Download C274/SQP368 Biology NATIONAL QUALIFICATIONS

C274/SQP368
Biology
Time: 2 hours 30 minutes
Advanced Higher
(Revised)
Specimen Question Paper
for use in and after 2012
NATIONAL
QUALIFICATIONS
SECTION A—Questions 1–25 (25 marks)
Instructions for completion of Section A are given on Page two.
SECTION B (65 marks)
The answer to each question should be written in ink in the answer book provided. Any additional
paper (if used) should be placed inside the front cover of the answer book.
Rough work should be scored through.
All questions should be attempted. Candidates should note that Question 11 contains a choice.
[C274/SQP368]
1
©
Read carefully
1 Check that the answer sheet provided is for Biology Advanced Higher (Section A).
2 For this section of the examination you must use an HB pencil and, where necessary, an eraser.
3 Check that the answer sheet you have been given has your name, date of birth, SCN (Scottish
Candidate Number) and Centre Name printed on it.
Do not change any of these details.
4 If any of this information is wrong, tell the Invigilator immediately.
5 If this information is correct, print your name and seat number in the boxes provided.
6 The answer to each question is either A, B, C or D. Decide what your answer is, then, using
your pencil, put a horizontal line in the space provided (see sample question below).
7 There is only one correct answer to each question.
8 Any rough working should be done on the question paper or the rough working sheet, not on
your answer sheet.
9 At the end of the examination, put the answer sheet for Section A inside the front cover of
the answer book.
Sample Question
Which of the following molecules contains six carbon atoms?
AGlucose
B
Pyruvic acid
C
Ribulose bisphosphate
D Acetyl coenzyme A
The correct answer is A—Glucose.
horizontal line (see below).
A
B
C
The answer A has been clearly marked in pencil with a
D
Changing an answer
If you decide to change your answer, carefully erase your first answer and using your pencil, fill in the
answer you want. The answer below has been changed to D.
A
[C274/SQP368] 2
B
C
D
Page two
SECTION A
All questions in this section should be attempted.
Answers should be given on the separate answer sheet provided.
1. Which of the following diagrams illustrates a
peptide bond?
A
B
O
C
C
H
N
H
C
D
H
C
H
H
C
C
C
H
N
N
H
H
H
Positively charged
H
C
Table 1
H
N
4. Table 1 shows the charge of amino acids at a
certain pH.
Negatively charged
arginine
tyrosine
lysine
cysteine
histidine
glutamate
aspartate
Table 2 shows the number of each amino acid
in a protein.
H
Table 2
2. A hydrophobic amino acid has an R group
that is
Amino acid
Number
arginine
13
lysine
19
Cnot polar
histidine
2
Dpolar.
tyrosine
7
cysteine
2
glutamate
20
aspartate
9
Anegatively charged
Bpositively charged
3.A buffered solution of four amino acids
was applied to the midline of a strip of
electrophoresis gel. The result of running the
gel is shown below.
_
+
1
2
3
4
Assuming that each amino acid carries a
single positive or negative charge, what is the
protein’s net charge at this pH?
A+34
B+4
C−4
Which of the amino acids was at its isoelectric
point?
A1
B2
C3
D4
[C274/SQP368] 3
Page three
D−38
5. The diagram below shows how phosphate
is used to modify the conformation of an
enzyme, phosphorylase, and so change its
activity.
ATP
7. The sodium-potassium pump spans the plasma
membrane. Various processes involved in the
active transport of sodium and potassium ions
take place either inside the cell (intracellular) or
outside the cell (extracellular).
ADP
Which line in the table correctly applies to the
binding of potassium ions?
Y
X
Binding location
of potassium ions
Conformation of
transport protein
A
intracellular
not phosphorylated
B
extracellular
phosphorylated
C
intracellular
phosphorylated
D
extracellular
not phosphorylated
Z
O H
O Pi
W
H2O
Pi
Which line in the table correctly identifies the
labels?
Kinase
Phosphatase Phosphorylase
A
Y
Z
W
B
W
Y
Z
C
X
Y
W
D
Y
W
Z
8. The Na,K–ATPase moves ions in the ratio
3 sodium : 2 potassium. 5000 of these ions
are pumped across the membrane every ten
seconds. The number of potassium ions
moved across in one second is
A200
B500
C2000
D3000.
6. The diagram below shows the distribution of
protein molecules in a cell membrane.
Protein 
Protein 
phospholipid
bilayer
Protein 
Protein 
9. The contribution of aquaporins (AQPs) to
osmosis was studied by measuring the rate of
movement of radioactive water across a plasma
membrane. Rates were measured in either
isotonic or hypertonic external solution when
the pores were either open or closed. Results
are shown in the table.
External
solution
Open AQPs
Closed AQPs
Isotonic
2·5
1·0
Hypertonic
20·0
1·8
Protein 
Which line in the table correctly identifies a
peripheral and an integral membrane protein?
Peripheral
membrane protein
Integral membrane
protein
A
1
5
B
2
1
C
3
4
D
5
2
[C274/SQP368] 4
Rate of water movement
(units s−1)
Which of the following is the dependent
variable in the experiment?
A
External solution
B
Radioactivity of water
C
Rate of water movement
DAquaporins
Page four
10. To which group of signalling molecules does
the sex hormone testosterone belong?
A
Extracellular hydrophilic
B
Extracellular hydrophobic
C
Peptide hormones
12. Which of the following diagrams correctly
represents the sequence of phases in the cell
cycle?
A
B
G2
M
G1
G1
S
M
S
DNeurotransmitters
11. The diagram below represents a well in an
immunoassay kit testing a blood sample from
a person who may have been exposed to a
virus. The substrate has been broken down
to form a coloured product, so the result is
positive.
C
G2
D
S
G2
M
G2
G1
M
S
G1
1
2
5
3
6
13. Animal cells growing in culture are found to
spend 20% of their time in the G2 phase of the
cell cycle. G2 lasts for 4 hours.
If cells spend 12% of their time in the M
phase, how long does this last?
4
plastic well
Which line in the table correctly identifies the
roles of numbered components?
4
3
2
1
B
5
2
3
6
C
4
6
5
2
D
6
3
2
5
[C274/SQP368] 5
2 hours 4 minutes
B
2 hours 12 minutes
C
2 hours 24 minutes
D
2 hours 40 minutes
14.Which type of ion is pumped
membranes by bacteriorhodopsin?
Antigen Antibody Enzyme Substrate
A
A
across
ASodium
BPotassium
CChloride
DHydrogen
15. Which of the following would be true if a
population’s gene pool remained unaltered for
many generations?
Page five
A
Genetic drift had occurred.
B
Mating was random.
C
Migration was common.
D
Certain alleles had a selective advantage.
16. Which line in the table correctly describes
cells in meiosis?
Stage
Chromosome
complement
Number of
cells
A
Meiosis I
haploid
4
B
Meiosis I
diploid
2
C
Meiosis II
haploid
4
D
Meiosis II
diploid
2
20. A quadrat with sides 50 cm long was used to
estimate the density of a plant species in two
areas X and Y. Five random samples were
taken in each of the two areas and the results
are given below.
17. The crossing over that generates new allele
combinations in meiosis I occurs between
Asister
chromatids
chromosomes
of
Number of plants
Quadrat
number
Area X
Area Y
1
27
15
2
19
16
3
39
42
4
19
31
5
11
16
homologous
B non-sister chromatids of homologous
chromosomes
The mean density per square metre in each of
the two areas is
Area X
Area Y
C sister chromatids of non‑homologous
chromosomes
A
23
24
D
B
46
48
C
92
96
D
115
120
non-sister chromatids of non‑homologous
chromosomes.
18. Which of the following is not a source of
DNA during horizontal gene transfer in
bacteria?
AVirus
BPlasmids
C
Bacterial cells
DGametes
19. In birds, females are heterogametic. The
gene for feather-barring in chickens is sexlinked and the allele for barred feathers is
dominant to the allele for non-barred feathers.
What ratio of offspring would be expected
when a non-barred male is crossed with a
barred female?
A
1 barred male : 1 non-barred female
B
1 non-barred male : 1 non-barred female
C
1 barred female : 1 barred male
D
1 non-barred male : 1 barred female
[C274/SQP368] 6
Page six
21. Anolis lizards are found on Caribbean islands.
They feed on prey of various sizes.
22.All viruses
surrounding
Histogram 1 shows the range of prey length
eaten by Anolis marmoratus on the island of
Jarabacoa, where there are five other Anolis
species.
Histogram 2 shows the range of prey length
eaten by Anolis marmoratus on the island of
Marie Galante, where it is the only Anolis
species.
Percentage of diet
Histogram 1: Jarabacoa Island
A
DNA or RNA
B
DNA and RNA
C
DNA only
D
RNA only.
of
a
protein
coat
23. Reverse transcriptase catalyses the production
of
A
DNA from DNA
B
mRNA from DNA
C
tRNA from mRNA
D
DNA from RNA.
26–30
21–25
16–20
11–15
6–10
0–5
24. Which line in the table correctly describes
imprinting?
Within
critical
period
Reversible
Learning
process
A
Yes
No
Rapid
B
No
Yes
Slow
C
No
No
Rapid
D
Yes
Yes
Slow
Prey length (mm)
46–50
41–45
36–40
31–35
26–30
21–25
16–20
11–15
6–10
Percentage of diet
Histogram 2: Marie Galante Island
0–5
consist
25. The formula N = MC/R is used to estimate
population size using mark and recapture data.
N = population estimate
Prey length (mm)
M=number first captured, marked
and released
Which of the following statements could
explain the different range of prey sizes eaten
by Anolis marmoratus on the two islands?
A Larger numbers of prey are found on
Marie Galante.
B
Anolis
marmoratus
occupies
fundamental niche on Jarabacoa.
C
Anolis marmoratus occupies its realised
niche on Marie Galante.
its
D Resource partitioning takes place on
Jarabacoa.
C = total number in second capture
R = number marked in second capture
In a survey to estimate a woodlouse
population, the following data were obtained:
Woodlice captured, marked and released = 80
Marked woodlice in second capture
= 24
Unmarked woodlice in second capture = 96
The estimated population of the woodlice was
A200
B320
C400
D
[C274/SQP368] 7
Page seven
3 840.
[END OF SECTION A]
Candidates are reminded that the answer sheet MUST be returned INSIDE the
front cover of the answer book.
[C274/SQP368] 8
Page eight
SECTION B
All questions in this section should be attempted.
All answers must be written clearly and legibly in ink.
1. Bovine spongiform encephalopathy (BSE) and variant Creutzfeldt-Jakob disease (vCJD)
are examples of fatal brain disease that can pass from one species to another. The nature
of the infectious agent is as yet unidentified but, in both diseases, a protein known as
PrPSC accumulates in brain tissue.
It has been shown that PrPSC is an altered form of the normal membrane protein
PrPC. Both molecules have the same primary structure (PrP) but they differ in how the
PrP protein folds. Molecules of PrPSC have a lower proportion of α-helix and a higher
proportion of β-sheets.
Proteins are normally broken down after a certain length of time by intracellular enzymes.
However, the increased β-sheet content makes PrPSC more resistant to enzymatic
breakdown, which leads to its accumulation.
PrP
normal
folding
abnormal
folding
PrPC
PrPSC
can be broken
down by enzymes
resistant to
breakdown
A substance capable of breaking β-sheets (β-breaker) was tested to find out if it could
make PrPSC more susceptible to the intracellular enzymes. PrPSC samples from mice and
humans were each incubated for 48 hours with different concentrations of β-breaker and
the percentage of PrPSC remaining after digestion was determined. The results are shown
in the Figure on Page ten.
In the same study, mouse PrPSC was further analysed to determine if any change in the
proportions of α-helix and β-sheet had occurred. The results are shown in Table 1.
In a second study, mice were treated with infectious material containing (i) PrPSC and
(ii) a 1:1 mixture of PrPSC and β-breaker. Quantities of PrPSC were equivalent in both
treatments. Table 2 shows the mean time to onset of symptoms of brain disease in the two
groups. Different concentrations of PrPSC were prepared by diluting stock solutions.
[Question 1 continues on Page ten
[C274/SQP368] 9
Page nine
Question 1 (continued)
Figure: The effect of β-breaker on mouse and human PrPSC
% PrPSC present after 48 hours
150
100
Mouse PrPSC
Human PrPSC
50
0
0:1
1:1
10:1
100:1
Ratio of β-breaker to PrPSC
1000:1
Table 1: Proportions of secondary structure in mouse PrPSC
before and after 48 hours incubation with β-breaker.
Proportions of secondary structure (%)
Secondary structure
Before incubation
After incubation
α-helix
18
27
β-sheet
36
9
Table 2: T
ime to onset of symptoms for mice treated with
PrPSC with and without β-breaker.
Time to onset of symptoms (days)
PrPSC concentration
(relative to stock solution)
PrPSC
PrPSC +
β-breaker
1 × 10−2
129
143
1 × 10−3
145
159
1 × 10−4
173
185
[C274/SQP368] 10
Page ten
Marks
Question 1 (continued)
(a) (i)PrPC and PrPSC have the same primary structure. Why should they be expected
to fold in an identical way?
(ii) Hydrogen bonds are important in the folding of proteins. Name two other
types of interaction involved in protein folding.
1
1
(b) Refer to the data in the Figure. The error bars represent confidence intervals.
(i) What do the confidence intervals show about the data collected?
1
(ii) How are confidence intervals used in the analysis of results?
1
(iii) Draw two conclusions about the effect of β-breaker on human and mouse PrPC
and PrPSC.
2
(c) Refer to the information in Table 1.
(i) Explain why the PrPSC protein should be more susceptible to breakdown by
intracellular enzymes after incubation.
2
(ii) How would a control showing the proportions of secondary structure in PrPC
help in judging the success of the β-breaker on PrPSC?
1
(d) Refer to the information in Table 2.
(i) Describe how to produce a 1 × 10−4 concentration of PrPSC by serial dilution
from the 1 × 10−2 solution.
1
(ii) What is the effect of changing the concentration on the onset of brain disease?
1
(iii) Calculate the greatest percentage improvement achieved in this experiment by
using β-breaker.
1
(iv) Suggest one alteration to the treatment that might improve the performance of
the β-breaker.
1
(13)
[C274/SQP368] 11
Page eleven
Marks
2. Gamma-aminobutyric acid (GABA) is a neurotransmitter that functions as a signalling
molecule in the central nervous system. GABA binds to a receptor protein located in
the plasma membrane of target cells as shown in Figure 1. Binding of a GABA molecule
opens a channel that allows chloride ions (Cl−) to enter the cell.
Figure 1
Figure 2
−
Cl
modulatory site
}
Drug
present
plasma
membrane
of target
cell
Chlorine ion movement
through channel (units)
GABA
binding
site
Drug
absent
GABA concentration (log units)
Cl−
Benzodiazepines are sedative drugs that bind to the receptor protein and increase its
affinity for GABA. These drugs act as allosteric modulators by binding at a site that is
distinct from the GABA-binding site. Figure 2, above, shows the movement of chloride
ions through the channel as GABA is increased with and without the drug being present.
(a) Use the information provided to explain why the GABA receptor is described as a
ligand-gated channel.
2
(b) What term describes the action of a membrane receptor in which signal binding
brings about an effect in the cytoplasm?
1
(c)
(i) How does the information in Figure 2 show that the affinity of the receptor for
GABA has been increased by the benzodiazepine?
1
(ii) Why does the affinity of the receptor for GABA increase when the drug binds
to the modulatory site?
1
(iii) What effect will chloride ion influx have on the membrane potential of the nerve
cell?
1
3. Discuss the advantages of a pilot study in the development of a biological investigation.
[C274/SQP368] 12
Page twelve
(6)
(4)
Marks
4. When insulin attaches to its receptor in the plasma membrane of fat cells and muscle cells,
GLUT 4 glucose transporter proteins in the cytoplasm are recruited into the membrane to
take in glucose. Type 2 diabetes is associated with insulin resistance in which cells are less
able to respond to insulin in this way.
GLUT 4 concentration relative
to control UT leg (%)
A recent study concluded that moderate strength training increases the GLUT 4 content
of muscle tissue in those with type 2 diabetes. Individuals taking part all did strength
training on one leg (T leg) for six weeks while the other leg was left untrained (UT leg).
The subjects either had type 2 diabetes or did not. At the end of the training, muscle
biopsies (samples) were taken from the trained and untrained legs and compared for
GLUT 4 protein content. The results are shown in the Graph below.
140
*
120
100
80
* shows significant
difference
UT leg
60
T leg
40
20
0
Non diabetic
Diabetic
Diabetic status
(a) In this investigation, what would be a suitable null hypothesis?
(b) What comparisons did the researchers make in order to conclude that moderate
strength training increases the GLUT 4 content of muscle tissue only in those with
type 2 diabetes?
1
2
(3)
[C274/SQP368] 13
Page thirteen
Marks
5. Thyroxine is a hormone that acts as a regulator of metabolic rate in most tissues.
Thyroxine causes an increase in metabolic rate by binding to specific receptors
located within the nucleus of a target cell. Hyperthyroidism is a condition caused by
overproduction of thyroxine. The figure shows the average change in metabolic rate of
individuals with hyperthyroidism who were treated over a twenty week period with a drug
(carbimazole). The drug decreases the synthesis of thyroxine from the thyroid gland.
40
Change in metabolic rate (%)
30
20
10
0
−10
−20
0
4
8
12
16
20
Time (weeks)
(a) What property of thyroxine allows it to cross the membrane of cells?
1
(b) Describe how thyroxine binding to its receptor affects transcription of genes that
influence metabolic rate.
2
(c)
(i) How many weeks of drug treatment were required to reach normal metabolic
rate?
1
(ii) How do the data support the conclusion that the thyroid gland has large stores
of thyroxine?
1
(iii) Why have changes in metabolic rate been presented as percentages?
1
(6)
[C274/SQP368] 14
Page fourteen
6. Rod cells and cone cells are photoreceptors in vertebrate eyes. Membranes in these cells
contain rhodopsin, a protein molecule that has a light-absorbing component. Rhodopsin
generates a nerve impulse when light is absorbed.
Marks
light-absorbing
component
protein
membrane
(a) Name the light-absorbing component of rhodopsin.
1
(b) Explain how the absorption of a photon by rhodopsin leads to the generation of a
nerve impulse.
2
(c) Give one feature of the photoreceptor system in rods that allows these cells to
function in low light intensity.
1
(d) Cone cells are sensitive to different wavelengths of light. Name the component of
rhodopsin that varies in cone cells.
1
(5)
3
7.(a) Describe the features of chromosomes in homologous pairs.
(b)In Drosophila, the genes for wing length (W), eye colour (E), body colour (B) and
presence of bristles (P) are linked.
The table below gives the frequency of recombination obtained in crosses involving
different pairs of the linked genes.
Gene pair in cross
Wing length × Eye colour
12%
Wing length × Body colour
18%
Wing length × Presence of bristles
15%
Eye colour
6%
× Body colour
Body colour × Presence of bristles
Frequency of recombination
3%
Using the letters W, E, B, and P to identify them, give the sequence in which the
genes would be arranged on the chromosome.
1
(4)
[C274/SQP368] 15
Page fifteen
Marks
8. The capercaillie (Tetrao urogallus) is the largest member of the grouse family. It is a
lekking species found in some old Caledonian pine forests.
Males are larger and noisier than females, particularly in spring when their displays make
them very conspicuous.
(a) What term is used to describe the condition, shown in the capercaillie, where there are
distinct differences between males and females?
1
(b) State one function of the male displays in spring.
1
(c) How do females benefit from being inconspicuous?
1
(d) Explain what is meant by “lekking species”.
2
(5)
[C274/SQP368] 16
Page sixteen
Marks
9. The figure shows the life cycle of the macroparasitic flatworm called Schistosoma
japonicum. The flatworm can live for many years within a host. In humans, if untreated, it
causes the disease schistosomiasis (bilharzia) that can be fatal.
Figure: Life cycle of Schistosoma japonicum
PRIMARY HOST
Within the body,
larvae migrate
to the liver as
they mature
Adult parasites
pass eggs into
large intestine
Eggs hatch
in water
Asexual
stage in snail
First free-living
stage
Second free-living
stage can
penetrate skin
SECONDARY HOST
(a)
(i) Explain why the snail may not be described as a vector.
1
(ii) Suggest a feature of the parasite’s life cycle that can lead to a higher rate of
transmission.
1
(iii) Name the phylum to which parasitic flatworms such as Schistosoma belong.
1
(b) Parasites living inside a host will be exposed to attack by the host’s immune system.
(i) Describe one way in which parasites may overcome the immune response of
their hosts.
1
(ii) Predict, in terms of the Red Queen hypothesis, how a host species would be
expected to respond to the parasite adaptation.
2
(6)
[C274/SQP368] 17
Page seventeen
Marks
10. In 1953, the viral disease myxomatosis was introduced to the UK to control rabbit
populations. The virus is spread by the rabbit flea. In successive years from 1952, young
rabbits removed from the wild population were tested against the virus and their response
noted, as shown in the Table.
Year in which
Number of
rabbits were
epidemics
selected from
previously
wild population
suffered by
for investigation the population
Response to the virus
(% of population)
Death
Moderate
symptoms
Mild
symptoms
1952
0
93
5
2
1953
2
95
5
0
1954
3
93
5
2
1955
4
61
26
13
1956
5
75
14
11
1957
6
54
16
30
(a) Use the data to show that rabbits were developing resistance over the period of the
study.
2
(b) Why was it important always to test the rabbits against the original strain of virus?
1
(3)
11.Answer either A or B
A. Discuss the concept of niche and how it applies to parasites.
(10)
OR
B. Discuss some of the challenges involved in the treatment and control of parasites.
[END OF SPECIMEN QUESTION PAPER]
[C274/SQP368] 18
Page eighteen
(10)
C274/SQP368
Biology
Advanced Higher (Revised)
Specimen Marking Instructions
for use in and after 2012
[C274/SQP368]
19
NATIONAL
QUALIFICATIONS
©
Section A
1A
2C
3B
4C
5D
6B
7B
8A
9C
10B
11D
12B
13C
14D
15B
16C
17B
18D
19A
20C
21D
22A
23D
24A
25C
[C274/SQP368] 20
Page two
Section B
Marks
Question 1
(a)
(i) They have the same sequence/order of amino acids
OR Idea showing understanding of primary structure.
1K
(ii) Disulfide (bridges), ionic (bonds), Van der Waals (attractions), hydrophobic
interactions.
1K
(b)
(i) Variability around mean.
1K
(ii) To judge if result is due to chance.
1K
(iii) As (ratio of) breaker to PrPSC increases and
– the % PrPSC decreases (in mice)
–there is a bigger % PrPSC decrease in mice than in humans OR it works
better/has a greater effect in mice than in humans
–(in mice the %) decrease gets less/the breaker gets less effective at higher
concentrations.
Any two
2P
(i) Beta sheets resist breakdown. 1
(c)
(Results show) a decrease in beta sheet content (36 to 9%). 1
(ii) Would show how close PrPSC is to normal structure after treatment.
(d)
(i) Add one (unit) of 10−2 to 9 (units) of diluent to get 10−3;
2P
1P
and repeat with 1 of this (+ 9).
OR Add 1 unit of 1 x 10−2 (concentration) to 99 units of diluent.
1K
(ii) The lower the concentration (of PrPSC) the greater the delay in the onset of
symptoms.
1P
(iii) 11% (for 10−2).
1P
(iv) Use more breaker.
Incubate PrPSC with breaker (before injection)/incubate longer/use different
lengths of incubation.
Reduce the concentration of PrPSC.
Any one
1P
(13)
U1 3K 7P
U3 2K 1P
U3 b(i) and c(ii) and d(iv)
Prac links d(i)
[C274/SQP368] 21
Page three
Marks
Question 2
(a) GABA is a ligand/substance that can bind to protein.
The channel is a protein that opens/gates in response to GABA/ligand binding.
Chloride passes through the protein when GABA is bound.
Any two
2K
(b)Transduction.
1K
(c)
(i) Chloride movement is (generally) greater at any GABA conc. if drug present.
1P
(ii) (Binding at the modulatory site) changes the conformation of the GABA site.
1K
(iii) (Make the cell more negative inside so) increase the membrane potential.
1P
(6)
Discuss the advantages of building a pilot study into the development of a biological
investigation.
(4)
U1 4K 2P
Question 3
1. Can be used to develop a new protocol.
2. Allows practice of established protocol.
3. Ensures appropriate range of values for independent variable.
4. Avoids results for dependent variable going “off the scale”.
5. Allows the number of repeat measurements required to be estimated.
6. Checks whether results can be produced in suitable time frame.
U3 4K
Question 4
1P
(a) The training will have no effect on GLUT 4 content of muscle.
(b) ND UT is baseline GLUT 4 (100%) and training does not produce significant
increase.
2
(1K 1P)
D UT is (significantly) lower GLUT 4 than baseline and exercise generates
significant increase.
OR 1 mark for noting controls are ND v D subjects and T v UT legs.
3 U3 (1K 2P)
[C274/SQP368] 22
Page four
(3)
Marks
Question 5
1K
(a)Hydrophobic.
(b) Thyroxine receptor protein is blocking transcription.
OR Thyroxine binding removes repression of genes. 1
More NaKATPase in membrane so more energy expenditure/higher metabolic rate. 1
2K
(c)
(i) 8 weeks.
1P
(ii) People need to be treated for several weeks before metabolic rate reaches normal.
1P
(iii) Starting metabolic rate is different for each individual.
1P
U1 3K 2P
U3 1P
(6)
Question 6
(a)Retinal.
1K
(b) Excited rhodopsin activates G protein which in turn activates many enzyme molecules.
Enzyme molecules cause closure of ion channels/catalyse the removal of molecules
that keep channels open.
Inward leakage of positive ions / Na+ and Ca+ is halted so membrane potential
increases.
Hyperpolarisation/increasing charge stimulates nerve impulse.
Any two
2K
(c) Wide range of wavelengths absorbed OR High degree of amplification from single
photon.
1K
(d)Opsin.
1K
U1 5K
(5)
Question 7
(a) Same size.
Same genes in same loci.
Same position of centromeres.
Inherited from different parents.
Alleles for a gene may be different on the two chromosomes.
5  = 3; 3 or 4  = 2; 1 or 2  =1
3K
(b) WEPB or BPEW
1P
U2 3K 1P
[C274/SQP368] 23
Page five
(4)
Marks
Question 8
(a) Sexual dimorphism.
1K
(b) To attract females/let females select “best” male.
1K
(c) Camouflage decreases chances of predation.
OR Improves survival chances of young.
1K
(d) In breeding season, males gather in leks/communal display areas. 1
Display enables female choice. 1
2K
(5)
U2 5K
Question 9
(a)
(i) Does not itself actively transmit parasite to another species.
1K
(ii) Waterborne dispersal stage.
1K
(iii)Platyhelminthes.
1K
(b)
(i) Mimic host antigens to evade detection.
OR Modify host immune response to reduce chances of destruction.
OR Antigenic variation allows rapid evolution to overcome host immune cell
clonal selection.
2K
(ii) Red Queen hypothesis = coevolution/“arms race”. 1
1K
Host adaptation to counter parasite. 1
(6)
U2 6K
Question 10
(a) Trend: Death rate decreases as the prior exposure to epidemics increases.
Supporting data: % deaths drop from 93 to 54% after 6 epidemics.
Trend: proportion with moderate/mild symptoms increases with exposure.
Supporting data: 5  26% after 3 epidemics.
OR Proportion with mild symptoms increases with exposure 2  30%.
2P
(b) To be sure that the response was not to a mutated virus.
OR To be sure that the rabbit response was not due to other factors.
U2 3P
[C274/SQP368] 24
1P
Page six
(3)
Marks
Question 11
Answer either A or B
A
Discuss the concept of niche and how it applies to parasites.
1. Define as multi-dimensional summary of tolerances and requirements.
2. Fundamental defined.
3. Realised defined.
4. Interspecific competition when niches overlap.
5. Interspecific competition can lead to competitive exclusion/local extinction.
6. When realized niches differ, species can co-exist by resource partitioning.
7. Symbionts explained.
8. Idea of cost/benefit of the relationship between parasite and host.
9. Parasites have narrow niche – explained in relation to host specificity.
10. Degenerate idea in relation to adaptations, ie loss rather than gain.
11. Endo- and ectoparasite distinction.
12. Definitive host defined.
13. Intermediate host.
14. Vector’s role in lifecycle of parasites.
15.Example.
16.Example.
(10)
Any ten
B
Discuss some of the challenges involved in the treatment and control of parasites.
1. Parasites are difficult to culture in the laboratory.
2. Rapid antigen change.
3. Should be reflected in the design of vaccines.
4. Host and parasite metabolism are similar.
5. Makes it difficult to develop drugs . . .
6. . . . that only target the parasite/do not harm hosts.
7. Limited number of control strategies.
8. Improvement of sanitation.
9. Vector control.
10. Parasites spread most rapidly in areas where control is difficult.
11. Example – eg overcrowded refugee camps, sea lice in salmon farming.
12. Parasites more abundant in tropical climates/developing countries.
13. Improvements in parasite control reduce child mortality.
14. Result in improvements in child development and intelligence.
15. Individuals have more resources for growth and development.
(10)
Any ten
U2 10K
[END OF SPECIMEN MARKING INSTRUCTIONS]
[C274/SQP368] 25
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