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Questions on handling data: units, scales, rates and uncertainty
Units, prefixes and standard form; where significant figures come from; ratios and magnification; percentage change against percentage difference; rate as a gradient, including a tangent to a curve; logarithmic scales for population and pH; and uncertainty in a measurement and how it propagates through a calculation.
15 original questions · 49 marks · the handling data: units, scales, rates and uncertainty notes · Scientific method and quantitative biology
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A potato chip is weighed at 5.00 g, left in a sucrose solution for an hour and weighed again at 4.60 g, on a balance reading to plus or minus 0.01 g. Calculate the percentage change in mass, and calculate the percentage uncertainty in that percentage change.
Mark scheme
- M1 change in mass = 4.60 − 5.00 = −0.40 g, and because the two masses were subtracted the absolute uncertainties add to give plus or minus 0.02 g
- A1 percentage change = −0.40 ÷ 5.00 × 100 = −8.0 per cent, the sign showing that the chip lost mass
- M1 percentage uncertainty in the change = 0.02 ÷ 0.40 × 100 = 5.0 per cent, and in the initial mass = 0.01 ÷ 5.00 × 100 = 0.20 per cent
- A1 the two were divided, so the percentage uncertainties add: about 5.2 per cent
A bacterial culture is counted every hour for ten hours and the counts are plotted with the logarithm to base 10 of the count on the vertical axis. Explain why a logarithmic axis is used here, and explain what a straight line on such a plot tells you.
Mark scheme
- B1 a population that multiplies repeatedly spans several factors of ten, so on a linear axis the early counts lie flattened against the baseline where nothing can be read from them
- B1 on a logarithmic axis equal distances represent equal multiplications, so the steps 1, 10, 100 and 1000 are evenly spaced and every hour is as readable as every other
- B1 each hour multiplies the population by the same factor and therefore adds the same amount to the logarithm, so constant growth plots as a straight line
- B1 a straight line is therefore evidence that the growth factor never changed over that period, which is what exponential growth means, and a rise of 3 on the axis is a multiplication of the population by 1000
The table shows the time an enzyme-catalysed reaction took to reach a colour change at five temperatures. Calculate the mean time to the colour change at 50 °C, saying what you do with the third reading and why, and then calculate the rate of reaction at 40 °C in s⁻¹.
Temperature / °C Repeat 1 / s Repeat 2 / s Repeat 3 / s 20 96 94 95 30 62 60 61 40 40 41 39 50 51 49 122 60 88 90 89 Mark scheme
- M1 identifies 122 s as an anomaly, because it is more than twice either of the other two readings at that temperature while every other set agrees within 2 s
- A1 excludes it and gives the mean of 51 and 49, which is 50 s
- M1 rate = 1 ÷ mean time; the mean at 40 °C is (40 + 41 + 39) ÷ 3 = 40 s
- A1 1 ÷ 40 = 0.025 s⁻¹
A treated group of seedlings has a mean height of 24 cm and a control group has a mean height of 20 cm. Calculate the percentage by which the treated group is taller than the control group, and calculate the percentage by which the control group is shorter than the treated group, using each one as its own reference value.
Mark scheme
- M1 difference between the means = 24 − 20 = 4 cm
- A1 treated relative to control: 4 ÷ 20 × 100 = 20% taller
- M1 control relative to treated: 4 ÷ 24 × 100, using a different reference value
- A1 = 17% shorter
A student measures an electron micrograph with a ruler. Its scale bar, labelled 500 nm, comes to 20 mm, and a ribosome in the same image comes to 1.0 mm across. Calculate the magnification the image was printed at, and calculate the actual diameter of the ribosome in nanometres.
Mark scheme
- M1 converts the scale bar's page length to the same unit as its labelled length: 20 mm = 20 000 000 nm
- M1 magnification = image size ÷ actual size = 20 000 000 ÷ 500 = ×40 000, with no unit
- M1 converts the ribosome's image size to the same units: 1.0 mm = 1 000 000 nm, then actual = image ÷ magnification
- A1 1 000 000 ÷ 40 000 = 25 nm, matching the expected size of a ribosome
A student collects 24 cm³ of gas in 20.0 s and writes the rate as 1.2 cm³ s⁻¹. Explain why the answer is given to two significant figures, and explain why counting the decimal places in the measurements is the wrong guide.
Mark scheme
- B1 an answer carries the same number of significant figures as the least precise measurement that went into it, and the volume of 24 cm³ carries only two
- B1 significant figures record how finely the instrument measured, so writing more digits claims a precision the syringe never had
- B1 decimal places depend only on the unit chosen: 24 cm³ written as 0.024 dm³ has three decimal places and still two significant figures, because changing the unit cannot improve the measurement
A tangent is drawn at the start of a curve of volume of oxygen collected against time. The tangent passes through the points (0 s, 0 cm³) and (30 s, 18.0 cm³). Calculate the initial rate of oxygen production and give its unit.
Mark scheme
- M1 the rate is the gradient of the tangent, which is the rise divided by the run, with both values read off the axes rather than counted in squares of paper
- M1 rise = 18.0 − 0 = 18.0 cm³ and run = 30 − 0 = 30 s
- A1 initial rate = 18.0 ÷ 30 = 0.60 cm³ s⁻¹
An osmosis investigation reports a percentage change in the mass of a potato chip carrying a percentage uncertainty of over 30 per cent, even though the balance reads to plus or minus 0.01 g. Suggest why the uncertainty is so large, and suggest one change to the method that would reduce it.
Mark scheme
- B1 the percentage change rests on the difference between two similar masses, and subtracting keeps the whole of both absolute uncertainties while leaving only a small value behind
- B1 so the percentage uncertainty in that small difference is large even though each individual weighing was good to a fraction of one per cent
- B1 leaving the chips immersed for longer, or using a solution further from the potato's own water potential, produces a larger change in mass and shrinks the percentage uncertainty without changing the balance
An enzyme-catalysed reaction is timed to a colour change at five temperatures, three repeats each; one repeat at 50 °C (122 s, against 51 s and 49 s) was left out of the mean as anomalous. The times fall from 20 °C to 40 °C and then rise again by 60 °C. Explain what this shows about the rate of the reaction, explain how the rate column is obtained from the times and why its unit is s⁻¹, and explain why the anomalous reading should be repeated rather than simply deleted.
Temperature / °C Mean time / s Rate / s⁻¹ 20 95 0.0105 30 61 0.0164 40 40 0.0250 50 50 0.0200 60 89 0.0112 Mark scheme
- B1 a shorter time means a faster reaction, so the rate rises to its highest measured value at 40 °C and falls after it, so the optimum lies between 30 °C and 50 °C, probably near 40 °C
- B1 each rate is 1 ÷ the mean time, an amount of change per second, so its unit is s⁻¹: 1 ÷ 40 s = 0.0250 s⁻¹ at 40 °C against 1 ÷ 95 s = 0.0105 s⁻¹ at 20 °C, which is why the rates run the opposite way up from the times
- B1 deleting a reading assumes it was a mistake; repeating it tests that assumption, and if the repeat agrees with the original the result is real and the explanation has to account for it
State the number of nanometres in one micrometre, the number of micrometres in one millimetre, and the unit that a ratio, such as a magnification, is given in.
Mark scheme
- B1 1 µm = 1000 nm
- B1 1 mm = 1000 µm
- B1 a ratio has no unit at all, because the units of the two quantities cancel when one is divided by the other
A student divides a total colour-change time by exactly 6 replicate tubes to calculate a mean, and wonders whether the 6 limits how many significant figures the mean can be given. Explain why the count of 6 tubes does not limit the significant figures of the mean, and explain what does.
Mark scheme
- B1 6 is an exact, counted number rather than a measurement, so it carries no uncertainty and imposes no limit on the number of significant figures of a result
- B1 the number of significant figures is instead set by the least precise instrument used to take the actual measurements that were averaged, such as the resolution of the stopwatch
- B1 the mean should therefore be rounded to match the significant figures of the timings themselves, not the number of readings that were combined
Compare percentage change with the symmetric percentage difference between two separate values, referring to what each is calculated from and the question each one answers.
Mark scheme
- B1 percentage change tracks a single quantity over time and divides by its initial value, keeping the sign of the change, whereas percentage difference compares two separate values measured at the same time and divides by their mean
- B1 percentage change answers 'how much did this one thing change', whereas a percentage difference or comparison answers 'how do these two separate values compare'
- B1 which value is used as the reference must be stated for a comparison between two values, because comparing a treated value to a control gives a different percentage from comparing the control to the treated
A student calculates the resting heart rate of a healthy adult human from their data and obtains an answer of 400 beats per minute. Suggest why this result should immediately be recognised as an error, and suggest one likely source of the mistake in the calculation.
Mark scheme
- B1 a healthy adult resting heart rate is around 60 to 100 beats per minute, so 400 is several times too large to be plausible for this organism
- B1 an order-of-magnitude check like this catches an error without redoing the whole calculation
- B1 a likely source is a unit not converted, such as working in seconds where minutes were needed, or a value multiplied where it should have been divided
A student times how long it takes a catalase-soaked filter-paper disc to float in hydrogen peroxide solution at five different pH values. Explain how the student should calculate a rate from each time, and explain why the times themselves should not be plotted directly as if they were the rate.
Mark scheme
- B1 rate is calculated as 1 ÷ t, in a unit such as per second, because the same fixed amount of reaction, the disc floating, has happened in every trial, so a shorter time means a faster reaction
- B1 the times themselves are not the rate: plotting time directly gives a graph the wrong way up compared with a graph of rate, since a longer time would appear to mean more when it actually means slower
- B1 the correct graph, of 1/t against pH, rises where the reaction is faster and falls where it is slower, matching how a rate is normally interpreted
State what is meant by the resolution of a measuring instrument, and state the uncertainty carried by a single reading taken from a millimetre rule.
Mark scheme
- B1 resolution is the smallest change an instrument can register, such as 1 mm on a metre rule or 0.01 g on a two-decimal-place balance
- B1 a single reading off a scale is taken as half the smallest division, so plus or minus 0.5 mm
Practise handling data: units, scales, rates and uncertainty one question at a time
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