KiLoEleMeNt
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Phenotype: The observable physical or bio-chemical characteristics of an organism , as determined by both genetic makeup and environmental influences.
Genotype: the genetic constitution of an individual, especially as distinguished from the phenotype; the whole of the genes in an individual or group
An understanding of plant breeding requires a basic understanding of the Hardy-Weinberg law. To illiterate the values of the Hardy-Weinberg law, ask yourself a question, like "If purple bud color is a dominant trait, why do some of the offspring of my purple bud strain have green buds?" Or "I have been selecting Indica mothers and cross-breeding them with mostly Indica male plants, but I have some sativa leafs why?" These Questions can be easily answered by developing an understanding of the Hardy-Weinberg law and the factors that disrupt genetic equilibrium.
The first of these questions reflects a very common misconception: that the dominant allele will always have the highest frequency in a population and the recessive allele will always have 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 always eventually 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 characteristics.
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 givin 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 knows 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 the number of times an allele for a particular trait occurs compared to the total number of alleles for that trait in the population. Gene frequency is calculated by dividing the number of a specific type of allele by the total number of alleles in the gene pool
The Hardy-Weinberg model of genetic equilibrium describes a theoretical situation in which there is no change in the pool. At equilibrium there can be no change or evolution.
Let's consider a population whose pool contains the alleles B and b
Assign the letter p to the dominant allele B and the letter q to the frequency of the recessive allele b. We know that the sum of all the alleles must equal 100%, so:
p + q = 100%
This can also be expressed as
p + q = 1
All the possible combinations of the members of a 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.
So now imagine that, hypothetically, you have grown a population of 1000 "black domina" plants from seeds obtained from a well known seed Bank. In that population 360 plants Emmitt a skunk smell, while the remaining 640 plants Emmitt a fruity smell you contact the seed Bank and ask them which smell is dominant in this particular strain. They tell you the breeder selected for a fruity smell and the skunk smell is a recessive genotype. You can call this recessive genotype "vv" and use the formula above to answer the following questions.
According to the Hardy-Weinberg law
Q.what is the frequency of "vv" genotype?
A. Since 360 out of 1000 plants have the "vv" genotype then 36% is the frequency of "vv" in this population of 'black domina'
Q. what is the frequency of the "v" allele?
A. The frequency of the "vv" allele is 36%. Since q2 is the percentage of homogeneous recessive individuals, and q is the frequency of the recessive allele in a population the following must be true: q2 = 0.36 | (q x q) = 0.36 | q = 0.6 Thus the frequency of the "v" allele is 60% !?! what is the frequency of the "V" allele A. Since the value of 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 the "V" allele is 40%
Q. What is the frequency of the genotypes 'VV'? A. Given what we know, the following must be true: VV = p2 | V = 0.4 = p | (p x p) = p2 | (0.4 x 0.4) = p2 | 0.6 = p2 | VV = 0.16 The frequency of "VV" is 16% VV = 16% | vv = 0.36 | VV + Vv + vv = 1 | 0.16 + Vv + o.36 = 1 | 0.52 + Vv = 1 | Vv = 1 - 0.52 | Vv = 0.48 or 48% / alternatively, Vv is 2pq, therefore Vv= 2pq |2pq = 2 x p x q | 2pq = 2 x 0.4 x 0.6 | 2pq = 0.48 or 48%
The frequencies of V and v (p and q ) will remain unchanged, generation after generation as long as the five following statements are true: 1. the population is big enough 2. There are no mutations, 3. There are no preferences; for example, a VV males 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, much more basic calculations when breedeing. The important thing to note here is the five conditions for equilibrium. Earlier we asked the question "I have been breeding Indica mothers and cross-breeding them with mostly Indica male plants and I have some sativa leafs why? The Hardy-Weinberg equilibrium tells us that outside genetics may have been introduced into the breeding program. Since the mostly Indica male plants are only mostly Indica and not pure Indica, you can expect to discover some sativa characteristics in the offspring, including sativa leaf traits.
Now some of you may be 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 have been told that a trait, such as potency, is homozygous dominant, heterozygous or homozygous recessive. However, you will 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 do what is called a test cross.
Determining the phenotype of a plant is fairly straight forward. You look at the plant and you see, smell, feel, or taste its phenotype. Determining the genotype can not 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 buds 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 visual characteristics but some, like bud taste, are phenotypes that can not be observed by the naked eye and are experienced instead of through the other senses. For example looking at a mostly sativa species like a skunk plant you will notice that the leafs are pale green. In a population of these skunk plant you may notice that a few have dark green leaves. This suggests that this skunks skunk strains leaf color is not true breeding, meaning that the leaf trait must be heterozygous because homozygous dominant and homozygous recessive traits are true breeding,. Some of the skunks pale green leaf traits will probably be homozygous dominant in this population. You may also be asking the question "could the pale green trait be the homozygous recessive trait and the dark green heterozygous trait?" Since a completely homozygous recessive population (bb) would not contain the alleles (B) for heterozygous expression (Bb) or for homozygous dominant expression (BB), it is impossible for the traits to be heterozygous (Bb) or homozygous dominant (BB) to exist in a population that is completely homozygous recessive 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 populationI is heterozygous for that trait (Bb) then that specific trait can be considered unstable, not true breeding, or will not breed true. If the trait for Bb or BB cannot exist in a bb population for that trait, then bb is the only trait you will dicover in that population. Hence, Bb is true breeding. If there is a variation in the trait, and the Hardy-Weinberg equilibrium law has not been broken, the trait must be heterozygous. In our skunk example there were only a few dark green leaves. This means that the leafs are homozygous recessive and the pale green leaves are heterozygous and may possibly be homozygous dominant too. You may also notice the bud is golden on most 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 is homozygous recessive (bb). So if a plant displays a recessive trait in its phenotype, it's genotype must be homozygous recessive genotype. This leaves you with an additional question to answer as well: are the golden buds or the pale green leaf color traits homozygous dominant (BB) or heterozygous (Bb)? You can not be completely certain of your inferences until you have completed a test cross.
A test cross is preformed 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 a plant of the opposite sex that is homozygous recessive for the same trait.
This brings to an important rule: If any offspring from a test cross display the recessive trait, the genotype of the parent with the dominant trait must be heterozygous and homozygous.
In our example our unknown genotype is either BB or Bb. The silver bud genotype is bb. We'll put this information into a mathematical series known as punnett squares ( this looks and is filled out exactly like a times table chart or battleship grid) We start by entering the known genotypes. We do these calculations for two parents that will breed we know that our recessive trait is bb and the other is either BB or Bb, so we'll use B? For the time being. Our next step is to fill in the boxes with what we can calculate it should look something like this
In this particular chart The first row of offspring Bb and Bb will have the dominant trait of golden bud. The second row can either contain Bb or bb offspring. This will either lead to offspring that will produce more golden bud (Bb) or silver bud (bb). The first possible outcome (where ? = B ) would give us golden bud (Bb) offspring. The second possible outcome where ( ? = b) would give us silver bud (bb) offspring. We can also predict what the frequency will be. Outcome where ? = B: Bb + Bb + Bb + Bb = 4Bb 100% golden bud. Outcome 2, where ? = b: Bb +Bb + bb + bb = 2Bb + 2Bb 50% golden 50% silver bud
Recall Homozygous dominant: BB = golden bud, Heterozygous Bb = Golden bud, Homozygous Recessive: bb = silver bud
To determine the identity of B? We used another cannabis plant of the opposite sex that was homozygous recessive (bb) for the same trait. Outcome 2 tells us that: both parents must have at least one b trait each to exhibit silver bud in the phenotype of the offspring. If any silver bud is produced in the offspring than the mystery parent (B?) Must be heterozygous (Bb). It cannot be homozygous dominant (BB)
So if a golden bud parent is crossed with a silver bud parent and produces only golden bud, then the golden bud parent must be homozygous dominant for that trait. If any silver bud offspring is produced, then the golden bud parent must be heterozygous for this trait. To summarize, the guidelines for performing a test cross is to determine the genotype of a plant exhibiting a domitrait are: 1. the plant with a dominant trait should always be crossed with a plant with a recessive trait. 2. If any offspring display recessive trait, the unknown genotype is heterozygous. 3. If all the offspring display the dominant trait the unknown genotype is homozygous dominant.
The main reasoning behind performing a test cross are. 1. When you breed plants you want to continue a trait, like hight, taste, smell, ect. 2. When you want to continue that trait you must know if it is homozygous dominant heterozygous or homozygous recessive. 3. You can only determine this with certainty by performing a test cross I should mention that, as a breeder you should be dealing with a large population in order to be certain of the results. The more plants you work with, the more reliable the results
TO BE CONTINUED
Genotype: the genetic constitution of an individual, especially as distinguished from the phenotype; the whole of the genes in an individual or group
An understanding of plant breeding requires a basic understanding of the Hardy-Weinberg law. To illiterate the values of the Hardy-Weinberg law, ask yourself a question, like "If purple bud color is a dominant trait, why do some of the offspring of my purple bud strain have green buds?" Or "I have been selecting Indica mothers and cross-breeding them with mostly Indica male plants, but I have some sativa leafs why?" These Questions can be easily answered by developing an understanding of the Hardy-Weinberg law and the factors that disrupt genetic equilibrium.
The first of these questions reflects a very common misconception: that the dominant allele will always have the highest frequency in a population and the recessive allele will always have 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 always eventually 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 characteristics.
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 givin 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 knows 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 the number of times an allele for a particular trait occurs compared to the total number of alleles for that trait in the population. Gene frequency is calculated by dividing the number of a specific type of allele by the total number of alleles in the gene pool
The Hardy-Weinberg model of genetic equilibrium describes a theoretical situation in which there is no change in the pool. At equilibrium there can be no change or evolution.
Let's consider a population whose pool contains the alleles B and b
Assign the letter p to the dominant allele B and the letter q to the frequency of the recessive allele b. We know that the sum of all the alleles must equal 100%, so:
p + q = 100%
This can also be expressed as
p + q = 1
All the possible combinations of the members of a 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.
So now imagine that, hypothetically, you have grown a population of 1000 "black domina" plants from seeds obtained from a well known seed Bank. In that population 360 plants Emmitt a skunk smell, while the remaining 640 plants Emmitt a fruity smell you contact the seed Bank and ask them which smell is dominant in this particular strain. They tell you the breeder selected for a fruity smell and the skunk smell is a recessive genotype. You can call this recessive genotype "vv" and use the formula above to answer the following questions.
According to the Hardy-Weinberg law
Q.what is the frequency of "vv" genotype?
A. Since 360 out of 1000 plants have the "vv" genotype then 36% is the frequency of "vv" in this population of 'black domina'
Q. what is the frequency of the "v" allele?
A. The frequency of the "vv" allele is 36%. Since q2 is the percentage of homogeneous recessive individuals, and q is the frequency of the recessive allele in a population the following must be true: q2 = 0.36 | (q x q) = 0.36 | q = 0.6 Thus the frequency of the "v" allele is 60% !?! what is the frequency of the "V" allele A. Since the value of 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 the "V" allele is 40%
Q. What is the frequency of the genotypes 'VV'? A. Given what we know, the following must be true: VV = p2 | V = 0.4 = p | (p x p) = p2 | (0.4 x 0.4) = p2 | 0.6 = p2 | VV = 0.16 The frequency of "VV" is 16% VV = 16% | vv = 0.36 | VV + Vv + vv = 1 | 0.16 + Vv + o.36 = 1 | 0.52 + Vv = 1 | Vv = 1 - 0.52 | Vv = 0.48 or 48% / alternatively, Vv is 2pq, therefore Vv= 2pq |2pq = 2 x p x q | 2pq = 2 x 0.4 x 0.6 | 2pq = 0.48 or 48%
The frequencies of V and v (p and q ) will remain unchanged, generation after generation as long as the five following statements are true: 1. the population is big enough 2. There are no mutations, 3. There are no preferences; for example, a VV males 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, much more basic calculations when breedeing. The important thing to note here is the five conditions for equilibrium. Earlier we asked the question "I have been breeding Indica mothers and cross-breeding them with mostly Indica male plants and I have some sativa leafs why? The Hardy-Weinberg equilibrium tells us that outside genetics may have been introduced into the breeding program. Since the mostly Indica male plants are only mostly Indica and not pure Indica, you can expect to discover some sativa characteristics in the offspring, including sativa leaf traits.
Now some of you may be 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 have been told that a trait, such as potency, is homozygous dominant, heterozygous or homozygous recessive. However, you will 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 do what is called a test cross.
Determining the phenotype of a plant is fairly straight forward. You look at the plant and you see, smell, feel, or taste its phenotype. Determining the genotype can not 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 buds 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 visual characteristics but some, like bud taste, are phenotypes that can not be observed by the naked eye and are experienced instead of through the other senses. For example looking at a mostly sativa species like a skunk plant you will notice that the leafs are pale green. In a population of these skunk plant you may notice that a few have dark green leaves. This suggests that this skunks skunk strains leaf color is not true breeding, meaning that the leaf trait must be heterozygous because homozygous dominant and homozygous recessive traits are true breeding,. Some of the skunks pale green leaf traits will probably be homozygous dominant in this population. You may also be asking the question "could the pale green trait be the homozygous recessive trait and the dark green heterozygous trait?" Since a completely homozygous recessive population (bb) would not contain the alleles (B) for heterozygous expression (Bb) or for homozygous dominant expression (BB), it is impossible for the traits to be heterozygous (Bb) or homozygous dominant (BB) to exist in a population that is completely homozygous recessive 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 populationI is heterozygous for that trait (Bb) then that specific trait can be considered unstable, not true breeding, or will not breed true. If the trait for Bb or BB cannot exist in a bb population for that trait, then bb is the only trait you will dicover in that population. Hence, Bb is true breeding. If there is a variation in the trait, and the Hardy-Weinberg equilibrium law has not been broken, the trait must be heterozygous. In our skunk example there were only a few dark green leaves. This means that the leafs are homozygous recessive and the pale green leaves are heterozygous and may possibly be homozygous dominant too. You may also notice the bud is golden on most 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 is homozygous recessive (bb). So if a plant displays a recessive trait in its phenotype, it's genotype must be homozygous recessive genotype. This leaves you with an additional question to answer as well: are the golden buds or the pale green leaf color traits homozygous dominant (BB) or heterozygous (Bb)? You can not be completely certain of your inferences until you have completed a test cross.
A test cross is preformed 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 a plant of the opposite sex that is homozygous recessive for the same trait.
This brings to an important rule: If any offspring from a test cross display the recessive trait, the genotype of the parent with the dominant trait must be heterozygous and homozygous.
In our example our unknown genotype is either BB or Bb. The silver bud genotype is bb. We'll put this information into a mathematical series known as punnett squares ( this looks and is filled out exactly like a times table chart or battleship grid) We start by entering the known genotypes. We do these calculations for two parents that will breed we know that our recessive trait is bb and the other is either BB or Bb, so we'll use B? For the time being. Our next step is to fill in the boxes with what we can calculate it should look something like this
In this particular chart The first row of offspring Bb and Bb will have the dominant trait of golden bud. The second row can either contain Bb or bb offspring. This will either lead to offspring that will produce more golden bud (Bb) or silver bud (bb). The first possible outcome (where ? = B ) would give us golden bud (Bb) offspring. The second possible outcome where ( ? = b) would give us silver bud (bb) offspring. We can also predict what the frequency will be. Outcome where ? = B: Bb + Bb + Bb + Bb = 4Bb 100% golden bud. Outcome 2, where ? = b: Bb +Bb + bb + bb = 2Bb + 2Bb 50% golden 50% silver bud
Recall Homozygous dominant: BB = golden bud, Heterozygous Bb = Golden bud, Homozygous Recessive: bb = silver bud
To determine the identity of B? We used another cannabis plant of the opposite sex that was homozygous recessive (bb) for the same trait. Outcome 2 tells us that: both parents must have at least one b trait each to exhibit silver bud in the phenotype of the offspring. If any silver bud is produced in the offspring than the mystery parent (B?) Must be heterozygous (Bb). It cannot be homozygous dominant (BB)
So if a golden bud parent is crossed with a silver bud parent and produces only golden bud, then the golden bud parent must be homozygous dominant for that trait. If any silver bud offspring is produced, then the golden bud parent must be heterozygous for this trait. To summarize, the guidelines for performing a test cross is to determine the genotype of a plant exhibiting a domitrait are: 1. the plant with a dominant trait should always be crossed with a plant with a recessive trait. 2. If any offspring display recessive trait, the unknown genotype is heterozygous. 3. If all the offspring display the dominant trait the unknown genotype is homozygous dominant.
The main reasoning behind performing a test cross are. 1. When you breed plants you want to continue a trait, like hight, taste, smell, ect. 2. When you want to continue that trait you must know if it is homozygous dominant heterozygous or homozygous recessive. 3. You can only determine this with certainty by performing a test cross I should mention that, as a breeder you should be dealing with a large population in order to be certain of the results. The more plants you work with, the more reliable the results
TO BE CONTINUED