Essential Maths Skills. for AS/A-level. Biology Answers. Dan Foulder

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1 Essential Maths Skills for AS/A-level Biology Answers Dan Foulder

2 1 Arithmetic and numerical computation Appropriate units in calculations Guided questions (p.8) = 15 g As there are 1 mm 3 in 1 cm 3, divide the volume in mm 3 by 1: 65 1 =.65 cm 3 3 The unit of the rate should be volume per time, so cm 3 min 1 would be an appropriate unit. (Any other answer with sensible volume and time components is acceptable.) Practice questions (p.8) 4 1 dm = 1 cm, so 1 dm 3 = 1 cm 1 cm 1 cm = 1 cm 3 Therefore, to convert 3.6 dm 3 to cm 3, multiply by 1: = 36 cm m 3 is a large volume, which equals 1 million cm 3. It is too large to use conveniently in most biological situations. 6 It would be difficult to use a measuring cylinder to measure accurately such a small volume as 1 mm 3, which is equivalent to only.1 cm 3. 7 a To describe the amount of energy flow per unit of woodland or grassland area each year, kj m year 1 would be an appropriate unit (as would any other unit made up of sensible energy, area and time components). b In an aquatic ecosystem, it would be necessary to measure the amount of energy flow in a volume of water rather than over an area of land, so the area component of the unit in a would need to be changed to a volume component. Expressions in decimal and standard form Guided questions (p.9) = 1 = 1gs 1 Step 1: after the first dilution, the volume has changed from 1 cm 3 to 1 cm 3, so the concentration becomes one-tenth of the original. Dilution factor = 1 =.1 = Step : after the second dilution, the concentration becomes one-hundredth of the 1 original, so the dilution factor is 1 =.1 = 1 1. Step 3: after the third dilution, the dilution factor relative to the original sample is = 1 =

3 Practice questions (p.1) 3.3 mm = mm 4 1 m = 1 mm, so 5 m = 5 mm = mm 5 Using the standard form expressions from questions 3 and 4, = = ( 5) 9 So the whale is approximately times larger than the ribosome = = bacteria in 1 cm 3 of water. 7 There are four dilutions, and after each dilution the concentration becomes of what it was previously, so the final dilution factor is 1 = 1 4. After the final dilution, 1 cm 3 contained 15 bacteria. Therefore, in the original sample, 1 cm 3 contained = = bacteria. Ratios, fractions and percentages Guided questions (p.14) 1 Step 1: the surface area to volume ratio is 4 : 8. = 1 1 Step : the highest common factor of 4 and 8 is 8, so divide both numbers by 8 to get 4 8 = 3 and 8 8 = 1. Therefore 4 : 8 = 3 : 1. This is the simplest form of the surface area to volume ratio of the cube. mass of product The percentage yield is mass of substrate = 45 1% = 33.6% 134 Practice questions (p.14) mass of product 3 The percentage yield is mass of substrate = 6 1% = 35.3% 17 4 The surface area to volume ratio is 16 : 1 = 4 : 3. 5 To see which ratio is largest, convert each ratio to the form number : 1 by dividing both numbers in the ratio by the second number. Thus, A : 3 =.667 : 1 B 15 : 8 = : 1 C 7 : = 3.5 : 1 Then compare them by looking at the first number in each. So ratio C is the largest. 6 The percentage of adenine must equal the percentage of thymine, so 4% of the bases are adenine. The percentages of all four bases must add up to 1: % adenine + % thymine + % cytosine + % guanine = 1

4 That is, % cytosine + % guanine = 1 Rearranging gives % cytosine + % guanine = = 5 But the percentages of cytosine and guanine must be equal, so each accounts for 5% = 6% Therefore 6% of the bases are guanine. 7 Let T denote the dominant (all) allele and t the recessive allele. Then the genotype of the heterozygous tall pea plant is Tt, and a short pea plant has genotype tt. The gametes are T, t from the tall parent and t, t from the short parent. Table A.1 Genetic cross T t t Tt tt t Tt tt The offspring genotype ratio is Tt : tt = : = 1 : 1 So the offspring phenotype ratio is 1 tall : 1 short. 8 Let the dominant allele for larger dorsal fin be D, and let the dominant allele for yellow scales be Y. Then heterozygotes with yellow scales and larger dorsal fin have genotype Dd Yy. So both parents are Dd Yy The gametes can be DY, Dy, dy or dy from each parent. Table A. Genetic cross DY Dy dy dy DY DD YY DD Yy Dd YY Dd Yy Dy DD Yy DD yy Dd Yy Dd yy dy Dd YY Dd Yy dd YY dd Yy dy Dd Yy Dd yy dd Yy dd yy Hence the offspring are expected to be distributed as follows: 9 with larger dorsal fin and yellow scales (DD YY, DD Yy, Dd YY, Dd Yy) 3 with larger dorsal fin and blue scales (DD yy, Dd yy) 3 with smaller dorsal fin and yellow scales (dd YY, dd Yy) 1 with smaller dorsal fin and blue scales (dd yy) Therefore the phenotype ratio is 9 : 3 : 3 : 1. Estimating results Guided questions (p.15) 1 Step 1: 478 g 5 g to the nearest hundred. 18 s s to the nearest ten. Step : estimate for the rate of reaction is 5 = 5 g s 1. 3

5 Step 1: 3.7 mm 4 mm to the nearest whole number. Step : estimate of volume is 4 mm 4 mm 4 mm = 64 mm 3. Practice questions (p.16) 3 Rounding both numbers to the nearest million (1 6 ): = = bacteria 4 Rounding both numbers to the nearest ten: 43 1 = 43 flowers per km 5 Rounding the probability.53 to one decimal place: 17.5 = 85 rabbits are expected to have white coats. 6 Rounding 4.54 cm up to 5 cm, a rough estimate of the volume of the cube would be 5 cm 5 cm 5 cm = 5 3 cm = 15 cm 3. 7 One of the numbers has been rounded incorrectly: 16 should be rounded upwards to, not downwards to 1. Power, exponential and logarithmic functions Guided questions (p.18) 1 From calculator: 15 3 = 3375 cm 3 [H + ] = , so ph = log[ ] = 1.5 Practice questions (p.18) 3 From calculator: 365 = m 4 From calculator: 3 9 = zooplankton 5 In this case r =.5, t = and P = 1. So after hours the bacterial population is.5 A = 1 e = e 1 = 6 6 [H + ] =.1, so ph = log[.1] = 1 7 a Since n = 3 for a human, n must represent the haploid number, i.e. the number of homologous pairs of chromosomes in an organism. b The diploid number is 8, so n = 8 = 4 for a fruit fly. Therefore the number of different combinations of chromosomes is 4 = 16. c For each gamete, the number of different combinations of chromosomes due to independent assortment is n. So, following random fertilisation of two gametes, the number of possible combinations of chromosomes is n n (which can also be written as 4 n or n ). 4

6 Handling data Significant figures Guided questions (p.) 1 The first significant figure is 1, the second significant figure is, and the digit 6 tells us to round up. So.168 =.11 ( s.f.) The trailing zero is not significant, so the number has two significant figures (8 and 9). Practice questions (p.) cm 3 = 19 cm 3 ( s.f.) M =.98 M (3 s.f.) has 4 significant figures (look only at the number multiplying the power of 1). 6 The length measurement was given to 3 s.f., so the volume should be rounded to 3 s.f., giving 1 8 cm 3. 7 To find the volume of juice produced per gram of fruit, divide the volume of juice by the mass of the fruit: 54 cm 3.3 g =.6956 cm 3 g 1 The mass was measured to 5 s.f. and the volume was measured to s.f. Therefore the answer should be rounded to s.f. (the least accurate measurement):.6956 =.7 cm 3 g 1 ( s.f.) Arithmetic means Guided questions (p.) cm 7 days = 19.9 cm per day = = 7.5 beats per minute 1 Practice questions (p.3) = 3.4cm 5 4 a Total rainfall = = 16 mm Mean rainfall over 1 days = 16 1 = 1.6 mm b Mean rainfall over days = 16 =.8 mm 5

7 5 The means (obtained by dividing each total by 3) are rounded to the nearest whole number. Table A.3 External solute potential (kpa) Total change in mass (%) Mean change in mass (%) The means (obtained by dividing each total by 3) are rounded to the nearest whole number. Table A.4 Hydrogen peroxide concentration (% stock solution) Total time (seconds) Mean time (seconds) a Table A.5 Temperature ( C) Total transmission (%) Mean transmission (%) b The mean value at C is lower than the mean value at 3 C, which goes against the overall trend of the higher the temperature, the lower the transmission. c Measurement number 5 at C is anomalously low compared to the other results. To remove its adverse effect on the mean, this outlier could be discarded before calculating the mean value for C (in which case the total should be divided by 4 instead of 5). 6

8 Interpreting tables and diagrams Guided questions (p.6) 1 The mean values for the different age groups are as follows: Table A.6 Age (years) Mean vital capacity (litres) a Looking for either the highest value overall or the highest mean value, the 39 age group has the highest vital capacity. b As age increases, the vital capacity increases, reaching a peak for the 39 age group. After this the vital capacity decreases with age. c Calculating the difference between the highest and lowest values in each group, we get the following: Table A.7 Age (years) Range of vital capacities (litres) So the 4 59 age group shows the greatest variation of vital capacities. The time between two successive R peaks is approximately.96. =.76 seconds 1 In one second there are.76 cycles, so in one minute there are cycles. 1 Therefore the heart rate is 6 = 78.9 beats per minute..76 The time between atrial and ventricular contractions is the time between the P and R points: P R interval =..5 =.15 seconds Practice questions (p.7) 3 The means are obtained by adding up the data values at each ph and dividing by 3. They are rounded to the nearest second. 7

9 Table A.8 Time taken for reaction to occur (seconds) ph 1 3 Mean Here are some mistakes the student has made: The units of the independent variable (distance from lamp) are written in the body of the table rather than in the heading. The mean is given in a separate column, not under the dependent variable, so it doesn t have a proper heading or units. The entries of the table don t have a consistent number of decimal places. 5 The peaks are not evenly spaced, indicating an irregular heart beat. 6 a Stage 1 b Stage 3 c Stage 5 d If after treatment the patient s GFR moved up to stage, which corresponds to a range of 6 89, then before treatment the GFR must have been 4 ml/min/1.73 m lower, i.e. in the range 49. Therefore the lowest GFR that the patient could have before treatment is ml/min/1.73 m, and this corresponds to stage 4. Simple probability Guided questions (p.9) 1 Step 1: the parents are of genotypes RR and RW. The possible gametes are R, R from the red parent and R, W from the pink parent. Step : from the genetic cross (Table.15), we see that: offspring genotypes: 5% RR, 5% RW offspring phenotypes: 5% red, 5% pink, % white Step 3: of the 3 offspring plants, we would expect 16 to have red flowers and 16 to have pink flowers (and none with white flowers). Step 1: = 16 Step : = 8, which is the expected number that each 1 in the ratio 9 : 3 : 3 : 1 represents. Step 3: Table A.9 Phenotype Expected number Orange, white-tipped 9 8 = 7 Orange, black-tipped 3 8 = 4 Yellow, white-tipped 3 8 = 4 Yellow, black-tipped 1 8 = 8 8

10 Practice questions (p.31) 3 5% =.5. The expected number of boys is 6.5 = 3. 4 The probability of a seed being orange is 1.54 =.46. So in a sample of 763 seeds, the expected number of orange seeds is = 351 (rounded to the nearest whole number). 5 a = = 3, which is the expected number that each 1 in the ratio 1 : : 1 represents. Therefore, among the offspring, we expect the following number of each genotype: Table A.1 Genotype Expected number Homozygous dominant 3 1 = 3 Heterozygous 3 = 6 Homozygous recessive 3 1 = 3 b The genetic cross gives the expected proportion, and hence number, of each outcome, but the actual distribution of offspring could deviate from this prediction due to chance = = 4, which is the expected number that each 1 in the ratio 9 : 3 : 3 : 1 represents. Table A.11 Phenotype Expected number Yellow and round 9 4 = 36 Green and round 3 4 = 1 Yellow and wrinkled 3 4 = 1 Green and wrinkled 1 4 = , because previous offspring do not the affect the probability of the next offspring having long fur. 8 No. Even though the table shows the rate of a heart attack being 1.9 per thousand in the UK, the chance of actually experiencing a heart attack is affected by a number of risk factors, including age, weight, genetics, lifestyle etc. Depending on their particular circumstances, the probability of a random person having a heart attack may be greater or less than Principles of sampling Guided questions (p.34) 1 a Step 1: of the 1 quadrats used, 4 contained non-zero numbers of U. dioica. Step : so the species frequency of U. dioica is 4 1% = 4% 1 9

11 b Step 1: the total number of U. dioica found in the different quadrats was = 7 Step : the total area of quadrats used was 1.5 m =.5 m Step 3: therefore the species density of U. dioica was 7.5 =.8 per square metre Number in first sample = 6 Number in second sample = 3 Number marked in second sample = 1 population size = = total number in first sample total number in second sample = 37 Practice questions (p.35) number marked in second sample 3 a Fraxinus excelsior occurred in 4 out of the 8 quadrats So species frequency is 4 8 1% = 5%. b Total number of Fraxinus excelsior = = 17 Total area of quadrats = 8 1 m = 8 m Species density = 17 8 =.1 F. excelsior per square metre. 4 a percentage cover = number of squares covered 13 1 = 1 = 5 total number of squares 5 So the percentage cover of H. helix was 5%. number of squares covered b Rearranging the formula percentage cover = 1 gives total number of squares number of squares covered = percentage cover 1 total number of squares Substitute in the given values for R. magellanicus to get number of squares covered by R. magellanicus = = 9 5 a Substituting the given values into the Lincoln index formula: total number in first sample total number in second sample population size = number marked in second sample = = = b When using mark and recapture together with the Lincoln index to estimate population size, we are assuming that no individuals die, are born, immigrate into the population or emigrate out of the population. If any of these events occur during the interval between the two samples, the Lincoln index will no longer provide an accurate estimate of the population. 1

12 6 The total number of individual organisms in the area is N = = 157 Make a table to calculate the values needed for Simpson s diversity index. Table A.1 Species Number (n) n N n N Titanus giganteus Theraphosa blondi Damon diadema To calculate n, add together the numbers in the last column: N =.435 Then D = 1 n N = =.565 Mean, median and mode Guided questions (p.37) 1 Step 1: in ascending order, the data values are 19 cm, cm, 3 cm, 4 cm, 35 cm, 45 cm Step : there are 6 data points, so the middle two are the 3rd and 4th. These are 3 cm and 4 cm. So the median is (3 + 4) = 3.5 cm Using the tally chart (Table.3), we get a frequency table for the data set. Table A.13 Mass of seed (g) Frequency The data value 3 g occurs 15 times and is the most common value. Therefore the mode is 3 g. 11

13 Practice questions (p.37) 3 In ascending order the data values are 4.1, 4.7, 4.9, 5., 5.4, 5.9, 6.1 There are 7 values. The middle one is the 4th value, which is 5.. Therefore the median red blood cell count is 5. million cells per microlitre. 4 In ascending order the data values are 6, 75, 76, 8, 89, 99 There are 6 data points, so the middle two are the 3rd and 4th. These are 76 and 8. The median is (76 + 8) = 78 beats per minute. 5 The largest number of Crematogaster were caught, so Crematogaster is the modal organism caught. 6 There is no mode, as none of the values are repeated. 7 a AB occurs most often, so the modal blood group is AB. b The data is non-numerical, so using the arithmetic mean does not make sense. Nor can the data points be put in a linear sequence to calculate a median value. Scatter diagrams Guided questions (p.4) 1 As you move from left to right in the scatter diagram, the data points tend to get higher. Therefore this data shows a positive correlation between limpet shell height and distance from the tide line. As the wavelength of light increases, the plant height tends to decrease. Therefore this data shows a negative correlation between wavelength of light and plant height. Practice questions (p.41) 3 There is a positive correlation between the number of cigarettes smoked a day and the incidence of lung cancer. 4 There is no obvious pattern in the data. So there appears to be no correlation between the drug dose and the number of allergic reactions. 5 There is a positive correlation between partial pressure of oxygen and oxygen saturation of haemoglobin up to about 7 mm Hg partial pressure. However, for partial pressures greater than 7 mm Hg, the curve plateaus and so there is no longer a correlation. 6 The scatter diagram indicates a positive correlation between the number of cars on the roads and the incidence of asthma. However, this does not show conclusively that the rise in the number of cars is causing the increase in asthma cases, as there are many other factors that could affect the number of asthma cases. 1

14 Order of magnitude Guided questions (p.45) 1 Step 1: converting both lengths to µm: 1 cm = µm Step : substituting the two values into the magnification equation: image size magnification = object size = 5 = 4 So the magnification is 4. Step 1: rearranging the magnification equation: image size magnification = object size image size = magnification object size Step : substituting the two given values into the rearranged equation: image size = 4.4 cm = 16 cm Practice questions (p.45) 3 Converting both lengths to nm: 4 cm = nm Substituting the two values into the magnification equation: image size magnification = object size = = So the magnification is times. 4 Rearranging the magnification equation gives image size object size = magnification Substituting the given values into the rearranged equation: object size = 9 cm 4.5 = cm 5 Rearranging the magnification equation gives image size = object size magnification = 7 µm 19 = 13 3 µm = 1.33 cm 13

15 6 Rearranging the magnification equation gives image size = object size magnification = 3 nm = nm = 3 cm 7 A magnification less than 1 means that the image is smaller than the original object. Statistical tests Practice questions (p.53) 1 We use the chi-squared test. Null hypothesis: there is no significant difference between the observed and expected results. Alternative hypothesis: there is a significant difference between the observed and expected results. There are = 6 offspring in total. Splitting 6 in a 1 : : 1 ratio gives 65 red, 13 purple and 65 blue offspring. These are the expected results. Make a table to calculate the chi-squared value. Table A.14 o e o e (o e) e (o e) Red Purple Blue (o e) χ = =.1 e Degrees of freedom = 3 1 = From Table.7, the critical value at.5 significance level is Because.1 < 5.99, accept the null hypothesis and reject the alternative hypothesis. So we conclude that there is no significant difference between the observed and expected results. Null hypothesis: there is no significant difference between the numbers of zooplankton in the high-salinity sample and in the low-salinity sample. Alternative hypothesis: there is a significant difference between the numbers of zooplankton in the high-salinity sample and in the low-salinity sample. To apply the t-test, we need the mean (x) and variance (s ) of each data set. mean of high salinity sample = =

16 mean of low salinity sample = = 56.5 We use tables to do the calculations of the variances. Table A.15 High-salinity sample Number of zooplankton in 1 cm 3 of high-salinity water (x) ( x x) =887.5 Deviation from mean (x x) Squared deviation from mean (x x) Variance of high salinity sample = Table A.16 Low-salinity sample Number of zooplankton in 1 cm 3 of low-salinity water (x) ( x x) =44 = Deviation from mean (x x) Squared deviation from mean (x x) Variance of low salinity sample = 44 8 = 53 t = = = x x s 1 1 n 1 s + n = 6. 15

17 Degrees of freedom = (8 1) + (8 1) = 14 From Table.3, the critical value at.5 significance level is.145. Because 6. >.145, reject the null hypothesis and accept the alternative hypothesis. So we conclude that there is a significant difference between the numbers of zooplankton in the high-salinity water and the low-salinity water. Measures of dispersion Guided questions (p.56) 1 Step 1: highest value is 191 mm. Lowest value is 17 mm. Step : so the range is = 64 mm. Step 1: Table A.17 Mean values at the different times Time (minutes) Mean x Step : Table A.18 Deviations from the mean Deviation from mean (x x) Time (minutes) Step 3: Table A.19 Squared deviations from the mean Squared deviation from mean (x x) Time (minutes)

18 Step 4: Table A. Variances Time (minutes) Squared deviation from mean (x x) Σ(x x) Variance s Step 5: Table A.1 Standard deviations Time (minutes) Standard deviation s The standard deviations are generally quite small, except for the one at 5 minutes. This is due to the anomalously large value of the third measurement (). Practice questions (p.57) 3 Largest mass = 1.5 kg. Smallest mass =.6 kg. Range = =.9 kg. 4 Greatest height = 11 m. Smallest height = 5 m. Range = 11 5 = 49 m. This is a large range, because one of the tree heights (5 m) is anomalously low. 5 The mean number of rabbits is x = Table A. =51 Year Number of rabbits Deviation from mean (x x) Squared deviation from mean (x x)

19 Summing the values in the last row: (x x) = 133 s Σ x x = ( ) = 133 = 19 n 7 So the standard deviation is s = 19 = Uncertainties in measurements Guided questions (p.59) 1 apparatus margin of error apparatus percentage error = 1% quantity measured =.5 1% 1.1 =.495% observed value expected value percentage error = 1% expected value = % = 4 1% = 1% 19 So the percentage error is 1%. Practice questions (p.59) 3 4 observed value expected value percentage error = 1% expected value = 1% 54 = 6 1% = 11% 54 Therefore the percentage error is 11%. apparatus margin of error apparatus percentage error = 1% quantity measured =.1 1% = 1.4% 7. 5 The volume measured is larger, so since we are dividing by it, the percentage error will turn out to be smaller. 6 apparatus percentage error = apparatus margin of error quantity measured 1% =.5 1% = 1.7%.3 18

20 This is a large percentage error, which indicates that the syringe is not suitable for use in this investigation. A piece of equipment that can measure a smaller volume more accurately should be used. 7 It is likely that experimental error will lead to a lower actual yield than the theoretical yield. This means that the observed value minus the expected value will be negative and so the percentage error calculation will give a negative number. 19

21 3 Algebra Algebraic equations Guided questions (p.6) 1 Replacing the quantities in the right-hand side of the equation by their given values: CO produced RQ = O consumed = 19 7 =.7 Step 1: Table A.3 Species Number (n) n N Daisy 34 Bramble 9 Buttercup =.5 =.175 =.135 Total 68 = N Step : n = N =.4 Step 3: Therefore D = 1 n N = 1.4 =.6

22 Practice questions (p.63) 3 Cardiac output = stroke volume heart rate 3 = 65cm per beat 8 beats per min = 5 cm 3 min 1 4 Water potential ψ = ψ s + ψ p = + ( ) = kpa 5 number of cells undergoing mitosis mitotic index = = 14 total number of cells in sample 35 =.4 6 proportion of polymorphic gene loci = number of polymorphic gene loci total number of loci = 4 48 = N = = 9 Species Number of individuals n(n 1) Deer (19 1) = 34 Rabbits (65 1) = 416 Shrews 8 8 (8 1) = 56 n(n 1) = = 4558 Therefore d = N( N 1) nn ( 1) 9 (9 1) = 4558 = 1.84 Changing the subject of an equation Guided questions (p.65) image size 1 Step 1: magnification = object size Step : image size = magnification object size Step 3: image size = magnification object size = μm = μm = μm This can also be expressed as 1 cm. 1

23 Step 1: rearranging the original equation gives photosynthetic efficiency energy falling on plant = energy incorporated into products of photosynthesis Step : energy incorporated into products of photosynthesis energy falling on plant = photosynthetic efficiency Step 3: substituting the numbers gives energyfallingonplant =. 1 kj m yr = kj m yr 1 3 area = πr Dividing through by π gives area = r π Then take the square root of both sides to get r on its own: area = r π Therefore r = area π = 78 π = 4.98 cm Practice questions (p.66) 4 Adding R to both sides of the equation net primary productivity (NPP) = gross primary productivity (GPP) respiration (R) gives NPP + R = GPP. Then, subtracting NPP from both sides leads to R = GPP NPP Substituting the numbers gives R = = kj m yr 1 5 Dividing both sides of the equation by heart rate gives cardiac output stroke volume = heart rate Therefore, for this person, stroke volume = 43 7 = 61.4 cm3 per beat 6 Multiplying both sides of the equation by distance moved by solvent front gives distance moved by substance = Rf distance moved by solvent front =.43 7 cm = 3.1 cm

24 7 Multiplying both sides of the equation by time and then dividing through by rate of reaction gives time = = concentration of product rate of reaction 6.5 cm dm cm dm s = 9.3 s So the time taken is 9.3 seconds. 8 In this example q =.5, so p = 1.5 =.75 The frequency of heterozygotes is the term pq in the Hardy Weinberg equation. Rearranging the equation to make pq the subject, we get pq = 1 p q = =.375 Alternatively, we can directly calculate pq =.75.5 =.375 Therefore the frequency of heterozygotes in the population is.375. Logarithms Guided questions (p.68) 1 The graph is declining (falling from left to right), so as the ph increases, the concentration of hydrogen ions decreases. Because the graph is a straight line, this decrease is exponential. The graph is flat from 34 to 38 hours, indicating no growth, so the stationary phase begins at 34 hours. Practice questions (p.69) 3 Bacterial populations can grow very quickly. A logarithmic scale allows a small initial population to be clearly represented on the same graph as later populations which may be several orders of magnitude larger. 4 a The graph peaks at around hours, when the population is between and bacteria. b Growth was fastest between 1 and 13 hours, as over this period the graph was steepest. 7 The graph with the linear scale shows a period of very fast growth before plateauing for a while and then declining. So graph C best represents this behaviour on a logarithmic scale, because it consists of a rising straight line (exponential growth) followed by a plateau and then a falling shape. 3

25 4 Graphs Bar charts, histograms and line graphs Guided questions (p.74) 1 Step 1: a frequency table for the data is as follows. Table A.4 Height (cm) Frequency Steps and 3: this gives the following histogram. Frequency Height (cm) Figure A.1 Number of penguins Figure A Year 4

26 Practice questions (p.75) 3 As the blood groups are distinct, non-overlapping categories, a bar chart is suitable. Number of people A B AB O Blood group Figure A.3 4 It would be clearest to represent the data on a line graph of the mean time against temperature. Mean time taken for methylene blue to decolourise (s) Temperature ( C) Figure A.4 5 A histogram is an appropriate graph to use for this data. Frequency Dry mass (g) Figure A.5 5

27 6 The horizontal axis does not have a linear scale (i.e. the numbers marked on the axis don t increase by the same amount from each one to the next) or an origin. The vertical axis does not have a unit. y = mx + c Guided questions (p.79) 1 Step 1: in this equation, m =.5 and c = 9. As m is negative, the graph is a straight line with negative gradient. Step : At x =, y = = 9 At x = 1, y = = 4 Step 3: Figure A.6 Step 1: y = 3x + 1 At x =, y = = 1 At x =, y = = 7 Step : ensure that the x-axis runs from to and that the y-axis goes up to at least Figure A.7 6

28 Practice questions (p.8) y = x + 3 At x =, y = + 3 = 3 At x = 5, y = = Figure A y = x + 9 At x = 5, y = = 14 At x = 1, y = = Figure A.9 5 y = x + 15 At x =, y = + 15 = 15 At x = 1, y = = Figure A.1 7

29 6 C represents y = x + 3: it has a higher y-intercept and a less steep gradient. A represents y = 3x + 1: it has the lowest y-intercept of the three. B represents y = 4x + 3: it has the same y-intercept as C but a steeper gradient. Determining the intercept of a graph Guided questions (p.84) 1 First extend the line to the x-axis: Change in mass of onion (%) Concentration of NaCl (M) Figure A.11 The extended graph has x-intercept at 1.5, so there will be no change in mass at a sodium chloride concentration of 1.5 M. Steps 1 and : Rate of oxygen production (cm 3 min 1 ) Light intensity (lux) Figure A.1 Step 3: the graph has x-intercept at approximately 39, so the compensation point is 39 lux. 8

30 Practice questions (p.85) 3 Join successive data points with straight lines: Rate of oxygen production (cm 3 min 1 ) Light intensity (lux) Figure A.13 The compensation point is the point at which no oxygen is produced, i.e. y =, so it corresponds to the graph s x-axis intercept. This is at approximately 83, so the compensation point is 83 lux. 4 No, the compensation point cannot be determined from Figure 4.. The origin of the y-axis is 1 rather than, so it isn t possible to identify the x-intercept where y, the rate of carbon dioxide uptake, is zero, which is the compensation point. 5 No, extrapolating backwards to determine an axis intercept is not always accurate. By extrapolating, you are assuming that the trend in the data will continue to where y =. However, the shape of the graph could change between the last recorded value and zero. 6 Solute potential (kpa) Change in mass of onion (%) Figure A

31 The water potential inside the cell equals the water potential outside the cell when the change in mass is zero, i.e. at the x-intercept. From the graph, this occurs at kpa external solute potential. Calculating the rate of change for a linear relationship Guided questions (p.89) 1 Step 1: Tapeworm length (cm) Time (hours) Figure A.15 Step : change in y = Step 3: change in x = 11 change in y gradient of line = change in x = = =.5 The unit of the rate of change is the length unit (cm) divided by the time unit (hour), i.e. cm hour 1. Therefore the rate of growth of the tapeworm is.5 cm hour 1. 3

32 Step 1: Maximum flight speed (km h 1 ) Mass (kg) Figure A.16 Step : change in y = Step 3: change in x = change in y gradient of line = change in x = =.5.5 = 5 The unit of the rate of change is the speed unit (km hr 1 ) divided by the mass unit (kg), i.e. km hr 1 kg 1. Therefore the rate of decrease of maximum flight speed with body mass is 5 km hr 1 kg 1. Practice questions (p.9) 3 change in y gradient of line = change in x 4 1 = = 8.1 =

33 The unit of the rate of change is the unit of the rate of reaction (g min 1 ) divided by the unit of concentration (mol dm 3 ), i.e. g dm 3 mol 1 min 1. 4 Therefore the rate of change is 13.3 g dm 3 mol 1 min 1. change in y gradient of line = change in x = 6 ( 9) ( ) = 35 =.175 The unit of the rate of change is the unit of the change in onion mass (%) divided by the unit of the solute potential (kpa), i.e. % kpa 1. Hence the rate of change is.175 % kpa 1. 5 a i From to 8 minutes, change in y rate of change = change in x = =.39 cm 3 min 1 ii From 8 to 1 minutes, the rate of water uptake is cm 3 min 1 because the line is flat and has zero gradient. b At a warmer temperature, the rate of water loss would be expected to increase. Measuring the rate of change at a point on a curve Guided questions (p.95) 1 Step 1: Concentration (M dm 3 ) Time (hours) Figure A.17 3

34 Step : gradient of tangent = change in change in = = 9 =. Therefore the rate of change in concentration at 44 hours is. M dm 3 h 1. Step 1: y x Oxygen saturation (%) Partial pressure of oxygen (mm Hg) Figure A Step : gradient of tangent = = 58 5 = 1.1 Therefore the rate of change in concentration at 44 mm Hg is 1.1 % (mm Hg) 1. 33

35 Practice questions (p.96) 3 Mass of product (g) Figure A gradient of tangent = 1 3 = 8 9 = 3.11 So the rate of reaction at 7 minutes is 3.11 g min 1. Time (minutes) 4 Number of birds Figure A. 46 gradient of tangent = = 6 4 = So the rate of change of the population in 199 was approximately 6.5 birds year Year 34

36 5 a Oxygen saturation (%) Partial pressure of oxygen (mm Hg) Figure A.1 For humans (lower curve and triangle): 81 1 gradient of tangent = = 71 6 =.73 So the rate of change in oxygen saturation at 8 mm Hg is.73 % (mm Hg) 1 for human haemoglobin. For llamas (upper curve and triangle): 1 47 gradient of tangent = = = 1.51 So the rate of change in oxygen saturation at 8 mm Hg is 1.51 % (mm Hg) 1 for llama haemoglobin. b As lamas live in an environment with low partial pressure of oxygen, their haemoglobin has a higher affinity for oxygen, so the oxygen dissociation curve is shifted to the left of the curve for a human. The rate calculated in a is higher in a human because 8 mm Hg is already past the steepest part of the llama s oxygen dissociation curve. 35

37 5 Geometry and trigonometry Circumferences, surface areas and volumes of regular shapes Guided questions (p.99) 1 Step 1: height of rectangle = π.6 = 1.885cm Step : length of rectangle = 9 cm surface area of curved side of cylinder = area of the rectangle Step 3: area of each circular end = πr = length of rectangle height of rectangle = 9 cm cm = 17. cm.6 = π =.83 cm area of both ends =.83 cm =.57 cm Therefore total surface area of cylinder = 17. cm +.57 cm = 17.6 cm Step 4: volume of cylinder = area of one circular end length Radius r = 6 = 3 µm =.83 9 =.5 cm 3 volumeofsphere = πr = π 3 = µm surface area = 4πr = 4 π 3 = µm 3 Step 1: for cube A: surface area = = 15 volume = = 15 Step : surface area : volume = 15 : 15 = 6 : 5 Step 3: for cube B: surface area = = 54 volume = = 7 36

38 Step 4: surface area : volume = 54 : 7 = : 1 Step 5: by expressing the ratio for cube A in the form 6 : 1 = 1. : 1, we can see that 5 cube B has the greater surface area to volume ratio. Practice questions (p.11) 4 Cell Z surface area = = 15 µm Cell Y surface area = (6 1) + (3 1) + (3 6) = = 54 μm So cell Z has the larger surface area. 5 Area of larger circle = π 3 = 8.3 cm Area of smaller disc = π.5 =.785 cm Area of clear ring = area of larger circle area of disc = = 7.5 cm 6 The section of aorta has the shape of a cylinder. volume of cylinder = area of one circular end length = π = 49.5 cm 3 7 Such species increase their surface area and reduce their volume by being very flat in shape, thus yielding a greater surface area to volume ratio. This ensures that diffusion pathways are short enough to provide sufficiently rapid gas exchange by diffusion through the body surface. 8 Surface area of a sphere = 4πr Surface area of original droplet = 4 π 15 = 87 µm Surface area of each new droplet = 4 π 8 = 84. µm Total surface area of new droplets = 84 5 = 41 µm After being split up into smaller droplets, the total surface area has increased considerably. The greater surface area helps to increase hydrolysis of the lipids by enzymes such as lipase. 37

39 Exam-style questions AS and A-level questions 1 a i Radius of a bead is r = cm = 1 cm Surface area of a spherical bead = 4πr = 4 π 1 = 1.57 cm marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer ii Volume of a spherical bead = 4 3 π r3 4 = π 1 3 = 4.19 cm 3 3 marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer iii Surface area : volume = 1.57 : 4.19 = 3 : 1 1 mark awarded for the correct answer iv 54 is approximately 5; 1.57 is approximately 13 So an estimate of the total surface area of these beads is 5 13 = 65 cm. marks awarded: 1 mark for the correct estimation; 1 mark for the correct answer b 3 marks awarded, as below: At x =.5, y = = 6.5 (1 mark) At x = 3, y = = 19 (1 mark) Flow rate (cm 3 min 1 ) Diameter of bead (cm) Figure A. 1 mark for the correctly drawn graph 38

40 c observed value expected value percentage error = 1% expected value = 1% 15 = 1% 15 = 13.3% So the percentage error is 13.3%. marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer a i mean length = 6 = = 1.7 μm 1 mark awarded for the correct answer ii 4 marks awarded, as below: Range = 5.9 = 4.1 µm 1 mark for calculating the range Make a table to calculate the standard deviation. Table A.5 A B C D E F Length of mitochondrion, x (µm) (x x) (x x) mark for completing the bottom two rows in the table The variance is Σ x x s = ( ) n = =.165 So the standard deviation is s =.165 = mark for calculating the standard deviation The range is large and the standard deviation is also quite large. This large spread is due to the exceptionally high value of D. 1 mark for a suitable comment on the values obtained 39

41 iii 3 marks awarded, as below: As there is an exceptionally high value for D (1 mark) and the range and standard deviation are both relatively large (1 mark), the mean is probably not the best average to use in this situation. The median may be a more useful average (1 mark). b In terms of µm, the drawing size of 4 cm is µm, so image size magnification = object size = = marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer 3 a 6 marks awarded, as below: Assuming that the flatworm is shaped like a rectangular prism: surface area = (14 3) + (3 1) + (14 1) = = 118 mm (1 mark) volume = = 4 mm 3 (1 mark) Therefore, for the flatworm, surface area : volume = 118 : 4 = 59 : 1 (1 mark) Assuming that the leech is shaped like a cylinder: surface area of curved side = circumference of circular end length of cylinder = π 6.5 = 4.84 cm area of each circular end= π = 3.14 cm Area of both ends = 3.14 cm = 6.8 cm Therefore total surface area = 4.84 cm cm = 47.1 cm (1 mark) volume = area of one circular end length = =.4 cm 3 (1 mark) Therefore, for the flatworm, surface area : volume = 47.1 :.4 (1 mark) 4

42 b Expressing both ratios in the form <a value> : 1, we have.81 : 1 for the flatworm.31 : 1 for the leech So the flatworm has the greater surface area to volume ratio. 1 mark awarded for the correct answer c marks awarded, as below: As the leech has a smaller surface area to volume ratio, its diffusion pathways are longer (1 mark), so diffusion by itself isn t fast enough to meet the leech s metabolic requirements, and a specialised circulatory system is required (1 mark). 4 a marks awarded, as below: Radius of a red blood cell is r = 7 µm = 3.5 µm Area of one circular side = πr = π 3.5 = µm (1 mark) Volume of blood cell = µm µm = µm 3 (1 mark) b marks awarded, as below: Each haemoglobin molecule can carry four oxygen molecules (1 mark), so each red blood cell can carry = oxygen molecules (1 mark). c i 5 = mark awarded for the correct answer ii As a human has red blood cells and each red blood cell can carry oxygen molecules, the total number of oxygen molecules that can be carried at any one time in a human body is ( ) ( ) =.8 1 marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer iii marks awarded, as below: At any one time, many of the red blood cells will not be carrying oxygen (1 mark), as they will be returning to the lungs from respiring tissues (1 mark). d i volume of a spherical nucleus= 4 r 3 π 3 = 4 6 π 3 = μm 3 marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer ii marks awarded, as below: A typical nucleus has a volume that is greater than the volume of a red blood cell. (1 mark) A red blood cell requires space inside to carry large amounts of oxyhaemoglobin. (1 mark) 3 41

43 A-level only questions 1 a The growth appears to be exponential, so on a logarithmic scale the graph would be a straight line. 1 mark awarded for the correct answer b Substituting P = 1, r = 1 and t = 5 into the equation: A = 1 e 1 5 = = So the population after 5 hours is bacteria. marks awarded: 1 mark for the correct substitution; 1 mark for the correct answer a 8 marks awarded, as below: Let R denote the allele for red flowers and Y the allele for yellow flowers. F1 generation: Parents RR YY Gametes: R only from one parent, Y only from the other parent Table A.6 Genetic cross R Y RY Offspring genotypes: 1% RY Offspring phenotypes: 1% orange flowers 1 mark for both genotype and phenotype F generation: Parents : RY RY Gametes: R, Y from both parents Table A.7 Genetic cross R Y R RR RY Y RY YY Offspring genotypes: 5% RR, 5% RY, 5% YY Offspring phenotypes: 5% red, 5% orange, 5% yellow 1 mark for both genotype and phenotype To determine whether the observed results differ significantly from the expected results, we perform a chi-squared test. Null hypothesis: there is no significant difference between the observed and expected results. Alternative hypothesis: there is a significant difference between the observed and expected results. 1 mark for noting the hypotheses 4

44 Total number of offspring = = 4 Splitting 4 in a 1 : : 1 ratio gives 51 red, 1 orange and 51 yellow offspring. These are the expected results. Make a table to calculate the chi-squared value. Table A.8 (o e) o e o e (o e) e Red Orange 1 1 Yellow mark for completing each of the second and fifth columns in the table (o e) χ = =.68 e 1 mark for the chi-square calculation Degrees of freedom = 3 1 = From Table.7, the critical value at.5 significance level is mark for identifying the correct level of significance Because.68 < 5.99, accept the null hypothesis and reject the alternative hypothesis. So we conclude that there is no significant difference between the observed and expected results. 1 mark for the final answer b area covered =.5 6% = =.15 cm marks awarded: 1 mark for the correct calculation; 1 mark for the correct answer 3 a 3 marks awarded, as below: Radius of a cylindrical potato core is r =.7 cm =.35 cm surface area of curved side = circumference of circular end length of cylinder = π.35 3 area of each circular end = π.35 = cm (1 mark) =.385 cm area of both ends =.385 cm =.77 cm Therefore total surface area = cm +.77 cm = 7.37 cm (1 mark) 43

45 volume = area of one circular end length = π.35 3 b i = 1.15 cm 3 (1 mark) percentage change in mass = final mass initial mass initial mass 1 Table A.9 Solute potential of external solution (kpa) Initial mass (g) Final mass (g) Percentage change in mass (%) marks awarded if all entries in the percentage change column are correct; 1 mark awarded if five of the entries are correct ii As the potato cores had different initial masses, just calculating the change in mass in grams would not have allowed the results from different solute potentials to be compared meaningfully. 1 mark awarded for the correct answer iii Change in mass (%) Figure A.3 Solute potential (kpa) 15 3 marks awarded: 1 mark for correct axes labels; 1 mark for use of linear scales; 1 mark for correctly plotting points and drawing line of best fit iv We need to find the value of x (solute potential) at which the y value (change in mass) is zero; this is the x-axis intercept. The line in iii crosses the x-axis at 5. Therefore the solute potential at which no change in mass occurred is 5 kpa. marks awarded: 1 mark for the correct answer; 1 mark for the explanation of how to reach the answer 44

46 v marks awarded, as below: The potato cells would be turgid at any solute potential higher than 5 kpa (1 mark) and plasmolysed at any solute potential lower than 5 kpa (1 mark). vi marks awarded, as below: Animal cells would burst upon gaining water (1 mark), so no gain in mass would be observed in hypotonic solutions (1 mark). vii marks awarded, as below: At any point where the solute potential is below 5 kpa, the cells are plasmolysed and then the pressure potential is zero (1 mark), which means that the water potential of the cells equals the solute potential (1 mark). 4 a 3 marks awarded, as below: In the Hardy Weinberg equation p + pq + q = 1, p and q are the frequencies of the dominant and recessive alleles. Here we have q = =.1744 (1 mark) So q =.1744 =.418 p + q = 1, so p = 1 q = =.58 (1 mark) The frequency of heterozygotes in the population is pq = =.49 (1 mark) b 3 marks awarded, as below: 44 mice out of 86 gives an actual frequency of.51 (1 mark). The frequency given by the Hardy Weinberg equation is an expected, or theoretical, frequency (1 mark). A discrepancy between the actual and expected frequencies could be due to chance or one of the assumptions of the Hardy Weinberg equation not being met (1 mark). c 3 marks awarded, as below: There appears to be a negative correlation between the drug dose and the effect of the deactivated allele (1 mark). This would suggest that the drug is mitigating the effects of the deactivated allele (1 mark). However, further investigation and statistical analysis of the results are needed to determine whether the correlation is significant (1 mark). 5 a i The energy lost from the second trophic level is the difference between the energy flowing into the primary consumers and the energy flowing out of the primary consumers into the secondary consumers and decomposers: = 491 kjm yr 1 1 mark awarded for the correct answer ii The gross primary productivity of the producers is the total energy flowing out of the producers: = 6 kjm yr 1 1 mark awarded for the correct answer 45

47 iii marks awarded, as below: The photosynthetic efficiency of the producers is % iv v = 1% 6 (1 mark) 1 = 3% (1 mark) The trophic efficiency of the primary consumers is 147 1% = 1.3% mark awarded for the correct answer The trophic efficiency of the secondary consumers is 54 1% = 17.3% mark awarded for the correct answer b marks awarded, as below: Secondary consumers are carnivores and can digest their high-protein diet more efficiently, resulting in higher tropic efficiency (1 mark). The primary consumers are herbivores, whose digestion of plant material is less efficient due to the high cellulose content (1 mark). 46