These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Useful relationships: clock rate = differences ÷ time since a dated split; time = differences ÷ rate. A calculator is assumed. All species, sequences and data sets are fictional.
Questions 1 and 2 use the tree below.
Question 1 (multiple choice · foundation)
Which node represents the most recent common ancestor of species B and species C?
- (A) Node 1
- (B) Node 2
- (C) Node 3
- (D) Node 4
Answer and explanation
Answer: (B). Trace back from B and from C. The two paths first meet at node 2. That node is their most recent common ancestor.
- (A) Node 1 is a common ancestor of B and C, but not the most recent one. It is the ancestor of all five species.
- (C) Node 3 is the most recent common ancestor of A and B only. C does not branch from it.
- (D) Node 4 is the ancestor of D and E. Neither B nor C descends from it.
Question 2 (multiple choice · core)
Which statement is best supported by the tree?
- (A) C is more closely related to D than to A, because C and D are drawn next to each other.
- (B) C shares a more recent common ancestor with B than with D.
- (C) A has changed least since the root, because it is at the top of the tree.
- (D) E is more distantly related to C than D is, because E is farther from C on the page.
Answer and explanation
Answer: (B). C and B meet at node 2. C and D meet only at node 1, the root, which is older. So C is more closely related to B (and equally to A) than to D.
- (A) judges by tip position. Rotate the branches at node 1 and C would be far from D. Relatedness depends on nodes, not on printing order.
- (C) The order of the tips says nothing about how much each species has changed. All five are modern species.
- (D) D and E both meet C at node 1, so they are equally related to C. Distance on the page is irrelevant.
Question 3 (multiple choice · core)
Which statement about cladograms and phylogenetic trees is correct?
- (A) Only a phylogenetic tree can show how long ago two lineages split, using dates from fossils or a molecular clock.
- (B) The branch lengths of a cladogram show how much genetic change happened in each lineage.
- (C) Phylogenetic trees are built only from fossils, and cladograms only from DNA sequences.
- (D) A phylogenetic tree is a proven fact, whereas a cladogram is only a hypothesis.
Answer and explanation
Answer: (A). A phylogenetic tree places nodes against a scale of time or change. That scale must be calibrated, either with dated fossils or with a molecular clock.
- (B) A cladogram shows branching order only. Its branch lengths are drawn for neatness and carry no information.
- (C) Either kind of diagram can be built from body features of living or fossil species, or from DNA or protein sequences.
- (D) Both are hypotheses. Both can be tested and revised when new evidence appears.
Question 4 (constructed response · core)
Four fictional species of desert beetle (F, G, H and I) are compared with outgroup V, a related insect that is not a beetle.
| Character | V | F | G | H | I |
|---|---|---|---|---|---|
| 1. Six legs | 1 | 1 | 1 | 1 | 1 |
| 2. Hardened forewings | 0 | 1 | 1 | 1 | 1 |
| 3. Hind wings lost (cannot fly) | 0 | 0 | 1 | 1 | 1 |
| 4. Waxy white coating | 0 | 0 | 0 | 1 | 1 |
| 5. Forewings fused together | 0 | 0 | 0 | 0 | 1 |
(a) Explain why V is used as the outgroup and what its character states tell you. (b) Identify the two characters that cannot be used to define any clade on this cladogram, and explain why each does not help. (c) Describe or draw the cladogram, marking where characters 2, 3 and 4 appear. (d) A student says “I is the most evolved species because it has the most derived characters.” Evaluate this claim.
Model answer and suggested Marlbridge rubric
(a) V is less closely related to the beetles than they are to each other. Character states found in V are taken to be ancestral; states found only in some beetles are derived.
(b) Character 1 (six legs) is present in the outgroup and in every beetle, so it is a shared ancestral character: it cannot separate any group. Character 5 (fused forewings) is found only in I, so it tells you about I’s own branch but does not link I with any other species.
(c) The tree is (F, (G, (H, I))), with V branching off first.
- Character 2 is marked on the branch leading to all four beetles.
- Character 3 is marked on the branch leading to G, H and I (after F has branched off).
- Character 4 is marked on the branch leading to H and I, which are sister species.
- (Character 5 may be shown on I’s own branch.)
(d) The claim is wrong. All four beetle lineages have been evolving for the same length of time since their common ancestor. F may have changed in features not scored here. I simply has more of the particular changes chosen for this table.
| Point | What earns it |
|---|---|
| 1 | Outgroup is less closely related and its states are taken as ancestral |
| 1 | Character 1 rejected as shared ancestral and character 5 rejected as unique to one species |
| 1 | Correct branching order (F, (G, (H, I))) with V outside |
| 1 | Characters 2, 3 and 4 placed on the correct branches |
| 1 | Claim rejected with a valid reason (equal time evolving, or table covers only some traits) |
Question 5 (calculation · core)
The amino acid sequence of the same 120-residue protein was compared in three fictional rodents. Fossils show that the lineages of species F and species G split 20 million years ago. F and G differ at 10 positions. F and a third species, H, differ at 6 positions.
(a) Calculate the clock rate, in amino acid differences per million years. (b) Estimate when the F and H lineages split. (c) A fourth species, N, differs from F at 25 positions. Estimate when F and N split, and give the percentage of positions that differ. (d) State two reasons why the estimate in (c) is less certain than the estimate in (b).
Worked solution and suggested Marlbridge rubric
(a) Rate = 10 differences ÷ 20 million years = 0.50 differences per million years (of separation, counting both branches).
(b) Time = 6 ÷ 0.50 = 12 million years ago. H is more closely related to F than G is.
(c) Time = 25 ÷ 0.50 = 50 million years ago. Percentage = 25 ÷ 120 × 100 = 20.8%.
(d) Any two:
- It lies far beyond the calibrated split (50 against 20 million years), so any error in the rate is magnified.
- Over long times a position can change twice, or change back, so the count underestimates the changes and the date may be too young.
- The rate may differ between lineages or eras, and the older the split, the more time there is for such differences to add up. | Point | What earns it | |—|—| | 1 | Rate 0.50 per million years with units | | 1 | 12 million years ago | | 1 | 50 million years ago and 20.8% | | 2 | Two distinct, valid limitations (1 each) |
Common error: dividing time by differences gives 2 million years per difference, the reciprocal of the rate. It still works (6 × 2 = 12) if the units are kept.
Question 6 (data analysis · core)
Four fictional songbird species were compared. Species W and Z both have a red throat patch; X and Y do not. A 24-base section of the same gene was sequenced:
| Species | Sequence (positions 1–24) |
|---|---|
| W | ATGGCTAAGTTCCGATACGGTCTA |
| X | ATGGCTAAGTTACGATACGGTCTG |
| Y | ATGCCTGAGTTCCGTTACAGTCTA |
| Z | ATGCCTGAGATCCGTTTCAGTCAA |
(a) Count the differences between W and each of the other species. (b) The other counts are X–Y 6, X–Z 9 and Y–Z 3. Use all the counts to state which species are sister pairs. (c) Explain why the throat patch and the sequence data suggest different trees, and which evidence is more reliable. (d) State one limitation of this sequence evidence.
Model answer and suggested Marlbridge rubric
(a) W–X: 2 differences (positions 12 and 24). W–Y: 4 (positions 4, 7, 15, 19). W–Z: 7 (positions 4, 7, 10, 15, 17, 19, 23).
(b) The smallest counts are W–X (2) and Y–Z (3). So W and X are sister species, and Y and Z are sister species: the tree is ((W, X), (Y, Z)).
(c) The shared throat patch would group W with Z. But W and Z have 7 differences, more than W has with X or Y. The likeliest explanation is that the red patch evolved twice, for example because females in both lineages preferred bright throats: this is convergent evolution, and the trait is analogous. Sequence data are more reliable because they provide many independent characters (here 24 positions, in practice thousands) and are less likely to become similar by chance than a single visible trait.
(d) Any one: only 24 bases from one gene were compared; one gene can give a different tree from the rest of the genome; a few differences could be sequencing errors; no outgroup was included, so the root (and the ancestral state of the patch) is not known.
| Point | What earns it |
|---|---|
| 1 | All three W counts correct |
| 1 | Sister pairs (W, X) and (Y, Z) justified by the smallest counts |
| 1 | Conflict explained as convergent evolution of the throat patch |
| 1 | Molecular evidence preferred, with a reason (many characters or less affected by convergence) |
| 1 | A valid limitation |
Question 7 (argumentation · stretch)
Four fictional snail species live in four lakes: Amber, Birch, Cedar and Delta. Amber and Cedar contain shell-crushing crabs; Birch and Delta do not. Snails in Amber and Cedar have thick shells, and those in Birch and Delta have thin shells. An older cladogram, based on shell features, grouped (Amber, Cedar) and (Birch, Delta). A new study compared 4,000 genes and grouped (Amber, Birch) and (Cedar, Delta).
(a) Explain how the new tree can be correct even though the thick-shelled snails look alike. (b) Explain why the older cladogram was a reasonable hypothesis at the time, and why it has now been revised. (c) Propose one further investigation that could test the new tree, and state the result that would support it.
Model answer and suggested Marlbridge rubric
(a) Crabs select for thick shells, because thin-shelled snails are crushed and leave fewer offspring. If the same selection pressure acted separately in Amber and Cedar, thick shells could have evolved twice. The resemblance would then be convergent (analogous), not inherited from a shared thick-shelled ancestor. The genome data show the true pattern of shared ancestry.
(b) Shell features were the evidence available, and a shared feature is often a sign of common ancestry, so the grouping was reasonable. A tree is a hypothesis that predicts what new evidence should show. The genome data, with far more characters, did not match the prediction, so the hypothesis was revised.
(c) Any one, with the supporting result:
- Identify the genetic changes that cause thick shells in each lake. If the thick shells are caused by different mutations in Amber and Cedar snails, that supports two separate origins and the new tree.
- Sequence further, unlinked genes and include an outgroup. The new tree is supported if they also group Amber with Birch and Cedar with Delta.
| Point | What earns it |
|---|---|
| 1 | Crab predation identified as the selection pressure for thick shells |
| 1 | Thick shells evolved independently (convergent / analogous) |
| 1 | Older tree reasonable because based on available shared traits and revised because new evidence contradicted its prediction |
| 1 | A workable investigation |
| 1 | The result that would support the new tree |
How did you do?
- Q1 or Q2 wrong: re-read “Reading a tree” in the study guide, especially the parts on nodes and rotation.
- Q3 wrong: compare the two panels of Figure 1 in the study guide.
- Q4 incomplete: rework Worked example 1 and Figure 2; check you used the outgroup to decide which states are derived.
- Q5 wrong: rework Worked example 2 (b) and (c); keep the units on the rate.
- Q6 or Q7 incomplete: re-read the “Evidence” section and the whale example.
Then tick off the topic checklist.