Heredity

What this unit covers

The topics below follow the published Biology course framework for Unit 5. This unit is worth 8–11% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Meiosis & Sources of Genetic Variation14 min · 3 objectivesExplain how meiosis halves the chromosome number to produce four haploid gametes · Describe how crossing over and independent assortment generate genetic variation · Distinguish meiosis I from meiosis II and contrast meiosis with mitosis
• Mendelian Genetics & Probability15 min · 3 objectivesApply the laws of segregation and independent assortment to predict offspring · Use Punnett squares and probability rules to solve mono- and dihybrid crosses · Explain how a test cross reveals the genotype of a dominant-phenotype individual
• Non-Mendelian Inheritance15 min · 3 objectivesDistinguish incomplete dominance from codominance in heterozygotes · Predict blood-type offspring from the multiple-allele, codominant ABO system · Explain how polygenic traits and sex-linked genes deviate from simple dominance
• Chromosomal Inheritance & the Environment13 min · 3 objectivesRelate the chromosomal theory of inheritance to linked genes and recombination frequency · Explain how nondisjunction produces aneuploidy such as trisomy 21 · Describe how the environment interacts with genotype to shape phenotype
• Dihybrid Crosses & the Chi-Square Test15 min · 3 objectivesPredict dihybrid and trihybrid ratios efficiently using the forked-line (probability) method · State a null hypothesis and compute expected counts for a 9:3:3:1 cross · Carry out a complete chi-square goodness-of-fit test and decide whether to reject the null hypothesis
• Linkage, Recombination & Gene Mapping14 min · 3 objectivesExplain why linked genes deviate from independent assortment and how the parent’s cis/trans arrangement sets the recombinant classes · Calculate recombination frequency from testcross data and convert it to map units (centimorgans) · Construct a linear gene map by ordering three genes from their pairwise recombination frequencies

Formulas in Unit 5

Every term in Unit 5

All 28 terms we publish for Heredity, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Recombination frequency and map units

The percentage of recombinant offspring estimates the distance between two linked genes; 1% recombination is one map unit.

Law of segregation

The two alleles of a gene separate into different gametes during meiosis I, so each gamete carries one.

Law of independent assortment

Alleles of genes on different chromosomes assort independently. Linked genes on the same chromosome violate this in proportion to their closeness.

Genotype vs phenotype

Genotype is the allele combination; phenotype is the observable trait, which depends on genotype and environment together.

Test cross

Crossing an unknown dominant phenotype with a homozygous recessive. Any recessive offspring reveals the unknown parent was heterozygous.

Incomplete dominance

Heterozygote shows an intermediate phenotype — red × white snapdragons give pink, because one functional allele makes half the pigment.

Codominance

Both alleles are fully expressed in the heterozygote, as in AB blood type where both antigens appear.

Multiple alleles

More than two alleles exist in the population even though an individual carries only two, as with the ABO gene's I^A, I^B and i.

Polygenic inheritance

Several genes contribute additively to one trait, producing continuous variation and a bell-shaped distribution — height, skin color.

Pleiotropy

One gene affects several apparently unrelated traits, as in sickle cell where a single substitution alters cell shape, oxygen carriage and malaria resistance.

Epistasis

One gene masks the expression of another — in Labradors, the E gene controls whether the B gene's pigment gets deposited at all.

Sex-linked inheritance

Genes on the X chromosome. Males, being XY, express every X allele, so recessive conditions like hemophilia are far commoner in males.

Nondisjunction

Chromosomes fail to separate in meiosis, producing gametes with extra or missing chromosomes — the cause of trisomy 21.

Linked genes

Genes close together on one chromosome are inherited together more often than chance predicts; recombination frequency measures the distance between them.

Chi-square test

χ² = Σ(observed − expected)²/expected. Compared to a critical value at df = categories − 1; p < 0.05 rejects the null hypothesis that deviation is due to chance.

Cytoplasmic inheritance

Mitochondrial and chloroplast genes pass through the egg only, so these traits show strictly maternal inheritance and do not follow Mendelian ratios.

Environmental effect on phenotype

The same genotype can give different phenotypes in different conditions — hydrangea color with soil pH, coat color in Himalayan rabbits with temperature.

Monohybrid cross

A cross tracking one gene. Heterozygote × heterozygote gives the 3:1 phenotypic and 1:2:1 genotypic ratio.

Dihybrid cross

A cross tracking two genes. Two heterozygotes give 9:3:3:1 if the genes assort independently — a departure from that ratio suggests linkage.

Punnett square and probability

The multiplication rule gives the chance of independent events together; the addition rule gives the chance of either of two mutually exclusive outcomes.

Pedigree analysis

Autosomal recessive traits can skip generations and appear in both sexes equally; X-linked recessive traits appear far more often in males.

Carrier

A heterozygote for a recessive condition who does not show it but can pass the allele on. Two carriers have a 1 in 4 chance of an affected child.

Wild type vs mutant

The wild type is the most common phenotype in a natural population; a mutant allele is any variant from it, not necessarily a harmful one.

Barr body and X-inactivation

One X chromosome condenses in each female cell, chosen at random early in development. This makes calico cat coat patterns possible.

Degrees of freedom in chi-square

Number of phenotypic categories minus one. For a monohybrid cross with two categories, df = 1 and the critical value at p = 0.05 is 3.84.

Interpreting a chi-square result

If χ² is below the critical value, deviation from expectation is attributable to chance and the null hypothesis stands. Above it, something other than chance is acting.

Mean, standard deviation, standard error

The mean is the average, standard deviation measures spread within a sample, and standard error estimates how precisely the sample mean estimates the population mean.

Interpreting error bars

Error bars of ±2 SE that do not overlap indicate a statistically significant difference between two means; overlapping bars do not.

What examiners penalize here

• On free-response questions about "how sexual reproduction increases variation," name all three mechanisms explicitly: **crossing over** (prophase I), **independent assortment** (metaphase I), and **random fertilization**. Listing the mechanism *and* its stage is what earns the point.
• For "what fraction of offspring…" questions, do **not** draw a 16-box dihybrid grid. Solve each gene as a small monohybrid cross (¾ or ¼) and multiply. It is faster and far less error-prone under time pressure.
• For sex-linked pedigree questions, remember the two shortcuts: a son inherits his single X-linked allele from his **mother**, and X-linked recessive traits are **more common in males** because their one X has no partner to mask a recessive allele.
• The AP throughline for this unit is **phenotype = genotype + environment**. When a question shows one genotype producing different phenotypes (fur color, flower color, height), point to an environmental factor acting on gene expression — not to a mutation or a genotype change.
• The single most common χ² mistake on the AP exam is miscounting degrees of freedom. **df = number of phenotype categories − 1**, never the sample size. Four categories (9:3:3:1) → df = 3; two categories (3:1 or 1:1) → df = 1. Pick the wrong row of the table and a correct χ² still yields the wrong conclusion.
• Expect a question where the directly measured distance between the two outer genes is **smaller** than the sum of the inner intervals. The cause is **double crossovers**: a second crossover flips the middle marker back but leaves the outer alleles in their parental arrangement, so those offspring are never scored as outer-gene recombinants — long spans underestimate true distance.

Practice Biology

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 8 units of AP Biology

1. Unit 1 · Chemistry of Life
2. Unit 2 · Cell Structure & Function
3. Unit 3 · Cellular Energetics
4. Unit 4 · Cell Communication & Cycle
5. Unit 5 · Heredity
6. Unit 6 · Gene Expression & Regulation
7. Unit 7 · Natural Selection
8. Unit 8 · Ecology