Taxonomy & Classification

Question 1

(a) Proportional distances $p = D/L$, $L=200$: (4 — half mark each, rounded up)

Award 4 marks for all six correct; deduct proportionally.

(b) Jukes–Cantor (6)

• Why correction needed (2): Observed differences saturate — the same site can mutate more than once (back-mutations, multiple hits). Raw $p$ therefore underestimates the true number of substitutions; the JC model estimates the true count $d$ that $p$ hides. (1 for saturation/multiple hits, 1 for underestimation.)

• W–X: $p=0.1$. $d = -\tfrac34\ln(1-\tfrac43(0.1)) = -0.75\ln(0.86667) = -0.75(-0.143101) = 0.10733 \approx \mathbf{0.107}$ (2)

• W–Z: $p=0.33$. $d = -0.75\ln(1-\tfrac43(0.33)) = -0.75\ln(0.56) = -0.75(-0.579818) = 0.43486 \approx \mathbf{0.435}$ (2)

Note $d>p$ in both cases, confirming the correction inflates distance (larger effect for larger $p$).

(c) Topology from p-distances (6)

• Smallest distance = W–X ($p=0.10$) → W and X pair first. (2)
• Next smallest = Y–Z ($p=0.12$) → Y and Z pair. (2)
• Cross distances (W/X to Y/Z ≈ 0.29–0.33) are all much larger, so {W,X} and {Y,Z} are separate clades. Newick: ((W,X),(Y,Z)). (2) Accept an explicit outgroup rooting if justified (e.g. Z is most divergent from W, so may sit basal), but the two-clade split is the required answer.

(d) Overturning morphology / Woese (6)

• Morphology can mislead: convergent evolution produces similar forms in unrelated lineages, and simple organisms (prokaryotes) offer few morphological characters. Molecular sequences give many independent, quantifiable characters directly reflecting genetic ancestry. (3)
• Woese (2): Comparing 16S/18S rRNA sequences, Woese found that "bacteria" were actually two deeply divergent groups — Bacteria and Archaea — as different from each other as either is from eukaryotes. (2 for rRNA + Archaea split.)
• This forced the three-domain system (Bacteria, Archaea, Eukarya), replacing the old prokaryote/eukaryote or five-kingdom view. (1)

Question 2

(a) Domain/kingdom assignment (6 — 0.5 each cell, 6 total incl. distinguishing feature)

Decisive feature (2): E. coli has peptidoglycan (murein) in its cell wall and ester-linked membrane lipids; Methanococcus (Archaea) lacks peptidoglycan and has ether-linked lipids. (Either character accepted; peptidoglycan is the classic marker.)

(b) Dichotomous key (8 — 1 mark per correct, single-character couplet; full 8 for a valid complete key)

1a  Cell has a true membrane-bound nucleus ...................... go to 3
1b  No membrane-bound nucleus (prokaryote) ..................... go to 2

2a  Cell wall contains peptidoglycan ................. Escherichia coli
2b  Cell wall lacks peptidoglycan; methanogen ..... Methanococcus jannaschii

3a  Single-celled, no cell wall, moves by pseudopodia ... Amoeba proteus
3b  Multicellular or with a cell wall .......................... go to 4

4a  Cell wall of chitin; heterotroph (absorptive) ........... Rhizopus
4b  Cell wall of cellulose OR no wall .......................... go to 5

5a  Autotroph, cellulose wall, produces cones/seeds ........... Pinus
5b  Heterotroph, no cell wall, nervous system .......... Panthera leo

Marks: each couplet uses one verifiable character and correctly partitions the set.

(c) Decision-tree depth & lower bound (6)

• Each yes/no test gives 1 bit of information, splitting the candidate set into (at most) two parts. After $k$ tests, at most $2^k$ distinct outcomes (leaves) are reachable. (2)
• To distinguish $n$ organisms we need at least $n$ distinct leaves, so $2^k \ge n \Rightarrow k \ge \log_2 n \Rightarrow k \ge \lceil\log_2 n\rceil$ (integer tests). This is the information-theoretic lower bound. (2)
• For $n=6$: $\log_2 6 = 2.585$, so $\lceil 2.585\rceil = \mathbf{3}$ tests minimum (best case, balanced tree). (2)

Question 3

(a) Binomial nomenclature rules (any 4, ×1) (4) + hierarchy (2)

1. Two-part Latin(ised) name: Genus + species.
2. Genus capitalised, species epithet lowercase: Panthera leo.
3. Whole name italicised (or underlined when handwritten).
4. Genus may be abbreviated after first use: P. leo.
5. Author/date may follow (e.g. Panthera leo Linnaeus, 1758).

Hierarchy for the lion (2): Domain Eukarya → Kingdom Animalia → Phylum Chordata → Class Mammalia → Order Carnivora → Family Felidae → Genus Panthera → Species leo.

(b) Domain determination (6)

• Domain = Archaea (2). Three justifying characters (1 each ×3 = but weighted 4):

  1. No nucleus → prokaryote (rules out Eukarya).
  2. No peptidoglycan in wall → rules out Bacteria (which have it).
  3. Ether-linked membrane lipids + extremophile (thermoacidophile) → hallmark of Archaea.

• Why old systems failed (2): Two-kingdom (Plant/Animal) had no place for microbes at all; the five-kingdom (Monera) lumped all prokaryotes together, hiding the deep Bacteria–Archaea split revealed only by molecular (rRNA) data.

(c) Combinatorial proof + max genera (6)

• Tree proof (3): Each species has a unique ordered 8-rank path (domain,…,species). Ranks nest strictly: a genus belongs to exactly one family, a family to one order, etc. Thus every node except the root (domain-level) has exactly one parent, and no cycles exist → the structure satisfies the definition of a rooted tree (single root = top rank, species = leaves). (3)

• Max distinct genera (3): 12 species, each genus holds ≤ 3 species. To maximise the number of genera, minimise species per genus. But genera must partition the 12 species, and the count is maximised by putting as few as possible per genus — the minimum is 1 species per genus, giving 12 genera. If the intended reading requires "as full as allowed" is not imposed, max = 12. If the constraint is read as maximising while each must be ≤3 (no minimum), the maximum number of genera = 12 (each monotypic). (Award full marks for 12 with the partition argument; if a student instead computes the minimum number of genera = $\lceil 12/3\rceil = 4$, award 2 for correct interpretation of the packing bound.)

[
  {"claim":"JC distance W-X (p=0.1) ~ 0.107",
   "code":"p=Rational(1,10); d=-Rational(3,4)*ln(1-Rational(4,3)*p); result=abs(float(d)-0.107)<0.001"},
  {"claim":"JC distance W-Z (p=0.33) ~ 0.435",
   "code":"p=Rational(33,100); d=-Rational(3,4)*ln(1-Rational(4,3)*p); result=abs(float(d)-0.435)<0.001"},
  {"claim":"Min yes/no tests to distinguish 6 items = 3",
   "code":"import math; result=math.ceil(math.log(6,2))==3"},
  {"claim":"Minimum genera packing 12 species at <=3 each = 4",
   "code":"import math; result=math.ceil(12/3)==4"},
  {"claim":"Smallest p-distances are W-X(0.10) and Y-Z(0.12)",
   "code":"pairs={'WX':20,'WY':60,'WZ':66,'XY':58,'XZ':64,'YZ':24}; ps={k:v/200 for k,v in pairs.items()}; s=sorted(ps,key=ps.get); result=(s[0]=='WX' and s[1]=='YZ')"}
]