Plant Systematics: Nomenclature, Classification, APG, Major Families, Cladistics and DNA Barcoding

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.

1. A species was described in 1885 from three specimens, all cited in the original publication, but the author never designated a holotype. A revising taxonomist now selects one of those three specimens as the single name-bearing type. Under the ICN, the selected specimen is a

   1. neotype, because it is designated long after the original description was first published
   2. isotype, because it belongs to the set of specimens the original author examined
   3. lectotype, because it is chosen later from original material when no holotype exists
   4. epitype, because it is chosen to settle how the original name should be interpreted
   Show answer

   Answer: C — lectotype, because it is chosen later from original material when no holotype exists

   A lectotype is a specimen chosen after publication from the original material (here the three cited syntypes) when no holotype was designated or it is missing. Neotype is the tempting answer because it is also chosen later, but it is allowed only when all original material is lost, which is not the case here. An isotype is a duplicate of a holotype, and an epitype supplements a type that is itself too ambiguous to identify the taxon.

2. Which of the following species names would be rejected under the ICN as a tautonym?

   1. Cajanus cajan, because its epithet is derived from the genus name
   2. Linaria linaria, because its epithet exactly repeats the genus name
   3. Mangifera indica, because its epithet refers to a geographical region
   4. Solanum lycopersicum, because its epithet was formerly a genus name
   Show answer

   Answer: B — *Linaria linaria*, because its epithet exactly repeats the genus name

   Botany forbids a specific epithet that exactly repeats the genus name, so Linaria linaria is a tautonym and is not permitted (zoology, by contrast, accepts names such as Naja naja). Cajanus cajan is the tempting choice, but its epithet differs in spelling from the genus name, so it is not a tautonym and is a correct name. Geographical epithets such as indica and epithets drawn from former genus names such as lycopersicum are legitimate.

3. Which of the following statements about botanical nomenclature under the ICN are correct?

   1. An isotype is a specimen chosen by a later author to replace a holotype that has been destroyed
   2. For most vascular plants, priority of names starts from Linnaeus’s Species Plantarum of 1753
   3. Gramineae is an illegitimate family name that must always be replaced by Poaceae
   4. A new plant species published in 2020 may be validated by a description or diagnosis written in English
   Show answer

   Answer: B — For most vascular plants, priority of names starts from Linnaeus’s *Species Plantarum* of 1753; D — A new plant species published in 2020 may be validated by a description or diagnosis written in English

   Priority for most plants runs from 1 May 1753, and since 1 January 2012 a Latin or an English description or diagnosis validates a new name. An isotype is merely a duplicate of the holotype from the same gathering; a specimen chosen to replace a lost type is a lectotype (from original material) or a neotype (if none survives). Gramineae is one of eight conserved alternative family names and may be used alongside Poaceae.

4. Which classification system placed the gymnosperms between the dicotyledons and the monocotyledons, a position now regarded as its chief flaw?

   1. Engler and Prantl’s system in Die Natürlichen Pflanzenfamilien
   2. Linnaeus’s sexual system, based mainly on the stamens
   3. Cronquist’s system of Magnoliopsida and Liliopsida
   4. Bentham and Hooker’s system in Genera Plantarum
   Show answer

   Answer: D — Bentham and Hooker’s system in *Genera Plantarum*

   Bentham and Hooker’s natural system (Genera Plantarum, 1862–1883) inserted Gymnospermae between the dicotyledons and the monocotyledons, separating two groups of flowering plants. Engler and Prantl is the tempting choice because it is the other great older system, but it placed the gymnosperms before all angiosperms; its weakness was treating the catkin-bearing Amentiferae as primitive. Linnaeus’s artificial system and Cronquist’s phylogenetic system did not make this placement.

5. Linnaeus’s sexual system is classed as an artificial system chiefly because it

   1. sorted plants mainly by stamen number, union and length, so unrelated plants sharing a stamen count fell into one class
   2. grouped plants by their inferred evolutionary descent, which could not be tested without any fossil evidence at that time
   3. relied on the overall similarity of very many characters, which made its classes too broad to be useful in practice
   4. treated the simple, unisexual flowers of catkin-bearing trees as primitive and placed them at the start
   Show answer

   Answer: A — sorted plants mainly by stamen number, union and length, so unrelated plants sharing a stamen count fell into one class

   An artificial system uses one or a few convenient characters; Linnaeus classed plants by stamen number, union and relative length, which put unrelated plants together and separated close relatives. Grouping by descent defines phylogenetic systems, overall similarity defines natural systems such as Bentham and Hooker’s, and treating catkin-bearers (Amentiferae) as primitive was Engler and Prantl’s idea, the tempting confusion.

6. In APG IV, which pair of families named in the syllabus belongs to the campanulid clade of the asterids?

   1. Lamiaceae and Verbenaceae
   2. Apiaceae and Asteraceae
   3. Apocynaceae and Solanaceae
   4. Myrtaceae and Apiaceae
   Show answer

   Answer: B — Apiaceae and Asteraceae

   Asterales (Asteraceae) and Apiales (Apiaceae) are campanulids. Lamiaceae and Verbenaceae (Lamiales), Apocynaceae (Gentianales) and Solanaceae (Solanales) are lamiids, the other asterid clade, which makes them the tempting distractors. Myrtaceae (Myrtales) is not an asterid at all but a rosid, so the pair with Apiaceae fails on one member.

7. A tree with quadrangular young twigs, large opposite leaves, small flowers in large panicles and a superior ovary with a terminal style is identified as Tectona grandis (teak). In the APG IV classification it is placed in

   1. Verbenaceae, because a terminal style on an unlobed ovary is still the diagnostic feature of that family
   2. Acanthaceae, together with the mangrove genus Avicennia removed from the old Verbenaceae
   3. Lamiaceae, which on molecular evidence received Tectona, Vitex, Gmelina and other former Verbenaceae
   4. Apocynaceae, because its opposite leaves and superior ovary match the gentianalean flower plan
   Show answer

   Answer: C — Lamiaceae, which on molecular evidence received *Tectona*, *Vitex*, *Gmelina* and other former Verbenaceae

   Molecular phylogenies showed that the woody genera of the old Verbenaceae (Tectona, Vitex, Clerodendrum, Gmelina, Premna) nest within Lamiaceae, so APG places teak there although it keeps the terminal style once used to define Verbenaceae. The Verbenaceae answer is the trap: it is right for the older morphological systems, not for APG IV. Only Avicennia went to Acanthaceae, and Apocynaceae is excluded by the absence of latex, contorted corolla and paired free ovaries.

8. A specimen has milky latex, opposite simple leaves, a corolla with contorted aestivation, five epipetalous stamens, and two carpels whose ovaries are free but whose styles join into a single stigma head; the fruit is a pair of follicles with tufted seeds. The specimen belongs to

   1. Euphorbiaceae, since latex together with a superior ovary identifies the family
   2. Solanaceae, since a bicarpellary superior ovary with epipetalous stamens marks it
   3. Lamiaceae, since opposite leaves and an ovary that separates into parts identify it
   4. Apocynaceae, since paired free ovaries united only at the stigma head mark it
   Show answer

   Answer: D — Apocynaceae, since paired free ovaries united only at the stigma head mark it

   Latex, a contorted corolla and two carpels free at the ovaries but united by the style head, ripening into paired follicles with comose seeds, is the Apocynaceae plan. Euphorbiaceae is the tempting choice because of the latex, but its flowers are unisexual with a tricarpellary syncarpous ovary that forms a capsule. Solanaceae lacks latex and has a syncarpous ovary with an oblique septum, and Lamiaceae lacks latex and has a four-lobed ovary with a gynobasic style forming nutlets.

9. Which of the following family–character pairings are correct for typical members of the family?

   1. Asteraceae: syngenesious anthers and an inferior ovary with a single basal ovule
   2. Myrtaceae: a superior ovary and a fixed number of five stamens
   3. Poaceae: spikelets, a perianth reduced to lodicules, and a caryopsis
   4. Orchidaceae: a superior ovary with axile placentation and free stamens
   Show answer

   Answer: A — Asteraceae: syngenesious anthers and an inferior ovary with a single basal ovule; C — Poaceae: spikelets, a perianth reduced to lodicules, and a caryopsis

   Asteraceae florets have united anthers and an inferior unilocular ovary with one basal ovule, and grasses bear florets in spikelets, with lodicules for a perianth and a caryopsis for a fruit. Myrtaceae has an inferior ovary and numerous stamens, and orchids have an inferior unilocular ovary with parietal placentation and stamens fused with the style into a column.

10. Which of the following features are found in a typical member of Solanaceae such as brinjal?

    1. A deeply four-lobed ovary with a gynobasic style, ripening into four nutlets
    2. A bicarpellary ovary whose septum lies oblique to the median plane of the flower
    3. An inferior ovary that ripens into a cremocarp splitting into two mericarps
    4. Bicollateral vascular bundles, with phloem on both sides of the xylem
    Show answer

    Answer: B — A bicarpellary ovary whose septum lies oblique to the median plane of the flower; D — Bicollateral vascular bundles, with phloem on both sides of the xylem

    Solanaceae has obliquely placed carpels with swollen axile placentae, and its stems have bicollateral bundles (internal phloem). A four-lobed ovary with a gynobasic style and nutlets belongs to Lamiaceae, and an inferior ovary forming a cremocarp belongs to Apiaceae.

11. Add the number of fertile stamens in one flower of Solanum melongena (Solanaceae), the number of stamens in one flower of Ocimum tenuiflorum (Lamiaceae) and the number of fertile anthers in one flower of Vanda (Orchidaceae, a monandrous orchid). The total is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 10

    Solanum has five epipetalous stamens, Ocimum has four didynamous stamens, and Vanda, like most orchids, has a single fertile anther on its column: 5 + 4 + 1 = 10. Counting six orchid stamens to match the six tepals, or two for Lamiaceae (true only of Salvia), are the usual slips.

12. On the rooted tree (O,(A,(B,(C,D)))), where O is the outgroup, four unordered two-state characters are scored as follows (states listed for O, A, B, C and D in that order). Character 1: 0, 0, 1, 1, 1. Character 2: 0, 1, 0, 1, 0. Character 3: 0, 0, 0, 1, 1. Character 4: 0, 1, 1, 0, 1. The minimum total number of state changes (tree length) needed on this tree is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 6

    Character 1: one change on the branch to (B,(C,D)) = 1. Character 2: state 1 arises separately in A and in C = 2. Character 3: one change on the branch to (C,D) = 1. Character 4: the (C,D) node is ambiguous and C differs from D, costing 1; B and D share 1 while the outgroup is 0, so one more change is needed, for example 0 to 1 at the base of the ingroup and 1 to 0 in C, giving 2. Tree length = 1 + 2 + 1 + 2 = 6 (checked by exhaustive reconstruction of all ancestral states). The minimum conceivable is 4, so the ensemble CI is 4/6.

13. The number of distinct unrooted, fully bifurcating trees for n labelled taxa is (2n − 5)!!, the product of all odd numbers from 1 up to 2n − 5. How many unrooted trees are possible for 6 taxa?

    Numerical answer — type the value.

    Show answer

    Answer: 105

    For n = 6, 2n − 5 = 7, so the count is 7!! = 1 × 3 × 5 × 7 = 105. Using the rooted formula (2n − 3)!! = 9!! = 945 is the usual slip; rooting multiplies the count because the root can sit on any of the 2n − 3 branches of an unrooted tree (105 × 9 = 945).

14. Which of the following groups is paraphyletic?

    1. The dicotyledons in the traditional sense, which exclude the monocots
    2. The eudicots, defined by tricolpate pollen
    3. The angiosperms, defined by carpels and double fertilisation
    4. Engler’s Amentiferae, catkin-bearing trees drawn from several unrelated lineages
    Show answer

    Answer: A — The dicotyledons in the traditional sense, which exclude the monocots

    The monocots arose from within the old dicot assemblage, so ‘dicotyledons’ contains a common ancestor but not all of its descendants: paraphyletic. Amentiferae is the tempting choice, but grouping unrelated lineages by a convergent feature (catkins) makes it polyphyletic, not paraphyletic. Eudicots and angiosperms are both clades.

15. In a study of four species P, Q, R and S with a closely related outgroup, P, Q and the outgroup have opposite leaves, whereas R and S have alternate leaves. What do the opposite leaves of P and Q show about their relationship?

    1. They are a synapomorphy, so P and Q should be grouped together as a clade
    2. They are an autapomorphy of P that has been inherited unchanged by Q
    3. They are a homoplasy, showing that P and Q acquired opposite leaves independently
    4. They are a symplesiomorphy, so they give no evidence that P and Q form a clade
    Show answer

    Answer: D — They are a symplesiomorphy, so they give no evidence that P and Q form a clade

    Outgroup comparison makes opposite leaves the ancestral state, so its presence in P and Q is a shared ancestral character (symplesiomorphy) and cannot group them; the alternate leaves of R and S, a shared derived state, are the potential synapomorphy uniting R and S. Calling the opposite leaves a synapomorphy is the classic error of grouping by overall similarity. Nothing indicates independent origin, so homoplasy is unsupported, and a state shared by two taxa cannot be an autapomorphy.

16. Why was the mitochondrial COI gene, the standard animal barcode, not adopted as the core DNA barcode for land plants?

    1. Plant cells lack mitochondria in the green tissues used for routine sampling
    2. Plant mitochondrial genes change so slowly in sequence that closely related species share them
    3. In plants the COI gene has been transferred to the nucleus, where it now occurs in many copies
    4. Plant COI is inherited from both parents and so gives mixed sequences in every hybrid
    Show answer

    Answer: B — Plant mitochondrial genes change so slowly in sequence that closely related species share them

    Land-plant mitochondrial genomes rearrange often but change very slowly in nucleotide sequence, so COI varies too little between congeneric species to identify them; hence the plastid rbcL + matK core barcode. Plant cells do have mitochondria, the plant COI gene (cox1) is still encoded in the mitochondrion, and mitochondria are usually inherited maternally, so the other options are false. The nuclear-transfer option is tempting because some mitochondrial genes have moved to the nucleus during plant evolution, but cox1 is not among them.

17. Which of the following statements about DNA barcoding of plants are correct?

    1. Barcoding discriminates species best when variation within species exceeds that between species
    2. The core plant barcode recommended by the CBOL Plant Working Group combines the plastid regions rbcL and matK
    3. A plastid barcode readily identifies both parents of an interspecific hybrid in most angiosperms
    4. Barcoding can identify a powdered herbal drug whose morphology is no longer visible
    Show answer

    Answer: B — The core plant barcode recommended by the CBOL Plant Working Group combines the plastid regions rbcL and matK; D — Barcoding can identify a powdered herbal drug whose morphology is no longer visible

    rbcL + matK is the two-locus core barcode (2009), and because only DNA is needed, processed material such as powders can be identified. Plastids are inherited maternally in most angiosperms, so a plastid barcode reveals only the mother of a hybrid. Identification needs the reverse of the within-versus-between statement: a barcode gap in which divergence between species exceeds that within them.

18. Which of the following pairs of herbarium acronym and location is correct?

    1. MH: the Madras Herbarium of the Botanical Survey of India, located in Chennai
    2. DD: the herbarium of the Botanical Survey of India, Northern Regional Centre
    3. CAL: the Central National Herbarium, located in Howrah near Kolkata
    4. LWG: the herbarium of the Forest Research Institute in Dehradun
    Show answer

    Answer: C — CAL: the Central National Herbarium, located in Howrah near Kolkata

    CAL is the Central National Herbarium of the BSI at Howrah. MH is the tempting choice because of its name, but the Madras Herbarium is at Coimbatore. DD belongs to the Forest Research Institute, Dehradun (the BSI Northern Regional Centre’s herbarium is BSD), and LWG is the herbarium of the National Botanical Research Institute (NBRI), Lucknow.