Download EP02/AH/02 Biology

AH
National
Qualications
EXEMPLAR PAPER ONLY
EP02/AH/02
Biology
Section 1 — Questions
Date — Not applicable
Duration — 2 hours 30 minutes
Instructions for the completion of Section 1 are given on Page 02 of your question and answer
booklet EP02/AH/01.
Record your answers on the answer grid on Page 03 of your question and answer booklet.
Before leaving the examination room you must give your question and answer booklet to the
Invigilator; if you do not, you may lose all the marks for this paper.
©
*EP02AH02*
SECTION 1 — 25 marks
Attempt ALL questions
1. The diagram below shows a haemocytometer counting chamber containing animal cells. The
depth of the chamber is 0·01 cm.
0·1 cm
0·1cm
central
square
animal
cell
The concentration of animal cells, based on the cell count from the central square, is
A
2·0 × 104 cells per cm3
B
2·0 × 105 cells per cm3
C
2·0 × 106 cells per cm3
D
2·0 × 107 cells per cm3.
Page 02
2. The following diagram shows a small polypeptide integrated into a membrane.
membrane
er
et Tyr Asn
Thr S Gly P r
o Il e P h e M
Which row in the table below classifies amino acids in this polypeptide?
polar
non-polar
A
Thr
Pro
B
Ile
Tyr
C
Asn
Ser
D
Phe
Gly
3. In the post-translational modification of a protein, which of the following enzymes would
remove a phosphate?
A
proteinase
B
ATPase
C
phosphatase
D
kinase
Page 03
The graph below shows the changes in the activity of enzymes that synthesise tryptophan
and utilise lactose in a cell after the addition of tryptophan and lactose.
Activity of enzymes
4.
lactose utilisation
enzymes
tryptophan synthesising
enzymes
0
5
10
tryptophan and
lactose added
15
20
Time (minutes)
What valid conclusion may be made from the graph?
5.
A
Addition of lactose acts as a negative enzyme modulator.
B
Addition of tryptophan acts as a positive enzyme modulator.
C
Enzyme induction is occurring in lactose utilisation enzymes.
D
Enzyme induction is occurring in tryptophan synthesising enzymes.
Which row in the table below describes the charges on the two components of nucleosomes?
DNA
histone proteins
A
negative
negative
B
positive
negative
C
positive
positive
D
negative
positive
Page 04
6. Colorimetry was used to produce the standard curve below.
0·7
Absorbance
0·6
0·5
0·4
0·3
0·2
0·1
0
0
1
2
3
4
5
6
Protein concentration (µg/cm3)
In an experiment to extract soluble protein from potato tubers, 25 g tissue was ground with
50 cm3 of buffer and centrifuged. The volume of extract produced was 65 cm3.
When 1 cm3 of extract was tested, the absorbance was found to be 0·5.
What is the protein content of the potato tissue in μg/g fresh tissue?
A
3·9
B
9·0
C
11·7
D
13·5
Page 05
7. The following diagram shows cotransport (symport) of sodium ions (Na+) and glucose into a
cell lining the gut.
outside cell
Na-binding
site
Na+ glucose
glucose-binding
site
Na+
glucose
inside cell
Which row in the table below represents the relative concentrations of glucose and Na+ on
the two sides of the plasma membrane when cotransport occurs?
sodium
glucose
outside cell
inside cell
outside cell
inside cell
A
high
low
low
high
B
high
low
high
low
C
low
high
low
high
D
low
high
high
low
8. Which row in the table below describes features of rod cells in humans?
function in low
light intensity
contain different
forms of opsin
A
no
yes
B
yes
yes
C
no
no
D
yes
no
Page 06
9.
10.
Which of the following statements describes how genes that increase metabolic rate are
activated by a hydrophobic signalling molecule?
A
Thyroxine binds to a receptor protein on DNA and stops it inhibiting transcription.
B
Testosterone binds to a receptor protein on DNA and stops it inhibiting transcription.
C
Thyroxine binds to a receptor protein in the cytosol and the complex regulates
transcription.
D
Testosterone binds to a receptor protein in the cytosol and the complex regulates
transcription.
Which of the following diagrams best represents the sequence of phases involved in the cell
cycle?
A
C
11.
G2
M
G1
B
G1
S
S
M
G2
S
M
D G2
G2
G1
M
Which of the following would not be a substrate for caspases?
A
DNA
B
actin
C
histone
D
tubulin
Page 07
S
G1
12. Some characteristics and properties of four proteins are shown below.
protein
symbol
molecular
mass
type
fibrinogen
Fb
330 000
fibrous
haemoglobin
Hb
68 000
globular
albumin
Alb
65 000
globular
myoglobin
My
17 000
globular
The solubility of proteins in salt solutions
Fb
Hb
My
Increasing solubility
Alb
0
2
4
6
8
10
Salt concentration (units)
Which conclusion about the solubility of proteins is valid from this information?
A
Protein solubility increases as salt concentration decreases.
B
Globular proteins are more soluble than fibrous proteins.
C
Fibrous proteins are more soluble than globular proteins at low salt concentrations.
D
Solubility of proteins is not related to molecular mass.
Page 08
13. Dicrocoelium dendriticum is a flatworm parasite of grazing vertebrates such as sheep and
cattle.
Which row in the table below shows the phyla to which these species belong?
Dicrocoelium
cattle/sheep
A
Nematoda
Chordata
B
Platyhelminthes
Arthropoda
C
Nematoda
Arthropoda
D
Platyhelminthes
Chordata
14. Northern elephant seals, intensely hunted in the 19th century, have significantly less genetic
variation than southern elephant seals that were hunted less in the same period.
This reduced genetic diversity is most likely a result of
A
sexual selection
B
genetic drift
Cmutation
D
natural selection.
Page 09
15. Eggs from leopard geckos kept in breeding cages were collected and incubated at two
temperatures over five breeding seasons. When each new gecko hatched, its gender was
noted. The graph below shows how temperature affected gender in the population.
Percentage of males in
the population
100
30°C
32.5°C
80
60
40
20
0
1
2
3
4
Breeding season
5
How many females would be present in a population of 500 leopard geckos after four
seasons at 32·5°C?
A150
B200
C300
D350
Page 10
16. Red-green colour deficiency is X-linked. The diagram below shows a family tree in which
this condition occurs.
unaffected male
affected male
unaffected female
affected female
P
R
S
Q
T
U
V
W
Which of the following individuals passed on the allele responsible for red-green colour
deficiency in individual W?
A
T only
B
U only
C
T and U
D
T and P
Page 11
Number of eggs laid per year
17. The graph below shows the fecundity (reproductive output) of three different strains of
white Leghorn domestic hens in relation to age.
260
240
220
200
strain 1
strain 2
strain 3
180
160
140
120
100
80
60
40
20
0
1
2
3
4
5
6
7
8
9 10 11
Age of hen (years)
Which of the following conclusions can be supported from the information shown?
A
Young hens lay more eggs during their lifetime than older hens.
B
Fecundity decreases faster with age in birds that lay more eggs early in life.
C
The lifelong reproductive output of all three strains is approximately equal.
D
Fecundity in later life is independent of fecundity in earlier years.
18. An organism’s parental investment strategy can be classified as k-selected or r-selected.
Which row in the table below describes the characteristics of k-selected organisms
compared to r-selected organisms?
number of offspring
produced
size of offspring
produced
A
larger
larger
B
larger
smaller
C
smaller
larger
D
smaller
smaller
Page 12
19. The black belly stripe of great tit males (Parus major) is an important stimulus in territorial
and courtship displays. Males with broader stripes make better parents and are more
attractive to prospective mates. The width of the stripe is therefore correlated with male
quality.
This stripe functions as
A
an honest signal
B
a sign stimulus
C
a fixed action pattern
D
an imprinting stimulus.
20. Which row in the table below describes the ecological niche of a parasite?
niche
host specificity
A
wide
high
B
wide
low
C
narrow
high
D
narrow
low
21. The virulence of an infectious organism is defined as the case fatality risk (CFR). CFR can be
represented as the percentage of infections that result in death. The table below shows the
numbers of people infected by the “bird flu” virus (H5N1) and the numbers who died from it
over a five year period.
year
2004
2005
2006
2007
2008
total
infections
of H5N1
46
98
115
88
44
number
dying from
H5N1
infection
32
43
79
59
33
In which year was H5N1 most virulent?
A2004
B2006
C2007
D2008
Page 13
22. The diagram below shows the life cycle of a parasitic worm that causes a disease in humans.
adult worms
(about 1 cm)
cercaria
(about 300 µm)
free-swimming
stage
egg
(about 140 µm)
miracidia
(about 180 µm)
free-swimming stage
Which row in the table below describes the roles of the other species in this parasite’s life
cycle?
definitive host
intermediate host
snail is a vector
A
snail
human
no
B
human
snail
no
C
snail
human
yes
D
human
snail
yes
23. Which of the following is a non-specific immune response to a parasite?
A
Apoptosis induced by T lymphocytes
B
Presentation of antigens by phagocytes
C
Apoptosis induced by natural killer cells
D
Production of antibody by B lymphocyte clone
Page 14
24. For a species of butterfly, the duration of its flight periods and the week of its first sighting
were recorded from 1976 to 1998.
Graph 1: Duration of flight period
Duration (weeks)
20
15
10
5
1980
Year
1990
2000
Graph 2: Week of first sighting
Week in year
20
15
10
5
1980
Year
1990
2000
Which row in the table below identifies the behaviour trends shown in the graphs?
flight period
first sighting
A
shorter
earlier
B
longer
earlier
C
longer
later
D
shorter
later
Page 15
Percentage of time
one individual is vigilant
25. The Arctic fox is a predator of barnacle geese. To avoid predation, geese periodically look
up from grazing to scan for foxes. In a study of this vigilance behaviour, different flock sizes
of geese were monitored for ten minutes when a cardboard model of an Arctic fox was
placed 100 m away. The percentage of the time per individual spent on vigilance behaviour
was recorded for different flock sizes, as shown in the graph below.
50
40
30
20
10
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Number of geese in flock
Which of the following would also have to be done to test the hypothesis that vigilance
behaviour in response to the Arctic fox decreases as flock size increases? Repeat the work
A
and calculate the average results
B
using a recording time of 20 minutes
C
to get data for the missing flock sizes
D
with no model fox present.
[END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2
OF YOUR QUESTION AND ANSWER BOOKLET]
Page 16
AH
FOR OFFICIAL USE
National
Qualications
EXEMPLAR PAPER ONLY
EP02/AH/01
Mark
Biology
Section 1 — Answer Grid
and Section 2
Date — Not applicable
*EP02AH01*
Duration — 2 hours 30 minutes
Fill in these boxes and read what is printed below.
Full name of centre
Town
Forename(s)
Date of birth
Day
Month
Surname
Year
Number of seat
Scottish candidate number
Total marks — 90
SECTION 1 — 25 marks
Attempt ALL questions.
Instructions for completion of Section 1 are given on Page 02.
SECTION 2 — 65 marks
Attempt ALL questions.
Write your answers clearly in the spaces provided in this booklet. Additional space for answers
and rough work is provided at the end of this booklet. If you use this space you must clearly
identify the question number you are attempting. Any rough work must be written in this
booklet. You should score through your rough work when you have written your final copy.
Use blue or black ink.
Before leaving the examination room you must give this booklet to the Invigilator; if you do not
you may lose all the marks for this paper.
©
*EP02AH0101*
SECTION 1 — 25 marks
The questions for Section 1 are contained in the question paper EP02/AH/02.
Read these and record your answers on the answer grid on Page 03 opposite.
Use blue or black ink. Do NOT use gel pens or pencil.
1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in
the appropriate bubble (see sample question below).
2. There is only one correct answer to each question.
3. Any rough working should be done on the additional space for answers and rough work at
the end of this booklet.
Sample Question
The thigh bone is called the
Ahumerus
Bfemur
Ctibia
Dfibula.
The correct answer is B — femur. The answer B bubble has been clearly filled in (see below).
A
B
C
D
Changing an answer
If you decide to change your answer, cancel your first answer by putting a cross through it (see
below) and fill in the answer you want. The answer below has been changed to D.
A
B
C
D
If you then decide to change back to an answer you have already scored out, put a tick (3) to the
right of the answer you want, as shown below:
A
B
C
A
D
B
C
or
*EP02AH0102*
Page 02
D
SECTION 1 — Answer Grid
*OBJ25AD1*
A
B
C
D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
*EP02AH0103*
Page 03
SECTION 2 — 65 marks
Attempt ALL questions
It should be noted that question 11 contains a choice.
1. Some species of Daphnia (water fleas) are able to develop their head spines and tail spines
as structural defences against predators such as fish. These structures can increase in
length in response to kairomones, chemicals in water where the fish occur.
One species, Daphnia lumholtzi, occurs naturally in freshwater habitats in Africa, Asia and
Australia. It has now spread throughout North America, first appearing in lakes in the south
in 1990 and reaching more northern and western lakes within four years. It is thought to
have been introduced when lakes were stocked with African fish species.
Figure 1: Illustration of Daphnia lumholtzi before and after exposure to kairomones
head spine
body length
tail spine
before exposure
after exposure
The successful spread of D. lumholtzi has been attributed to its ability to develop defensive
spines. To investigate the relevance of this feature to Daphnia survival, laboratory
experiments were carried out to compare the population dynamics of D. lumholtzi with
Daphnia pulicaria, the most widely distributed American species.
All the experiments were conducted under standard conditions of temperature (20°C) and
light in identical plastic tanks. The culture medium was based on minerals and phosphate
buffer made up in water of a very high purity. Daphnia were fed with green algae in
quantities that maintained constant food availability. The density of each species was the
same at the start and populations were left for several days before sampling began.
Figure 2 shows the population changes observed from the first day of sampling in
experiments set up as below:
Experiment A: Single species alone without predators
Experiment B: Two species together without predators
Experiment C: Two species together with fish predators.
*EP02AH0104*
Page 04
1. (continued)
Figure 3 shows the results of measuring the lengths of head spines and tail spines for the
two species in culture medium either containing or lacking kairomones.
Figure 2: Population changes in Experiments A, B and C
Density of adults (individuals 1−1 )
60
50
40
Experiment A: D. pulicaria
30
Experiment A: D. lumholtzi
Experiment B: D. pulicaria
20
10
Experiment B: D. lumholtzi
0
Density of adults
(individuals 1−1 )
1
5
9 13 17 21 25 29 33 37 41 45
Time (days)
20
Experiment C: D. lumholtzi
10
0
Experiment C: D. pulicaria
1
5
9 13 17 21 25
Time (days)
Figure 3: Relative lengths of spines before and after exposure to kairomones
Relative spine length
(% of body length)
100
head spine
80
tail spine
60
40
20
0
Control Kairomone Control Kairomone
D. lumholtzi
D. pulicaria
*EP02AH0105*
Page 05
1. (continued)
MARKS
(a) Refer to Figure 2.
(i) Use the data at Day 41 to demonstrate that competition is a
negative interaction for both species.
2
(ii) Suggest how long it takes for spine formation to affect predator
behaviour. Justify your answer.
2
Number of days
Justification
(b) Refer to Figure 3.
(i) Suggest the defence used by D. pulicaria against fish predation.
1
(ii) For D. lumholtzi with a mean body length of 1·6 mm, calculate the
difference in length between the head spines in the control and
kairomone cultures.
1
Space for calculation and working
mm
*EP02AH0106*
Page 06
DO NOT
WRITE IN
THIS
MARGIN
1. (continued)
MARKS
(c) Explain why the researchers believed the formation of head spines to be
critical in the survival of D. lumholtzi in competition with D. pulicaria.
*EP02AH0107*
Page 07
2
DO NOT
WRITE IN
THIS
MARGIN
MARKS
2. The table below shows data comparing some stages in the purification of an
enzyme from a tissue sample. Total protein and enzyme activity are measured
at the end of each stage.
stage
1
liquidised tissue
2
precipitation by salt
3
iso-electric separation
4
affinity chromatography
total protein
(mg)
enzyme activity
(units)
10 000
2 000 000
3000
1 500 000
500
500 000
30
42 000
(a) (i) Calculate the percentage of the original protein that had been
removed by the end of the process.
1
Space for calculation and working
%
(ii) Enzyme purity can be calculated from these values as the activity
per mg of protein.
Calculate by how many times the enzyme purity increased by the
end of stage 4.
2
Space for calculation and working
(b) Explain the principle of iso-electric separation.
*EP02AH0108*
Page 08
2
DO NOT
WRITE IN
THIS
MARGIN
2. (continued)
MARKS
(c) In affinity chromatography, a ligand specific to the enzyme was bonded
to beads in a burette.
Explain how this method can improve purity.
*EP02AH0109*
Page 09
2
DO NOT
WRITE IN
THIS
MARGIN
MARKS
3. Describe the transport of sodium and potassium ions across the plasma
membrane.
*EP02AH0110*
Page 10
5
DO NOT
WRITE IN
THIS
MARGIN
MARKS
4. Myoglobin and haemoglobin are oxygen-carrying proteins. Myoglobin has one
polypeptide chain and is found in muscle. Haemoglobin has four polypeptide
chains and is found in red blood cells. The tertiary structures of the
myoglobin and the haemoglobin chains are very similar. Each chain has one
binding site for oxygen.
The proportion of binding sites occupied by oxygen is known as saturation.
saturation =
number of oxygen binding sites occupied
total number of oxygen binding sites
The graph below shows the binding of oxygen to haemoglobin and myoglobin
as the available oxygen is increased.
1∙00
myoglobin
haemoglobin
Saturation
0∙75
0∙50
0∙25
0
0
10
20
30
40
50
O2 pressure (units)
(a)(i)Use the data to compare the saturation of myoglobin and
haemoglobin between 0 and 30 units.
1
(ii)Explain how the information shows that quaternary structure
affects the binding of oxygen to haemoglobin.
2
*EP02AH0111*
Page 11
DO NOT
WRITE IN
THIS
MARGIN
MARKS
4. (continued)
(b) Use the formula to calculate the change in the number of oxygen
molecules bound to haemoglobin as the oxygen pressure is reduced from
30 to 20 units.
1
Space for calculation and working
(c) Haem groups are an example of non-polypeptide components present in
proteins.
State the term that describes such components.
*EP02AH0112*
Page 12
1
DO NOT
WRITE IN
THIS
MARGIN
5. Starch consists of glucose molecules in long coiled chains, periodically joined at branching
points. In the iodine test for starch, the iodine lies within the coils and the solution has a
blue/black colour.
coiled chain of glucose
iodine
The enzyme alpha amylase (α-amylase) hydrolyses starch into maltose and dextrins (short
chains of glucose molecules). Samples taken from an α-amylase/starch mixture, when
reacted with iodine solution, will show changes in colour as hydrolysis proceeds.
Different reaction products contribute different colours, as shown below.
starch
(blue/black)
dextrins
(red/purple)
maltose
(colourless)
Within an investigation to test the effect of an inhibitor on the rate of amylase activity,
researchers needed to develop a quick way to measure starch concentrations. They
produced a colorimetric method based on the starch-iodine colour change. Part of the
experiment to find the best wavelength is outlined below.
Amylase buffered at pH7 was incubated with different reaction mixtures at optimum
temperature. After 30 minutes, hydrochloric acid was used to stop any enzyme action and
then the iodine solution was added to produce the colour. Absorbance was measured
across a range of wavelengths for different reaction mixtures, as shown in the graph below.
Absorbance
2.5
starch 2g/l
2.0
2g/l starch with maltose
starch 1g/l
1.5
reference blank
1.0
0.5
0
400
450
500
550
600
Wavelength (nm)
650
700
*EP02AH0113*
Page 13
5. (continued)
MARKS
(a) State the term that describes an experiment, like the one outlined, that
is developing a technique within a larger investigation.
1
(b) State what would be left out of the reference blank so it functions as a
control.
1
(c) The researchers concluded that 580 nm would be the best wavelength for
quantifying amylase activity by this method.
(i) Explain how they arrived at this conclusion.
1
(ii) The results show that the presence of the reaction product maltose
is not a confounding variable. State what is meant by the term
confounding variable.
1
(iii) Explain how the results show that the method will be valid in
investigating the effect of amylase inhibitor on the rate of starch
breakdown.
1
*EP02AH0114*
Page 14
DO NOT
WRITE IN
THIS
MARGIN
MARKS
6. The hormone insulin is involved in the control of blood glucose concentration.
(a) Describe how insulin stimulates the uptake of glucose into cells.
2
(b) Research has shown that fatty (adipose) tissue secretes a number of
signalling molecules that regulate a variety of metabolic processes. One
of these molecules, adiponectin, is thought to increase the sensitivity of
cells to the hormone insulin.
Table 1 below shows the results of a study that compared the
concentration of adiponectin in subjects having type 2 diabetes with
non-diabetic subjects.
Table 1
subjects
average plasma
adiponectin concentration
(μg cm−3 ± SE)
type 2 diabetes
6·6 ± 0·4
non-diabetics
7·9 ± 0·5
Use the information to explain the relationship between type 2 diabetes
and the average plasma concentration of adiponectin.
*EP02AH0115*
Page 15
2
DO NOT
WRITE IN
THIS
MARGIN
6. (continued)
MARKS
(c) Table 2 below shows the results of a second study that measured changes
in adiponectin following treatment of individuals at risk of developing
type 2 diabetes.
Table 2
treatment
average increase in
adiponectin concentration
(μg cm−3 ± SE)
drug treatment
0·83 ± 0·05
lifestyle changes
0·23 ± 0·05
none
0·10 ± 0·05
Explain how the data in Table 2 confirm that both treatments were
effective in increasing adiponectin concentration.
1
(d) Both studies used human subjects. For this type of research:
(i) give one important ethical consideration;
1
(ii) explain why a large sample size is required to produce valid
conclusions.
1
*EP02AH0116*
Page 16
DO NOT
WRITE IN
THIS
MARGIN
MARKS
7. The protein p53 plays an important role in controlling cell division. The
diagram below represents how the activation of p53 can result in arrest of the
cell cycle.
Mdm2
Mdm2
p53
kinase
p53 inactive
p53
p53 active
production of
protein p21
activated
p21
Cdk
p21
Cdk
Cdk inhibited
(a) Explain why binding of p21 protein to cyclin dependent kinase (Cdk)
prevents the cell cycle from progressing.
2
(b)State one trigger that would stimulate the activation of p53.
1
(c) State one other fate, apart from arrest of the cell cycle, of a cell that has
had p53 activated.
1
(d) Explain why it is important that the cell cycle is controlled in a
multicellular organism.
2
*EP02AH0117*
Page 17
DO NOT
WRITE IN
THIS
MARGIN
8. Courtship behaviour in the dung beetle Onthophagus sagittarius was
investigated in the laboratory.
MARKS
Beetles were paired by randomly selecting males and females, and the pairs
were placed in breeding chambers. Mating success in relation to the
frequency of courtship behaviour was recorded for large and small males.
Probability of mating
small males
large males
Frequency of courtship behaviour
(a) From the results, describe how female choice changes in relation to male
size.
1
(b) State the term used to indicate the time from introduction until first
courtship during the experiment.
1
(c) Male and female dung beetles can be distinguished by their horns.
State the term used to describe this difference in appearance.
(d) Females in many species are relatively inconspicuous. Suggest why this is
of importance to them.
*EP02AH0118*
Page 18
1
1
DO NOT
WRITE IN
THIS
MARGIN
MARKS
9. (a) The spread of a buttercup plant, Ranunculus repens, from an established
flowerbed into a nearby disturbed area, is shown in the illustration
below.
flowers
runner
flowerbed
disturbed area
(i) Explain what is meant by the term hermaphroditic as it applies to
plants such as the buttercup.
1
(ii) Explain how asexual reproduction is of advantage to the buttercup
in the colonisation of this disturbed area.
1
(b) State the term that describes the mode of asexual reproduction in
arthropods where offspring arise from unfertilised eggs.
1
*EP02AH0119*
Page 19
DO NOT
WRITE IN
THIS
MARGIN
10. The parasite–stress hypothesis states that children who contract parasites
have reduced cognitive abilities (on average). Some research has shown that
parasites alone account for 67% of the worldwide variation in intelligence (IQ)
and that there is a significant negative correlation between the prevalence of
parasitic infection in a country and the mean IQ of its population.
MARKS
To test this hypothesis, researchers examined data from each of the states in
the USA. Results from their study are shown in the scatterplot below.
Correlation between mean US state IQ and
relative values for infectious disease stress
106
104
Mean IQ
102
100
98
96
94
−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5
Parasite stress (relative scale)
3
(a) Explain why parasitic infections might result in lower intelligence in a
developing individual.
2
(b) Give one reason why the results in the scatterplot are thought to have
low reliability.
1
(c) As a criticism of the parasite–stress hypothesis, it has been argued that
“correlation does not necessarily indicate causation”.
With reference to the scatterplot, explain the phrase “correlation does
not necessarily indicate causation”.
*EP02AH0120*
Page 20
2
DO NOT
WRITE IN
THIS
MARGIN
MARKS
11. Answer either A or B.
A Discuss meiosis under the following headings:
(i) the sequence of events;
7
(ii) the origin of genetically variable gametes.
3
OR
B
Discuss the process of evolution under the following headings:
(i) importance of mutation in enabling evolution;
6
(ii) factors increasing the rate of evolution.
4
Labelled diagrams may be used where appropriate.
[END OF EXEMPLAR QUESTION PAPER]
*EP02AH0121*
Page 21
DO NOT
WRITE IN
THIS
MARGIN
MARKS
ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK
*EP02AH0122*
Page 22
DO NOT
WRITE IN
THIS
MARGIN
MARKS
ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK
*EP02AH0123*
Page 23
DO NOT
WRITE IN
THIS
MARGIN
AH
National
Qualications
EXEMPLAR PAPER ONLY
EP02/AH/01
Biology
Marking Instructions
These Marking Instructions have been provided to show how SQA would mark this
Exemplar Question Paper.
The information in this publication may be reproduced to support SQA qualifications only
on a non-commercial basis. If it is to be used for any other purpose, written permission
must be obtained from SQA’s Marketing team on [email protected].
Where the publication includes materials from sources other than SQA (ie secondary
copyright), this material should only be reproduced for the purposes of examination or
assessment. If it needs to be reproduced for any other purpose it is the user’s
responsibility to obtain the necessary copyright clearance.
©
General Marking Principles for Advanced Higher Biology
This information is provided to help you understand the general principles you must apply when
marking candidate responses to questions in this paper. These principles must be read in
conjunction with the detailed Marking Instructions, which identify the key features required in
candidate responses.
(a)
Marks for each candidate response must always be assigned in line with these general
marking principles and the detailed Marking Instructions for this assessment.
(b)
Marking should always be positive. This means that, for each candidate response, marks
are accumulated for the demonstration of relevant skills, knowledge and understanding:
they are not deducted from a maximum on the basis of errors or omissions.
(c)
Half marks may not be awarded.
(d)
Where a candidate makes an error at an early stage in a multi-stage calculation, credit
should normally be given for correct follow-on working in subsequent stages, unless the
error significantly reduces the complexity of the remaining stages. The same principle
should be applied in questions which require several stages of non-mathematical
reasoning.
(e)
Unless a numerical question specifically requires evidence of working to be shown, full
marks should be awarded for a correct final answer (including units if required) on its own.
(f)
Larger mark allocations may be fully accessed whether responses are provided in
continuous prose, linked statements or a series of discrete developed points.
(g)
In the detailed Marking Instructions, if a word is underlined then it is essential; if a word is
(bracketed) then it is not essential.
(h)
In the detailed Marking Instructions, words separated by / are alternatives.
(i)
If two answers are given where one is correct and the other is incorrect, no marks are
awarded.
(j)
Where the candidate is instructed to choose one question to answer but instead answers
both questions, both responses should be marked and the better mark awarded.
(k)
The assessment is of skills, knowledge and understanding in Biology, so marks should be
awarded for a valid response, even if the response is not presented in the format
expected.
For example, if the response is correct but is not presented in the table as requested, or if
it is circled rather than underlined as requested, give the mark.
(l)
Unless otherwise required by the question, use of abbreviations (eg DNA, ATP) or chemical
formulae (eg CO2, H20) are acceptable alternatives to naming.
(m) Content that is outwith the Course assessment specification should be given credit if used
appropriately, eg metaphase of meiosis.
(n)
If a numerical answer is required and units are not given in the stem of the question or in
the answer space, candidates must supply the units to gain the mark. If units are required
on more than one occasion, candidates should not be penalised repeatedly.
(o)
Incorrect spelling is given. Sound out the word(s):
•
•
•
if the correct term is recognisable then give the mark
if the word can easily be confused with another biological term then do not give the
mark, eg ureter and urethra
if the word is a mixture of other biological terms then do not give the mark, eg
mellum, melebrum, amniosynthesis.
Page 02
(p)
When presenting data:
•
•
•
•
(q)
if a candidate provides two graphs or charts in response to one question (eg one in the
question and another at the end of the booklet), mark both and give the higher mark
for marking purposes no distinction is made between bar charts (used to show
discontinuous features, have descriptions on the x-axis and have separate columns) and
histograms (used to show continuous features, have ranges of numbers on the x-axis
and have contiguous columns)
other than in the case of bar charts/histograms, if the question asks for a particular
type of graph or chart and the wrong type is given, then do not give the mark(s) for
this
where provided, marks may still be awarded for correctly labelling the axes, plotting
the points, joining the points either with straight lines or curves (best fit rarely used),
etc. The relevant mark should not be awarded if the graph uses less than 50% of the
axes; if the x and y data are transposed; if 0 is plotted when no data for this is given
(ie candidates should only plot the data given).
Marks are awarded only for a valid response to the question asked. For example, in
response to questions that ask candidates to:
•
•
•
•
•
•
•
•
calculate, they must determine a number from given facts, figures or information;
compare, they must demonstrate knowledge and understanding of the similarities
and/or differences between things;
describe, they must provide a statement or structure of characteristics and/or
features;
evaluate, they must make a judgement based on criteria;
explain, they must relate cause and effect and/or make relationships between things
clear;
identify, name, give, or state, they need only name or present in brief form;
predict, they must suggest what may happen based on available information;
suggest, they must apply their knowledge and understanding of Biology to a new
situation. A number of responses are acceptable: marks will be awarded for any
suggestions that are supported by knowledge and understanding of Biology.
Page 03
Detailed Marking Instructions for each question
SECTION 1
Question Response
Mark
1
B
1
2
A
1
3
C
1
4
C
1
5
D
1
6
C
1
7
A
1
8
D
1
9
A
1
10
B
1
11
A
1
12
A
1
13
D
1
14
B
1
15
A
1
16
D
1
17
B
1
18
C
1
19
A
1
20
C
1
21
D
1
22
B
1
23
C
1
24
B
1
25
D
1
Page 04
SECTION 2
Question
1
a
Max
mark
Expected response
i Population of both species higher
when separate/in Experiment A
OR
equivalent for Experiment B (1)
Additional guidance
2
Comparison of Experiment A with B
Any day in range 9-13
Comparison of the population
change in both species
D. lumholtzi 25 vs 11
OR
D. pulicaria 37 vs 22 (1)
1
a
ii 9-13 days (1)
(density of)
D. lumholtzi > D. pulicaria
OR
D. lumholtzi increasing D. pulicaria
decreasing (1)
2
1
b
i Increase in length of tail spine
1
1
b
ii 0·40 mm
1
1
c
2
(When both are competing) in
presence of fish / predators /
kairomones / Experiment C (1)
D. lumholtzi increases head spine as
well as tail spine
Increase = grow, produce, make
Both species increase tail spines and
D. lumholtzi also increases head
spine / D. pulicaria does not
increase head spine (1)
2
a
i 99·7%
1
2
a
ii 7 times
2
Specific activities
Stage 1: 200
Stage 4: 1400
2
Amino acid OK instead of protein
1 mark for one correct specific
activity
difference is 1200 units
2
b
Isoelectric point is pH where a
protein has no net/overall charge
(1)
Electrophoresis = electric field
In a buffer, proteins at their
isoelectric point will +
not move in an electric field
OR
(become insoluble and) settle out /
precipitate (1)
2
c
The ligand binds the enzyme (1)
The enzyme is held in the burette
while the other proteins pass
through / wash out (1)
Page 05
2
Question
3
1.
2.
3
4.
5.
6.
7.
4
a
Max
mark
Expected response
Protein that spans the
membrane
Works against concentration
gradient/by active transport
ATP provides phosphate
Phosphate attaches to
pump/protein/protein
phosphorylated
Phosphorylation/
dephosphorylation alters
conformation/shape of protein
OR description
Different conformations have
different affinity for
sodium/potassium
(3) Sodium ions (pumped) out
of cell and (2) potassium in
i (For the increase in O2 pressure
0-30 units)
Myoglobin increases to 0·975 and
5
Additional guidance
Any five
Not channel or carrier
Diagrams might be useful here but
must be annotated correctly
1
Haemoglobin increases to 0·50
4
a
ii Curves differ/binding differs
AND
Tertiary (structure) the
same/similar (1)
2
Or illustrate with correct values
Not changed proportions
Only Hb has quaternary (1)
4
b
1 (less)
1
4
c
Prosthetic groups
1
5
a
Pilot study
1
5
b
Starch
1
5
c
i This gives maximum absorbance for
the starch reaction and lowest for
blank
OR
Maximum difference between
reaction and blank/control
1
5
c
ii (Other than the independent
variable) a factor that may affect
dependent variable/results
1
5
c
iii (Aim is to monitor rates of reaction
and) method can distinguish
between starch concentrations
1
Page 06
Question
6
a
Max
mark
Expected response
Insulin binding triggers recruitment
of glucose transporters/GLUT 4 (1)
Additional guidance
2
to (plasma) membrane of
fat/muscle/target cells (1)
6
b
Adiponectin levels are lower in type
2 diabetics / higher in non-diabetics
(1)
2
Type 2 diabetes is associated with
decreased sensitivity to insulin /
loss of receptor function (1)
Increases are (statistically)
significant
OR
use error data to show the ranges
do not overlap
1
6
c
6
d
i Informed consent / right to
withdraw data / confidentiality
1
6
d
ii It captures / represents the
variation in the population
OR
large number of confounding
variables
OR
other factors (such as body weight)
may have an effect
1
7
a
(p21 binding to Cdk) prevents Cdks
phosphorylating proteins (1)
2
(Proteins) that then stimulate the
cell cycle (1)
OR
unphosphorylated protein inhibits
cycle/stops cycle at the checkpoint
7
b
DNA damage
OR
death signals from lymphocytes
1
7
c
Apoptosis/programmed cell death /
DNA damage repaired / cell
continues through cell cycle
1
7
d
Uncontrolled decrease in the rate
of the cell cycle may lead to
degenerative disease (1)
2
Uncontrolled increase in the rate of
the cell cycle may lead to tumour
formation (1)
Page 07
Need idea that unphosphorylated
protein inhibits the cell cycle
Rb = protein
Question
Max
mark
Expected response
8
a
At low courtship frequencies large
males are chosen more often / have
a higher probability of mating but
this reverses at higher frequencies
OR
It changes from large males to small
males as courtship rates increase
1
8
b
Latency
1
8
c
Sexual dimorphism
1
8
d
Less easy for predators to see them
OR
May be nesting/laying eggs (1)
OR
survival chances of the young
increase (1)
1
9
a
i Has both male and female parts/sex
organs
1
9
a
ii Clones/offspring are genetically
identical to parent/already adapted
to habitat
OR
new plants sustained by
runner/parent until established
OR
rapid spread/colonisation/rapid
population increase
1
9
b
Parthenogenesis/parthenogenic
1
10
a
Combating parasite/infection uses
energy/resources (1)
2
Additional guidance
Not less predation
Less brain/cognitive development /
damaged brain (1)
10
b
eg some samples long way from
best-fit line
1
eg small/no samples for some
infection levels
eg some obvious
discrepancies/inconsistencies such
as at stress 0
10
c
IQ increases as parasitic infection
decreases (1)
The relationship does not mean that
level of parasite stress causes IQ
Page 08
2
not “small sample size”
Question
Max
mark
Expected response
Additional guidance
OR
another variable could account for
differences in IQ
OR
eg temperature (parasite abundance
higher in tropics)
/nutrition/literacy/education/GDP
or other economic indicator (1)
11
A
i
1.
Occurs in gamete mother cells
+ in sex organs
2. Two cycles of nuclear division
OR
State meiosis I following by II
Meiosis I
3. Nuclear membrane breaks
down and spindle forms
4. Homologous chromosomes
associate in pairs
5. Homologous pairs line up along
equator
6. Homologous chromosomes
segregate/pulled apart
7. New nuclear membrane forms
and cytoplasm divides
8. Two cells with half the number
of chromosomes/one set of
chromosomes/one copy of
genome
Meiosis II
9. Two new spindles form, one in
each cell
10. Chromosomes line up singly on
equator
11. Chromatids separate/are
pulled apart
12. New nuclear membranes form
and cytoplasm divides
13. To give 4 haploid cells/gametes
7
Any seven
11
A
ii
14. Crossing over occurs at
chiasmata
15. Chromatids break and rejoin
16. This process shuffles sections
of DNA between the
homologous pairs and
recombining alleles
17. Crossing over separate linked
genes
18. Chromosomes move apart
irrespective of their maternal
or paternal origin
OR
Chromosomes show
Independent Assortment
3
Any three
Page 09
Question
11
B
i
1.
2.
3.
4.
5.
6.
7.
8.
9.
11
B
Max
mark
Expected response
ii
Evolution is the change in
genetic composition of
population over time
Mutation gives rise to new
sequences of DNA/novel alleles
Results in new
variants/variation in traits/new
phenotypes
Mutation can be harmful,
neutral or beneficial
Beneficial mutations increases
“fitness”
Individuals compete for limited
resources . . .
. . . individuals with favourable
alleles more likely to survive
and breed
These (favoured) alleles
become more frequent in
subsequent generations
Definition of fitness in relation
to fecundity
10.
11.
12.
13.
High selection pressure
Shorter generation time
Warmer environments
Sharing of beneficial DNA
sequences between lineages
14./ 15. eg of how . . .
• through sexual reproduction
OR
• horizontal gene transfer (as
in X-bacteria)
Additional guidance
6
Any six
4
Any four
[END OF EXEMPLAR MARKING INSTRUCTIONS]
Page 10