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A-Level · Biology

AB4 Genetic Information, Variation and Relationships

AQA 7402/7405/7408 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about the structure of DNA.

(a)State the three components of a single DNA nucleotide.(3)

(b)Describe how individual nucleotides join together to form a polynucleotide strand.(2)

(c)Explain why DNA is a chemically stable molecule, well suited to storing genetic information over long periods of time.(3)

(d)A section of DNA contains 5,200,000 bases in total. 30% of these bases are adenine. Calculate the number of guanine bases in this section of DNA.(2)

(Total for Question 1 is 10 marks)

This question is about semi-conservative DNA replication.

(a)Name the enzyme that unwinds the DNA double helix at the start of replication and state the type of bond it breaks.(2)

(b)Describe the role of DNA polymerase in semi-conservative replication.(3)

(c)Meselson and Stahl grew bacteria in a medium containing only heavy nitrogen (15N) so that all their DNA became heavy. The bacteria were then transferred to a medium containing only light nitrogen (14N) and allowed exactly one round of DNA replication. Predict, with a reason, the density of DNA that would be found after this one round of replication if DNA replication is semi-conservative.(3)

(Total for Question 2 is 8 marks)

This question is about the genetic code and protein synthesis. Use the following codon table extract where needed: AUG = methionine (start), CCA = proline, CGA = arginine, CAU = histidine, UUU = phenylalanine.

(a)State three features of the genetic code.(3)

(b)Describe the process of transcription in a eukaryotic cell, from RNA polymerase binding to DNA through to the production of mature mRNA.(4)

(c)A gene, containing no introns, has a coding sequence of 1,020 base pairs. Calculate the maximum number of amino acids that could be coded for by this gene, excluding the stop codon.(2)

(d)Which of the following correctly describes the relationship between a codon and an anticodon?
A) A codon is a triplet on tRNA; an anticodon is a triplet on mRNA.
B) A codon is a triplet on mRNA; an anticodon is a complementary triplet on tRNA.
C) A codon and an anticodon are both found on the same tRNA molecule.
D) A codon codes for a start signal only; an anticodon codes for amino acids.(1)

A) A codon is a triplet on tRNA; an anticodon is a triplet on mRNA.
B) A codon is a triplet on mRNA; an anticodon is a complementary triplet on tRNA.
C) A codon and an anticodon are both found on the same tRNA molecule.
D) A codon codes for a start signal only; an anticodon codes for amino acids.

(Total for Question 3 is 10 marks)

The original (coding strand) DNA sequence for part of a gene is: ATG CCA TTT

(a)Give the mRNA sequence transcribed from this coding strand and, using the codon table in Question 3, determine the sequence of amino acids it codes for.(2)

(b)A substitution mutation changes the 4th DNA base from C to G, giving the coding strand ATG CGA TTT. State the type of mutation this is and, using the codon table, give the new amino acid sequence.(3)

(c)A second mutation deletes the 4th base (C) from the original DNA, giving the coding strand ATGCATTT (with the reading frame shifted from that point on). Explain why this type of mutation is likely to have a greater effect on the polypeptide than the substitution in part (b).(3)

(d)The complete, non-mutated gene is 999 base pairs long and contains no introns. Calculate how many amino acids the complete polypeptide would contain, assuming the sequence includes exactly one stop codon.(2)

(Total for Question 4 is 10 marks)

This question is about meiosis and genetic variation.

(a)State the number of chromosomes present in a human somatic (body) cell and in a human gamete.(2)

(b)Describe how independent assortment during meiosis I contributes to genetic variation.(3)

(c)Calculate the number of genetically different gamete combinations (due to independent assortment alone) that can be produced by a species with a diploid number of 8.(2)

(d)Explain how crossing over during prophase I further increases genetic variation, beyond that produced by independent assortment alone.(3)

(Total for Question 5 is 10 marks)

Explain how meiosis and random fertilisation increase genetic variation within a species, and explain why this genetic variation is important for the survival of the species.

(Total for Question 6 is 6 marks)

A group of students investigated the biodiversity of a rocky shore habitat using a 0.25 m² quadrat placed at 10 random points along a 20 m transect. They counted the number of individuals of each species present. Their results are shown below.
Species A (limpet): 24 individuals
Species B (barnacle): 56 individuals
Species C (dog whelk): 8 individuals
Species D (periwinkle): 12 individuals
Total (N) = 100 individuals

(a)Describe how the students could have used random sampling to choose the position of each quadrat along the transect, ensuring their method avoided bias.(3)

(b)Simpson's Index of Diversity can be calculated using the formula:
D = 1 - [ sum( n/N )² ]
where n = number of individuals of one species and N = total number of individuals of all species.
Calculate the Index of Diversity, D, for this habitat using the data given. Give your answer to 3 significant figures.(4)

(c)A second, more disturbed area of the same shore was found to have an Index of Diversity of 0.31. Using this value and your answer to part (b), explain which of the two areas has the higher biodiversity and suggest one possible reason for the difference.(3)

(d)State two variables, other than species number and abundance, that the students should have kept the same when comparing the two areas to make the comparison valid.(2)

(Total for Question 7 is 12 marks)

This question is about species, taxonomy and evidence of evolutionary relationships.

(a)The taxonomic hierarchy used to classify organisms, from largest to smallest group, is: Domain, Kingdom, ___, Class, Order, Family, Genus, Species. Name the missing group.(1)

(b)State the biological definition of a species.(2)

(c)Explain why courtship behaviour between two organisms is used as evidence that they belong to the same species.(3)

(d)Scientists compare the base sequence of a specific gene (for example, the gene coding for cytochrome c) in different species to assess how closely related they are. Explain how the number of differences in the base sequence between two species can be used to determine how recently they diverged from a common ancestor.(3)

(Total for Question 8 is 9 marks)

In a population of 500 ladybirds, a single gene controls wing case colour, with allele M (melanic/black) dominant to allele m (red). Genotyping of the population gave the following results: MM = 20, Mm = 160, mm = 320.

(a)Define the term 'gene pool'.(1)

(b)Calculate the frequency of the m allele in this population.(3)

(c)Suggest and explain one factor, other than mutation, that could reduce the genetic diversity of this ladybird population in the future.(3)

(d)Explain why natural selection acting on this population could not occur without genetic diversity.(2)

(Total for Question 9 is 9 marks)

As part of an investigation into diversity, students studied whether the height above the low water mark on a rocky shore is correlated with the mean shell length of periwinkles at six sites.
Site 1: height 0.5 m, mean shell length 28 mm
Site 2: height 1.0 m, mean shell length 25 mm
Site 3: height 1.5 m, mean shell length 22 mm
Site 4: height 2.0 m, mean shell length 19 mm
Site 5: height 2.5 m, mean shell length 15 mm
Site 6: height 3.0 m, mean shell length 12 mm

(a)State what is meant by a negative correlation.(1)

(b)Rank the height and shell length data (rank 1 = smallest value in each set), calculate d² for each site, and hence calculate Spearman's rank correlation coefficient, rs, using the formula:
rs = 1 - [ 6 x sum(d²) / (n(n² - 1)) ](5)

(c)The critical value of rs for n = 6 at the 5% (p = 0.05) significance level is 0.886. Using this critical value and your answer to part (b) (or ft your value), explain what conclusion can be drawn about the relationship between height on the shore and mean periwinkle shell length.(3)

(Total for Question 10 is 9 marks)

Students hypothesised that a species of snail was randomly distributed across a woodland floor. They divided the woodland into 4 equal-sized zones and counted the number of snails found in each zone:
Zone 1: 8, Zone 2: 22, Zone 3: 15, Zone 4: 15 (Total = 60)

(a)State the null hypothesis for this investigation.(1)

(b)Explain why the chi-squared test, rather than the Spearman's rank correlation test, is an appropriate statistical test to use with this data.(2)

(c)Calculate the expected number of snails per zone if they were randomly distributed, and complete the chi-squared calculation to find the value of chi-squared (x²). Show your working.(4)

(d)The critical value of chi-squared at the 5% significance level for 3 degrees of freedom is 7.815. Using this value and your answer to part (c) (or ft your value), state and explain the conclusion that should be drawn.(2)

(Total for Question 11 is 9 marks)

This question links genetic diversity, natural selection and speciation.

(a)Compare allopatric and sympatric speciation, giving the type of isolating mechanism associated with each.(4)

(b)Explain, using ideas about natural selection, how two geographically isolated populations of the same original species could eventually become two separate species.(5)

(c)A population of 2,000 insects is founded by just 5 individuals that colonise a remote island (a founder event). Explain, using this figure, why the new island population is likely to show reduced genetic diversity compared with the original mainland population.(2)

(Total for Question 12 is 11 marks)

Mark scheme · AB4 Genetic Information, Variation and Relationships

Question 1

(a) B1 a phosphate group
(a) B1 a (deoxyribose) pentose sugar
(a) B1 a nitrogen-containing (organic) base
(a) Answer: Phosphate group, deoxyribose sugar, nitrogen-containing base
(b) B1 a condensation reaction occurs (releasing a water molecule)
(b) B1 forms a phosphodiester bond between the phosphate group of one nucleotide and the sugar of the next
(b) Answer: Condensation reaction forms a phosphodiester bond between adjacent nucleotides.
(c) B1 hydrogen bonds form between specific complementary base pairs (A with T, C with G), holding the two strands together
(c) B1 the double helix is a large, coiled molecule that protects the bases (which carry the genetic code) on the inside of the molecule from chemical/physical damage
(c) B1 the sugar-phosphate backbone is held together by strong covalent (phosphodiester) bonds, so the molecule does not break down easily
(c) Answer: DNA is stable due to complementary H-bonding, a protective coiled double-helix structure, and a strong covalent backbone.
(d) M1 recognises %A = %T = 30%, so %G = %C = (100 - 60) / 2 = 20% oe
(d) A1 1,040,000 (guanine bases) cao
(d) Answer: 1,040,000 guanine bases

Question 2

(a) B1 DNA helicase
(a) B1 breaks hydrogen bonds (between complementary base pairs)
(a) Answer: DNA helicase; breaks hydrogen bonds between base pairs.
(b) B1 aligns free (activated) nucleotides against the exposed bases on each template strand by complementary base pairing
(b) B1 catalyses formation of (phosphodiester) bonds between adjacent nucleotides, building the new sugar-phosphate backbone
(b) B1 works in one direction only, so synthesis on one template strand (the lagging strand) is discontinuous, e.g. as Okazaki fragments
(b) Answer: DNA polymerase adds complementary nucleotides and joins them via phosphodiester bonds, working directionally.
(c) B1 DNA of intermediate density would be found (a single band, between the heavy and light bands)
(c) B1 because each new DNA molecule contains one original (heavy, 15N) strand and one newly made (light, 14N) strand
(c) B1 this differs from the conservative model, which would predict two separate bands (one fully heavy, one fully light) instead of one intermediate band
(c) Answer: A single band of intermediate density DNA, because each molecule is a hybrid of one old heavy strand and one new light strand.

Question 3

(a) B1 it is a triplet code (three bases code for one amino acid)
(a) B1 it is degenerate (more than one codon can code for the same amino acid)
(a) B1 it is non-overlapping (each base is only read as part of one codon); accept universal (the same in all living organisms)
(a) Answer: Triplet, degenerate, non-overlapping (accept universal).
(b) B1 RNA polymerase binds to a promoter region on DNA and hydrogen bonds break, unwinding/exposing the template strand
(b) B1 free RNA nucleotides align with exposed bases on the template strand by complementary base pairing (A pairs with U)
(b) B1 RNA polymerase joins adjacent RNA nucleotides via phosphodiester bonds, forming pre-mRNA
(b) B1 pre-mRNA is spliced: introns (non-coding sections) are removed and exons (coding sections) joined to form mature mRNA, which leaves the nucleus via a nuclear pore
(b) Answer: RNA polymerase unwinds DNA, builds pre-mRNA by complementary base pairing, then splicing removes introns to give mature mRNA.
(c) M1 1020 / 3 = 340 (codons)
(c) A1 339 (amino acids) cao
(c) Answer: 339 amino acids
(d) B1 B cao
(d) Answer: B

Question 4

(a) M1 mRNA: AUG CCA UUU
(a) A1 methionine - proline - phenylalanine ft from mRNA
(a) Answer: AUG CCA UUU; methionine - proline - phenylalanine
(b) B1 substitution (point mutation)
(b) B1 ft new mRNA/amino acid sequence: AUG CGA UUU giving methionine - arginine - phenylalanine
(b) B1 described as a missense mutation because only the amino acid at that one position changes; the rest of the sequence (reading frame) is unaffected
(b) Answer: Substitution (missense); methionine - arginine - phenylalanine.
(c) B1 deletion of a base causes a frameshift mutation - all codons downstream of the deletion point are read differently
(c) B1 this changes most/all of the amino acid sequence after the mutation point, not just one amino acid, potentially altering the tertiary structure of the whole protein
(c) B1 this could produce a premature stop codon or a completely non-functional protein, whereas the substitution changed only one amino acid
(c) Answer: The deletion causes a frameshift affecting every codon downstream, unlike the substitution which affects only one codon.
(d) M1 999 / 3 = 333 (codons)
(d) A1 332 (amino acids) cao
(d) Answer: 332 amino acids

Question 5

(a) B1 46 (23 pairs) in a somatic cell
(a) B1 23 in a gamete
(a) Answer: 46 in a somatic cell; 23 in a gamete.
(b) B1 homologous pairs of chromosomes line up in a random orientation at the equator during metaphase I
(b) B1 each pair consists of one maternal and one paternal chromosome, and either can end up in either daughter cell independently of how other pairs orientate
(b) B1 this creates many different possible combinations of maternal and paternal chromosomes in the resulting gametes
(b) Answer: Random orientation of homologous pairs at metaphase I produces many chromosome combinations in gametes.
(c) M1 haploid number n = 8 / 2 = 4
(c) A1 2⁴ = 16 cao
(c) Answer: 16 different gamete combinations
(d) B1 non-sister chromatids of homologous chromosomes pair up (forming a bivalent) and become twisted around each other at points called chiasmata
(d) B1 sections of chromatids (alleles) are exchanged/swapped between the non-sister chromatids
(d) B1 this produces new/recombinant combinations of alleles on a chromosome not present in either parent, increasing variation beyond that from independent assortment alone (which only shuffles whole chromosomes)
(d) Answer: Crossing over exchanges sections between non-sister chromatids, creating new allele combinations on chromosomes.

Question 6

Level 3 (5-6): A detailed and coherent explanation that links specific mechanisms (independent assortment, crossing over, random fertilisation) to genetic variation, and clearly explains the importance of this variation for the survival of a species facing a changing environment. Appropriate scientific terminology is used accurately throughout.
Level 2 (3-4): The explanation includes several relevant mechanisms of variation and makes some link to the importance of variation for survival, but the explanation may lack detail, full accuracy or coherence in places.
Level 1 (1-2): Basic or isolated statements that meiosis and/or fertilisation cause variation, with little or no explanation of why this variation is important. Scientific detail is limited.
Level 0 (0): No relevant content.
Indicative content:
- Independent assortment of homologous chromosomes during meiosis I produces many combinations of maternal and paternal chromosomes in gametes.
- Crossing over (chiasmata formation) between non-sister chromatids in prophase I creates new combinations of alleles on chromosomes (recombinants).
- Meiosis therefore produces genetically different (non-identical) gametes.
- Fertilisation is random: any one of millions of genetically different gametes from one parent can fuse with any one of millions from the other parent.
- This creates a very large number of possible genetically different offspring (zygotes).
- Genetic variation means some individuals in a population may possess alleles/characteristics better suited to a changing environment.
- Variation provides the raw material for natural selection; without variation a population cannot evolve or adapt.
- If the environment changes (for example a new disease or climate change), a genetically varied population is more likely to contain individuals that survive and reproduce, reducing the risk of extinction of the species.

Question 7

(a) B1 divide the transect into a numbered grid/sections (using a tape measure marked at regular intervals)
(a) B1 use a random number generator/table to generate coordinate or position numbers
(a) B1 place the quadrat at the positions given by the random numbers, rather than the students choosing where to place it, avoiding the students unconsciously selecting areas with more or fewer visible organisms
(a) Answer: Number the transect, generate random positions, place quadrats only at those positions.
(b) M1 calculates n/N for each species (0.24, 0.56, 0.08, 0.12)
(b) M1 squares each proportion and sums them: 0.0576 + 0.3136 + 0.0064 + 0.0144 = 0.392
(b) dM1 subtracts sum from 1: D = 1 - 0.392
(b) A1 D = 0.608 (3 sf) cao
(b) Answer: D = 0.608
(c) B1 ft the first area (D = 0.608, or candidate's value from (b)) has the higher biodiversity, because a value of D closer to 1 indicates greater species diversity/a more even distribution of species
(c) B1 a valid suggested reason, e.g. the disturbed area may have been affected by trampling/pollution, or a dominant coloniser species may be outcompeting others, reducing the number of species or making one species very dominant
(c) B1 further valid development, e.g. lower diversity reduces the community's stability/resilience to further environmental change
(c) Answer: The first area has higher biodiversity (D closer to 1); disturbance likely reduced species evenness/richness in the second area.
(d) B1 quadrat size (area sampled)
(d) B1 number of sample points/quadrats used (sampling effort); accept time of year, tide height or time spent searching
(d) Answer: Quadrat size and number of samples (accept time of year/tide height).

Question 8

(a) B1 Phylum
(a) Answer: Phylum
(b) B1 a group of organisms with similar characteristics/morphology (and genetics)
(b) B1 that can interbreed to produce fertile offspring
(b) Answer: A group of similar organisms able to interbreed and produce fertile offspring.
(c) B1 courtship behaviour is species-specific: it involves a stereotyped, recognisable sequence of behaviours unique to a species
(c) B1 if two organisms recognise and respond correctly to each other's courtship behaviour (and successfully mate), this suggests they are of the same species (able to interbreed)
(c) B1 courtship behaviour also ensures individuals identify a mate of the correct species, preventing hybridisation or wasted reproductive effort
(c) Answer: Courtship behaviour is species-specific, so mutual recognition indicates the organisms are the same, interbreeding species.
(d) B1 the fewer the differences in the DNA base sequence (or the amino acid sequence of the coded protein) between two species, the more closely related they are
(d) B1 mutations accumulate in DNA over time at a roughly constant/measurable rate, so more differences indicate a greater length of time since the two species diverged from a common ancestor
(d) B1 this evidence can be used to construct or confirm phylogenetic trees showing evolutionary relationships, supporting or independent of evidence from physical characteristics
(d) Answer: More sequence differences indicate more time (and mutations) since divergence, allowing construction of phylogenetic trees.

Question 9

(a) B1 all the alleles of all the genes present in a population at a given time
(a) Answer: The complete set of alleles present in a population at a given time.
(b) M1 total number of alleles = 500 x 2 = 1000
(b) M1 number of m alleles = (320 x 2) + (160 x 1) = 800
(b) A1 frequency = 800 / 1000 = 0.8 cao
(b) Answer: Allele frequency of m = 0.8 (80%)
(c) B1 identifies a valid factor, e.g. a genetic bottleneck (a dramatic reduction in population size, due to disease or habitat loss) or a founder effect (a small number of individuals colonising a new isolated habitat)
(c) B1 explains that only a small, potentially non-representative, sample of the alleles present in the original gene pool would survive/be passed on
(c) B1 so some alleles may be lost completely (allele frequencies change), reducing overall genetic diversity and increasing the risk of extinction/reducing the ability to adapt to future environmental change
(c) Answer: A genetic bottleneck or founder effect would reduce diversity because only a small, unrepresentative sample of alleles would survive/found the new population.
(d) B1 natural selection depends on differences (variation) between individuals in their alleles/phenotypes, giving some individuals a selective (survival/reproductive) advantage over others in a given environment
(d) B1 without genetic diversity, all individuals would be genetically identical/equally suited (or unsuited) to the environment, so there would be nothing for selection pressures to act on and allele frequencies could not change over time
(d) Answer: Without variation there is no differential survival/reproduction for selection to act on, so allele frequencies cannot change.

Question 10

(a) B1 as one variable increases, the other variable decreases
(a) Answer: As one variable increases, the other decreases.
(b) M1 correctly ranks height data as 1, 2, 3, 4, 5, 6 for sites 1 to 6 (already in ascending order)
(b) M1 correctly ranks shell length data as 6, 5, 4, 3, 2, 1 for sites 1 to 6 (values decrease as height increases)
(b) M1 calculates d values (-5, -3, -1, 1, 3, 5) and d² values (25, 9, 1, 1, 9, 25), sum d² = 70
(b) M1 substitutes correctly: rs = 1 - (6 x 70) / (6 x 35) = 1 - 420/210
(b) A1 rs = -1 cao
(b) Answer: rs = -1
(c) B1 ft the magnitude of the calculated rs (1) is greater than the critical value (0.886)
(c) B1 so the correlation is statistically significant (unlikely to be due to chance) at the 5%/p=0.05 level
(c) B1 there is a significant negative correlation between height on the shore and mean shell length (shell length decreases as height above the low water mark increases); accept reject the null hypothesis
(c) Answer: The correlation is significant: shell length decreases significantly as shore height increases.

Question 11

(a) B1 there is no significant difference between the observed and expected number of snails in each zone (the snails are randomly distributed across the woodland floor)
(a) Answer: There is no significant difference between observed and expected snail numbers per zone.
(b) B1 chi-squared tests for association/difference between observed and expected frequencies of categoric data, whereas Spearman's rank tests for correlation between two variables that can be ranked
(b) B1 the data here are counts/frequencies in discrete categories (zones), not two measured variables to be correlated, so chi-squared is the appropriate test
(b) Answer: Chi-squared suits categoric count data being tested against expected frequencies, not correlation between two ranked variables.
(c) M1 expected number per zone = 60 / 4 = 15
(c) M1 (O - E)² calculated correctly for each zone: 49, 49, 0, 0
(c) M1 (O - E)² / E summed correctly: 49/15 + 49/15 + 0 + 0 = 6.53
(c) A1 x² = 6.53 (accept 6.5 to 6.54) cao
(c) Answer: x² = 6.53
(d) B1 ft the calculated value of x² (6.53) is less than the critical value (7.815)
(d) B1 so there is no significant difference between the observed and expected distribution (the null hypothesis is accepted, any difference is due to chance): the snails are randomly distributed across the woodland floor
(d) Answer: x² < critical value, so the null hypothesis is accepted; snails are randomly distributed.

Question 12

(a) B1 allopatric speciation occurs when populations are separated by a physical/geographical barrier (e.g. a river, mountain range or ocean)
(a) B1 geographical isolation prevents gene flow between the two populations
(a) B1 sympatric speciation occurs without geographical separation; populations become reproductively isolated while occupying the same area (e.g. behavioural, mechanical, seasonal or ecological isolation)
(a) B1 reproductive isolation prevents interbreeding/gene flow between the two populations even though they occupy the same area
(a) Answer: Allopatric = geographical isolation; sympatric = reproductive isolation without geographical separation.
(b) B1 mutation introduces new alleles into each population independently (different mutations may arise in each population)
(b) B1 within each population, the environment/selection pressures may differ (e.g. different climate, food availability or predators)
(b) B1 individuals with alleles advantageous in their local environment are more likely to survive and reproduce (differential reproductive success)
(b) B1 these advantageous alleles are passed on to offspring, so allele frequencies in each population gradually change, over many generations, in different directions
(b) B1 eventually the genetic differences between the two populations become so great that, even if reunited, individuals could no longer interbreed successfully to produce fertile offspring, so they are now two separate species
(b) Answer: Independent mutation and differing selection pressures cause divergent allele frequency changes until reproductive isolation results in two species.
(c) B1 only a small number of individuals (5) founded the new population, so only a small, possibly unrepresentative, sample of the alleles present in the original mainland gene pool would have been carried to the island
(c) B1 as the population grows it does so from this limited allele pool, so the genetic diversity of the island population will be lower than that of the mainland population (some mainland alleles may be completely absent from the island population)
(c) Answer: A founder population of 5 carries only a small, unrepresentative sample of mainland alleles, reducing diversity in the descendant population.

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