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This is the marking scheme of the book A2 Biology unit six workbook! This will help the students to check the work they ...
Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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Marking schemes for Advanced Level Biology
A2 Biology with Stafford Unit Six: Practical Workbook
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
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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Copyright © Stafford Valentine Redden Unauthorized duplication contravenes applicable laws. Typeset & layouts by : Mohamed Sobir Cover designed by : Mohamed Sobir Printed in Maldives by : Copier Repair Published by : Author publisher All rights reserved. No part of this publication may be Reproduced, stored in a database or retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, 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 Departm ent (Biology) Villa International School, Male’, Republic of Maldives Em ail:
[email protected] Mobile: +960 7765507
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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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 th e h abitat. (3) b. Survey the area to identify any plants that may contain toxic chemicals or ir r itants, 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 centip edes, scor p ions or bees. Carry long forceps and gloves to handle such hazards. Pr otective cloth ing could also h elp. Wearing shoes could help to prevent from cuts on the feet and legs from sharp stones or th or ns. (2)
c. Quadrat X is too small and will enclose only a very small number of p lants , giving a sample size that would be too small. Quadrat Z is too lar ge for th e ar ea and ver y few quadrats could be placed, without much overlap. At least 10 quadrats of size Y could be p laced into the habitat without any over lap and th e quadr at Y is lar ge 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 oth er. Select p airs of r andom numbers and use it as coordinates to place the q uadr at at th ese randomly selected sites within the sampling ar ea. (2)
ii. Grid on next page. Your results may differ. Quadrat numbe r
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) Me an =
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 / 16cm 2 = 0.1125 stars per cm 2 (2)
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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[Total 15 marks] A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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SAQ 2. a. The pe rce ntage cove r of the lichens can be measured to provide a re liable quantitative measure of abundance. Place a quadrat of 15 cm x 15cm , containing 25 smalle r 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 ke pt
constant or monitored. These factors may be Height from ground, Diameter of tree, Spe cie s of tree, Exposure to light, wind or te mperature.
(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 e nz yme 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, le ading
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 subje ctive measurements. Some squares are only partly covered and results in inaccurate measure me nt 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)
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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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 o C , the standard deviation increases, indicating that the data from repeated measurements is furthe r 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. Be yond 5 0 o C there is a decrease in the hatch rate of brine shrimps. (2)
[Total 11 marks] SAQ 4 a) Te mperature/ C o
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 re duced 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 re ason that the y do not ge rminate. (2)
c) The seeds may be germinated in Petri dishes with the cover just placed on to it without
se aling 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 se e dling 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] A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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SAQ 5 (a) Restriction Endonucle ase s
(1)
(b) The analysis of the DNA will require DNA Fingerprinting, which requires a relative ly
large quantity of DNA. The sample obtained from the window sill is too small and the PC R is used to increase the quantity of DNA very rapidly . (1)
(c) D NA Nucleotides and DNA Polymerase enz yme
(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 sequence s that e nable it to bind with human DNA fragments only, then only human DNA fragments will be re plicated. (2)
(d) D NA is double stranded and the 3’ end of each strand has a different sequence of base s. 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 D NA. (1)
e) The amplification factor = 2n, where n is the number of replication cycles =
2 12
= 4096 DNA fragments
(2) [Total 12 marks]
SAQ 6. a) Obtain DNA samples from many animals of the same species and carry o ut D NA
profiles for each. Select animals which have greatest differences in the DNA banding patterns. (1) b) The selective breeding of genetically diverse organisms increases heteroz ygosity 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 multiplie d by 0.3 4nm will give an accurate calculation of the length of a D NA fragment.
(2)
ii)
(2) A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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iii) D istance moved by the fragment depends on the length of the fragment. Short fragme nts 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 bacte ria 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 z one.
(2)
c) The area of the clear z one around Ampicillin is the largest.
(1)
d) The concentration of each antibiotic must be the same in all the discs to ensure that the
re sults are reliable. Otherwise the rate of diffusion into the nutrient agar will diffe r and influence the reliability of the re sults. (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 os motic shock. (3)
(Total 9 marks)
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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Marking scheme for A2 Biology with Stafford, Unit Six Practical Workbook. Book available at
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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 dm 3 of oxygen (8.5+8+8)÷3 mm on Y axis = 0.82 dm 3
(2)
Note: it is a good practice to measure the tidal volume three times (as shown on the graph) and then find the me an tidal volume .
b) breathing rate = (numbe r of breaths ÷ time ) x 60 = (5.5 ÷ 29) x 60 = 11.4 breaths per minute
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
(2)
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c) at rest
Minute ventilation rate = bre athing rate x tidal volume = 11.4 x 0.82 = 9.3 dm 3 min - 1 D uring exercise Minute ventilation rate = bre athing rate x tidal volume = 3 1.6 x 1.5 = 47.4 dm 3 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 exe rcise . D uring exercise, the tissues (specially muscles) have an increased demand for oxyge n for re spiration 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 de te cte d by chemoreceptors in the carotid body and medulla. Respiratory centre in the me dulla stimulates the intercostals muscles and diaphragm to increase the breathing rate and de pth (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
e nsures that any change in volume of gases in the tube is only due to oxygen being used up for re spiration. (1)
(b) The 1 cm3 syringe is used to reset the apparatus by readjusting the liquid leve l in the manometer U tube. This allows multiple readings to be taken without having to reinstall the e ntire apparatus. (1)
(c) Setup the apparatus as shown in the diagram and allow the seeds to respire. The se e ds use
up oxygen the liquid in the manometer U tube will begin to fall. Note the distance move d by the liquid in the U tube. The rate of oxygen uptake by the seeds is found by the following e quation. rate of oxyge n uptake = (π r 2 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 take n for the liquid to move.
d) The antiseptic will prevent the growth of bacteria on the seed. If the bacteria are allowe d to grow along with the seeds then the bacteria will respire and the readings will be unre liable .
(2) (Total 7 marks) A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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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 le arning because it brings about a change in behaviour . (1) (ii) ignore irre le vant stimuli/pre vents unnecessary behaviour; uninterrupte d fe e ding; e nergy not waste d/save d; re turns if stimulus change s;
(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 affe cte d; action potential/impulse not regenerated; (3) (d) no impulse to/ no depolarisation of motor end plate/ neuromuscular junction; muscle s re lax/fail to contract; (2)
(Total 8 marks) SAQ12. (a) The correlation coefficient is used to determine whether there is a signi ficant association between two measured variables. Height and weight are two measured variable s.
(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 (0.3 8) the null hypothe sis must be re je cte d and we must conclude that there is a significant
re lationship between the he ight and weight of the boys. (3) Other variables that can influence height and weight ration must be considere d whe n 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(15 2 – 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 grove s and the measured yie ld of orange s in each grove . (1) A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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(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 hive s 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 expecte d number of se e ds.
b) done in the table . (c) X axis Phenotype;
Y axis number of individuals; Paired bar graph;
(d) degrees of fre e dom = (r-1) (c -1)
= (4-1) (2-1) = 3 r – is the numbe r of rows of data for observed and expected values c – is the numbe r 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 diffe r e nce be tween the obse rve d and e xpe cted 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) D one in the table.
(4)
b) There is no significant difference between the observed and expected number of exits from e ach aperture by the mouse .
(d) degrees of fre e dom = 3 (e) Since the calculated Chi2 value (0.95) is greater than the critical value (7.82), at 3 degre e s 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 e ach aperture .
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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SAQ 16 (a) Range of light intensities (class inte rvals) 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 D e ciduous C oniferous II I IIII IIII IIII I III II IIII I II I
Frequency D eciduous C oniferous 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 diffe re nce be tween the mean light inte nsitie s 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 0.85 0r 9.0 Le af length / cm (class inte rvals) 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
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(4)
a. Suggest a suitable null hypothesis for this experiment.
(1)
...................................................................................................................................................................... ............................................................................................................................................................ b. State the number of degrees of freedom for the data.
(1)
............................................................................................................ ...................................................... c. The formula used for the t test is given below.
Where,
Calculate the value of t. Show your working.
(3)
d. A statistical table showed that the critical value at the 5 % level was 2.00. What does this tell you about the difference between the mean leaf lengths from sites A and B? (1) ............................................................................................................................. ..................................... ............................................................................................................................. ..................................... .................................................................................................................................................................. ............................................................................................................................. ..................................... ............................................................................................................................. ..................................... .................................................................................................................................................................. ............................................................................................................................. ..................................... .................................................................................................................................................................. .................................................................................................................................................................. [Total 14 marks]
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SAQ18 A new fertilizer [‘super-grow’] has been developed. During testing, super-grow was applied to 7 plots of a wheat field. A further 8 plots were left as a control. The numbers of ‘giant’ wheat plants per metre of row in each plot were recorded. The results are shown in the table below. Without super-grow With super-grow
Number of ‘Giant’ plants per metre of row 6.2 8.5 12.3 13.2 14.1 15.8 19.4 21.0 3.9 8.4 9.1 11.8 12.4 14.9 18.2
(a)
Suggest why a Mann-Whitney U test is appropriate to determine whether super-grow increases the number of ‘Giant’ wheat plants. (2) .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ............................................................................................................................................................... (b) State a suitable null hypothesis for this investigation. (1) .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. (c) The U values for the two sets of data were calculated and found to be: U1 = 37 U2 = 19 A statistical table showed that the critical value is 11. What can you conclude from this investigation? Explain your answer. (3) .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ..................................................................................................................................................................
(d) Prepare a table of the raw data and organise it in such a way that the median number of ‘Giant’ plants in each field can be identified. (4)
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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(e) Use the data in your table to present the information in a suitable graphical form. (3)
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(Total 13 marks)
SAQ 19 A study was carried out to see if male birds in territories provided with extra food allocate more
of their time to singing. The results of the study are shown in the table below.
a) Use the data in the table to present the data in a suitable graphical form.
(3)
(b) State the null hypothesis for this investigation.
(1)
(c) The student decided to apply the wilcoxon signed rank test to the paired data. This statistical test uses the differences between paired samples and is equivalent to a t-test for paired data. The calculations produced W values of 3 and 33. i) Use the table of critical values in the notes for W test and determine the critical value for this set of data. (1) ……………………………………………………………………………………………………………. ii) Use the information to draw conclusions. ........................................................................................................................................................ ........................................................................................................................................................ ............................................................................................................................................ (3) (Total 7 marks)
A2 Biology with Stafford. Unit Six: Practical Workbook answers / Unit Six paper BIO8
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