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NEET (UG) · revision cheat sheet · neetlogic.in

Botany

17 chapters · 45 topics · ~180 marks · 33 solved previous-year questions

Class 11

1. The Living World

Characteristics of living organisms: Growth, reproduction and metabolism all have exceptions — mules and sterile worker bees are alive but never reproduce — so NCERT treats consciousness, the ability to sense and respond to the environment, as the one property that defines life without exception. Watch for options that call reproduction a universal criterion of a living organism.
Taxonomy, nomenclature and taxonomic categories: The rank between kingdom and class is named differently by convention — phylum for animals, division for plants — and NEET tests this exact word swap more often than the rest of the hierarchy combined. Keep taxon (an actual group, e.g. Insecta) distinct from category (the rank itself, e.g. Class).

Which one of the following statements refers to reductionist biology?

Ans: Physico-chemical approach to study and understand living organisms·Reductionist biology explains living systems by breaking them into their physico-chemical parts — the approach described in Statement 1.

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2. Biological Classification

Five-kingdom classification: Whittaker's five kingdoms are separated on five criteria — cell structure, body organisation, mode of nutrition, reproduction and phylogenetic relationships — and mode of respiration is not one of them, though options often slip it in. Viruses, viroids and lichens are deliberately left out of all five kingdoms because they don't fit a cellular definition of life.
Monera, Protista and Fungi: Match fungal class to its sexual spore, not just to a genus name: zygospores or oospores in Phycomycetes, ascospores in an ascus for Ascomycetes, basidiospores on a basidium for Basidiomycetes — and Deuteromycetes are grouped separately for exactly one reason, that no sexual stage is known for them at all.
Viruses, viroids and lichens: The one fact worth knowing cold: a viroid is infectious RNA with no protein coat at all, unlike a virus, which always packages its nucleic acid inside a protein capsid — that missing coat is the whole reason Diener's discovery was notable, and it's the detail options most often drop or reverse.

Each of the following characteristics represents a kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organisation. A. Multicellular heterotrophs with a cell wall made of chitin. B. Heterotrophs with tissue/organ/organ-system level of body organisation. C. Prokaryotes with a cell wall made of polysaccharides and amino acids. D. Eukaryotic autotrophs with tissue/organ level of body organisation. E. Eukaryotes with cellular body organisation. Choose the correct answer from the options given below:

Ans: C, E, A, D, B·Monera (prokaryotes with peptidoglycan walls, C) → Protista (unicellular eukaryotes, E) → Fungi (multicellular with chitin walls, A) → Plantae (tissue/organ level autotrophs, D) → Animalia (organ-system level heterotrophs, B). So C, E, A, D, B.

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Which of the following organisms cannot fix nitrogen? A. Azotobacter · B. Oscillatoria · C. Anabaena · D. Volvox · E. Nostoc Choose the correct answer from the options given below:

Ans: D only·Volvox is a green alga with no nitrogen-fixing ability. Azotobacter is a free-living fixer, and Oscillatoria, Anabaena and Nostoc are nitrogen-fixing cyanobacteria.

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3. Plant Kingdom

Algae: NEET draws heavily on one table — Chlorophyceae stores true starch, Phaeophyceae stores it as laminarin or mannitol, Rhodophyceae as floridean starch, each with its own matching pigment set — and wrong options routinely swap the storage product or pigment between two of the three classes.
Bryophytes and pteridophytes: One distinction organises this whole comparison: the gametophyte is the dominant, independent generation in bryophytes, while in pteridophytes the sporophyte is dominant and independent and the gametophyte (prothallus) is reduced to a small, short-lived stage.
Gymnosperms and angiosperms; alternation of generations: The gymnosperm fact NEET returns to again and again: their endosperm forms before fertilisation, as part of the haploid female gametophyte — the opposite of angiosperms, where the endosperm is triploid precisely because it is a product of fertilisation itself, via triple fusion.

The correct sequence of events in the life cycle of bryophytes is: A. Fusion of antherozoid with egg B. Attachment of gametophyte to the substratum C. Reduction division to produce haploid spores D. Formation of sporophyte E. Release of antherozoids into water Choose the correct answer from the options given below:

Ans: B, E, A, D, C·The gametophyte attaches to the substratum (B), releases antherozoids into water (E), an antherozoid fuses with the egg (A), the sporophyte develops on the gametophyte (D), and meiosis in the capsule produces haploid spores (C). So B, E, A, D, C.

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In bryophytes, the gemmae help in which one of the following?

Ans: Asexual reproduction·Gemmae are green, multicellular buds formed in gemma cups on the thallus of liverworts such as Marchantia; they detach and grow into new plants — asexual reproduction.

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4. Morphology of Flowering Plants

Root, stem and leaf: modifications: The classic mix-up is thorn versus spine: a thorn (Bougainvillea, Citrus) is a hardened, axillary branch — a stem structure — while a spine (Opuntia, Acacia) is a modified leaf or stipule, and NEET relies on their shared "sharp defensive structure" look to test whether you actually know the difference in origin.
Inflorescence, flower and floral formula: Whether the main axis keeps growing decides everything: a racemose inflorescence has an indefinite, still-growing axis and opens acropetally (oldest flower at the base), while a cymose one ends in a terminal flower that stops the axis, so it opens basipetally instead — options frequently swap which order goes with which type.
Fruit and seed; families: Not every fruit is simply "a ripened ovary": apple, pear and cashew are false fruits whose fleshy, edible part develops from the thalamus (receptacle) rather than the ovary, which is exactly the example set NEET uses to check this distinction.

In the seeds of cereals, the outer covering of the endosperm separates the embryo by a protein-rich layer called:

Ans: Aleurone layer·In cereal grains the endosperm is separated from the embryo by the aleurone layer, a protein-rich layer that also secretes enzymes during germination.

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Given below are two statements: Statement I: In a floral formula, the symbol % stands for the zygomorphic nature of the flower, and G with a line above it stands for an inferior ovary. Statement II: In a floral formula, the symbol ⊕ stands for the actinomorphic nature of the flower, and G with a line below it stands for a superior ovary. In the light of the above statements, choose the correct answer from the options given below:

Ans: Statement I is incorrect but Statement II is correct·In floral formulae % denotes zygomorphy and ⊕ actinomorphy; a line below G marks a superior ovary and a line above G an inferior ovary. Statement I gets the ovary marking backwards; Statement II is correct.

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5. Anatomy of Flowering Plants

Tissues and tissue systems: Living versus dead at maturity is the axis NEET tests: collenchyma stays alive and can keep dividing, giving flexible support to organs still growing, while sclerenchyma is dead and uniformly lignified, giving rigid support wherever growth has already stopped — options like to swap "living" and "dead" between the two.
Anatomy of dicot and monocot root, stem, leaf: The anatomical fact that predicts almost everything else asked here: dicot stems have open vascular bundles (cambium present, so secondary growth is possible) arranged in a ring, while monocot stems have closed bundles (no cambium, no secondary growth) scattered through the ground tissue.

Find the statement that is not correct with regard to the structure of the monocot stem.

Ans: The hypodermis is parenchymatous.·In a monocot stem the hypodermis is sclerenchymatous, not parenchymatous — Statement 1 is wrong. Scattered, conjoint, closed bundles and absent phloem parenchyma are all correct.

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Given below are two statements: Statement I: Parenchyma is living but collenchyma is dead tissue. Statement II: Gymnosperms lack xylem vessels but the presence of xylem vessels is characteristic of angiosperms. In the light of the above statements, choose the correct answer from the options given below:

Ans: Statement I is false but Statement II is true·Collenchyma is a living tissue (Statement I false). Vessels are characteristic of angiosperms and absent from gymnosperms, whose xylem has only tracheids (Statement II true).

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6. Cell: The Unit of Life

Cell theory, prokaryotic and eukaryotic cells: "No membrane-bound organelles" does not mean "no ribosomes" — prokaryotic cells still have 70S ribosomes, since a ribosome is never membrane-bound to begin with; what prokaryotes actually lack is the endomembrane system and DNA-containing double-membrane organelles.
Cell membrane, cell wall and endomembrane system: Keep wall and membrane separated by function, not just position: the cell wall is freely (fully) permeable and never itself controls what enters or leaves, while it's the plasma membrane underneath that is selectively permeable — crediting the wall with selective transport is the standard trap.
Mitochondria, plastids, ribosomes, cytoskeleton, nucleus: Mitochondria and plastids are the two organelles carrying their own circular DNA and their own 70S ribosomes — the same size class as a bacterial ribosome, not the 80S kind on the rest of the eukaryotic cell's machinery — which is exactly the evidence the endosymbiotic theory is built on.

A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is:

Ans: Mesosome·The mesosome, an infolding of the prokaryotic plasma membrane, helps in cell-wall formation, DNA replication and its distribution, and in respiration.

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Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver them to intracellular targets and outside the cell. Reason (R): Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus. In the light of the above statements, choose the correct answer from the options given below:

Ans: Both A and R are true and R is the correct explanation of A·Packaging and delivering ER products is the Golgi's principal job (A true), and material does arrive in vesicles at the cis face and leave, modified, from the trans face (R true) — R describes how A is achieved.

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7. Cell Cycle and Cell Division

Phases of the cell cycle: S phase doubles the DNA content but not the chromosome number — each chromosome simply gains a second sister chromatid, so the count only changes once anaphase splits the centromeres. Resist calling G1 and G2 empty "gaps": both are long, active periods of growth and synthesis, not pauses.
Mitosis: Mitosis isn't restricted to diploid cells the way meiosis is — plenty of haploid cells, in fungi, algae and gametophyte tissue, divide mitotically too, since mitosis only has to copy whatever chromosome number is already present, never halve it.
Meiosis and its significance: Only meiosis I actually reduces the chromosome number, by separating homologous pairs; meiosis II is mechanically just mitosis run on haploid cells, separating sister chromatids without reducing anything further. Treating both divisions as equally "reductional" is the recurring error.

What is the main function of the spindle fibres during mitosis?

Ans: To separate the chromosomes·Spindle fibres attach to kinetochores and pull the sister chromatids apart to opposite poles during anaphase.

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Following are the stages of cell division: A. Gap 2 phase · B. Cytokinesis · C. Synthesis phase · D. Karyokinesis · E. Gap 1 phase Choose the correct sequence of stages from the options given below:

Ans: E-C-A-D-B·The cell cycle runs

G_{1} \to S \to G_{2} \to M

, and M phase is karyokinesis followed by cytokinesis: E, C, A, D, B.

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8. Photosynthesis in Higher Plants

Light reaction: photosystems, electron transport, photophosphorylation: Cyclic photophosphorylation runs on Photosystem I alone, cycling electrons back to the same photosystem — so it produces ATP only, with no NADPH and no oxygen evolved, since splitting water is strictly a Photosystem II job.
Calvin cycle, C4 pathway, photorespiration: C4 plants don't skip the Calvin cycle — they still run it, just relocated to the bundle sheath cells, after PEP carboxylase in the mesophyll (which cannot bind oxygen) concentrates CO2 there and starves RuBisCO of the oxygen that would otherwise trigger photorespiration.
Factors affecting photosynthesis: Blackman's law of limiting factors is the lens for every graph question here: the factor nearest its minimum sets the rate, so a light-saturation plateau means light has stopped being limiting and something else — usually CO2 or temperature — now is, not that photosynthesis has maxed out for good.

Which of the following statements about RuBisCO is true?

Ans: It catalyses the carboxylation of RuBP·RuBisCO catalyses the carboxylation of RuBP in the Calvin cycle. It is active in light, has a higher affinity for

C O_{2}

than for

O_{2}

, and has nothing to do with photolysis of water.

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Match List I with List II.

List I	List II
A. Chlorophyll a	I. Yellow-green
B. Chlorophyll b	II. Yellow
C. Xanthophylls	III. Blue-green
D. Carotenoids	IV. Yellow to yellow-orange

Choose the option with all correct matches:

Ans: A-III, B-I, C-II, D-IV·On a chromatogram chlorophyll a is blue-green (III), chlorophyll b yellow-green (I), xanthophylls yellow (II) and carotenoids yellow to yellow-orange (IV).

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9. Respiration in Plants

Glycolysis and fermentation: Glycolysis happens in the cytoplasm, not the mitochondrion, and runs identically whether the cell goes on to respire aerobically or ferment. Fermentation itself adds no extra ATP beyond glycolysis's own net 2 — its only job is regenerating NAD+ so glycolysis can keep going.
Krebs cycle and electron transport chain: FADH2 hands its electrons to the transport chain at Complex II, skipping Complex I entirely — which is exactly why it yields fewer ATP than NADH (2 versus 3, in the standard accounting) even though both are reduced coenzymes generated by the same turn of the cycle.
Respiratory quotient and amphibolic pathway: RQ reveals what's being burned: carbohydrate respiration gives RQ = 1, fat gives RQ below 1 (oxidising its extra hydrogen needs more O2 than the CO2 released), and organic acids give RQ above 1. Respiration is also amphibolic, not purely catabolic — its intermediates are routinely drawn off to build fats and amino acids too.

Complex II of the mitochondrial electron transport chain is also known as:

Ans: Succinate dehydrogenase·Complex II of the electron transport chain is succinate dehydrogenase, which passes electrons from

FAD H_{2}

to ubiquinone. Complex I is NADH dehydrogenase, III the cytochrome bc

_{1}

complex and IV cytochrome c oxidase.

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Identify the step in the tricarboxylic acid cycle which does not involve oxidation of the substrate:

Ans: Succinyl-CoA → Succinic acid·Succinyl-CoA → succinic acid is a substrate-level phosphorylation that releases CoA and makes GTP; no hydrogen is removed. The other three steps are dehydrogenations (oxidations).

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10. Plant Growth and Development

Phases of growth and differentiation: Growth plotted against time gives a sigmoid curve, not a straight line — a slow lag phase, a rapid exponential phase, then a stationary phase as the growth rate itself declines — and NEET's graph questions usually just want the segment named correctly.
Plant growth regulators: auxin, gibberellin, cytokinin, ethylene, ABA: Auxin's apical dominance is a genuine paradox worth remembering: the same hormone that drives elongation at the shoot apex simultaneously suppresses the lateral buds below it, so it's removing the apex — not adding more auxin — that finally lets the laterals grow.

Which one of the following phytohormones promotes nutrient mobilisation, which helps in the delay of leaf senescence in plants?

Ans: Cytokinin·Cytokinins promote nutrient mobilisation and so delay leaf senescence; ethylene and abscisic acid promote senescence.

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Read the following statements on plant growth and development. (A) Parthenocarpy can be induced by auxins. (B) Plant growth regulators can be involved in the promotion as well as the inhibition of growth. (C) Dedifferentiation is a pre-requisite for redifferentiation. (D) Abscisic acid is a plant growth promoter. (E) Apical dominance promotes the growth of lateral buds. Choose the option with all correct statements:

Ans: A, B, C only·Auxins induce parthenocarpy (A); growth regulators can promote or inhibit growth (B); cells must dedifferentiate before they can redifferentiate (C). Abscisic acid is a growth inhibitor, and apical dominance suppresses lateral buds. Hence A, B and C.

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Class 12

11. Sexual Reproduction in Flowering Plants

Flower structure, microsporogenesis and megasporogenesis: Microsporogenesis and megasporogenesis are not symmetric: meiosis in the microspore mother cell gives four functional pollen grains, but meiosis in the megaspore mother cell gives four megaspores of which normally only one — usually the chalazal one — survives to form the embryo sac.
Pollination and double fertilisation: "Triple fusion" names one single fusion event involving three haploid nuclei — one male gamete plus both polar nuclei — not three separate fusions, and it's this event, not syngamy, that produces the triploid primary endosperm nucleus.
Post-fertilisation: embryo, seed, fruit; apomixis: Apomixis produces a seed without any fertilisation at all, often by a nucellar cell developing directly into an embryo — and it matters commercially because it would let a hybrid variety breed true, passing on its hybrid vigour indefinitely instead of segregating out in the very next generation.

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. Reason (R): The presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells. In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Both A and R are true and R is the correct explanation of A·Tapetal cells are multinucleate with dense cytoplasm (A true), and this polyploid, multinucleate state supports their role of nourishing the developing microspores (R true, and it explains A).

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How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?

Ans: 1 meiosis and 3 mitosis·One meiosis of the megaspore mother cell gives four megaspores; the surviving one undergoes three mitotic divisions to form the 8-nucleate embryo sac. So 1 meiosis and 3 mitoses.

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12. Principles of Inheritance and Variation

Mendel's laws and deviations: Don't let "both parental traits show up" collapse into "blended appearance": incomplete dominance genuinely blends into an intermediate phenotype (pink Mirabilis), while codominance keeps both parental phenotypes fully and separately visible at once (AB blood group) — NEET options trade directly on that difference.
Chromosomal theory and linkage: Genes on the same chromosome recombine less often than the 50% independent assortment would predict, and less often still the closer together they sit — the extreme case NCERT names is male Drosophila, where crossing over does not happen at all.
Sex determination, mutation, genetic disorders: Don't default to the mammalian pattern: in the ZW system that governs birds (and some fish and moths), it's the female who is heterogametic and determines the offspring's sex — the reverse of the XY system, where that role belongs to the human male.

What is the pattern of inheritance for a polygenic trait?

Ans: Non-Mendelian inheritance pattern·Polygenic traits such as skin colour and height are controlled by several genes with additive effects and show continuous variation — a non-Mendelian pattern.

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Genes R and Y follow independent assortment. If RRYY produces round yellow seeds and rryy produces wrinkled green seeds, what will be the phenotypic ratio of the

F_{2}

generation?

Ans: Phenotypic ratio 9 : 3 : 3 : 1·Independent assortment of two genes in a dihybrid cross gives the classical 9 : 3 : 3 : 1 phenotypic ratio in the

F_{2} (

round yellow : round green : wrinkled yellow : wrinkled green).

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13. Molecular Basis of Inheritance

DNA structure and replication: Meselson and Stahl's density-gradient experiment on E. coli, tracking 15N and 14N through generations, is the direct proof that replication is semiconservative — each new DNA molecule keeps exactly one parental strand — and NEET likes to ask which generation produces which density band.
Transcription, genetic code, translation: The genetic code is degenerate — most amino acids answer to more than one codon — but not evenly: methionine (AUG) and tryptophan (UGG) are the two exceptions with only a single codon each, which is exactly why AUG can double as both the start signal and an amino-acid codon without ambiguity.
Gene regulation: lac operon; Human Genome Project, DNA fingerprinting: The lac operon is off by default — the repressor sits on the operator and blocks transcription — and lactose itself (as allolactose) is the inducer that removes the repressor, switching on one polycistronic mRNA for all three structural genes at once, unlike any eukaryotic gene.

Histones are enriched with:

Ans: Lysine and arginine·Histones are rich in the basic amino acids lysine and arginine; their positive charge binds the negatively charged DNA.

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Match List I with List II.

List I	List II
A. Alfred Hershey and Martha Chase	I. Streptococcus pneumoniae
B. Euchromatin	II. Densely packed and dark-stained
C. Frederick Griffith	III. Loosely packed and light-stained
D. Heterochromatin	IV. Confirmation of DNA as genetic material

Choose the correct answer from the options given below:

Ans: A-IV, B-III, C-I, D-II·Hershey and Chase confirmed DNA as the genetic material (IV); euchromatin is loosely packed and lightly stained (III); Griffith worked with Streptococcus pneumoniae (I); heterochromatin is densely packed and dark (II).

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14. Microbes in Human Welfare

Microbes in household and industrial products: Beyond curd and bread, know the mechanism behind at least one product: statins from Monascus purpureus lower blood cholesterol by competitively inhibiting the enzyme that synthesises it — the kind of specific, mechanistic fact NEET prefers over a bare organism-product pairing.
Sewage treatment, biogas, biocontrol and biofertilisers: Higher BOD means more polluted water, not cleaner — it measures how much oxygen bacteria would consume breaking down the organic matter present, so secondary (biological) treatment exists specifically to bring that number down before effluent is released.

Streptokinase, produced by the bacterium Streptococcus, is used for:

Ans: Removing clots from blood vessels·Streptokinase, from Streptococcus, is a 'clot buster' given to patients after a heart attack to dissolve blood clots in vessels.

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Which of the following is an example of a non-distilled alcoholic beverage produced by yeast?

Ans: Beer·Beer is fermented by yeast but not distilled. Whisky, brandy and rum are distilled spirits.

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15. Organisms and Populations

Organism and its environment; adaptations: Regulation isn't always physiological: plenty of conformers — reptiles, most famously — keep their body temperature workable purely through behaviour, basking in the sun or retreating to shade, without ever regulating internally the way a true homeotherm does.
Population attributes, growth models and interactions: Exponential growth (dN/dt = rN) describes a population with no resource ceiling and produces an ever-steepening J-shaped curve that no real population sustains for long; logistic growth (dN/dt = rN(K-N)/K) is the realistic, resource-limited S-shaped curve NEET actually expects you to identify.

Epiphytes growing on a mango branch are an example of which of the following?

Ans: Commensalism·An epiphyte uses the mango branch only for support and takes nothing from it — the orchid benefits and the tree is unaffected, which is commensalism.

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Which one of the following equations represents the Verhulst–Pearl logistic growth of a population?

Ans:

\frac{d N}{d t} = r N (\frac{K - N}{K})

·The Verhulst–Pearl logistic equation is

\frac{d N}{d t} = r N (\frac{K - N}{K})

, growth slowing as the population

N

approaches the carrying capacity

K

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16. Ecosystem

Structure, productivity and decomposition: Net primary productivity is what's left after the plant's own respiration is subtracted from gross primary productivity — and it's this leftover, not GPP, that is actually available as food to every heterotroph in the ecosystem.
Energy flow, ecological pyramids: Pyramids of numbers and biomass can both turn upside down — a single large tree supporting swarms of insects, for instance — but the pyramid of energy cannot: every trophic transfer loses energy as heat, so it stays upright without exception.
Nutrient cycling: Sort each cycle by where its reservoir sits: carbon, nitrogen and oxygen are gaseous cycles with the atmosphere as reservoir, while phosphorus and sulphur are sedimentary, drawing from the earth's crust — phosphorus in particular has essentially no gaseous phase at all, unlike the other three.

Which of the following is the unit of productivity of an ecosystem?

Ans:

(kcal m^{- 2}) y r^{- 1}

·Productivity is the rate of biomass production per unit area per unit time, expressed in

g m^{- 2} y r^{- 1}

(kcal m^{- 2}) y r^{- 1}

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Given below are two statements: Statement I: The primary source of energy in an ecosystem is solar energy. Statement II: The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP). In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Statement I is correct but Statement II is incorrect·Solar energy is the primary source for almost every ecosystem (I correct). The rate of organic-matter production during photosynthesis is the gross primary productivity; NPP is what remains after respiratory loss, so II is incorrect.

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17. Biodiversity and Conservation

Levels and patterns of biodiversity: The species-area slope isn't a fixed number: within a region it comes out to roughly 0.1-0.2 fairly regardless of the taxonomic group studied, but measured across whole continents that same slope steepens sharply to somewhere between 0.6 and 1.2.
Loss of biodiversity and its conservation: Of the "Evil Quartet," habitat loss and fragmentation is the single largest driver of biodiversity loss, ahead of over-exploitation, invasive species and co-extinction — and a region only counts as a biodiversity hotspot when it combines high endemism with being under serious threat, not richness alone.

Which one of the following is an example of ex-situ conservation?

Ans: Zoos and botanical gardens·Zoos, botanical gardens, seed banks and cryopreservation keep species outside their natural habitat — ex-situ conservation. National parks, sanctuaries and other protected areas are in-situ.

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Match List I with List II.

List I	List II
A. The Evil Quartet	I. Cryopreservation
B. Ex-situ conservation	II. Alien species invasion
C. Lantana camara	III. Causes of biodiversity losses
D. Dodo	IV. Extinction

Choose the option with all correct matches:

Ans: A-III, B-I, C-II, D-IV·The Evil Quartet names the causes of biodiversity loss (III); cryopreservation is an ex-situ method (I); Lantana camara is an invasive alien species (II); the dodo is the classic extinction (IV).

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