Lesson 24: Genetics Extensions (HL) HL ONLY

This HL extension covers Theme D — genetics extensions (HL only): gene linkage and recombination in depth, polygenic inheritance, and the molecular toolkit of gene technology and bioinformatics that HL papers use as application contexts.

Linkage groups and mapping chromosomes

Genes on the same chromosome form a linkage group — a species has as many linkage groups as chromosome types (humans 23, *Drosophila* 4). Linked genes do not assort independently: parental allele combinations dominate offspring, and recombinants arise only when crossing over exchanges segments between homologous chromosomes in prophase I of meiosis. The further apart two loci are, the more likely a crossover falls between them, so recombination frequency (RF) maps distance: RF% ≈ map units/centimorgans (Lesson 8's drill). Limits: RF saturates at 50% (indistinguishable from unlinked), and double crossovers between distant loci are missed — hence maps are built from short intervals summed in threes (three-point test crosses). Worked reasoning: test-cross counts 1,000 offspring: parental classes 430 and 420, single-crossover classes 60 and 55, other singles 20 and 12, double crossovers 2 and 1. RF for the outer pair ≈ (60+55+20+12+2+1)/1000 = 15%; for the inner intervals 7.7% and 3.5%, consistent with a gene order deduced because double crossovers restore the middle gene to parental combination — the standard deduction examiners love. Chi-square tests whether a cross deviates from independent assortment (df based on phenotypic classes) before linkage is claimed.

Polygenic inheritance and continuous variation

Many traits (height, skin colour, wheat grain colour) are controlled by several genes with additive effects, giving continuous variation — a normal distribution rather than discrete classes. With two additive genes the F₂ phenotypic classes run 1:4:6:4:1 (0–4 "doses" of the contributing allele); three genes give seven classes. Environmental effects smooth the distribution further (nutrition on height). Distinguish carefully from multiple alleles (one gene, several versions) and from epistasis (one gene masking another — e.g. coat colour in Labradors, where the E locus decides whether the B locus pigment is deposited: a classic 9:3:4 ratio). HL exam crosses mix these: read whether the phenotype classes are discrete (single-gene/epistasis) or graded (polygenic) before calculating anything.

Gene technology and bioinformatics in practice

DNA profiling revisited: STR loci are amplified by PCR and separated by electrophoresis; match probabilities multiply across loci (with population-frequency caveats — Lesson 6). Sequencing: Sanger (dideoxy chain-termination) reads hundreds of bases; modern high-throughput methods sequence whole genomes — the Human Genome Project (2003) found ~20,000 protein-coding genes, far fewer than expected, shifting attention to regulation, splicing and the proteome (Lesson 5). Bioinformatics: sequence databases (GenBank) and alignment tools (BLAST) let any researcher compare a new sequence against all known ones; phylogenetic analyses (Lesson 12) and barcode identification rest on this infrastructure. Gene therapy and editing: vectors (modified viruses, lipid nanoparticles) deliver working alleles to somatic cells (e.g. for SCID, some retinal dystrophies); CRISPR–Cas9 editing uses a guide RNA to target a nuclease — somatic editing is in clinical use (sickle-cell therapy approved in 2023–24), germline editing remains prohibited after the 2018 CRISPR-babies case, an ethics landmark examiners cite. GM crops and pharmacogenomics extend Lesson 19's evaluation framework: weigh measured benefits (yield, reduced insecticide) against assessed risks (gene flow, resistance evolution) with evidence, and note that risk assessment differs from hazard identification.

Population genetics (HL awareness)

Allele frequencies in a gene pool change by selection, drift, migration and mutation (Lesson 11). The Hardy–Weinberg equilibrium (p² + 2pq + q² = 1) models the null state — no evolution — under conditions (random mating, no selection, no migration, large population, no mutation) that real populations violate in informative ways. Worked use: if 1 in 2,500 newborns has cystic fibrosis (q² = 0.0004), then q = 0.02, p ≈ 0.98, and carrier frequency 2pq ≈ 0.039 ≈ 1 in 26 — the calculation behind genetic-counselling figures for Northern European populations. Deviations from Hardy–Weinberg proportions in samples (tested by chi-square) signal that one of the assumptions fails — evidence of evolution in action.

Nature of Science / TOK link

Genome-scale data changed genetics from a hypothesis-led to a data-led science; bioinformatics conclusions are only as good as databases and algorithms — a new layer of mediation between scientist and organism. TOK: a BLAST "match" is a statement of statistical similarity (an E-value), not identity of function — inference at scale still requires the same scepticism as a single gel.

Common mistakes (from examiner reports)

• Treating 50% recombination as proof genes are on different chromosomes (distant linked genes also approach 50%).
• Confusing epistasis ratios (9:3:4) with dihybrid (9:3:3:1) and polygenic distributions.
• Hardy–Weinberg: using p² for the recessive phenotype (it is q²) or forgetting allele frequencies sum to 1.
• Describing CRISPR as "adding genes" generically — it edits at a targeted sequence guided by RNA.

IB Exam Experience

Paper 1 style. 1. Two genes show a recombination frequency of 50%. The best interpretation is… A. they are adjacent on a chromosome B. they are unlinked, or so far apart on one chromosome that they assort as if unlinked C. they are sex-linked D. crossing over never occurs 2. In a population in Hardy–Weinberg equilibrium, 9% are homozygous recessive. The heterozygous frequency is… A. 9% B. 21% C. 42% D. 70% 3. Which ratio suggests epistasis rather than a standard dihybrid cross? A. 9:3:3:1 B. 9:3:4 C. 1:2:1 D. 3:1 Answer key. 1 — B. 2 — C (q = 0.3, p = 0.7, 2pq = 0.42). 3 — B. Paper 2 style. A test cross of a plant heterozygous for two linked genes gives 480 offspring: 218 and 212 of the parental phenotypes, 36 and 34 of the recombinant phenotypes. (a) Calculate the recombination frequency and state the distance between the loci. [2] (b) Explain why recombinant offspring are produced at all. [2] (c) State the phenotypic ratio expected in this test cross if the genes were unlinked. [1] *Marking notes.* (a) Recombinants = 70/480 = 14.6% ≈ 15 map units (cM) apart [2]. (b) Crossing over between homologous chromosomes in prophase I of meiosis exchanges alleles between the loci [1]; producing new allele combinations in some gametes [1]. (c) 1:1:1:1 [1]. IA idea. A Hardy–Weinberg investigation using published population data (e.g. ABO or MN blood-group frequencies, or disease-carrier frequencies) — test fit to equilibrium with chi-square and discuss which assumptions the population violates.

Extended data: a three-point test cross, fully worked

A *Drosophila* test cross for three linked genes (a, b, c) produced 1,000 offspring. Phenotype counts (test-cross progeny reveal the heterozygote parent's gametes): parental types 420 and 405; one recombinant pair 45 and 40; a second pair 55 and 28; the rarest pair (double crossovers) 4 and 3. Step 1 — identify parentals (the two largest classes) and double crossovers (the two smallest). Step 2 — find the gene order: compare a double-crossover gamete with its parental counterpart; the gene whose allele has *switched* relative to the other two is the middle gene (double crossovers swap only the middle segment). Suppose that comparison shows gene b switched: order is a—b—c. Step 3 — distances: a–b RF = (single crossovers in the a–b interval + doubles) ÷ total = (45+40+4+3)/1000 = 9.2%; b–c RF = (55+28+4+3)/1000 = 9.0%. Map: a —9.2— b —9.0— c. Step 4 — sanity check: the a–c distance measured directly (all non-parentals = 175/1000 = 17.5%) is slightly less than 9.2 + 9.0 = 18.2%, because double crossovers between a and c restore the parental combination for the outer markers and are missed without the middle gene — which is *why* three-point crosses exist. Examiners award this reasoning in stages; write it in stages.

Evaluating gene therapy claims — a framework you can reuse

When a paper or an exam presents a gene-therapy breakthrough, run four questions. (1) Target: is the edit somatic (affects only the patient) or germline (heritable)? — the ethical and regulatory line runs exactly here. (2) Delivery: how does the therapeutic DNA reach the right cells, and what fraction actually get it (viral vectors raise immune and insertional risks; ex vivo editing of blood stem cells, as in the approved sickle-cell therapy, sidesteps delivery by treating cells outside the body and returning them)? (3) Durability: does the treated cell population persist (stem-cell targets do; short-lived cells need re-treatment)? (4) Evidence: is the claim from a trial with controls and follow-up, or from a single case? The 2018 germline-editing scandal failed every test — no medical need (sperm washing already prevented transmission), mosaicism risk uncontrolled, consent invalid — and is the standard counterexample. This framework converts a news story into IB-grade evaluation, and it generalises to stem cell and GM claims throughout the course.

One more application: pedigree meets population — the royal haemophilia calculation

Queen Victoria was a carrier of haemophilia (a new mutation, most likely — Lesson 7's mechanisms reaching into history); her descendants carried the allele into the Spanish and Russian royal families. The population-genetics frame sharpens the pedigree story: for an X-linked recessive condition with male incidence q, roughly q of males are affected and 2q(1−q) ≈ 2q of females are carriers (for haemophilia A, q ≈ 1/5,000 males, so ~1 in 2,500 women is a carrier). Contrast an autosomal recessive at the same incidence and the carrier frequency would be far higher (~1 in 35 for cystic fibrosis) — the geometry of X-linkage concentrates risk in males while keeping the carrier pool comparatively small. Expect exam hybrids of exactly this kind: half pedigree, half Hardy–Weinberg, solved by keeping the two models' assumptions visibly separate until the final line.

Examiner's lens: final drills (genetics HL)

• Map-distance hygiene: recombination frequency can never exceed 50%; a measured 48% means "far apart or effectively unlinked", and only a third marker between them can resolve the order — the practical moral of the three-point cross.
• Hardy–Weinberg in one breath: p + q = 1 (alleles) and p² + 2pq + q² = 1 (genotypes). Recessive phenotype frequency is q² — take the square root *first*, then compute carriers as 2pq.
• Polygenic signature: a bell-shaped distribution of phenotypes in the F₂ and classes counted in binomial proportions (1:4:6:4:1 for two genes) — if the data are graded, do not force a Mendelian ratio onto them.
• Self-test: a population has 16% homozygous recessive individuals and mating is random with no selection. What percentage are carriers? (q = 0.4, p = 0.6, 2pq = 48%.)