Genetics and Genomics: Cell Division, Mendel, Linkage, Organelle Genes, Epigenetics, Genome Change and Omics
2. Mendelian inheritance, gene interaction and the chi-square test
Mendel's law of segregation says the two alleles of a gene separate into different gametes; the law of independent assortment says alleles of genes on different chromosomes (or far apart on one) assort independently. A monohybrid F2 gives 3:1 (1:2:1 by genotype), a dihybrid 9:3:3:1 and a dihybrid test cross 1:1:1:1. With n independent heterozygous genes there are 2ⁿ kinds of gamete and 3ⁿ F2 genotypes, and probabilities multiply: from AaBbCc selfed, all three dominant phenotypes appear in (3/4)³ = 27/64 of the progeny. Incomplete dominance (red × white snapdragon gives pink) turns 3:1 into 1:2:1, and interaction between two genes modifies 9:3:3:1:
| F2 ratio | Interaction | Logic |
|---|---|---|
| 9:7 | Complementary genes (duplicate recessive epistasis) | A dominant allele of both genes is needed, as for sweet pea flower colour |
| 9:3:4 | Recessive epistasis | aa masks the second gene |
| 12:3:1 | Dominant epistasis | A_ masks the second gene, as for summer squash fruit colour |
| 15:1 | Duplicate dominant genes | Either dominant allele suffices, as for shepherd's purse capsule shape |
| 13:3 | Dominant suppression | A_ suppresses the trait, so only aaB_ shows it |
The chi-square test asks whether a deviation from an expected ratio is larger than chance would give: χ² = Σ(O − E)²/E, with degrees of freedom = number of classes − 1 when the ratio comes from the hypothesis. Critical values at P = 0.05 are 3.84 (1 df), 5.99 (2 df) and 7.81 (3 df). Worked example: an F2 of 290 purple and 110 white plants tested against 3:1 has expected values 300 and 100, so χ² = 100/300 + 100/100 = 1.33 < 3.84, and the 3:1 hypothesis is not rejected. Not rejecting means the data are consistent with the hypothesis, not that they prove it.
3. Linkage, recombination and genetic mapping
Genes close together on a chromosome are linked and stay together more often than independent assortment predicts; crossing over at pachytene recombines them. The recombination frequency (RF) = recombinant progeny ÷ total, most easily read from a test cross, and 1% RF = 1 map unit (centimorgan, cM). RF cannot exceed 50%, because even many crossovers between distant genes yield at most as many recombinants as independent assortment. Phase matters: in coupling (AB/ab) the parental classes are AB and ab, in repulsion (Ab/aB) they are Ab and aB, and in either case the two most frequent test-cross classes are parental.
A three-point test cross maps and orders three genes at once. The two largest classes are parental and the two smallest are double crossovers (DCO); the gene that has switched in the DCO relative to the parental combination is the middle gene. Each interval's distance = (single crossovers in that interval + DCO) ÷ total × 100. The coefficient of coincidence (CoC) = observed DCO ÷ expected DCO, where expected DCO = RF₁ × RF₂ × total, and interference I = 1 − CoC; positive interference means one crossover lowers the chance of another nearby. Over long intervals RF underestimates map distance because double crossovers go unseen, which mapping functions such as Haldane's correct.
Worked example. ABC/abc × abc/abc gives 1,000 progeny: ABC 405, abc 395, Abc 45, aBC 55, ABc 42, abC 48, AbC 6, aBc 4. The DCO classes AbC and aBc differ from the parentals only at B, so B is in the middle. A–B = (45 + 55 + 10)/1000 = 11 cM; B–C = (42 + 48 + 10)/1000 = 10 cM; so A–C = 21 cM. Expected DCO = 0.11 × 0.10 × 1000 = 11, observed 10, so CoC = 0.91 and I = 0.09.
4. Extrachromosomal inheritance
Plastid and mitochondrial genes do not follow Mendel: their inheritance is non-Mendelian, usually maternal, and reciprocal crosses give different results. In most angiosperms plastids pass through the egg. In Mirabilis jalapa, Correns found that seeds from flowers on white branches give white seedlings, those from green branches give green seedlings and those from variegated branches give green, white and variegated seedlings, whatever the pollen. Plastid transmission is biparental in some genera (Pelargonium, Oenothera) and largely paternal in many conifers. Mixed (heteroplasmic) organelle populations sort out during cell divisions, producing variegated sectors.
Cytoplasmic male sterility (CMS) is caused by rearranged mitochondrial genes, often chimeric open reading frames, that disrupt pollen development; in the Texas (T) cytoplasm of maize the mitochondrial gene T-urf13 also made plants susceptible to southern corn leaf blight. Nuclear restorer-of-fertility (Rf) genes, many encoding pentatricopeptide-repeat (PPR) proteins that act on mitochondrial transcripts, suppress the sterility, and CMS with Rf underpins hybrid seed production. Do not confuse maternal inheritance (the gene is in the cytoplasm) with a maternal effect (the mother's nuclear genotype decides the offspring phenotype, as in snail shell coiling).
5. Epigenetics and gene silencing
Epigenetics is heritable change in gene activity without change in DNA sequence. Plants methylate cytosine in three contexts: CG, maintained by MET1; CHG (H = A, T or C), maintained by CMT3 together with H3K9 dimethylation; and CHH, maintained by CMT2 and by RNA-directed DNA methylation (RdDM). In RdDM, RNA polymerase IV transcripts are copied by RDR2 into dsRNA and cut by DCL3 into 24-nt siRNAs; these load AGO4, which pairs with RNA polymerase V transcripts and recruits the methyltransferase DRM2. Histone marks add a second layer: H3K27me3, laid down by Polycomb (PRC2) complexes, silences FLC after vernalisation. Epigenetic states can be inherited as epialleles, such as the peloric form of Linaria vulgaris, caused by methylation of the Lcyc gene, and they underlie paramutation and genomic imprinting in endosperm.
Gene silencing is transcriptional (TGS) when promoter methylation and heterochromatin stop transcription, and post-transcriptional (PTGS) when the mRNA is made but destroyed. Transgenes are prone to both, especially when present as multiple copies or inverted repeats or when very highly expressed. In the classic co-suppression of petunia, extra copies of the chalcone synthase gene, meant to deepen flower colour, gave white or variegated flowers because the transgene and the endogenous gene were silenced together; nuclear run-on assays showed that transcription continued while the mRNA was lost, the signature of PTGS. PTGS is RNA interference: double-stranded RNA is diced into siRNAs, one strand guides an Argonaute protein to cleave matching RNA, and RNA-dependent RNA polymerase (RDR6) amplifies the signal, which can spread through the plant. Viruses fight back with suppressors such as the tombusvirus P19 protein, which sequesters siRNA duplexes.
| Feature | miRNA | siRNA |
|---|---|---|
| Origin | MIR genes transcribed by RNA polymerase II into single-stranded fold-back precursors with imperfect stems | Long, perfectly paired dsRNA from inverted-repeat transgenes, viruses, RDR activity or transposons |
| Processing | DCL1 in the nucleus; 21-nt duplex 2′-O-methylated by HEN1 | DCL4 (21 nt), DCL2 (22 nt) or DCL3 (24 nt) |
| Targets | Other genes, often transcription factors: miR156 and SPL, miR165/166 and HD-ZIP III, miR172 and AP2 | Usually the sequence they came from: the transgene, virus or transposon itself |
| Action | AGO1 cleaves near-perfectly matched targets; some translational repression | Cleavage (21–22 nt, AGO1) or DNA methylation through RdDM (24 nt, AGO4) |
6. Genome organisation, gene expression, mutation and repair
Genome size varies enormously without matching complexity (the C-value paradox): Arabidopsis thaliana has about 135 Mb, hexaploid bread wheat roughly 16–17 Gb. The difference lies mostly in repetitive DNA, above all LTR retrotransposons, which make up most of large cereal genomes, together with satellite repeats at centromeres, rDNA arrays at nucleolar organisers and telomeres (TTTAGGG repeats in Arabidopsis). Plant genome evolution is dominated by polyploidy and whole-genome duplication followed by gene loss, by tandem and segmental duplications that create gene families, and by transposon bursts. Organelle genomes are small and gene-dense: plastid DNA (typically 120–160 kb) has two inverted repeats separating large and small single-copy regions, while plant mitochondrial genomes are large, variable and rich in recombining repeats. A nuclear protein-coding gene has a promoter (TATA box and upstream cis-elements bound by transcription factors) and exons and introns; it is transcribed by RNA polymerase II, and its pre-mRNA is capped, spliced (often alternatively) and polyadenylated. Plants also have RNA polymerases IV and V, which serve silencing.
Mutations are base substitutions (transitions within purines or within pyrimidines; transversions between them), small insertions or deletions that may shift the reading frame, or larger rearrangements; in coding sequence they are silent, missense or nonsense. EMS ethylates guanine to O⁶-ethylguanine, which pairs with thymine, giving almost only G:C→A:T transitions, which is why it is the mutagen of TILLING populations; ionising radiation breaks DNA and causes deletions and rearrangements; UV makes cyclobutane pyrimidine dimers and 6-4 photoproducts. Repair pathways are photoreactivation by photolyase (light-dependent), base excision and nucleotide excision repair, mismatch repair, and for double-strand breaks non-homologous end joining (fast, error-prone, the route exploited by CRISPR knockouts) and homologous recombination (accurate, template-dependent).
7. Chromosomal aberrations and transposons
Euploidy changes whole sets. Autopolyploids carry multiple sets of one genome (triploid banana, autotetraploid potato) and can form multivalents; allopolyploids combine different genomes that pair as bivalents. Bread wheat, Triticum aestivum, is an allohexaploid, 2n = 6x = 42 (AABBDD); Brassica napus (AACC, 2n = 38) arose from B. rapa (AA, 2n = 20) × B. oleracea (CC, 2n = 18). The F1 of such a cross is sterile because its chromosomes lack partners; doubling the number with colchicine restores pairing. Aneuploidy changes single chromosomes: nullisomic 2n − 2, monosomic 2n − 1, trisomic 2n + 1, tetrasomic 2n + 2, double monosomic 2n − 1 − 1. A diploid can have as many primary trisomics as it has chromosome pairs (12 in Datura stramonium), and polyploids such as wheat tolerate nullisomics, which is how genes are assigned to chromosomes.
| Aberration | Meiotic signature | Genetic consequence |
|---|---|---|
| Deletion | Unpaired loop in the normal homologue | Pseudodominance of recessive alleles on the normal homologue; often lethal when homozygous |
| Duplication | Unpaired loop in the duplicated homologue | Extra gene dose; raw material for new genes; often arises by unequal crossing over |
| Paracentric inversion | Inversion loop; a crossover inside it gives a dicentric bridge and an acentric fragment at anaphase I | Recombinant gametes are unbalanced, so recombination appears suppressed |
| Pericentric inversion | Loop includes the centromere; no bridge | Crossover products carry duplications and deficiencies; recombination appears suppressed |
| Reciprocal translocation | Cross-shaped quadrivalent at pachytene; ring or chain at metaphase I | Only alternate segregation gives balanced gametes, so heterozygotes are semi-sterile (about half the pollen aborts); genes on the two chromosomes appear linked |
Transposable elements were discovered by Barbara McClintock in maize through the Ac/Ds system: Ac (Activator) is autonomous and encodes transposase; Ds (Dissociation) is a non-autonomous derivative that moves only when Ac is present, and its insertion and excision produce spotted kernels. Class II DNA transposons (Ac/Ds, Mutator) move by cut-and-paste, have terminal inverted repeats and leave target-site duplications. Class I retrotransposons copy themselves through RNA and reverse transcriptase: LTR types (Ty1/copia and Ty3/gypsy) dominate plant genomes, and non-LTR LINEs and SINEs also occur. Most elements are kept silent by DNA methylation and siRNAs; stress or tissue culture can wake them, as the rice retrotransposon Tos17 is activated in culture and used for insertional mutagenesis.
8. Model organisms, omics and data analysis
| Organism | Why it is used |
|---|---|
| Arabidopsis thaliana | Small genome (about 135 Mb, 2n = 10), life cycle of about 6–8 weeks, selfing, floral-dip transformation, sequence-indexed T-DNA insertion lines |
| Rice (Oryza sativa) | Cereal and monocot model, compact genome, 2n = 24, efficient transformation, Tos17 and T-DNA mutant collections |
| Medicago truncatula and Lotus japonicus | Legume models for nodulation and mycorrhiza |
| Maize (Zea mays) | Classical genetics, transposons, heterosis and C4 photosynthesis |
| Physcomitrium patens (formerly Physcomitrella) | Moss with a haploid-dominant life cycle and efficient homologous recombination for gene targeting |
| Chlamydomonas reinhardtii | Unicellular alga for photosynthesis, flagella and chloroplast transformation |
| Nicotiana benthamiana | Transient expression by agroinfiltration and virus-induced gene silencing |
Forward genetics goes from a mutant phenotype to the gene; reverse genetics starts from a gene and disrupts it (T-DNA or transposon insertion, TILLING, RNAi, VIGS, CRISPR) to see the phenotype. Transcriptomics measures all transcripts, now by RNA-seq: reads are mapped to the genome, counts are normalised for gene length and sequencing depth (RPKM/FPKM or TPM), and differentially expressed genes are called by fold change and adjusted P-value. Proteomics separates proteins by 2-D gel electrophoresis (isoelectric focusing, then SDS-PAGE) or liquid chromatography and identifies them by mass spectrometry: in bottom-up proteomics trypsin cuts after lysine and arginine, and peptide masses (MALDI-TOF fingerprinting) or fragment spectra (LC-MS/MS) are matched to sequence databases. Metabolomics profiles small molecules by GC-MS (volatile or derivatised compounds), LC-MS and NMR, targeted or untargeted.
Bioinformatics stores and compares the data: databases such as GenBank, UniProt, the Protein Data Bank and TAIR for Arabidopsis; global (Needleman–Wunsch) and local (Smith–Waterman) alignment scored with substitution matrices such as BLOSUM62; and the fast heuristic local search BLAST, whose E-value is the number of hits of at least that score expected by chance in a database of that size, so lower is more significant. Statistics turns data into conclusions: mean, variance and standard error; t-tests and ANOVA for means; chi-square for counts; correlation and regression; clustering and principal component analysis for expression patterns; QTL mapping and GWAS to link markers to traits. Testing thousands of genes at once demands multiple-testing correction: 20,000 tests of true null hypotheses at P < 0.05 give about 1,000 false positives. The Bonferroni threshold divides α by the number of tests, and the Benjamini–Hochberg method controls the false discovery rate instead.
Key takeaways
- CDK–cyclin activity and checkpoints drive the cell cycle; plant cells use a preprophase band and a phragmoplast-built cell plate. Meiosis I is reductional; crossing over happens at pachytene; DNA content goes 2C (G1), 4C (prophase I), 2C (metaphase II), 1C (spore).
- Monohybrid 3:1, dihybrid 9:3:3:1, test cross 1:1:1:1; probabilities multiply; gene interaction gives 9:7, 9:3:4, 12:3:1, 15:1 and 13:3; χ² = Σ(O − E)²/E with df = classes − 1, critical 3.84 at 1 df.
- RF = recombinants ÷ total, 1% = 1 cM, never above 50%; in a three-point cross the double-crossover class reveals the middle gene, CoC = observed ÷ expected DCO and I = 1 − CoC.
- Organelle genes are usually maternal, so reciprocal crosses differ; CMS is mitochondrial and nuclear Rf genes restore it. Methylation is kept by MET1 (CG), CMT3 (CHG) and RdDM or CMT2 (CHH); PTGS destroys mRNA that is still transcribed; miRNAs (DCL1, MIR hairpins) act on other genes, siRNAs on their own source.
- Repeats and polyploidy shape genome size; EMS gives G:C→A:T transitions; NHEJ is error-prone, homologous recombination accurate. Count aneuploids from 2n (wheat: nullisomic 40, trisomic 43); a paracentric inversion crossover gives a bridge and fragment; translocation heterozygotes are semi-sterile; Ds moves only with Ac.
- Each model organism serves a purpose; RNA-seq counts are normalised for length and depth, proteins are identified from tryptic peptides by mass spectrometry, a low BLAST E-value is significant, and thousands of tests need Bonferroni or false-discovery-rate correction.
Practice questions (18)
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.
During which substage of meiotic prophase I is synapsis complete and crossing over taking place between non-sister chromatids?
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Answer: B — Pachytene
At pachytene the synaptonemal complex is complete along the bivalents and crossing over occurs. Zygotene is tempting because synapsis begins there, but it is not yet complete. Chiasmata, the visible result of crossing over, become evident at diplotene as the synaptonemal complex dissolves, and they terminalise during diakinesis.Plants heterozygous for three independently assorting genes, each showing complete dominance (AaBbCc), are self-pollinated and 1,280 progeny are raised. How many progeny are expected to show the dominant phenotype for exactly one of the three traits?
Numerical answer — type the value.
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Answer: 180
For each gene P(dominant) = 3/4 and P(recessive) = 1/4. Exactly one dominant can happen in 3 ways (which gene is dominant), each with probability (3/4)(1/4)(1/4) = 3/64, so the total is 9/64. Expected number = 9/64 × 1280 = 180. Forgetting the factor of 3 gives 60; the full set of phenotype classes is 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 out of 64, so all dominant is 540, exactly two dominant is also 540 and all recessive is 20.Two pure-breeding white-flowered lines of a plant are crossed. The F1 has purple flowers, and the F2 segregates 9 purple : 7 white. Which explanation fits these results?
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Answer: D — Two genes, each of which must carry a dominant allele for pigment to form
9:7 is complementary gene action: only A_B_ (9/16) is purple, because each gene controls a step of the pigment pathway and both steps must work; the two white parents were AAbb and aaBB. One gene with incomplete dominance would give 1:2:1 (here 2 purple : 2 white), not 9:7. Either dominant allele sufficing gives 15:1, and dominant epistasis gives 12:3:1.An F2 population contains 380 tall and 100 dwarf plants. Calculate χ² for the hypothesis of a 3:1 ratio, to two decimal places.
Numerical answer — type the value.
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Answer: 4.44
Total = 480, so expected = 360 tall and 120 dwarf. χ² = (380 − 360)²/360 + (100 − 120)²/120 = 400/360 + 400/120 = 1.11 + 3.33 = 4.44. With 1 degree of freedom the 5% critical value is 3.84, so the 3:1 hypothesis is rejected at P = 0.05. Dividing by the observed instead of the expected values gives 5.05, a common slip.A plant heterozygous for three linked genes, with the wild-type alleles of all three on one homologue and the recessive alleles a, b and c on the other, is test-crossed to a b c / a b c. The 1,000 progeny are listed below, alleles written in the order a, b, c and + meaning wild type:
+ + + 356; a b c 352; + b c 47; a + + 45; + b + 97; a + c 95; + + c 4; a b + 4.
What is the coefficient of interference, to one decimal place?Numerical answer — type the value.
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Answer: 0.6
Parentals are + + + and a b c; the rarest classes, + + c and a b + (8 in all), are double crossovers. They differ from the parentals only at c, so c is the middle gene and the order is a–c–b. a–c recombinants: + b c, a + +, plus DCO = 47 + 45 + 8 = 100, so 10 cM. c–b recombinants: + b +, a + c, plus DCO = 97 + 95 + 8 = 200, so 20 cM. Expected DCO = 0.10 × 0.20 × 1000 = 20; observed 8; CoC = 8/20 = 0.4; interference = 1 − 0.4 = 0.6. Assuming the written order a–b–c misidentifies the classes and gives a wrong answer.A dihybrid that received Ab from one parent and aB from the other is test-crossed. The progeny are Ab 205, aB 195, AB 52 and ab 48. Which conclusion is correct?
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Answer: A — The genes are 20 cM apart, and the recombinants are the AB (52) and ab (48) classes
The dihybrid is in repulsion (Ab/aB), so Ab and aB are parental and AB and ab are recombinant. RF = (52 + 48)/500 = 0.20, so 20 cM. Calling Ab and aB recombinant is the classic slip of assuming coupling phase. Map distance is the recombinant percentage, not the parental one, and linked genes still produce all four classes, only in unequal numbers.A cytoplasmic male-sterile line (S cytoplasm, rf rf) is pollinated by a line homozygous for a single dominant restorer gene Rf that acts sporophytically (the genotype of the plant bearing the anthers decides fertility). The F1 is self-pollinated. What is expected in the F2?
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Answer: B — Male-fertile and male-sterile plants in 3:1, all carrying S cytoplasm
Mitochondria pass through the egg, so the F1 and all F2 plants carry S cytoplasm. The F1 is Rf rf; selfing gives Rf Rf : Rf rf : rf rf = 1 : 2 : 1, and with sporophytic restoration any plant carrying Rf is fertile, so 3 fertile : 1 sterile. The all-sterile option ignores the nuclear restorer, and pollen does not transmit the restorer's cytoplasm. With gametophytic restoration only Rf pollen would function and the F2 would be all fertile.Which of the following statements about epigenetic regulation in plants are correct?
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Answer: A — CHH methylation is maintained largely by RNA-directed DNA methylation involving 24-nt siRNAs and DRM2; B — Vernalisation silences FLC through H3K27 trimethylation deposited by Polycomb complexes
Asymmetric CHH sites have no template on the new strand, so they are re-methylated by RdDM (Pol IV, RDR2, DCL3, 24-nt siRNA, AGO4, Pol V, DRM2) and by CMT2. Polycomb-deposited H3K27me3 accumulates at FLC during cold and keeps it silent. MET1 maintains CG, not CHH, methylation. An epiallele has the same DNA sequence and differs only in its epigenetic state, which is why epialleles such as peloric Linaria can revert.Which feature distinguishes a plant miRNA from a typical siRNA?
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Answer: C — It is processed by DCL1 from a fold-back precursor transcribed from its own gene
A miRNA is encoded by its own MIR gene, transcribed by Pol II into an imperfect hairpin and excised by DCL1; it then regulates other genes in trans. Every other option describes siRNAs: long perfect dsRNA from RDR activity, self-targeting RdDM through AGO4, and 24-nt heterochromatic siRNAs from Pol IV transcripts cut by DCL3.Petunia plants carrying extra copies of the chalcone synthase gene have white flowers. Which observation shows that the silencing is post-transcriptional rather than transcriptional?
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Answer: A — Run-on assays show both genes still transcribed in petal nuclei while their mRNAs are scarce
PTGS means the gene is transcribed but its mRNA is degraded, so the decisive evidence is continued transcription (nuclear run-on) together with low steady-state mRNA. Promoter methylation with no transcription is the signature of TGS. Recovery after the transgene segregates away happens with either mechanism, so it does not discriminate, and the downstream-enzyme observation says nothing about how chalcone synthase is silenced.In a plant heterozygous for a paracentric inversion, a single crossover within the inversion loop produces, at anaphase I,
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Answer: C — a bridge formed by a dicentric chromatid, together with an acentric fragment
In a paracentric inversion the centromere lies outside the loop, so a crossover inside it joins two centromeres on one chromatid (a dicentric bridge as the centromeres move apart) and leaves an acentric fragment that is lost. Duplication–deficiency chromatids without a bridge are the result in a pericentric inversion, the tempting near-miss. The quadrivalent belongs to a reciprocal translocation, and an unpaired loop with pseudodominance to a deletion.Bread wheat is an allohexaploid with 2n = 6x = 42. In a nulli-tetrasomic line, both copies of chromosome 5A are missing and chromosome 5B is present in four copies. How many chromosomes does a root-tip cell of this line contain?
Numerical answer — type the value.
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Answer: 42
Start from 42. Nullisomy for 5A removes two chromosomes (40), and tetrasomy for 5B adds two extra copies of 5B (42). The count is unchanged, though the genome is unbalanced; such lines are viable because the homoeologous 5B compensates for 5A, and they are used to assign genes to chromosomes. Counting only the loss gives 40; counting four added copies gives 44.The F1 of Brassica rapa (AA, 2n = 20) × Brassica oleracea (CC, 2n = 18) is highly sterile, but the same hybrid becomes fertile after colchicine treatment. Why?
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Answer: D — Its 19 chromosomes lack homologues and form univalents; doubling gives each a partner
The F1 receives 10 A and 9 C chromosomes (2n = 19), none with a homologue, so they form univalents at meiosis and segregate at random into unbalanced, mostly inviable gametes. Doubling to 2n = 38 (AACC, as in B. napus) gives every chromosome a partner, so bivalents form and gametes are balanced. The F1 is an amphihaploid, not a triploid; colchicine blocks spindle formation rather than repairing DNA; and doubling does not change the cytoplasm.Which statement about the maize Ds element is correct?
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Answer: B — It moves only when Ac is present in the genome to supply transposase
Ds (Dissociation) is a non-autonomous class II DNA element, typically a deletion derivative of Ac that has lost a functional transposase gene but keeps the terminal inverted repeats; it transposes only when Ac (Activator) supplies transposase. The roles are reversed in the transposase option: Ac encodes it. Ds is a cut-and-paste DNA transposon, not a retrotransposon, and not a centromeric satellite.Which of the following statements about model organisms and omics methods are correct?
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Answer: A — Physcomitrium patens is used for gene targeting because homologous recombination is efficient in it; B — In bottom-up proteomics, trypsin cuts proteins on the carboxyl side of lysine and arginine before mass spectrometry; C — TPM and FPKM normalise RNA-seq read counts for gene length and sequencing depth
The moss Physcomitrium integrates DNA by homologous recombination at high frequency, which makes targeted knockouts easy. Trypsin cleaves after lysine and arginine, giving predictable peptides for MALDI-TOF fingerprinting or LC-MS/MS. TPM and FPKM correct read counts for transcript length and library size. GC-MS needs volatile (or derivatised) small molecules; intact proteins are far too large and non-volatile for gas chromatography.An RNA-seq study tests 25,000 genes for differential expression. To keep the family-wise error rate at 0.05 using the Bonferroni correction, the per-gene P-value threshold is x × 10⁻⁶. What is x?
Numerical answer — type the value.
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Answer: 2
Bonferroni divides the overall α by the number of tests: 0.05/25,000 = 2 × 10⁻⁶, so x = 2. Without correction, 25,000 true-null tests at P < 0.05 would yield about 1,250 false positives. Multiplying instead of dividing, or using 0.01, are the usual slips.A BLAST search returns a database hit with an E-value of 10⁻⁴⁰. This means that
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Answer: C — about 10⁻⁴⁰ hits this good are expected by chance in a database of this size
The E-value is the number of alignments with at least this score expected by chance when searching a database of this size; 10⁻⁴⁰ means a chance hit is essentially impossible, so the match is highly significant and very likely reflects homology. It is not a percentage identity or a position, and it is not literally the probability of homology, which also depends on biology the statistics do not model.Which of the following statements about mutagens and DNA repair in plants are correct?
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Answer: A — EMS produces mainly G:C→A:T transitions because O⁶-ethylguanine pairs with thymine; B — Cyclobutane pyrimidine dimers caused by UV can be reversed directly by photolyase in the light; C — Non-homologous end joining repairs double-strand breaks without a template and often leaves small insertions or deletions
EMS ethylates guanine at O⁶; the adduct pairs with T, so after replication G:C becomes A:T. Photolyase uses blue light to split cyclobutane pyrimidine dimers directly (photoreactivation). NHEJ ligates broken ends without a template and often leaves small indels, which is why CRISPR–Cas9 knockouts work. Gamma rays are ionising radiation that break DNA strands, causing deletions and rearrangements; ethylation is the action of EMS.