Genetics and Evolutionary Biology — Mendel to Population Genetics
1. Mendelian inheritance, gene interaction and complementation
Segregation gives a 3:1 phenotypic (1:2:1 genotypic) F₂ for one gene and independent assortment gives 9:3:3:1 for two unlinked genes; a test cross of the dihybrid to the double recessive gives 1:1:1:1. Departures from these ratios are information. Incomplete dominance and codominance turn 3:1 into 1:2:1; a recessive lethal turns it into 2:1. Whether observed counts fit an expected ratio is judged by the chi-square test, χ² = Σ(O − E)²/E with degrees of freedom = classes − 1; at p = 0.05 the critical values are 3.84 (1 df), 5.99 (2 df) and 7.81 (3 df), and a χ² below the critical value means the deviation can be put down to chance.
| F₂ ratio | Interaction | Logic |
|---|---|---|
| 9:7 | Complementary genes (duplicate recessive epistasis) | Both dominant alleles are needed; either recessive homozygote blocks the pathway |
| 9:3:4 | Recessive epistasis | aa masks whatever the second gene does |
| 12:3:1 | Dominant epistasis | A dominant allele of one gene masks the other gene |
| 13:3 | Dominant suppression (inhibitory gene) | One dominant allele suppresses expression of the other |
| 15:1 | Duplicate dominant genes | Either dominant allele alone gives the phenotype |
| 9:6:1 | Duplicate genes with cumulative effect | One dominant allele gives an intermediate, two give the full phenotype |
2. Linkage, recombination, chromosome mapping and extra-chromosomal inheritance
Genes on the same chromosome are inherited together unless a crossover between them in meiosis produces recombinant gametes. Recombination frequency RF = recombinant progeny/total progeny, and 1% RF is one map unit, a centimorgan. RF cannot exceed 50%, because at that point independent assortment and linkage look identical; distant genes are mapped by adding shorter intervals. In a test cross of AB/ab with genes 12 cM apart, 12% of progeny are recombinant, split equally as Ab and aB — 6% each. In a three-point cross the parental classes are the most frequent and the double crossovers the least; the gene that switches position between the parental and double-crossover classes is the one in the middle.
Extra-chromosomal inheritance does not follow Mendel because the genes are in mitochondria or plastids, which in most plants and animals come through the egg. Leaf variegation in Mirabilis jalapa follows the plastids of the maternal branch, and mitochondrial disorders pass from an affected mother to all her children. It must not be confused with a maternal effect, where a nuclear gene of the mother decides the offspring’s phenotype: shell coiling in the snail Limnaea is set by the mother’s genotype (dextral D dominant), so a dd mother has only sinistral progeny whatever their own genotype, and the ratio appears one generation late.
3. Microbial genetics, gene mapping, horizontal gene transfer and transposable elements
| Process | What carries the DNA | Key facts |
|---|---|---|
| Transformation | Naked DNA taken up by a competent cell | Griffith’s pneumococcus experiment; Avery, MacLeod and McCarty showed the principle is DNA; DNase-sensitive |
| Transduction | A bacteriophage | Generalised (P1, P22): any host gene packaged by mistake; specialised (λ): only genes beside the prophage site, such as gal and bio; cotransduction frequency maps genes close together |
| Conjugation | Cell-to-cell contact through a sex pilus | F⁺ × F⁻ transfers the F plasmid and makes the recipient F⁺; an Hfr has F integrated and transfers chromosomal genes in a fixed order from oriT, rarely the whole F, so recipients stay F⁻; interrupted mating maps genes by minutes of entry; F′ carries a piece of chromosome |
These three are the mechanisms of horizontal gene transfer, the movement of genes between organisms other than parent to offspring, and they explain how antibiotic resistance and pathogenicity islands spread across species. Transposable elements move within and between genomes. Bacterial insertion sequences carry only a transposase between inverted repeats; composite transposons such as Tn10 carry extra genes between two IS elements. Transposition may be conservative (cut-and-paste) or replicative, and insertion creates a short target-site duplication. In eukaryotes McClintock’s Ac/Ds elements of maize were the first described (Ds moves only when Ac supplies transposase), and retrotransposons — LINEs such as L1 and SINEs such as Alu — move through an RNA intermediate and reverse transcriptase.
4. Chromosomal variation, sex determination, genetic disorders and epigenetics
- Structural variation: deletions (cri-du-chat, 5p−), duplications, inversions — paracentric (not including the centromere) or pericentric (including it), both of which suppress recoverable recombinants in heterozygotes — and translocations, reciprocal (the Philadelphia chromosome t(9;22) making BCR–ABL in chronic myeloid leukaemia) or Robertsonian (fusion of two acrocentrics, the cause of familial Down syndrome).
- Numerical variation: aneuploidy from non-disjunction — trisomy 21 (Down), 47,XXY (Klinefelter), 45,X (Turner), trisomy 13 and 18 — and euploid changes such as polyploidy, common and useful in plants.
- Sex determination: XX/XY with a dominant SRY gene on the Y in mammals; in Drosophila the X-to-autosome ratio decides (1.0 female, 0.5 male) and the Y is not male-determining; XX/XO in grasshoppers; ZZ male and ZW female in birds; haplodiploidy in bees (unfertilised eggs become males); temperature in many reptiles. Dosage compensation equalises X-linked expression — X inactivation into a Barr body by XIST RNA in female mammals, doubled transcription of the single X in male Drosophila.
- Genetic disorders by pattern: autosomal recessive (cystic fibrosis, sickle-cell anaemia, phenylketonuria), autosomal dominant (Huntington disease, an expanded CAG repeat), X-linked recessive (haemophilia A, Duchenne muscular dystrophy, red–green colour blindness — affected sons of carrier mothers, no father-to-son transmission) and mitochondrial (maternal only).
- Epigenetics is heritable change in gene expression without change in DNA sequence: methylation of cytosine in CpG dinucleotides (usually silencing), histone acetylation (which neutralises lysine charge and opens chromatin) and deacetylation or certain methylations (which close it), genomic imprinting — expression from only the maternal or only the paternal copy, so that loss of the paternal 15q11–q13 region gives Prader–Willi syndrome and loss of the maternal one Angelman syndrome — and X inactivation.
5. Population genetics, selection, drift and speciation
In a large, randomly mating population with no selection, mutation or migration, allele frequencies p and q stay constant and genotype frequencies settle in one generation at p² : 2pq : q² — the Hardy–Weinberg equilibrium. It is used backwards: if 1 in 2500 people show a recessive disorder, q² = 1/2500, q = 0.02, p = 0.98 and the carrier frequency is 2pq = 2 × 0.98 × 0.02 = 0.0392, about 1 in 25 — fifty times the disease frequency. Each assumption, broken, is an evolutionary force. Selection changes frequencies according to fitness; against a rare recessive it is very slow, because most copies hide in heterozygotes. Mutation supplies variation, migration (gene flow) homogenises populations, and genetic drift changes frequencies at random, strongest in small populations — bottlenecks and founder effects — and ends in fixation or loss. A new neutral allele is fixed with a probability equal to its initial frequency, 1/(2N) in a diploid population.
Adaptive evolution is change driven by selection; the neutral theory (Kimura) holds that most molecular differences between species are selectively neutral and were fixed by drift, which is why molecular clocks tick. A dN/dS ratio above 1 in a protein-coding gene points to positive selection, below 1 to purifying selection. A species, under the biological species concept, is a group of interbreeding populations reproductively isolated from others — by prezygotic barriers (habitat, timing, behaviour, mechanical, gametic) or postzygotic ones (hybrid inviability, sterility as in the mule, hybrid breakdown). Speciation is allopatric when a geographic barrier splits a population, sympatric when isolation arises within one area — instantly by polyploidy in plants — and parapatric along a cline.
Key takeaways
- 9:7 complementary, 9:3:4 recessive epistasis, 12:3:1 dominant epistasis, 13:3 suppression, 15:1 duplicate dominant; χ² at 1 df must be under 3.84 to accept the fit at p = 0.05.
- 1% recombination = 1 cM, never above 50%; coincidence = observed/expected double crossovers, interference = 1 − coincidence; the middle gene is the one that switches in the double crossovers.
- Maternal inheritance (organelle genes) differs from a maternal effect (the mother’s nuclear genotype sets the phenotype, as in Limnaea coiling).
- Generalised transduction moves any gene, specialised only genes beside the prophage; Hfr transfers chromosomal genes in fixed order and recipients stay F⁻.
- Hardy–Weinberg: q from √(affected), carriers 2pq; drift is strongest in small populations and a neutral allele fixes with probability equal to its frequency.
Practice questions (12)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A dihybrid F₂ showing a 9:7 phenotypic ratio indicates
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Answer: A — complementary gene action
In 9:7 only the 9 A_B_ individuals show the phenotype; A_bb, aaB_ and aabb (3 + 3 + 1 = 7) do not, so both dominant alleles are needed — two genes acting in one pathway. Dominant epistasis gives 12:3:1 and duplicate dominant genes 15:1.Fruit colour in a squash gives an F₂ of 12 white : 3 yellow : 1 green. The interaction is
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Answer: A — dominant epistasis
A dominant allele W of one gene gives white whatever the second gene is (9 W_Y_ + 3 W_yy = 12); only ww plants show the second gene, yellow for Y_ and green for yy. Recessive epistasis would give 9:3:4, and dominant suppression folds the classes into 13:3.Three linked genes lie in the order a–b–c, with a–b = 10 map units and b–c = 20 map units. In a three-point test cross of 1000 progeny, 12 double-crossover progeny are observed. The interference is ______.
Numerical answer — type the value.
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Answer: 0.4
Expected double crossovers = 0.10 × 0.20 × 1000 = 20. Coefficient of coincidence = 12/20 = 0.6, and interference = 1 − 0.6 = 0.4. Reporting 0.6 gives the coincidence, not the interference; adding the distances (0.30 × 1000) confuses single with double crossovers.An autosomal recessive disorder affects 1 in 2500 individuals in a population at Hardy–Weinberg equilibrium. The frequency of heterozygous carriers is ______ (round off to four decimal places).
Numerical answer — type the value.
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Answer: 0.0392
q² = 1/2500, so q = 1/50 = 0.02 and p = 0.98; carriers = 2pq = 2 × 0.98 × 0.02 = 0.0392. Using q = 1/2500 directly, without the square root, gives about 0.0008 — the commonest mistake. The carrier frequency is roughly fifty times the disease frequency, which is why selection against rare recessives is slow.Specialised transduction by phage λ differs from generalised transduction in that it
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Answer: A — transfers only bacterial genes lying next to the prophage integration site
Specialised transducing particles arise from faulty excision of an integrated prophage, which takes adjacent host DNA such as gal or bio with it. Generalised transducers such as P1 package random host fragments, so any gene can move. Pilus contact is conjugation and naked-DNA uptake is transformation.In a cross between an Hfr strain and an F⁻ strain of E. coli, which of the following are generally true?
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Answer: A — Chromosomal genes are transferred in a fixed linear order.; B — Most recipients remain F⁻.; C — Interrupting mating at intervals maps genes by their time of entry.
Transfer starts inside the integrated F at oriT and proceeds along the chromosome, so genes enter in order and time of entry in minutes maps them. The rest of F would enter last, after about 100 minutes, and the mating bridge nearly always breaks first — so recipients stay F⁻ and do not become Hfr.In the snail Limnaea, dextral coiling (D) is dominant to sinistral (d) and coiling is determined by maternal effect. A dd female is crossed with a DD male. The F₁ snails are
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Answer: A — all Dd and all sinistral
Every F₁ snail is Dd, but its coiling was set by the gene products its dd mother put into the egg, so all are sinistral. When these F₁ snails reproduce, their own Dd genotype takes over and all F₂ are dextral; the 3:1 segregation appears only in the F₃. Choosing 'all dextral' reads the offspring’s own genotype, which is exactly what a maternal effect does not do.A monohybrid F₂ of 400 plants shows 290 with the dominant and 110 with the recessive phenotype. The χ² value for the fit to a 3:1 ratio is ______ (round off to two decimal places).
Numerical answer — type the value.
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Answer: 1.33
Expected counts are 300 and 100. χ² = (290 − 300)²/300 + (110 − 100)²/100 = 100/300 + 100/100 = 0.33 + 1.00 = 1.33. With 1 degree of freedom this is below 3.84, so the data fit 3:1 at p = 0.05. Dividing by the observed instead of the expected counts is the usual slip.The karyotype 47,XXY corresponds to
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Answer: A — Klinefelter syndrome
An extra X in a male gives Klinefelter syndrome — the presence of a Y with SRY makes the individual male, and one of the two X chromosomes is inactivated as a Barr body. Turner syndrome is 45,X, Down syndrome is trisomy 21, and cri-du-chat is a deletion of 5p.Genetic drift is expected to change allele frequencies most rapidly in
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Answer: A — a small, isolated population
Drift is sampling error between generations, and its variance scales as pq/(2N), so it is largest when N is small, as after a bottleneck or in a founder population. Gene flow from a large neighbour and a very large population both damp random change, and selection is a directed force, not drift.Which of the following are epigenetic mechanisms of gene regulation?
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Answer: A — Methylation of cytosine in CpG dinucleotides; B — Acetylation of lysines on histone tails; C — Genomic imprinting
Epigenetic changes alter expression heritably without altering the DNA sequence: CpG methylation, histone modifications and imprinting — itself maintained by allele-specific methylation — all qualify. A point mutation changes the sequence itself, which is the definition of a genetic rather than an epigenetic change.Genes A and B are 12 map units apart. A dihybrid AB/ab (in coupling) is test-crossed with ab/ab. The percentage of progeny with the Ab phenotype is ______.
Numerical answer — type the value.
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Answer: 6
12 map units means 12% recombinant gametes, shared equally between the two reciprocal recombinant classes Ab and aB, so each is 6%. The parental classes AB and ab are 44% each. Reporting 12 forgets that the recombinants are split between two classes.