The Cannabis Breeder's Bible by Greg Green Read Free Book Online Page A

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.
     
b
b
B
?
     
    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.
     
b
b
B
Bb
bb
?
?b
?b
    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:
     
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 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.
     
B
b
B
BB
Bb
b
Bb
bb*
* Take special note of this offspring and compare it with the parents.
    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