Analytical Approaches in Genetics
High-Yield Summary
- Test cross: cross an individual of unknown genotype (showing dominant phenotype) with a homozygous recessive (pp). If ALL offspring show dominant phenotype → unknown was PP. If ~half show recessive phenotype → unknown was Pp.
- Sex-linked crosses: males are XY, hemizygous for X-linked genes — whatever allele is on their single X is expressed (dominant or recessive), since there's no second X to mask it. This is why X-linked recessive disorders (e.g. hemophilia) are more common in males.
- Females need 2 copies of an X-linked recessive allele to be affected; males need only 1. A carrier female (heterozygous) × unaffected male never produces an affected daughter, only affected sons.
- Gene mapping: crossing over during prophase I creates recombinant chromosomes. Recombination frequency reflects physical distance between two genes — farther apart = higher frequency. 1 map unit (centimorgan) = 1% recombination frequency. Recombination frequency caps at 50% (behaves like independent assortment).
- Hardy-Weinberg: allele frequencies p + q = 1; genotype frequencies p² + 2pq + q² = 1 (p² = homozygous dominant, 2pq = heterozygous, q² = homozygous recessive). Models a population where evolution is NOT occurring.
- 5 Hardy-Weinberg conditions: large population, random mating, no mutation, no migration, no natural selection. Deviation from predicted p²/2pq/q² signals at least one condition is violated → evolution is occurring.
Hardy-Weinberg Equations
p + q = 1; p² + 2pq + q² = 1
- p = frequency of the dominant allele
- q = frequency of the recessive allele
- p² = frequency of homozygous dominant individuals
- 2pq = frequency of heterozygous individuals
- q² = frequency of homozygous recessive individuals
- Worked example: observed 0.49 YY, 0.42 Yy, 0.09 yy → p = 0.7, q = 0.3 → predicted p²=0.49, 2pq=0.42, q²=0.09 matches observed exactly → population is in equilibrium (not evolving).
5 Hardy-Weinberg Equilibrium Conditions
| Condition | Why It Matters |
|---|---|
| Large population | Minimizes genetic drift (random fluctuation) |
| Random mating | No genotype/phenotype-based pairing bias |
| No mutation | No new alleles introduced |
| No migration | No gene flow in or out |
| No natural selection | All genotypes have equal survival/reproduction |
Key Terms
- Test cross
- Crossing an individual of unknown genotype with a homozygous recessive to reveal whether it's homozygous dominant or heterozygous.
- Hemizygous
- Carrying only one copy of a gene (males, for X-linked genes) — that single allele is always expressed.
- Recombination frequency
- The probability two genes are separated by a crossover event; increases with physical distance between loci.
- Centimorgan (map unit)
- A unit of genetic distance equal to a 1% recombination frequency.
Common MCAT Trap
- A test cross tester must be homozygous RECESSIVE (pp), not homozygous dominant — only a recessive tester makes the unknown parent's contribution unambiguous.
- Recombination frequency cannot exceed 50% — at 50%, genes behave as if unlinked/independently assorting, even if technically on the same chromosome.
- p and q are ALLELE frequencies; p², 2pq, q² are GENOTYPE frequencies — don't plug an allele frequency into a genotype-frequency slot or vice versa.
- Hardy-Weinberg equilibrium means evolution is NOT happening — it's the null model. A population matching p²/2pq/q² predictions is NOT evolving; a mismatch signals it IS.
Quick Recall
A purple-flowered plant of unknown genotype is crossed with a white-flowered (pp) plant, producing only purple offspring. What was the unknown parent's genotype?
Why are X-linked recessive disorders more common in males than females?
If two genes have an 8% recombination frequency, how far apart are they in map units?
A population's observed genotype frequencies exactly match p², 2pq, and q² predictions. What does this tell you?