HL Paper 2
The Chinese soft-shelled turtle, Pelodiscus sinensis, lives in salt water marshes. The turtle can live under water and out of water.
These turtles have fully developed lungs and kidneys, however, many microvilli have been discovered in the mouth of P. sinensis. A study was undertaken to test the hypothesis that oxygen uptake and urea excretion can simultaneously occur in the mouth.
Initial experiments involved collecting nitrogen excretion data from P. sinensis. The turtle urinates both in water and out of water. When in water it allows waste products to be washed out of its mouth. When out of water it regularly dips its head into shallow water to wash its mouth. The table shows the mean rates of ammonia and urea excretion from the mouth and kidney over six days.
It was noted that during long periods out of water, turtles rhythmically moved their mouths to take in water from a shallow source and then discharge it. Changes in the dissolved oxygen and the quantity of accumulated urea in the rinse water discharged by the turtles were monitored over time as shown in this graph.
In order to test whether a urea transporter was present in the mouth tissues of the turtles, phloretin (a known inhibitor of membrane proteins that transport urea) was added to the water in which a further set of turtles submerged their heads. The results of that treatment are shown.
Further research was conducted to determine where mRNA expression of a urea transporter gene might be occurring in P. sinensis. Gel electrophoresis was used to analyse different tissue samples for mRNA activity.
Expression of the urea transporter gene by cells in the turtle’s mouth was assessed by measuring mRNA activity. Turtles were kept out of water for 24 hours and then injected with either a salt solution that matched the salt concentration of the turtle, dissolved ammonia or urea, followed by another 24 hours out of water.
Deduce whether the excretion of ammonia or urea changes more when a turtle emerges from water.
Compare and contrast the changes in urea excretion in the mouth with the changes in urea excretion in the kidney when a turtle emerges from the water.
Describe the trends shown by the graph for dissolved oxygen in water discharged from the mouth.
Suggest reasons for these trends in dissolved oxygen.
Deduce with a reason whether a urea transporter is present in the mouth of P. sinensis.
Outline the additional evidence provided by the gel electrophoresis results shown above.
Identify which of these turtle groups represent the control, giving a reason for your answer.
Suggest a reason for the greater expression of the gene for the urea transporter after an injection with dissolved ammonia than an injection of urea.
The salt marshes where these turtles live periodically dry up to small pools. Discuss the problems that this will cause for nitrogen excretion in the turtles and how their behaviour might overcome the problems.
Markscheme
a. urea
b. for both mouth and kidney
c. percentage change/change in μmol day−1 g−1 greater with urea/other acceptable numerical comparison
a. both higher/increased on emergence from/with turtle out of water
b. both increased by 0.66 «μmol−1 g−1 when turtle emerges from water»
c. % increase is higher in kidney / kidney 940% versus mouth 73/75% / increase is higher proportionately higher in kidney / kidney x10 versus mouth nearly double/x1.73
d. urea excretion by mouth greater than kidney out of water «despite larger % increase in kidney excretion»
decrease «when head is submerged» and increase when head is out of water
a. oxygen absorbed from water/exchanged for urea when head dipped in water«so oxygen concentration decreases»
b. lungs cannot be used with head in water / can «only» be used with head out of water
c. oxygen from water «in mouth» used in «aerobic cell» respiration
d. oxygen from air dissolves in water when head out of water «so oxygen concentration increases»
a. urea transporter is present
b. less urea «excreted»/ lower rate «of urea excretion» / excretion almost zero when phloretin/inhibitor was present
a. mRNA only in mouth and tongue/in mouth and tongue but not esophagus intestine kidney or bladder
b. bands / lines indicate mRNA for/expression of urea transporter gene
c. urea transporter gene expressed / urea transporters in mouth/tongue / not expressed/made in esophagus/intestine/kidneys/bladder
d. mRNA/transcription/gene expression/urea transporters higher in tongue/more in tongue «than mouth»
salt solution is control because it does not contain a nitrogenous/excretory waste product / it matches the salt concentration of the turtle / the turtle’s body already contains salt / because the turtle lives in salt water/salt marshes / because nothing has been altered
a. ammonia is «highly» toxic/harmful
b. ammonia is more toxic than urea/converse
c. ammonia converted to urea
d. urea concentration raised «by injecting ammonia»
e. difference between ammonia and urea «possibly» not «statistically» significant
Problems:
a. urea becomes more concentrated «in small pools» / lower concentration gradient «between tongue/mouth and water»
b. less water available for urine production/excretion by kidney
OR
less water in ponds for mouth rinsing/more competition for pools (to use for mouth rinsing)
Behaviour to overcome problems:
c. «still able to» dip mouth into/mouth rinse in water/pools
d. «still able to» excrete urea «though the mouth» in the small pools
e. more conversion of ammonia to urea/urea excretion rather than ammonia
f. more urea transporters/expression of urea transporter gene
g. urea excreted «in mouth/via microvilli» by active transport/using ATP
h. excretion with little/no loss of water
Examiners report
Isolated communities in rural Finland, Hungary and some of the Scottish islands have a high incidence of red-green colour blindness. Describe the inheritance of red-green colour blindness.
Outline the causes of variation in one example of continuous variation in humans.
Explain how evolution occurs and which factors can cause the process to be rapid.
Markscheme
a. sex linked/gene is on the X chromosome ✔
b. allele «for red-green colour blindness» is recessive/colour blindness is recessive trait/disorder ✔
c. heterozygous females are unaffected/carriers ✔
d. XB denotes normal allele and Xb denotes colour blindness allele ✔
e. more frequent in males because they only have one X chromosome ✔
f. 50% chance of colour blindness in sons whose mother is heterozygous/XBXb ✔
Accept any other letter for the alleles.
Award mpb, mpc, mpd and mpf if these points are clearly made on a Punnett grid.
a. height/skin colour/other valid example ✔
b. with continuous variation any level of the variable/phenotype is possible/OWTTE ✔
c. polygenic inheritance/combined effect of more than one gene on the trait ✔
d. additive effects on the trait of alleles of multiple genes/OWTTE ✔
e. histogram showing effects of alleles of multiple genes ✔
f. environment «may» also affect the trait/sunlight affects skin colour/other example ✔
g. normal distribution curve drawn or described to show typical pattern with continuous variation ✔
a. evolution is a change in the heritable characteristics «of a species» ✔
b. natural selection «causes evolution» ✔
c. overpopulation/over-reproduction/more offspring «than the environment can support» ✔
d. competition «for resources/mates» ✔
e. variation in population/species ✔
f. mutation/meiosis/sexual reproduction contributes to variation ✔
g. adaptation increases chance of survival ✔
h. reproduction/offspring produced «by the better adapted/by those that survive» ✔
i. characteristics passed to offspring by reproduction/variation is heritable ✔
j. allele frequencies/number of organisms carrying a gene changes/gene pool changes ✔
k. environmental change stimulates/triggers/speeds up natural selection/evolution ✔
l. increase in rainfall/introduction of antibiotic/pollution on tree trunks/other valid example of environmental change/new selection pressure ✔
m. artificial selection/selective breeding can speed up evolution ✔
Mark points can be awarded if explained using an example.
Examiners report
A broad bean is the seed of a species, Vicia faba, in the Fabaceae, a family of flowering plants. This family contains many species that are used as sources of food.
On the diagram, label the testa and the radicle.
An experiment was done to test the hypothesis that temperature affects the rate of germination of the broad bean. Outline two factors apart from temperature that should be controlled in this experiment.
State the genus of the broad bean.
Broad beans are rich in starch and cellulose. Compare and contrast the structure of starch and cellulose.
Once the germinated bean grows above the ground, state the process used by the bean in the production of starch.
Markscheme
a. testa labelled ✔
b. radicle labelled ✔ (must point to the bottom half of the embryo or the tip).
a. same amount/type of soil/substrate / pH of soil ✔ e.g.: heater
b. same amount of water / humidity ✔
c. oxygen/aeration ✔
d. same measurement of germination / time ✔ e.g.: emergence of radicle
e. same number/source/age of seeds ✔
Accept discussion of light only in as much as it relates to temperature.
Vicia ✔
No mark if the species name is included.
Mark can be awarded if the genus is not capitalized.
a. both polymers of glucose molecules / polysaccharides
OR
both form 1,4 glycosidic bond ✔
b. starch is formed by alpha glucose while cellulose is formed by beta glucose
OR
in starch C1 hydroxyl groups are found in same plane while in cellulose on different planes
OR
in cellulose, alternatively the beta glucose needs to be placed upside-down in order to have C1 hydroxyl groups on the same plane
OR
two types of starch (amylose and amylopectin) but one type of cellulose ✔
One similarity and one difference.
Comparison to cellulose needed.
photosynthesis
Do not accept condensation or polymerization; if list of processes given, mark the first answer.
Examiners report
Many candidates did not correctly label the testa and in many cases the radicle was labelled too high up, pointing at the hypocotyl.
Several answers considered how light is needed for germination without any reference to its relationship to temperature.
This question presented some confusion to Spanish speaking candidates since the word "género" is the same for "genus" and "gender", so some candidates write "male/female" in their answers.
Candidates seemed to find it difficult to find similarities and differences in structure between starch and cellulose, and many wrote about their functions, or simply mentioned that both were made of "glucose" without referencing that they are both polymers. There was limited recognition that there are two forms of starch: amylose and amylopectin that differ in their degree of branching.
The land snail Cepaea nemoralis is very common in North America and in Europe.
The base colour of its shell varies between brown, pink and yellow, and also in its intensity. Some shells are unbanded, but most show one to five bands of different width on top of the shell base colour.
The genes coding for the shell base colour and the presence of banding have linked loci. Allele C causes a dark pink shell and is dominant to allele c, which codes for a dark yellow shell. Allele B codes for the absence of bands and is dominant to allele b, which codes for the presence of bands.
C. nemoralis is a mollusc. Identify two external features that distinguish this snail from an arthropod.
State the major chemical component of the shell in molluscs.
Distinguish between sex-linked genes and genes with linked loci.
Outline how it can be shown that the genes for shell base colour (Cc) and presence or absence of bands (Bb) are linked.
Markscheme
a. unsegmented body (whereas arthropods are segmented)
b. shell (versus exoskeleton in arthropods)
c. muscular foot (which arthropods do not have)
d. no jointed appendages/jointed legs (whereas arthropods have jointed legs/appendages)
e. slimy/mucus-covered / arthropod is not slimy
Do not award marks for any answers after the first two given.
calcium carbonate/CaCO3
Do not accept carbon or calcium – the mass of oxygen in calcium carbonate is greater than both these and the chemical component is the compound not its constituent elements.
sex-linked genes are on sex/X chromosome(s)/on chromosome 23 whereas genes with linked loci are on the same autosome/chromosome
a. perform a cross/test cross
b. (if) double heterozygotes/CcBb are crossed with double homozygous recessives/ccbb
OR
Punnett square/genetic diagram showing CcBb crossed with ccbb
OR
c. (then) expected ratio (for unlinked genes) is 1:1:1:1
d. (if) double heterozygotes/CcBb are crossed together
OR
Punnett square showing CcBb crossed with CcBb
OR
e. (then) expected ratio (for unlinked genes) is 9:3:3:1
f. no/fewer than expected recombinants if genes are linked
OR
fewer pink banded/yellow unbanded if the genes are linked
OR
linked genes are expressed together more often than expected
g. use chi-square test (for significance of difference)
h. linked genes are on the same chromosome/diagram showing this
Examiners report
Boreal forests stretch across Canada, Russia and Scandinavia. This northern ecosystem accounts for 29 % of the world’s forest areas. The long, cold winters favour tall evergreen trees with either needles or scale-like leaves. These trees are wind-pollinated and their seeds are not enclosed in a fruit. The photograph shows a typical boreal forest in winter.
Identify the dominant plant phylum in the boreal forest.
In some areas there are gaps in the boreal forest where trees fail to grow and peat tends to accumulate. Suggest reasons for this.
An increase in global temperatures poses a critical threat to boreal forests. Explain the consequences of climate change to this northern ecosystem.
Suggest one advantage for the evergreen trees of the boreal forest being pollinated by wind.
Discuss the advantages of the production of seeds enclosed in fruit.
The boreal forests are situated close to the north pole and even in summer the intensity of sunlight is lower than at the equator. Sketch a graph showing the effect of light on the rate of photosynthesis, labelling the axes.
In some boreal species, Rubisco is down-regulated during the winter months. Describe the role of Rubisco in photosynthesis.
Markscheme
coniferophyta/conifer/coniferous/gymnosperms/pinophyta ✔
a. waterlogged soils/poor drainage
OR
acidic soil
OR
anaerobic conditions/soil ✔
b. organic matter not «fully» decomposed «leading to peat formation»
OR
decomposers/saprotrophs less active/fewer in cold «temperatures» ✔
a. higher temperatures so more transpiration/droughts/dehydration/water shortage ✔
b. more forest fires ✔
c. more/new pests/diseases because of the changed conditions ✔
d. competition from trees/plants «that colonize/spread to boreal forests» ✔
e. trees/«named» organisms «of boreal forests» not adapted to warmer conditions
OR
trees/«named» organisms migrate/change their distribution due to warmer conditions ✔
f. trees die so loss of habitat for animals ✔
g. faster decomposition/nutrient cycling «so conditions in the ecosystem change» ✔
h. standing water/floods due to more snow/permafrost melting ✔
animals/insects/mutualistic «relationships» not needed «for pollination»
OR
pollen not eaten by animals/insects ✔
a. seeds are protected «inside the fruit» ✔
b. seed dispersal by fruits ✔
c. example of a strategy for seed dispersal by a fruit ✔
d. dispersal reduces competition/spreads seeds away from parent plant/to colonize new areas ✔
For mpc suitable strategies are dispersal by wind, by animals ingesting/carrying away succulent fruits, by animals being attracted to colourful/sweet/tasty fruits, by animals burying nuts, by burrs or other hooked fruits sticking to animals and by self-explosion.
a. x-axis labelled as light intensity/amount of light AND y-axis labelled as rate of photosynthesis/rate of oxygen release/rate of carbon dioxide uptake ✔
b. curve starting at/slightly to the right of the x-axis origin and rising rapidly and then more slowly and plateauing but never dropping ✔
a. carbon fixation/fixes carbon dioxide/carboxylation
OR
rubisco is used in the Calvin cycle/light independent stage ✔
b. carbon dioxide linked to RuBP/ribulose bisphosphate «by rubisco» ✔
c. glycerate 3-phosphate/glycerate phosphate produced «by rubisco» ✔
Examiners report
The black-legged tick (Ixodes scapularis) is an arthropod which sucks blood from humans and other mammals. It is encountered mainly in wooded and semi-wooded areas. Some ticks can be infected by the bacterium Borrelia burgdorferi. When a tick bites a human, the bacterium is often introduced, causing Lyme disease. Lyme disease is a public health problem in North America and, if left untreated, can cause important neurological impairment. The diagram represents the two-year life cycle of a tick.
[Source: Cary Institute of Ecosystem Studies / Leslie Tumblety]
Scientists fear that global warming will change the distribution range of ticks.
The graphs show the developmental stages of ticks throughout seasons in a densely human-populated area of south-eastern Canada, surrounded by woods (circled on the map). Values are already established for 2000 and are predicted for 2080.
[Source: reprinted from International Journal for Parasitology, 36(1), N.H. Ogden, A. Maarouf, I.K. Barker, M. Bigras-Poulin,
L.R. Lindsay, M.G. Morshed, C.J. O’Callaghan, F. Ramay, D. Waltner-Toews, D.F. Charron, Climate change
and the potential for range expansion of the Lyme disease vector Ixodes scapularis in Canada,
63–70, Copyright (2006), with permission from Elsevier]
White-footed mice (Peromyscus leucopus) in eastern North America’s wooded areas often host B. burgdorferi bacteria. To determine whether bacterial transmission from mice to tick nymphs could be prevented, mice were vaccinated with antigens from Lyme disease-causing B. burgdorferi. Scientists captured wild mice at two different sites in the woods once a month, over 4 months. Each time, they measured the levels of antibodies to B. burgdorferi present in the captured and re-captured mice, inoculated all of them, and released them into the woods. The control group was not vaccinated with B. burgdorferi antigen.
[Source: Copyright (2004) National Academy of Sciences, U.S.A. An ecological approach to preventing human infection:
Vaccinating wild mouse reservoirs intervenes in the Lyme disease cycle, Jean I. Tsao, J. Timothy Wootton, Jonas Bunikis,
Maria Gabriela Luna, Durland Fish, Alan G. Barbour, Proceedings of the National Academy of Sciences
Dec 2004, 101 (52) 18159–18164; DOI: 10.1073/pnas.0405763102]
The summer after vaccination, the prevalence of B. burgdorferi infection in tick nymphs collected on mice from the two sites was measured.
[Source: Copyright (2004) National Academy of Sciences, U.S.A. An ecological approach to preventing human infection:
Vaccinating wild mouse reservoirs intervenes in the Lyme disease cycle, Jean I. Tsao, J. Timothy Wootton, Jonas Bunikis,
Maria Gabriela Luna, Durland Fish, Alan G. Barbour, Proceedings of the National Academy of Sciences
Dec 2004, 101 (52) 18159–18164; DOI: 10.1073/pnas.0405763102]
State the domain into which ticks are classified.
Using information from the text, identify one possible simple treatment for Lyme disease.
Identify the month when small birds had the greatest chance of being infected by B. burgdorferi bacteria in the year 2000 and the month when they would be most likely to become infected according to the 2080 predictions.
2000:
2080:
Using the life cycle diagram and the graph for the year 2000, analyse the distribution of adult ticks throughout the different seasons.
Evaluate the effect of the change in distribution of the different life stages of ticks on the spread of Lyme disease in south-eastern Canada.
State the reason for performing the experiment in the months of May to August.
Suggest possible reasons for the observed pattern of presence of antibodies in vaccinated mice.
Analyse the data on the state of infection of tick nymphs with B. burgdorferi in control and vaccinated mice.
Using all the data, discuss whether inoculating mice with the antigen to B. burgdorferi could be an effective method of controlling the spread of Lyme disease.
Markscheme
eukaryote ✔
Accept eukaryotes.
antibiotics / named antibiotics ✔
«2000» August AND «2080» July ✔
Both required.
a. adults present through autumn and winter «according to the life cycle diagram»
OR
some adults «must» survive winter «despite graph suggesting zero» ✔
b. adults peak in October «& November»/in autumn/between September and December ✔
c. adults die after laying eggs in winter/beginning of spring ✔
d. smaller peak/10 % versus 55 % peak/smaller numbers of adults in April/spring ✔
e. adults absent from June to September/summer ✔
Each mark point, requires month or season.
Jan - Mar = winter
Apr - Jun = spring
Jul - Sep = summer
Oct - Dec = autumn = fall
Do not accept that there are the lowest number or no adults in winter.
a. nymphs present through most of year/longer period/from March to November/through spring and summer «so more risk of infection» ✔
b. more adults in winter/in January/February so more risk of infection then ✔
c. infection will be possible through more/most months of/throughout the year ✔
d. Lyme disease likely to/will increase ✔
because nymphs are present/numbers of nymphs rise «in these months»
OR
build up immunity/antibodies in mice before nymphs «peak» ✔
Ignore references to larvae.
a. low antibody level initially as mice not previously exposed to antigen/bacteria ✔
b. vaccination causes antibody production/development of immunity ✔
c. increased proportion of mice have been vaccinated in each successive month ✔
d. second vaccination/booster shot increases antibody level/speeds up antibody production ✔
e. memory cells produced so greater/faster antibody production ✔
f. many/rising numbers of nymphs which may spread the bacteria/antigens to mice ✔
Ignore any references to non-vaccinated/control mice – this means that no marks are awarded for them because the question is about vaccinated mice, but there is no penalty for including this information in an answer.
a. at Site 1 there is little/no significant difference in the proportion of infected nymphs/numbers of infected and uninfected nymphs collected from both control and vaccinated mice ✔
b. at Site 2 the proportion of infected nymphs is lower in those collected from vaccinated than control mice
OR
at Site 2 «significantly» more nymphs are not infected from vaccinated than control mice ✔ For mpb and mpd, accept converse answers that give the proportions/percentages of uninfected nymphs rather than infected.
c. at both sites there are fewer infected than uninfected nymphs in those collected from both vaccinated and control mice ✔
d. proportion of infected nymphs is lower at Site 1 than Site 2 in nymphs collected from both control and vaccinated mice
OR
22 % of control mice and 23 % of vaccinated mice with infected nymphs at Site 1 AND 39 % of control mice and 29 % of vaccinated mice with infected nymphs at Site 2 ✔ Percentages are required for the second alternative of mpd.
Accept “ticks” instead of “tick nymphs” or “nymphs”
Do not accept quoting of untransformed numerical data.
a. Site 2 suggests that vaccination could reduce «nymph» infection rate «so method might be effective» ✔
b. Site 1 suggests that vaccination does not reduce «nymph» infection rate «so method probably not effective» ✔
c. effective «to some extent» as vaccination increases antibodies/immunity in mice ✔
d. high antibody levels needed/ many mice need to be vaccinated «for the method to be effective» ✔
e. some nymphs are still infected / «absolute» numbers «rather than proportions» of infected nymphs are similar in those collected from control and vaccinated mice ✔
f. there are other hosts/mammals/birds ✔
g. difficult/expensive «to vaccinate many small mammals/mice»
OR
cheaper to use protective clothing/tick repellant/avoid wooded areas/other method ✔
Examiners report
Variation in genetically inherited characteristics is the basis for evolution.
Outline the inheritance of blood types in the ABO blood system in humans.
Explain how genetic variation between the individuals in a species can be generated.
Outline the use of analogous and homologous traits in natural classification.
Markscheme
a. i and IA and IB are alleles (of the blood group gene) ✔
b. IA is dominant and i is recessive / IB is dominant and i is recessive ✔
c. Group O (only) with ii and Group A with IAi or IAIA / Group B with IBi or IBIB
d. IA and IB are co-dominant so Group AB with IAIB ✔
e. one allele/copy of the gene inherited from each parent/Punnett square showing this ✔
Disallow mpa if allele notation is incorrect (such as using different letters for different alleles) but allow other mps.
The notation Ii can be accepted instead of i.
Reject blood groups or types being dominant/recessive – it should be alleles.
a. mutation (in genes/DNA generates variation)
b. base substitution / change to base sequence of gene / makes single nucleotide polymorphisms /SNPs ✔
c. new alleles formed / different alleles of gene / multiple alleles ✔
d. radiation/mutagenic chemicals/mutagens cause/increase the chance of mutation ✔
e. meiosis (generates variation) ✔
f. recombination/new combinations of genes/alleles produced by crossing over ✔
g. independent assortment/random orientation of (pairs of homologous) chromosomes/bivalents ✔
h. gametes/chromosomes/DNA/genes from two parents combined (in sexual reproduction)
i. random fertilization (increases genetic variation) ✔
j. in reproductively/geographically isolated populations natural selection may differ ✔
k. in small/isolated populations gene pools change/evolution occurs due to natural selection/genetic drift ✔
l. disruptive selection can cause different varieties/variants/types/phenotypes to diverge ✔ transfer of genes (between bacteria) in plasmids ✔
For mpf it is not enough just to state ‘crossing over’.
Do not accept types of mutation other than substitution for mpb.
Analogous
a. different evolutionary origin/do not share (recent) common ancestor (despite similarities of function) ✔
b. arise by convergent evolution ✔
c. classification based on analogous traits brings together dissimilar species/is artificial ✔
Homologous
d. similar (internal) structures/pentadactyl limb/other example of homologous structures due to common ancestry ✔
e. different uses/functions ✔
f. arise by adaptive radiation/divergent evolution ✔
g. (natural) classification is based on homologous traits (not analogous) ✔
h. classification based on homologous traits has predictive values/matches evolutionary history ✔
Examiners report
For a well-prepared candidate this was a very routine exercise, but many candidates were not so prepared. A common fault was to confuse blood groups with the alleles that determine the groups, either in notation for alleles, or overall. Commonly, answers did not distinguish between gene and allele. Some candidates focused too much on the consequences of blood groups (including rhesus +ve/−ve) for blood transfusion which was not relevant to the question.
More marks were scored here, with most candidates at least mentioning mutation and some of the events of meiosis. As this was an explain question it was not enough merely to name a process, such as crossing over, some indication of how it generates variation was expected. A common misunderstanding was to think that geographic isolation in itself generates variation.
This was well-known by stronger candidates. Weaker candidates tended to omit significant ideas or get analogous and homologous traits muddled up.
The images show parts of plants belonging to two different phyla.
State the phylum of plant X and of plant Y.
X:
Y:
Some plant families, such as the figwort family, have been reclassified on the basis of evidence from cladistics. Explain the methods that have been used recently to reclassify groups of plants.
Successful sexual reproduction in flowering plants depends on several essential processes. Outline the role of pollination and seed dispersal.
Pollination:
Seed dispersal:
Markscheme
X: Filicinophyta ✔
Y: Coniferophyta/Conifera/Gymnosperms ✔
a. «previous» classification used to be based on the appearance/structures of the plant/leaves/flowers/seeds/analogy/phenotype ✔
b. «modern cladistics uses» RNA/DNA nucleotide/base sequencing/amino acid sequencing/homology ✔
c. DNA mutation occurs at a relatively constant rate allowing estimation of when species diverged ✔
d. a shared/common derived characteristic places organisms in the same clade ✔
e. the number of changes in sequences indicates distance from common ancestor
OR
the fewer the differences «in sequences» means the closer the relationship ✔
pollination:
transfer/dispersal/movement of pollen from anther/stamen to stigma
OR
transfer/dispersal/movement of pollen between plants/flowers prior to/allowing fertilization ✔
seed dispersal:
«strategy of» distribution of seeds so that new plants have space/nutrients to develop/avoid competition/colonize new habitats ✔
Accept any other valid role.
Accept OWTTE. Answers must be about the role.
Examiners report
This question requiring students to identify plant phyla from images was successfully answered by most. It was more common for students to correctly identify Plant Y as a member of the Coniferophyta phylum with a greater number being unable to identify Plant X as a member of the Filicinophyta phylum.
Many students were able to identify molecular sequencing as a method used to re-classify plants. Fewer were clear on the notion of original classification being mistakenly based on analogous features.
Most students were able to outline the role of seed dispersal in minimizing competition between parent and offspring. Slightly fewer were successful in correctly outlining the role of pollination in sexual reproduction with a number being unable to correctly draw upon the necessary vocabulary.
Describe what is shown in a cladogram.
Outline how variation in organisms of the same species could lead to natural selection.
Markscheme
a. similarities/differences between organisms/species/clades ✔ mpa and mpd concern actual characteristics, not evolutionary relationships.
b. «probable» evolutionary relationships/closeness/common ancestry/phylogeny ✔ mpb concerns such relationships
c. divergence/splits/speciation/branches/nodes ✔ mpc concerns the structure branching of the cladogram
d. relative similarity/differences between base sequence/amino acid sequence ✔
a. survival of the better adapted/fittest ✔
b. more reproduction of better adapted/fittest/individuals with favorable variations ✔
c. genes for favorable variations/adaptations passed on to offspring ✔ Accept answers in the converse.
d. competition for resources/more offspring produced than the environment can support/a struggle for existence ✔ Accept answers in the converse.
Examiners report
Outline the roles of helicase and ligase in DNA replication.
Explain how natural selection can lead to speciation.
Outline the features of ecosystems that make them sustainable.
Markscheme
helicase:
a. unwinds/uncoils the DNA «double helix» ✔
b. breaks hydrogen bonds «between bases» ✔
c. separates the «two» strands/unzips the DNA/creates replication fork ✔
ligase:
d. seals nicks/forms a continuous «sugar-phosphate» backbone/strand ✔
e. makes sugar-phosphate bonds/covalent bonds between adjacent nucleotides ✔
f. after «RNA» primers are removed/where an «RNA» primer was replaced by DNA ✔
g. «helps to» join Okazaki fragments ✔
a. variation is required for natural selection/evolution/variation in species/populations ✔
b. mutation/meiosis/sexual reproduction is a source of variation ✔
c. competition/more offspring than the environment can support ✔
d. adaptations make individuals suited to their environment/way of life ✔
e. survival of better adapted «individuals)/survival of fittest/converse ✔
f. inheritance of traits/passing on genes of better adapted «individuals»
OR
reproduction/more reproduction of better adapted/fittest «individuals» ✔
g. speciation is formation of a new species/splitting of a species/one population becoming a separate species ✔
h. reproductive isolation of separated populations ✔
i. geographic isolation «of populations can lead to speciation» ✔
j. temporal/behavioral isolation «of populations can lead to speciation» ✔
k. disruptive selection/differences in selection «between populations can lead to speciation» ✔
l. gradual divergence of populations due to natural selection/due to differences in environment ✔
m. changes in the gene pools «of separated populations»/separation of gene pools ✔
n. interbreeding becomes impossible/no fertile offspring «so speciation has happened» ✔
a. recycling of nutrients/elements/components/materials ✔
b. carbon/nitrogen/another example of recycled nutrient/element ✔
c. decomposers/saprotrophs break down organic matter/release «inorganic» nutrients ✔
d. energy supplied by the sun
OR
energy cannot be recycled «so ongoing supply is needed»
OR
energy is lost from ecosystems as heat ✔
e. energy flow along food chains/through food web/through trophic levels ✔
f. photosynthesis/autotrophs make foods/trap energy
OR
autotrophs supply the food that supports primary consumers ✔
g. oxygen «for aerobic respiration» released by autotrophs/photosynthesis/plants ✔
h. carbon dioxide «for photosynthesis» released by respiration ✔
i. populations limited by food supply/predator-prey/interactions/competition
OR
populations regulated by negative feedback
OR
fewer/less of each successive trophic level «along the food chain»/OWTTE ✔
j. supplies of water from rainfall/precipitation/rivers/water cycle ✔
Examiners report
This was generally well answered, with most candidates knowing at least something of the roles of these two enzymes. Most candidates knew that ligase connects Okazaki fragments but some claimed that it creates hydrogen bonds between nucleotides on template and the new strand. Many candidates did not distinguish between unwinding of DNA and separating the strands. Two details that should be more widely known are that helicase separates the two strands of a DNA molecule by encouraging the breakage of hydrogen bonds between bases and that ligase seals nicks by making sugar phosphate bonds.
Most candidates think they understand evolution by natural selection but many do not. Here the focus was on speciation - the splitting of a species into two or more species. Often answers described the evolution of one species over time, rather than speciation itself. An idea central to natural selection that was frequently missing from an answer is adaptation or fitness. Often traits were referred to as ‘favourable’ and therefore likely to lead to survival and reproduction but there is a circularity of argument there. Survival depends on traits fitting the environment, hence being an adaptation to it. The mostly common ideas seen in answers were differential survival and reproduction, due to differences in traits. A common fault was to confuse individuals and species and to refer to a whole species surviving and reproducing more successfully than another species.
There were some vague answers to this question but also some impressive ones that explained ecological processes including nutrient recycling, energy flow and regulation of population sizes.
The biological insights of Mendel and Darwin in the 19th century remain important to this day.
Discuss the role of genes and chromosomes in determining individual and shared character features of the members of a species.
Outline the process of speciation.
Describe, using one example, how homologous structures provide evidence for evolution.
Markscheme
Genes
a. mutation changes genes/causes genetic differences
b. genes can have more than one allele/multiple alleles
OR
alleles are different forms/versions of a gene
c. different alleles «of a gene» give different characters
OR
variation in alleles between individuals
d. eye colour/other example of «alleles of» a gene affecting a character
e. alleles may be dominant or recessive
OR
dominant alleles determine trait even if recessive allele is present
f. both alleles influence the characteristic with codominance
OR
reference to polygenic inheritance
g. all members of a species are genetically similar/have shared genes
OR
certain genes expressed in all members of a species
h. reference to epigenetics/methylation/acetylation / not all genes are expressed «in an individual»
i. genes are inherited from parents/passed on to offspring/passed from generation to generation
Chromosomes
j. same locus/same position of genes
OR
same sequence of genes/same genes on each chromosome «in a species»
k. same number of chromosomes «in a species»/all humans have 46 chromosomes/differences in chromosome number between species
l. some individuals have an extra chromosome/Down syndrome/other example of aneuploidy
OR
polyploidy divides a species/creates a new species
m. X and Y/sex chromosomes determine the sex/gender of an individual
n. meiosis/independent assortment/fertilization/sexual reproduction give new combinations «of chromosomes/genes»
a. speciation is the splitting of a species «into two species»
b. reproductive isolation/lack of interbreeding
c. isolation due to geography/«reproductive» behavior/«reproductive» timing
d. polyploidy can cause isolation
e. gene pools separated
f. differences in/disruptive selection cause traits/gene pools to change/diverge
g. gradualism / speciation/changes accumulating over long periods
h. punctuated equilibrium / speciation/changes over a short time period
a. similar structure but different function «in homologous structures»
b. pentadactyl limbs/limb with five digits/toes / other example
c. similar bone structure/example of similarity of bones «in pentadactyl limbs» but different uses/functions
d. two examples of use of pentadactyl limb by a vertebrate group
e. suggests a common ancestor «and evolutionary divergence»
f. process called adaptive radiation
Examiners report
Cell biologists play an important role in research into disease, fertility, evolution and many other areas of science.
Describe the origin of eukaryotic cells according to the endosymbiotic theory.
Cell biologists play an important role in research into disease, fertility, evolution and many
other areas of science.
Compare and contrast the processes of spermatogenesis and oogenesis.
Cell biologists play an important role in research into disease, fertility, evolution and many
other areas of science.
Outline the evidence for evolution provided by selective breeding.
Markscheme
a. mitochondria and chloroplasts are similar to prokaryotes
b. «host» cell took in another cell by endocytosis/by engulfing «in a vesicle»
Allow “taking in” in place of “engulfing”
c. but did not digest the cell/kept the «ingested» cell alive
OR
symbiotic/mutualistic relationship «between engulfed and host cell»
d. chloroplasts and mitochondria were once independent/free-living «organisms»
e. DNA «loop» in chloroplast/mitochondrion
f. division/binary fission of chloroplast/mitochondrion
g. double membrane around chloroplast/mitochondrion
h. 70s ribosomes «in chloroplast/mitochondrion»
Award up to [2] for evidence from mpe to mph
[Max 4 Marks]
a. both result in haploid cells/gametes
b. both involve mitosis at the start/in the «germinal» epithelium
c. both have cell growth «before meiosis»
d. both involve «two divisions of» meiosis
e. both involve differentiation to produce a gamete
f. both are stimulated by hormones
OR
spermatogenesis stimulated by testosterone and oogenesis stimulated by FSH
A table is not required but both statements in one row of the table must either be explicitly stated or clearly implied to award the mark
[Max 8 Marks]
a. crop plants/domesticated animals/livestock produced by selective breeding
b. specific example of a domesticated animal/crop plant and the wild species from which it
was developed
OR
specific example of a domesticated animal/crop plant and the features in it which have been improved «compared with the wild species»
For example dogs have been developed from wolves
c. artificial selection/crossing selected varieties/eliminating undesirable varieties
d. «selective breeding/artificial selection can cause» significant/rapid change over time/from the original wild species
e. «changes due to selective breeding/artificial selection» shows natural selection can cause change/evolution «in a species»
[Max 3 Marks]
Examiners report
The image shows a food web.
Using the food web, identify a detritivore.
Using the food web, identify a saprotroph.
State the name of the domain to which birds, such as the Elf owl, belong.
Outline the energy flow through this food web.
Markscheme
earthworm/woodlouse
bacteria/fungi
Do not accept protozoans or nematodes as they are consumers.
eukaryote/eukaryota/eukarya
a. light energy of Sun is converted by plant/autotroph to chemical energy «in carbon compounds through photosynthesis»
b. detritivores/saprotrophs decay plant material «that accumulates in the soil» to obtain energy OWTTE
c. consumers release energy from the carbon compounds by cell respiration energy lost as heat
d. energy is used by organisms for metabolism
e. energy is transferred between organisms/trophic levels through the food chains/web
For mp e, accept specific example such as energy is transferred from primary to secondary consumer etc.
f. energy is lost at each trophic level «so lengths of food chains/web are restricted»
OR
approximately 80/90 % of energy is lost «between trophic levels»
Vice versa
Award mark points that refer to the specific organisms from this food web.
Examiners report
State two features that are found only in mammals.
Birds, bats and humans are all vertebrates with pentadactyl limbs. Birds and bats use their forelimbs to fly whereas humans can use them to lift and manipulate objects. Outline how the bird, bat and human forelimb can be used to illustrate the concept of homologous structures.
Markscheme
a. hair/fur
b. mammary glands/breasts/(secretion of) milk/lactation
c. sweat glands
d. lungs with alveoli
e. placenta (in most mammals)
f. fetus develops in uterus/gives birth to live young
Do not award marks for any answers after the first two given.
a. (homologous structures have evolved) from a common ancestor
b. divergent evolution/adaptive radiation
c. similarities in forelimb bones (in birds, bats and humans) / description of the similarities in bones
d. different uses/functions
Do not accept (positive) correlation.
Examiners report
List two causes of variation within a gene pool.
Describe how variation contributes to evolution by natural selection.
Outline what is required for speciation to occur.
Markscheme
a. sexual reproduction / random fertilization / meiosis
b. mutation
No mark for crossing over unqualified.
Reject natural selection/evolution as causes of variation.
a. (variation is) different phenotypes/differences between individuals in a population/species
b. struggle/competition for survival
c. some individuals have advantageous characteristics/are better adapted/have greater chance of survival/reproduction (than others)
d. favourable alleles/genetic variations passed on/inherited by offspring/next generation
Reject “pass on phenotypes”.
a. divided species/gene pool / part of species/gene pool becomes separated / species splits into separate populations
b. reproductive isolation / lack of interbreeding
Mark point b refers to a lack of interbreeding between separated populations in a species, not the lack of interbreeding after speciation.
c. may be due temporal/behavioural/geographic isolation
d. different natural selection/different selective pressures