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.