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combinations of the members of a population would equal:
p 2 + 2pq + q 2
    Where:
p = frequency of the dominant allele in a population
q = frequency of the recessive allele in a population
p 2 = percentage of homozygous dominant individuals
q 2 = percentage of heterozygous recessive individuals
2pq = percentage of heterozygous individuals
    Imagine that you have grown a population of 1,000 ‘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 dominant in this particular strain. Hypothetically, they tell you that 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.
     
    Question: According to the Hardy-Weinberg law, what is the frequency of the ‘vv’ genotype?
     
    Answer: Since 360 out of the 1,000 plants have the ‘vv’ genotype, then 36% is the frequency of ‘vv’ in this population of ‘Black Domina.’
     
    Question: According to the Hardy-Weinberg law, what is the frequency of the ‘v’ allele?
     
    WILD POLLINATION
    Cannabis pollination and seed production.

    Answer: The frequency of the ‘vv’ allele is 36%. Since q 2 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:
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