In Johannsen's experiments with beans, selection for heavier seeds shifted the mean of a mixed population but had no effect when practised within a single pure line. Why?

All plants of a pure line share one genotype, so the seed-weight differences among them are environmental

Too few seeds were weighed in each generation of the pure line for selection to have any detectable effect

Bean is cross-pollinated, so outcrossing with other lines erased the effect of selection every generation

All plants of a pure line share one genotype, so the seed-weight differences among them are environmental

Seed weight in bean is set by one dominant gene that masks recessive alleles within the line

In a cytoplasmic-genetic male sterility system, the A line has sterile (S) cytoplasm and an rf rf nucleus, the B line has normal (N) cytoplasm and an rf rf nucleus, and the R line carries Rf Rf. Which cross is made each season to multiply seed of the A line itself?

A line as the female parent pollinated by the B line

A line as the female parent pollinated by the R line

B line as the female parent pollinated by the A line

R line as the female parent pollinated by the B line

A male-sterile maize plant is pollinated by a fertile inbred line Q, and every F1 plant is male sterile. The F1 and each later generation are again pollinated by Q. After six such backcrosses, when the nuclear genome is almost entirely that of Q, every plant is still male sterile. Which explanation fits these results?

A sterility-inducing cytoplasm passed through the egg, with Q lacking a nuclear restorer allele

A recessive nuclear male-sterility allele, with Q homozygous for the dominant fertile allele

A sterility-inducing cytoplasm passed through the egg, with Q lacking a nuclear restorer allele

A sterility-inducing cytoplasm passed through the egg, with Q itself homozygous for a dominant nuclear restorer allele

A dominant nuclear male-sterility allele carried heterozygously by the original sterile plant

Which of the following statements about molecular markers and marker-assisted selection are correct?

An SSR marker scored in an F2 population distinguishes the heterozygote from both homozygotes, giving a 1:2:1 ratio

Background selection with markers spread over the genome speeds recovery of the recurrent-parent genome in a backcross programme

An SSR marker scored in an F2 population distinguishes the heterozygote from both homozygotes, giving a 1:2:1 ratio

A RAPD band present in one parent and absent in the other is codominant, so heterozygotes look different from band-bearing homozygotes

Background selection with markers spread over the genome speeds recovery of the recurrent-parent genome in a backcross programme

A marker 25 cM from a gene is as reliable for foreground selection as one 1 cM away, since both lie on the same chromosome

Barley (Hordeum vulgare) pollinated with Hordeum bulbosum forms embryos that, when rescued and cultured, grow into haploid barley plants. What makes these plants haploid?

The bulbosum chromosomes are eliminated from the embryo during its early cell divisions

Fertilization never occurs, because bulbosum pollen tubes cannot reach the barley ovule

The bulbosum chromosomes are eliminated from the embryo during its early cell divisions

A meiotic division in the hybrid zygote halves the chromosome number before mitosis begins

Colchicine in the embryo-rescue medium removes one chromosome set from every cell

Tobacco pith explants are cultured on media that differ only in the ratio of auxin to cytokinin. What is expected on a medium with a high cytokinin-to-auxin ratio?

Shoot buds form on the explant while the formation of roots is suppressed

Roots form freely from the explant while the formation of shoots is suppressed

Undifferentiated callus proliferates without forming any organs at all

Somatic embryos form directly, each one with its own shoot and root poles

Shoot buds form on the explant while the formation of roots is suppressed

Which of the following statements about plant tissue culture techniques are correct?

Plants regenerated from a cultured apical dome are often free of viruses present in the mother plant

Artificial seeds are made by encapsulating somatic embryos in sodium alginate and hardening the beads in calcium chloride

A somatic embryo is joined to the explant tissue by vascular strands, in the same way as an adventitious shoot

Plants regenerated from a cultured apical dome are often free of viruses present in the mother plant

Somaclonal variation is a temporary change that disappears once the regenerated plant is propagated by seed

Artificial seeds are made by encapsulating somatic embryos in sodium alginate and hardening the beads in calcium chloride

In a binary-vector system for Agrobacterium-mediated transformation, which component must lie on the same plasmid as the gene to be transferred, rather than being supplied in trans by the helper Ti plasmid?

The left and right 25-bp border repeats, which mark the two ends of the transferred DNA

The virA and virG genes, which sense wound phenolics and switch on the vir regulon

The virD1 and virD2 genes, whose products nick the T-DNA and pilot it into the cell

The opine catabolism genes, which let the bacterium feed on compounds the tumour makes

The left and right 25-bp border repeats, which mark the two ends of the transferred DNA

Which of the following are features of plastid (chloroplast) transformation as practised in tobacco?

The transgene integrates into the plastome by homologous recombination between plastid sequences flanking it in the vector

Repeated rounds of regeneration on spectinomycin are needed to replace every wild-type plastome copy and reach homoplasmy

The transgene integrates into the plastome by homologous recombination between plastid sequences flanking it in the vector

Repeated rounds of regeneration on spectinomycin are needed to replace every wild-type plastome copy and reach homoplasmy

The foreign protein receives eukaryotic N-linked glycosylation, as it would in the endoplasmic reticulum

The transgene spreads readily through pollen because plastids are inherited from the male parent in most angiosperms

An sgRNA has a 20-nucleotide spacer that exactly matches a chosen 20-bp protospacer in a gene. On the protospacer-bearing strand, the three nucleotides immediately 3′ of the protospacer are 5′-TCA-3′. When this guide is used with Streptococcus pyogenes Cas9 (SpCas9), which recognizes a 5′-NGG-3′ PAM, what outcome is expected at this site and why?

No double-strand break forms, because the sequence next to the protospacer is not a 5′-NGG-3′ PAM, which SpCas9 requires before it will cut

A precise cut forms, because perfect pairing of the 20-nucleotide spacer with the target is by itself sufficient for cleavage

A cut forms but only on the strand paired with the guide, because the RuvC domain is inactive without a purine-rich PAM

A cut forms far from the site, because in the absence of a PAM Cas9 cleaves 17 base pairs downstream instead

A construct has the arrangement promoter–loxP–stop–loxP–GUS, where the stop cassette blocks GUS expression. Cre recombinase is expressed from a heat-shock promoter so that only scattered cells excise the stop. Why does each GUS-positive region appear as a clonal sector of blue cells rather than as single isolated cells?

Once the stop is excised in a cell the change is permanent and is inherited by all its mitotic descendants

Cre protein diffuses from each expressing cell into its neighbours and excises the stop in all of them at once

Excision at the loxP sites reverses spontaneously, so blue cells keep re-appearing in the same patch

Once the stop is excised in a cell the change is permanent and is inherited by all its mitotic descendants

Heat shock synchronizes the cell cycle, so every cell in a region divides and turns blue together

Seeds are treated with ethyl methanesulfonate (EMS), grown to the M1, and the M1 plants are selfed to raise the M2. Why are recessive point mutations screened in the M2 rather than in the M1?

A new recessive mutation is heterozygous or chimeric in the M1 and turns homozygous only after selfing, in the M2

EMS keeps on acting on the plant's DNA throughout the whole of the M1 generation and finally stops only once the M2 has been formed

M1 plants are sterile, so no phenotype of any kind can be scored until the M2 restores fertility

Dominant mutations appear only in the M2, while recessive ones are already visible in the M1

Plant Breeding, Genetic Modification and Genome Editing: Selection, Heterosis, Markers, Tissue Culture, Transgenics and CRISPR

This chapter answers Section XL2.6 of XL2 Botany, one of the five optional sections of the Life Sciences (XL) paper: “Plant Breeding, Genetic Modification, Genome Editing”. The heading names selection, hybridization and heterosis; male sterility; genetic maps and molecular markers; embryo rescue; haploids and doubled haploids; plant tissue culture (micropropagation, embryo culture and in vitro regeneration, somatic embryogenesis, artificial seed, cryopreservation, somaclonal variation, somatic cell hybridization); marker-assisted selection; direct and vector-mediated gene transfer and the generation of transgenic plants; CRISPR/Cas9 and the Cre-Lox system to generate chimeras; plastid transformation; and chemical mutagenesis. For each tool the chapter asks one question: what does it let a breeder see, fix or move that ordinary crossing cannot? The numericals are heterosis percentages, homozygosity after selfing, backcross recovery, doubled-haploid and marker-linkage arithmetic, and the segregation of transgenes.

1. Breeding systems and selection

Breeding is the directed change of a crop's heritable make-up, and the method follows the mating system. Self-pollinated crops (rice, wheat, chickpea) exist as near-homozygous lines; cross-pollinated crops (maize, pearl millet) are heterozygous populations that suffer inbreeding depression when selfed; clonal crops (potato, sugarcane, banana) fix any genotype, heterozygous or not, by vegetative propagation. Every scheme has three steps: create variation (introduction, hybridization, mutation, transgenesis), select within it, and test the selections over locations and seasons before release.

In self-pollinators, mass selection bulks seed of many superior plants; pure-line selection picks single plants and keeps their progenies apart. Johannsen's bean experiments showed why selection within a pure line fails: all its variation is environmental. After a cross, the pedigree method selects plants and keeps records from F2 onward; the bulk method and single-seed descent advance the population with little selection until it is nearly homozygous; the backcross method moves one or a few genes from a donor into an elite recurrent parent. Each backcross halves the donor share, so with BC1 as the first backcross generation the recurrent-parent genome in BCn is 1 − (1/2)ⁿ⁺¹: 75% in BC1, 87.5% in BC2, 93.75% in BC3. Cross-pollinators are improved by mass and recurrent selection, synthetic and composite varieties, and hybrids.

Selfing arithmetic. Selfing Aa gives 1/4 AA, 1/2 Aa and 1/4 aa, so heterozygosity halves every generation. Counting F1 as the first generation, the heterozygous fraction in Fn is (1/2)ⁿ⁻¹ and the homozygous fraction 1 − (1/2)ⁿ⁻¹. Independent loci multiply: the fraction homozygous at all k loci is [1 − (1/2)ⁿ⁻¹]ᵏ. Worked example: in F6 with 10 segregating loci, (31/32)¹⁰ = 0.728, so about 73% of plants are homozygous at all ten. An F1 heterozygous at n loci forms 2ⁿ kinds of gamete and gives 3ⁿ genotypes in F2, of which 2ⁿ are homozygous.

⚠️ Exam trap

Count generations carefully: F2 has had one round of selfing, so its heterozygosity is 1/2, not 1/4. Most slips in these questions are one generation off.

2. Hybridization, heterosis and male sterility

Heterosis is the superiority of an F1 over its parents, measured three ways: mid-parent heterosis = (F1 − MP)/MP × 100, where MP is the parental mean; better-parent heterosis (heterobeltiosis) = (F1 − BP)/BP × 100; and standard heterosis against the best commercial check, the figure that decides whether a hybrid is worth releasing. Worked example: parents yield 40 and 50 q/ha and the F1 63 q/ha. MP = 45, so mid-parent heterosis = 18/45 × 100 = 40%, while better-parent heterosis = 13/50 × 100 = 26%. The dominance hypothesis explains heterosis as favourable dominant alleles from each parent masking the other's deleterious recessives; the overdominance hypothesis credits the heterozygote itself (Aa superior to AA); epistasis adds to both. Vigour falls in F2 as genotypes segregate, so hybrid seed must be produced afresh each season.

Male sterility: a female parent that sheds no viable pollen makes hybrid seed production possible on a field scale
Type	Basis	How it is used
Genetic (GMS)	Recessive nuclear ms alleles	msms × Msms gives 1 sterile : 1 fertile, so fertile plants must be rogued from the female rows
Cytoplasmic (CMS)	Chimeric open reading frames in the mitochondrial genome, inherited through the egg	On its own, useful only where seed is not the harvested product
Cytoplasmic-genetic (CGMS)	Sterile cytoplasm (S) plus a nuclear fertility restorer (Rf)	Three lines: A (S, rf rf) sterile; B (N, rf rf) maintainer; R (Rf Rf) restorer
Environment-sensitive GMS	Nuclear genes whose effect depends on temperature (TGMS) or day length (PGMS)	Sterile in one season and self-fertile in another: two-line hybrids, as in rice
Transgenic	Barnase, a ribonuclease, expressed in the tapetum blocks pollen development	A barstar line, whose protein inhibits barnase, restores fertility in the F1

In the three-line system, A × B gives A-line seed again (S cytoplasm from the mother, rf rf nucleus from both parents), and A × R gives the fertile commercial hybrid (S, Rf rf), which matters wherever grain is the product.

📖 Uniform cytoplasm is a risk

By 1970 most hybrid maize in the USA carried Texas (T) male-sterile cytoplasm. Its mitochondrial protein URF13 makes the plant sensitive to the T-toxin of Cochliobolus heterostrophus race T, and southern corn leaf blight swept the crop. Breeders now diversify sterile cytoplasms.

3. Genetic maps, molecular markers and marker-assisted selection

A linkage map orders loci by recombination frequency: over short intervals 1% recombination is about 1 centimorgan (cM). Over longer intervals double crossovers hide recombinants, so mapping functions convert recombination frequency to distance (Haldane's assumes no interference, Kosambi's allows some). A physical map counts base pairs, and cM per megabase varies along a chromosome, being low near centromeres.

Marker	Basis	Inheritance	Note
RFLP	Restriction digestion, Southern blot and a labelled probe	Codominant	Reliable but slow; needs much DNA
RAPD	PCR with single short arbitrary primers (often 10-mers)	Dominant	Quick, poorly reproducible
AFLP	Restriction, adapter ligation and selective PCR	Mostly dominant	Many loci per reaction
SSR (microsatellite)	PCR across tandem repeats such as (CA)n whose length varies	Codominant	Multi-allelic and locus-specific; the workhorse of selection
SNP	Single-base differences read by arrays or sequencing	Codominant	Most abundant; biallelic; high throughput

A dominant marker scores only presence or absence, so an F2 gives 3:1 and the heterozygote looks like the band-bearing homozygote; a codominant marker shows both alleles and gives 1:2:1. Marker-assisted selection (MAS) selects the genotype, so it works on seedlings, off-season, on recessive alleles and on traits costly to screen. In a backcross programme, foreground selection tracks the target gene, recombinant selection with flanking markers trims the donor segment around it (linkage drag), and background selection picks plants carrying the most recurrent-parent markers, reaching in two or three backcrosses what would otherwise take six. Pyramiding stacks genes: Improved Samba Mahsuri rice carries the bacterial blight resistance genes Xa21, xa13 and xa5, and Swarna-Sub1 carries SUB1A for submergence tolerance.

Marker arithmetic. If a marker lies at recombination fraction r from the gene, a plant selected for the donor marker allele received from a heterozygous parent lacks the donor gene allele with probability r: at 4 cM, 4 of every 100 selected plants are false positives. With markers on both sides at r₁ and r₂, only a double crossover separates both from the gene, about r₁ × r₂ without interference: 0.04 × 0.05 = 0.002, or 2 in 1,000.

4. Haploids, doubled haploids and embryo rescue

A haploid sporophyte carries the gametic chromosome number. Routes to it: androgenesis by anther or isolated microspore culture (haploid embryos were first raised from anthers of Datura innoxia by Guha and Maheshwari in 1964); gynogenesis from unfertilized ovules or ovaries (onion, sugar beet); chromosome elimination after wide crosses, where the alien chromosomes are lost from the early embryo (barley × Hordeum bulbosum leaves haploid barley, wheat × maize leaves haploid wheat); and in vivo inducer lines, such as maize lines descended from Stock 6 or Arabidopsis lines with an altered centromeric histone CENH3. Colchicine binds tubulin, blocks the spindle and doubles the chromosome number, giving a fertile doubled haploid (DH) homozygous at every locus in one step instead of six or more generations of selfing.

DH arithmetic. An F1 heterozygous at n unlinked loci gives 2ⁿ DH genotypes in equal frequency, so the wanted homozygote appears at (1/2)ⁿ against (1/4)ⁿ in an F2. For 4 loci that is 1/16 against 1/256, sixteen times fewer plants to grow. For two linked loci at recombination fraction r, each recombinant DH class is r/2 and each parental class (1 − r)/2. Because every line breeds true and can be replicated, DH populations are also ideal for mapping.

Embryo rescue saves hybrid embryos that form and then abort, usually because the endosperm fails in a wide cross. The immature embryo is excised and grown on a nutrient medium, younger embryos needing more sucrose and a higher osmotic strength. It recovers interspecific hybrids and early-ripening and seedless (stenospermocarpic) grapes, and shortens breeding cycles by bypassing seed dormancy. Culture of mature embryos is the simpler, related technique.

5. Plant tissue culture

Tissue culture rests on totipotency, the capacity of a living plant cell to regenerate a whole plant. The standard medium is that of Murashige and Skoog (MS, 1962): mineral salts, vitamins, sucrose as the carbon source and agar. Skoog and Miller's tobacco pith experiments (1957) set the rule for regeneration: a high auxin:cytokinin ratio gives roots, a high cytokinin:auxin ratio gives shoots, and a balance gives undifferentiated callus. Regeneration runs through organogenesis (shoots, then roots) or somatic embryogenesis, directly from the explant or indirectly through callus.

Technique	What is done	Use or principle
Micropropagation	Axillary or adventitious shoots multiplied through stock-plant preparation, establishment, multiplication, rooting and acclimatization	True-to-type clones in large numbers (banana, sugarcane, orchids)
Meristem culture	The apical dome, a few tenths of a millimetre, is cultured, often after heat therapy	Virus-free plants: the dome has no differentiated vascular tissue and carries little or no virus
Somatic embryogenesis	Bipolar embryos (shoot and root poles) from somatic cells, with no vascular link to the explant; in carrot 2,4-D induces them and an auxin-free medium lets them mature	Mass clonal propagation; source of artificial seed
Artificial seed	A somatic embryo or shoot bud mixed with sodium alginate is dropped into calcium chloride, forming a calcium alginate bead	Sowable, storable clonal propagules
Cryopreservation	Storage at −196 °C in liquid nitrogen after cryoprotectants (DMSO, glycerol, sucrose), vitrification or encapsulation-dehydration	Long-term germplasm storage, including species with recalcitrant seed; the enemy is intracellular ice
Somaclonal variation	Heritable variation among regenerants, from chromosome changes, transposon activation and altered DNA methylation; rises with time in callus	A source of new variants, and a threat to clonal fidelity
Somatic hybridization	Protoplasts released with cellulase and pectinase in an osmoticum such as mannitol, fused with PEG or an electric pulse, hybrids selected by complementation	Combines sexually incompatible species; cybrids (nucleus of one parent, mixed cytoplasm) transfer CMS and chloroplast traits

⚠️ Exam trap

A somatic embryo is genetically identical to its explant, whereas a zygotic embryo is a new recombinant. Somaclonal variation is heritable, unlike the passing abnormalities that culture conditions cause and that vanish in the next generation.

6. Gene transfer and generation of transgenic plants

Vector-mediated transfer uses Agrobacterium tumefaciens, which naturally inserts the T-DNA of its Ti plasmid into plant nuclear DNA. Phenolics from wounded cells, such as acetosyringone, are sensed by VirA, which phosphorylates VirG, the activator of the other vir genes. VirD1 and VirD2 nick the 25-bp border repeats; VirD2 stays covalently bound to the 5′ end of the single-stranded T-strand, VirE2 coats it, and a type IV secretion system delivers it into the plant cell. Only the borders act in cis, so in a binary vector the gene of interest and a selectable marker lie between borders on a small plasmid that replicates in E. coli and Agrobacterium, while a disarmed helper Ti plasmid, its tumour genes deleted, supplies the vir proteins in trans.

Direct transfer needs no bacterium: particle bombardment (gold or tungsten particles coated with DNA, usable on any tissue and on plastids), PEG-mediated or electroporation-mediated uptake into protoplasts, microinjection and silicon carbide whiskers. Direct methods more often give multiple, rearranged copies.

Generating a transgenic plant: co-cultivate explants with Agrobacterium (or bombard them); kill the bacteria with cefotaxime or carbenicillin; select transformed cells with a marker (nptII for kanamycin, hpt for hygromycin, bar for phosphinothricin); regenerate T0 plants; confirm by PCR, by Southern blotting for copy number and by RT-qPCR or western blotting for expression, with GUS or GFP as reporters. A T0 plant is hemizygous, so one insert selfed gives 3 resistant : 1 sensitive in T1, and two unlinked inserts give 15 : 1. A homozygous line is found in T2 as one whose progeny no longer segregate. Common constitutive promoters are CaMV 35S in dicots and maize ubiquitin in monocots.

ℹ️ Note

Marker genes can be removed after selection, either by co-transforming with two separate T-DNAs that later segregate apart or by Cre-lox excision.

7. Genome editing, Cre-lox chimeras, plastid transformation and chemical mutagenesis

In CRISPR/Cas9, the Streptococcus pyogenes Cas9 nuclease is guided by a single guide RNA whose 20-nucleotide spacer pairs with the target DNA. Cutting needs a PAM, 5′-NGG-3′, immediately 3′ of the protospacer on the non-target strand; the HNH domain cuts the strand paired with the guide and RuvC the other, leaving a blunt double-strand break 3 bp upstream of the PAM. Error-prone non-homologous end joining leaves small insertions or deletions that usually shift the reading frame and knock the gene out; homology-directed repair from a donor template makes precise changes; base editors (a Cas9 nickase fused to a deaminase, for C•G to T•A or A•T to G•C) and prime editors change bases without a double-strand break. The Cas9 transgene can then be segregated away, leaving an edited plant with no foreign DNA. Edits are grouped as SDN-1 (no template), SDN-2 (short template) and SDN-3 (a whole gene inserted).

Cre-lox. Cre recombinase, from bacteriophage P1, recombines two 34-bp loxP sites (two 13-bp inverted repeats around an 8-bp asymmetric spacer that gives the site a direction). Sites in the same orientation on one molecule excise the DNA between them; opposite orientation inverts it; sites on two molecules allow integration or exchange. To make a chimera, Cre is expressed in only some cells, from a heat-shock or tissue-specific promoter, and a reporter built as promoter–loxP–stop–loxP–GUS switches on only where the stop is excised. Excision is permanent and passes to every daughter cell, so each event marks a clonal sector in an otherwise unchanged plant: this is how cell lineages in meristems are traced. The same system removes selectable markers; FLP/FRT from the yeast 2-micron plasmid is the analogue.

Plastid transformation. DNA, usually delivered by bombardment, carries the gene flanked by plastid sequences and enters the plastome by homologous recombination at a chosen site. The marker aadA gives spectinomycin and streptomycin resistance, and because a leaf cell holds thousands of plastome copies, repeated rounds of regeneration on selection are needed to reach homoplasmy. Advantages: very high expression, polycistronic operons, no position effect or silencing, and maternal plastid inheritance in most angiosperms, which limits transgene escape through pollen. Limits: no eukaryotic glycosylation, and routine success in few species, tobacco being the first.

Chemical mutagenesis. EMS (ethyl methanesulfonate) ethylates guanine at O6; O6-ethylguanine pairs with thymine, giving mostly G:C to A:T transitions. Sodium azide (much used in barley) and MNU are other mutagens. Treated seed grows into the M1, which is chimeric because the seed embryo is multicellular, and in which recessive mutations stay hidden; recessive mutants are screened in the M2. TILLING pairs an EMS population with a screen for point mutations in a chosen gene, a non-transgenic route to reverse genetics.

Key takeaways

The method follows the mating system: pure lines, pedigree and backcross for selfers; recurrent selection and hybrids for outcrossers. Heterozygosity halves with each selfing, (1/2)ⁿ⁻¹ in Fn, and BCn recovers 1 − (1/2)ⁿ⁺¹ of the recurrent parent.
Heterosis is measured against the mid-parent, the better parent or a standard check. In CGMS, A × B maintains the sterile line and A × R makes the fertile hybrid.
Codominant markers (RFLP, SSR, SNP) read heterozygotes and dominant ones (RAPD, mostly AFLP) do not. A marker at recombination fraction r misses the gene with probability r; flanking markers fail only on a double crossover, about r₁r₂.
Doubled haploids give complete homozygosity in one step, the wanted genotype appearing at (1/2)ⁿ rather than (1/4)ⁿ as in F2. Embryo rescue saves wide-cross embryos whose endosperm fails.
High auxin:cytokinin gives roots and the reverse gives shoots; somatic embryos are bipolar and clonal; artificial seeds are calcium alginate beads; somaclonal variation is heritable.
Binary vectors need only the T-DNA borders in cis, and a single insert segregates 3:1 in T1. Cas9 needs an NGG PAM; Cre on directly repeated loxP sites excises, marking clonal sectors; plastid transgenes integrate by homologous recombination; EMS gives G:C to A:T transitions, screened in M2.

Practice questions (17)

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.

In Johannsen's experiments with beans, selection for heavier seeds shifted the mean of a mixed population but had no effect when practised within a single pure line. Why?
1. Too few seeds were weighed in each generation of the pure line for selection to have any detectable effect
2. Bean is cross-pollinated, so outcrossing with other lines erased the effect of selection every generation
3. All plants of a pure line share one genotype, so the seed-weight differences among them are environmental
4. Seed weight in bean is set by one dominant gene that masks recessive alleles within the line
Show answer
Answer: C — All plants of a pure line share one genotype, so the seed-weight differences among them are environmental
A pure line is the progeny of one homozygous, self-pollinated plant, so all its members carry the same genotype; seed-weight differences within it are environmental and are not transmitted, and selection among them changes nothing. Bean is self-pollinated, so outcrossing did not undo selection. Sample size was not the issue, since the mixed population, which contained genetically different lines, did respond.
In a cytoplasmic-genetic male sterility system, the A line has sterile (S) cytoplasm and an rf rf nucleus, the B line has normal (N) cytoplasm and an rf rf nucleus, and the R line carries Rf Rf. Which cross is made each season to multiply seed of the A line itself?
1. A line as the female parent pollinated by the B line
2. A line as the female parent pollinated by the R line
3. B line as the female parent pollinated by the A line
4. R line as the female parent pollinated by the B line
Show answer
Answer: A — A line as the female parent pollinated by the B line
A × B gives progeny with S cytoplasm from the A-line mother and an rf rf nucleus from both parents, which is the A line again, male sterile. A × R gives S Rf rf plants that are fertile: that is the commercial hybrid, not A-line seed. The A line sheds no viable pollen, so it cannot be a pollen parent, and an R-line mother would pass on its own cytoplasm rather than S cytoplasm.
A male-sterile maize plant is pollinated by a fertile inbred line Q, and every F1 plant is male sterile. The F1 and each later generation are again pollinated by Q. After six such backcrosses, when the nuclear genome is almost entirely that of Q, every plant is still male sterile. Which explanation fits these results?
1. A recessive nuclear male-sterility allele, with Q homozygous for the dominant fertile allele
2. A sterility-inducing cytoplasm passed through the egg, with Q lacking a nuclear restorer allele
3. A sterility-inducing cytoplasm passed through the egg, with Q itself homozygous for a dominant nuclear restorer allele
4. A dominant nuclear male-sterility allele carried heterozygously by the original sterile plant
Show answer
Answer: B — A sterility-inducing cytoplasm passed through the egg, with Q lacking a nuclear restorer allele
Repeated backcrossing replaces the nucleus with Q's but keeps the cytoplasm of the original mother, because cytoplasm passes through the egg. Sterility that persists after the nucleus has become Q's therefore lies in the cytoplasm, and Q must lack a restorer (it acts as a maintainer). A recessive nuclear allele would give a fertile F1 when Q carries the dominant fertile allele, and a restorer in Q would likewise make the F1 fertile. A heterozygous dominant sterility allele would give about half fertile progeny, not an all-sterile F1.
Two inbred lines of pearl millet yield 60 g and 80 g of grain per plant, and their F1 yields 91 g per plant. Calculate the better-parent heterosis (heterobeltiosis) of the F1, in per cent, to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 13.75
Better parent BP = 80 g. Heterobeltiosis = (F1 − BP)/BP × 100 = (91 − 80)/80 × 100 = 11/80 × 100 = 13.75%. Using the mid-parent value (60 + 80)/2 = 70 g instead gives (91 − 70)/70 × 100 = 30%, which is mid-parent heterosis, not the quantity asked.
An F1 wheat plant is heterozygous at three unlinked loci. Its descendants are advanced by single-seed descent (selfing, no selection) to the F5 generation, counting the F1 as the first generation. What percentage of F5 plants is expected to be homozygous at all three loci? Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 82.4
F5 has had four rounds of selfing (F1 to F2, F2 to F3, F3 to F4, F4 to F5), so heterozygosity at each locus is (1/2)⁴ = 1/16 and homozygosity 15/16. For three independent loci, (15/16)³ = 3375/4096 = 0.8240, i.e. 82.4%. Counting five rounds of selfing gives (31/32)³ = 90.9%, the usual off-by-one slip.
Which of the following statements about molecular markers and marker-assisted selection are correct?
1. An SSR marker scored in an F2 population distinguishes the heterozygote from both homozygotes, giving a 1:2:1 ratio
2. A RAPD band present in one parent and absent in the other is codominant, so heterozygotes look different from band-bearing homozygotes
3. Background selection with markers spread over the genome speeds recovery of the recurrent-parent genome in a backcross programme
4. A marker 25 cM from a gene is as reliable for foreground selection as one 1 cM away, since both lie on the same chromosome
Show answer
Answer: A — An SSR marker scored in an F2 population distinguishes the heterozygote from both homozygotes, giving a 1:2:1 ratio; C — Background selection with markers spread over the genome speeds recovery of the recurrent-parent genome in a backcross programme
SSRs are codominant: both allele sizes are seen, so the F2 shows 1:2:1. Background selection picks backcross plants with the most recurrent-parent alleles and saves several generations. A RAPD band is a dominant marker: heterozygotes and band-bearing homozygotes look alike. Reliability falls with distance, because a marker 25 cM away recombines away from the gene in about a quarter of meioses (somewhat less after correcting for double crossovers), against about 1% at 1 cM.
In a backcross programme, the donor allele of a disease-resistance gene is linked in coupling to an SSR marker allele 6 cM away. The F1 is backcrossed to the susceptible recurrent parent, and 400 BC1 plants carrying the donor marker allele are selected without any disease test. Taking the map distance as equal to the recombination percentage, how many of these 400 plants are expected to lack the donor resistance allele?
Numerical answer — type the value.
Show answer
Answer: 24
Each selected BC1 plant received the donor marker allele in a gamete from the F1. It lacks the linked donor resistance allele only if a crossover occurred between marker and gene in that meiosis, with probability r = 0.06. Expected false positives = 0.06 × 400 = 24. Halving r because only half the BC1 carries the marker is the usual slip: the 400 plants were already selected for the marker.
An F1 rice plant is heterozygous at five unlinked loci, each with one favourable allele. A breeder raises 640 doubled-haploid lines from its anthers. How many of these lines are expected to be homozygous for the favourable allele at all five loci?
Numerical answer — type the value.
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Answer: 20
Each DH line is homozygous at every locus and carries either allele with probability 1/2 at each. The chance of all five favourable alleles is (1/2)⁵ = 1/32, so 640/32 = 20 lines. From 640 F2 plants the expectation would be 640 × (1/4)⁵ = 0.625 plant, which is why doubled haploids make rare combinations accessible.
Barley (Hordeum vulgare) pollinated with Hordeum bulbosum forms embryos that, when rescued and cultured, grow into haploid barley plants. What makes these plants haploid?
1. Fertilization never occurs, because bulbosum pollen tubes cannot reach the barley ovule
2. The bulbosum chromosomes are eliminated from the embryo during its early cell divisions
3. A meiotic division in the hybrid zygote halves the chromosome number before mitosis begins
4. Colchicine in the embryo-rescue medium removes one chromosome set from every cell
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Answer: B — The bulbosum chromosomes are eliminated from the embryo during its early cell divisions
Fertilization does occur and a hybrid zygote forms, but the H. bulbosum chromosomes are lost from the young embryo in its early divisions, leaving only the barley set; embryo rescue is needed because the endosperm fails. The haploids therefore arise by chromosome elimination, not by failure of fertilization. Zygotes do not undergo meiosis, and colchicine doubles chromosome number rather than removing a set.
Tobacco pith explants are cultured on media that differ only in the ratio of auxin to cytokinin. What is expected on a medium with a high cytokinin-to-auxin ratio?
1. Roots form freely from the explant while the formation of shoots is suppressed
2. Undifferentiated callus proliferates without forming any organs at all
3. Somatic embryos form directly, each one with its own shoot and root poles
4. Shoot buds form on the explant while the formation of roots is suppressed
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Answer: D — Shoot buds form on the explant while the formation of roots is suppressed
Skoog and Miller showed that the auxin:cytokinin balance decides the organ: high cytokinin relative to auxin gives shoots, high auxin relative to cytokinin gives roots, and an intermediate balance gives callus. Root formation is the outcome of the opposite ratio, the commonest confusion. Somatic embryos need a different regime, typically induction by 2,4-D followed by an auxin-free medium.
Which of the following statements about plant tissue culture techniques are correct?
1. A somatic embryo is joined to the explant tissue by vascular strands, in the same way as an adventitious shoot
2. Plants regenerated from a cultured apical dome are often free of viruses present in the mother plant
3. Somaclonal variation is a temporary change that disappears once the regenerated plant is propagated by seed
4. Artificial seeds are made by encapsulating somatic embryos in sodium alginate and hardening the beads in calcium chloride
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Answer: B — Plants regenerated from a cultured apical dome are often free of viruses present in the mother plant; D — Artificial seeds are made by encapsulating somatic embryos in sodium alginate and hardening the beads in calcium chloride
The apical dome has no differentiated vascular tissue and little or no virus, so meristem culture (often after heat therapy) yields virus-free plants. Calcium ions cross-link alginate into a gel bead around the embryo, giving an artificial seed. A somatic embryo is bipolar and has no vascular connection with the explant, which distinguishes it from an adventitious shoot. Somaclonal variation is heritable, arising from chromosomal, transposon and methylation changes.
In a binary-vector system for Agrobacterium-mediated transformation, which component must lie on the same plasmid as the gene to be transferred, rather than being supplied in trans by the helper Ti plasmid?
1. The virA and virG genes, which sense wound phenolics and switch on the vir regulon
2. The virD1 and virD2 genes, whose products nick the T-DNA and pilot it into the cell
3. The opine catabolism genes, which let the bacterium feed on compounds the tumour makes
4. The left and right 25-bp border repeats, which mark the two ends of the transferred DNA
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Answer: D — The left and right 25-bp border repeats, which mark the two ends of the transferred DNA
Only the border repeats act in cis: the VirD1/VirD2 complex recognizes and nicks them wherever they are, so the gene of interest must sit between borders on the binary plasmid. All vir proteins, including VirD1 and VirD2, act in trans and can come from the disarmed helper Ti plasmid. Opine catabolism genes are not needed for transfer at all.
A T0 tobacco plant carries two unlinked, single-copy T-DNA inserts, each with a functional kanamycin-resistance gene. It is selfed and 480 T1 seedlings are grown on kanamycin. How many seedlings are expected to be kanamycin-sensitive?
Numerical answer — type the value.
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Answer: 30
Each insert is hemizygous in T0, so a seedling lacks a given insert with probability 1/4. It is sensitive only if it lacks both: 1/4 × 1/4 = 1/16. Sensitive seedlings = 480/16 = 30, the 15 resistant : 1 sensitive ratio. With a single insert the ratio would be 3:1, giving 120.
Which of the following are features of plastid (chloroplast) transformation as practised in tobacco?
1. The transgene integrates into the plastome by homologous recombination between plastid sequences flanking it in the vector
2. Repeated rounds of regeneration on spectinomycin are needed to replace every wild-type plastome copy and reach homoplasmy
3. The foreign protein receives eukaryotic N-linked glycosylation, as it would in the endoplasmic reticulum
4. The transgene spreads readily through pollen because plastids are inherited from the male parent in most angiosperms
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Answer: A — The transgene integrates into the plastome by homologous recombination between plastid sequences flanking it in the vector; B — Repeated rounds of regeneration on spectinomycin are needed to replace every wild-type plastome copy and reach homoplasmy
Plastid vectors carry plastid DNA on both sides of the transgene, which integrates by homologous recombination at a defined site, and because a cell has thousands of plastome copies, homoplasmy needs repeated selection with aadA on spectinomycin. Plastids have prokaryote-like expression machinery and no ER-type N-glycosylation. Plastids are maternally inherited in most angiosperms, so pollen rarely carries the transgene; that containment is one of the method's advantages.
An sgRNA has a 20-nucleotide spacer that exactly matches a chosen 20-bp protospacer in a gene. On the protospacer-bearing strand, the three nucleotides immediately 3′ of the protospacer are 5′-TCA-3′. When this guide is used with Streptococcus pyogenes Cas9 (SpCas9), which recognizes a 5′-NGG-3′ PAM, what outcome is expected at this site and why?
1. No double-strand break forms, because the sequence next to the protospacer is not a 5′-NGG-3′ PAM, which SpCas9 requires before it will cut
2. A precise cut forms, because perfect pairing of the 20-nucleotide spacer with the target is by itself sufficient for cleavage
3. A cut forms but only on the strand paired with the guide, because the RuvC domain is inactive without a purine-rich PAM
4. A cut forms far from the site, because in the absence of a PAM Cas9 cleaves 17 base pairs downstream instead
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Answer: A — No double-strand break forms, because the sequence next to the protospacer is not a 5′-NGG-3′ PAM, which SpCas9 requires before it will cut
SpCas9 interrogates DNA for its PAM first and only then unwinds and pairs the guide; with 5′-TCA-3′ next to the protospacer there is no 5′-NGG-3′ PAM, so no cleavage occurs however well the spacer matches. Perfect spacer pairing alone is therefore not sufficient. Cas9 makes a blunt break (both HNH and RuvC acting) about 3 bp from the PAM, not a single-strand nick far downstream.
A construct has the arrangement promoter–loxP–stop–loxP–GUS, where the stop cassette blocks GUS expression. Cre recombinase is expressed from a heat-shock promoter so that only scattered cells excise the stop. Why does each GUS-positive region appear as a clonal sector of blue cells rather than as single isolated cells?
1. Cre protein diffuses from each expressing cell into its neighbours and excises the stop in all of them at once
2. Excision at the loxP sites reverses spontaneously, so blue cells keep re-appearing in the same patch
3. Once the stop is excised in a cell the change is permanent and is inherited by all its mitotic descendants
4. Heat shock synchronizes the cell cycle, so every cell in a region divides and turns blue together
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Answer: C — Once the stop is excised in a cell the change is permanent and is inherited by all its mitotic descendants
Cre excision between directly repeated loxP sites deletes the stop irreversibly (the product has a single loxP and cannot recombine back), so a cell that has switched GUS on passes the switched-on gene to every daughter by mitosis, building a clonal sector. Cre acts within the cell that makes it and does not diffuse into neighbours, and heat shock induces expression rather than synchronizing division; this permanence is exactly what makes the system a lineage tracer.
Seeds are treated with ethyl methanesulfonate (EMS), grown to the M1, and the M1 plants are selfed to raise the M2. Why are recessive point mutations screened in the M2 rather than in the M1?
1. A new recessive mutation is heterozygous or chimeric in the M1 and turns homozygous only after selfing, in the M2
2. EMS keeps on acting on the plant's DNA throughout the whole of the M1 generation and finally stops only once the M2 has been formed
3. M1 plants are sterile, so no phenotype of any kind can be scored until the M2 restores fertility
4. Dominant mutations appear only in the M2, while recessive ones are already visible in the M1
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Answer: A — A new recessive mutation is heterozygous or chimeric in the M1 and turns homozygous only after selfing, in the M2
A mutation induced in one strand of one cell of the multicellular seed embryo gives an M1 plant that is heterozygous and often chimeric for it; a recessive allele is masked by the wild-type allele and shows no phenotype. Selfing the M1 lets the allele become homozygous in a quarter of the M2, where it is finally visible. EMS does not keep acting across generations, M1 plants are generally fertile, and recessive alleles are hidden, not visible, in the M1.