Unit 5 · Heredity
Unit 5 · Heredity
Unit 5: Heredity accounts for 8% to 11% of the AP Biology exam score in Section I (multiple choice), and its mechanisms also appear in Section II free-response questions.
Weighting is the College Board CED range for the current exam. Higher-weight units repay proportionally more review time; use this share to size your effort before drilling the mechanisms below.
Heredity problems become unstable when homologs, sister chromatids, alleles, gametes, genotypes, and phenotypes are treated as interchangeable. A chromosome diagram describes physical behavior, a probability model describes possible outcomes, and offspring data test whether that model fits the observed inheritance pattern.
You will use chromosome bookkeeping and conditional probability to connect meiosis to inheritance while keeping linkage, nondisjunction, environment, and statistical evidence in their correct layers.
The decision that organizes this unit
Which chromosome copies and alleles can enter each gamete under the stated meiotic event, and what observations would distinguish the resulting inheritance model?
Mechanism route and repair branches
- Main route: Identify homologous chromosomes, sister chromatids, loci, and parental allele arrangements before following a division.
- Main route: Track replication separately from homolog separation and sister-chromatid separation so chromosome and DNA-copy counts remain consistent.
- Diagnostic cue: DNA replication is said to double chromosome number. Wrong branch: Chromatid copies and chromosomes are counted as the same object. Repair: Count chromosomes by centromeres and DNA copies by chromatids, then state which structures separate at each meiotic division.
- Main route: List possible gametes under independent assortment, linkage, crossing over, or nondisjunction before combining parental probabilities.
- Diagnostic cue: Linked loci are multiplied as independent probabilities. Wrong branch: Independent assortment is applied without checking chromosome location. Repair: Use parental arrangement and recombination evidence; only use the product rule after the relevant events are justified as independent.
- Main route: Translate gamete combinations into offspring genotypes, then use the stated dominance, interaction, or environmental relation to predict phenotype.
- Diagnostic cue: A genotype is claimed to determine one phenotype in every environment. Wrong branch: Gene expression and environmental effects disappear from the model. Repair: State the genotype-dependent capacity and then include the environmental condition or interaction that changes the measured phenotype.
- Main route: Compare observed and expected counts with an appropriate model, preserving degrees of freedom and the limits of the statistical conclusion.
Load-bearing representation lab
One crossover between nonsister chromatids changes allele combinations while leaving the centromere inventory intact. The tetrad shows four chromatids: parental AB and ab remain, while Ab and aB carry the exchanged segments. Thus exactly two products are recombinant and two are parental after this single exchange between the marked loci. Line style encodes product class: solid marks parental products and dashed marks recombinant products, so the styles must not be interpreted as chromosome breakage or different ploidy. Students often mark all four chromatids recombinant because all four homologous products are drawn together, or they place the exchange between sister chromatids and expect new allele combinations. Count chromatids first, locate one centromere on each product, and then read the allele order on both sides of the exchange. The figure establishes product classes from one crossover; it does not state the population recombination frequency, which also depends on how often crossovers occur and whether multiple exchanges restore parental combinations. A warranted answer names the two reciprocal recombinant types and preserves the two parental chromatids.
The two reaction norms cross between cool and warm environments, so genotype rank reverses. Genotype B has the higher trait value in the cool condition, whereas genotype A has the higher value in the warm condition. That crossing is evidence of a genotype-by-environment interaction: the effect of changing environment is not the same for both genotypes. It is not enough to say that genes and environment both matter, because parallel lines could show additive effects without an interaction. Trace each labeled genotype across the two conditions, compare the direction and size of its change, and then compare ranks within each environment. The common mistake selects the highest endpoint and declares one genotype universally superior, ignoring the reversal at the other endpoint. The graph also does not establish fitness unless the measured trait is linked to reproductive contribution in the stated environment. A precise response names the trait value as the dependent measure, describes the crossing pattern, and limits the conclusion to these genotypes and conditions rather than treating warm and cool as a universal environmental scale.
