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A-Level · Biology

AB10 Biology: Genetics, Populations and Evolution

AQA 7402 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A gardener grows a variety of runner bean with a diploid chromosome number of 2n = 22 (n = 11 pairs of homologous chromosomes). Like all organisms that reproduce sexually, the genetic variation seen among the offspring of two runner bean plants arises largely from events that occur during meiosis.

(a)Explain how crossing over during meiosis I can produce gametes with new combinations of alleles that were not present in either parent plant.(2)

(b)Explain how independent assortment of homologous chromosomes during meiosis I increases the genetic variation of the gametes produced.(2)

(c)Ignoring any effect of crossing over, calculate the number of genetically different gamete combinations that independent assortment alone could produce in this runner bean plant.(2)

(Total for Question 1 is 6 marks)

A researcher at Fenwick Down nature reserve is investigating the inheritance of shell banding pattern in the grove snail, Cepaea nemoralis. For the purposes of this question, assume shell pattern is controlled by a single gene with two alleles: U (unbanded shell, dominant) and u (banded shell, recessive). The researcher crosses two snails that are both heterozygous (Uu) and rears 200 offspring to hatching.

(a)Complete a genetic diagram to show the expected genotype ratio of the offspring from this cross, and state the expected phenotype ratio.(2)

(b)State a suitable null hypothesis for a chi-squared test on these data.(1)

(c)Using the 3:1 ratio and the total sample of 200 offspring, calculate the expected number of unbanded offspring and the expected number of banded offspring.(2)

(d)The researcher actually observed 155 unbanded and 45 banded offspring. Using the formula chi² = sum[ (O - E)² / E ], calculate the chi-squared value for these data. Show your working.(3)

(e)The critical value of chi-squared at the 5% (p = 0.05) significance level with 1 degree of freedom is 3.84. Compare your calculated value with this critical value and state a conclusion about the null hypothesis.(2)

(Total for Question 2 is 10 marks)

In humans, red-green colour blindness is caused by a recessive allele (n) located on the X chromosome; the dominant allele (N) gives normal colour vision. A woman who is a known carrier of the n allele has normal colour vision. She has children with a man who has normal colour vision and does not carry the allele.

(a)Using the symbols given, state the genotypes of the mother and the father.(2)

(b)Complete a genetic diagram for this cross and use it to determine the probability that a son born to this couple will be colour blind.(3)

(c)Explain, in terms of the chromosomal location of the allele, why X-linked recessive conditions such as colour blindness are much more common in males than in females.(3)

(d)Explain, using ideas about dominance, why natural selection acting against the colour-blindness phenotype is unlikely to remove the n allele completely from the population's gene pool.(3)

(Total for Question 3 is 11 marks)

The human ABO blood group gene has three alleles: I^A and I^B, which are codominant with each other, and I^O, which is recessive to both. A father who is heterozygous for blood group A and a mother who is heterozygous for blood group B have children together.

(a)State the genotype of a person with blood group O.(1)

(b)Complete a genetic diagram for this cross and use it to determine the probability that a child of this couple has blood group O.(3)

(c)Explain what is meant by the term codominance, using the I^A and I^B alleles as your example.(2)

(d)Explain what is meant by the term multiple alleles, and explain why the ABO gene can have three alleles in the population's gene pool even though any one person can carry only two of them.(2)

(Total for Question 4 is 8 marks)

In a breed of dog, coat colour is controlled by two genes on different chromosomes. Gene 1 has alleles B (black pigment, dominant) and b (brown pigment, recessive). Gene 2 has alleles E (allows pigment to be deposited in the coat, dominant) and e (prevents pigment being deposited, so the coat appears yellow regardless of genotype at gene 1, recessive). Two dogs that are both double heterozygotes (BbEe) are crossed.

(a)Define the term epistasis.(2)

(b)Complete a dihybrid genetic diagram for the cross BbEe x BbEe and determine the genotype ratio of the offspring, before considering the effect of epistasis.(3)

(c)Given that the ee genotype prevents pigment deposition regardless of the genotype at the B gene, state the phenotypic ratio of black : brown : yellow puppies expected from this cross.(1)

(d)Explain why the phenotypic ratio in part (c) differs from the standard 9:3:3:1 ratio usually expected from a dihybrid cross.(2)

(Total for Question 5 is 8 marks)

A population of two-spot ladybirds, Adalia bipunctata, living at Fenwick Down nature reserve shows two colour forms. For the purposes of this question, assume colour is controlled by a single gene with two alleles: M (melanic, black form, dominant) and m (red form, recessive). A survey of 1000 ladybirds in this population found that 360 were the red form.

(a)State the Hardy-Weinberg equations that link allele frequencies (p and q) and genotype frequencies in a population.(1)

(b)Calculate the frequency of the m allele (q) and the frequency of the M allele (p) in this population.(3)

(c)Using your values of p and q, calculate the number of ladybirds in this population of 1000 expected to be homozygous dominant (MM) and the number expected to be heterozygous (Mm).(3)

(d)State two conditions that must be met for a population to be in Hardy-Weinberg equilibrium.(2)

(e)A follow-up survey twenty years later found that q had risen to 0.75. Over this period, average local temperatures at Fenwick Down have risen due to climate change. Suggest, using ideas about thermal melanism and natural selection, why the allele frequencies in this population might have changed over this time.(3)

(Total for Question 6 is 12 marks)

Three scenarios below describe natural selection acting on continuously variable traits in different populations. For each scenario, name the type of natural selection taking place and describe its expected effect on the distribution of the trait in the population.

(a)Scenario 1: In a population of human babies, data collected over many years show that babies born with a birth mass close to the population mean (around 3.2 kg) have consistently had the highest survival rate to one year old, while babies with much lower or much higher birth mass have a lower survival rate. This pattern has remained stable over many decades. Name the type of selection and describe its effect on the distribution of birth mass in the population.(3)

(b)Scenario 2: In a population of a moth species living near a town, most individuals historically had pale-coloured wings, closely matching the colour of the lichen-covered tree bark on which they rested, making them well camouflaged from predatory birds. After nearby factories began releasing soot that darkened the tree bark over several decades, the proportion of dark-coloured (melanic) moths in the population increased substantially, generation after generation. Name the type of selection and describe its effect on the distribution of wing colour in the population.(3)

(c)Scenario 3: A population of finches on an island feeds only on seeds from two plant species: one produces small, soft seeds and the other produces large, hard seeds; no plants producing seeds of intermediate hardness grow on the island. Finches with small, shallow beaks can only feed efficiently on the soft seeds, and finches with large, deep beaks can only feed efficiently on the hard seeds; finches with intermediate beak depth feed inefficiently on both food sources. Name the type of selection and predict its effect on the distribution of beak depth in the population over time.(3)

(Total for Question 7 is 9 marks)

Genetic drift is the random change in allele frequencies from one generation to the next, due to chance events rather than natural selection.

(a)Define the term genetic drift.(2)

(b)Explain why genetic drift tends to have a much greater effect on allele frequencies in small populations than in large populations.(3)

(c)A small group of 15 grove snails (Cepaea nemoralis) from a mainland population is accidentally transported to a previously snail-free island, where they establish a new, isolated population. By chance, this founding group includes an unusually high proportion of yellow-shelled (bb) individuals compared with the mainland population. Explain, using this example, what is meant by the founder effect.(3)

(d)Suggest one event, other than colonising a new habitat, that could cause a population to undergo a genetic bottleneck, and explain the likely effect of this bottleneck on the genetic diversity of the population.(3)

(Total for Question 8 is 11 marks)

Two related concepts describe how new species can arise: allopatric speciation and sympatric speciation.

(a)State what is meant by the biological species concept.(2)

(b)A single population of freshwater fish living in a large lake becomes divided into two separate populations when a change in the river system creates a physical barrier between two smaller lakes. Explain how this could eventually lead to allopatric speciation.(5)

(c)A population of a flowering plant species growing in a single meadow, with no geographical barrier, gives rise to some offspring with a doubled chromosome number (polyploid) due to an error during meiosis. Explain how this could lead to sympatric speciation within the same meadow.(4)

(Total for Question 9 is 11 marks)

Evaluate the relative importance of natural selection and genetic drift in causing changes in allele frequencies within populations, and explain how the size of a population affects the balance between these two processes.

(Total for Question 10 is 6 marks)

Priya is investigating the population size of Cepaea nemoralis snails in a section of hedgerow at Fenwick Down using the mark-release-recapture method. She catches, marks and releases a first sample of snails, then returns 48 hours later to take a second sample. Her results are shown below.
Number of snails marked and released in the first sample (n1): 84
Number of snails caught in the second sample (n2): 96
Number of marked snails recaptured in the second sample (m2): 21
The Lincoln index formula is: estimated population size (N) = (n1 x n2) / m2

(a)Name this method of estimating population size.(1)

(b)Suggest three features that a suitable marking method for these snails must have, for this technique to give a valid estimate of population size.(3)

(c)Explain why Priya waited 48 hours between releasing the first sample and taking the second sample.(2)

(d)Using the Lincoln index formula given, calculate Priya's estimate of the total population size of snails in this section of hedgerow. Show your working.(3)

(e)State three assumptions that must be true for the Lincoln index to give a valid, reliable estimate of population size.(3)

(f)The following year, a new road is built alongside the hedgerow, splitting the snail population into two smaller, isolated sub-populations with no gene flow between them. Using ideas about genetic drift, explain how this fragmentation could affect the genetic diversity of the snail population at Fenwick Down over subsequent generations.(4)

(g)Define the term genetic diversity.(2)

(Total for Question 11 is 18 marks)

A population's gene pool is the complete set of alleles present in all the individuals of that population at a given time.

(a)Explain what is meant by the term gene pool.(2)

(b)Explain why genetic variation within a population's gene pool is essential for natural selection to bring about evolutionary change.(3)

(Total for Question 12 is 5 marks)

Mark scheme · AB10 Biology: Genetics, Populations and Evolution

Question 1

(a) B1 non-sister chromatids of a homologous pair become physically joined at a chiasma (during prophase I) and exchange section(s) of DNA/alleles
(a) B1 this produces recombinant chromatids carrying new combinations of alleles that differ from either parental chromosome
(a) Answer: Crossing over exchanges sections of DNA between non-sister chromatids of homologous chromosomes at chiasmata, producing recombinant chromosomes with new allele combinations.
(b) B1 each homologous pair orientates/lines up randomly and independently of every other pair at the equator during metaphase I
(b) B1 so maternal and paternal chromosomes are distributed to daughter cells in many different possible combinations
(b) Answer: Random, independent orientation of each homologous pair at metaphase I means maternal and paternal chromosomes end up distributed in many different combinations in the resulting gametes.
(c) M1 uses 2ⁿ with n = 11 (the haploid number)
(c) A1 2048 cao
(c) Answer: 2048

Question 2

(a) M1 correct gametes (U, u) from each parent combined correctly to give offspring genotypes UU : Uu : uU : uu = 1 : 2 : 1
(a) A1 expected phenotype ratio of unbanded : banded = 3 : 1
(a) Answer: 3 : 1 (unbanded : banded)
(b) B1 there is no significant difference between the observed numbers of unbanded and banded offspring and the numbers expected from a 3:1 ratio oe
(b) Answer: There is no significant difference between the observed and expected numbers of unbanded and banded offspring.
(c) M1 200 x 3/4 = 150 (unbanded)
(c) A1 200 x 1/4 = 50 (banded), both correct
(c) Answer: Expected unbanded = 150; expected banded = 50.
(d) M1 (155 - 150)² / 150 = 25/150 (= 0.17)
(d) M1 (45 - 50)² / 50 = 25/50 (= 0.5)
(d) A1 chi² = 0.67 (2 dp) (allow 0.667 or 2/3) cao
(d) Answer: chi² = 0.67 (awrt)
(e) M1 the calculated chi-squared value (0.67, or ecf from part (d)) is less than the critical value of 3.84
(e) A1 the null hypothesis is accepted; there is no significant difference between observed and expected numbers, so the data are consistent with a 3:1 ratio and simple monohybrid inheritance with complete dominance
(e) Answer: 0.67 < 3.84, so the null hypothesis is accepted; the data are consistent with a 3:1 ratio.

Question 3

(a) B1 mother: X^N Xⁿ
(a) B1 father: X^N Y
(a) Answer: Mother X^N Xⁿ; father X^N Y.
(b) M1 correct gametes shown: mother X^N, Xⁿ; father X^N, Y
(b) M1 correct offspring genotypes shown: X^N X^N, X^N Xⁿ, X^N Y, Xⁿ Y, each with probability 1/4
(b) A1 probability a son is colour blind = 1/2 oe (of the two possible male genotypes, X^N Y and Xⁿ Y, each is equally likely)
(b) Answer: 1/2 (50%)
(c) M1 males are hemizygous for X-linked genes/have only one X chromosome, paired with a Y chromosome that does not carry this gene locus
(c) M1 so a male carrying a single copy of the recessive allele always shows the recessive phenotype, as there is no second allele to mask it
(c) M1 females have two X chromosomes, so a female needs two copies of the recessive allele (homozygous) to show the condition; a female with only one copy is an unaffected carrier
(c) Answer: Males have only one X chromosome, so a single recessive allele is always expressed; females have two X chromosomes, so the recessive allele can be masked in heterozygotes, making the condition much rarer in females.
(d) M1 natural selection can only act on the phenotype produced (in females, the homozygous recessive genotype), not directly on the allele itself
(d) M1 in heterozygous females the recessive allele is masked by the dominant allele, so carriers are unaffected and are not selected against
(d) M1 the allele therefore persists, hidden, in the gene pool within unaffected carriers, maintaining it at low frequency rather than being eliminated
(d) Answer: Selection cannot act directly on a masked recessive allele; because heterozygous carriers are unaffected, the allele is protected from selection and persists at low frequency in the gene pool rather than being eliminated.

Question 4

(a) B1 I^O I^O (or ii) oe
(a) Answer: I^O I^O
(b) M1 correct gametes: father I^A, I^O; mother I^B, I^O
(b) M1 correct offspring genotypes: I^A I^B, I^A I^O, I^B I^O, I^O I^O, each with probability 1/4
(b) A1 probability of blood group O = 1/4 cao
(b) Answer: 1/4 (25%)
(c) B1 codominance is when two different alleles of a gene are both expressed in the phenotype of a heterozygote, with neither allele masking the other
(c) B1 an individual with genotype I^A I^B expresses both the A and B antigens on their red blood cells (blood group AB), rather than one allele masking the other
(c) Answer: Codominance means both alleles in a heterozygote are expressed in the phenotype; an I^AI^B individual shows both A and B antigens (blood group AB) rather than one allele being masked.
(d) B1 multiple alleles means more than two alleles of a gene exist within a population's gene pool, even though a diploid individual can carry only two of them at any one locus
(d) B1 each allele arose from a separate mutation event (at different times/in different individuals), so the extra allelic variants can all persist and spread through the population's gene pool
(d) Answer: Multiple alleles means more than two versions of a gene exist across the whole population, even though each diploid individual can only carry two (one per homologous chromosome); the extra alleles arose from separate mutation events and can co-exist in the gene pool.

Question 5

(a) B1 epistasis is where the allele(s) of one gene mask(s) or affect(s) the expression of the allele(s) of a different gene (at a different locus)
(a) B1 so the phenotype produced by one gene depends on the genotype at the other gene
(a) Answer: Epistasis is the interaction between genes at different loci, in which the allele(s) of one gene mask(s) or affect(s) the expression of the allele(s) of another gene.
(b) M1 correct gametes from each parent: BE, Be, bE, be
(b) M1 correct 4 x 4 Punnett square combination of gametes
(b) A1 genotype/phenotype ratio (before epistasis) = 9 B_E_ : 3 B_ee : 3 bbE_ : 1 bbee
(b) Answer: 9 : 3 : 3 : 1 (B_E_ : B_ee : bbE_ : bbee)
(c) B1 9 black : 3 brown : 4 yellow
(c) Answer: 9 black : 3 brown : 4 yellow
(d) M1 the ee genotype is epistatic to the B gene, masking whichever pigment would otherwise be produced, so both B_ee and bbee puppies appear yellow
(d) M1 this combines the B_ee (3) and bbee (1) classes into a single yellow phenotype class of 4, giving 9:3:4 instead of 9:3:3:1
(d) Answer: Because ee is epistatic to the B gene, both B_ee and bbee genotypes (3 + 1 = 4 parts) appear yellow instead of forming two separate classes, merging two classes into one and changing the ratio from 9:3:3:1 to 9:3:4.

Question 6

(a) B1 p + q = 1 and p² + 2pq + q² = 1
(a) Answer: p + q = 1; p² + 2pq + q² = 1
(b) M1 q² = 360/1000 = 0.36
(b) M1 q = √0.36 = 0.6
(b) A1 p = 1 - q = 0.4, both values correct
(b) Answer: q = 0.6; p = 0.4
(c) M1 p² x 1000 = 0.4² x 1000 = 0.16 x 1000 = 160 (MM)
(c) M1 2pq x 1000 = 2 x 0.4 x 0.6 x 1000 = 0.48 x 1000 = 480 (Mm)
(c) A1 both values correct (160 and 480) cao
(c) Answer: MM = 160; Mm = 480
(d) B1 any one of: no mutation occurring
(d) B1 any other one of: no natural selection (all genotypes have equal survival and reproductive success); mating is random; no migration into or out of the population (no gene flow); the population is very large (so no genetic drift)
(d) Answer: Any two of: no mutation; no selection; random mating; no migration/gene flow; very large population (no genetic drift).
(e) M1 melanic (M_) ladybirds absorb more solar radiation/heat than red (mm) ladybirds, which is an advantage for warming up in cool conditions but a disadvantage (risk of overheating) in warm conditions
(e) M1 as local temperatures have risen, melanic ladybirds are now more likely to overheat, reducing their survival/reproductive success relative to red ladybirds
(e) M1 red (mm) individuals are therefore more likely to survive and reproduce in the warmer climate, increasing the frequency of the m allele (and reducing the frequency of M) in the gene pool over successive generations (directional selection)
(e) Answer: Rising local temperatures mean melanic ladybirds are now more likely to overheat, so red (mm) ladybirds are more likely to survive and reproduce, increasing the frequency of the m allele over time (directional selection favouring the red form).

Question 7

(a) B1 stabilising selection
(a) B1 individuals with the mean/intermediate value of the trait are favoured (have highest fitness/survival) and those at both extremes are selected against
(a) B1 over time this reduces the variance/spread of the distribution around the mean, without shifting the mean itself
(a) Answer: Stabilising selection; the distribution becomes narrower (reduced variance) around the same mean, as both extremes are selected against.
(b) B1 directional selection
(b) B1 pale moths became more conspicuous to predators against the darkened bark and were preferentially eaten, while melanic moths were now better camouflaged and more likely to survive and reproduce
(b) B1 the mean phenotype/allele frequency shifts progressively towards one extreme (darker colouration) over successive generations, and the whole distribution curve moves in that direction
(b) Answer: Directional selection; the distribution shifts progressively towards the darker (melanic) phenotype over generations as the mean moves towards one extreme.
(c) B1 disruptive selection
(c) B1 individuals with intermediate beak depth are selected against (lower fitness, feeding inefficiently on both food sources), while both extremes (shallow and deep beaks) are favoured
(c) B1 over time this produces a bimodal distribution (two separate peaks, around small and large beak depths) with a reduced frequency of the intermediate phenotype, increasing overall variance
(c) Answer: Disruptive selection; the distribution becomes bimodal, with two peaks corresponding to shallow and deep beak depths and a trough at the intermediate depth, increasing overall variance.

Question 8

(a) B1 a random change in allele frequency in a population's gene pool from one generation to the next
(a) B1 caused by chance (random) sampling of alleles/gametes during reproduction, rather than by natural selection
(a) Answer: Genetic drift is a random (chance) change in allele frequencies in a population's gene pool between generations, unrelated to natural selection.
(b) M1 the alleles present in each generation are effectively a random sample of gametes/alleles taken from the previous generation's gene pool
(b) M1 in a small population this sample is small, so chance/sampling error can cause a relatively large change in allele frequency from one generation to the next
(b) M1 in a large population the sample is large, so random chance fluctuations tend to average out and allele frequencies change relatively little by chance alone
(b) Answer: Each generation's alleles are effectively a random sample from the previous gene pool; small samples (small populations) are more strongly affected by chance/sampling error, whereas in large populations random fluctuations tend to average out.
(c) M1 the founder effect is when a new population is established by a small number of individuals from a larger original population
(c) M1 the founding individuals carry only a small, non-representative sample of the alleles present in the original population's gene pool, purely by chance
(c) M1 so the new (island) population's allele frequencies differ from the original (mainland) population from the outset, here having a higher-than-average frequency of the b allele/yellow shell phenotype
(c) Answer: The founder effect occurs when a new population is started by a small number of individuals whose alleles are, by chance, not representative of the original population's gene pool; here the island population has a higher-than-average frequency of the b allele purely by chance, not because of selection.
(d) M1 any valid suggestion, e.g. a disease epidemic, natural disaster (flood, fire) or severe winter that kills most of the population, leaving only a small number of survivors
(d) M1 the surviving individuals carry only a small, by-chance sample of the alleles that were present in the gene pool before the bottleneck, so many alleles are likely to be lost
(d) M1 this reduces the overall genetic diversity of the population, and allele frequencies in the survivors' descendants may differ substantially from those in the original (pre-bottleneck) population
(d) Answer: For example, a disease epidemic could kill most of the population; the small number of survivors carry only a chance sample of the original alleles, reducing genetic diversity and shifting allele frequencies away from those before the bottleneck.

Question 9

(a) B1 a species is a group of organisms with similar morphology/physiology/behaviour that share a common gene pool
(a) B1 and that can interbreed to produce fertile offspring, but cannot normally interbreed successfully with members of other such groups
(a) Answer: A species is a group of organisms, similar in structure, physiology and behaviour, that can interbreed to produce fertile offspring, but cannot normally do so with members of other such groups.
(b) M1 the physical (geographical) barrier prevents gene flow/interbreeding between the two separated populations
(b) M1 different mutations arise independently in each population over time, and/or genetic drift acts independently on each (smaller) population
(b) M1 the two lakes may impose different selection pressures (e.g. different predators, food sources or water conditions), so natural selection favours different alleles/phenotypes in each population
(b) M1 allele frequencies in the two populations therefore diverge (become increasingly different) over many generations
(b) M1 eventually the genetic differences between the two populations become so great that, even if the fish were reunited, they could no longer interbreed to produce fertile offspring; reproductive isolation has occurred and two new species now exist
(b) Answer: The geographic barrier stops gene flow between the populations; independent mutation, drift and different selection pressures in each lake cause allele frequencies to diverge over many generations, until the two populations can no longer interbreed successfully and have become reproductively isolated, separate species.
(c) M1 an error in meiosis (non-disjunction) produces gametes with an unreduced/doubled chromosome number; fusion of such gametes produces polyploid offspring with a different chromosome number from the original diploid population
(c) M1 the polyploid's chromosomes cannot pair up correctly at meiosis if it is crossed with a normal diploid plant, so any hybrid offspring would be sterile; the polyploid is reproductively isolated from the original population
(c) M1 the polyploid individuals can still interbreed successfully with each other, or self-fertilise (common in plants), so a new, reproductively isolated population can establish
(c) M1 because this occurs within the same location/habitat as the original population, with no geographical separation required, this is sympatric speciation; a new species can arise in a single generation
(c) Answer: Non-disjunction during meiosis produces a polyploid plant with a different chromosome number from the original population; this prevents fertile hybrid offspring with the normal diploid plants, but polyploid individuals can interbreed with each other or self-fertilise, so a new, reproductively isolated species can form instantly in the same location (sympatric speciation).

Question 10

Level 3 (5-6): A clear, logically structured evaluation that explains how natural selection changes allele frequency through consistent, non-random differences in survival/reproduction linked to phenotype, and how genetic drift changes allele frequency through random chance, explicitly linking population size to the relative importance of each process, and reaches a reasoned overall conclusion. Appropriate scientific terminology is used accurately throughout.
Level 2 (3-4): A reasonably clear explanation of how both natural selection and genetic drift can change allele frequencies, with some attempt to link this to population size, but the comparison/evaluation is only partially developed or somewhat one-sided.
Level 1 (1-2): Basic, mostly descriptive statements about natural selection and/or genetic drift, with little or no reference to population size or comparison between the two processes.
Level 0 (0): No relevant content.
Indicative content:
- Natural selection changes allele frequency because phenotypes with higher fitness survive/reproduce more successfully than others; it is a non-random, consistent process related to phenotype and the environment.
- Genetic drift changes allele frequency through random sampling of gametes/alleles between generations; it is unrelated to phenotype or fitness.
- In small populations, each generation is effectively a small sample of the gene pool, so random sampling error can cause large, unpredictable changes in allele frequency (drift is proportionally much stronger).
- In small populations, drift can cause harmful or neutral alleles to become fixed, and beneficial alleles to be lost, purely by chance, potentially overriding the direction selection would otherwise favour.
- Bottlenecks and founder events are examples of drift with especially large effects in small, isolated populations.
- In large populations, random sampling effects tend to average out across many individuals, so drift causes only small changes in allele frequency.
- In large populations, natural selection more reliably and consistently increases the frequency of advantageous alleles over generations, because chance fluctuations are relatively insignificant.
- Both processes can act simultaneously in any population; their relative importance depends on population size (and the strength of any selection pressure).
- A valid overall conclusion, e.g. genetic drift is most important in determining allele frequency change in small or isolated populations, while natural selection is the more dominant and predictable process in large populations.

Question 11

(a) B1 mark-release-recapture (Lincoln index method)
(a) Answer: Mark-release-recapture (Lincoln index)
(b) B1 any one of: the mark must not be toxic/must not harm the snail or affect its health
(b) B1 any other one of: the mark must not be so visible/conspicuous that it increases the snail's risk of predation (e.g. by making it easier for birds to see)
(b) B1 any other one of: the mark must not wash off, fade or be shed before the second sample is taken (durable enough for the study period), and must not affect the snail's normal behaviour or movement
(b) Answer: For example: the marking method must not be toxic or harm the snails; must not increase predation risk by being too visible; and must be durable enough not to wash off before the second sample.
(c) B1 to allow enough time for the marked snails to redistribute themselves randomly/remix thoroughly throughout the whole population
(c) B1 so that the second sample taken is representative of the whole population, with marked and unmarked individuals equally likely to be caught
(c) Answer: To allow the marked snails time to disperse and mix randomly back into the whole population, so that the second sample is representative of the population as a whole.
(d) M1 substitutes correctly: N = (84 x 96) / 21
(d) M1 84 x 96 = 8064
(d) A1 N = 8064 / 21 = 384 (snails) cao
(d) Answer: 384 snails
(e) B1 any one of: the population is 'closed', i.e. there is no significant immigration, emigration, birth or death in the population between the two samples
(e) B1 any other one of: marked individuals mix/redistribute completely at random throughout the population before the second sample is taken
(e) B1 any other one of: every individual (marked or unmarked) has an equal chance of being caught in the second sample; the marks are not lost and do not affect survival or catchability
(e) Answer: Any three of: the population is closed (no immigration, emigration, births or deaths during the study); marked individuals mix randomly with the rest of the population before recapture; every individual has an equal chance of being caught; the marks are retained and do not affect survival/catchability.
(f) M1 each of the two new sub-populations is smaller than the original population, so allele frequencies in each are more susceptible to large random change by genetic drift between generations
(f) M1 by chance, some alleles could be lost completely from one or both sub-populations (reducing genetic diversity within each), while others could increase in frequency (become fixed) purely by chance
(f) M1 because the two sub-populations are now isolated (no gene flow/interbreeding between them), drift is likely to act independently in each one
(f) M1 so, over time, the allele frequencies of the two sub-populations are likely to diverge from each other (and from the original population), even without any difference in natural selection between the two sides of the road
(f) Answer: Both resulting sub-populations are smaller than the original, so genetic drift is likely to have a stronger effect, potentially causing loss of alleles or chance fixation of others. Because the sub-populations are isolated from each other, drift will act independently in each, so their allele frequencies are likely to diverge over subsequent generations, even without any difference in selection.
(g) B1 the number/variety of different alleles (of each gene) present within the gene pool of a population or species
(g) B1 a greater variety of alleles corresponds to greater genetic diversity
(g) Answer: Genetic diversity is the number/variety of different alleles present within the gene pool of a population or species.

Question 12

(a) B1 the gene pool is the complete range of alleles (of every gene) present in a population at a given time
(a) B1 allele frequencies within it can be described as the proportion of each allele present
(a) Answer: The gene pool is the total collection of alleles (of all genes) present in the individuals of a population at a given time.
(b) M1 mutation is the ultimate source of new alleles/new genetic variation within a gene pool
(b) M1 if there is no variation in a heritable trait, all individuals have the same alleles/phenotype, so there can be no difference in survival or reproductive success between individuals
(b) M1 without variation, allele frequencies cannot change over generations, so the population cannot adapt/evolve in response to a changing environment, increasing its risk of extinction
(b) Answer: Mutation is the ultimate source of genetic variation; without variation in a trait, no phenotype has an advantage over another, so natural selection cannot favour some individuals over others, allele frequencies cannot change, and the population cannot evolve to adapt to a changing environment.

Biology: Genetics, Populations and Evolution - Worksheets, Questions and Revision