Outline how hydrogen bonds form in water.

Describe the processes that cause water to move from the roots of plants to their leaves.

Explain the role of the kidney in osmoregulation.

a. water (molecules) are polar/dipolar/have partially positive and negative poles/have δ+ and δ-;

b. attraction/bonding between positive and negative (poles);

c. hydrogen bond formed between hydrogen and oxygen; Reject if H and O in same molecule.

d. bond/attraction between different water molecules/intermolecular;

Marks can be awarded in an annotated diagram.

Reject answers stating or implying that there are whole positive or negative charges for mpa.

a. water moved/transported in xylem vessels;

b. transported under tension/suction/pulled up (in xylem vessels);

c. transpiration/loss of water (vapour) generates pulling forces/low pressure/tension;

d. tension/pull generated when water evaporates from cell walls (in mesophyll);

e. transpiration is loss of water vapour from leaf (surface)/stomata;

f. cohesivity/cohesion in water due to hydrogen bonding/attractions between water molecules;

g. cohesion/WTTE so chain/column of water (molecules) doesn’t break/remains continuous;

h. transpiration stream is a column of/flow of water in xylem from roots to leaves;

Do not award marks for absorption of water by roots.

a. osmoregulation is regulation of water and solute/salt balance/solute concentrations;

b. nephron (is the functional unit of the kidney/osmoregulates);

c. ultrafiltration in glomerulus / glomerular filtrate collected by Bowman’s capsule;

d. loop of Henle establishes/maintains hypertonic conditions in medulla;

e. osmosis/reabsorption of water (from filtrate) in the collecting duct;

f. brain/hypothalamus monitors blood solute concentration / pituitary secretes ADH;

g. ADH secreted when solute concentration of blood is too high/hypertonic/when dehydrated;

h. ADH increases permeability of collecting duct to water;

i. ADH causes more aquaporins (in membranes of collecting duct wall cells);

j. more water reabsorbed resulting in more concentrated/hypertonic urine/less volume of urine;

k. less/no ADH secreted when solute concentration (of blood) is too low/hypotonic;

l. less water reabsorbed resulting in dilute/hypotonic urine/large volume of urine;

Reject ‘water balance’ and ‘water concentration’ for mpa.

Answers were mostly good here, with many candidates scoring full marks. A very common misconception was that hydrogen bonds are strong – a single hydrogen bond is a weak interaction and it is only because there are many hydrogen bonds in water that they collectively exert large cohesive forces. A few candidates thought that hydrogen bonds were within rather than between water molecules.

This was answered quite well. There were some traditional areas of confusion, with cohesion and adhesion either muddled up or treated as the same thing. Some candidates thought high pressure caused by roots and low pressure caused by leaves could exist at the same time in xylem. Few responses referred to water being transported in the vessels of xylem. In some answers water was said to evaporate from stomata, instead of the moist, blotting-paper-like cells walls of the mesophyll. Some answers included details of how water is absorbed into roots, which was outside the scope of the question. Despite these common faults, many candidates described clearly how tension is generated in xylem and how cohesive columns of water can be pulled up to leaves.

There was some excellent knowledge of kidney function, frequently going way beyond the question’s requirements. In some cases, candidates just described all processes occurring in the kidney instead of actually answering the question. This generally led to the loss of the extra clarity mark.

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.

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

Identify the week and year in which the first cases were recorded in the suburbs.

Week:

Year:

Based on the graph, compare and contrast the progress of the epidemic in the suburbs and rural areas.

Suggest two reasons for the overall decline in the epidemic after week 51.

Compare and contrast the data for Conakry with the data for the three suburbs.

Suggest reasons for the high percentage of fatal cases at Ebola treatment centres.

Based on these data, outline the evidence that T-705 has potential to be used as a treatment for EVD.

Explain how vaccination can lead to the production of B cells specific to the Ebola virus.

Suggest possible reasons for the difficulty of preventing or controlling a viral epidemic such as the 2014 EVD epidemic in a remote rural region.

week 34 AND 2014 ✔

both needed

a. start of epidemic/first cases in rural areas
OR
epidemic spread to suburbs later ✔

b. higher maximum number of cases/greater increase in rural areas
OR
converse for suburbs ✔

c. increase came earlier in rural areas «than suburbs»
OR
number of cases peaked earlier in rural areas
OR
more cases in rural areas «than suburbs» in 2014 ✔

d. decrease came earlier in rural areas «than suburbs»
OR
decreasing in rural areas but not in suburbs in 2015/by end of study period
OR
more cases in suburbs than rural areas in 2015 ✔

e. «large» fluctuations in both ✔

a. «overall decline due to» fewer cases in rural areas ✔

Answers relating to people who died from the disease or develop immunity to it:
b. fewer cases due to deaths of people who had the disease/people recovering
OR
more people vaccinated/became immune/made antibodies/were not vulnerable to infection ✔

Answers relating to health care workers or availability of resources:
c. more doctors/nurses/medical equipment/treatment centers/hospitals/spending/aid/NGOs ✔

Answers relating to medical techniques used to tackle the epidemic:
d. better treatments/infection control/hygiene/quarantine/new vaccine/new antiviral drugs ✔

Answers relating to the public and patients:
e. education/better awareness/avoidance of infection/taking precautions/vaccination accepted ✔

Answers relating to reservoirs of infection:
f. fewer infected people «who could spread infection»/fewer bats/less contact with bats ✔

differences:
a. Conakry has more cases than any of the suburbs
OR
more cases in total in the suburbs than in Conakry ✔

b. more male cases in Conakry whereas more female cases in suburbs ✔

c. higher «% of» fatal cases at Ebola treatment centers in suburbs than in Conakry ✔

similarity:
d. in both Conakry and suburbs «% of» fatal cases in treatment centers is higher than outside ✔

a. most serious cases are in/are taken to treatment centers
OR
treatment centers are set up where there are most cases/most serious cases ✔

b. long time/distance to travel between contracting disease and arrival at treatment center
OR
travel to treatment center weakens/upsets/harms the patient ✔

c. Ebola is a virulent disease/Ebola virus mutated «to become virulent»
OR
little known about Ebola/new disease so treatments not yet developed ✔

d. no/not enough vaccine/antiviral drug available «in 2014/15»
OR
antibiotics do not work against viral diseases ✔

e. secondary infections/Ebola patients infected with other diseases/other Ebola strains
OR
ineffective hygiene/cleaning/sterilization/use of contaminated equipment/disposal of corpses ✔

f. small number of staff relative to patients/treatment centers overcrowded/swamped with patients
OR
insufficient equipment/supplies for large number of patients/with the rapid rise in patients ✔

g. better reporting at Ebola centers/deaths due to Ebola not reported in rural areas ✔

a. cells not killed/few cells killed «even at high concentrations» ✔

b. «T-705» effective/viruses reduced/viruses killed at 100 μM
OR
«T-705» very effective/viruses much reduced/nearly all viruses killed at 1000 μM ✔

c. virus concentration decreases as T-705 concentration increases ✔

d. drug has «high» potential for treatment «at high enough concentration» ✔

a. vaccine contains Ebola antigens ✔

b. vaccine «could» contain weakened/attenuated/dead/killed form of «Ebola» virus/virus genetically modified to express an Ebola/viral protein ✔

c. phagocyte/macrophage engulfs the antigen/presents the antigen to T cell ✔

d. antigen recognized by «specific» T cells/binds to T cells ✔

e. «activated» T cells activate «specific) B cells ✔

f. «activated» B cells make the antibodies «against Ebola» ✔

g. B cells divide forming «clone of» plasma cells/producing more B cells specific to Ebola ✔

a. poor transport infrastructure/poor communication/bad roads/difficult access/no maps/support slow arriving/scattered population ✔

b. poor education/understanding of disease amongst health workers/local population
OR
continued contact with infected people / other example of unsafe actions ✔

c. more sources of infection such as bats/difficult to find sources of infection ✔

d. lack of/limited access to medical care/doctors/health care workers ✔

e. lack of/no access to/unaffordability of treatment centers/medicalsupplies/equipment/antivirals/drugs/vaccine/treatments ✔

f. refusal/reluctance in local population to be vaccinated
OR
difficult to find/reach everyone to vaccinate them/repeat the vaccination ✔

g. migration of people spreads the infection ✔

h.poor sanitation/lack of clean water ✔

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

Nearly all candidates read from the graph the week and year in which the first cases were recorded in the suburbs.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

The skill required in this question was picking significant trends out of data with a great deal of noise. Only one similarity was accepted – the large week-to-week fluctuation in number of cases. The differences between the rural areas and the suburbs were that Ebola epidemic started earlier in rural areas, rose to higher peaks and started declining earlier.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

A wide range of possible reasons for the decline in the epidemic was accepted here and most candidates scored both marks. The commonest type of misunderstanding was that there might be few people left to become infected. Given that the total number of cases was only in the thousands and the population of Guinea must be in the millions, that explanation for the decline was implausible. As this was a suggest question and candidates are not required to have specific knowledge of the Ebola epidemic, any reasonable answer was accepted even if it was not historically true.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

This was another question where the skill was in picking out significant trends. Differences between individual suburbs were not significant so all the answers accepted were either similarities or differences between the capital city and the suburbs generally. Most but not all candidates made two or more statements about the data that were significant enough to score marks. Some students treated percentages as though they were absolute numbers and for example added the percentage of fatal cases in and outside treatment centres together, which does not give a meaningful total.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

This was a more difficult question. We might expect fewer fatalities in percentage terms at treatment centres than elsewhere, but the data showed that there were more. Not all candidates realised that the figures given indicated the percentage of people diagnosed with Ebola that died, not the percentage of the population as a whole. As in 1(c) a wide variety of answers was accepted. Some of the answers related to Ebola having a high mortality rate generally, but to score three marks it was necessary to find reasons for treatment centres in particular having high rates of death. One suggestion commonly given but not accepted was that patients became infected after arrival at treatment centres – this would not necessarily give a high fatality rate or percentage and also uninfected people are unlikely to be brought to treatment centres. Some students thought that the more infected people a patient came into contact with, the more infected they would be. This ignores the potential for the Ebola virus to multiply exponentially once inside a patient. Material shortages were only accepted as a reason if this was combined with the idea that treatment centres were swamped with patients.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

This was generally very well answered, with candidates correctly concluding that the drug had potential as a treatment because at a high enough dose it reduced the number of viruses without harming a significant number of cells.

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

Vaccination is a major focus on 11.1 of the programme so it was perhaps surprising that answers here were not better. Given the current outbreaks of measles and other diseases that can be prevented by vaccination, this topic should be given greater prominence in some schools. There were a few impressive answers giving the details of the steps that lead up to production of antibodies against a specific pathogen, but less than a quarter of candidates scored all three marks. There were many misconceptions. The terms ‘resistant’ and ‘immune’ have different meanings in biology, which should be carefully distinguished. Some candidates think that a vaccine contains a small amount of the pathogen. Would those candidates be happy to be infected with Ebola viruses as long as it was only a few? There was too much talk of cells ‘remembering’ the antigen or the virus. Memory cell is perhaps an unfortunate term. They are only memory cells in the sense that if an antigen binds to the antibodies they display, after cell multiplication more of those antibodies are produced and they will bind to the same antigens on the surface of an invading pathogen. This is just a series of molecular processes, not true memory. 

This was a timely question with Ebola epidemics again causing great problems in parts of Africa. Students were expected to have had some practice at analysing data relating to an epidemic because of this Skill in the programme: Analysis of epidemiological data related to vaccination programmes. It was assumed that candidates would understand terms such as ‘case’, ‘fatal case’, ‘infect’ and ‘treat’.

This was a third question where a very wide range of answers were accepted. Even so, not all candidates scored both marks. The commonest weakness was vague answers such as “shortage of resources”. The best answers gave common-sense reasons for special difficulties in controlling an epidemic in the population of a remote rural region.

Estimate the difference between the highest and lowest mean body temperatures.

Compare and contrast the changes in mean ambient and body temperatures during 2012.

Explain the change in heart rate during the period of hibernation.

Distinguish between the changes in porosity of the bones in humans and bears as age increases.

The life expectancy of a human at the time of the study was 80 years. Estimate the porosity of the bones of the individual who was approximately 32 years old.

The researchers assessed age as a proportion of normal life span, rather than in years. Suggest one reason for this.

Describe what is happening to the bone during hibernation.

Suggest how the graph would differ for a human during a long period of inactivity.

Calculate the percentage increase in the mean concentration of osteocalcin from pre-hibernation to hibernation.

A hypothesis has been proposed that an increase in parathyroid hormone concentration causes an increase in osteocalcin in bears. Evaluate the evidence for this hypothesis provided by the data.

Discuss how helpful these studies of bears can be in developing an understanding of osteoporosis in humans.

5 °C;

Units required. Accept answers in the range 4.5 to 5.5 °C.

  Accept one similarity:

1. both rise and then fall / both fall with hibernation and rise with activity
2. both reach minimum during hibernation and maximum during activity;
3. both lowest in January/February / both rise from January/February;
   
   Accept one difference:
   
   
4. one peak of ambient temp but body temp has two peaks / OWTTE;
5. body temp remains maximal for longer/plateaus whereas ambient peaks;
6. body temperature is always higher than ambient temperature;
7. ambient range is greater than body temperature range / OWTTE;

1. decreased/slower heart rate because bears less active/use less energy;
2. less (cell) respiration / lower (rate of) metabolism;
3. less oxygen/glucose required / less CO₂ produced/needing to be removed;
4. less muscle contraction/muscles require less blood;
5. conserves energy;

porosity increased in humans and decreased in bears;

Both needed.

6 %;

Accept answers in the range 6.0 to 6.5 %.
Percentage sign required.

1. to allow comparison of bears and humans;
2. bears have a different life span to humans / bears do not live to 80 years;
3. because they age at different rates;

1. resorption/breaking down occurs and formation/rebuilding occurs;
2. at similar rates / more resorption at most times;
3. no/little (overall) change (in bone mass);
4. lag between bone resorption rising and formation rising / OWTTE;
5. bone resorption rising towards end as formation dropping / OWTTE;

1. more resorption than formation;
2. PICP/bone formation (always) lower (than in bears);
3. ICTP/bone resorption (always) higher (than in bears);
4. ICTP above PICP by a greater amount in humans (than in bears);

250 % (Allow 240 to 260 %)

1. (hypothesis supported by)
   positive/direct correlation/direct relationship (between parathyroid hormone and osteocalcin)
   OR
   osteocalcin rises as parathyroid hormone rises/vice versa;
2. no evidence for causal link / causal link cannot be assumed / correlation does not prove causation;
   OR
   no evidence that parathyroid hormone causes change in osteocalcin;
   OR
   other factors may cause change in osteocalcin;

  Accept one reason for the studies being helpful:

1. (helps us understand how)) bears avoid osteoporosis;
2. bone structure of bears and humans is similar / both are mammals;
3. suggests that hormones/osteocalcin/parathyroid hormone might be a (preventative) treatment;
   
   Accept one reason for the studies not being helpful:
4. humans do not hibernate / are not inactive for long periods;
5. humans live for much longer;

70 % of candidates were able to read two temperatures from the graph and subtract the higher from the lower correctly.

Many candidates realized that they were expected to give similarities and differences between the curves, but most struggled to do this clearly. Rather than make eclectic statements about the data that may be correct but are not significant, the aim with a question such as this is to make comments that would allow someone who has not been shown the curves to sketch them. Some students confused maxima or minima with increases and decreases, so for example stating that ambient body temperature increased in July, when it reached a maximum in July and stopped increasing.

Many candidates mentioned lack of activity as a reason for lower heart rate and some went on to mention reduced respiration rates or other aspects of physiology. Other candidates failed to offer an explanation and instead just described the data. A few claimed wrongly that bone porosity was directly proportional to age, in which case all those of a certain age would have the same degree of osteoporosis, which is plainly not the case.

Well answered with most candidates referring to the positive and negative correlations.

This question was very unusual in that the discrimination index was negative — stronger candidates were slightly less likely to answer correctly than weaker candidates, for reasons that are not obvious! Only a minority gave the expected answer of 6 %. A very common answer was 7 %, which is reached by calculating the age proportion correctly as 0.4 but using the trend line rather than the data point for the individual who was 32 years old. Careful reading of the question was needed to avoid this mistake.

Most candidates were successful here, using the argument that humans and bears have different life spans.

In this question candidates were expected to deduce the changes to bone from the concentrations of the markers. Given that neither of them reaches zero, we can assume that bone formation and resorption both continue throughout the hibernation period. Few candidates made this point clearly. The other 'best' answer was to comment on relative overall amounts of formation and resorption, either by stating that they are approximately equal or that there is slightly more resorption.

Candidates were expected to suggest a difference between the bears and humans in the concentration of one or both of the markers, that would result in loss of bone mass. A common fault was to predict changes in the markers during a period of inactivity in humans, but changes over time were not the issue — it was differences between humans and bears that were relevant to the question.

A very small proportion of candidates correctly calculated the percentage difference. There were many different incorrect answers, with 350 % being the most popular, which is the amount of osteocalcin during hibernation as a percentage of the amount pre-hibernation.

This type of question is increasingly well answered with more candidates understanding that two variables being positively correlated does not prove a causation. The situation would have been different if an experiment had been done with the levels of parathyroid hormone as the independent variable controlled by the researchers.

Answers here were very varied. The best included the idea that an understanding of how bears maintain bone mass despite inactivity during hibernation might lead to preventative treatments for humans. Because this was a discuss question, some counterargument was expected, based on differences between bears and humans.

Identify the dark structure indicated by I.

Identify the protein producing the thick filament in the dark band indicated by II.

Identify the structure indicated by III.

Discuss whether the tissue shown in the micrograph consists of cells or not.

Explain how calcium is involved in muscle contraction.

nucleus ✔

myosin ✔

muscle fibre/muscle cell ✔

Reject myofibril because it would be much narrower – diameter 

a. «muscle fibres are» multinucleate/contain many nuclei «whereas cells are expected to have only one/so muscle fibers are an exception to the cell theory» ✔

b. one cell membrane/sarcolemma enclosing a whole muscle fibre «as expected for cells» ✔

c. very large/much larger/longer/than most cells ✔

d. muscle fibres formed by fusion of cells/are syncytia ✔

a. action potential/nerve impulse causes release of calcium ✔

b. from sarcoplasmic reticulum/specialized endoplasmic reticulum ✔

c. binds to troponin ✔

d. causes tropomyosin to move/be removed «from binding sites» ✔

e. exposes myosin-binding sites on actin/allows myosin «heads» to bind to actin ✔

Explain how circulation of the blood to the lungs and to other systems is separated in humans and what the advantages of this separation are.

Distinguish between the composition of the blood of the renal artery and the blood of the renal vein.

a. double circulation / pulmonary and systemic circulations 

b. heart is a double pump / heart has separate pumps for lungs and other systems / left and right sides of heart are separate / no hole in heart (after birth)

c. deoxygenated blood pumped to the lungs and oxygenated to other organs/tissues/whole body (apart from lungs)

d. each side of the heart has an atrium and a ventricle

e. left ventricle/side pumps blood to the systems/tissues and right ventricle/side pumps blood to the lungs

f. left atrium receives blood from the lungs and right atrium receives blood from systems/tissues

g. left ventricle pumps blood via the aorta and right ventricle pumps blood via the pulmonary artery

h. left atrium receives blood via the pulmonary vein and right atrium receives blood via the vena cava

i. lungs require lower pressure blood / high pressure blood would damage lungs

j. high pressure required to pump blood to all systems/tissues apart from lungs

k. pressure of blood returning from lungs not high enough to continue to tissues / blood has to be pumped again after returning from lungs

l. oxygenated blood and deoxygenated blood kept separate / all tissues receive blood with high oxygen content/saturation

Points may be earned using an annotated diagram.

a. less urea/excretory waste products/creatinine in renal vein

b. less oxygen in the renal vein

c. more carbon dioxide in renal vein

d. less glucose in renal vein

e. concentration of sodium ions/chloride ions/pH at normal level in the renal vein whereas it is variable in renal artery

f. solute concentration/osmolarity/water balance at normal level in the renal vein whereas it is variable in renal artery

Allow answers in a table format. For all these mark points accept the converse as long as it is clear whether the artery or vein has the higher amount.

Answers relating to volume and pressure are not relevant to the question.

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.

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 ✔

State the role of plasma cells in the immune system.

Describe the production of hybridoma cells.

State one possible use of hybridoma cells.

produce/secrete antibodies

a. antigen injected into mouse/mammal/host

    Accept animal

b. B cells/B lymphocytes/plasma cells «obtained/extracted from host»

c. fusion «of plasma cell» with myeloma cell/tumour cell

d. division «of hybridoma cells» to produce a clone

[Max 2 Marks]

produce monoclonal antibodies
OR
diagnosis of diseases/malaria/cancer/HIV
OR
treatment of rabies
OR
blood and tissue typing
OR
pregnancy testing
OR
targeting of cancer cells «with a chemotherapy drug»
OR
treatment of infection if too late for vaccination/successful immune response

Only accept the first use of hybridoma cells given in the answer

Not treatment of malaria

Outline how the properties of water make it an effective coolant for the body.

Describe how changes in weather conditions affect the transport and loss of water in plants.

Explain how water balance is restored in mammals when they are dehydrated.

1. hydrogen bonds hold water molecules together/make water molecules cohere;
2. evaporation requires breaking of hydrogen bonds / heat needed to break hydrogen bonds
3. water has a high heat of vaporization/high latent heat;
4. evaporation of water/sweat removes heat from/cools the skin/body;

1. water (vapor) lost by transpiration/through stomata;
2. transpiration/loss of water from leaves causes transport of water (in xylem);
   Temperature:
3. faster/more water loss/transpiration/transport in hotter weather;
4. more heat for evaporation;
   Humidity:
5. slower/less water loss/transpiration/transport in more humid weather;
6. faster diffusion of water (vapor) out of the leaf/through the stomata with low humidity outside;
   OR
   no evaporation if air is saturated with water vapor/with 100 % humidity;
   Wind:
7. faster/more water loss/transpiration/transport in windy/windier weather;
8. wind/air movement carries away water vapor from around the leaf/stomata;
9. high winds can cause stomatal closure and so reduce transpiration;
   Drought:
10. drought causes stomata to close so reduces loss/transport;

1. thirst;
2. more water drunk / more water reabsorbed from feces (in the colon/large intestine);
3. osmoreceptors in the hypothalamus detect dehydration/high solute concentration in blood;
4. ADH secreted;
5. by the pituitary gland;
6. ADH signals to collecting duct/DCT (cells) to increase permeability to water;
7. more aquaporins (in plasma membranes of collecting duct/distal convoluted tubule cells);
8. more water reabsorbed from filtrate (in collecting ducts/distal convoluted tubules);
9. reabsorption by osmosis / reabsorption due to medulla being hypertonic;
10. reabsorbed water passes into the blood/reduces the solute concentration of blood;
11. smaller volume/more concentrated/hypertonic urine formed;
12. less sweating;

This question revealed widespread misunderstanding of the properties of water and changes of state. Most candidates though that water's high specific heat capacity explains its role as a coolant and that sweat cools the body by taking heat from it to raise its temperature. Few candidates referred to the requirement for heat to break hydrogen bonds as water evaporates. To be able to understand biological processes properly, a firm grounding in physics and chemistry is needed, but all too often it was lacking.

More marks were scored here, but again there was widespread misunderstanding of the forces that cause water to evaporate and diffuse out of leaves, and of how conditions such as humidity, temperature and wind can influence the process. Few candidates mentioned concentration gradients between air spaces in the leaf and the atmosphere outside the leaf. There was a tendency to get trends the wrong way round, for example by suggesting that transpiration increases during rainfall because plants have plenty of water and therefore choose to open their stomata more widely.

This was poorly answered by many candidates. The discrimination index was very high but the mean mark was only 1.5 out of 7. Often candidates did not get beyond the basic ideas of thirst and drinking water to rehydrate. The best candidates gave concise but detailed accounts of the roles of the hypothalamus, pituitary gland, collecting duct and aquaporins.

Outline reasons for the therapeutic use of stem cells.

Describe how monoclonal antibodies are produced.

Explain the role of the electron transport chain in the generation of ATP by cell respiration.

1. unspecialized/undifferentiated stem cells can divide/differentiate along different pathways;
2. (stem cells are accessible as they) come from embryos/bone marrow/umbilical cord blood/adult tissue;
3. (stem cells) can regenerate/repair/regrow diseased/damaged tissues in people;
4. valid specific example;
5. drugs can be tested on stem cells (in laboratories to see if they are harmful);

1. mice/rabbit/small mammal injected with one type of antigen;
2. cells from the spleen/antibody-producing cells are removed;
3. plasma cells that produce antibodies (are used);
4. myeloma/tumor cells that divide endlessly (are used);
5. fusion of plasma cells with tumor/myeloma cells / fusion produces hybridoma cells;
6. selection of hybridoma cells / medium used that only allows growth of hybridoma cells;
7. fused cells/hybridoma cells are cultured/grown in tissue culture/grown in a fermenter;
8. (hybridoma) cells divide endlessly and produce the desired antibodies;

1. electron transport chain performs chemiosmosis / chemiosmosis generates ATP;
2. receives energy/electrons from oxidation reactions/from Krebs cycle/glycolysis;
3. receives electrons from reduced NAD/NADH/reduced FAD/FADH;
4. energy released as electrons pass from carrier to carrier (in the chain);
5. release of energy (from electron flow) coupled to proton pumping;
6. protons pumped into intermembrane space;
7. creates proton gradient;
8. protons diffuse back/move down the concentration gradient (across membrane);
9. protons pass through ATP synthase;
10. protons return to the matrix;
11. flow of protons provides energy for generating ATP;
12. electrons transferred to oxygen at end of electron transport chain;

Most candidates knew something of the therapeutic uses of stem cells, including differentiation for specific roles. There was a tendency for over-optimism over what can be fixed using stem cells. For example, stem cells are not a treatment for most cancers. The best answers stuck to well-established procedures such as the treatments for leukaemia using stem cells from bone marrow.

Production of monoclonal antibodies was not widely understood and the mean mark was below 1 (out of 5). Many answers described the normal immune response by the body that results in production of antibodies, rather than the production of hybridoma cells that allow large-scale antibody manufacture.

For well-prepared candidates this question posed no difficulty and there were some excellent detailed accounts of chemiosmosis. The discrimination index was the highest for any question on the paper, indicating that there was no room for lucky guesses about the biology here!

Outline four different processes, with examples, that allow substances to pass through the plasma membrane.

Humans need to balance water and solute concentrations and also excrete nitrogenous wastes. Explain how the different parts of the kidney carry out these processes.

Describe adaptations in mammals living in desert ecosystems to maintain osmolarity in their bodies.

1. simple/passive diffusion down a concentration gradient / from high concentration to low concentration (without the use of channels/proteins); (e.g., CO₂ / O₂ / H₂O / steroid hormones)
2. osmosis is the diffusion of water from an area of high water potential / low solute concentration to low water potential / high solute concentration;
3. facilitated diffusion is passive transport/diffusion through a protein channel; (e.g., glucose)
4. active transport requires energy/ATP to move the molecules through a protein channel (e.g., Na-K pump / sodium potassium pump) against a concentration gradient/from low solute concentration to high concentration;
5. endocytosis is the infolding of membranes to form a vesicle and take in a large molecule; (e.g., macrophages engulfing pathogens)
6. exocytosis is the fusion of vesicles with membranes to release a large molecule; (e.g. neurotransmitters)

1. humans are osmoregulators/maintain the internal concentrations of the blood/osmolarity within specific/ limited range / OWTTE;
2. glomerulus / Bowman’s capsule (in the nephron) carry out ultrafiltration;
3. proximal convoluted tubule selectively reabsorbs glucose/solute/salts/amino acids;
4. loop of Henle maintains hypertonic conditions in the medulla/absorbs salts (by active transport);
5. loop of Henle reabsorbs water (by osmosis);
6. (osmoreceptors in the hypothalamus) cause production of ADH if the blood is too concentrated / person is dehydrated / OWTTE;
7. ADH causes more uptake of water/increases permeability in the collecting duct;
8. resulting in a more concentrated urine / lower volume of urine;
9. excess amino acids are broken down producing nitrogenous waste / ammonia / urea as a result;
10. ammonia is toxic and is converted into non-toxic urea;
11. urea is eliminated in the urine;

Marks can be awarded to clearly annotated diagrams.

1. behavioural adaptations to avoid over-heating / hiding in burrows/out of sun during hot period of day / active at cooler times of the day/nocturnal animals / panting;
2. adaptations for heat exchange such as large ears;
3. may have longer loop of Henle (to reabsorb more water);
4. may produce more ADH (according to osmotic concentrations of the blood) / produce concentrated urine / lower volume of urine;
5. camel humps that store fat that releases (metabolic) water when broken down;
6. reduced sweat;
7. any other valid adaptation; (e.g., light coloured coats)

This question in section B was the second most commonly selected, but in performance, tended to do more poorly than the other questions.

Most candidates could list the processes that allow passage across the plasma membrane, but many answers lacked some of the necessary elements, most commonly leaving out examples (e.g. an example for gradient).

Generally, there were good accounts of the functioning of the nephron, but some elements were missing such as which substances are reabsorbed in the proximal tubule and which are absorbed in the loop of Henle, although permeability was mentioned. Discussion about the role of ADH was well done. The discussion of the management of nitrogenous wastes was least well done in this question.

Long loops of Henle was the most common desert adaptation discussed. There were common misconceptions about camels’ humps being water storage organs rather than the production of metabolic water.