AQA A-level Biology · 3.7

Genetics, populations, evolution and ecosystems

Revision questions with answers for 3.7 Genetics, populations, evolution and ecosystems in AQA A-level Biology (spec 7402). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

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Question 1 · Spec 3.7.1 · Inheritance terms · Recall
Define genotype, phenotype, homozygous and heterozygous.
Answer:
Genotype is an organism's genetic constitution. Phenotype is its expressed characteristics, resulting from genotype and environment. A homozygote has identical alleles at a locus, while a heterozygote has different alleles. State allele symbols clearly in genetic diagrams.
Question 2 · Spec 3.7.1 · Allele interactions · Recall
Distinguish dominant, recessive and codominant alleles.
Answer:
A dominant allele affects phenotype in a heterozygote. A recessive allele's effect is masked by a dominant allele at that locus. With codominance, both alleles are expressed in the heterozygote. Dominance describes expression and does not imply that an allele is common or beneficial.
Question 3 · Spec 3.7.1 · Multiple alleles · Application
How can a gene have multiple alleles while a diploid individual carries only two? Use ABO blood groups.
Answer:
More than two alleles can occur in the population, but an individual has two copies at an autosomal locus. In ABO, I^A and I^B are codominant and both dominate i. An I^Ai x I^Bi cross can produce A, B, AB or O offspring in equal expected proportions.
Question 4 · Spec 3.7.1 · Monohybrid cross · Application
For Aa x Aa with complete dominance, what genotype and phenotype ratios are expected?
Answer:
Gametes carry A or a with equal probability. Expected genotypes are 1 AA : 2 Aa : 1 aa. Expected phenotypes are 3 dominant : 1 recessive. These are probabilities, so a small family need not match the ratio exactly.
Question 5 · Spec 3.7.1 · Dihybrid cross · Application
When does AaBb x AaBb give a 9:3:3:1 phenotypic ratio?
Answer:
It is expected when the two loci assort independently, each shows complete dominance and there is no epistasis or differential survival. Each parent produces AB, Ab, aB and ab gametes in equal proportions. State these assumptions before applying the ratio.
Question 6 · Spec 3.7.1 · Test cross · Application
How can a test cross help identify an unknown dominant-phenotype genotype?
Answer:
Cross the individual with a homozygous recessive organism. At one locus, a heterozygote gives an expected 1:1 dominant-to-recessive ratio. A homozygous dominant parent gives only dominant offspring. Larger offspring numbers make the inference more reliable.
Question 7 · Spec 3.7.1 · Sex linkage · Application
A carrier mother X^AX^a and unaffected father X^AY have children. What outcomes are expected for an X-linked recessive condition?
Answer:
Daughters are expected to be half X^AX^A and half X^AX^a, both unaffected under this model. Sons are expected to be half X^AY unaffected and half X^aY affected. Each son receives his X from his mother and Y from his father.
Question 8 · Spec 3.7.1 · Autosomal linkage · Recall
Why can two autosomal genes fail to assort independently?
Answer:
Genes on the same chromosome are linked and tend to be inherited together. Crossing over can separate their alleles. The closer their loci, the less likely a crossover occurs between them, so parental combinations may greatly outnumber recombinant combinations.
Question 9 · Spec 3.7.1 · Linked-cross data · Analysis
In a test cross, AB and ab offspring greatly outnumber Ab and aB. What does this suggest?
Answer:
It suggests the heterozygous parent carried linked alleles in the AB/ab arrangement. AB and ab are parental combinations. Ab and aB can result from crossing over. Check the stated genetic model and sample size before concluding that linkage explains the departure from independent assortment.
Question 10 · Spec 3.7.1 · Epistasis · Application
What is epistasis, and how should an unfamiliar epistatic cross be solved?
Answer:
Epistasis occurs when one gene modifies or masks the expression of another gene at a different locus. Use the supplied biochemical pathway or phenotype rules to group genotypes after constructing the cross. Derive the resulting ratio from those rules rather than assuming 9:3:3:1.
Question 11 · Spec 3.7.1, MS 1.9 · Chi-squared test · Calculation
What does a chi-squared goodness-of-fit test assess, and how is it calculated?
Answer:
It tests whether differences between observed and expected category frequencies are consistent with chance under a stated model. Calculate chi-squared = sum[(O - E)²/E]. Use counts, mutually exclusive categories, adequate expected frequencies and the appropriate degrees of freedom.
Question 12 · Spec 3.7.1, MS 1.9 · Chi-squared example · Calculation
A cross gives 70 dominant and 30 recessive offspring, with a 3:1 expectation. Calculate chi-squared.
Answer:
Expected counts are 75 and 25. Chi-squared = (70 - 75)²/75 + (30 - 25)²/25 = 25/75 + 25/25 = 1.33. For two categories with no parameters estimated from the observations, degrees of freedom = 2 - 1 = 1.
Question 13 · Spec 3.7.1, MS 1.9 · Statistical conclusion · Analysis
For chi-squared = 1.33 and a supplied 5% critical value of 3.84, what conclusion is justified?
Answer:
The calculated value is below the critical value, so do not reject the null hypothesis. The difference from the expected 3:1 ratio is not statistically significant at the 5% level. This result is compatible with the model but does not prove that the model is correct.
Question 14 · Spec 3.7.2 · Populations · Recall
What is a population in biological terms?
Answer:
A population is a group of organisms of the same species occupying a particular area at a particular time and potentially able to interbreed. A species may contain several populations with different environmental conditions, allele frequencies or degrees of isolation.
Question 15 · Spec 3.7.2 · Gene pool · Recall
Define gene pool and allele frequency.
Answer:
A gene pool contains all the alleles present in a population. An allele frequency is the proportion of all copies of a gene represented by a particular allele. Frequencies may change through selection, genetic drift, mutation or movement of alleles between populations.
Question 16 · Spec 3.7.2, MS 0.3 · Allele-frequency calculation · Calculation
Among 100 diploid individuals, 36 are AA, 48 Aa and 16 aa. Calculate the frequency of A.
Answer:
There are 200 allele copies. A copies = (2 x 36) + 48 = 120. Frequency of A = 120/200 = 0.60. Frequency of a is 0.40. A heterozygote contributes one copy of each allele.
Question 17 · Spec 3.7.2 · Hardy-Weinberg equations · Recall
State the Hardy-Weinberg equations for a gene with two alleles.
Answer:
p + q = 1 describes allele frequencies. p² + 2pq + q² = 1 describes expected genotype frequencies under Hardy-Weinberg conditions. p² is AA, 2pq is Aa and q² is aa when p represents A and q represents a.
Question 18 · Spec 3.7.2 · Hardy-Weinberg conditions · Recall
Which conditions are assumed by the Hardy-Weinberg model?
Answer:
A sufficiently large population with random mating, no mutation, no migration and no natural selection at the locus. These conditions allow allele frequencies to remain constant between generations. Departures can change genotype frequencies, allele frequencies or both.
Question 19 · Spec 3.7.2, MS 2.4 · Hardy-Weinberg application · Calculation
A recessive phenotype occurs in 9% of a Hardy-Weinberg population. Calculate allele and carrier frequencies.
Answer:
q² = 0.09, so q = 0.30 and p = 0.70. Heterozygous carrier frequency = 2pq = 2 x 0.70 x 0.30 = 0.42, or 42%. Do not treat the recessive phenotype frequency itself as q.
Question 20 · Spec 3.7.2, MS 2.4 · Population prediction · Calculation
Using the preceding allele frequencies, how many heterozygotes are expected in a population of 2500?
Answer:
Expected heterozygotes = 2pq x population size = 0.42 x 2500 = 1050. Expected AA individuals = 0.49 x 2500 = 1225 and aa individuals = 0.09 x 2500 = 225. The expected counts total 2500.
Question 21 · Spec 3.7.3 · Sources of variation · Recall
How do genes and environment contribute to phenotypic variation?
Answer:
Different alleles and allele combinations produce genetic variation. Environmental factors can change how a genotype is expressed. Many characteristics reflect both influences. Mutation introduces new alleles, while meiosis and fertilisation generate new combinations of existing alleles.
Question 22 · Spec 3.7.3 · Selection and evolution · Application
How does differential reproductive success produce evolution?
Answer:
A selection pressure makes some heritable phenotypes more successful than others. Individuals carrying advantageous alleles leave more surviving offspring, so these alleles increase in frequency. Evolution is a change in allele frequencies in a population over generations.
Question 23 · Spec 3.7.3 · Selection patterns · Recall
Compare stabilising, directional and disruptive selection.
Answer:
Stabilising selection favours intermediate phenotypes and reduces variation. Directional selection favours one extreme and shifts the mean. Disruptive selection favours both extremes over intermediates and can increase distinct phenotypic groups. The pattern depends on the selective conditions.
Question 24 · Spec 3.7.3 · Genetic drift · Application
Why can genetic drift have a large effect in small populations?
Answer:
Chance differences in survival or reproduction can change allele frequencies independently of selective advantage. A small number of individuals represents a larger fraction of a small gene pool, so random events can cause substantial changes, including loss or fixation of alleles.
Question 25 · Spec 3.7.3 · Allopatric speciation · Application
Explain how geographical separation can lead to speciation.
Answer:
A physical barrier reduces gene flow. Mutation, selection and drift change the separated populations' gene pools independently. Genetic differences can accumulate until members can no longer interbreed successfully to produce fertile offspring. Reproductive isolation establishes separate species.
Question 26 · Spec 3.7.3 · Sympatric speciation · Application
How can speciation occur without a geographical barrier?
Answer:
Gene flow can be reduced within the same area by reproductive separation, such as different breeding times or mate preferences. In plants, chromosome changes may rapidly create incompatible groups. Subsequent selection and drift can increase differences until fertile interbreeding no longer occurs.
Question 27 · Spec 3.7.3 · Evidence for speciation · Analysis
What evidence is needed before concluding that two diverging populations are separate species?
Answer:
Differences in appearance or allele frequency alone are insufficient under the biological species definition. Evidence of reproductive isolation, such as failure to mate successfully or to produce fertile offspring, is needed. Consider whether observed separation would persist if populations met.
Question 28 · Spec 3.7.4 · Community and ecosystem · Recall
Distinguish population, community, habitat and ecosystem.
Answer:
A population consists of one species in an area. A community includes populations of different species. A habitat is the place where an organism lives. An ecosystem includes the community and its non-living environment, with interactions between them.
Question 29 · Spec 3.7.4 · Ecological niche · Recall
What is a species' ecological niche?
Answer:
A niche describes how a species uses resources and interacts with its abiotic and biotic environment. It includes requirements and roles such as food use, tolerances and relationships with other species. Adaptations enable occupation of particular niches within a habitat.
Question 30 · Spec 3.7.4 · Carrying capacity · Recall
What is carrying capacity and why can it change?
Answer:
Carrying capacity is the population size an environment can support under prevailing conditions. Resources, abiotic factors, competition, predation and disease influence it. Seasonal or long-term environmental changes can therefore alter the supported population size.
Question 31 · Spec 3.7.4 · Abiotic factors · Application
How can an abiotic factor limit a population's abundance or distribution?
Answer:
Conditions such as temperature, water availability, light or pH may affect survival, growth and reproduction. Abundance often declines outside a species' tolerance range. Use the stated physiological mechanism and consider interacting factors when explaining field patterns.
Question 32 · Spec 3.7.4 · Competition · Recall
Compare intraspecific and interspecific competition.
Answer:
Intraspecific competition occurs within a species and interspecific competition between species. Both arise when resources are limiting. Their effects can reduce growth, survival or reproduction. Strong overlap in resource use can restrict distribution or prevent long-term coexistence under particular conditions.
Question 33 · Spec 3.7.4 · Predator-prey relationships · Application
Why may predator population peaks follow prey population peaks?
Answer:
More prey can increase predator survival and reproduction. Rising predator numbers then increase predation, reducing prey. Reduced prey availability can subsequently reduce predator numbers. Time lags produce offset cycles, although real ecosystems also involve other prey, predators and environmental factors.
Question 34 · Spec 3.7.4, AT k · Random quadrats · Practical
How should quadrats be placed to estimate abundance without deliberate selection bias?
Answer:
Define the study area, assign coordinates and use random numbers to choose positions. Apply a consistent boundary rule for organisms at quadrat edges. Use enough quadrats to represent habitat variation and record the same abundance measure in each.
Question 35 · Spec 3.7.4, MS 2.3 · Population estimate · Calculation
Twenty 0.25 m² quadrats contain 150 plants in total. Estimate the number in a 400 m² habitat.
Answer:
Total sampled area = 20 x 0.25 = 5.0 m². Density = 150/5.0 = 30 plants m⁻². Estimated population = 30 x 400 = 12,000 plants. The estimate assumes representative sampling and a suitable counting method.
Question 36 · Spec 3.7.4, AT k · Cover and frequency · Recall
Distinguish percentage cover and frequency as measures of plant abundance.
Answer:
Percentage cover estimates the fraction of ground occupied by a species. Frequency is the percentage of sampled quadrats containing that species. Frequency depends on quadrat size, while cover estimates can be subjective. Keep the method consistent when comparing sites.
Question 37 · Spec 3.7.4, AT k · Transects · Practical
When is a belt transect useful and how can it be used?
Answer:
Use one to investigate distribution along an environmental gradient. Place quadrats along a line in a continuous or systematic belt and record abundance and environmental measurements. Repeat transects improve representation and help distinguish a consistent trend from local variation.
Question 38 · Spec 3.7.4, RP12 · Required practical 12 · Practical
How could light intensity effects on a plant species' distribution be investigated?
Answer:
Sample quadrats across a light gradient and record abundance with light readings at comparable positions and times. Repeat across several transects or sites. Standardise quadrat area and recording methods, and measure possible confounders such as soil moisture and pH.
Question 39 · Spec 3.7.4, RP12 · Required practical 12 · Analysis
A plant's abundance correlates negatively with light intensity. What conclusions and limitations should be stated?
Answer:
Abundance tends to be lower at higher measured light intensity in this sample. A suitable correlation test can assess evidence for association. The study alone does not establish that light caused the distribution, because moisture, competition or other variables may also vary.
Question 40 · Spec 3.7.4, MS 2.3 · Mark-release-rec apture · Calculation
First capture marks 40 animals. A second sample contains 50 animals, including 10 marked ones. Estimate population size.
Answer:
Estimated population = first marked sample x second sample/marked recaptures = (40 x 50)/10 = 200 animals. The estimate depends on marked animals being representative and the marked proportion in the second sample reflecting that in the population.
Question 41 · Spec 3.7.4 · Recapture assumptions · Analysis
What assumptions are needed for mark-release-recapture?
Answer:
The population is approximately closed between samples. Marks remain visible, do not harm animals or alter capture probability, and are recorded correctly. Marked animals mix sufficiently with the population before recapture. All individuals have a reasonably comparable chance of capture.
Question 42 · Spec 3.7.4 · Primary succession · Recall
What is primary succession and how does it begin?
Answer:
Primary succession develops on a surface initially lacking developed soil. Pioneer species colonise and tolerate harsh conditions. Their growth and decomposition contribute organic material and soil development, changing conditions so that additional species can establish.
Question 43 · Spec 3.7.4 · Successional change · Application
How can organisms alter conditions during succession?
Answer:
Organic material accumulates, soil develops, water retention changes and nutrients become more available. Larger plants may create shade and alter microclimate. These changes can favour new species and make conditions less suitable for earlier colonists, changing community composition and biodiversity.
Question 44 · Spec 3.7.4 · Climax community · Recall
What is a climax community, and why does conservation sometimes interrupt succession?
Answer:
A climax community is a relatively stable late-stage community under prevailing conditions. Management such as grazing or cutting can prevent later stages from replacing a valued habitat. Maintaining an earlier stage may conserve species adapted to that stage.
Question 45 · Spec 3.7.4 · Conservation decisions · Analysis
How should a habitat-conservation proposal be evaluated?
Answer:
Use evidence about species, habitat condition and likely management effects. Weigh ecological benefits against land use, cost and local needs. Consider study quality, uncertainty and conflicting evidence. A reasoned conclusion should connect the proposed action to maintaining species or habitat features.
Question 46 · Spec 3.7.4 · Sustainability · Application
How can human resource use be made sustainable while conserving ecosystems?
Answer:
Harvest or use resources at rates compatible with replenishment and ecosystem functioning. Monitor population and habitat changes, protect key areas and adjust management when evidence shows decline. Evaluate social and economic needs alongside long-term conservation outcomes.
Question 47 · Spec 3.7.4, MS 1.9 · Student's t-test · Analysis
When may a Student's t-test be appropriate in an ecological investigation?
Answer:
It can test evidence for a difference between two means, using quantitative measurements and the version suited to the sampling design. Check assumptions, including independence for an unpaired test and appropriate distributional conditions. Use the supplied formula, degrees of freedom and critical values.
Question 48 · Spec 3.7.4, RP12, MS 1.9 · Correlation test · Analysis
When could a correlation coefficient be used, and what can it show?
Answer:
Use paired quantitative data to test association between variables, such as light and abundance. Select a suitable coefficient and follow the supplied method. Spearman's uses ranks and assesses a monotonic association. A significant association establishes evidence of correlation, while causal interpretation needs further evidence.
Question 49 · Spec 3.7.1, 3.7.4, MS 1.9 · Statistical interpretation · Analysis
How should a p-value and a non-significant result be interpreted?
Answer:
A p-value describes how unusual data at least this extreme would be if the null model were true. A small p-value can justify rejecting that model at a chosen threshold. A non-significant result may reflect limited power or sample size and does not prove there is no biological effect.
Question 50 · Spec 3.7.1, 3.7.2, 3.7.3, 3.7.4, 4.2 · Synoptic link · Synoptic
How could an essay link inheritance, evolution and ecology when explaining changes in a population?
Answer:
Explain how alleles produce heritable phenotypic differences, how ecological conditions create selection pressures, and how differential reproduction changes allele frequencies. Use genetic drift and isolation where relevant. Develop the mechanism and its significance for the population using evidence from several specification areas.

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