Natural Selection: Unit Diagnostic (Biology Unit 7)

What this diagnostic is for

Use this before or just after revising Unit 7: one short question per topic, two for Hardy–Weinberg equilibrium.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your result does not predict an exam score; treat it as a map of what to revisit.

Calculator and data. A calculator is allowed on both exam sections. You need square roots; every other number is given. All organisms and data are fictional.

How to work. Answer everything first, with a short reason for each choice. A right answer for the wrong reason is still a gap.

Question 1 (multiple choice · 7.1)

A mouse population produces far more young than its habitat’s seeds can feed. Tooth shape varies and is inherited. What does Darwin’s theory predict?

(A) Each mouse reshapes its teeth by gnawing, and its young inherit the new shape.
(B) Mice whose inherited tooth shape helps them gather seeds breed more, so that shape spreads over generations.
(C) Seeds are shared equally, so every mouse breeds equally.
(D) The mice produce fewer young because the population needs to fit the food supply.

Answer and explanation

Answer: (B). Limited food means competition; mice with a heritable advantage leave more offspring.

(A) is inheritance of acquired traits: wear does not change genes.
(C) leaves out competition, the start of Darwin’s argument.
(D) gives evolution a goal.

If you missed this: “Darwin’s argument, step by step”, Topic 7.1 study guide.

Question 2 (multiple choice · 7.2)

A vole has two forms of hemoglobin. Form H loads oxygen well in thin air; form L releases oxygen to muscles more easily. H is commoner high on the mountain, L in the valley. Which is best supported?

(A) High altitude causes voles to switch from making L to making H.
(B) H is better overall and will soon replace L everywhere.
(C) Variation in one molecule affects fitness, and which form is favoured depends on the environment.
(D) Both forms carry oxygen, so the difference cannot affect fitness.

Answer and explanation

Answer: (C). Each form helps in one setting, so the same variation gives different fitness in different places.

(A) treats an inherited difference as a response.
(B) forgets the valley, where L is favoured.
(D) ignores how a molecule’s function affects fitness.

If you missed this: “Variation in molecules affects fitness”, Topic 7.2 study guide.

Question 3 (multiple choice · 7.3)

For 20 generations, a breeder uses only the longest-finned fish as parents. Compared with the wild population, the bred line most likely has:

(A) longer fins on average, and less genetic variation at fin-length genes.
(B) longer fins on average, and more variation, because selection creates new alleles.
(C) fish that grow longer fins during life because their parents did.
(D) no change, because only natural environments select.

Answer and explanation

Answer: (A). Choosing parents shifts the mean; alleles for short fins are not passed on, so variation falls.

(B) reverses the effect: selection sorts alleles, it does not make them.
(C) confuses inheritance with growth.
(D) is wrong: humans can be the selecting force.

If you missed this: “What selection does to a population”, Topic 7.3 study guide.

Question 4 (multiple choice · 7.4)

Five birds reach an empty island. Three carry one copy of allele b; none is bb. Mainland frequency of b is 0.02. What is its frequency among the founders?

(A) 0.02
(B) 0.15
(C) 0.30
(D) 0.60

Answer and explanation

Answer: (C). Five diploid birds carry 10 copies; 3 ÷ 10 = 0.30, by chance far above the mainland value: a founder effect.

(A) assumes a small sample matches the mainland.
(B) divides by 20; (D) divides by 5 birds instead of 10 copies.

If you missed this: “The founder effect”, Topic 7.4 study guide.

Question 5 (multiple choice · 7.5)

In a population in Hardy–Weinberg equilibrium, 64% show the dominant phenotype. What is the frequency of the recessive allele?

(A) 0.20
(B) 0.36
(C) 0.40
(D) 0.60

Answer and explanation

Answer: (D). Only the recessive phenotype gives q² directly: q² = 1 − 0.64 = 0.36, so q = √0.36 = 0.60.

(A) treats 0.64 as p², but the dominant phenotype also includes heterozygotes.
(B) is q², not q. (C) is p.

If you missed this: “From alleles to genotypes: p² + 2pq + q² = 1”, Topic 7.5 study guide.

Question 6 (short answer · 7.5)

A sample of 1000 grass plants gives AA 200, Aa 700 and aa 100.

(a) Calculate p and q by counting alleles. (b) Calculate the expected number of each genotype under Hardy–Weinberg equilibrium. (c) Identify the condition most likely broken, and justify your answer. (d) A student finds q as √(100 ÷ 1000). Explain why this is wrong here.

Answer and explanation

(a) 2000 alleles. A: (2 × 200) + 700 = 1100, so p = 0.55; q = 0.45.

(b) AA 0.55² × 1000 = 302.5; Aa 2 × 0.55 × 0.45 × 1000 = 495; aa 0.45² × 1000 = 202.5.

(c) Heterozygotes far exceed expectation (700 against 495). Most likely natural selection favouring heterozygotes (non-random mating of unlike plants would also fit).

(d) q = √(q²) assumes equilibrium, which fails here, so √0.10 = 0.316 is wrong. Counting alleles always works.

If you missed this: “Two routes to allele frequencies”, Topic 7.5 study guide.

Question 7 (multiple choice · 7.6)

Species on a volcanic island chain resemble species on the nearest mainland more than species in similar habitats on distant islands. What does this best support?

(A) Similar habitats always produce identical species.
(B) The island species descend from mainland colonists and changed after arriving.
(C) The island and distant-island species share the most recent common ancestor.
(D) Each island animal changed during its life to match the mainland species.

Answer and explanation

Answer: (B). Biogeography: where species live reflects where their ancestors came from.

(A) is contradicted: similar distant habitats hold different species.
(C) reverses the evidence.
(D) describes individuals, not populations.

If you missed this: “Biogeography: where species live”, Topic 7.6 study guide.

Question 8 (multiple choice · 7.7)

Membrane-bound organelles, linear chromosomes and genes with introns occur in plants, animals and fungi, but not bacteria. Why is this evidence of common ancestry of eukaryotes?

(A) All living things need these features.
(B) It shows animals evolved from plants, and fungi from animals.
(C) Similar environments produced them separately in each group.
(D) Each is complex and unlikely to arise twice in the same form, so one shared ancestor best explains them.

Answer and explanation

Answer: (D). Complex features shared by all eukaryotes, and absent from prokaryotes, point to one ancestor.

(A) is false: bacteria lack them.
(B) treats ancestry as a ladder; living groups share an ancestor.
(C) is convergence, which cannot explain identical detailed structures everywhere.

If you missed this: “Why shared features point to one ancestor”, Topic 7.7 study guide.

Question 9 (multiple choice · 7.8)

Which observation is the strongest evidence that a crop pest is evolving now?

(A) The pest looks like fossils of its relatives.
(B) After a wet summer, adults were larger than usual.
(C) Over 10 years, a resistance allele rose from 0.05 to 0.40 in sprayed fields but stayed near 0.05 in unsprayed fields.
(D) The pest population doubled in one year.

Answer and explanation

Answer: (C). Evolution is a change in allele frequencies; the unsprayed fields are a comparison.

(A) concerns past ancestry. (B) may be a response to food. (D) is a change in numbers, not alleles.

If you missed this: “Evidence 3: resistance to the chemicals we use”, Topic 7.8 study guide.

Question 10 (multiple choice · 7.9)

On a cladogram, legless lizard L sits inside a clade whose other members have legs. The outgroup has legs. What does this best support?

(A) L descends from legged ancestors; leg loss is a derived character.
(B) L is the oldest species in the clade, because it lacks legs.
(C) L must be most closely related to other legless reptiles.
(D) L cannot belong to the clade, because it lacks the shared trait.

Answer and explanation

Answer: (A). Traits can be gained or lost; since legs are ancestral, their loss is derived.

(B) confuses “simpler” with “older”; all living tips are equally recent.
(C) forgets that leglessness can evolve more than once.
(D) ignores trait loss.

If you missed this: “Building a tree from traits”, Topic 7.9 study guide.

Question 11 (short answer · 7.10)

Rising mountains split a salamander population into northern (N) and southern (S) groups two million years ago. In a laboratory, 88% of eggs hatch from N × N or S × S matings and 76% from N × S. The hybrids grow normally, but almost none reproduce.

(a) Name the type of speciation and explain how the split allowed divergence. (b) Identify and classify the barrier shown by the data. (c) Are N and S separate species under the biological species concept?

Answer and explanation

(a) Allopatric: geographic separation. Without gene flow, mutation, drift and different selection changed each group separately.

(b) Mainly hybrid sterility, a post-zygotic barrier. The lower hatch rate (76% against 88%) adds reduced hybrid viability, also post-zygotic.

(c) Yes: they cannot produce fertile offspring, so no genes pass between them.

If you missed this: “Reproductive barriers”, Topic 7.10 study guide.

Question 12 (multiple choice · 7.11)

Reef A was replanted from fragments of a few coral colonies. Reef B grew from larvae of many parents. A heatwave strikes both. Which reef is more likely to keep living coral?

(A) Reef A, because its identical colonies are equally well adapted.
(B) Reef B, because it more likely contains colonies with heat-tolerant alleles.
(C) Neither, because heat is abiotic and affects all coral alike.
(D) Reef B, because its corals will develop heat tolerance when they need it.

Answer and explanation

Answer: (B). More genetic diversity means a greater chance that some individuals withstand a new pressure.

(A) is the risk of low diversity: if one colony is susceptible, so are its copies.
(C) ignores diversity. (D) has the right reef for the wrong reason: tolerance is not a response to need.

If you missed this: “High diversity: a resilient population”, Topic 7.11 study guide.

Question 13 (multiple choice · 7.12)

Which result would most weaken the RNA world hypothesis?

(A) RNA is less chemically stable than DNA.
(B) The ribosome’s site that joins amino acids is made of RNA.
(C) Modern cells store genetic information in DNA.
(D) After a long search, no RNA can copy even a short RNA template unless a protein is added.

Answer and explanation

Answer: (D). The hypothesis assumes genetically encoded proteins were not needed as catalysts; this result challenges that.

(A) fits the hypothesis: it explains why DNA later took over.
(B) supports it. (C) is expected; the hypothesis concerns the earliest genetic material.

If you missed this: “The RNA world hypothesis”, Topic 7.12 study guide.

Your next step

A short answer missed in any part counts as missed.

Topic	Question(s)	If you missed it, read
7.1 Introduction to natural selection	1	Study guide 7.1
7.2 Natural selection	2	Study guide 7.2
7.3 Artificial selection	3	Study guide 7.3
7.4 Population genetics	4	Study guide 7.4
7.5 Hardy–Weinberg equilibrium	5, 6	Study guide 7.5
7.6 Evidence of evolution	7	Study guide 7.6
7.7 Common ancestry	8	Study guide 7.7
7.8 Continuing evolution	9	Study guide 7.8
7.9 Phylogeny	10	Study guide 7.9
7.10 Speciation	11	Study guide 7.10
7.11 Variations in populations	12	Study guide 7.11
7.12 Origins of life on Earth	13	Study guide 7.12

How to use your result

Wrong answers first: read the named section, then that topic’s practice set.
Lucky guesses count as misses if your reason did not match the explanation.
Missed 1, 3, 9 or 12? Check you never let an individual “adapt” or “need” a trait. Start with Topic 7.1.
Missed 4, 5 or 6? Practise counting alleles, two per individual, in Topic 7.4.
No misses? Go to the mixed unit review.