MC

Notes on Mendelian Genetics: Crosses and Probability

Genetic Crosses: Monohybrid, Dihybrid, and Test Crosses

Best Practices for Conducting Genetic Crosses

  • List Parental Genotypes First: The initial step for any genetic cross is to clearly identify the genotypes of the parents involved.
  • Determine Gamete Genotypes: Once parental genotypes are known, ascertain the types of gametes each parent can produce. These gametes will be the components placed on the axes of a Punnett square.
    • Gametes are Haploid: Remember that gametes contain only one allele for each gene, reflecting the haploid state.
  • Mendel's Law of Segregation: This fundamental law states that during gamete formation, the two alleles for a heritable character in an individual (e.g., Gg) will separate (segregate) from each other. Each gamete receives only one of the two alleles (either G or g), transitioning from a diploid, heterozygous state to haploid gametes of two types.

Monohybrid Crosses: Inheritance of a Single Trait

Example: Sparty's Color (Green vs. White)

  • Phenotypes: Green (dominant, represented by G) and White (recessive, represented by g).
  • Parental Cross (P Generation): True-breeding green Sparty crossed with a true-breeding white Sparty.
    • Genotypes: GG imes gg
    • Gametes:
      • Green parent (GG) produces only G gametes.
      • White parent (gg) produces only g gametes.
  • First Filial Generation (F1):
    • Genotype: Gg
    • Phenotype: All offspring are Green (heterozygous, expressing the dominant trait).
  • Second Filial Generation (F2) from F1 Self-Cross: A cross between two F1 individuals (Gg imes Gg).
    • Gametes for each F1 parent:
      • Each Gg parent has a 50\% chance of producing a G gamete and a 50\% chance of producing a g gamete.
    • Punnett Square Setup: Parent 1's gametes (G, g) on one axis, Parent 2's gametes (G, g) on the other.
    • F2 Offspring Genotypes:
      • GG: 1/4 probability
      • Gg: 1/2 probability (1/4 + 1/4 from two combinations)
      • gg: 1/4 probability
    • F2 Genotypic Ratio: 1 ext{ (homozygous dominant)} : 2 ext{ (heterozygous)} : 1 ext{ (homozygous recessive)} (1:2:1)
    • F2 Phenotypes:
      • Green (dominant phenotype, from GG and Gg genotypes): 3/4 probability
      • White (recessive phenotype, from gg genotype): 1/4 probability
    • F2 Phenotypic Ratio: 3 ext{ (dominant)} : 1 ext{ (recessive)}

Dihybrid Crosses: Inheritance of Two Independent Traits

  • Parental Generation (P): True-breeding round yellow seeds (dominant for both) crossed with true-breeding wrinkled green seeds (recessive for both).
    • Genotypes: RRYY imes rryy
    • Gametes:
      • Round Yellow parent (RRYY) produces only RY gametes.
      • Wrinkled Green parent (rryy) produces only ry gametes.
  • First Filial Generation (F1):
    • Genotype: RrYy
    • Phenotype: All offspring are Round Yellow seeds (heterozygous for both traits).
  • Determining F1 Gamete Genotypes (for F2 generation):
    • The FOIL method (First, Outer, Inner, Last) can be used for a dihybrid heterozygote (RrYy):
      • First alleles: RY
      • Outer alleles: Ry
      • Inner alleles: rY
      • Last alleles: ry
    • Each gamete will contain one allele for each gene (e.g., one R/r and one Y/y).
  • Independent Assortment: Since the genes for seed shape (R/r) and seed color (Y/y) are on different chromosomes (or sufficiently far apart on the same chromosome to act as if unlinked), these four gamete types (RY, Ry, rY, ry) are produced in equal proportions (1/4 each).
  • Second Filial Generation (F2) from F1 Self-Cross: RrYy imes RrYy
    • Punnett Square: A 4 imes 4 Punnett Square is needed, with the four gamete types (RY, Ry, rY, ry) from each parent on the axes. This can be a very tedious process.
    • F2 Genotypic Ratio (example): 1:2:1:2:4:2:1:2:1 (This reflects the combinations of homozygous and heterozygous states for both genes).
    • F2 Phenotypic Ratio: 9 ext{ (dominant for both)} : 3 ext{ (dominant for one, recessive for other)} : 3 ext{ (recessive for one, dominant for other)} : 1 ext{ (recessive for both)}
      • e.g., 9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green

Independent Assortment Explained

  • Definition: Mendel's Law of Independent Assortment states that the alleles of one gene segregate independently of the alleles of another gene during gamete formation. This typically occurs when genes are on different non-homologous chromosomes.
  • Mechanism (Meiosis I):
    • During Anaphase I of meiosis, homologous chromosomes separate.
    • Two possible orientations of homologous chromosome pairs at the metaphase plate are equally likely:
      • Coupling-like arrangement: If G and A go together to one pole, and g and a go to the other. This would lead to GA and ga gametes.
      • Repulsion-like arrangement: If G and a go together to one pole, and g and A go to the other. This would lead to Ga and gA gametes.
    • Because both arrangements are equally probable, all four possible gamete combinations (GA, Ga, gA, ga) are produced in equal frequency (1/4 each) from a dihybrid heterozygote (GgAa).
    • Note on Linkage (future topic): When genes are linked (on the same chromosome and close together), this independent assortment pattern is altered. However, for current problems, we assume independent assortment.

Test Crosses: Revealing Unknown Genotypes

  • Purpose: To determine the genotype of an individual that exhibits a dominant phenotype (e.g., Round pea plant), but whose genotype (homozygous dominant RR or heterozygous Rr) is unknown.
  • Procedure: The individual of unknown genotype and dominant phenotype is crossed with a homozygous recessive individual.
    • Example: A mystery Round pea plant is crossed with a pure-breeding Wrinkled pea plant (rr).
      • The homozygous recessive parent (rr) can only produce r gametes.
  • Interpreting Results:
    • Scenario 1: Mystery parent is heterozygous (Rr).
      • Cross: Rr imes rr
      • Offspring: 1/2 ext{ Round } (Rr) : 1/2 ext{ Wrinkled } (rr)
      • If the offspring show both dominant and recessive phenotypes in a 1:1 ratio, the mystery parent must have been heterozygous.
    • Scenario 2: Mystery parent is homozygous dominant (RR).
      • Cross: RR imes rr
      • Offspring: All Round (Rr)
      • If all offspring exhibit the dominant phenotype, the mystery parent must have been homozygous dominant.
  • Principle: The appearance of any recessive offspring immediately indicates that the mystery parent carried the recessive allele, thus being heterozygous. If no recessive offspring appear among a sufficiently large sample, it strongly suggests the mystery parent is homozygous dominant.

Probability in Genetics

  • Key Idea: For independent genes, calculate probabilities for each gene separately, then combine them using probability rules.
1. Addition Rule ('OR')
  • When to Use: Calculates the probability that one of two or more mutually exclusive events will occur. You add their individual probabilities.
  • Formula: P(A ext{ or } B) = P(A) + P(B)
  • Example: The probability of having an XY child OR an XX child is P(XY) + P(XX) = 1/2 + 1/2 = 1. (Assumes normal sex determination and equal likelihood).
2. Multiplication Rule ('AND')
  • When to Use: Calculates the probability that two or more independent events will occur together. You multiply their individual probabilities.
  • Formula: P(A ext{ and } B) = P(A) imes P(B)
  • Example 1: Gamete formation from a dihybrid heterozygote (AaBb).
    • P( ext{gamete gets } A) = 1/2
    • P( ext{gamete gets } B) = 1/2
    • P( ext{gamete is } AB) = P(A) imes P(B) = (1/2) imes (1/2) = 1/4
    • Similarly, P(Ab) = 1/4, P(aB) = 1/4, P(ab) = 1/4. The sum of all possible gamete probabilities is 1/4 + 1/4 + 1/4 + 1/4 = 1.
  • Example 2: Determining offspring genotype for multiple independent genes.
    • Consider a cross: GgBbDd imes GgbbDd
    • Question: What is the probability that an offspring will have the genotype GgBbDd?
    • Step 1: Calculate probability for each gene locus independently.
      • For gene 'G': cross Gg imes Gg
        ightarrow P(Gg) = 1/2
      • For gene 'B': cross Bb imes bb
        ightarrow P(Bb) = 1/2
      • For gene 'D': cross Dd imes Dd
        ightarrow P(Dd) = 1/2
    • Step 2: Multiply the individual probabilities.
      • P(GgBbDd) = P(Gg) imes P(Bb) imes P(Dd) = (1/2) imes (1/2) imes (1/2) = 1/8 or 0.125

Conditional Probability

  • When to Use: Applied when specific information about an outcome is already known after a cross has been made. This new information reduces the sample space of possible outcomes.
  • Key Idea: It's the probability of an event occurring given that another event has already occurred. You eliminate possibilities inconsistent with the given information.
  • Example: In a cross between two heterozygous plants (Aa imes Aa) for seed color (yellow dominant, green recessive), what is the probability that a yellow-seeded offspring is heterozygous (Aa)?
    • Outcomes from Aa imes Aa:
      • Genotypes: 1/4 AA, 1/2 Aa, 1/4 aa
      • Phenotypes: 3/4 Yellow (AA, Aa), 1/4 Green (aa)
    • Given Information: The offspring has a dominant (yellow-seeded) phenotype. This means we know the offspring is not aa (green).
    • Reduced Sample Space: The possible genotypes for a yellow-seeded offspring are AA, Aa, Aa
      • This represents 1 homozygous dominant and 2 heterozygotes out of a total of 3 possible dominant phenotypes.
    • Conditional Probability Calculation: Out of these 3 yellow-seeded possibilities, 2 are heterozygous (Aa).
      • Therefore, the probability that a yellow-seeded offspring is heterozygous is 2/3.