Natural Selection: Mixed Unit Review (Biology Unit 7)

Before you start

This review joins the twelve topics of Unit 7. Do the unit diagnostic first and close its gaps.

Big ideas of the unit

Selection needs three things: heritable variation, competition for limited resources and unequal reproductive success. Fitness is offspring left, not survival alone (Topic 7.1).
A trait is only “good” in an environment. Selection acts on phenotypes, down to single molecules, and changes direction when the environment does (Topic 7.2).
Humans can be the selecting force. Artificial selection shifts traits quickly but uses up variation (Topic 7.3).
Chance also changes populations. Mutation adds alleles; drift (bottlenecks, founders) acts most strongly in small populations; gene flow makes populations more alike (Topic 7.4).
Hardy–Weinberg is the “nothing is happening” model. Count alleles, predict p², 2pq and q², and read any clear mismatch as a broken condition (Topic 7.5).
Independent evidence agrees. Dated rocks, fossils, homologous and vestigial structures, biogeography and sequences point the same way (Topic 7.6), including to one ancestor of all eukaryotes (Topic 7.7).
Evolution is still happening. Resistance to drugs and pesticides and new pathogens are selection we can watch (Topic 7.8), and genetic diversity decides who withstands it (Topic 7.11).
Trees and species are the long-term result. Trees are testable hypotheses (Topic 7.9); without gene flow, populations diverge into species (Topic 7.10); life arose between about 3.9 and 3.5 billion years ago, perhaps in an RNA world (Topic 7.12).

Key relationships and methods

Idea	Rule or method	Where it appears
Allele frequency	copies of the allele ÷ (2 × number of individuals)	any genotype count
Hardy–Weinberg	p + q = 1; genotypes p², 2pq, q²; q = √(q²) only if in equilibrium	null hypothesis, carriers
Relative fitness	success of a phenotype ÷ success of the best phenotype	survival or offspring data
Selection or drift?	consistent direction, repeated in replicates, large population → selection; random, small population → drift	allele-frequency data
Molecular clock	rate = differences ÷ time of a dated split; time = differences ÷ rate	phylogenetic trees
Dating	lower layers older; ash beds bracket fossils; carbon-14 (half-life 5,700 years) only for recent remains	fossils
Species	biological species concept: interbreed and produce viable, fertile offspring	speciation, barriers
Diversity	more alleles → more likely some individuals withstand a new pressure	resilience, resistance

These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All organisms and data are fictional.

Question 1 (multiple choice · mixed)

Two desert plants on different continents both have thick water-storing stems and spines. DNA shows each is more closely related to leafy, spineless plants on its own continent than to the other. Which conclusion is best supported?

(A) The stems evolved separately under similar selection in dry habitats: convergent evolution, so the stems are not evidence of close relationship.
(B) The stems are homologous, so the two plants share a recent common ancestor.
(C) The DNA must be wrong, because the plants look so alike.
(D) Each plant grew a thick stem during its life because it lacked water.

Answer and explanation

Answer: (A). Similar pressures produced similar phenotypes in separate lineages. Shared DNA reflects ancestry better than one shared adaptation.

(B) assumes similarity always means homology.
(C) reverses the evidence: many sequence positions outweigh one body feature.
(D) is a response during life, not evolution.

Question 2 (multiple choice · mixed)

A flood leaves 25 plants of a once-large population: 14 AA, 10 Aa, 1 aa. The survivors mate at random and the population grows large again. What frequency of aa is expected in the next generation?

(A) 0.040
(B) 0.058
(C) 0.24
(D) 0.36

Answer and explanation

Answer: (B). Count alleles among survivors: a = 10 + 2 = 12 of 50, so q = 0.24. Random mating gives q² = 0.24² = 0.0576 ≈ 0.058.

(A) is aa among the 25 survivors (1 ÷ 25), not the prediction after random mating.
(C) is q, not q².
(D) is 2pq ≈ 0.36, the heterozygotes.

Question 3 (multiple choice · mixed)

A grain crop was bred for 50 years for high yield in irrigated fields. Breeders now need drought tolerance and find the allele they want only in a wild relative. Which explanation is best supported?

(A) Selection under irrigation narrowed variation; drought-tolerance alleles gave no advantage there and were lost, but the wild population kept them.
(B) Drought caused the wild plants to make the tolerance allele.
(C) Artificial selection creates new alleles, but only for the selected trait.
(D) The crop and its wild relative are different species, so they never shared alleles.

Answer and explanation

Answer: (A). Artificial selection and small breeding groups reduce diversity; alleles that were not useful under the breeders’ conditions can be lost.

(B) treats mutation as a response to need.
(C) is wrong: selection sorts alleles.
(D) ignores that the crop came from wild ancestors.

Question 4 (constructed response · mixed)

Volcanic ash turns a pale beach dark. A lizard there has one colour gene with incomplete dominance: LL pale, Ll medium, ll dark. Hawks hunt the lizards.

	LL	Ll	ll
Adults before the breeding season	144	192	64
Adults that survived to breed	36	96	48

On two other ash-covered beaches, the frequency of L fell by similar amounts. On a nearby pale beach, it stayed at about 0.60.

(a) Show that the starting population matches Hardy–Weinberg proportions. (b) Calculate the survival fraction and relative fitness of each phenotype. (c) Calculate the frequency of L among the breeders. (d) Explain the change using natural selection. (e) A student says the change is genetic drift. Evaluate the claim. (f) Predict the genotype frequencies of the offspring if the breeders mate at random.

Worked solution and suggested Marlbridge rubric

(a) 400 lizards, 800 alleles. L = (2 × 144) + 192 = 480, so p = 0.60, q = 0.40. Expected: 0.36 × 400 = 144, 0.48 × 400 = 192, 0.16 × 400 = 64. Observed equals expected.

(b) Survival: LL 36 ÷ 144 = 0.25; Ll 96 ÷ 192 = 0.50; ll 48 ÷ 64 = 0.75. Relative fitness: LL 0.33, Ll 0.67, ll 1.00.

(c) 180 breeders, 360 alleles. L = (2 × 36) + 96 = 168. p = 168 ÷ 360 = 0.467 (down from 0.60).

(d) Colour varies and is heritable. On dark ash, paler lizards stand out, so hawks catch more of them; darker lizards breed more and pass on l. The ash changed the direction of selection.

(e) Not supported. 400 is not tiny, and the same large change happened on two other ash beaches but not on the pale beach. Drift is random in direction, so it would not repeat like this.

(f) p = 0.467, q = 0.533: LL 0.467² = 0.218; Ll 2 × 0.467 × 0.533 = 0.498; ll 0.533² = 0.284.

Point	What earns it
1	p = 0.60 and expected numbers equal to observed
1	Survival fractions and relative fitness 0.33, 0.67, 1.00
1	p = 0.467 among breeders
1	Selection explained: heritable colour, contrast with background, differential survival and reproduction
1	Drift rejected using population size, replicate beaches and the pale-beach comparison
1	Offspring frequencies 0.218, 0.498, 0.284

Total: 6 points. Accept answers carried forward from (c).

Question 5 (constructed response · mixed)

A lava field split the range of a ground squirrel into north (N) and south (S) populations. Radiometric dating gives the lava an age of 3.0 million years. A 1,500-base gene differs by 24 bases between N and S, and by 40 bases between each of them and a related species, E. Today a narrow strip of forest has grown across the lava, and N and S meet there. Hybrids are healthy, but hybrid females raise on average 30% as many young as pure females.

(a) Explain why carbon-14 cannot date the lava. (b) Calculate the clock rate and estimate when E split from the N–S lineage. (c) Describe the tree, with dates at the nodes, and state whether it is a cladogram or a phylogenetic tree. (d) Name the type of speciation that began, and evaluate whether N and S are now separate species. (e) Explain why this tree is a hypothesis, and give one result that would revise it.

Worked solution and suggested Marlbridge rubric

(a) Lava contains no carbon taken in by a living thing, and 3.0 million years is about 526 half-lives of 5,700 years, so no measurable carbon-14 would remain.

(b) Rate = 24 ÷ 3.0 = 8 differences per million years. E: 40 ÷ 8 = 5.0 million years ago.

(c) N and S are sister groups, joined at a node dated 3.0 million years ago. E branches off at an earlier node, 5.0 million years ago. It shows time calibrated by a dated event and a clock, so it is a phylogenetic tree.

(d) Allopatric: the lava cut gene flow. Under the biological species concept, N and S are not yet fully separate: hybrids are viable and partly fertile, so some genes still pass between them. Reduced hybrid fertility is a post-zygotic barrier; speciation is under way but incomplete.

(e) It rests on one gene, one calibration and a steady rate. Many more genes showing S closer to E than to N would revise it.

Point	What earns it
1	Carbon-14 rejected: no biological carbon and far too many half-lives
1	Rate of 8 per million years
1	E split 5.0 million years ago
1	Tree described with both node dates and called a phylogenetic tree
1	Allopatric, with reduced hybrid fertility as an incomplete post-zygotic barrier and a conclusion that uses the species concept
1	Tree as a hypothesis, with a result that would change it

Total: 6 points.

Question 6 (constructed response · mixed)

Researchers compare a 1,500-base segment of a ribosomal RNA (rRNA) gene, found in every living cell, between yeast Y and other organisms. The values are illustrative, not measured.

Compared with yeast Y	Second yeast	Plant	Animal	Archaeon	Bacterium
Bases identical / %	97	85	86	66	62

(a) Explain why a gene found in every cell is useful for comparing very distantly related organisms. (b) Describe the pattern, and explain how it supports the common ancestry of eukaryotes. (c) A student concludes: “The archaeon result proves that eukaryotes evolved from archaea.” Evaluate this conclusion. (d) In the ribosome, the site that joins amino acids is made of rRNA. Explain how this supports the RNA world hypothesis, naming one of its assumptions.

Worked solution and suggested Marlbridge rubric

(a) Every organism has it, so any two can be compared. Its job is essential, so most changes are harmful and it changes slowly; differences stay countable over billions of years.

(b) The eukaryotes are most similar to yeast Y: the second yeast 97%, then the animal 86% and plant 85%. The prokaryotes are far less similar (66%, 62%). This fits one eukaryotic ancestor, matching shared features such as membrane-bound organelles, linear chromosomes and introns.

(c) Not proved. 66% against 62% is a small gap, from one gene and one organism each. The data fit archaea being closer relatives, but a tree is a hypothesis; many genes and organisms are needed.

(d) RNA, not protein, catalyses a reaction central to all life. That fits the assumption that genetically encoded proteins were not needed as catalysts: RNA could both store information and catalyse reactions.

Point	What earns it
1	Universal gene that changes slowly because its function is essential
1	Pattern described with values and linked to a shared eukaryotic ancestor
1	Claim evaluated: small difference, one gene, tree is a hypothesis
1	RNA catalysis linked to the assumption that encoded proteins were not needed as catalysts

Total: 4 points.

Question 7 (constructed response · mixed)

Two farms grow the same crop and spray the same insecticide. Resistance comes from a recessive allele r: at the dose used, most rr insects survive spraying, but most RR and Rr insects die. Insects also fly in from the land around both farms. Farm B leaves 20% of its field unsprayed as a refuge. The table shows the frequency of r.

Year	0	2	4	6	8
Farm A (sprayed everywhere)	0.01	0.04	0.15	0.42	0.78
Farm B (with refuge)	0.01	0.02	0.04	0.07	0.11

(a) Describe and compare the two trends, with values. (b) Explain why r was present before any spraying, and why it rises on farm A. (c) Assuming Hardy–Weinberg proportions, calculate the percentage of resistant (rr) insects on each farm in year 8, and the percentage of r alleles carried by heterozygotes on farm B. (d) Use your answers to explain how the refuge slows the evolution of resistance. (e) Propose an investigation to test whether refuge size affects the rate of resistance evolution.

Worked solution and suggested Marlbridge rubric

(a) Both start at 0.01. Farm A rises slowly, then fast: +0.03, +0.11, +0.27, +0.36 per two years, reaching 0.78. Farm B rises slowly and steadily to 0.11, about one-seventh of farm A.

(b) Mutation is random and occurs without the insecticide, so a few r alleles were already present. Spraying kills most susceptible insects; rr survivors breed and pass on r, so its frequency rises. The rise speeds up as more r alleles meet in rr insects.

(c) Farm A: 0.78² = 60.8% rr. Farm B: 0.11² = 1.2% rr. On farm B, heterozygotes (2 × 0.11 × 0.89 = 19.6% of insects) carry 0.89, about 89%, of all r alleles.

(d) The refuge keeps many susceptible insects carrying R. They mate with survivors from the sprayed area (gene flow), so most r alleles sit in heterozygotes, which mostly die when sprayed. Few rr insects form, so r rises slowly.

(e) Independent variable: refuge size, for example 0%, 10%, 20% and 40% of a plot. Dependent variable: r frequency, or survival of a test dose, each generation. Control the starting r frequency, insect number, insecticide dose, crop and number of generations. Use at least three replicate plots per refuge size.

Point	What earns it
1	Both trends described with values and compared
1	r present by random mutation before spraying, then selected
1	60.8% and 1.2% rr
1	About 89% of r alleles in heterozygotes on farm B
1	Refuge explained: gene flow of susceptible alleles, heterozygotes killed
1	Investigation with independent and dependent variables, controls and replication

Total: 6 points.

How did you do?

Add up your points: 3 for the multiple-choice questions and 22 for the constructed responses. Use the total only to compare with your own next attempt; it does not predict an exam score.

Lost points on calculations (Questions 2, 4, 7(c)): count alleles first, then use p², 2pq and q², in Topic 7.5.
Mixed up selection and drift (Question 4(e)): reread Topic 7.4.
Lost points on trees and dating (Questions 1, 5, 6): rework Topic 7.9 and Topic 7.6.
Want a quick recheck? Retake the unit diagnostic.

Then tick off each topic checklist: