Genetics Crosses, Probability & Pedigrees – Detailed Study Notes

Monohybrid Cross Basics

  • True-breeding parents (P generation) are homozygous for opposite alleles.
    • Example used in class: red stem (dominant) × green stem (recessive).
  • Symbol convention (for Mendelian complete dominance)
    • Dominant allele: capital letter (e.g. R for red stem).
    • Recessive allele: lowercase letter (e.g. r for green stem).
  • F₁ generation from RR \times rr is 100 % heterozygous Rr and all show the dominant phenotype.
  • Self-cross of F₁ (Rr \times Rr) is the classic monohybrid cross.
    • Genotypic ratio: 1\;RR : 2\;Rr : 1\;rr.
    • Phenotypic ratio: 3 dominant : 1 recessive.
  • DO NOT write heterozygotes as R + or mixed capitalization—use standard two-letter symbols.

Genotypic vs. Phenotypic Ratios (quick reference)

  • Homozygous dominant × homozygous recessive ⇒ F₁ all heterozygous.
    • Phenotypic ratio: 4:0 (all dominant).
    • Genotypic ratio: 0:4:0 (all heterozygous).
  • Heterozygous × heterozygous (test yourself frequently!)
    • Phenotype 3:1, genotype 1:2:1.
  • Homozygous dominant × heterozygous
    • Genotype 2\;RR : 2\;Rr : 0\;rr → simplify 1:1:0.
    • Phenotype 4 dominant : 0 recessive.
  • Heterozygous × homozygous recessive (classic test-cross)
    • Expect 1 dominant : 1 recessive phenotype.
    • Genotype 1\;Rr : 1\;rr.

Test-Cross Logic

  • Purpose: reveal the unknown genotype of an individual showing the dominant trait.
  • Cross the unknown with a homozygous recessive (rr).
  • Outcomes
    • 100 \% dominant progeny ⇒ tester was RR.
    • 50 \% dominant : 50 \% recessive ⇒ tester was Rr.

Probability & Product Rule

  • Independent events: multiply individual probabilities.
    • Example: chance of two heads when flipping two coins
      \frac12 \times \frac12 = \frac14.
  • Genetics example: heterozygous × heterozygous → probability F₂ is recessive =\frac14.
    • Probability that two children in a row are recessive = \frac14 \times \frac14 = \frac1{16}.
  • Important: probabilities describe possible outcomes, not guarantees. Once a child is born the probability for that individual collapses to 0 or 1 (e.g., sex of an existing child is now 100 % known).

Common Crosses Worth Memorising

  • Rr \times Rr → 3:1 phenotype, 1:2:1 genotype (monohybrid).
  • AaBb \times AaBb (dihybrid heterozygous for both) → phenotype 9:3:3:1 (see below).
  • Aa \times aa (test-cross) → 1:1 phenotype.
  • Homozygous × heterozygous for same allele yields 100 \% dominant phenotype.

Dihybrid Cross & Product Rule (Branching Method)

  • Consider two independent genes, e.g. stature (T/t) and flower colour (P/p).
    • T = tall dominant, t = short recessive.
    • P = purple dominant, p = white recessive.
  • TtPp \times TtPp.
  • Solve each gene separately (two simultaneous monohybrids):
    • Tall : short = 3:1.
    • Purple : white = 3:1.
  • Combine with product rule:
    • Tall & purple \frac34 \times \frac34 = \frac{9}{16}.
    • Tall & white \frac34 \times \frac14 = \frac{3}{16}.
    • Short & purple \frac14 \times \frac34 = \frac{3}{16}.
    • Short & white \frac14 \times \frac14 = \frac1{16}.
  • Produces the canonical 9:3:3:1 phenotypic ratio—memorise.
  • Genotypic ratios for dihybrids are complex (16 classes); exam will focus on phenotype.

Pedigree Chart Conventions

  • Square = male, circle = female.
  • Horizontal line joins mating pair; vertical line drops to offspring.
  • Left-to-right within a sibship = birth order (oldest left).
  • Shaded symbol = individual expresses the tracked trait (often a disorder).
    • Unshaded = phenotypically normal for that trait.
  • Generations in rows: P → F₁ → F₂…

Example Pedigree (Albinism in Instructor’s Family)

  • Gene: pigmentation control, alleles N (normal) vs n (albino, recessive).
  • Grandparents each Nn (carriers) → three children.
    • Two normal (unknown genotype), one albino (nn).
  • Albino son (nn) marries carrier (Nn) → children have 50 % chance albino (Punnett nn \times Nn).
  • Calculations demonstrated:
    • Probability albino child = \frac12.
    • Three consecutive albino sons = (\frac12)^3 = \frac1{8}.
    • Probability of three albino sons who are also male = (\frac12\;\text{albino})^3 \times (\frac12\;\text{male})^3 = \frac1{64}.
  • Key takeaway: carrier status of phenotypically normal relatives often unknown unless a test cross (or modern genetic test) reveals it.

Mendel’s Laws (review)

  1. Law of Segregation – two alleles separate during gamete formation; each gamete gets one.
  2. Law of Independent Assortment – alleles of different genes assort independently (true for genes on different chromosomes or distant on same chromosome).

Non-Mendelian Single-Gene Patterns

Incomplete Dominance

  • Heterozygote shows an intermediate (blended) phenotype.
  • Example: snapdragon flower colour.
    • Alleles written with superscripts: C^{R} (red) and C^{W} (white).
    • Cross C^{R}C^{R} \times C^{W}C^{W} → F₁ C^{R}C^{W} (pink).
    • Heterozygote phenotypic ratio in F₂ = 1\;\text{red} : 2\;\text{pink} : 1\;\text{white}.
  • Because no allele dominates, upper- vs. lower-case letters are NOT used.

Codominance

  • Both alleles fully and distinctly expressed in heterozygote (no blending).
  • Classic examples
    • Roan cattle: red hairs and white hairs intermixed.
    • ABO human blood group: I^{A} and I^{B} codominant (not covered deeply in transcript but analogous).
  • Notation identical to incomplete dominance, but phenotype shows side-by-side expression (e.g. red-and-white patches).

Tips & Connections

  • Memorise key ratios but be able to derive them with Punnett squares.
  • Always distinguish genotype (allele combination) from phenotype (observable trait).
  • Use product rule for multi-gene or multi-event probability; order irrelevant.
  • Pedigree analysis relies on understanding of dominance vs. recessiveness and test crosses.
  • Non-Mendelian patterns (incomplete dominance, codominance) violate the simple dominant-recessive phenotypic ratios—watch allele notation.
  • Real-world relevance: human single-gene disorders (e.g., cystic fibrosis, albinism) often recessive; carriers can be identified via pedigrees or molecular tests.
  • Ethical implications: genetic counselling uses these calculations to inform prospective parents of risks.

Formula Cheat-Sheet

  • Monohybrid heterozygous cross genotype: 1:2:1.
  • Monohybrid heterozygous cross phenotype (complete dominance): 3:1.
  • Dihybrid heterozygous cross phenotype: 9:3:3:1.
  • Product rule example: P(\text{event A and B}) = P(A) \times P(B), valid when events independent.
  • Fraction conversions: \frac14 = 25\%\quad \frac34 = 75\%\quad \frac1{16} = 6.25\%.

Practice Recommendations

  • Drill Punnett squares until set-up is automatic (don’t “constantly draw” but be able to when stuck).
  • Create mini-pedigrees and label each individual’s possible genotype.
  • Challenge problems: compute probability of specific child orders (e.g., first two girls both recessive).
  • When unsure, quickly sketch a square—better to spend 30 s drawing than miss a ratio.