Stafford answers a2 practicals

Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at http://www.amazon.com/gp/aw/s/ref=is_s_?k=AS+A2+Biology+with+stafford

Marking schemes for
Advanced Level Biology

A2 Biology with Stafford
Unit Six: Practical Workbook

www.staffordeducationalservices.com
Paper reference: 6BIO8

The book is available at the following link
http://www.amazon.com/gp/aw/s/ref=is_s_?k=AS+A2+Biology+with+stafford

A copy of this marking scheme and many other resources can be downloaded from the
following link
http://www.facebook.com/groups/biologywithstafford/

A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8

Page 1


Copyright © Stafford Valentine Redden
Unauthorized duplication contravenes applicable laws.
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Cover designed by
: Mohamed Sobir
Printed in Maldives by
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Published by
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All rights reserved.
No part of this publication may be
Reproduced, stored in a database or
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form or by any means, electronic,
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or otherwise, without the prior
written permission of the author

ISBN: 978-81-910705-3-8

Any other queries on any aspect of the practical paper can be clarified on my FB group
http://www.facebook.com/groups/biologywithstafford/

Cheers and all the best….
Stafford Valentine Redden
(M.A; M.Sc.; M.Ed.; (Ph.D))
Head of Department (Biology)
Villa International School,
Male’, Republic of Maldives
Email: staffordv@yahoo.com
Mobile: +960 7765507


Page 2


SAQ 1. a. Random Quadrat Sampling. This is because there are too many individuals to
count in the entire habitat. Random sampling will help to eliminate bias and provide a
good representation of the habitat.
(3)
b. Survey the area to identify any plants that may contain toxic chemicals or irritants,
like stinging nettle. Make students aware of these plants and ask them to avoid contact
with such plants. Also look out for stinging animals, like centipedes, scorpions or bees.
Carry long forceps and gloves to handle such hazards. Protective clothing could also
help. Wearing shoes could help to prevent from cuts on the feet and legs from sharp stones
or thorns.
(2)

c. Quadrat X is too small and will enclose only a very small number of plants, giving a
sample size that would be too small. Quadrat Z is too large for the area and very few
quadrats could be placed, without much overlap. At least 10 quadrats of size Y could be
placed into the habitat without any overlap and the quadrat Y is large enough to
enclose a sizeable number of plants. So, quadrat Y is most suitable.
(3)
d. i. Tie two measuring tapes along the edges of the sampling area at right angles to each
other. Select pairs of random numbers and use it as coordinates to place the quadrat at
these randomly selected sites within the sampling area.
(2)

ii. Grid on next page. Your results may differ.
Quadrat number

Number of stars / quadrat

1 (2,10)
2 (4,9)
3 (2,7)
4 (6,7)
5 (2,5)
6 (4,5)
7 (6,7)
8 (5,4)
9 (3,3)
10(4,2)
Mean =

2
2
2
2
4
1
2
2
0
1
1.8

Number of circles /
quadrat
3
3
3
0
0
3
1
1
1
1
1.6

iii. Mean number of organisms / area of quadrat = 1.8 / 16cm2 = 0.1125 stars per cm2


(2)

Page 3


[Total 15 marks]

Page 4


SAQ 2. a. The percentage cover of the lichens can be measured to provide a reliable

quantitative measure of abundance. Place a quadrat of 15 cm x 15cm , containing 25 smaller
squares within it, at a fixed position on the trunk and count the number of squares covered by
the lichens.
The percentage cover =
(number of squares covered by lichens/total number of squares in the quadrat) x 100

(2)
b. Repeating the experiment on few trees (6 to 8) provides a large enough sample to minimise
the effect of anomalous results and gives us a means to assess the variation and reliability of
the data. A large sample is also necessary to perform certain statistical tests.
(2)

c.i. In order to compare the results for each tree, the confounding variables must be kept

constant or monitored. These factors may be Height from ground, Diameter of tree, Species of
tree, Exposure to light, wind or temperature.
(2)

ii. The texture (smooth or rough bark) can affect the ability of the lichens to attach or hold on

to the bark. The water content of the bark could also affect the growth of lichens. Warm moist
conditions are ideal for the growth of most lichens. The pH of the bark can affect enzyme
activity in photosynthesis and respiration of the lichens. Most lichens are intolerant to acidic
conditions and are used as good indicators of acid rain or pollution.
(2)

d. The Light intensity, Temperature and Wind speed may differ in the different trees, leading

to differences in the rate of photosynthesis and water content of lichen cells.
The percentage cover used as a measure of abundance of lichens is dependent on subjective
measurements. Some squares are only partly covered and results in inaccurate measurement
of abundance.
(2)

e.
Distance from Light intensity /
the ground / m arbitrary units
0.5
0.4
1.0
0.36
1.5
0.33
2.0
0.24
2.5
0.19
3.5
0.17

Tree one
14
16
19
18
12
11

Abundance (%)
Tree two Tree three
13
14
12
11
16
15
14
14
13
9
9
6

Mean
13.7
13.0
16.7
15.3
11.3
8.7

(3)
[Total 13 marks]
SAQ3. a)
Temperature
/ oC
5
10
20
30
40
50
60

Number of eggs
per beaker
100
100
100
100
100
100
100

Number of larvae per beaker
Trial 1
Trial 2 Trial 3
Mean
26
32
39
45
60
67
32

27
36
38
50
63
79
12

29
35
39
52
75
72
9

27.3
34.3
38.3
49.0
66.0
72.7
17.7

Standard
deviation
1.5
2.0
0.6
3.6
7.9
6.0
(3)


Page 5


b) Plot the data into a suitable graph.

(3)

c) The range bars are relatively short for temperatures below 30 oC. This indicates low variance

in the repeated measurements and indicates high reliability. At temperatures above 40 oC, the
standard deviation increases, indicating that the data from repeated measurements is further
away from the mean. This indicates lower reliability.
(3)

d) As temperature increases from 5 to 50 oC, the hatch rate of brine shrimp increases. Beyond
50 oC there is a decrease in the hatch rate of brine shrimps.
(2)

[Total 11 marks]
SAQ 4 a)
Temperature/ oC
20
25
30
35
40
45

Number of seeds germinated per dish
Trial 1 Trial 2 Trial 3
Mean
6
5
6
5.7
5
4
3
4.0
3
2
3
2.7
2
3
1
2.0
0
1
0
0.3
0
0
1
0.3

(3)
b) The Petri dishes are sealed around their circumference, so oxygen availability may be

reduced and this can inhibit the germination of seeds. So temperature is not the only factor
which is influencing germination. Also the space inside the Petri dishes may not be enough to
allow the seedlings to grow freely. Some of the seeds may be non-viable and that may be the
reason that they do not germinate.
(2)

c) The seeds may be germinated in Petri dishes with the cover just placed on to it without

sealing it with cello tape. This will allow oxygen to freely enter the Petri dish and facilitate
germination. Using larger deeper Petri dishes can solve the problems of space for seedling
growth. The seeds can be X-rayed to check for the presence and condition of the embryo. This
will help to select seeds that are more likely to be viable.
(3)

[Total 8 marks]

Page 6


SAQ 5 (a) Restriction Endonucleases

(1)

(b) The analysis of the DNA will require DNA Fingerprinting, which requires a relatively
large quantity of DNA. The sample obtained from the window sill is too small and the PCR is
used to increase the quantity of DNA very rapidly.
(1)

(c) DNA Nucleotides and DNA Polymerase enzyme

(2)

(c) The primers are single stranded short chains of nucleotides which are complementary to the
bases at the 3’ end of each parent DNA strand. If the primers contain base sequences that
enable it to bind with human DNA fragments only, then only human DNA fragments will be
replicated.
(2)

(d) DNA is double stranded and the 3’ end of each strand has a different sequence of bases. So,
a different primer will be needed to bind to each strand of the DNA fragment.
(2)

(e) (i) Suspect 3. Because every dark band on the DNA profile from the window sill sample
matches the bands on the DNA profile of suspect 3.
(1)
(ii) To make a reliable and valid comparison, all DNA sample must be treated in the same way

and cut with the same restriction enzymes. Any differences or similarities in the banding
patterns will then be attributed to similarities or differences in the molecular structure of the
DNA.
(1)

e) The amplification factor = 2n, where n is the number of replication cycles
= 212

= 4096 DNA fragments

(2)
[Total 12 marks]

SAQ 6. a) Obtain DNA samples from many animals of the same species and carry out DNA

profiles for each. Select animals which have greatest differences in the DNA banding patterns.
(1)
b) The selective breeding of genetically diverse organisms increases heterozygosity in the
population and increases the chances of survival of the species in a changing environment. The
increased genetic diversity will allow the tigers to adapt to different environments and also
increase the chances of successful reintroduction programmes in the wild.
(3)

ii) The DNA is passed unchanged from parent to offspring and holds a reliable record of our
ancestry. The recording of information in stud books can involve human error.
(2)
c) (i) The length of a nucleotide is fixed (0.34nm). So, the number of base pairs multiplied by
0.34nm will give an accurate calculation of the length of a DNA fragment.

(2)

ii)

(2)

Page 7


iii) Distance moved by the fragment depends on the length of the fragment. Short fragments
move fastest and largest fragment move slowest through the gel.
(1)
iv) 5

(2)
[Total 13 marks]

SAQ7. (a)
Antibiotic
Streptomycin
Penicillin
Tetracycline
Ampicillin

Diameter of clear zone / mm

(2)
b) The antibiotic diffuses into the nutrient medium and inhibits the growth of bacteria or kills
bacteria, providing a clear area with no bacterial colonies. If the antibiotic is ineffective in
killing the bacteria then there will be no clear zone.
(2)

c) The area of the clear zone around Ampicillin is the largest.

(1)

d) The concentration of each antibiotic must be the same in all the discs to ensure that the
results are reliable. Otherwise the rate of diffusion into the nutrient agar will differ and
influence the reliability of the results.
(2)

[Total 7 marks]
SAQ8 a) The lag phase for the control setup of species A lasts for 6 hours, where as, the lag

phase for the control setup of species B lasts for 4 hours. The growth rate of Species A only is
higher than the growth rate for species B only, between 9 hours to 21 hours. The growth of
species A stops (at 18.5 hours) 3 hours before the growth for species B stops (at 21.5 hours).

(3)

b) Antibiotics are a group of chemicals, which kill bacteria or stop them from reproducing.
Bactericidal antibiotics: kill bacteria.
Bacteriostatic antibiotics: stop bacteria from reproducing.

(3)

c) Penicillin has no effect during lag phase. It starts having effect once bacteria are in in the

growth phase. This is because penicillin only affects dividing cells. It interferes with the
formation of peptidoglycan in the bacterial cell wall. Cell walls cannot for and cells that are
dividing will lack cell walls. These cells will then absorb water rapidly and die from osmotic
shock.
(3)

(Total 9 marks)


Page 8


SAQ9 A spirometer was used to compare a person's breathing at rest and during exercise. The results
are shown in the graphs below.

a) 10 mm on Y axis of the graph = 1 dm3 of oxygen
(8.5+8+8)÷3 mm on Y axis = 0.82 dm3

(2)

Note: it is a good practice to measure the tidal volume three times (as shown on the graph) and
then find the mean tidal volume.

b) breathing rate = (number of breaths ÷ time) x 60

= (5.5 ÷ 29) x 60 = 11.4 breaths per minute


(2)

Page 9


c) at rest

Minute ventilation rate = breathing rate x tidal volume
= 11.4 x 0.82 = 9.3 dm3 min-1
During exercise
Minute ventilation rate = breathing rate x tidal volume
= 31.6 x 1.5 = 47.4 dm3 min-1

(3)
d) after exercise, the breathing rate is 2.8 times more than at rest and the tidal volume is 1.8
times more after exercise. The rate of oxygen consumption is also greater following exercise.
During exercise, the tissues (specially muscles) have an increased demand for oxygen for
respiration to provide ATP at an increased rate for muscle contraction. The pH of the blood
decreases due to increased carbon dioxide and lactic acid production. The low pH is detected
by chemoreceptors in the carotid body and medulla. Respiratory centre in the medulla
stimulates the intercostals muscles and diaphragm to increase the breathing rate and depth
(tidal volume).
(4)
(Total 11 marks)

SAQ10 The diagram below shows a simple respirometer, which can be used to measure, the uptake of
oxygen by germinating seeds.

(a) The potassium hydroxide solution absorbs carbon dioxide released during respiration. This

ensures that any change in volume of gases in the tube is only due to oxygen being used up for
respiration.
(1)

(b) The 1 cm3 syringe is used to reset the apparatus by readjusting the liquid level in the
manometer U tube. This allows multiple readings to be taken without having to reinstall the
entire apparatus.
(1)

(c) Setup the apparatus as shown in the diagram and allow the seeds to respire. The seeds use

up oxygen the liquid in the manometer U tube will begin to fall. Note the distance moved by
the liquid in the U tube. The rate of oxygen uptake by the seeds is found by the following
equation.
rate of oxygen uptake = (π r2 l ) / t
where, π = 3.14 ; r= radius of manometer U tube; l = length moved by the liquid in the
manometer U tube; t = time taken for the liquid to move.

d) The antiseptic will prevent the growth of bacteria on the seed. If the bacteria are allowed to
grow along with the seeds then the bacteria will respire and the readings will be unreliable.

(2) (Total 7 marks)

Page 10


SAQ11 The table below shows the data obtained when a snail is subjected to a sudden touch stimulus.
Number of times Exposed Time
taken
for
to the Stimulus
tentacles to be fully reextended / s
1
180
2
170
3
165
4
156
5
130
6
110
7
95
8
180
9
63
10
50
(a) (i) Habituation is a type of learning because it brings about a change in behaviour.
(1)
(ii) ignore irrelevant stimuli/prevents unnecessary behaviour;
uninterrupted feeding;
energy not wasted/saved;
returns if stimulus changes;

(2)

(b) stimulate the muscle directly/change the stimulus (following habituation);

muscle will
contract (allow response will still occur); (accept measure pH; build-up of lactate if fatigued) (2)

(c) lack of transmitter substance/depletion of Ach; (accept reference to inhibitory synapse)
depolarisation affected; action potential/impulse not regenerated;
(3)
(d) no impulse to/ no depolarisation of motor end plate/ neuromuscular

junction; muscles

relax/fail to contract;

(2)

(Total 8 marks)
SAQ12. (a) The correlation coefficient is used to determine whether there is a significant
association between two measured variables. Height and weight are two measured variables.

(1)
(4)
(1)

(b) Scatter diagram.
c) There is no significant relationship between the height and weight of boys.
(d) i. Since the calculated coefficient correlation (0.558) is greater than the critical value (0.38)
the null hypothesis must be rejected and we must conclude that there is a significant
relationship between the height and weight of the boys.
(3)
Other variables that can influence height and weight ration must be considered when
making a conclusion. For example the age of the boys can also influence the results and must
be considered or all boys of the same age should be included in this study.
e) degrees of freedom = number of pairs of measurements - 1
= 20 -1
= 19
(1) (Total 11 marks)

SAQ13. a) Scatter diagram.
b) r = 1 – ((1071) / 15(152 – 1))

(4)

= 1 – ((1071) / 15 (224))
= 1 – 0.31875
= 0.68125

(2)

(c) There is no significant relationship between the number of hives in different groves
and the measured yield of oranges in each grove.
(1)

Page 11


(d) Since the calculated correlation coefficient (0.68125) is greater than the critical value (0.361)

(on page 58 in the book), at the 5% significance level, the null hypothesis must be rejected and it
must be concluded that there is a significant relationship between the number of bee hives in
different groves and the measured yield of oranges in each grove.
(3)

(Total 10 marks)
SAQ14

(O – E)2
PHENOTYPE
GRAIN

OF OBSERVED
NUMBER
(O)
Yellow and Smooth
53
Yellow and Wrinkled
20
White and Smooth
17
White and Wrinkled
10
100
∑

EXPECTED
NUMBER
(E)

E

56.25
18.75
18.75
6.25
100

0.188
0.083
0.163
2.25
2.684

(a) There is no significant relationship between the observed number of seeds and the expected
number of seeds.

b) done in the table.
(c) X axis Phenotype;

Y axis number of individuals;
Paired bar graph;

(d) degrees of freedom = (r-1) (c-1)

= (4-1) (2-1)
=3
r – is the number of rows of data for observed and expected values
c – is the number of columns of data for observed and expected values.

(e) Since the calculated 2 value (2.68) is lesser than the critical value (7.81), at p=0.05, the null
hypothesis must be accepted and we must conclude that there is no significant difference
between the observed and expected number of seeds for each phenotype.

SAQ15.
Aperture
A
B
C
D
∑

OBSERVED
NUMBER
(O)
20
24
18
22
84

EXPECTED
NUMBER
(E)

(O – E)2 ÷ E

21
21
21
21
84

0.047
0.428
0.428
0.047
0.95

a) Done in the table.

(4)

b) There is no significant difference between the observed and expected number of exits from
each aperture by the mouse.

(d) degrees of freedom = 3
(e) Since the calculated Chi2 value (0.95) is greater than the critical value (7.82), at 3 degrees of
freedom and 5% significance level, the null hypothesis must be rejected and we must conclude
that there is a significant difference between then observed and expected number of exits from
each aperture.


Page 12


SAQ 16 (a)
Range of light intensities
(class intervals)
9.1 – 9.5
9.6 – 10.0
10.1 – 10.5
10.6 – 11.0
11.1 – 11.5
11.6 – 12.0
12.1 – 12.5

Tally
Deciduous
Coniferous
II
I
IIII
IIII
IIII I
III
II
IIII
I
II
I

Frequency
Deciduous
Coniferous
0
2
1
4
4
6
3
2
4
1
2
0
1
0

(3)
(4)

(b) Bar Graph type histogram.

(c) The data for deciduous woodland shows a wider range and hence a lower variability.

(1)

(d) Since the calculated t value (3.42) is greater than the critical value (2.05), at p = 0.05, the

null hypothesis must be rejected and we must conclude that there is a significant difference
between the mean light intensities reaching the ground in two different woodlands. (3)

(Total 11 marks)
SAQ17. (a) 13.2 – 5.7 = 7.5
7.5/9 = 0.83 0r round off to 0.9
Leaf length / cm
(class intervals)
5.7 – 6.5
6.6 – 7.4
7.5 – 8.3
8.4 – 9.2
9.3 – 10.1
10.2 – 11.0
11.1 – 11.9
12.0 – 12.8
12.9 – 13.7

Tally
Shaded
Sunlit
II
IIII
III
III
IIII IIII IIII
II
III
IIII IIII
I
IIII IIII
II
IIII
I

Frequency
Shaded
Sunlit
0
2
0
5
0
3
3
14
2
3
10
1
9
2
5
0
1
0


Page 13


b. Bar graph type histogram. Same as the one in the SAQ 16. X axis is leaf length in cm. Y
axis frequency.
(4)

c. There is no significant difference between the length of leaves growing in sun lit and
shaded areas.
(1)
d. Degrees of freedom = 60 – 2 = 58.

(1)

e. Calculate the value of t. Show your working.

t = (11.03 – 8.45) / √((1.29/30)+(1.68/30))
= 2.85 /√(0.043+0.056)
= 2.85 / √0.099
= 2.85 / 0.3146
= 9.05

(3)

f. Since the calculated t value (9.05) is greater than the critical value (2.00), at p = 0.05, the

null hypothesis must be rejected and we must conclude that there is a significant difference
between the mean leaf length at site A and B.
(1)

[Total 14 marks]

SAQ18
(a)
This test is used to determine whether the difference between the MEDIAN values for
the two sets of data is significant or not. The two sets of data do not show a normal
distribution.
(2)

There is no significant difference between the number of ‘Giant’ plants grown with super
grow and without super grow.
(1)

(b)

Since the calculated U value (19) is greater than the critical value (11), the null
hypothesis must be Accepted and we must conclude that there is no significant difference
between the number of ‘Giant’ plants grown with super grow and without super grow.
(3)

(c)

(d)
Without super-grow
With super-grow

Median Number of ‘Giant’ plants per metre of
row
13.65
Mean of 13.2 and 14.1 (the two middle
values)
11.8
Middle value

(4)
(e) Bar graph. Two bars with space between the bars and between the first bar and the y axis.

X axis: treatment given to plants. Y axis: Median Number of ‘Giant’ plants per metre of row.
(3) (Total 13 marks)


Page 14


SAQ 19

a)

(3)
(b) There is no significant difference in the songs per hour sung by birds provided with food
and the birds not provided with food.
(1)
(c) i) 6 (from the table on page 75)

(1)

ii) Since the calculated W value (3) is LESS than the critical value (6), the null hypothesis must
be Rejected and we must conclude that there is a significant difference between the duration of
songs sung by birds provided with food and birds not provided with food.
(3)

(Total 7 marks)


Page 15

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