Chapter 1-7 Genetics Introduction (Probability Rules & Mendelian Inheritance)

Probability Rules and Mendelian Genetics: Comprehensive Notes

• Key ideas from the lecture

  • Probability rules used in genetics
  • Multiplication rule: to find the probability that several independent events all occur, multiply their probabilities: P(A \, \text{and}\, B) = P(A) \cdot P(B) (assuming independence, which is typically justified by the law of independent assortment for genes on different chromosomes).
  • Addition rule: to find the probability that one of several mutually exclusive events occurs, add their probabilities: P(A \cup B) = P(A) + P(B) \quad\text{(when A and B are mutually exclusive)}. This is used for an "or" scenario (one genotype or another) when the alternatives cannot happen at the same time.
  • Law of independent assortment
  • Genes on different chromosomes assort independently during meiosis, so the genotype probability for one gene can be treated separately from another gene.
  • If genes are not independently assorting (e.g., linked on the same chromosome), then independence is violated and probabilities must account for linkage.
  • Distinction between genotype and phenotype
  • Genotype: the specific alleles an individual carries (e.g., PP, Pp, pp for a single gene).
  • Phenotype: the observable trait (dominant vs recessive) which can be determined by genotype (e.g., a dominant phenotype can be produced by PP or Pp).
  • Mendel’s core ideas (from the Mendelian cross discussion)
  • Pure-breeding (true-breeding) lines produce offspring that consistently look like the parent when crossed with each other.
  • Law of segregation: alleles separate during gamete formation and then unite at fertilization, producing the observed phenotypic/genotypic ratios.
  • Dominant vs recessive alleles: dominant phenotype appears in heterozygotes; recessive phenotype appears only in homozygous recessives.
  • Law of independent assortment: when considering two traits on different chromosomes, the inheritance of one trait’s alleles is independent of the other trait’s alleles.

• First practice problem (three traits: p, r, t)

  • Scenario: Determine probability of offspring genotypes across three traits with specific parental genotypes.
  • Trait p (alleles P/p)
  • Parents: Pp × pp (heterozygous × homozygous recessive)
  • Probability to obtain little p from parent 1: P(p) = \tfrac{1}{2}
  • Probability to obtain little p from parent 2: P(p) = 1 (pp parent can only contribute p)
  • Offspring genotype for the p-trait that is recessive (pp): P(pp) = \tfrac{1}{2} \times 1 = \tfrac{1}{2}
  • Trait r (alleles R/r)
  • Parents: Rr × RR (heterozygous × homozygous dominant)
  • Probability of heterozygous offspring (Rr): P(Rr) = \tfrac{1}{2}
  • Probability of homozygous dominant (RR): P(RR) = \tfrac{1}{2}
  • Trait t (alleles T/t)
  • Parents: Tt × tt (heterozygous × homozygous recessive)
  • Probability to obtain recessive tt: P(tt) = \tfrac{1}{2}
  • Combining per-trait probabilities
  • The problem targeted offspring matching: p is recessive (pp), r is either heterozygous (Rr) or homozygous dominant (RR), and t is recessive (tt).
  • For ppRrtt: P(pp) \cdot P(Rr) \cdot P(tt) = \tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}
  • For ppRRtt: P(pp) \cdot P(RR) \cdot P(tt) = \tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}
  • These two genotypes are mutually exclusive, so add their probabilities:
    P(\text{either } ppRrtt \text{ or } ppRRtt) = \tfrac{1}{8} + \tfrac{1}{8} = \tfrac{1}{4}
  • Takeaway from approach
  • Break a multi-gene problem into per-gene problems using the law of independent assortment.
  • Use the multiplication rule for the probability of all required alleles appearing (and/or all genotype criteria being met).
  • Use the addition rule to combine mutually exclusive genotype possibilities that yield the same target outcome.

• Second practice problem (dominant phenotype for three traits: P, R, T)

  • Parental genotypes (as described in the lecture):
  • One parent is PpRrTt, the other is ppRRtt (homozygous recessive for p, homozygous dominant for R, homozygous recessive for T).
  • Target phenotype: dominant for P, dominant for R, and dominant for T (P-, R-, T- phenotypes).
  • Probability for each trait (phenotype) separately
  • P phenotype (dominant): cross Pp × pp yields at least one P with probability
    P(P-) = 1 - P(pp) = 1 - \tfrac{1}{2} = \tfrac{1}{2}
  • R phenotype (dominant): cross Rr × RR yields dominant phenotype in all offspring (RR or Rr), so
    P(R-) = 1
  • T phenotype (dominant): cross Tt × tt yields at least one T with probability
    P(T-) = 1 - P(tt) = 1 - \tfrac{1}{2} = \tfrac{1}{2}
  • Assuming independence among genes (independent assortment), the probability of all three dominant phenotypes is:
    P(P- \, \text{and}\, R- \, \text{and}\, T-) = P(P-) \cdot P(R-) \cdot P(T-) = \tfrac{1}{2} \cdot 1 \cdot \tfrac{1}{2} = \tfrac{1}{4}
  • Note on genotype vs phenotype
  • If you instead want the specific genotype that yields the dominant phenotype for all three traits, there are multiple genotype combinations (e.g., for P: Pp or PP; for R: RR or Rr; for T: Tt or TT depending on parental contributions). The lecture focused on the phenotype, which is often the practical question in Mendelian crosses.

• Mendel’s key findings summarized

  • Pure-breeding parents produce uniform F1 offspring for the trait being studied.
  • The F1 offspring are typically all intermediate in terms of dominance in their genotype, but phenotypically resemble one parent for a single trait.
  • Law of segregation: each parent contributes one allele per trait to the offspring, and alleles segregate during gamete formation.
  • Law of independent assortment (for genes on different chromosomes): the combination of alleles for one trait is independent of the combination for another trait in the formation of gametes.
  • The F2 generation shows segregation and a characteristic ratio for a single trait: phenotypic ratio of dominant to recessive is 3:1; genotypic ratio is typically 1:2:1 (for a heterozygous cross like Aa × Aa).
  • For two traits (dihybrid cross) with independent assortment, the classic phenotypic ratio is 9:3:3:1 in the F2 generation:
  • 9: dominant for both traits (AB)
  • 3: dominant for the first trait, recessive for the second (A_bb)
  • 3: recessive for the first trait, dominant for the second (aaB_)
  • 1: recessive for both traits (aabb)
  • Practical implications and caveats
  • Individuals that look the same (phenotype) may have different underlying genotypes (genotype diversity within phenotypically similar individuals).
  • The model assumes no linkage between genes; if genes are on the same chromosome and close together, linkage reduces independent assortment and changes expected ratios.
  • Real-world relevance
  • These principles underpin approaches in breeding, genetics education, and foundational population genetics models. They also illustrate how probabilities scale when considering multiple genes and how small probability events combine (multiplication) or are aggregated (addition).

• Quick recap of why this matters

  • Using probability rules allows us to predict offspring outcomes without enumerating every possible Punnett square, which becomes impractical for more complex crosses.
  • Understanding when to apply multiplication vs addition helps avoid common mistakes in combining probabilities across multiple genes.
  • The concepts of dominance, segregation, and independent assortment lay the groundwork for more advanced topics in genetics, such as linkage, recombination, and population genetics modeling.

• If you want more practice

  • Try a dihybrid cross with different dominant/recessive alleles across two traits and verify the 9:3:3:1 phenotypic ratio.
  • Create small problems where one gene is linked to another and explore how linkage alters the independent assortment assumption.

• Quick conceptual connections to earlier material

  • The per-gene probability approach relies on treating each gene separately because of independent assortment, a foundational principle introduced earlier.
  • The transition from single-trait ratios (3:1) to multi-trait ratios (9:3:3:1) mirrors moving from monohybrid to dihybrid analyses, reinforcing the same probabilistic framework.

• Practical implications discussed in class

  • If you observe deviations from expected Mendelian ratios in real populations, it can suggest linkage or other genetic mechanisms (epistasis, incomplete penetrance, etc.), prompting deeper investigations beyond simple Mendelian models.

• Final takeaway

  • For multi-lactor genetic problems, break the problem into manageable pieces: compute per-gene genotype/phenotype probabilities, multiply for independent genes, and add for mutually exclusive genotype options that satisfy a given condition.