AQA A-level Biology · 3.4

Genetic information, variation and relationships between organisms

50 revision questions with answers for 3.4 Genetic information, variation and relationships between organisms 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.

All 50 questions and answers

Question 1 · Spec 3.4.1 · Chromosomes · Recall
Compare the organisation of DNA in prokaryotes and eukaryotic nuclei.
Answer:
Prokaryotic chromosomal DNA is usually short, circular and free in the cytoplasm, without associated histones in the AQA model. Nuclear eukaryotic DNA is long, linear and associated with histone proteins. A DNA molecule and its associated proteins form a chromosome.
Question 2 · Spec 3.4.1 · Organelle DNA · Recall
How does mitochondrial and chloroplast DNA compare with nuclear DNA?
Answer:
Mitochondria and chloroplasts contain short, circular DNA molecules that are not associated with histones in the specification model. Nuclear DNA is linear and histone-associated. Similarities between organelle and prokaryotic DNA contribute to evidence of evolutionary relationships.
Question 3 · Spec 3.4.1 · Genes and loci · Recall
Define a gene, an allele and a locus.
Answer:
A gene is a DNA base sequence coding for a polypeptide's amino acid sequence or for a functional RNA. An allele is an alternative form of a gene. A locus is the fixed position of a gene on a particular DNA molecule.
Question 4 · Spec 3.4.1 · Genetic code · Recall
What do universal, degenerate and non-overlapping mean when describing the genetic code?
Answer:
Universal means the same triplets generally specify the same amino acids across organisms. Degenerate means more than one triplet may specify an amino acid. Non-overlapping means each base is read as part of only one triplet in a given reading frame.
Question 5 · Spec 3.4.1 · Non-coding DNA · Recall
Distinguish exons, introns and non-coding repeats.
Answer:
Exons are sequences retained in mature RNA and, for protein-coding regions, specify amino acid sequence. Introns are intervening sequences removed from pre-mRNA. Non-coding repeated sequences can occur between genes. Much eukaryotic nuclear DNA does not code for polypeptides.
Question 6 · Spec 3.4.2 · Genome and proteome · Application
What are the genome and proteome, and why can cells with the same genome differ?
Answer:
The genome is the complete genetic material of a cell or organism. The proteome is the full range of proteins a cell can produce. Different cells express different genes, so the proteins present and their relative amounts can differ despite a shared genome.
Question 7 · Spec 3.4.2 · RNA types · Recall
Compare mRNA and tRNA in protein synthesis.
Answer:
mRNA carries a sequence of codons copied from DNA to a ribosome. tRNA folds into a specific shape, carries an amino acid at one end and has an anticodon that pairs with a complementary mRNA codon.
Question 8 · Spec 3.4.2 · Transcription · Recall
How does transcription produce an RNA copy of a gene?
Answer:
DNA strands separate locally. Complementary RNA nucleotides pair with exposed bases on the template strand, using uracil opposite adenine. RNA polymerase joins adjacent RNA nucleotides. In eukaryotes, the initial transcript is pre-mRNA.
Question 9 · Spec 3.4.2 · RNA splicing · Recall
What happens during splicing, and how does prokaryotic transcription differ?
Answer:
Introns are removed from eukaryotic pre-mRNA and exons are joined to form mature mRNA. This mRNA can leave the nucleus for translation. In prokaryotes, transcription produces mRNA directly without this eukaryotic pre-mRNA processing stage.
Question 10 · Spec 3.4.2 · Translation · Recall
How is a polypeptide assembled during translation?
Answer:
A ribosome binds mRNA. tRNAs with complementary anticodons bring specific amino acids. The ribosome holds them in position for peptide-bond formation and moves along the mRNA. ATP supplies energy for processes including amino acid attachment to tRNA. Translation ends at a stop codon.
Question 11 · Spec 3.4.2 · Sequence application · Application
A DNA template sequence is TAC GGA. What is the complementary mRNA sequence, and how would its amino acids be identified?
Answer:
The mRNA sequence is AUG CCU, using A-U and C-G pairing. Read an mRNA codon table to identify the amino acids. A DNA coding strand would instead match the mRNA sequence except that DNA has thymine in place of uracil.
Question 12 · Spec 3.4.1, 3.4.2, MS 0.3 · Coding calculations · Calculation
A polypeptide contains 120 amino acids. What is the minimum number of coding DNA bases for those amino acids?
Answer:
Each amino acid requires a triplet, so 120 x 3 = 360 bases on one DNA strand. A stop codon adds another three bases if included. Introns and regulatory sequences can make the full eukaryotic gene region longer.
Question 13 · Spec 3.4.3 · Gene mutation · Recall
How do substitutions and deletions create genetic variation?
Answer:
A substitution replaces a base and may alter one encoded amino acid. A single-base deletion changes the reading frame downstream, potentially altering many amino acids. Mutations change DNA sequence and can create new alleles that may be inherited.
Question 14 · Spec 3.4.3 · Degeneracy and mutation · Application
Why can a base substitution leave a protein unchanged?
Answer:
The genetic code is degenerate, so the changed triplet may code for the same amino acid. A substitution in some non-coding sequences may also leave the amino acid sequence unchanged. The effect depends on the site and nature of the mutation.
Question 15 · Spec 3.4.3 · Mutagens · Recall
What are mutagenic agents, and do they create mutations because organisms need them?
Answer:
Mutagenic agents, such as ionising radiation or certain chemicals, increase mutation rate. Mutations also occur spontaneously during DNA replication. Their occurrence is random with respect to an organism's needs; selection determines which inherited variants become more common.
Question 16 · Spec 3.4.3 · Non-disjunction · Recall
How can non-disjunction change chromosome number?
Answer:
A homologous pair or sister chromatids fail to separate during meiosis. Resulting gametes may contain an extra chromosome or lack one. Fertilisation involving such a gamete can produce a zygote with an abnormal chromosome number.
Question 17 · Spec 3.4.3 · Meiosis overview · Recall
How does meiosis reduce chromosome number?
Answer:
One DNA replication is followed by two nuclear divisions. Homologous chromosomes separate in the first division and sister chromatids in the second. A diploid cell usually produces four genetically different haploid cells, each with one chromosome from each homologous pair.
Question 18 · Spec 3.4.3 · Crossing over · Recall
How does crossing over increase genetic variation?
Answer:
Homologous chromosomes pair and non-sister chromatids exchange corresponding sections of DNA. This produces new combinations of alleles along chromatids. It rearranges existing alleles and can create genetic combinations different from those in either parent chromosome.
Question 19 · Spec 3.4.3 · Independent segregation · Recall
How does independent segregation produce different gametes?
Answer:
Each homologous pair can orient in either direction during the first meiotic division. The orientation of one pair is independent of other pairs. Maternal and paternal homologues therefore enter daughter cells in different combinations.
Question 20 · Spec 3.4.3, MS 0.5 · Chromosome combinations · Calculation
An organism has four homologous chromosome pairs. How many gamete chromosome combinations are possible without crossing over?
Answer:
The number is 2^n, where n is the number of pairs. For n = 4, there are 16 combinations. Random fertilisation between two such parents gives 16 x 16 = 256 possible combinations of whole parental chromosomes.
Question 21 · Spec 3.4.3 · Random fertilisation · Application
How does random fertilisation add to variation produced by meiosis?
Answer:
Any one of many genetically different male gametes may fuse with any one of many genetically different female gametes. Their combination produces a diploid zygote with a new combination of alleles. Fertilisation combines variation rather than creating new alleles by itself.
Question 22 · Spec 3.4.3, 3.2.2 · Comparing divisions · Recall
Compare the genetic outcomes of mitosis and meiosis.
Answer:
Mitosis normally produces two cells with the same chromosome number and genetic information as the parent cell. Meiosis normally produces four haploid cells with different allele combinations through crossing over and independent segregation. Mutation can introduce additional differences in either process.
Question 23 · Spec 3.4.4 · Genetic diversity · Recall
What is genetic diversity, and why does it matter to natural selection?
Answer:
Genetic diversity is the number of different alleles of genes in a population. It contributes to heritable variation in phenotype. When conditions impose selection, some variants may have greater reproductive success, enabling the population's allele frequencies to change.
Question 24 · Spec 3.4.4 · Natural selection · Application
Explain how an advantageous allele can increase in frequency.
Answer:
Mutation creates alleles and individuals vary. A selection pressure makes some phenotypes more successful. Their carriers survive and reproduce more, passing the advantageous allele to more offspring. Over generations, its frequency rises in the population.
Question 25 · Spec 3.4.4 · Antibiotic resistance · Application
Explain the evolution of antibiotic resistance in a bacterial population.
Answer:
A random mutation may confer resistance before exposure. Antibiotic treatment kills susceptible bacteria while resistant bacteria survive. Survivors reproduce and pass resistance alleles to descendants. The frequency of resistance rises. The antibiotic selects resistant variants rather than directing a useful mutation.
Question 26 · Spec 3.4.4 · Directional selection · Recall
What happens to a phenotype distribution under directional selection?
Answer:
One extreme phenotype has a selective advantage. Associated alleles increase in frequency, shifting the population mean towards that extreme. Antibiotic resistance illustrates selection favouring variants with greater resistance when an antibiotic is present.
Question 27 · Spec 3.4.4 · Stabilising selection · Application
Explain stabilising selection using human birth mass.
Answer:
Intermediate birth masses have historically had higher survival than very low or very high masses. Selection against both extremes maintains an intermediate optimum and reduces variation around it. This is a population pattern influenced by medical and environmental conditions.
Question 28 · Spec 3.4.4 · Adaptation types · Recall
Distinguish anatomical, physiological and behavioural adaptations.
Answer:
Anatomical adaptations involve structure, such as a thick insulating covering. Physiological adaptations involve functioning, such as producing concentrated urine. Behavioural adaptations involve actions, such as avoiding daytime heat. Heritable adaptations can increase reproductive success in a particular environment.
Question 29 · Spec 3.4.4, RP6 · Required practical 6 · Practical
Which aseptic techniques reduce contamination during an antimicrobial investigation?
Answer:
Disinfect the work surface, use sterile equipment and media, minimise time containers are open and avoid touching sterile surfaces. Use an appropriate approved microorganism and incubation conditions. Secure plates according to the laboratory protocol and do not reopen cultures after incubation.
Question 30 · Spec 3.4.4, RP6 · Required practical 6 · Practical
How can antimicrobial substances be compared using agar plates?
Answer:
Spread the same microbial suspension evenly on agar. Apply equal-sized discs carrying controlled amounts of each substance and include a solvent-only control. Keep agar, inoculum, incubation time and temperature consistent. Repeat plates and compare inhibition zones using a defined measurement method.
Question 31 · Spec 3.4.4, RP6, MS 4.1 · Required practical 6 · Calculation
An inhibition zone has diameter 18 mm, including a 6 mm disc. Calculate the clear area outside the disc.
Answer:
Clear area = pi x 9² - pi x 3² = 72pi, approximately 226 mm². State whether a reported diameter or area includes the disc. Use the same definition for every treatment.
Question 32 · Spec 3.4.4, RP6 · Required practical 6 · Analysis
Why does a larger inhibition zone not always prove that one antimicrobial is more effective in patients?
Answer:
Zone size also depends on diffusion through agar, concentration, disc loading and microbial susceptibility. Laboratory conditions differ from conditions in the body. Compare controlled experiments and use clinical evidence before drawing conclusions about treatment effectiveness.
Question 33 · Spec 3.4.5 · Species · Recall
What is the biological species definition, and where can it be difficult to apply?
Answer:
Members of a species can interbreed to produce fertile offspring. The definition can be difficult to apply to asexual organisms, fossils and geographically separated populations whose ability to interbreed cannot be tested easily.
Question 34 · Spec 3.4.5 · Courtship · Application
How can courtship behaviour support species recognition?
Answer:
Species-specific signals help individuals recognise a suitable mate of the same species and coordinate mating. Failure to respond to another species' signals can reduce interbreeding and help maintain reproductive separation.
Question 35 · Spec 3.4.5 · Taxonomic hierarchy · Recall
Put the major taxonomic ranks in order from broadest to most specific.
Answer:
Domain, kingdom, phylum, class, order, family, genus and species. Each taxon fits within a broader group in the hierarchy. A phylogenetic classification aims to reflect evolutionary origins and relationships.
Question 36 · Spec 3.4.5 · Binomial naming · Recall
How should a scientific binomial name be written?
Answer:
Use the genus name followed by the species name, for example Homo sapiens. Capitalise the genus, use a lower-case species name and italicise both when typed. Universal naming reduces ambiguity caused by different common names.
Question 37 · Spec 3.4.5 · Phylogenetic evidence · Analysis
How have DNA sequencing and immunology helped classification?
Answer:
Molecular comparisons provide evidence of inherited similarities that may be obscured by outward appearance. DNA sequences and immunological similarity between proteins can help infer common ancestry and revise classifications when they conflict with older groupings.
Question 38 · Spec 3.4.7 · Sequence comparison · Analysis
What can similar DNA or amino acid sequences suggest about evolutionary relationships?
Answer:
Fewer sequence differences usually suggest a more recent common ancestor when comparing the same homologous region or protein. Use comparable sequence lengths and multiple lines of evidence. Differences can accumulate at different rates, so counts alone have limitations.
Question 39 · Spec 3.4.5 · Immunological comparison · Application
How can an antigen-antibody comparison provide evidence about relatedness?
Answer:
Antibodies raised against one species' protein may also bind similar proteins in another species. Greater binding or precipitation under controlled conditions can suggest greater protein similarity and closer evolutionary relationships. The inference depends on comparing equivalent proteins and methods.
Question 40 · Spec 3.4.6 · Species richness · Recall
What is species richness, and why is it insufficient on its own to describe diversity?
Answer:
Species richness is the number of species present in a community. It ignores how individuals are distributed among those species. Two communities can have equal richness but different diversity if one is dominated by a single species.
Question 41 · Spec 3.4.6 · Index of diversity · Recall
State AQA's index of diversity equation and define its symbols.
Answer:
d = N(N - 1)/sum[n(n - 1)]. N is the total number of organisms of all species, and n is the number belonging to each species. Calculate n(n - 1) for each species, then sum these values.
Question 42 · Spec 3.4.6, MS 2.3 · Diversity calculation · Calculation
A sample contains 5, 3 and 2 individuals of three species. Calculate d.
Answer:
N = 10, so N(N - 1) = 90. The denominator is (5 x 4) + (3 x 2) + (2 x 1) = 28. Therefore d = 90/28 = 3.21 to three significant figures.
Question 43 · Spec 3.4.6 · Interpreting diversity · Analysis
How should values of AQA's index of diversity be compared?
Answer:
For this form of the index, a higher value indicates greater diversity, reflecting richness and relative abundance. Compare samples collected with similar effort and methods. Different published diversity indices use different formulae and scales, so identify the formula before interpreting values.
Question 44 · Spec 3.4.6 · Farming and biodiversity · Application
Why can intensive farming reduce biodiversity?
Answer:
Removing hedges and other habitats reduces available niches. Monocultures reduce plant diversity, herbicides remove wild plants and pesticides can affect non-target organisms. Fertiliser enrichment may favour a few competitive species. Effects extend through food webs.
Question 45 · Spec 3.4.6 · Conservation and farming · Analysis
How can biodiversity conservation be balanced with food production?
Answer:
Habitat margins, hedgerows, mixed cropping and targeted pesticide use can support more species. Evaluate evidence for biodiversity gains alongside yield, cost and land requirements. A balanced conclusion should address both food supply and conservation using the data provided.
Question 46 · Spec 3.4.7 · Measuring genetic diversity · Analysis
Why can direct DNA comparison be more informative than comparing appearance alone?
Answer:
Appearance is affected by environment, gene interactions and the alleles chosen for measurement. DNA comparisons reveal sequence variation directly, including variation without an obvious visible effect. The value of a comparison still depends on representative sampling and the regions examined.
Question 47 · Spec 3.4.7 · Sampling variation · Practical
How should a sample be collected to investigate variation within a species?
Answer:
Define the population and use a method that gives individuals an appropriate random chance of selection. Use a sufficiently large sample and a consistent measurement method. Random selection reduces selection bias and helps make the sample representative.
Question 48 · Spec 3.4.7, MS 1.10 · Mean and standard deviation · Analysis
Two samples have the same mean leaf length but different standard deviations. What can you conclude?
Answer:
The sample with the larger standard deviation has more spread in leaf lengths around its mean. The means alone conceal that difference. Standard deviation describes variability in the data, while uncertainty in the estimated mean also depends on sample size.
Question 49 · Spec 3.4.7, MS 1.3, MS 1.7 · Displaying variation · Analysis
When should a bar chart, histogram or scatter graph be used?
Answer:
Use a bar chart for separate categories, with gaps between bars. Use a histogram for continuous data grouped into intervals, with adjoining bars and frequency density where widths differ. Use a scatter graph to examine association between two quantitative variables.
Question 50 · Spec 3.4.3, 3.4.4 · Synoptic link · Synoptic
How do mutation, meiosis, fertilisation and selection have different roles in evolution?
Answer:
Mutation creates new alleles. Meiosis reshuffles existing alleles through crossing over and independent segregation. Random fertilisation combines gametes. Selection changes allele frequencies through differences in reproductive success. Together these processes generate and act on heritable variation.

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