Unit 5 Genotype vs. phenotype Tap / Space to flip
Unit 5 Genotype = genetic makeup; phenotype = observable trait. Phenotype is genotype + environment.
Unit 5 · 8–11% of exam
Heredity
Meiosis, Mendelian and non-Mendelian inheritance, linkage, pedigrees, chi-square. See the meiosis priority deep dive for the full breakdown.
Must-know content
- Meiosis covered in detail on the priority page — reductional vs. equational, sources of variation, vs. mitosis.
- Mendelian genetics:
- Law of segregation — alleles separate during gamete formation.
- Law of independent assortment — alleles of different genes assort independently (different chromosomes, or far apart on the same chromosome).
- Punnett squares: monohybrid (3:1), dihybrid (9:3:3:1), test cross (1:1).
- Non-Mendelian inheritance:
- Incomplete dominance (RR × WW → pink Rr).
- Codominance (AB blood type).
- Multiple alleles (ABO).
- Sex-linked (X-linked recessive — hemophilia, color blindness).
- Pleiotropy — one gene affects many traits.
- Polygenic — many genes additive on one continuous trait.
- Epistasis — one gene's expression masks another.
- Linked genes: recombination frequency (RF) ≤ 50%; RF ≈ map distance in centimorgans.
- Pedigrees: recognize autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive.
- Chi-square goodness-of-fit: compare observed to expected ratios; reject H₀ if χ² > critical value at p = 0.05.
- Environmental effects on phenotype (e.g., Himalayan rabbit fur color depends on temperature).
Example questions
MCQ A heterozygous black guinea pig (Bb) is crossed with a homozygous white (bb). What fraction of offspring are expected to be white? (A) 0 (B) 1/4 (C) 1/2 (D) 3/4
Answer: C. Bb × bb yields 1/2 Bb (black) and 1/2 bb (white).
Long FRQ A dihybrid cross predicts 9:3:3:1. Observed: 152:42:50:16 (total = 260). Calculate χ² and determine if the data fit the expected ratio. Critical value at df = 3, p = 0.05 is 7.815.
Answer:
Expected counts: 9/16 × 260 = 146.25; 3/16 = 48.75; 3/16 = 48.75; 1/16 = 16.25.
χ² = (152−146.25)²/146.25 + (42−48.75)²/48.75 + (50−48.75)²/48.75 + (16−16.25)²/16.25
≈ 0.226 + 0.935 + 0.032 + 0.004 ≈ 1.20Since 1.20 < 7.815, fail to reject H₀. The observed data fit the 9:3:3:1 ratio.
MCQ In humans, hemophilia is X-linked recessive. A carrier mother (X^H X^h) and an unaffected father (X^H Y) have a son. What's the probability the son has hemophilia? (A) 0 (B) 1/4 of all offspring (C) 1/2 of sons (D) All sons
Answer: C. Sons inherit Y from dad and either X^H or X^h from mom; half receive the affected X^h.
Drill flashcards
Unit 5 Homozygous Tap / Space to flip
Unit 5 Two identical alleles for a given gene (e.g., AA or aa).
Unit 5 Heterozygous Tap / Space to flip
Unit 5 Two different alleles for a given gene (e.g., Aa).
Unit 5 Test cross Tap / Space to flip
Unit 5 Cross with a homozygous recessive individual to determine an unknown genotype.
Unit 5 Codominance Tap / Space to flip
Unit 5 Both alleles are fully expressed in the heterozygote (e.g., AB blood type).
Unit 5 Incomplete dominance Tap / Space to flip
Unit 5 Heterozygote shows an intermediate phenotype between the two homozygotes (e.g., RR × WW → Rr pink).
Unit 5 Sex-linked Tap / Space to flip
Unit 5 Gene located on a sex chromosome (almost always X). X-linked recessive traits affect more males.
Unit 5 Linked genes Tap / Space to flip
Unit 5 Genes on the same chromosome that tend to inherit together. RF measures their map distance.