The hardy-wineburg model of genetic equilibrium

This is a run down on the hardy-wineburg law..

An understanding of plant breeding requires a basic understanding of the hardy-wineburg law.. to illustrate the value of this law, ask yourself a question, like " if purple bud color is a dominant trait, why do some of the offspring have green bud?" Or " I have been selecting indica mother's and cross-breeding them with mostly indica male plants but I have some sativa leaves. Why?" These questions can be easily answered by developing an understanding of the hardy-wineburg law and the factors that can disrupt genetic equilibrium..
The first of these questions, reflects a very common misconception that the dominant allele of a trait will always have the highest frequency in a population and the recessive allele will always show the lowest frequency.. this is not always the case. A dominant trait will not necessarily spread to a whole population, nor will a recessive trait die out..

Gene frequencies can occur in high or low ratios, regardless of how the allele is expressed. The allele can also change, depending on certain conditions. It is these changes in gene frequencies over time that result in different plant characteristic's

A genetic population is basically a group of individuals of the same species(cannabis indica or cannabis sativa) or strain ( skunk #1 or master kush) in a given area whose members can breed with one another. This means that they must share a common group of genes. This common group of genes is locally known as the gene pool.. the gene pool contains the alleles for all of the traits in the entire population..for a step in evolution- a new plant species,strain or trait - to occur, some of the gene frequencies must change. The gene frequency of an allele refers to the number of alleles for that trait in a population.. gene frequency is calculated by dividing the number of specific type of allele by the total number of alleles in the gene pool..

Genetic equilibrium theory and applications

The hardy-wineburg model of equilibrium describes a theoretical situation in which there is no change in the gene pool.. At equilibrium there can be no change or evolution.
Let's consider a population whose gene pool contains the alleles B and b.
Assign the letter p to the frequency of the dominant allele B and the letter q to the frequency of recessive allele b. We know the sum of all the alleles are 100% so:

p+q=100%
This can also be expressed as :
p+q=1
And all the random possible combinations of the members of the population would equal: p2+2pq+q2

WHERE:
p = frequency of the dominant allele in a population
q=frequency of the recessive allele in a population
p2=percentage of homozygous dominant individuals
q2=percentage of heterozygous recessive individuals
2pq=percentage of heterozygous individuals

Imagine that you have grown a population of 1000 ' black domina' cannabis plants from seeds obtained from a well known seed bank. In that population, 360 plants emit a skunky smell, while the remaining 640 plants emit a fruity smell.
You contact the seed bank and ask them which smell is the dominant in this particular strian. Hypothetically, they tell you that the breeder selected for fruity smell and the skunk smell is the recessive genotype. You can call this recessive genotype 'vv' and use the formula above to answer the following questions..

QUESTION: according to the hardy-wineburg law, what is the frequency of the 'vv' genotype?
ANSWER: since 360 out of a 1000 plants have the 'vv' genotype, then 36% is the frequency of 'vv' in this population of 'black domina'
QUESTION: according the hardy-wineburg law, what is the frequency of the 'v'allele?
ANSWER: the frequency of 'vv' is 36% . Since q2 is the percentage of homozygous recessive individuals, and q is the frequency of the recessive allele in a population, the following must also be true:
q2=0.36
(q×q)=0.36
q=0.6
Thus , the frequency of 'v'allele is 60%

Since q=0.6 we can solve for p.
p+q=1
p+0.6=1
p=1-0.6
P=0.4
The frequency of 'V' allele is 40%

QUESTION: according to the hardy-wineburg law , what is the frequency of the genotypes 'VV'and 'Vv'
ANSWER: given what we know the following must be true:
VV=p2
V=0.4=p
(p ×p)=p2
(0.4 × 0.4)=p2
0.16=p2
VV=0.16
The frequency of the genotype 'VV' is 16%
VV=0.16
vv=0.36
VV + Vv +vv =1
0.16+ Vv + 0.36= 1
0.52 + Vv =1
Vv= 1 - 0.52
Vv = 0.48 or 48%
Or alternatively'Vv' is 2pq, therefore:
Vv= 2pq
2pq= 2 × p × q
2pq= 2 × 0.4 × 0.6
2pq = 0.48 or 48%
The frequency of V and v (p and q) will remain unchanged, generation after generation, as long as the following five statements are true
1. The population is large enough
2. There are no mutations
3. There are no preferences, for example A VV male does not prefer a vv female by its nature.
4. No other outside population exchanges genes with this population
5. Natural selection does not favor any specific gene

The equation p2 + 2pq + q2 can be used to calculate the different frequencies. Although this equation is important to know about, we make use of other more basic calculations when breeding. The important thing to note here is the five conditions for equilibrium.
Earlier the question was asked" I've been selecting indica mother's and cross-breeding them with mostly indica males but some of the plants have sativa leaves. Why?'' the hardy-wineburg equilibrium tells us that outside genetics may have been introduced into the breeding program.. since the mostly indica male plants are mostly indica and not pure indica, you can expect to discover some sativa characteristic's in the offspring,, including the sativa leaf trait..

Please don't post on this thread until I've completed it...

I'll be going over test crosses tomorrow..

Again!! Please don't post until I'm finished.. this is going to take several days for me to write this up...

THE TEST CROSS

Some of you maybe asking the question, " how do I know if a trait such as bud color is homozygous dominant(BB) , heterozygous(Bb) or homozygous recessive(Bb)?

If you have been given seeds or a clone you may of been told that a trait, such as potency, is homozygous dominant, heterozygous or homozygous recessive.. however you'll want to establish this yourself, especially if you intend to use those specific traits in a future breeding plan.. to do this, you will have to preform what is called a test cross..

Determining the phenotype of a plant is pretty straightforward.. you look at the plant and you see, smell, feel or taste it's phenotype. Determining the genotype cannot be achieved through simple observation alone.
Generally speaking, there are three possible genotypes for each plant trait.. for example, if golden bud is dominant and silver bud is recessive, the possible genotypes are:

Homozygous dominant:. BB= golden bud
Heterozygous: Bb = golden bud
Homozygous recessive: bb = silver bud

The golden and silver bud colors are the phenotypes..BB,Bb and bb denote the genotypes. Because B is the dominant allele, Bb would appear golden and not silver. Most phenotypes are visible characteristic's but some, like bud taste, are phenotypes that can't be observed by the naked eye and are experienced instead through the other senses..
For example, looking at a mostly sativa like a skunk plant you will notice that the leaves are pale green. In a population of these skunk plants you may notice that a few have dark green leaves.. this suggests that this skunk strain's leaf color is not true breeding (tbl) meaning that the leaf trait must be heterozygous dominant in this population.

Not finished.. please no posting until I'm done writing this up.. thank you..

THE TEST CROSS ( continued)

You maybe asking the question: " could the pale green trait be the homozygous recessive trait and the dark green trait heterozygous?" Since a completely homozygous recessive population (bb) would not contain the allele (B) for heterozygous expression (Bb) or for homozygous dominant(BB) to exist in a population that is completely homozygous recessive(bb) for that trait. If a population is completely homozygous for that trait( bb or BB),then that specific trait can be considered stable, true breeding or ' will breed true'. If a population is heterozygous for that trait(Bb) then that specific trait can be considered unstable, not breeding true or ' will not breed true'.

If the trait for Bb or BB can not exist in a bb population for that trait, then bb is the only trait you will discover in that population. Hence, bb is true breeding. If there is a variation in the trait, and the hardy-wineburg law of equilibrium has not been broken, the trait must be heterozygous. In our skunk example there were only a few dark leaves. This means that the dark green leaves are homozygous recessive and the pale green leaves are heterozygous and may possibly be homozygous dominant too.
You may also notice that the bud is golden on most of the plants. This also suggests that the golden bud color is a dominant trait.. if buds on only a few of the plants are silver, this suggests that the silver trait is recessive. You know the only genotype that produces the recessive trait in its phenotype, it's genotype must be homozygous recessive. A plant that displays a recessive trait in its phenotype always has a homozygous recessive genotype. But this leaves you with an additional question to answer as well: Are the golden bud or pale green leaf color traits homozygous dominant(BB) or heterozygous (Bb)? You cannot be completely certain of any of you inferences until you have completed the test cross..

A test cross is performed by breeding a plant with an unknown dominant genotype(BB or Bb) with a plant that is homozygous recessive(bb) for the same trait. For this test you will need another cannabis plant of opposite sex that is homozygous recessive(bb) from the same trait.

This brings us to an important rule: if any offspring from a test cross displays the recessive trait, the genotype of the parent with the dominant trait must be heterozygous and not homozygous..

( Please no posting until I'm finished.. expect several days) thank you

1.mutation: A mutation is a change in genetic material, which can give rise to heritable variations in the offspring. Exposure to radiation can cause a genetic mutation, for example. In this case the result would be a mutation of the plants generic code that would be transferred to its offspring. The effect is equivalent to a migration of foreign genetic material being introduced into the population. There are other factors that can cause mutation. Essentially a mutation is the result of DNA repair failure at the cellular level. Anything that causes DNA repair to fail can result in a mutation.
2. Gene mutation:
Over time, a population will reach equilibrium that will be maintained as long as no other genetic material migrates into the population. When new genetic material is introduced from another population, this is called introgression. During the process of introgression many new traits can arise in the original population, resulting in a shift of equilibrium.
3. Genetic drift:
If a population is small, equilibrium is more easily violated, because of a slight change in the number of alleles results in a significant change in genetic frequency. Even by chance alone certain traits can be eliminated from the population and the frequency of alleles can drift toward higher or lower values. Genetic drift is actually an evolutionary force that alters a population and demonstrates that the hardy-wineburg law of equilibrium cannot hold true over an indefinite period of time.
4. Non-random mating:
External or internal factors may influence a population to a point at which mating is no longer random. For example, if some female flowers develop earlier than others they will be able to gather pollen earlier then the rest. If some of the males release pollen early and then stop producing pollen. The mating between these early males and females is not random, and could result in late-flowering females ending up as a sinsemilla crop. This means the late flowering won't be able to make their contribution to the gene pool in future generations. Equilibrium will not be maintained.
5. Natural selection:
With regards to natural selection, the environment and other factors can cause certain plants to produce a greater or smaller number of offspring. Some plants may have traits that make them less immune to disease, for example, meaning that when the population is exposed to disease, less of their offspring will survive to pass on genetic material, while others may produce more seeds or exhibit a greater degree of immunity, resulting in a greater number of offspring surviving to contribute genetic material to the population..


( Please no posting until I'm done writing this up.. expect several days)

How to true breed a strain:

Breeding cannabis strains is about manipulating gene frequencies. Most strains sold by reputable breeders through seed banks are very uniform in growth. This means that the breeder has attempted to lock certain genes down so that the genotypes of those traits are homozygous.
Imagine that a breeder has two strains: master kush and silver Haze. The breeder lists a few traits that they particularly like(denoted by *).

This means they want to create a plant that is homozygous for the following traits and call it something like silver kush.
SILVER KUSH
Pale green leaf
Hashy smell
Silver flowers
Short plants

All the genetics needed are contained in the gene pools for master kush and silver Haze. The breeder could simply mix both populations and hope for the best or try and save time , space ,and money by calculating the genotype for each trait and using the results to create an IBL.
The first thing the breeder must do is to understand the genotype of each trait that will be featured in ideal "SILVER KUSH" strain. In order to do this genotype of each parent strain for that same trait must be understood. Since there are four traits that the breeder is trying to isolate, and 4×2= 8, alleles make up the genotypes for these phenotype expressions and must be made known to the breeder.
Let's take the pale green leaf trait of silver Haze for starters. The breeder will grow out as many silver Haze plants as possible, noting if any plants in the population display other leaf colors. If they do not, the breeder can assume that the trait is heterozygous(Ss) and must be locked down through selective breeding.. let's look closely at the parents for a moment.

|. |. S. | SS. |
| S. |. SS.|. SS |
|. S. |. SS |. SS. |

In both parents were SS there wouldn't be any variation in the population for this trait. I would already be locked down and would always breed true without any variations.

|. |. S. |. s. |
|. S. |. SS |. Ss. |
|. S. | SS |. Ss. |

With one SS parent and one Ss parent the breeder would produce a 50/50 population. One group being homozygous(SS) and the other heterozygous(Ss) .

In both parents were SS, the breeder would have a 25% percent SS, 50%percent Ss, and a 25% percent ss. Even though the Gene frequencies can be predicted, the breeder will not know with certainty whether the pale green leaf trait is dominant or recessive until the perform a test cross. By running several test crosses the breeder can isolate the plant that is either SS or ss and eliminate any Ss from the group. Once the genotype has been isolated and the population reduced to contain only plants with the same genotype, the breeding program can begin in earnest. Remember that the success of any cannabis breeding program hinges on the breeder maintaining accurate records about the parent plants and their descendants so that they can control gene frequencies.

( Not complete, no posting until I'm down with this write up please expect several day)

Hate to be the first one to post against your wishes, but it's been a month and a half. I want more!!!

1) Hardy-Weinberg

2) Not sure how this is relevant to selective breeding. Harry-Weinberg is population genetics theory. It allows population geneticists to identify when allele frequencies deviate from equilibrium, suggesting some external factor, like natural selection, is in play. As a breeder, that external factor is you. You’re applying selection, you’re limiting the population size, thus violating Hardy-Weinberg.

Also, in the vast majority of situations, traits under selection are neither simple dominant or recessive, rather they are complex traits with many loci responsible for the phenotype. For a simple example, hybrids are rarely indica short or sativa tall, they run a spectrum from short to tall. Understanding Mendelian genetics isn’t exactly useful in these situations because it’s extremely difficult to identify how many loci underlie the trait.