GENERAL CERTIFICATE OF EDUCATION
TYSTYSGRIF ADDYSG GYFFREDINOL
2009 - 2010
HUMAN BIOLOGY
SPECIMEN QUESTION PAPERS
SPECIMEN MARKING SCHEMES
GCE HUMAN BIOLOGY Specimen Assessment Materials 3
Contents
Page
Question Papers
Unit 1
Biology/Human Biology
5
Unit 2
Human Biology
21
Unit 4
Human Biology
37
Unit 5
Biology/Human Biology
53
Unit 1
Biology/Human Biology
71
Unit 2
Human Biology
79
Unit 4
Biology/Human Biology
87
Unit 5
Biology/Human Biology
95
Mark Schemes
GCE HUMAN BIOLOGY Specimen Assessment Materials 5
CYD-BWYLLGOR ADDYSG CYMRU
Tystysgrif Addysg Gyffredinol
Uwch Gyfrannol/Uwch
WELSH JOINT EDUCATION COMMITTEE
General Certificate of Education
Advanced Subsidiary/Advanced
BIOLOGY/HUMAN BIOLOGY
UNIT 1
SPECIMEN PAPER
(1 hour 30 minutes)
INSTRUCTIONS TO CANDIDATES
Write your name, centre number and candidate number in the spaces at the top of this page.
Answer all questions.
Write your answers in the spaces provided in this booklet.
INFORMATION FOR CANDIDATES
The number of marks is given in brackets at the end of each question or part-question.
You are reminded of the need for good English and orderly presentation in your answers.
You are reminded that assessment will take into account the quality of written communication
used in your answers.
No certificate will be awarded to a candidate detected in any unfair practice during the
examination.
GCE HUMAN BIOLOGY Specimen Assessment Materials 6
1.
Complete the table below by placing ticks in the appropriate column(s). Each row may have
one or more ticks.
Monosaccharide
Disaccharide
Structural
Polysaccharide
Storage
Polysaccharide
Glucose is an example of
Cellulose is an example of
Amylose is an example of
Maltose is an example of
Insoluble
Deoxyribose is an example of
(Total 6 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 7
2.
The diagram below shows a section of DNA molecule.
(a)
Name the parts labelled A and B.
[1]
A ............................................
B ............................................
(b)
What type of bonding is shown at E?
[1]
..............................................................................................................................
(c)
(i)
What type of nitrogenous base is guanine?
[1]
..............................................................................................................................
(ii)
Name the bases C and D.
[2]
C ............................................
D ............................................
(d)
On the diagram above, draw a ring around a single nucleotide unit.
[1]
GCE HUMAN BIOLOGY Specimen Assessment Materials 8
(e)
A large sample of DNA was analysed and it was found that guanine made up 28% of
the number of base molecules present. Calculate the number of thymine base
molecules present as a percentage.
You must show your working.
[3]
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(Total 9 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 9
3.
Beetroot cells contain a red pigment.
Cut cubes of beetroot were washed to remove the red pigment from damaged surface cells.
The cubes were placed in test tubes containing 10cm3 water, in waterbaths at a range of
temperatures.
The tubes were left at each temperature for 10 minutes.
The beetroot was removed from the tubes and the colour of the solution in the tube measured
on a colorimeter.
The darker the red colour of the solution, the greater the colorimeter reading.
The following results were obtained.
Temperature (ºC)
Colorimeter
Reading
(Arbitrary
units)
(i)
10
20
30
40
50
60
70
0
1
4
7
80
90
100
Describe the pattern of results from the above table.
[2]
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(ii)
Explain the results as fully as you can. Refer to cell membrane structure in your
answer.
[3]
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(iii)
Explain the difference in the results if you were investigating the effect of ethanol on
the permeability of the membrane.
[2]
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(Total 7 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 10
4.
In an experiment the enzyme catalase, obtained from liver tissue, was used to break down
hydrogen peroxide releasing oxygen. The experiment was carried out at selected pH values
and the volume of oxygen given off was measured at intervals. The results obtained were
plotted as shown below.
(a)
Which one of the four pH values would be nearest to the optimum for catalase?
[1]
..........................................
(b)
Select the letter which indicates:
(i)
The fastest rate of reaction.
[2]
..........................................
Give a reason for your choice. ................................................................
..................................................................................................................
(ii)
The least concentration of hydrogen peroxide.
[2]
..................................................................................................................
Give a reason for your choice ...............................................................
..................................................................................................................
(c)
Explain fully the results obtained at pH 4.4.
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[3]
GCE HUMAN BIOLOGY Specimen Assessment Materials 11
(d)
A control was set up at optimum pH conditions using boiled and cooled liver.
Suggest an explanation for the shape of the control curve shown on the above graph.
[1]
..............................................................................................................................
..............................................................................................................................
(e)
Outline how the experiment could be modified in order to investigate the effect of the
concentration of catalase on the rate of the reaction.
[2]
..............................................................................................................................
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(Total 11 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 12
5.
Milk can be made lactose-free by passing it down a column of the immobilised enzyme
lactase.
An experiment was carried out to determine the optimum size of alginate beads to use in this
process.
Three bead sizes were prepared and placed in columns. The same volume of milk was run
into each column at the same rate of flow.
The percentage product for each experiment was calculated.
The entire experiment was repeated a number of times.
Bead diameter (mm)
Mean percentage of product
(a)
(i)
2
4
6
98
84
70
Suggest the bead size which should be used in the process. Give a reason for
your answer.
[1]
..................................................................................................................
..................................................................................................................
(ii)
Give two reasons for the different results from the three bead sizes.
[2]
..................................................................................................................
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..................................................................................................................
(iii)
What would you expect to happen to the results if the flow rate was
decreased? Explain your answer.
[2]
..................................................................................................................
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(iv)
Which other factor should be kept constant during the experiment?
[1]
..................................................................................................................
(b)
Give two advantages of using immobilised enzymes in industrial processes.
[2]
..............................................................................................................................
..............................................................................................................................
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(Total 8 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 13
6.
The diagram represents the nucleus of an animal cell (2n = 6) at early prophase of
mitosis.
(a)
In the space below draw an annotated diagram to indicate what happens to this cell at
anaphase of mitosis.
[4]
GCE HUMAN BIOLOGY Specimen Assessment Materials 14
(b)
What events must occur during interphase before mitosis can take place?
[5]
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(Total 9 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 15
7.
When a potato tuber is cut open, the surfaces exposed to the air gradually turn brown due to
the production of dark brown coloured pigments from phenolic compounds (e.g. catechol).
The enzyme, polyphenol oxidase, catalyses this process.
The graph below shows the results of an experiment in which different concentrations of the
enzyme were added to tubes containing catechol solution. The tubes were kept at a constant
temperature and shaken periodically during the experiment.
The time taken for a standard brown colour to develop was recorded and from this the rate of
reaction was calculated, using
1
for the colour to develop.
time in minutes
A graph was then plotted of rate against enzyme concentration.
(a)
What general conclusion concerning enzyme action can be drawn from these results?
[1]
..............................................................................................................................
..............................................................................................................................
(b)
At what enzyme concentration would the standard colour be obtained after 18
minutes? Show your working.
[2]
..............................................................................................................................
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GCE HUMAN BIOLOGY Specimen Assessment Materials 16
(c)
Explain why the addition of lemon juice (citric acid) to the tubes stops any colour
change.
[1]
..............................................................................................................................
..............................................................................................................................
(d)
Explain what is likely to happen to the rate of colour change if the temperature of a
tube containing the enzyme polyphenol oxidase and catechol was gradually raised
from 10ºC to 100ºC.
[2]
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
(e)
Suggest what might happen to the rate of reaction if a competitive inhibitor was
added. Explain your answer.
[2]
..............................................................................................................................
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..............................................................................................................................
..............................................................................................................................
(f)
'All enzymes are proteins', describe how you would test to show that polyphenol
oxidase is a protein.
[2]
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
(Total 10 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 17
8.
Answer ONE of the following questions.
Any diagrams included in your answer must be fully annotated.
EITHER
(a)
Distinguish between primary, secondary, tertiary and quaternary
structure of proteins.
[10]
OR
(b)
With the aid of labelled drawings, compare and contrast a
mitochondrion and a chloroplast.
[10]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 18
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GCE HUMAN BIOLOGY Specimen Assessment Materials 19
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GCE HUMAN BIOLOGY Specimen Assessment Materials 21
WELSH JOINT EDUCATION COMMITTEE
General Certificate of Education
Advanced
CYD-BWYLLGOR ADDYSG CYMRU
Tystysgrif Addysg Gyffredinol
Uwch
HUMAN BIOLOGY
UNIT 2
SPECIMEN PAPER
(1 hour 30 minutes)
INSTRUCTIONS TO CANDIDATES
Write your centre number, name and candidate number in the spaces provided above.
Answer all questions.
Write your answers in the spaces provided in this booklet.
INFORMATION FOR CANDIDATES
The number of marks is given in brackets at the end of each question or part-question.
You are reminded of the need for good English and orderly presentation in your answers.
You are reminded that assessment will take into account the quality of written communication
used in your answers.
No certificate will be awarded to a candidate detected in any unfair practice during the
examination.
GCE HUMAN BIOLOGY Specimen Assessment Materials 22
1.
Complete the following table, using a tick (9) to indicate which type of immunity is shown
by each of the following circumstances.
[4]
Active
natural
Active
artificial
Passive
natural
Passive
artificial
Exposure to measles
Receiving MMR vaccine
Receiving anti-rabies injection
Transfer of antibodies from mother to
child in breast milk
(Total 4 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 23
2.
Before competing, athletes often train at high altitudes for several weeks after which time
their red blood cell count increases. The table shows these changes in a group of athletes.
(a)
(i)
Altitude/m
Number of red blood cells
/ 1012dm-3
0
5.0
6000
6.20
Calculate the percentage increase in red blood cells in the athletes after
several weeks at 6000 metres. Show your working.
[2]
............................................................................................................................
............................................................................................................................
(iii)
Explain the benefit of this increase in red blood cell count.
[3]
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
(b)
Name the form in which oxygen is transported in the red blood cells.
...........................................................................
[1]
GCE HUMAN BIOLOGY Specimen Assessment Materials 24
(c)
The following diagram shows blood pressure changes as blood travels through one
circuit of the circulatory system.
(i)
Explain fully the reasons for the alternating high and low blood pressure in
region A.
[4]
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
(ii)
Why does the blood pressure decrease so rapidly in region B.?
[1]
..................................................................................................................
..................................................................................................................
(d)
Why is it important that the blood pressure in the lung capillaries is lower than that in
the body capillaries?
[1]
..............................................................................................................................
..............................................................................................................................
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GCE HUMAN BIOLOGY Specimen Assessment Materials 25
(e)
There is a difference between the arterial pressures and the capillary pressures.
Explain, in relation to the functions of these vessels, why this is more efficient. [4]
..............................................................................................................................
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(f)
Give one reason to explain how a return flow to the heart is possible when the vein
pressures are so low.
[1]
..............................................................................................................................
..............................................................................................................................
(Total 17 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 26
3.
(a)
The graph shows the concentration of antibodies for the virus that causes measles in
the blood of a child following MMR vaccination. At point X the child comes into
contact with another child who is suffering from measles.
(i)
Explain what is happening during the first week after vaccination.
[2]
............................................................................................................................
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(ii)
Explain what is happening between weeks 1 and 3 after vaccination.
[2]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 27
(iii)
Explain the response after point X.
[3]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 28
(b)
Measles vaccine was first introduced in 1968. In 1988 the combined MMR vaccine
was introduced. A vaccination campaign for all school age children was conducted in
1994. For the vaccine to be effective in providing immunity for the whole
population, it is estimated that 90% of the children should be vaccinated.
The graph below shows the number of reported cases of measles and the uptake of the
vaccination in England and Wales from 1967 to 2000.
(i)
With reference to the above information and the graph, explain the change in
the number of reported cases of measles during this period.
[4]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 29
(ii)
Suggest why a vaccination campaign was introduced in 1994.
[1]
............................................................................................................................
............................................................................................................................
(iii)
Since 2000, a significant number of parents have refused to allow their
children to be given the MMR vaccine. In some areas, vaccine uptake has
reduced to only 70%. Suggest why fewer children are being vaccinated and
the likely effect on the number of cases of measles.
[4]
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(Total 16 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 30
4.
(a)
Describe the digestion of carbohydrate in the alimentary canal.
[6]
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(b)
Some people continually overproduce acid in the stomach. During acid reflux, the
acidic contents of the stomach move up into the oesophagus, causing irritation and
damage to the cells lining the oesophagus. Acid reflux is one cause of 'heartburn'.
A pH probe was used to discover whether a patient's 'heartburn' was caused by acid
reflux. This probe monitored the pH in the oesophagus from 8am to midnight. The
results are shown below.
GCE HUMAN BIOLOGY Specimen Assessment Materials 31
(i)
State two functions of the acid secreted in the stomach.
[1]
............................................................................................................................
............................................................................................................................
............................................................................................................................
(ii)
Why are cells lining the stomach not affected by acid in the same way as the
cells lining the oesophagus?
[1]
............................................................................................................................
(iii)
Suggest an explanation for the pH values observed in the oesophagus up to
an hour after a meal is eater.
[2]
............................................................................................................................
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(Total 10 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 32
5.
The diagram below, shows an alveolus and its blood supply. The numbers show the relative
amounts of oxygen and carbon dioxide in various positions.
(a)
(i)
Name the process by which gases enter and leave the blood from the
alveolus.
[1]
............................................................................................................................
(ii)
Suggest the value for the amount of carbon dioxide in the alveolus.
[1]
............................................................................................................................
(iii)
Which of the vessels R and S represents the arterial end of the capillary and
explain your choice..
[2]
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(b)
Why do the amounts of oxygen and carbon dioxide remain relatively constant in the
alveolus?
[1]
..............................................................................................................................
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(c)
Explain the function of the surfactant found in the alveolus.
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[1]
GCE HUMAN BIOLOGY Specimen Assessment Materials 33
(d)
The diagram below compares a normal spirometer trace with one from a trained
athlete and one from an asthma sufferer.
(i)
What feature shows the effect of training on the trace obtained from the
athlete?
[1]
............................................................................................................................
(ii)
What additional feature distinguishes the asthma sufferer's trace from the
other traces and explain the cause of this feature.
[2]
............................................................................................................................
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(e)
Carbon dioxide in the exhaled gases is removed by passing the gases through soda
lime or carbosorb before they are returned to the spirometer. Explain why the carbon
dioxide levels in the blood would rise if the carbon dioxide in the exhaled gases was
not removed.
[2]
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(f)
Explain why the spirometer cannot be used to measure the total capacity of the lungs.
[2]
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(Total 13 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 34
6.
Answer one of the following questions.
Any diagrams included in your answer must be fully annotated.
Either, (a)
Describe the short and long term effects of smoking on the respiratory
system.
[10]
Or
Describe how a knowledge of the life cycle of both parasite and vector is
important in controlling malaria.
[10]
(b)
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GCE HUMAN BIOLOGY Specimen Assessment Materials 35
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GCE HUMAN BIOLOGY Specimen Assessment Materials 36
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GCE HUMAN BIOLOGY Specimen Assessment Materials 37
WELSH JOINT EDUCATION COMMITTEE
General Certificate of Education
Advanced
CYD-BWYLLGOR ADDYSG CYMRU
Tystysgrif Addysg Gyffredinol
Uwch
HUMAN BIOLOGY
UNIT 4
SPECIMEN PAPER
(1 hour 45 minutes)
INSTRUCTIONS TO CANDIDATES
Write your centre number, name and candidate number in the spaces provided above.
Answer all questions.
Write your answers in the spaces provided in this booklet.
INFORMATION FOR CANDIDATES
The number of marks is given in brackets at the end of each question or part-question.
You are reminded of the need for good English and orderly presentation in your answers.
You are reminded that assessment will take into account the quality of written communication
used in your answers.
No certificate will be awarded to a candidate detected in any unfair practice during the
examination.
GCE HUMAN BIOLOGY Specimen Assessment Materials 38
1.
The diagram below represents a synapse in the nervous system.
(a)
When a nerve impulse reaches a synapse, synaptic vesicles fuse with the presynaptic
membrane and shed their contents into the synaptic cleft. State the general names
given to:
(i)
the contents of the synaptic vesicle;
[1]
........................................................................
(ii)
the shedding of contents into the synaptic cleft.
[1]
........................................................................
(b)
Explain fully how the shedding of the vesicle contents is caused.
[2]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 39
(c)
Intrinsic proteins act as channels across the post synaptic membrane.
(i)
What is the other main chemical constituent of the membrane?
[1]
........................................................................
(ii)
The diagram shows X passing through these channels. Name X.
[1]
........................................................................
(d)
Explain how synaptic transmission can be disrupted by using an enzyme inhibitor. [2]
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(e)
The chemical caffeine, which is present in tea, coffee and cola drinks, increases the
metabolic rate in presynaptic cells. Explain fully how strong coffee stimulates the
nervous system.
[2]
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(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 40
2.
The diagram shows the arrangement of protein filaments in a muscle fibril.
(a)
(b)
If the muscle fibre contracted, how would the appearance of regions P and Q
compare with the diagram?
[2]
(i)
P
..............................................
Q
..............................................
What is meant by the primary structure of a protein molecule?
[1]
............................................................................................................................
............................................................................................................................
(ii)
Name the specific, three dimensional, spiral shaped, secondary structure of
muscle fibril protein molecules.
[1]
............................................................................................................................
GCE HUMAN BIOLOGY Specimen Assessment Materials 41
(c)
The data below were collected during an experiment to investigate the effects of
training on respiration in human muscle.
In the investigation a group of trained athletes and a control group of non-athletes
exercised at different levels. The levels of exercise were expressed as rates of energy
expenditure in J min-1kg-1 of body mass. At each level, oxygen consumption and
lactic acid production were measured and converted into rates per kg of body mass.
The results are shown in the table below, the figures being the means of each group.
Rate of oxygen consumption/
cm3 min-1kg-1
Rate of lactic acid
production/mg min-1 kg-1
Rates of energy
expenditure/
J min-1kg-1
Athletes
Non-Athletes
600
30
29
0
0
800
40
39
0
0
1000
50
44
0
185
1200
57
45
85
350
1400
58
45
305
590
(i)
Athletes
Non-Athletes
Describe and compare the effects of increasing the levels of exercise on rates
of oxygen consumption in athletes and non-athletes. Suggest an explanation
for these differences.
[3]
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
(ii)
Describe and compare the effects of increasing the levels of exercise on rates
of lactic acid production in athletes and non-athletes. Suggest an explanation
for these differences.
[3]
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
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(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 42
3.
ATP is a compound that is described as the universal energy currency in living organisms.
(a)
(i)
Using the structure of ATP, explain what is meant by the phrase energy
currency.
[5]
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
(ii)
Give two examples of the use of ATP.
[1]
.........................................................................................
........................................................................................
(b)
A diagram summarising the role of the electron transport system in ATP production is
shown below.
This process takes place during both respiration and photosynthesis.
GCE HUMAN BIOLOGY Specimen Assessment Materials 43
State the electron donor (X) and the final electron acceptor (Y) during
(i)
respiration;
[2]
X .........................................
Y .........................................
(ii)
photosynthesis.
[2]
X .........................................
Y .........................................
(c)
As the electron is transferred along the electron transport chain, energy is made
available for ATP formation.
Explain how this energy is used to produce ATP.
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
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[4]
GCE HUMAN BIOLOGY Specimen Assessment Materials 44
(d)
Dinitrophenol is a chemical which was used in the manufacture of explosives during
the First World War but it poisoned some of the workers. If dinitrophenol is absorbed
into the body it reduces the quantity of ATP produced by the electron transport chain.
The electron transport chain continues but most of the energy released is in the form
of heat. Explain why people suffering from dinitrophenol poisoning experience
symptoms such as lack of energy and weight loss.
[4]
.........................................................................................................................................
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(Total 18 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 45
4.
Bacteria can be cultured in the laboratory on agar plates.
(a)
State three conditions which are necessary for growth of bacterial colonies on these
plates.
[2]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
(b)
Suggest how you could ensure that aseptic and sterile conditions are present during
the preparation of agar plates and the transfer of the bacteria from the culture medium
to the plates.
[3]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
(c)
Some substances are know to interfere with the action of antibiotics and are called
antagonist. Two strips of filter paper, one soaked with the antibiotic and the other
soaked with a suspected antagonist, are placed so that they cross at right angles on a
Petri dish. The Petri dish contains nutrient agar inoculated with a microorganism
sensitive to the antibiotic. Draw and label on the diagram the pattern of bacterial
growth you would expect to see after 48 hours incubation.
[2]
GCE HUMAN BIOLOGY Specimen Assessment Materials 46
(d)
Three students were carrying out an experiment in which bacteria were cultured in a
liquid. A sample of the bacteria was stained purple by the Gram stain.
(i)
What does the staining indicate about the structure of the bacterial cell wall?
[2]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(ii)
In order to monitor population growth, a number of different methods may be
used.
One student suggested using a viable count.
(i)
What assumption must be made when using this method?
[1]
...............................................................................................................
...............................................................................................................
(ii)
State one limitation of using this method.
[1]
...............................................................................................................
...............................................................................................................
(iii)
Another student used a total count method. He counted all the bacteria in the
field of view of the microscope.
Suggest why this method gave a higher estimate of the population than the
viable count.
[1]
............................................................................................................................
GCE HUMAN BIOLOGY Specimen Assessment Materials 47
(iv)
In both methods the original culture must be diluted before a count can be
made.
The diagrams below show how a dilution was carried out and the result of
incubating 1cm3 of the diluted sample on a nutrient agar plate for 24 hours.
Using this information, complete the dilution factors and calculate the
estimated total population in 1cm3 of the original culture.
Show your working.
[3]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(Total 15 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 48
5.
When kidneys fail, patients can lead a relatively normal life as the functions of the kidneys
are performed by dialysis, which is the main form of treatment available.
(a)
In haemodialysis, a connection is made to a blood vessel through which blood passes
out of the body into a machine. Once in the machine, the blood passes through a
system of tubes formed from an artificial membrane, which is bathed in a dialysis
solution, after which the blood is returned to the body. The principle of the machine
is shown in the simplified diagram below. Typically, haemodialysis sessions are
required three times a week.
Explain why blood cells would not be removed from the blood as it passes through
the machine.
[1]
...........................................................................................................................
...........................................................................................................................
GCE HUMAN BIOLOGY Specimen Assessment Materials 49
(b)
An alternative method of dialysis is peritoneal dialysis. In this, instead of the blood
being 'cleaned' using an artificial membrane outside the body, it is 'cleaned' inside the
body. It uses the peritoneum, the thin membrane that surrounds the organs of the
abdomen. The peritoneum is well supplied with blood vessels. This membrane has
two layers and the space between them is used as a reservoir for the dialysis fluid.
The principle is illustrated in the simplified diagram below (side view).
The fluid remains in the body for a few hours, after which it is drained out and
replaced with fresh dialysis fluid. This procedure needs to take place four times a
day.
(i)
Explain why the peritoneum is a suitable membrane for dialysis within the
body.
[3]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(ii)
Using the information provided and your own knowledge, suggest why
peritoneal dialysis needs to take place 4 times a day whereas haemodialysis
only needs to take place 3 times a week.
[3]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
GCE HUMAN BIOLOGY Specimen Assessment Materials 50
(iii)
Suggest how the dialysis fluid may be altered to allow more water to be
removed from the blood.
[1]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(iv)
If peritoneal dialysis is used for long-term treatment, the peritoneum becomes
thickened.
Suggest how and why this affects the efficiency of peritoneal dialysis.
[2]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(c)
One possible treatment for kidney failure is a kidney transplant. It is important to get
the best donor/recipient 'match' as possible in order to avoid severe rejection of the
new kidney. The best match will be found between identical twins, followed by
members of the immediate family.
(i)
Why is the best match between identical twins?
[1]
...........................................................................................................................
...........................................................................................................................
(ii)
Even with a good match, some degree of rejection is common. If rejection
occurs, the patient can be treated using immuno-suppressive drugs such as
anti-lymphocyte globulin or orthoclone K T-cell receptor 3 antibody.
Explain how and why the body might reject a new kidney.
[5]
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
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(Total 16 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 51
6.
Answer one of the following questions.
Any diagrams included in your answer must be fully annotated.
Either, (a)
Give an account of how the hormonal control of the kidney enables it to
function in osmoregulation in a mammal.
[11]
Or
Describe
(b)
(i)
(ii)
how the resting potential is maintained in the axon of a neurone;
how a nerve impulse is transmitted along a myelinated axon.
Graphs may be used to illustrate your answer.
[11]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 52
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(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 53
WELSH JOINT EDUCATION COMMITTEE
General Certificate of Education
Advanced
CYD-BWYLLGOR ADDYSG CYMRU
Tystysgrif Addysg Gyffredinol
Uwch
BIOLOGY/HUMAN BIOLOGY
UNIT 5
SPECIMEN PAPER
(1 hour 45 minutes)
INSTRUCTIONS TO CANDIDATES
Write your centre number, name and candidate number in the spaces provided above.
Answer all questions.
Write your answers in the spaces provided in this booklet.
INFORMATION FOR CANDIDATES
The number of marks is given in brackets at the end of each question or part-question.
You are reminded of the need for good English and orderly presentation in your answers.
You are reminded that assessment will take into account the quality of written communication
used in your answers.
No certificate will be awarded to a candidate detected in any unfair practice during the
examination.
GCE HUMAN BIOLOGY Specimen Assessment Materials 54
1.
The diagram shows the flow of energy through an ecosystem.
(a)
The energy flow through the herbivores can be expressed by the equation:
C1 = C2 + R2 + E2
(i)
Calculate the respiratory loss by the herbivores and write your answer in the
correct position on the diagram.
[1]
(ii)
Give the equation that expresses the energy flow through the carnivores. [1]
..................................................................................................................
(b)
The values given are for a small wood and are in kJm-2yr-1. The area of wood is
25 000 m2.
(i)
Calculate the total amount of energy 'expelled' by this ecosystem in one year,
showing your workings.
[2]
..................................................................................................................
..................................................................................................................
..................................................................................................................
(ii)
Describe what happens to this 'expelled' energy.
[2]
..................................................................................................................
..................................................................................................................
..................................................................................................................
(Total 6 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 55
2.
In Africa, Lake Nabugabob separated from Lake Victoria thousands of years ago. There are
five species of cichlid fish of the genus Haplochromis in Lake Nabugabob, each descended
from a different species in the main lake, Lake Victoria.
(a)
Suggest how analysis of DNA or proteins might be used to supply additional
evidence that the Lake Nabugabob fish have descended from ancestors in Lake
Victoria.
[2]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
(b)
Explain how the splitting of the fish population into Lake Nabugabob and Lake
Victoria populations has led to the formation of the separate species.
[6]
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(Total 8 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 56
3.
Duchenne muscular dystrophy (DMD) is a generative muscular disease. It is caused by a sexlinked recessive allele. A family pedigree showing inheritance is shown below.
(a)
Fully explain the term sex-linked recessive allele.
[4]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
(b)
Use the key below to answer the following questions.
KEY:
(i)
XN
Normal allele
Xn
Muscular dystrophy allele
Y
Male chromosome
The probable genotype of A, B, C and D is
[1]
…………………………………………….
(ii)
The genotype of E, F and G is
[1]
……………………………………………..
(iii)
Give the genotype of all the normal males.
…………………………………………….
[1]
GCE HUMAN BIOLOGY Specimen Assessment Materials 57
(c)
(d)
Complete the genetic diagram to show the mechanism by which E inherited the
disease.
[4]
Parents
Male
X
Female
Genotype
………….
………….
Gametes
…… ……
…… ……
Offspring Genotypes
………….
………….
………….
………….
Corresponding
Offspring Phenotypes
………….
………….
………….
………….
Discuss an argument for and an argument against the screening of apparently normal
individuals for genetic disorders.
[2]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
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(Total 13 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 58
4.
During protein synthesis, amino acids are carried by tRNA. Each amino acid is first joined to
its tRNA by an enzyme. The diagram below shows the stages in this process.
(a)
(i)
What is the name given to the region consisting of ABC?
[1]
.....................................................................
(ii)
The shape of the enzyme molecule partly depends upon different types of
bond. Name two of these bonds.
[2]
..................................................................................................................
..................................................................................................................
The diagram below represents a stage in the synthesis of the polypeptide chain.
(b)
Name
(i)
the organelle labelled C,
[1]
.......................................................................
(ii)
the structure labelled D,
.......................................................................
[1]
GCE HUMAN BIOLOGY Specimen Assessment Materials 59
(c)
The process shown arranges the correct sequence of amino acids in the peptide.
(i)
Name this process.
[1]
.......................................................................
(ii)
Explain what ensures that the correct amino acid is added.
[3]
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
(d)
After the stage shown, explain what happens to A and B.
[3]
.........................................................................................................................................
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.........................................................................................................................................
.........................................................................................................................................
(e)
It is important that this whole process is carried out correctly.
Explain why.
[2]
.........................................................................................................................................
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(Total 14 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 60
5.
The following diagram is a simplified representation of a technique used in genetic
engineering.
GCE HUMAN BIOLOGY Specimen Assessment Materials 61
(a)
Suggest why the same enzyme (enzyme A) is used to cut both the vector DNA and
the chromosomal DNA fragment.
[2]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
(b)
What is the function of enzyme B?
[1]
.........................................................................................................................................
.........................................................................................................................................
(c)
Describe two ways in which the chromosomal DNA fragment can be obtained.
[3]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
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(d)
Trials have been done to genetically modify cauliflowers to make their leaves much
greener than normal. 'Green genes' were introduced into cauliflower cells by either
attaching them to part of a cauliflower virus (which has been rendered harmless) or to
a bacterial plasmid. Suggest two reasons why there are concerns about this
technology.
[2]
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
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GCE HUMAN BIOLOGY Specimen Assessment Materials 62
(e)
Recent progress has been made towards effective treatment for cystic fibrosis. The
key problem remains developing an effective delivery system to introduce the
replacement gene into the cells. One delivery system is a virus which is known not to
cause disease in humans.
(i)
What are the two major components of a virus?
[1]
........................................................................
........................................................................
(ii)
Briefly describe another delivery system which could be used to introduce the
gene to the lung cells.
[3]
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
..................................................................................................................
(iii)
Which events must occur once the replacement gene is inside the lung cell for
the treatment to be successful?
[3]
..................................................................................................................
..................................................................................................................
..................................................................................................................
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(Total 15 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 63
6.
The cultivated banana is sterile and seedless. It reproduces asexually, producing shoots or
suckers that develop into new plants. It is extremely difficult to develop disease-resistant
varieties of this plant.
In the 1950s the main variety grown in banana plantations was Gros Michel. This variety was
totally wiped out by Panama disease, caused by a soil fungus (Fusarium oxysporum). The
disease spread rapidly from one plant to another and from plantation to plantation.
Following the devastation, a variety of banana that was resistant to Panama disease,
Cavendish, was planted and has been used ever since. However, it became apparent that
Cavendish was susceptible to the disease Sigatoka, caused by another fungus
(Mycosphaerella fijiensis). Only the use of massive amounts of fungicide spray is keeping
Sigatoka under control. It seems that as soon as a new fungicide is used that fungus develops
resistance to it.
(a)
(i)
State the genus of the fungus that causes Panama disease.
[1]
..........................................................................
(ii)
Explain why Panama disease spread so quickly through the banana
plantations and wiped out the Gros Michel variety in the 1950s.
[3]
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
(b)
Explain how the fungus that causes Sigatoka becomes resistant to fungicides.
[7]
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GCE HUMAN BIOLOGY Specimen Assessment Materials 64
(c)
Explain why it is so difficult to develop disease-resistant varieties of banana.
[3]
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(Total 14 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 65
8.
Answer one of the following questions.
Any diagrams included in your answer must be fully annotated.
Either, (a)
Give an account of the steps involved in cloning plants and mammals.
Describe one advantage and one disadvantage of these techniques.
or
Man has been accused of rape of the land and the sea and with agricultural
exploitation.
What are the consequences of:
(b)
(i)
(ii)
[10]
forest destruction
fishing.
What should be done to allow sustainability of both these resources?
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[10]
GCE HUMAN BIOLOGY Specimen Assessment Materials 66
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GCE HUMAN BIOLOGY Specimen Assessment Materials 67
MARK SCHEMES
GCE HUMAN BIOLOGY Specimen Assessment Materials 69
STRICTLY CONFIDENTIAL
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
MARK SCHEME
COVER SHEET
AS/A2 UNITS
NOTES
Assistants are asked to read and digest thoroughly all the information set out in the document
"Instructions for Examiners". It is essential for the smooth running of the examination that these
instructions are adhered to by all. Particular attention should be paid to the following instructions
regarding marking:
1.
The markscheme indicates the central information required to access the mark. However
points, particularly in extended prose answers, should be made in the correct context and
expanded sufficiently beyond a basic phrase or term in order to provide a complete and
coherent answer to the question asked including relevant explanation of the points being
made.
2.
Compliance with the mark scheme is required of all assistants in order to ensure
comparability. The mark scheme shows a variety of suitable correct answers to the questions.
Over-rigidity in its interpretation is not intended and it is accepted that points may be made in
a variety of different ways to include converse and counterpoints. Thus, except where terms
are specifically requested, all correct responses even if expressed using different words are
acceptable provided the points are explicit, unambiguous and made in the correct context. If
the meaning is unclear then the mark cannot be awarded.
3.
Marking is to be carried out with a red ballpoint or felt-tip pen. The ticks are to be placed on
the word or phrase which qualifies for the mark.
4.
Sub-totals in the margins must clearly be seen to be the sum of ticks within the script and all
questions should show evidence of marking. The total for the question should be written at
the end of the question. Transfer each checked total to the front cover.
GCE HUMAN BIOLOGY Specimen Assessment Materials 71
UNIT 1: BIOLOGY/HUMAN BIOLOGY
MARK SCHEME
1.
Monosaccharide
Glucose is an
example of
Disaccharide
Structural
Polysaccharide
Storage
Polysaccharide
[1]
9
Cellulose is an
example of
[1]
9
Amylose is an
example of
9
Maltose is an
example of
[1]
9
9
Insoluble
Deoxyribose is an
example of
[1]
9
9
[1]
[1]
(Total 6 Marks)
2.
(a)
A
B
Pentose sugar / deoxyribose (allow: 5 carbon sugar)
Phosphate
[1]
(b)
Hydrogen (Bonding) (not: H)
[1]
(c)
(i)
Purine
[1]
(ii)
C Adenine
D Thymine [allow a consequential error here]
[1]
[1]
Reject abbreviations A/T / consequential error if C is uracil.
(d)
Ring should cover any 1 base with 1 sugar and 1 phosphate (through dotted lines
i.e, bond)
[1]
(e)
If 28% G then C must be 28%
[1]
(28+28 = 56) / 100 − 56 = 44/(A+T = 44)
[1]
T = 22(%)
[1]
1 mark for each correct line.
(Total 9 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 72
3.
(i)
{very little increase/0-7 units} up to {40ºC/first 4 temperatures;
rapid rise at 50ºC/rise from 7 to 80 at 50ºC;
small rise from 80 to 100 units from 50 to 70ºC
higher temperature, darker solution/higher reading
(ii)
[2]
Membrane (proteins) are stable between 10 and 40ºC/at low temperatures;
(Appearance of the red pigment means that) the (tonoplast and) cell membrane has
been damaged between 40 and 50ºC/at high temperatures;
(meaning) pigment has leaked out of cells/sap vacuole/change in membrane
permeability/damaged membrane;
(iii)
proteins change shape/denature and come out of bilayer;
[3]
Dissolves phospholipids/destroys cell membrane structure;
[1]
(So) pigment leaks out.
(not: ref. to higher or lower readings)
[1]
(Total 7 Marks)
4.
(a)
6.5
[1]
(b)
(i)
A
[1]
the greatest volume of oxygen produced in unit time / reference to minutes [1]
(ii)
E
[1]
the greatest number of substrate / hydrogen peroxide molecules have been
broken down to produce the oxygen / all the enzyme active sites are
occupied.
[1]
(c)
A change in pH in the environment of the active site alters the ionic charge of the
acidic and basic groups of the amino acids
and causes the shape of the protein to change.
This may prevent the substrate molecules from making a close fit to the active site.
[3]
(d)
The slight reaction is due to the decomposition of hydrogen peroxide in air.
(e)
(The substrate concentration is constant)
Use the optimum pH value
[1]
Vary the concentration of catalase by dilution of minced liver / changing the mass of
the minced liver added.
[1]
[1]
(Total 11 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 73
5.
(a)
(i)
(ii)
2(mm)
(because) it gives the highest percentage of product
(not: converse)
[1]
smaller beads give a larger total area (exposed to the milk / substrate);
(not: smaller beads have large surface area)
OR
smaller beads will pack more closely (so there is more enzyme:substrate);
the milk will have longer contact / flow more slowly around small, close
packed beads;
OR
gives more time for the enzyme and substrate to come into contact
(iii)
(iv)
(b)
[2]
Percentage product would increase / get more product;
More time for enzyme-substrate complexes to occur / for reaction.
[2]
temperature / milk type / source (not: pH)
[1]
Can tolerate / more stable in a wider range of condition / temps / pH qualified;
Product not contaminated with enzyme / pure product / easily reused qualified;
Several enzymes (requiring different conditions) may be used together;
Enzymes easily added / removed = more control.
(allow: ref. to continuous production qualified,
(not: ref. to cost)
[2]
(Total 8 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 74
6.
(a)
(b)
12 single structures (chromosomes);
[1]
Separated from each other;
[1]
Labels: centrioles, spindle fibres, centromere, chromosomes
Max 2 for any 3 labels.
[2]
replication of DNA must take place;
Centrioles replicate;
Organelles e.g. mitochondria are produced;
Protein synthesis occurs;
RNA / nucleotide synthesis occurs;
High rate respiration / ATP synthesis is needed;
AVP. e.g. Two chromatids formed.
(Max. 5) (not: growth)
[5]
(Total 9 Marks)
7.
(a)
As the concentration increased the rate of reaction increased.
[1]
(b)
5.5; (allow : 5.6)
[1]
Fig. from graph (1.1 to 1.2)
[1]
(c)
Enzyme has a pH optimum / does not work in acid pH caused by lemon juice
(allow: denatured not: destroyed / inhibited)
[1]
(d)
Faster (colour change) at start as temperature increases;
[1]
No further colour change after enzyme denatured.
[1]
rate drops / stops if all active sites blocked;
[1]
as it competes with normal substrate for active sites.
[1]
Biuret test / NaOH or any alkali plus CuSO4;
[1]
Purple / violet ring / colour develops if protein.
(linked marks)
[1]
(e)
(f)
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 75
8.
(a)
A
Primary is (type, number and) sequence / order of amino acids
[1]
B
linked by peptide bonds
[1]
C
Secondary is (3D) folding / most common is (α) helix / β pleated sheet
[1]
D
formed by hydrogen bonds (between peptide bonds)
[1]
E
Tertiary is folding / twisting of a α helix / structure
[1]
F
(As shown by) globular proteins
[1]
G
To form very specific (3D) shapes
[1]
H
R groups project from α helix / side chains / part of active site
(i.e. function of R groups)
[1]
I
And may react to form bonds which stabilise 3D shape
(i.e. side chains)
[1]
J, K
Any two from, covalent / disulphide bridges; ionic / salt bridges;
hydrogen bonds; hydrophobic bonds.
[1]
L
Quaternary is two or more polypeptide / amino acid chains.
[1]
M
Joined by similar bonds between projecting R groups.
(i.e. named bond between R groups)
[1]
N
e.g. Haemoglobin.
[1]
O
Fibrous proteins have helices linked in strands.
(Maximum 10 from available 15)
[1]
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 76
(b)
A
Mitochondrion has double / inner and outer membrane.
[1]
B
with narrow fluid filled Inter-membrane space between.
[1]
C
cristae / stalked particles are infoldings of the inner membrane.
[1]
D
the interior contains an organic matrix.
[1]
E
DNA / ribosome present (in either mitochondrion or chloroplast).
[1]
F
and the organelles are self replicating (either mitochondrion or chloroplast).
[1]
G
The function is energy release / ATP production / aerobic respiration.
[1]
H
Chloroplast has double / inner and outer membrane.
[1]
I
interior of the chloroplast is the gelatinous stroma.
[1]
J
Flattened sacs / lamellae / thylakoids are present in the stroma.
[1]
K
usually (stacked to form) grana.
[1]
L
sites of photosynthetic pigments / chlorophyll
e.g. grana, thylakoids
[1]
M
Starch grains may be present.
[1]
N
Chloroplast function is photosynthesis.
[1]
O
Some attempt at genuine comparison.
(Maximum 10 from available 15)
[1]
Any 9 points plus point O.
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 77
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
General Certificate of Education
Tystysgrif Addysg Gyffredinol
MARK SPECIFICATION GRID
GCE BIOLOGY
UNIT TEST
Year of Examination ____Specimen______
Page__1____of_____1__
Session: summer/winter
Unit _BY1_
Assessment Objective
Paper
Other requirements
Total
Mark
Question
Number
Target Totals
AS BY1, BY2
A2 BY4, BY5
Specification
Reference
AO1
AO2
AO3
33
30
33
45
4
5
70
80
Synoptic
Quality of
written
com'cn.
N/A
9
9
9
1
1.1b
5
1
6
2
1.6 a&b
5
4
9
3
1.3a
3
4
7
9
4
1.4b
1
7
3
11
9
5
1.5a
3
4
1
8
6
1.7a
5
4
9
7
1.4c
3
7
10
8
1.1d/1.2a
7
3
10
32
34
Raw Totals:
4
70
9
9
GCE HUMAN BIOLOGY Specimen Assessment Materials 79
UNIT 2: HUMAN BIOLOGY
MARK SCHEME
1.
Active
natural
Exposure to measles
Active
artificial
Passive
natural
Passive
artificial
9
Receiving MMR vaccine
9
9
Receiving anti-rabies injection
Transfer of antibodies from
mother to child in breast milk
1 mark per row, any additional ticks, no mark.
9
[4]
(Total 4 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 80
2.
(a)
(i)
1.20
× 100
5.00
24
(-1 if wrong units given)
(ii)
[2]
low(er) oxygen levels at higher altitudes
more red blood cells, more haemoglobin
haemoglobin carries oxygen so more oxygen carried
to maintain / support (aerobic) respiration
maintain activities / named activities of person
(b)
oxyhaemoglobin
(c)
(i)
[3]
[1]
Contraction of ventricles causes a surge in blood pressure
This drops when the ventricles relax
But does not fall to zero because of the closing of the semi-lunar valves
The elastic recoil of the arteries maintains blood pressure.
[4]
(ii)
Arteries are dividing up (into arterioles) with a larger surface area / blood is
distributed in larger number of blood vessels with greater S.A. / greater
resistance due to larger S.A. / more friction / distance from heart greater. [1]
(d)
(Interstitial) fluid in the lungs must not build up as this reduces the efficiency of
gaseous exchange / fluid leaks out of capillaries and fills lungs.
[1]
(e)
(i)
The arteries function is transport
which is efficient at high speed, i.e. high pressure.
Capillary function is exchange
requiring slow speed
(longer contact with surroundings) and therefore lower pressure.
[4]
(f)
Any comment on the suction effect of relaxing heart / the massaging effects of
muscles / the prevention of backflow by valves.
[1]
(Total 17 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 81
3.
(a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
(iii)
(long) latent phase / incubation period (not: lag)
antigen is recognised
cloning / proliferation of B-lymphocytes
memory cells produced
(Any 2)
[2]
antibody production (not: rises)
by B-lymphocytes / by plasma cells
stimulated by T-lymphocytes
memory cells produced
(Any 2)
[2]
memory cells recognise antigen
rapid production of antibodies
with shorter latent phase / immediate response
high(er) level of antibodies (than with vaccination)
(Any 3)
[3]
as vaccinations increase the incidence decreases
fewer individuals to act as reservoir
particularly effective as exceed 90%
campaigns (1988, 1994) resulted in more vaccinations
Ref. natural variation in incidence
(Any 4)
[4]
measles starting to increase
wanted to eradicate measles
vaccination uptake dropping
(Any 1)
[1]
concerns about danger of vaccine / complacency
ref. to triple vaccine
adverse publicity / public perception of side effects / autism / bowel disease
increases reservoir
increase in number of measles cases
(Any 4)
[4]
(Total 16 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 82
4.
(a)
Digestion takes place in the mouth and duodenum/small intestine;
Amylase secretion in saliva and from pancreas
amylase converts starch to maltose;
maltase converts maltose to glucose;
sucrase converts sucrose to glucose and fructose;
digestion of disaccharides to monosaccharides occurs in or on epithelial cells;
ref. hydrolysis.
[Max. 6]
(b)
(i)
disinfectant/kills bacteria;
optimum pH/correct pH;
activation of enzymes/named enzyme;
ref. absorption of ions
(not: stops amylase action) Any 2.
[1]
(ii)
covered with mucus/alkaline layer.
[1]
(iii)
food (and acid) remain in stomach / acid contents being mixed
with food for 2-4 hours;
more acid production stimulated once food in stomach;
pH falls as more acid moves up into oesophagus.
[Max. 2]
(Total 10 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 83
5.
(a)
(i)
diffusion (not: gas exchange)
[1]
(ii)
5.1 kPa (unit needed)
[1]
(iii)
R because
low in oxygen/high in carbon dioxide
[1]
[1]
(b)
air in alveolus is (replaced) by ventilation/breathing
(not: refs. to diffusion/gas exchange)
(c)
stops alveoli clumping/sticking together/correct reference to surface tension/keep
open
[1]
(d)
(i)
Bigger vital capacity
(not: description; term needed)
(ii)
Flow rate decreased as air is expelled / (more difficult to breathe
out) / reference to time
(not: reference to height of peaks)
[1]
[1]
Constriction / inflammation (of lining) of the bronchii restricts flow
(not: reference to trachea)
(e)
[1]
high / increased levels of carbon dioxide in inhaled air;
Lower concentration gradient between blood and alveolar air;
Slower diffusion;
(f)
[1]
[Max. 2]
Total capacity includes residual volume / some air always remains in the lungs /
doesn't move.
[1]
Spirometer only measures movements of air in and out / the trace only
records movements.
[1]
(Total 13 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 84
6.
(a)
A.
smoke chemicals damages / destroys cilia
B.
and is irritant causing over-production of mucus
C.
mucus not removed from bronchi / bronchioles
D.
causing persistent coughing
E.
causes contraction of circular muscles reducing lumen
F.
accumulation of bacteria
G.
leads to infection of lungs and bronchitis
H.
damages alveoli; where walls break down / scar tissue forms
I.
so reduces area for gas exchange
J.
CO from smoke in blood
K.
means less oxygen carried in blood
L.
and reduction in physical activity
M.
tar / carcinogenic chemicals in cigarette smoke
N.
causes epithelial cells to divide uncontrollably
O.
growth of tumour which causes blockage usually of bronchi / often spreads
(Total 10 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 85
(b)
A
exploit weak points in transmission;
[1]
B
prevent mosquitoes biting;
[1]
C
by netting / clothing and repellants;
[1]
D
control mosquito population;
[1]
E
using insecticide / DDT for adults or insects;
[1]
F
release sterilised / infertile males;
[1]
G
fish to eat larvae;
[1]
H
bacteria to infect larvae;
[1]
I
drain / cover standing water / spray oil / detergent on water to kill larvae; [1]
J
very few points to attack parasite / only when 'free' in bloodstream;
[1]
K
preventative drugs effective only at this stage;
[1]
L
spends much of its time inside body cells;
[1]
M
cannot be targeted without damaging host;
[1]
N
parasite mutates / antigenic types;
[1]
O
Ref. vaccine as alternative e.g. vaccine difficult to develop because
of different antigenic types.
[1]
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 86
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
General Certificate of Education
Tystysgrif Addysg Gyffredinol
MARK SPECIFICATION GRID
GCE BIOLOGY
UNIT TEST
Year of Examination ____Specimen______
Page__1____of_____1__
Session: summer/winter
Unit _HB2_
Assessment Objective
Paper
Other requirements
Total
Mark
Question
Number
Target Totals
AS BY1, HB2
A2 HB4, BY5
Specification
Reference
AO1
AO2
AO3
33
30
33
45
4
5
70
80
Synoptic
Quality of
written
com'cn.
N/A
9
9
9
1
2.5
2
2
4
2
2.4
7
10
17
9
3
2.5
4
8
16
9
4
2.2
6
4
10
9
5
2.3
6
7
13
2.3 / 2.6
8
2
10
33
33
6a / 6b
Raw Totals:
4
4
70
9
GCE HUMAN BIOLOGY Specimen Assessment Materials 87
UNIT 4: HUMAN BIOLOGY
MARK SCHEME
1.
(a)
(i)
(ii)
(b)
(c)
(Neuro) transmitter substance.
(not: acetyl choline)
[1]
Exocytosis (secretion).
(not: active transport / diffusion)
[1]
As a result of depolarisation of the membrane / calcium ions diffuse / enter
into (the presynaptic cell) / and cause fusion of the vesicles with
the presynaptic membrane (2 from 3)
[2]
(i)
Phospholipid
[1]
(ii)
Sodium ions
[1]
(d)
(transmission) occurs via a transmitter substance / acetyl / choline which is broken
down by an enzyme
[1]
enzyme inhibitor blocks this breakdown (after the transmitter has served
its purpose).
[1]
(not: blocks receptors in post synaptic membrane)
(e)
Increased metabolism means increased mitochondrial activity / more ATP produced
(not: more ATP unqualified)
which stimulates more neurotransmitter synthesis / release.
[2]
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 88
2.
(a)
P - shorter
Q - same length
[1]
[1]
(b)
(i)
(number and) sequence of amino acids (in polypeptide chain)
[1]
(ii)
α - helix
[1]
(i)
Athletes have a higher rate of oxygen consumption (at all
levels)/converse/athletes have a maximum oxygen consumption which is
much higher (25% higher) than non-athletes;
[1]
(c)
There is only an increase of 1 from 1200 to 1400 J min-1 kg-1 / max oxygen
consumption is reached at 1400 J min-1 kg-1 /max 58, 45.
[1]
alternatively:
Non athletes have a lower overall increase in oxygen consumption over the
range of energy expenditures (or figs)/converse;
Max oxygen consumption is reached at 1200 J min-1 kg-1 ;
(i.e. comparison needed)
in athletes higher lung capacity;
better circulatory system/heart capacity;
greater number of mitochondria;
in non-athletes accept converse
(Any 1)
(not: more (developed) muscles)
(ii)
[1]
As exercise increases non-athletes produce more lactate than athletes.
Athletes have no lactate until 1200 J min-1 kg-1
max figure of 305 mg-1 min-1 kg
non-athletes have lactate appearing at 1000 mg min-1 kg-1
(much higher) figure of 590 mg min-1 kg-1
Any 2
[2]
in athletes more oxygen available so there is less anaerobic
respiration/converse
[1]
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 89
3.
(a)
(i)
sugar + base + 3 phosphates or correct names;
ATP to ADP + Pi releases energy/exergonic or description;
ADP + Pi to ATP needs energy/endergonic or description;
easily reversible;
transfers energy from place of release/one molecule to energyrequiring reactions;
provides energy in 'small packets'/figure e.g. 30.6 or 3.1.
[5]
(Max. 5)
(ii)
(b)
(i)
(ii)
(c)
two of synthesis/muscle contraction/active transport/other
e.g. nerve impulses/photosynthesis etc.;
(not: movement/growth)
[1]
X
Y
H (atom)/reduce NAD/reduced FAD;
oxygen; (not water)
[1]
X
Y
chlorophyll/photosystem I or II;
chlorophyll/NADP/photosystem I;
[1]
[1]
(electron) fuels proton pump;
across membrane/into intermembrane space;
H+/protons diffuse/flow down concentration gradient;
Through ATP synthetase/stalked particles (not: ATPase);
Membrane impermeable to protons.
(Max.4)
(d)
[4]
Why person lacking energy e.g. ATP used for muscle contraction/less absorption of
food.
Food not enough to meet metabolic demands of body/named;
i.e. food broken down but not making sufficient ATP
No excess food material to store;
therefore respiration using body tissues/fat/protein/food stores;
AVP e.g. heat production increases metabolic rate/body respires faster to produce
ATP.
[Max. 4]
(Total 18 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 90
4.
(a)
Suitable temperature (between 25-40);
Nutrients;
Correct pH (6-8);
Oxygen;
(b)
Water;
(not: ref. aseptic/sterile) (3 correct 2; 2 correct 1)
[2]
Autoclave / High temperatures and pressure/AVP;
[1]
Sterilise glass rod/loop etc by flaming;
Flame top of culture tube after removing lid;
Lift Petri dish lid at angle/short time open;
Bunsen to cause air flow.
(not: ref. to disinfect benches)
(c)
(d)
[2]
1 mark for clear zone around antibiotic clearly labelled;
[1]
1 mark for shape (horizontal 'bow-tie').
[1]
(i)
mostly peptidoglycan / thick murein layer;
carhohydrate / polysaccharide with amino acid side chains;
(ii)
no lipoprotein / lipopolysaccharide.
[2]
(I)
each cell produces 1 colony / cells separate
[1]
(II)
underestimate / doesn't allow for clumping.
[1]
(iii)
include dead bacteria
[1]
(iv)
calculation of dilution factor 10-2;
[1]
counting of colonies 7;
[1]
multiplication 700;
[1]
(allow: consequential error)
(Total 15 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 91
5.
(a)
too large to pass through (pores in) membrane
(b)
(i)
(c)
[1]
large surface area as lines whole abdomen
thin membrane so short diffusion distance;
dialysis fluid needs to be in close contact with blood (capillaries);
so that substances can pass from blood into dialysis fluid. (Any 3)
[3]
(ii)
dialysis fluid not continually replaced;
concentration gradient not maintained;
(once some movement of substances from blood has taken place) equilibrium
is reached;
less efficient removal of substances from blood.
(Any 3)
[3]
(iii)
increase solute / glucose / salt concentration / more negative water potential.
[1]
(iv)
less efficient;
further distance for substances to diffuse / slower diffusion.
[2]
(i)
they are genetically identical / clones
(not: similar)
[1]
(ii)
(cells of new kidney have) antigens
(cells of) new kidney recognised as foreign / not identical to those of
recipient;
T killer (lymphocytes) cause lysis of target cells / destroy kidney;
helper T cells are activated;
B lymphocytes / plasma cells;
produce antibodies.
(Any 5)
[5]
(Total 16 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 92
6.
(a)
Kidney Essay
A
Antidiuretic hormone
B
Is secreted by the (posterior) lobe of the pituitary
C
It is carried in the bloodstream to the (distal convoluted tubule and) collecting
duct
D
When the blood is more concentrated/low Ψ/high OP
E
Detected by (osmo) receptors
F
in hypothalamus
G
ADH levels are higher / ADH released
H
Higher ADH levels increase the permeability of the cells lining the DCT/CD
to water/explanation of water channels inserted into DCT membrane.
I
Water moves out of the DCT/CD by osmosis
J
Into the interstitial fluid where it is rapidly removed by the capillary network
K
This occurs because the medulla of the kidney has a high solute
concentration/low Ψ
L
Due to the countercurrent multiplier system operating in the Loop of Henlé
M
This conserves water and produces small volumes/
concentrated urine
N
When the blood is more dilute/high Ψ/low OP
O
ADH levels are lower so less water is reabsorbed
P
This allows more water to leave the tubule/collecting duct/results in large
volumes of dilute urine
(Total 11 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 93
(b)
A
inside about -60 to -70m V (compared with outside);
B
maintained by sodium (-potassium) pump;
C
[K+] higher inside / [NA+] higher outside;
D
membrane more permeable to K+;
E
K+ diffuse / leak / move out;
F
all or nothing / threshold value;
G
action potential occurs when membrane becomes more permeable to Na+;
H
Na+ flood in / sodium gates;
I
depolarisation;
J
inside about +40m V (compared with outside);
K
K+ then diffuse / move out and repolarise the membrane;
L
refractory period explained;
M
local currents stimulate next part of neurilemma / membrane/ axon;
N
myelin sheath increases distance over which local currents can act to bring
about depolarisation;
O
or speeds up transmission by jumping from node to node / saltatory
conclusion;
P
depolarisation only at nodes of Ranvier;
(Total 11 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 94
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
General Certificate of Education
Tystysgrif Addysg Gyffredinol
MARK SPECIFICATION GRID
GCE BIOLOGY
UNIT TEST
Year of Examination ____Specimen______
Page__1____of_____1__
Session: summer/winter
Unit _HB4_
Assessment Objective
Paper
Other requirements
Total
Mark
Question
Number
Target Totals
AS BY1, HB2
A2 HB4, BY5
Specification
Reference
AO1
AO2
AO3
33
30
33
45
4
5
Synoptic
Quality of
written
com'cn.
70
80
N/A
9
9
9
1
4.7
4
6
10
9
2
4.2
2
8
10
9
9
3
4.2/4.1
9
9
18
9
9
4
4.4
5
5
5
4.6
5
11
16
4.6/4.7
7
4
11
32
43
7a/b
Raw Totals:
5
5
9
15
80
9
9
9
GCE HUMAN BIOLOGY Specimen Assessment Materials 95
UNIT 5: BIOLOGY/HUMAN BIOLOGY
MARK SCHEME
1.
(a)
(b)
(i)
15576
[1]
(ii)
C2 = C3+R3+E3 / C3 = C2-(R3+E3)
[1]
(i)
total energy expelled per m2=972+3732+110+20+4834 (1)
total energy expelled = 4834 × 25000 (1) = 120,850,000 / 1.2085×108
[2]
(ii)
passes to decomposers / detritivores; respired / used / released by
decomposers; lost as heat
(Any 2) (not: eaten)
[2]
(Total 6 Marks)
2.
(a)
Reference to DNA / genetic fingerprinting / base sequencing / immunological
comparison;
Close similarities should be present between the species.
(i.e. technique and related result)
(b)
[2]
Populations physically / geographically separated (by a barrier);
Populations cannot interbreed;
Different mutations in each population;
Environmental conditions / habitats (or e.g.) differ;
Different selection pressures;
Leads to different gene pools / demes;
Leads to difference in physical characteristics / behaviour / appearance.
(i.e. change in genetics leads to change in phenotype)
Any 6 points in context.
[6]
(Total 8 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 96
3.
(a)
carried on the sex / X chromosome;
only expressed if both chromosomes carry the allele / gene in females / homozygous;
always expressed in the male, (since Y is not homologous);
allele = different from of the same gene;
[4]
(b)
(i)
XN Xn
[1]
(ii)
Xn Y
[1]
(iii)
XN Y
[1]
(c)
[1]
[1]
Accept 1:1:1:1, but only if linked to genotype / 1:1:1:1 ratio of all four genotypes. [2]
Genotypes 1, linked phenotypes 1.
(not: ref. infected male)
(d)
Increased worry as may not develop disease / insurance company may not provide
cover / confidentiality issues / affects lifestyle choices / prevents gene being passed
on / reduces incidence of gene in population / AVP.
[2]
(Total 13 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 97
4.
(a)
(b)
(c)
(d)
(i)
active site
[1]
(ii)
hydrogen / sulphur / peptide / ionic bond / van der Waals / disulphide bridges
(not: covalent)
[2]
(i)
Ribosome
[1]
(ii)
Anticodon
[1]
(i)
Translation
[1]
(ii)
Each type of tRNA has a specific linkage with one type of amino acid
[1]
tRNA joints to mRNA
[1]
by matching codon / anticodon
[1]
B moves into the vacant space to the right of A
and the addition of B's amino acid to the chain (removes the chain from A to B)
A is now free to move off and combine with another identical amino acid molecule
(i.e. reference to release).
B occupies position A was in / The ribosome moves one codon to the right / (mRNA
strand moves one codon to the left)
3 from 4.
[3]
(e)
The protein will only perform its function if the structure is correctly reproduced [1]
Active site altered or destroyed if amino acid sequence is changed
[1]
(not: makes the wrong protein)
(Total 14 marks)
5.
(a)
To give matching / the same (sticky) ends;
so that the vector / plasmid can join with the fragment;
Complementary bases. (not: codon)
(Any two)
[2]
(b)
Anneals / seals / splices sticky ends together.
[1]
(c)
Can be cut directly from host DNA / use of restriction enzymes.
[1]
RNA can be extracted / use of messenger RNA
and treated with reverse transcriptase
which turns RNA into DNA
i.e. Reverse transcriptase and a description of the process.
[2]
GCE HUMAN BIOLOGY Specimen Assessment Materials 98
(d)
Reversion of virus to disease causing form
Bacteria / virus could be toxic to humans or insects
Virus / bacterium could transfer to another species
Accept ref. ethical reasons qualified e.g. Why change colour of leaves when
there is no nutritional value?
Accept ref. increase cost of plants putting financial burden on farmers
Accept other valid points e.g. unknown future side effects.
(not: playing God / genes escaping)
(e)
(i)
protein and nucleic acid / DNA / RNA
(ii)
Use of liposomes;
[Max. 2]
[1]
Transfer of non CF gene (into plasmid);
Coated in (spheres of) phospholipids / lipid (bilayer);
Breathed into lung via aerosol / inhaler.
(iii)
[3]
Gets into nucleus;
Insertion into genome / chromosome;
Transcription / eq. (in context);
Translation / eq. (in context).
[3]
(Total 15 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 99
6.
(a)
(b)
(c)
(i)
Fusarium
[1]
(ii)
banana plants genetically identical / clones;
same susceptibility / lack of resistance to Panama disease;
no / little mutation;
close planting enabled easy spread.
(Any 3)
[3]
natural selection;
fungicide is selective agent / selective pressure;
variation between individuals;
chance / random mutation;
some individuals have selective advantage / better change of surviving;
survival of resistants / death of non-resistants (not: survival of the fittest);
mutation / ability to survive pesticide passed on to offspring;
increased allele frequency.
(Any 7)
[7]
normally achieved with breeding programmes;
(using) reservoir of genes / alleles;
bananas are sterile / seedless / reproduce asexually;
so no easy way to incorporate new genes / alleles.
(Any 3)
[3]
(Total 14 marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 100
7.
(a)
A
in plant tissue culture plants are grown from parts of a plant / cuttings /
suitable example
B
reference to term tissue culture / micropropagation
C
(explants) placed in sterile, (aerated) medium
D
cells divide (by mitosis)
E
to form callus
F
callus differentiates into plantlet
G
plantlets transplanted into (sterile) soil when big enough / eq. e.g. grown on
H
in animals a nucleus / DNA may be removed from diploid / somatic / udder /
body / undifferentiated cell
I
transferred to an exnucleated egg cell / egg cell with no nucleus
J
or an embryo split before differentiation/at early stage
K
the embryo allowed to develop in a surrogate (uterus)
L
reference to totipotent / stem cells / pre differentiation in either
M
advantages: maintain genetic stocks / gene library more quickly than seed
propagation / rapid increase in nos. / standardisation for transport / ref. to cost
qualified
N
disadvantages: in mammals unforeseen long term effects e.g. premature
ageing / gigantism
O
no variation / disease / entry of pathogens may be a problem
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 101
(b)
Forestry / Fishing
A
Habitat destruction / comment on destruction balanced ecosystems
B
Leading to reduction in biodiversity / owtte
C
Qual of above e.g. loss of potentially important therapeutic plants / reduction
gene pool
D
Tree removal leads to soil erosion / dust bowls
E
Causing increase in sedimentation / deposits
F
Plant removal contributes to climate change / example / ref. to global
warming
G
Fall in CO2 reduction / raising CO2 levels / O2
H
Reduction of fish population to below sustainable levels / numbers cannot be
maintained - will be further decline
I
Curtailment of breeding due to fewer mature fish
J
So no regeneration
K
Quotas could reduce catches while maintaining industry
L
Managed areas / farms to supply commercial timber / fish / plant renewable
resources
M
Reduction fleet size / number of days fishing reduce number of fish taken
N
Exclusion zones / protected areas allow population to rebuild
O
Restricted mesh size to allow young fish to escape and breed.
(Total 10 Marks)
GCE HUMAN BIOLOGY Specimen Assessment Materials 102
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
General Certificate of Education
Tystysgrif Addysg Gyffredinol
MARK SPECIFICATION GRID
GCE BIOLOGY
UNIT TEST
Year of Examination ____Specimen______
Page__1____of_____1__
Session: summer/winter
Unit _BY5_
Assessment Objective
Paper
Other requirements
Total
Mark
Question
Number
Target Totals
AS BY1, BY2
A2 BY4, BY5
Specification
Reference
AO1
AO2
AO3
33
30
33
45
4
5
70
80
Synoptic
Quality of
written
com'cn.
N/A
9
9
9
1
5.7
1
5
6
2
5.5
3
5
8
9
3
5.4
2
9
2
13
9
4
5.1
9
4
1
14
9
9
5
5.6
7
8
15
9
9
6
5.4/5.5
4
9
1
14
9
9
7
5.8 / 5.6
5
4
1
10
31
44
5
80
Raw Totals:
GCE Human Biology SAMs (2009-2010)/ED
30 July 2007
9