Skill Align QCE Biology - Free Online Pack 0 Paper 1
An independently authored QCE Biology practice paper with worked answers and question-specific marking guidance.
- Paper
- Paper 1 Combined Independent Practice Book
- Reading
- 5 minutes perusal
- Writing
- 90 minutes
- Assessment
- 49 marks
A QCAA-approved scientific or graphics calculator may be used. Notes and other unauthorised materials are not permitted.
Section 1 - Multiple-choice questions
Questions 1-20 are multiple choice. Select the best answer for each question. Each question is worth 1 mark.
Question 1
1 mark- Average the two branches of the key.
- Follow the branch of the key for toothed margins.
- Return the specimen to the smooth-margin group.
- Name a species without using later couplets.
Question 2
1 mark- All shore locations where any snail species has been observed.
- Its trophic level alone, with abiotic conditions excluded.
- The maximum number of snails the shore can support.
- Its resource use, required conditions and interactions with other species.
Question 3
1 mark- two parental strands joined end to end.
- only RNA nucleotides.
- one parental strand and one newly synthesised strand.
- four newly synthesised strands.
Question 4
1 mark- Individual plants develop inherited alleles because they need deeper roots.
- Every shallow-rooted plant mutates in the same direction.
- Plants with heritable deeper roots leave more offspring, changing allele frequencies over generations.
- The drought selects traits without differential reproduction.
Question 5
1 mark- 4 tussocks m⁻²
- 3 tussocks m⁻²
- 5 tussocks m⁻²
- 20 tussocks m⁻²
Question 6
1 mark- 1%
- 90%
- 1 000%
- 10%
Question 7
1 mark- random orientation of homologous chromosome pairs at metaphase I.
- replication of chromosomes after telophase II.
- separation of sister chromatids during mitosis only.
- translation of alleles by ribosomes.
Question 8
1 mark- directional selection for the lost allele.
- gene flow into the island.
- genetic drift following a population bottleneck.
- non-random mating creating a new species immediately.
Question 9
1 mark- Every lizard is captured on the first occasion.
- Marked lizards become easier to recapture.
- Births must exactly equal deaths during marking.
- The population is effectively closed between capture occasions.
Question 10
1 mark- They convert atmospheric nitrogen into nitrogen compounds usable within food webs.
- They release nitrogen gas by breaking down nitrate under all conditions.
- They convert glucose directly into atmospheric nitrogen.
- They prevent plants from absorbing ammonium ions.
Question 11
1 mark- More than two alternative alleles occur at the locus in the population.
- Each person inherits all three alleles.
- The alleles occur on three different chromosomes.
- Blood group is determined without genes.
Question 12
1 mark- artificial selection.
- convergent evolution within one individual.
- asexual reproduction.
- allopatric speciation.
Question 13
1 mark- Site Y must contain more individual organisms.
- Site X has exactly twice as many species.
- Site X has greater diversity according to this index.
- Site Y has greater evenness in every taxon.
Question 14
1 mark- Tree seedlings resprout immediately from an existing soil seed bank.
- Microorganisms and lichens colonise while organic material begins to accumulate.
- A mature forest appears before any colonising species.
- Weathering stops once the rock surface cools.
Question 15
1 mark- 5'-AUG-3'
- 5'-TAC-3'
- 3'-AUG-5'
- 5'-UAC-3'
Question 16
1 mark- 0.20.
- 0.50.
- 0.80.
- 1.00.
Question 17
1 mark- They ensure every quadrat contains the target species.
- They reduce deliberate placement bias.
- They remove natural variation among locations.
- They make replication unnecessary.
Question 18
1 mark- is the ecosystem's only primary producer.
- contains most of the ecosystem biomass.
- prevents all energy loss between trophic levels.
- has a disproportionately large effect on community structure.
Question 19
1 mark- They permanently separate all DNA strands at room temperature.
- They provide sequence-specific starting points for DNA polymerase.
- They translate amplified DNA into protein.
- They cut DNA at palindromic recognition sites.
Question 20
1 mark- A and B share a more recent common ancestor with each other than either does with C.
- Species A is the direct ancestor of both living species.
- The sequence differences establish an exact divergence date without further information.
- Species C must occupy a different habitat from A and B.
Section 2 - Short-response questions
Questions 21-26 are short response. Answer all parts and use the supplied evidence where required.
Question 21
4 marksEcologists capture 48 wallabies, mark and release them. One week later they capture 60 wallabies, including 12 marked animals.
Question 22
5 marksA mangrove community fixes 24 000 kJ m⁻² year⁻¹ as gross primary productivity. Producers respire 14 400 kJ m⁻² year⁻¹, and herbivores assimilate 960 kJ m⁻² year⁻¹.
Question 23
6 marksA coding DNA segment is 5'-ATG GCT TTT-3'. A variant is 5'-ATG GAT TTT-3'. Use the supplied codon reference.
Question 24
2 marksIn one orchid population, deeper nectar spurs receive more pollen from moths with longer feeding structures. Moths whose feeding structures match local spur depth obtain more nectar. Both traits vary heritably.
Question 25
5 marksAt Reef A, counts for four coral genera are 18, 18, 17 and 17. At Reef B, counts are 52, 8, 6 and 4. Use D = 1 - Σ[n(n - 1)]/[N(N - 1)].
Question 26
7 marksA coastal grass population varies in leaf wax thickness. During repeated salt-spray events, plants with thicker wax lose less water and produce more viable seeds. Wax thickness is heritable.
Worked Solutions And Marking Guide
General marking principles
- Award each listed mark independently when the required biological evidence is demonstrated.
- Accept scientifically accurate equivalent wording and logically equivalent calculations.
- Do not award an unsupported causal claim where the evidence establishes only an association.
Section 1 Question 1
Answer: Follow the branch of the key for toothed margins.
A dichotomous key is followed one observed alternative at a time until a taxon is identified.
Section 1 Question 2
Answer: Its resource use, required conditions and interactions with other species.
An ecological niche includes how a species uses resources and conditions and how it interacts with other organisms; it is not just a location or population limit.
Section 1 Question 3
Answer: one parental strand and one newly synthesised strand.
Each original strand templates a complementary new strand.
Section 1 Question 4
Answer: Plants with heritable deeper roots leave more offspring, changing allele frequencies over generations.
Selection acts on existing heritable variation through differences in reproductive success.
Section 1 Question 5
Answer: 4 tussocks m⁻²
The total of 20 tussocks divided by five square metres gives 4 tussocks per square metre.
Section 1 Question 6
Answer: 10%
Ecological efficiency is 1 800 divided by 18 000, multiplied by 100.
Section 1 Question 7
Answer: random orientation of homologous chromosome pairs at metaphase I.
Different homologous pairs orient independently at metaphase I, producing varied chromosome combinations in gametes.
Section 1 Question 8
Answer: genetic drift following a population bottleneck.
Chance survival in a severe population reduction can alter allele frequencies independently of fitness.
Section 1 Question 9
Answer: The population is effectively closed between capture occasions.
Immigration, emigration, births or deaths between samples can bias a capture-recapture estimate.
Section 1 Question 10
Answer: They convert atmospheric nitrogen into nitrogen compounds usable within food webs.
Nitrogen fixation introduces biologically available nitrogen by reducing atmospheric N₂.
Section 1 Question 11
Answer: More than two alternative alleles occur at the locus in the population.
A multiple-allele system has more than two allelic forms in the population, although each diploid individual carries at most two of them.
Section 1 Question 12
Answer: allopatric speciation.
Geographic isolation can reduce gene flow until reproductive isolation evolves.
Section 1 Question 13
Answer: Site X has greater diversity according to this index.
For the stated form of the index, the larger value indicates greater combined richness and evenness.
Section 1 Question 14
Answer: Microorganisms and lichens colonise while organic material begins to accumulate.
Primary succession begins without pre-existing soil, so early colonisers and weathering contribute to the gradual development of soil and biomass.
Section 1 Question 15
Answer: 5'-AUG-3'
RNA polymerase reads the DNA template 3' to 5' and synthesises complementary mRNA 5' to 3'.
Section 1 Question 16
Answer: 0.50.
Equal representative samples contribute equal numbers of allele copies, so the combined frequency is (0.80 + 0.20) / 2 = 0.50. Pooling populations redistributes existing alleles rather than creating new ones.
Section 1 Question 17
Answer: They reduce deliberate placement bias.
Random placement gives locations a known selection process and reduces observer choice bias.
Section 1 Question 18
Answer: has a disproportionately large effect on community structure.
A keystone species has an ecological effect that is large relative to its abundance.
Section 1 Question 19
Answer: They provide sequence-specific starting points for DNA polymerase.
Primers anneal to target-flanking sequences and supply free 3' ends for extension.
Section 1 Question 20
Answer: A and B share a more recent common ancestor with each other than either does with C.
Fewer differences in the same conserved sequence support closer relatedness, while ancestry direction and exact dates require additional evidence.
Section 2 Question 21
(a) - Substitution gives N = (48 × 60) / 12. - The estimated population size is 240 wallabies.
(b) - Marked wallabies should mix fully and have the same recapture probability as unmarked wallabies. - If marks make wallabies avoid traps, fewer marked animals are recaptured and the population is overestimated.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Substitution gives N = (48 × 60) / 12.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The estimated population size is 240 wallabies.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Marked wallabies should mix fully and have the same recapture probability as unmarked wallabies.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: If marks make wallabies avoid traps, fewer marked animals are recaptured and the population is overestimated.
Section 2 Question 22
(a) - Net primary productivity equals gross primary productivity minus producer respiration. - NPP = 24 000 - 14 400 = 9 600 kJ m⁻² year⁻¹.
(b) - Herbivore assimilation is (960 / 9 600) × 100 = 10%. - Herbivores do not ingest all producer biomass; some plant material dies, is inaccessible or is otherwise not consumed. - Some ingested material is not absorbed across the gut and is egested, so it is not assimilated.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Net primary productivity equals gross primary productivity minus producer respiration.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: NPP = 24 000 - 14 400 = 9 600 kJ m⁻² year⁻¹.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Herbivore assimilation is (960 / 9 600) × 100 = 10%.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Herbivores do not ingest all producer biomass; some plant material dies, is inaccessible or is otherwise not consumed.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Some ingested material is not absorbed across the gut and is egested, so it is not assimilated.
Section 2 Question 23
(a) - Reference mRNA is 5'-AUG GCU UUU-3'. - Variant mRNA is 5'-AUG GAU UUU-3'.
(b) - Reference codons specify Met-Ala-Phe. - The variant specifies Met-Asp-Phe, so alanine is replaced by aspartate.
(c) - The substitution may alter charge, folding or an active region, but its effect depends on the residue's structural context. - Protein activity and organism phenotype require functional evidence because some substitutions have little detectable effect.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Reference mRNA is 5'-AUG GCU UUU-3'.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Variant mRNA is 5'-AUG GAU UUU-3'.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Reference codons specify Met-Ala-Phe.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The variant specifies Met-Asp-Phe, so alanine is replaced by aspartate.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: The substitution may alter charge, folding or an active region, but its effect depends on the residue's structural context.
Part c (1 mark)
Awards one mark for accurately establishing this question-specific point: Protein activity and organism phenotype require functional evidence because some substitutions have little detectable effect.
Section 2 Question 24
(a) - Orchid reproductive success varies with the moth feeding trait, while moth feeding success varies with orchid spur depth. - Reciprocal selection can therefore change both populations across generations, producing coevolution.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Orchid reproductive success varies with the moth feeding trait, while moth feeding success varies with orchid spur depth.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Reciprocal selection can therefore change both populations across generations, producing coevolution.
Section 2 Question 25
(a) - For Reef A, N = 70 and Σn(n - 1) = 1 156. - D = 1 - 1 156/4 830 = 0.76 to two decimal places.
(b) - For Reef B, Σn(n - 1) = 2 750 and D = 1 - 2 750/4 830 = 0.43. - Reef A has the greater index because individuals are distributed more evenly among the four recorded genera. - The calculation describes the sampled genera and does not establish total reef biodiversity or the cause of the difference.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: For Reef A, N = 70 and Σn(n - 1) = 1 156.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: D = 1 - 1 156/4 830 = 0.76 to two decimal places.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: For Reef B, Σn(n - 1) = 2 750 and D = 1 - 2 750/4 830 = 0.43.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Reef A has the greater index because individuals are distributed more evenly among the four recorded genera.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The calculation describes the sampled genera and does not establish total reef biodiversity or the cause of the difference.
Section 2 Question 26
(a) - The starting population contains heritable variation in wax thickness. - Salt spray creates a selection pressure because thick-wax plants retain more water. - Those plants contribute more offspring, increasing the frequency of alleles associated with thicker wax.
(b) - Heritable wax variation and unequal viable-seed production support selection during salt-spray events. - No trait or allele-frequency observations across generations are supplied. - One population does not separate the salt effect from soil moisture, exposure or gene flow among sites. - The stated heritability does not show whether the same trait has a fitness advantage outside this coastal environment.
Detailed marking criteria
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: The starting population contains heritable variation in wax thickness.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Salt spray creates a selection pressure because thick-wax plants retain more water.
Part a (1 mark)
Awards one mark for accurately establishing this question-specific point: Those plants contribute more offspring, increasing the frequency of alleles associated with thicker wax.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: Heritable wax variation and unequal viable-seed production support selection during salt-spray events.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: No trait or allele-frequency observations across generations are supplied.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: One population does not separate the salt effect from soil moisture, exposure or gene flow among sites.
Part b (1 mark)
Awards one mark for accurately establishing this question-specific point: The stated heritability does not show whether the same trait has a fitness advantage outside this coastal environment.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3 Topic 1 - Biodiversity and populations | Q1, Q5, Q9, Q13, Q17, Q21, Q25 | 14 | ___ | Review classification, sampling, population estimates, diversity indices and evidence limits. |
| Unit 3 Topic 2 - Functioning ecosystems and succession | Q2, Q6, Q10, Q14, Q18, Q22 | 10 | ___ | Review energy transfer, biogeochemical cycling, species interactions, disturbance and succession. |
| Unit 4 Topic 1 - Genetics and heredity | Q3, Q7, Q11, Q15, Q19, Q23 | 11 | ___ | Review DNA, gene expression, meiosis, inheritance, biotechnology and quantitative genetic reasoning. |
| Unit 4 Topic 2 - Continuity of life on Earth | Q4, Q8, Q12, Q16, Q20, Q24, Q26 | 14 | ___ | Review mutation, natural selection, speciation, phylogeny and converging evolutionary evidence. |