effective size calculation in this example:
16.8/20 = 84%

Fun with numbers for sure. I posted these examplea to underscore that genetic drift is quantitatively measurable.

[/FONT]As you can see from the above the more the better, now if for example you took an example like Tomhills haze were he stated 5% were the great ones giving you 5/100 or 1/20, you need 20 to make sure you capture that 1.

https://www.icmag.com/ic/showthread.php?t=167789&highlight=drift&page=42

 

[zif]

The drift calculations, as you point out, show more is better. And it's tough, because ensuring just a given allele is preserved takes tons of plants. It's far worse given that we want entire ensembles of alleles!

As you can see from the above the more the better, now if for example you took an example like Tomhills haze were he stated 5% were the great ones giving you 5/100 or 1/20, you need 20 to make sure you capture that 1.

Your example is much trickier than this. You cannot count on getting a great thh with just 20 plants. Any given 20 plant grow, with 1/20 being 'great', has a 36% chance of not showing that rare 'great'! I'm not sure about you, but I think making sure you capture that greatness would mean having, what, a 90, 95, or maybe even 99% chance of finding the 1/20 type?

These levels of certainty require 45, 59, and 90 plants, respectively. But maybe we should account for females? Then we'd need roughly double the plants to be 'sure' to capture the 'great' type. The math is simple, but the implications are challenging!

 

[Yard dog]

The drift calculations, as you point out, show more is better. And it's tough, because ensuring just a given allele is preserved takes tons of plants. It's far worse given that we want entire ensembles of alleles!



Your example is much trickier than this. You cannot count on getting a great thh with just 20 plants. Any given 20 plant grow, with 1/20 being 'great', has a 36% chance of not showing that rare 'great'! I'm not sure about you, but I think making sure you capture that greatness would mean having, what, a 90, 95, or maybe even 99% chance of finding the 1/20 type?

These levels of certainty require 45, 59, and 90 plants, respectively. But maybe we should account for females? Then we'd need roughly double the plants to be 'sure' to capture the 'great' type. The math is simple, but the implications are challenging!

I'm struggling with the math! you have a 5% chance of finding the "great" plant with 20.. you have the same chance with 100 though I'd certainly give you bigger odds for finding it in 20.
Could you please tell me how you get the 99 per cent chance from 90 plants etc...

 

[Yard dog]

I presume it's to do with (.95) ^ (2)(50) = .00592 probability of elimination. (replacing the 50 with the rel number)?

 

[zif]

The easy way to figure this out is to use the binomial distribution. It's not so easy to get to how it does what it does, but it basically lets you move from asking questions like "what are the odds that this coin will come up heads?" to questions like "what are the odds that these 20 coins will have at least one come up heads?".

If you tell me there is a 1/20 chance that a thh will be awesome, I can use it to find out what the odds are that 10 seeds will have one of that awesome type.

It's really easy to see why maybe 20 is not enough to find that great plant, even though the odds any one plant are great are 1 in 20, though. Let's simplify the situation:

Imagine I give you a special pack with 2 regular seeds. Do you feel confident you will get a female? After all, each one has a 1/2 chance of being female, so you ought to jump at the chance to buy this 2 pack, right?

Do you see what's wrong there?

 

[MJPassion]

That's why 3 packs are better.

Good job explaining things zif.

 

[Betterhaff]

The probability of getting a female from one regular seed is 50%. The probability of getting 1 female from 2 seeds is 75%. The probability of getting 1 female from 3 seeds is 87.5%. And so on. But the key word is “probability”. From 3 seeds you could get all males (or all females).

 

[Yard dog]

The easy way to figure this out is to use the binomial distribution. It's not so easy to get to how it does what it does, but it basically lets you move from asking questions like "what are the odds that this coin will come up heads?" to questions like "what are the odds that these 20 coins will have at least one come up heads?".

If you tell me there is a 1/20 chance that a thh will be awesome, I can use it to find out what the odds are that 10 seeds will have one of that awesome type.

It's really easy to see why maybe 20 is not enough to find that great plant, even though the odds any one plant are great are 1 in 20, though. Let's simplify the situation:

Imagine I give you a special pack with 2 regular seeds. Do you feel confident you will get a female? After all, each one has a 1/2 chance of being female, so you ought to jump at the chance to buy this 2 pack, right?

Do you see what's wrong there?

Yes, I understand .... although with your example of 2 seeds or coins giving you a 0.5 chance or 50% in fact your chances are much higher to get 1 from a higher number although the 0.5 chance remains the same you have more chances with the higher number and therefore a greater chance at success..
I tried to calc it myself with the formula;

b(x; n, P) = nCx * Px * (1 – P)n – x


b = binomial probability
x = total number of “successes” (pass or fail, heads or tails etc.)
P = probability of a success on an individual trial
n = number of trials

but it was much simpler with the online calculator

Also correct me if I'm wrong but don't you have more chance at 1 in 20 than 5 in 100? as I understand you calculated for 1 in 90 etc to give you the 99%...

 

[Yard dog]

Also correct me if I'm wrong but don't you have more chance at 1 in 20 than 5 in 100? as I understand you calculated for 1 in 90 etc to give you the 99%...

It won't let me edit, but to clarify;

This is taken from the online calc and I've calculated for 1/20 and 5/100...

1.The probability of fewer than 1 = 36%
The probability of fewer than 5 = 44%

2.The probability of at most 1 = 74%
The probability of at most 5 = 62%

3.The probability of getting at least 1 = 64%
The probability of getting at least 5 = 56%

4.The probability of more than 1 = 26%
The probability of more than 5 = 38%

To me it stands to reason you that if you x4 your chances of looking for 1 you will find it easier, like where SamS suggests in that thread you're better at finding gold in a batch of 1000 than you are with 5 batches of 200 etc.....

the results for the 1/90 were;

1. .98%
2. 6%
3. 99%
4. 94%

Hope I've understood correctly.

 

[zif]

Yes, I understand .... although with your example of 2 seeds or coins giving you a 0.5 chance or 50% in fact your chances are much higher to get 1 from a higher number although the 0.5 chance remains the same you have more chances with the higher number and therefore a greater chance at success..

Right, and especially: just because you have 1/2 odds on any given seed, it does not mean that 2 seeds will give you the desired outcome. Exactly the same reasoning shows that you can't expect a great plant from 20 seeds with a 1/20 chance of great plants.

Also correct me if I'm wrong but don't you have more chance at 1 in 20 than 5 in 100? as I understand you calculated for 1 in 90 etc to give you the 99%...

To get a 99% chance, I was looking at what happens if you increase the number of trials (that is, the number of seeds you test), definitely not changing the odds for any given seed. The point was just that being confident you'll get the outcome depends on how confident you want to be, and if you want to be reasonably confident, you need far more seeds than you might think.

 

[zif]

The probability of getting a female from one regular seed is 50%. The probability of getting 1 female from 2 seeds is 75%. The probability of getting 1 female from 3 seeds is 87.5%. And so on. But the key word is “probability”. From 3 seeds you could get all males (or all females).

And with 10 seeds, you've got a 99.9% chance of 1 or more females. Makes the 'standard' pack amount seem pretty reasonable....

 

[zif]

To me it stands to reason you that if you x4 your chances of looking for 1 you will find it easier, like where SamS suggests in that thread you're better at finding gold in a batch of 1000 than you are with 5 batches of 200 etc.....

It looks like you've basically got it.

There is no probability argument that makes how you carve up your batches matter. Say you set aside 1000 seeds, you've just made a batch. I don't care how you decide to grow them out, you've already chosen the seeds and cannot change what you'll find. It's certainly possible that in practice it's easier to select when you are comparing plants at the same time and place, though.