Aussie Growers Thread
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Supa smoka
Well-Known Member
Yeah they obv couldnt get it all sorted before the big sale on tonight.
Supa smoka
Well-Known Member
Big raids in brizzy yesterday
forgetiwashere
Well-Known Member
really i havnt seen anyone have any trouble in any of the ones im following. watching a real nice one on ggg forums at the moment. pineapple pheno is killing it.
you sure the ones your watching are mosca's because female seeds have a cindy 99 bx aswell and i have heard a lot of people saying that its not a proper bx and that they have actually bred an auto into it wtf?
heres a shot of danksmith's c99
and heres one of his sun maiden
Supa smoka
Well-Known Member
Yeah which beans bucko
forgetiwashere
Well-Known Member
ggg's bright moments is up. nearly bought it got to the credit card details stage and changed my mind. i have about 8 thousand seeds already i really should just grow them all out before i spend any more money its getting out of hand lol
Supa smoka
Well-Known Member
A bit of info for those who are breeding.
We can often get get hung up on terminology and lose sight of what is really trying to be said. For this paper, when I discuss inbreeding, I'm talking about crossing individuals from the same generation and not backcrossing. I've haphazardly referred to this as generational inbreeding, although I'm not certain such terminology is considered accurate, haha. Also, in my F1 vs F2 discussion paper, I try to finely define terms such IBL, F1, and F2 from the perspective of a seed vendour and seed buyer. Those definitions won't apply here and I'll rely upon the most generic definitions of those terms for this discussion.
Your starting point of an inbreeding project can involve two parents that are related or two parents that are not. You could even start with a single parent and self it. In each case, we will arbitrarily assign the parents making up the starting point the P1 parents. In a typical inbreeding project, the progeny of the P1 parents will be called the F1 cross. When you cross individuals from an F1 generation together, you get an F2 generation. Cross the F2 generation and you get an F3 generation. The F3 generation gives rise to the F4 generation, and likewise, the F4 gives rise to the F5 generation. A similar inbreeding strategy can also be applied as a followup to a selfing or backcrossing project. We will first take a close look at how we can manipulate gene frequencies by solely working with generational inbreeding.
Lets say we want to stabilize the pineapple flavour of a special individual within a pine flavoured population. The genes controlling the pine flavour could be dominant or recessive. This fact can greatly influence the success of the project.
If the pineapple flavour is controlled by a dominant gene, there is a good chance the indivual will be heterozygous (Pp) where P symbolizes pineapple flavour and p symbolizes pine flavour. It can also safely be assumed that other individuals in the population are homozygous (pp) for the pine gene. Therefore our F1 cross will be:
F1 cross = Pp x pp = Pp + Pp + pp + pp
50% of the F1 generation will be pineapple flavoured and the frequency of the pineapple (P) gene will be 2/8 or 25%. Natually when selecting parents for the F2 generation, we would choose ones that were pineapple flavoured and therefore they would all be heterozygous (Pp). By being able to select both sets of parents, we call this a full sib cross. Again due to it's common simplicity I'll spare you the punnet square, we can determine the genetic combinations of our F2 population in our heads.
F2 (a) cross = Pp x Pp = PP + Pp + Pp + pp - typical mendelian phenotypic 3:1 and genotypic1:2:1 ratios.
75% of the F2 population will have pineapple flavour and our frequency of the P gene is now 4/8 or 50%. Now moving onto the F3 (a) generation gets a little harder to do in our heads. Again spotting the pine flavoured (pp) individuals should be easy and therefore removed from the breeding population. This leaves us with the PP + Pp + Pp individuals to make up the breeding population. We shorten it to PPPPpp to indicate the breeding population's genotype and frequency of the P gene. Since it can evenly be divided by 2, PPp just as accurately symbolizes the same genotype. Therefore the two parents become PPp x PPp. Each individual letter can represent the frequency of a single gamete (pollen or ovule) in the breeding population.
Fig 1: F3 (a) Cross
If we continued into the F5 generation using the same selective pressure, we would end up with 144PP + 72Pp + 9pp which translates into 96% of the population tasting like pineapple. The frequency of the pineapple gene would have risen to 80%.
b) But lets say that in all reality, that we can't determine flavour in the males, so we can only remove the pine (pp) flavoured individuals from the female parents. We call this a half-sib cross when we can't select our pollen source. So we would be doing two crosses Pp x Pp and Pp x pp. A shortcut is to combine the various genotypes into one and just write Pp x Pppp. I'll skip the punnet square on this one, but please feel free to do one yourself to be sure you understand what's happening. If you don't, it should become obvious when I go through the F3 cross in detail.
F2 (b) cross = Pp x Pppp = PP + Pp + Pp + Pp + Pp + pp + pp + pp
62.5% (5/
of the halfsib F2 population would be pineapple flavoured and the frequency of the P gene would be 6/16 or 37.5%. This is quite a decrease by simply not being able to remove the pp males from the breeding population. I'll carry this one more generation (F3 cross) in detail to show you the developing patterns. After removing the pine flavoured (pp) females, our female genepool would be PP+Pp+Pp+Pp+Pp = PPPPPPpppp = PPPpp. Without any selective pressure, our male genepool would remain PP+Pp+Pp+Pp+Pp+pp+pp+pp = PPPPPPpppppppppp = PPPppppp. Here's how the cross plays out.
As you can see, the F3 cross yields pineapple flavour in 75% (30/40) of the offspring. The frequency of the P gene has risen to 48.8% (39/80).
Lets look at the mathematical patterns developing. To recap, this F3 cross was PPPpp x PPPppppp. Lets rewrite it in a simpler fashion that expresses the ratio of each gene (or gamete). We would get 3P2p x 3P5p. If you note, when we add up the numerical value of each side of the cross and then multipy them (3+2)*(3+5) we get the number 40, which turns out is the same as the number of offspring created by the punnett square. Notice that when we multiply the two 3Ps, we get 9P, the same number of PP individuals from the punnett square? The same pattern holds for each combination, so what we have here is a simple way of calculating a punnet square outcome without actually drawing the punnet square. This can save alot of time when we get into complex combinations. So lets use the mathematical method of determining the results from the above F3 cross.
3P2p x 3P5p = (3*3)P + (3*5)Pp + (2*3)Pp + (2*5)pp = 9PP + 15Pp + 6Pp + 10pp = 9PP + 21Pp + 10pp
As you can see we came up with the same number as the punnet square without drawing all the lines. Now lets use the same formula to calculate the F4 and F5 generations. We will remove the 10pp from the female genepool and be left with 9PP + 21Pp. If we add up all the P's and p's, this works out to [2(9) + 21]P and 21p which translates to 39P21p. The male gene pool will work out to be [2(9)+21]P and [21+2(10)]p = 39P41p. Remember, each number infront of each gene simply represents the frequency of that particular gene.
F4 Cross:
39P21p x 39P41p = 39*39PP + 39*41Pp + 21*39Pp + 21*41pp = 1521PP + 1599Pp + 819Pp + 861pp
So now lets use the ratios we got from the F4 generation to calculate the gene frequencies of the parental genepools of the F5 cross. All we do is add together the frequency of each gene (gamete) and divide by the total of the ratio used in that genepool. Since the Pp is only half P, we divide this one in half. Therefore, the male parental genepool will be (.32+.25)/1 which equals .57P. Once we know P, we automatically know p since it is simply 1-P or .43. Therefore, the male parental genepool is .57P.43p. Now to determine the female gene frequencies, we need to subract the .18pp from the total numbers since they will be removed from the genepool. This is one way to do it. (.32+.25)/1-.18 = .57/.82 = .695P. Again 1-P=p so we end up with the female gene frequencies of .695P.305p. The F5 cross is as follows:
.695P.305p x .57P.43p = (.695*.57)PP + (.695*.43)Pp + (.303*.57)Pp + (.305*.43)pp = .40PP + .47Pp + .13pp
Finally in conclusion, after 4 generations of inbreeding where we only make our selections from the female population, we end up with an F5 population where 87% taste like pineapple. Plus the frequency of the pineapple gene in the F5 population is 63.5%. This is significantly less than if we were able to apply the selective pressure to both parental genepools. If you recall, in that situation we achieved 96% of the population tasting like pineapple. The frequency of the pineapple gene would have risen to 80%. This is just the case where the flavour gene is dominant, the situation when selecting for recessive traits is much nicer.
When the trait we want is recessive.
In this case, we will assign the symbol p to indicate pineapple flavour and P will indicate Pine flavour. If we find a single pineapple flavoured individual in a population of pine flavoured individuals, and the trait is recessive, then the individual must be homozygous (pp) for the trait. When choosing a mate to cross it with, there is a chance you could select a heterozygous individual, but it's more likely to use a homozygous dominant pine flavoured one so that is what we'll base this next model on. Therefore, our F1 cross will be:
F1 cross = pp x PP = Pp + Pp + Pp + Pp or just simply Pp since all the F1s are the same.
Just to maintain consistency, I will point out that none (0%) of the F1 cross will have pineapple flavour but the frequency of the p gene will be 50%. Now when we move onto the F2 population, our parents will both be Pp. Here is the F2 cross:
F2 cross = Pp x Pp = PP + Pp + Pp + pp
Since there was no selection in choosing the parents, the p gene frequency remained at 50%. However, 25% of the offspring will be pineapple flavoured. As has been shown previously, it is this reassortment within the F2 population that is key. Now we can spot the females that are homozygous (pp) for the pineapple flavour. If we can identify the pineapple flavoured males, then we will be finished with an F3 cross as follows:
F3(a) cross = pp x pp = pp + pp + pp + pp - all true breeding for the pineapple flavour, mission accomplished
But not so fast, many are unable to determine the flavour of a male plant and so therefore wouldn't be able to perform any selections on the male portion of the genepool. Again we are back in a half-sib breeding model. The male population's population be PP + Pp + Pp + pp which equals 4P4p which in turn can be simplified to Pp. The frequency of the p gene in the female genepool will remain 100% from now on in this model. In this case the F3 cross would be:
F3 (b) cross = pp x Pp = Pp + pp + Pp + pp
50% of the F3(b) generation would be pineapple flavoured and the frequency of the pineapple (p) gene has increased to 6/8 or 75%. We would select out the PP female but use both Pp and pp males in the F4 cross. From what we learned in the previous section we could designate our gene frequencies as the female breeding pool = .5p.5p and the male breeding pool as .25P.75p. You see where those came from? Remember that 6/8 or 75% were p from the F3(b) cross? Well the 75% simply becomes .75 when we convert to decimal form. And 1-p=P to arrive at the .25p. Hence our F4 cross is:
F4 (b) cross = .5p.5p x .25P.75p = (.5*.25)Pp + (.5*.75)pp + (.5*.25)Pp + (.5*.75)pp = .25Pp + .75pp or more simply Pp + pp + pp + pp
We'll skip to the F5 generation, see if you can figure where I get my gene frequencies.
F5 (b) cross = .5p.5p x .125P.875p = .125Pp + .875pp
So after doing half sib inbreeding for 4 generations, we achieve an F5 generation where 87.5% of the offspring will be pineapple flavoured and the frequency of the p gene will be 93.75%. Not bad at all, just as good as cubing, but the 100% we achieved with the previous full sib example was better and with two fewer generations (HINT!!).
Please keep in mind that these models assume that flavour is a trait controlled by a single gene or linked group of genes. Reality isn't as simple, but the principles mentioned here apply to more complex models as well. The main point to take from this is that the degree of selection we use can very much influence our success rate. And that selecting for dominant and recessive traits have some subtle differences
We can often get get hung up on terminology and lose sight of what is really trying to be said. For this paper, when I discuss inbreeding, I'm talking about crossing individuals from the same generation and not backcrossing. I've haphazardly referred to this as generational inbreeding, although I'm not certain such terminology is considered accurate, haha. Also, in my F1 vs F2 discussion paper, I try to finely define terms such IBL, F1, and F2 from the perspective of a seed vendour and seed buyer. Those definitions won't apply here and I'll rely upon the most generic definitions of those terms for this discussion.
Your starting point of an inbreeding project can involve two parents that are related or two parents that are not. You could even start with a single parent and self it. In each case, we will arbitrarily assign the parents making up the starting point the P1 parents. In a typical inbreeding project, the progeny of the P1 parents will be called the F1 cross. When you cross individuals from an F1 generation together, you get an F2 generation. Cross the F2 generation and you get an F3 generation. The F3 generation gives rise to the F4 generation, and likewise, the F4 gives rise to the F5 generation. A similar inbreeding strategy can also be applied as a followup to a selfing or backcrossing project. We will first take a close look at how we can manipulate gene frequencies by solely working with generational inbreeding.
Lets say we want to stabilize the pineapple flavour of a special individual within a pine flavoured population. The genes controlling the pine flavour could be dominant or recessive. This fact can greatly influence the success of the project.
If the pineapple flavour is controlled by a dominant gene, there is a good chance the indivual will be heterozygous (Pp) where P symbolizes pineapple flavour and p symbolizes pine flavour. It can also safely be assumed that other individuals in the population are homozygous (pp) for the pine gene. Therefore our F1 cross will be:
F1 cross = Pp x pp = Pp + Pp + pp + pp
50% of the F1 generation will be pineapple flavoured and the frequency of the pineapple (P) gene will be 2/8 or 25%. Natually when selecting parents for the F2 generation, we would choose ones that were pineapple flavoured and therefore they would all be heterozygous (Pp). By being able to select both sets of parents, we call this a full sib cross. Again due to it's common simplicity I'll spare you the punnet square, we can determine the genetic combinations of our F2 population in our heads.
F2 (a) cross = Pp x Pp = PP + Pp + Pp + pp - typical mendelian phenotypic 3:1 and genotypic1:2:1 ratios.
75% of the F2 population will have pineapple flavour and our frequency of the P gene is now 4/8 or 50%. Now moving onto the F3 (a) generation gets a little harder to do in our heads. Again spotting the pine flavoured (pp) individuals should be easy and therefore removed from the breeding population. This leaves us with the PP + Pp + Pp individuals to make up the breeding population. We shorten it to PPPPpp to indicate the breeding population's genotype and frequency of the P gene. Since it can evenly be divided by 2, PPp just as accurately symbolizes the same genotype. Therefore the two parents become PPp x PPp. Each individual letter can represent the frequency of a single gamete (pollen or ovule) in the breeding population.
Fig 1: F3 (a) Cross
b) But lets say that in all reality, that we can't determine flavour in the males, so we can only remove the pine (pp) flavoured individuals from the female parents. We call this a half-sib cross when we can't select our pollen source. So we would be doing two crosses Pp x Pp and Pp x pp. A shortcut is to combine the various genotypes into one and just write Pp x Pppp. I'll skip the punnet square on this one, but please feel free to do one yourself to be sure you understand what's happening. If you don't, it should become obvious when I go through the F3 cross in detail.
F2 (b) cross = Pp x Pppp = PP + Pp + Pp + Pp + Pp + pp + pp + pp
62.5% (5/
of the halfsib F2 population would be pineapple flavoured and the frequency of the P gene would be 6/16 or 37.5%. This is quite a decrease by simply not being able to remove the pp males from the breeding population. I'll carry this one more generation (F3 cross) in detail to show you the developing patterns. After removing the pine flavoured (pp) females, our female genepool would be PP+Pp+Pp+Pp+Pp = PPPPPPpppp = PPPpp. Without any selective pressure, our male genepool would remain PP+Pp+Pp+Pp+Pp+pp+pp+pp = PPPPPPpppppppppp = PPPppppp. Here's how the cross plays out.
Lets look at the mathematical patterns developing. To recap, this F3 cross was PPPpp x PPPppppp. Lets rewrite it in a simpler fashion that expresses the ratio of each gene (or gamete). We would get 3P2p x 3P5p. If you note, when we add up the numerical value of each side of the cross and then multipy them (3+2)*(3+5) we get the number 40, which turns out is the same as the number of offspring created by the punnett square. Notice that when we multiply the two 3Ps, we get 9P, the same number of PP individuals from the punnett square? The same pattern holds for each combination, so what we have here is a simple way of calculating a punnet square outcome without actually drawing the punnet square. This can save alot of time when we get into complex combinations. So lets use the mathematical method of determining the results from the above F3 cross.
3P2p x 3P5p = (3*3)P + (3*5)Pp + (2*3)Pp + (2*5)pp = 9PP + 15Pp + 6Pp + 10pp = 9PP + 21Pp + 10pp
As you can see we came up with the same number as the punnet square without drawing all the lines. Now lets use the same formula to calculate the F4 and F5 generations. We will remove the 10pp from the female genepool and be left with 9PP + 21Pp. If we add up all the P's and p's, this works out to [2(9) + 21]P and 21p which translates to 39P21p. The male gene pool will work out to be [2(9)+21]P and [21+2(10)]p = 39P41p. Remember, each number infront of each gene simply represents the frequency of that particular gene.
F4 Cross:
39P21p x 39P41p = 39*39PP + 39*41Pp + 21*39Pp + 21*41pp = 1521PP + 1599Pp + 819Pp + 861pp
= 1521PP + 2418Pp + 861pp and these add up to 4800
Therefore (1521+241/4800 or 82% will have pineapple flavouring and the frequency of the P gene will be 2(1521)+2418/2(4800) = 5460/9600 or 56.8%.
Can you imagine doing that with a punnet square? Even so, as you can see, the literal numbers are getting a little crazy and are becoming hard to follow. It may be easier to start working with gene frequencies in terms of decimals or simply percentages. Percentages are the easiest to follow but there is a trick or two to remember, so I'll stick with simple decimals. So lets move onto the F5 generation using decimals to indicate frequencies. First we have to calculate the gene frequencies of each parental genepool. If you recall, the F4 cross created the following genepool. 1521PP + 2418Pp + 861pp with a total of 4800. When we translate each ratio into a decimal we get 1521/4800PP + 2418/4800Pp + 861/4800pp = .32PP + .50Pp + .18pp after rounding to two decimal places. [Hint: Note that if we add up all our decimals, we get a total of 1. If they don't, a mistake was made.] Therefore (1521+241
/4800 or 82% will have pineapple flavouring and the frequency of the P gene will be 2(1521)+2418/2(4800) = 5460/9600 or 56.8%.
So now lets use the ratios we got from the F4 generation to calculate the gene frequencies of the parental genepools of the F5 cross. All we do is add together the frequency of each gene (gamete) and divide by the total of the ratio used in that genepool. Since the Pp is only half P, we divide this one in half. Therefore, the male parental genepool will be (.32+.25)/1 which equals .57P. Once we know P, we automatically know p since it is simply 1-P or .43. Therefore, the male parental genepool is .57P.43p. Now to determine the female gene frequencies, we need to subract the .18pp from the total numbers since they will be removed from the genepool. This is one way to do it. (.32+.25)/1-.18 = .57/.82 = .695P. Again 1-P=p so we end up with the female gene frequencies of .695P.305p. The F5 cross is as follows:
.695P.305p x .57P.43p = (.695*.57)PP + (.695*.43)Pp + (.303*.57)Pp + (.305*.43)pp = .40PP + .47Pp + .13pp
Finally in conclusion, after 4 generations of inbreeding where we only make our selections from the female population, we end up with an F5 population where 87% taste like pineapple. Plus the frequency of the pineapple gene in the F5 population is 63.5%. This is significantly less than if we were able to apply the selective pressure to both parental genepools. If you recall, in that situation we achieved 96% of the population tasting like pineapple. The frequency of the pineapple gene would have risen to 80%. This is just the case where the flavour gene is dominant, the situation when selecting for recessive traits is much nicer.
When the trait we want is recessive.
In this case, we will assign the symbol p to indicate pineapple flavour and P will indicate Pine flavour. If we find a single pineapple flavoured individual in a population of pine flavoured individuals, and the trait is recessive, then the individual must be homozygous (pp) for the trait. When choosing a mate to cross it with, there is a chance you could select a heterozygous individual, but it's more likely to use a homozygous dominant pine flavoured one so that is what we'll base this next model on. Therefore, our F1 cross will be:
F1 cross = pp x PP = Pp + Pp + Pp + Pp or just simply Pp since all the F1s are the same.
Just to maintain consistency, I will point out that none (0%) of the F1 cross will have pineapple flavour but the frequency of the p gene will be 50%. Now when we move onto the F2 population, our parents will both be Pp. Here is the F2 cross:
F2 cross = Pp x Pp = PP + Pp + Pp + pp
Since there was no selection in choosing the parents, the p gene frequency remained at 50%. However, 25% of the offspring will be pineapple flavoured. As has been shown previously, it is this reassortment within the F2 population that is key. Now we can spot the females that are homozygous (pp) for the pineapple flavour. If we can identify the pineapple flavoured males, then we will be finished with an F3 cross as follows:
F3(a) cross = pp x pp = pp + pp + pp + pp - all true breeding for the pineapple flavour, mission accomplished
But not so fast, many are unable to determine the flavour of a male plant and so therefore wouldn't be able to perform any selections on the male portion of the genepool. Again we are back in a half-sib breeding model. The male population's population be PP + Pp + Pp + pp which equals 4P4p which in turn can be simplified to Pp. The frequency of the p gene in the female genepool will remain 100% from now on in this model. In this case the F3 cross would be:
F3 (b) cross = pp x Pp = Pp + pp + Pp + pp
50% of the F3(b) generation would be pineapple flavoured and the frequency of the pineapple (p) gene has increased to 6/8 or 75%. We would select out the PP female but use both Pp and pp males in the F4 cross. From what we learned in the previous section we could designate our gene frequencies as the female breeding pool = .5p.5p and the male breeding pool as .25P.75p. You see where those came from? Remember that 6/8 or 75% were p from the F3(b) cross? Well the 75% simply becomes .75 when we convert to decimal form. And 1-p=P to arrive at the .25p. Hence our F4 cross is:
F4 (b) cross = .5p.5p x .25P.75p = (.5*.25)Pp + (.5*.75)pp + (.5*.25)Pp + (.5*.75)pp = .25Pp + .75pp or more simply Pp + pp + pp + pp
We'll skip to the F5 generation, see if you can figure where I get my gene frequencies.
F5 (b) cross = .5p.5p x .125P.875p = .125Pp + .875pp
So after doing half sib inbreeding for 4 generations, we achieve an F5 generation where 87.5% of the offspring will be pineapple flavoured and the frequency of the p gene will be 93.75%. Not bad at all, just as good as cubing, but the 100% we achieved with the previous full sib example was better and with two fewer generations (HINT!!).
Please keep in mind that these models assume that flavour is a trait controlled by a single gene or linked group of genes. Reality isn't as simple, but the principles mentioned here apply to more complex models as well. The main point to take from this is that the degree of selection we use can very much influence our success rate. And that selecting for dominant and recessive traits have some subtle differences
Supa smoka
Well-Known Member
I know what ya mean FIWH im uming and ahhing wether i should hit the checkout button
Buck123
Well-Known Member
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Supa smoka
Well-Known Member
They look nice
'ome Grown
Well-Known Member
less money on beans, more money on equipment
