The Cannabis Breeder's Bible (7 page)

q
²
= 0.36
(q x q) = 0.36
q = 0.6

Thus, the frequency of the ‘v’ allele is 60%.

Question:
According to the Hardy-Weinberg law, what is the frequency of the ‘V’ allele?

Answer:
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 the ‘V’ allele is 40%.

Question:
According to the Hardy-Weinberg law, what is the frequency of the genotypes ‘VV’ and ‘Vv’?

Answer:
Given what we know, the following must be true:

VV = p
²
V = 0.4 = p
(p x p) = p
²
(0.4 x 0.4) = p
²
0.16 = p
²
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 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 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 p
²
+ 2pq + q
²
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 we asked the question: “I have been selecting Indica mothers and crossbreeding them with mostly Indica male plants but I have some Sativa leaves. 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 the Sativa leaf trait.

THE TEST CROSS

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’ve 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 perform what is called a test cross.

Determining the phenotype of a plant is fairly straightforward. You look at the plant and you see, smell, feel or taste its 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:

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 visual characteristics 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 species 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, meaning that the leaf trait must be heterozygous because homozygous dominant and homozygous recessive traits are true breeding. Some of the Skunk’s 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 leaf the heterozygous trait?” Since a completely homozygous recessive population (bb) would not contain the allele (B) for heterozygous expression (Bb) or for homozygous dominant expression (BB), it is impossible for the traits for heterozygous (Bb) or 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 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 that you will discover in that population. Hence, bb is true breeding. If there is a variation in the trait, and the Hardy-Weinberg law of equilibrium 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 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 is homozygous recessive (bb). So if a plant displays a recessive trait in its phenotype, its 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 your inferences until you have completed a test cross.

A test cross is a 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 the opposite sex that is homozygous recessive (bb)
for the same trait.

This brings us 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 not 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.

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 the box in with what we can calculate.

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 1, where ? = B:

Bb + Bb + Bb + Bb = 4Bb
100% Golden Bud

Outcome 2, where ? = b:

Bb + Bb + bb + bb = 2 Bb + 2bb
50% Golden Bud and 50% Silver Bud

Recall:

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 then 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 that trait.

To summarize, the guidelines for performing a test cross to determine the genotype of a plant exhibiting a dominant trait are:

1. The plant with the dominant trait should always be crossed with a plant with the recessive trait.

2. If any offspring display the recessive trait, the unknown genotype is heterozygous.

3. If all the offspring display the dominant trait, the unknown genotype is homozygous dominant.

The main reasons behind performing a test cross are:

1. When you breed plants you want to continue a trait, like height, taste, smell, etc.

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.

We 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.

HARDY - WEINBERG LAW, PART 2

The question may arise: “How do I breed for several traits, like taste, smell, vigor and color?” To answer this question, you will need to learn more about the Hardy-Weinberg law of genetic equilibrium.

If you breed two plants that are heterozygous (Bb) for a trait, what will the offspring look like? The Punnett squares can help us determine the phenotypes, genotypes and gene frequencies of the offspring.

In this group, the resulting offspring will be:

1 BB - 25% of the offspring will be homozygous for the dominant allele (BB)
2 Bb - 50% will be heterozygous, like their parents (Bb)
1 bb - 25% will be homozygous for the recessive allele (bb)

Unlike their parents (Bb and Bb), 25% of offspring will express the recessive phenotype bb. So two parents that display Golden Bud but are both heterozygous (Bb) for that trait will produce offspring that exhibit the recessive Silver Bud trait, despite the fact that neither of the parents displays the phenotype for Silver Bud.

Understanding how recessive and dominant traits are passed down through the phenotype and genotype so that you can predict the outcome of a cross and lock down traits in future generations is really what breeding is all about.

When you breed a strain, how do you know that the traits you want to keep will actually be retained in the breeding process? This is where the test cross comes in. If you create seeds from a strain that you bought from a seed bank, how can you be sure that the offspring will exhibit the characteristics that you like? If the trait you wish to continue is homozygous dominant (BB) in both parent plants then there’s no way that you can produce a recessive genotype for that trait in the offspring, as illustrated in the Punnett square below.

It is impossible for the recessive trait to appear. And if both parents contain the recessive trait then they cannot produce the dominant trait.

In order to breed a trait properly you must know if it is homozygous, heterozygous or homozygous recessive so that you can
predict
the results before they happen.

MENDEL AND THE PEA EXPERIMENTS