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IB Biology A4.1: Evolution and speciation worked solutions

Worked solutions to every practice question in A4.1: Evolution and speciation of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which observation is an example of evolution in a population? (A) A runner's leg muscles enlarge with training. (B) The proportion of bacteria carrying an antibiotic resistance allele rises over 40 generations. (C) A plant grows taller in bright light. (D) A tanned person has darker skin after summer.
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B. A rise in the proportion of bacteria carrying a resistance allele is a change in heritable characteristics of a population over generations. A, C and D are changes in single organisms during a lifetime and are not written into the gametes.

Final answer

B

Question A2

[1 mark]
Which are evidence of common ancestry? I. Similar corresponding gene sequences. II. A shared limb bone plan. III. Features acquired by use are passed on. (A) Only I (B) Only II (C) I together with II (D) All three
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C. I together with II: sequence similarity and a shared structural plan are both evidence of common ancestry. III is the Lamarckian idea, which is not supported, because acquired features are not written into the gametes.

Final answer

C

Question A3

[1 mark]
Which pair of structures is homologous? (A) Bird and insect wings (B) Human arm, flipper (C) Octopus and human eyes (D) Moth and bat wings
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B. The human arm and the whale flipper share the humerus, radius, ulna and digits inherited from a common ancestor. A, C and D pair structures with similar functions or appearance but separate origins.

Final answer

B

Question A4

[1 mark]
The eye of an octopus and the eye of a human both form an image, but the common ancestor of the two animals had no such eye. What are these eyes? (A) Homologous (B) Analogous (C) Vestigial (D) Divergent evolution
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B. A similar function that arose independently makes the eyes analogous structures. Homologous structures are inherited from a common ancestor, vestigial structures are reduced and functionless, and divergent evolution describes homologous structures adapting differently.

Final answer

B

Question A5

[1 mark]
A protein is compared with the human version. Species P differs at 3 of 60 positions, Q at 4 of 100 and R at 6 of 200. Which is most similar to the human protein? (A) P (B) Q (C) Cannot be decided (D) R
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D. The proportions of positions that differ are 5%, 4% and 3%, so R is most similar even though it has the most differences. Always normalise for the number of positions compared.

Final answer

D

Question A6

[1 mark]
Which sequence correctly describes speciation by geographical separation? (A) Divergence, barrier, isolation, selection (B) Barrier, different selection, divergence, reproductive isolation (C) Isolation, barrier, divergence, selection (D) Selection, isolation, barrier, divergence
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B. A barrier first stops gene flow, different pressures then act on each side, allele frequencies diverge, and the populations end up unable to interbreed. A starts with divergence before there is a barrier, C and D reverse the order.

Final answer

B

Question A7

HL[1 mark]
Which statement describes sympatric speciation? (A) A new species forms behind a physical barrier. (B) A new species forms in one area without a physical barrier. (C) One species gradually changes without splitting. (D) A species is replaced after extinction.
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B. Sympatric speciation occurs in the same area, with isolation arising from, for example, breeding time or behaviour. A describes allopatric speciation, C is gradual change within a species, and D describes extinction.

Final answer

B

Question A8

HL[1 mark]
Females of one frog species approach only the calls of their own males. Which barrier is this? (A) Behavioural (B) Temporal (C) Hybrid sterility (D) Geographical
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A. A difference in courtship signals stops mating, so it is behavioural isolation before the zygote forms. Temporal isolation concerns timing, hybrid sterility acts after the zygote forms, and geographical isolation is a physical barrier.

Final answer

A

Question A9

HL[1 mark]
Which observation best indicates adaptive radiation? (A) A bird population doubles in size. (B) One finch species has a larger bill each year. (C) Several island species from one ancestor use different foods. (D) Many species vanish after a fire.
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C. Several species that descend from one recent ancestor, each using a different resource, is the pattern of adaptive radiation. A and D describe population growth or an extinction, and B describes one species with no splitting.

Final answer

C

Tutorial practice

Question B1

[3 marks]
Explain why the forelimbs of a human, a bat and a whale are evidence for common ancestry.
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The limbs have corresponding bones in the same arrangement, humerus, radius, ulna, wrist bones and digits, despite different functions (1). The shared plan is explained by inheritance from a common ancestor (1). Natural selection modified it in each lineage for different uses, which is divergent evolution (1). Similar function alone would not be evidence of shared descent.

Final answer

They share the same bones in the same order, humerus, radius, ulna, wrist bones and digits, although their functions differ. That plan was inherited from a common ancestor. Selection then modified the proportions of the bones for grasping, flying and swimming.

Question B2

[4 marks]
Compare and contrast homologous and analogous structures.
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Any four of: homologous structures share an origin in a common ancestor, analogous structures have separate origins (1); homologous structures may have different functions, analogous structures have similar functions (1); homologous structures arise by divergent evolution, analogous structures by convergent evolution (1); an example of homologous structures, such as the pentadactyl limbs of a human and a bat (1); an example of analogous structures, such as the wings of an insect and a bird (1). The comparison must pair each feature, not give two separate definitions.

Final answer

Homologous structures come from a common ancestor and may have different functions, while analogous structures arose independently and have similar functions. Homologous structures result from divergent evolution and analogous structures from convergent evolution. A human arm and a bat arm are homologous; an insect wing and a bird wing are analogous.

Question B3

[4 marks]
Sharks and dolphins have similar streamlined bodies, but they descend from different ancestors. Explain how this similarity arose.
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Sharks and dolphins evolved separately from different ancestors (1). Both faced a similar selection pressure, the need to move fast through water (1). In each lineage, individuals with a more streamlined shape survived and reproduced more, and passed on the alleles (1). Over generations the frequency of those alleles rose in both, which is convergent evolution producing analogous structures (1).

Final answer

Both lineages faced the same pressure: moving fast through water. In each, streamlined individuals swam faster, caught food or escaped predators, and left more offspring, so the alleles for a streamlined shape became more frequent. Selection acted on the variation in each lineage separately, so the similarity is analogous.

Question B4

[5 marks]
The table compares a 100-amino-acid protein from other primates with the human protein. The values are illustrative. Differences from the human protein out of 100 positions: chimpanzee 1, gorilla 3, orangutan 8, macaque 14. (a) State the species with the protein most similar to the human protein, and calculate the percentage of positions in the macaque protein that are identical to the human protein. [2] (b) Explain how the data support common ancestry. [3]
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(a) Chimpanzee, with 1 difference (1); 100-14=86 identical positions of 100, so 86% (1). (b) Differences arise from mutations that accumulate independently in each lineage after they separate (1); the similar sequences are inherited from a common ancestor (1); the fewer the differences, the less time has passed since the common ancestor, so the chimpanzee shares the most recent common ancestor with humans (1).

Final answer

(a) Chimpanzee; 86%. (b) Mutations change the sequence over time. A common ancestor passed on the same sequence, so fewer differences mean a more recent common ancestor, and the chimpanzee has the fewest.

Question B5

[4 marks]
Over 60 years a farmer bred cattle only from the best milk producers. Mean milk yield rose from 5000 to 8000 kg per cow per year. (a) Calculate the percentage increase in milk yield. [1] (b) Explain why this is an example of evolution, and state what it shows about the speed of heritable change. [3]
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(a) (8000-5000)\div5000\times100=60\% (1). (b) The cattle with high yield were selected as parents, and the yield trait is inherited, so the allele frequencies in the herd changed over generations, which is evolution (2); large heritable change can happen in a relatively short period when selection is strong (1).

Final answer

(a) A 60% increase. (b) The farmer chose the breeders and the yield is inherited, so the allele frequencies changed over generations. Heritable change can be large in a short time.

Question B6

[6 marks]
A glacier divided a valley into two parts, each with a population of the same ground beetle. After thousands of years the beetles from the two parts do not interbreed when brought together. Explain how two species could have formed.
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One ancestral population was divided by a geographical barrier (1). Gene flow stopped, so alleles were no longer exchanged (1). The two parts had different conditions, so different selection pressures acted (1). Beetles with alleles suited to each part survived and reproduced more, so the allele frequencies diverged (1). Mutations arose independently in each population (1). After many generations they could no longer produce fertile offspring together, so reproductive isolation persisted and two species existed (1).

Final answer

The glacier stopped gene flow. The two valleys differed in climate and food, so different inherited traits were selected, and allele frequencies diverged. Mutations arose separately. After many generations the beetles could not produce fertile offspring together, so reproductive isolation persisted and there were two species.

Question B7

[5 marks]
Fish from three lakes were bred in pairs of lakes, and their courtship was observed. The table gives the results. Pair 1 (lake A with B): courtship different, many offspring, fertile. Pair 2 (lake B with C): courtship similar, few offspring, sterile. Pair 3 (lake A with C): courtship different, no offspring. (a) Identify the pairs that provide evidence that speciation is complete, with a reason. [2] (b) Evaluate the claim that lakes A and B hold separate species because their courtship behaviour differs. [2] (c) Suggest one further test. [1]
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(a) Pairs 2 and 3 (1), because sterile hybrids or no offspring stop alleles passing between the populations (1). (b) The claim is not supported (1): pair 1 produces fertile offspring, so alleles can still flow even though courtship differs (1). (c) Any one: test the fertility of the second generation from pair 1; observe whether the populations interbreed under natural conditions (1).

Final answer

(a) Pairs 2 and 3. Pair 2 produces only sterile offspring and pair 3 produces none, so gene flow is blocked. (b) Not valid: pair 1 shows that they can still produce fertile offspring even though courtship differs, so the split is not complete. (c) Release them in one lake and test whether they interbreed in the wild.

Question B8

HL[4 marks]
Compare allopatric and sympatric speciation.
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Similarity, either one: both involve the splitting of a pre-existing species (1); both need reproductive isolation so that allele frequencies diverge (1). Difference: allopatric populations are separated by a geographical barrier, whereas sympatric populations live in the same area (1); sympatric isolation arises from temporal or behavioural differences or polyploidy, where allopatric begins with the barrier (1). Each point must be paired for both routes.

Final answer

Both involve a pre-existing species splitting into two and reproductive isolation. In allopatric speciation the populations are separated by a geographical barrier. In sympatric speciation they occupy the same area and isolation arises by other means, such as breeding at different times, behaviour or a change in chromosome number.

Question B9

HL[5 marks]
Two plant species were crossed. Of 200 flowers pollinated, the pollen tube reached the ovule in 60, 24 seeds germinated, 20 seedlings reached flowering and none of those 20 produced viable pollen. (a) Calculate the percentage of flowers in which the pollen tube reached the ovule. [1] (b) Identify the stage with the largest loss and state whether it is a pre-zygotic or post-zygotic barrier. [2] (c) Identify two post-zygotic barriers shown by the data. [2]
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(a) 60\div200\times100=30\% (1). (b) The largest loss is at the pollen-tube stage, where 140 of 200 flowers failed (1); this is before fertilisation, so pre-zygotic (1). (c) Two of the hybrid stages: only 24 of 60 fertilised seeds germinated, so hybrid inviability (1); none of the 20 flowering hybrids produced viable pollen, so hybrid sterility (1).

Final answer

(a) 30%. (b) Failure of pollen tubes to reach the ovule, which is before the zygote: pre-zygotic. (c) Failure of seeds to germinate and hybrid sterility.

Advanced practice

Question C1

[7 marks]
Researchers compared a 150-amino-acid protein and its gene in species M, N and O. M and N have identical proteins, but their gene sequences differ at 9 bases. M and N each differ from O at 6 amino acids. A student concludes that M and N are the same species and that O is distant from both. (a) State what the protein data show about M and N. [1] (b) Explain how the gene sequences can differ while the proteins are identical. [2] (c) Evaluate the student's conclusion that M and N are the same species. [2] (d) Explain why the data do not show which of M and N is closer to O, and state what extra evidence is needed. [2]
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(a) The proteins of M and N are identical in all 150 positions (1). (b) The genetic code is degenerate, so different codons can code for the same amino acid (1); a base substitution at the third position can therefore change the gene without changing the protein (1). (c) The conclusion is not supported: the gene sequences differ at 9 bases, so the species are not identical (1); one protein does not show whether they can interbreed (1). (d) The equal counts of 6 differences from O do not show whether M or N is closer to O (1); further evidence from other genes, or non-coding regions, from more than one sample would be needed (1).

Final answer

(a) Identical proteins. (b) The genetic code is degenerate, so several codons can specify one amino acid; changes at the third base can leave the protein unchanged. (c) Not valid: protein identity hides 9 base differences, and species are defined by interbreeding, not one protein. (d) M and N each differ from O by 6, so neither is shown closer to O; compare more genes, including non-coding sequence.

Question C2

HL[8 marks]
Plant species P has 14 chromosomes in its body cells (one set is 7) and plant species Q has 18 (one set is 9). They hybridize, and the sterile hybrid later doubles its chromosome number. The numbers are invented for this question. (a) Calculate the chromosome number of the first hybrid. [2] (b) Explain why the first hybrid is sterile. [2] (c) Deduce the chromosome number after doubling and the number of pairs. [2] (d) Explain why the doubled plant is a new species. [2]
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(a) Gametes of P carry 14\div2=7 and of Q carry 18\div2=9, so the hybrid has 7+9=16 chromosomes (2). (b) The chromosomes come from two species and have no homologous partners, so they cannot pair at meiosis and balanced gametes are not formed (2). (c) Doubling gives 2\times16=32 chromosomes, which is two sets of P and two sets of Q, forming 16 pairs (2). (d) With partners for every chromosome meiosis works, so the plant is fertile (1); crossing it with P or Q gives offspring with an odd number of sets, which are sterile, so it is reproductively isolated from both and is a new species (1).

Final answer

(a) 7 + 9 = 16. (b) The 16 chromosomes have no homologous partners, so they cannot pair at meiosis and no balanced gametes form. (c) 32 chromosomes: two sets from P and two from Q, 16 pairs. (d) Each chromosome pairs, so it is fertile; offspring with P or Q would have an odd number of sets and be sterile, so it is a new species.

Question C3

[7 marks]
Lizards live on two islands that separated 20 000 years ago. In choice trials, lizards were offered a partner from their own island or from the other island, and hybrid offspring were then bred together. Partner accepted from the same island 90 percent; from the other island 12 percent; second-generation hybrid offspring that are fertile 80 percent. (a) Describe the results. [2] (b) Explain how the separation could have produced these results. [2] (c) Evaluate the claim that the lizards on the two islands are now separate species. [3]
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(a) Acceptance is 90% within an island and 12% between islands (1); 80% of the hybrids' offspring are fertile (1). (b) The sea stopped gene flow (1); different conditions on each island led to different selection, so allele frequencies diverged (1). (c) Evaluation: the low acceptance between islands shows partial behavioural isolation (1); but 80% fertility means that alleles can still pass between the islands, so isolation is incomplete and speciation is not yet complete (1); a further test of whether lizards interbreed when released together would be needed (1).

Final answer

(a) Most lizards accept a partner from their own island, but few accept one from the other, and the second generation of hybrids is mostly fertile. (b) No gene flow after separation, different pressures, allele frequencies diverge. (c) Not complete: mating preference is partly lower, but hybrids are fertile, so alleles can still pass. Test: release together.

Question C4

HL[6 marks]
A crater lake formed 10 000 years ago with no outlet. It was colonised by one cichlid species. It now holds five cichlid species that differ in jaw shape and diet, with no barrier inside the lake. (a) State whether the speciation was allopatric or sympatric, with a reason. [2] (b) Explain how adaptive radiation produced the five species. [3] (c) State why this raises biodiversity. [1]
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(a) Sympatric speciation (1), because the five species live in one lake and there is no geographical barrier (1). (b) Any three: the ancestor found vacant niches with unused food (1); jaw variation was heritable and selected for different foods (1); fish feeding on the same food bred together, so allele frequencies diverged (1); competition between species fell because they used different resources (1). (c) Adaptive radiation produces several species from one ancestor, which is an increase in species number, not only in the number of fish (1).

Final answer

(a) Sympatric: the species share one lake with no barrier. (b) One ancestor entered a lake with unused food sources; variation in jaws was selected for different foods; groups specialised, mated within their group and diverged. (c) Because the number of species increased, not only the number of fish.

Connected questions

Question X1

Connected questions[6 marks]
A4.1 + D1.2 + D1.3. In a gene in species M and in species N, one codon is GAA in M and GAG in N. Both codons specify glutamic acid. All other codons are identical. (a) State the type of mutation that changed GAA to GAG. [1] (b) Explain why the protein is unchanged. [2] (c) Explain what this single difference suggests about the history of M and N. [2] (d) Suggest how to make the comparison more reliable. [1]
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(a) The base A has been replaced by G, a substitution (point) mutation (1). (b) Both codons specify the same amino acid, so the protein is unchanged (1), because the genetic code is degenerate: several codons specify one amino acid (1). (c) The difference arose by mutation after the lineages separated (1); a single base difference shows the DNA is not identical, so the protein comparison alone understates the divergence (1). (d) Comparing more genes gives a more reliable estimate of relationship (1).

Final answer

(a) A substitution mutation at the third base. (b) The genetic code is degenerate. (c) The proteins are identical, but the DNA differs, so the species are not identical and have been separate long enough for a mutation to arise. (d) More genes needed.

Question X2

Connected questions · HL[6 marks]
A4.1 + D3.1 + D2.1. Two plant species have 12 and 14 chromosomes in their body cells. A hybrid between them grows normally but sets no viable seed. (a) Calculate the chromosome number of the hybrid. [1] (b) Explain why the hybrid sets no viable seed. [2] (c) Explain why the hybrid still grows normally. [1] (d) Explain how chromosome doubling could produce a new species. [2]
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(a) Gametes carry 6 and 7 chromosomes, so the hybrid has 13 (1). (b) The chromosomes come from two species and have no homologous partners (1), so they cannot pair at meiosis I and balanced gametes are not produced (1). (c) The hybrid grows normally because mitosis does not need pairing of homologues (1). (d) Doubling the number to 26 gives 13 pairs, so each chromosome has a partner and meiosis can produce balanced gametes (1); the plant is fertile but isolated from both parents, so it is a new species (1).

Final answer

(a) 6 + 7 = 13. (b) The 13 chromosomes cannot pair as homologous pairs at meiosis, so gametes are unbalanced. (c) Mitosis does not need pairing, so the hybrid can grow. (d) Doubling to 26 gives 13 pairs, so meiosis works.

Question X3

Connected questions[6 marks]
A4.1 + A3.2 + D4.1. A student draws a clade that groups bats with birds because both have wings, and leaves humans outside it. (a) Explain why the grouping is not valid. [2] (b) State two features that support grouping bats with humans. [2] (c) Describe how sequence evidence could be used to test the grouping. [2]
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(a) The wings of bats and birds are analogous structures that arose independently (1), as a result of convergent evolution, because flight was favoured by selection in both lineages (1). (b) Two of: homologous forelimb bones, hair, mammary glands (2). (c) Comparing corresponding gene sequences from more than one gene (1); the bat sequences would be more similar to the human sequences than to the bird sequences, giving a clade of mammals (1).

Final answer

(a) Wings are analogous; flight evolved separately by convergent evolution under similar selection. (b) Bats and humans share homologous forelimb bones and hair. (c) Compare gene sequences from more than one gene; bats and humans would be more similar.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Evolution and speciation from the core ideas to full exam questions, with question types, worked examples and graded practice.

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