Question 1
1
PanI is a gene in cod fish that codes for an integral membrane protein called pantophysin. Two alleles of the gene, \(\operatorname{PanI}{ }^{\mathrm{A}}\) and \(\operatorname{PanI}{ }^{\mathrm{B}}\), code for versions of pantophysin, that differ by four amino acids in one region of the protein. Samples of cod fish were collected from 23 locations in the north Atlantic and were tested to find the proportions of PanI \({ }^{\mathrm{A}}\) and PanI \({ }^{\mathrm{B}}\) alleles in each population. The results are shown in pie charts, numbered 1-23, on the map below. The proportions of alleles in a population are called the allele frequencies. The frequency of an allele can vary from 0.0 to 1.0 . The light grey sectors of the pie charts show the allele frequency of PanI \({ }^{\mathrm{A}}\) and the black sectors show the allele frequency of PanI \({ }^{\mathrm{B}}\).
10 marks
Question 1(a)
1(a)
2 marks
Question 1(a)(i)
1(a)(i)
State the two populations with the highest \(\operatorname{PanI}^{\mathrm{B}}\) allele frequencies.
Mediumstructured1 marks
Answer
(population) 1 and (population) 2 (both needed)
Question 1(a)(ii)
1(a)(ii)
State the population in which the allele frequencies were closest to 0.5 .
Mediumstructured1 marks
Answer
(population) 11/7/7 and 11
Question 1(b)
1(b)
Deduce the allele frequencies of a population in which half of the cod fish had the genotype \(\operatorname{PanI} \mathrm{I}^{\mathrm{A}} \operatorname{PanI} \mathrm{I}^{\mathrm{A}}\), and half had the genotype \(\operatorname{PanI} \mathrm{I}^{\mathrm{A}} \operatorname{PanI} \mathrm{I}^{\mathrm{B}}\).
Mediumstructured1 marks
Answer
\(P a n \mathrm{I}^{\mathrm{A}} 0.75\) and \(P a n \mathrm{I}^{\mathrm{B}} 0.25 / 3 P a n \mathrm{I}^{\mathrm{A}}\) to \(1 P a n \mathrm{I}^{\mathrm{B}}\) Both must be correct for the mark to be awarded, accept frequencies in form of ratio.
Question 1(c)
1(c)
The graphs below show the latitude and the mean surface sea temperature in June of the sampling locations and the frequency of the \(\operatorname{PanI}^{\mathrm{A}}\) allele. State the relationship between
1 marks
Question 1(c)(i)
1(c)(i)
latitude and the frequency of the \(\operatorname{PanI}^{\mathrm{A}}\) allele.
Mediumstructured1 marks
Answer
greatest/great frequencies of \(\operatorname{PanI}^{\mathrm{A}}\) at lowest/low latitudes / a rapid drop in frequency at (60-65 degrees latitude) / lowest/low frequencies at highest/high latitudes Answers which describe/imply the correct step-wise relationship should get credit. Answers stating or implying a negative correlation alone should not get credit.
Question 1(c)(ii)
1(c)(ii)
mean surface sea temperature in June and the frequency of the PanI \({ }^{\mathrm{A}}\) allele.
Mediumstructured0 marks
Answer
lowest/low frequencies of \(\operatorname{PanI}^{\mathrm{A}}\) at lowest/low temperatures / a rapid increase in frequency at ( 8-10 degrees Celsius) / highest/high frequencies at warmest/warm temperatures Answers which describe/imply the correct step-wise relationship should get credit. Answers stating or implying a negative correlation alone should not get credit.
Question 1(d)
1(d)
Suggest how natural selection could have caused the relationships shown in the graphs.
Hardstructured2 marks
Answer
(cod with) \(\operatorname{PanI}^{\mathrm{A}}\) allele selected/favoured/better adapted to warmer water; (cod with) PanI \({ }^{\mathrm{B}}\) allele selected/favoured/better adapted to colder water; cod that survive can reproduce and pass alleles on to offspring; It takes a whole organism to reproduce in order to pass on the allele, hence we expect reference to the fish to gain this last marking point
Question 1(e)
1(e)
The sites close to Iceland, at a latitude of \(60-65^{\circ}\), had very varied allele frequencies, with both PanI \({ }^{\mathrm{A}}\) and PanI \({ }^{\mathrm{B}}\) occurring. The water at these sample sites was highly stratified, with much warmer water at the surface and much colder water below. Suggest reasons for both \(\operatorname{PanI}^{\mathrm{A}}\) and \(\operatorname{PanI}^{\mathrm{B}}\) alleles occurring at these sites.
Hardstructured2 marks
Answer
higher frequency of \(\operatorname{PanI}^{\mathrm{A}} / \operatorname{PanI} \mathrm{I}^{\mathrm{A}} \operatorname{PanI} \mathrm{I}^{\mathrm{A}}(\operatorname{cod})\) in warm (surface) water; higher frequency of \(\operatorname{PanI}^{\mathrm{B}} / \operatorname{PanI}^{\mathrm{B}} \operatorname{PanI}^{\mathrm{B}}\) (cod) in colder (deeper) water; interbreeding results in \(\operatorname{PanI}^{\mathrm{A}} \operatorname{PanI}^{\mathrm{B}} \operatorname{cod} /\) heterozygous cod;
Question 1
1
Several studies have been undertaken to determine whether there is an evolutionary explanation for menopause, the time when reproductive capacity stops in women. Two contemporary hunter-gatherer societies were studied. The graph shows what percentage of women survive to each of the ages given.
5 marks
Question 1(f)
1(f)
Explain the possible natural selection of menopause among humans during the hunter-gatherer period of their evolution.
Hardstructured2 marks
Answer
a. older/skilled women could provide more food when they no longer reproduce; b. this could provide more food for offspring/group leading to greater success for the group; c. (in competition with other groups) this group would survive to pass on the genes for menopause; d. allows younger women to use time/energy to reproduce and care for offspring;
Question 1
1
Obesity (excessive weight) is recognized as a global health problem and has been correlated with a large number of health issues, diseases and deaths. The increased consumption of fructose, now widely used as a sweetener, has been associated with the increase in obesity. In a study, mice were divided into four groups. Each group was given the same amount of food and either a soft drink with a different sweetener or water.
15 marks
Question 1(h)
1(h)
Using all of the data, evaluate the evidence that suggests the consumption of large amounts of fructose poses a risk to human health.
Harddata_response3 marks
Answer
evidence that fructose causes (body) fat accumulation/obesity; evidence that fructose is related to increased (blood) triglycerides which are correlated with obesity/coronary heart disease; evidence that fructose is related to reduced insulin sensitivity/diabetes; evidence that fructose is used in ribose synthesis but no clear evidence that fructose causes pancreatic cancer;
Question 1
1
The yeast Metschnikowia bicuspidata is a parasite of a species of zooplankton, Daphnia dentifera. Biologists monitored the infections of D. dentifera populations in a series of lakes in Indiana (USA). An increase in nitrogen compounds dissolved in the lakes causes the phytoplankton populations to increase. D. dentifera feed on phytoplankton. The graphs show the - relationship between nitrogen levels dissolved in the water and the size of the parasite epidemic in the D. dentifera population. - relationship between the size of the parasite epidemic in the D. dentifera population and the change in the resistance (established by comparing the infection of the D. dentifera populations before and after the epidemic).
0 marks
Question 1(c)
1(c)
Outline, according to the theory of natural selection, how increased size of the parasite epidemic in the D. dentifera will result in the evolution of increased resistance to the parasite. Predatory fish tend to eat infected D. dentifera more than uninfected D. dentifera.
Mediumdata_response2 marks
Answer
high epidemic sizes associated with higher populations of D. dentifera/Daphnia; the greater the population of D. dentifera/Daphnia, the greater the possibility of variation/mutations in population; some of the variations may include greater resistance to the parasite; (resistant strains) have a greater chance of reproducing/produce more offspring; (leads to) selection in favour of resistant strains / death of non resistant strains;
Question 1
1
The larval stage of the fly Eurosta solidaginis develops in the plant Solidago altissima. The larva secretes a chemical which causes plant tissue to grow around it forming a swelling called a gall. The gall provides the developing insect with protection from predators. The E. solidaginis fly is preyed upon by the parasitic wasp Eurytoma gigantea. The graph shows the relationship between gall diameter and the percentage of flies that avoid predation by E. gigantea.
0 marks
Question 1(c)
1(c)
Explain the concept of directional selection with respect to this example.
Mediumstructured2 marks
Answer
a. directional selection is when an extreme phenotype/characteristic is favoured b. flies that form small galls will be selectively predated c. over time, flies that produce small galls will become rarer OR mean gall size will increase OWTTE OWTTE - accept vice versa 2 max
Question 1
1
Mice (Mus musculus) have various defence systems against predators such as foxes (Vulpus vulpus) or stoats (Mustela erminea). The mice release specific alarm compounds when under threat that serve as danger signals for other mice. Predators also release scents that the mice can detect. In one study, mice were exposed to paper soaked in compounds taken from other mice, foxes or stoats in a test chamber. The scientists then measured the reaction of three different groups of neurons used in smelling: G1, G2 and G3, as shown in the diagram. The percentage of G1, G2 and G3 neurons responding to the mouse, fox and stoat compounds, as well as a control compound, are shown in the chart.
14 marks
Question 1(j)
1(j)
Discuss whether natural selection would favour the transgenic EBF-producing thale cress plants if they were released into the wild.
HardEssay2 marks
Answer
a. mutant aphids/varieties may be indifferent to/attracted to transgenic plants as these do not present a hazard «not favour» \(\boldsymbol{}\) b. initially/for limited time the plants would thrive as the aphids would be «largely» repelled and thus not eat the plants «so natural selection would favour them» c. over time/in a few generations, the aphids population become more resistant/more attracted/less repelled to EBF and return to feed on the plants so long-term benefit very limited «so natural selection would not favour them» \(\boldsymbol{}\) d. the aphids resistant to EBF would not respond to other aphid alarms and «likely» be more readily eaten by predators «so the long-term benefit to plants could be supported by natural selection» The answers must indicate whether natural selection would support or not for each statement.
Question 1
1
A group of nine semi-wild Asian elephants (Elephas maximus) from an elephant camp in Myanmar were trained to pull two ends of a single rope to get access to food in food trays located on a table behind a barrier. When both elephants pulled the rope, the table moved towards them making the food trays accessible. The task required two elephants to work together in cooperation for mutual benefit to obtain food. The graph shows the rate of cooperation between pairs of elephants in pulling the rope when there were two food trays, one at each end of the table (as shown in the diagram), or when there was one food tray placed in the centre of the table. The cooperation rate was the number of successful cooperative attempts divided by the number of all attempts in one session. The experiment was repeated for many sessions on successive days.
structured13 marks
Question 1(i)
1(i)
Evaluate the hypothesis the researchers put forward from their experiments that in some circumstances natural selection favours competitive behaviour.
HardEssay2 marks
Answer
a. when there is more food available the elephants are willing to share/cooperate OR when food is scarce, survival/competition is more beneficial than cooperation OR natural selection will favour the behaviour that gives elephants sufficient food; b. the elephants may not behave in the same way in the wild; OR controls are not specified/how much food was available was not given/gender of elephants/sizes/ages/whether or not the same elephants were used for the trials;
Question D2
D2
6 marks
Question D2(a)
D2(a)
State two assumptions that are made when the Hardy-Weinberg equation is used.
Mediumshort_answer2 marks
Answer
D2. (a) large population; randomly breeding population; no mutation; no selection (for an allele); no migration/immigration or emigration;
Question 2
2
0 marks
Question 2(b)
2(b)
0 marks
Question 2(b)(i)
2(b)(i)
In fruit flies (Drosophila melanogaster) the allele for long wings is dominant to the allele for short (vestigial) wings. In a laboratory population of 200 flies there were 168 with long wings. Using the Hardy-Weinberg equation, calculate the percentage of flies that would be heterozygous for this characteristic. Show your working.
Mediumcalculation2 marks
Answer
\(\left.\begin{array}{l}200-168=32 \text { flies had short wings } \\ \mathrm{q}^{2}=32 / 200=0.16 \\ \mathrm{q}=0.4 \\ \mathrm{p}=0.6 \\ 2 \mathrm{pq}=0.48 ;\end{array}\right\} \begin{aligned} & \text { Award [1] for correct working. } \\ & \text { (Looking for the logic of the } \\ & \text { working) }\end{aligned}\) percentage of heterozygotes =48 %; Award [1] for correct answer.
Question 2(b)(ii)
2(b)(ii)
State two assumptions made when using the Hardy-Weinberg equation to calculate allele frequencies.
Mediumshort_answer1 marks
Answer
large population/random mating/no selection/no migration/no mutation Award [1] for any two correct answers. Mark the first two answers given.
Question D2
D2
6 marks
Question D2(a)
D2(a)
2 marks
Question D2(a)(i)
D2(a)(i)
Define the term gene pool.
Easyshort_answer1 marks
Answer
D2. (a) (i) all the genes in an interbreeding population (at a certain time) ..... [1] Do not accept "species" instead of "population".
Question D2(b)
D2(b)
Describe sickle-cell anemia as an example of balanced polymorphism.
Hardshort_answer2 marks
Answer
heterozygotes have two alleles \(/ \mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{S}}\) which gives them (partial) resistance to malaria; heterozygotes \(/ \mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{S}}\) more likely to survive and reproduce than either homozygote \(/ \mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{A}}\) and \(\mathrm{Hb}^{\mathrm{S}} \mathrm{Hb}^{\mathrm{S}}\); \(\mathrm{Hb}^{\mathrm{A}} \mathrm{Hb}^{\mathrm{A}}\) may die from malaria \(/ \mathrm{Hb}^{\mathrm{S}} \mathrm{Hb}^{\mathrm{S}}\) are more likely to die from sickle-cell anemia; an equilibrium between allele frequency is established / lethal allele retained in population;
Question D2(c)
D2(c)
During a screening campaign of 281884 babies in Sao Paulo, Brazil, it was found that the frequency of the sickle-cell anemia allele was 0.02 . Calculate the expected number of babies not carrying the allele.
Mediumcalculation2 marks
Answer
Hardy-Weinberg equation (e.g. \(\mathrm{p}^{2}+2 \mathrm{pq}+\mathrm{q}^{2}=1\) ) / calculation of p (e.g. p=1-q=0.98 ) \(/ \mathrm{p}^{2}=0.9604\); calculation of total of babies without allele \(=\mathrm{N} \times \mathrm{p}^{2}\) or \(281884 \times 0.9604=270721\);