1 markA group of scientists is setting up an experiment to investigate the effect of exercise on participants’ lung function.
Which of the following variables should the scientists control?
1 | Age of the participants |
2 | Type of exercise completed |
3 | The amount of air the participant’s lungs can hold |
4 | Gender of the participants |
1, 2, 3 and 4
Only 1, 2 and 4
Only 2 and 3
Only 1
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1 markA student wants to investigate the effect of mineral nutrients on the growth of barley seedlings. They grow 5 seedlings in a control medium, and 5 seedlings in a nitrogen-deficient medium. After 4 weeks, they measure the length of the seedlings. The mean length of the seedlings in the control group was 29.82 cm.
| Length of barley seedling (cm) | |||||
1 | 2 | 3 | 4 | 5 | ||
Condition | Control | 30.1 | 29.5 | X | 31.2 | 30.3 |
Nitrogen-deficient medium | 19.9 | 18.9 | 16.1 | 17.0 | 18.4 | |
What was the length of control seedling 3 (X)? Give your answer to one decimal place.
28.0 cm
18.06 cm
28.0 mm
27.9 cm
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1 markWhich of the following statements describes Student’s t-test?
A statistical test of the significance of the difference between expected and observed results
A statistical test which can compare three or more groups
A statistical test which compares the means of two sets of data and assesses whether the difference could be due to chance
A statistical test used to establish the level of correlation between two sets of data
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1 markWhy is it important to identify outliers in experimental results?
1 | Abnormal results might suggest an error in the experiment methodology |
2 | Outliers could skew the conclusions of the experiment if they are not excluded |
3 | Outliers could impact the statistical analysis of results |
1, 2 and 3
Only 1 and 2
Only 2 and 3
Only 1
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1 markWhat is standard deviation?
An average value of a group of measurements.
A test for correlation between values.
Another name for an outlier value.
A measure of the spread of data around the average value.
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1 markAn extract of the bark of the Indian fig tree, Ficus racemosa, has antioxidant properties. It has been suggested that this extract could protect against damage to non-cancerous cells during radiotherapy.
Cultures of non-cancerous cells were exposed to increasing doses of radiation and the percentage of damaged cells was measured. The experiments were performed multiple times either with or without the bark extract.
The graph shows the results for the two sets of data. Each data point shows the mean ±1 standard deviation.
Which of the following statements, A to D, about the graph is correct?
The data for the ‘with extract’ group is more variable than the data for the ‘without extract’ group.
The error bars show the range of data at each radiation dose.
The error bars show whether the difference between the two sets of data is significant.
The χ2 (chi-squared) test can be used to determine whether there is a significant difference between the two sets of data.
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1 markCelery contains the enzyme catalase, which breaks down hydrogen peroxide into oxygen and water.
A student added liquidised celery to a solution of hydrogen peroxide and collected the oxygen given off by the reaction. The results are shown in the graph below.
Which of the following shows the rate of reaction at 30 s?
0.85 cm3 s-1
1.00 cm3 s-1
1.15 cm3 s-1
1.50 cm3 s-1
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1 markPeak expiratory flow (PEF) is a measure of the maximum rate at which a person can exhale.
The graph below shows the typical PEF values for men of different ages and heights.
Which of the following is the percentage increase from the PEF of a 20 year old man of 175 cm to the PEF of a 45 year old man of 183 cm?
19.4%
10.9%
12.3%
5.2 %
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1 markThe graph below shows data on the density of the plant species R. ponticum and on the number of species present in an area of UK woodland.
Which of the following provides a correct analysis of the graph?
The presence of R. ponticum has a negative impact on the number of other species due to competition for resources.
The largest range for the number of species present at any particular density of R. ponticum is 10.
The overall diversity is greatest at 1 log arbitrary unit of R. Ponticum.
There is a significant negative correlation between the density of R. Ponticum and the number of species present.
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1 markWhich of the following statements is true when drawing conclusions from experimental results?
Random and systematic errors reduce accuracy.
Taking many repeat measurements or having a large sample size to analyse will improve precision.
Conclusions cannot be drawn from data which is accurate but not precise.
An evaluation should be carried out after all the data has been collected and conclusions have been drawn from the results.
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1 markIn trained athletes the metabolism of lactate can occur at a fast enough rate to reduce the buildup of lactic acid. As a result trained athletes are able to maintain a lower lactate level even during periods when their heart rate is very high, compared to an untrained individual.
A study was carried out to compare the blood lactate levels to the heart rate of 5 individuals during exercise.
Individual | Blood Lactate / mol per L | Average VE / L per min | Heart Rate / min-1 |
1 | 1.8 | 60.1 | 156 |
2 | 2.8 | 83.1 | 163 |
3 | 4.9 | 79.8 | 148 |
4 | 5.0 | 88.9 | 152 |
5 | 9.5 | 148.0 | 184 |
Identify the most highly trained individual from the study.
1
2
4
5
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1 markThe graph below shows the growth of a tumour over time. The tumour can only be detected once it reaches a certain size, indicated on the graph by line X.
How many cells will a tumour contain before it will be detected?
81
3 486 784 401
9
1 x 109
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1 markSome researchers were studying the resilience of some crop species. In their investigation they measured the water content of four varieties of wheat (v1, v2, v3, and v4) at the end of a 10 day period without water.
Their results are shown in the graph below.
Which of the following statements about the data are valid?
The water content of all wheat varieties showed a significant decrease in water content over the 10 day period.
The smallest decrease in water content was shown in v4.
v2 shows an 80 % decrease in water content.
v4 would be the most successful crop for growth in dry conditions.
Only 1
Only 1, 2 and 3
Only 2 and 3
1, 2, 3 and 4
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1 markSome students wanted to study the impact of soil mineral ion concentration on the growth of plant seedlings over a 10 day period.
Select the option which contains a correct combination of variables to carry out a valid investigation.
| Independent Variable | Dependent Variable | Control Variable |
A. | Concentration of soil fertiliser | Number of leaves grown | Concentration of soil fertiliser |
B. | Concentration of soil fertiliser | Change in mass of seedlings | Sterilised soil for planting all seedlings |
C. | Volume of water added to soil | Change in mass of soil | Same volume of fertiliser and water used for each repeat |
D. | Growth of seedlings in mm | Concentration of soil fertiliser | Same volume of fertiliser and water used for each repeat |
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1 markGoat (Capra hircus) livestock are the main animal grazed in northern Africa. An increase in demand for meat has resulted in more intensive farming which has increased the degradation of rangelands. To reduce the impact of grazing, farmers have been supplementing the grazing with additional feed resources.
A study was conducted into the effect of the different additional feeds on the digestive systems of the goats.
The results below show the pH of a sample of digestive juice from the digestive system of goats after being fed 1 of 4 different diets:
C = control diet
OC = olive cake diet
CC = cactus cladodes diet
OC+CC - olive cake and cactus cladodes diet
The results for diets marked * are significantly different at p < 0.05 %.
What is meant by 'significantly different at p < 0.05 %'?
There is more than a 5 % probability that any difference in the results is due to chance.
There is less than a 5 % probability that any difference in the results is due to chance.
The null hypothesis can be accepted.
There is a significant difference between all of the values shown in the results.
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1 markThe water potential in potato tissue can be investigated by placing potato discs in boiling tubes containing sucrose solutions of different concentrations.
Results from an investigation such as this are shown below, using potato discs with a mass of 2.5g:
Concentration of sucrose solution (mol dm-3) | Change in mass of potato discs (%) |
0.00 | + 16.00 |
0.10 | + 9.50 |
0.20 | + 4.00 |
0.30 |
|
0.40 | - 6.50 |
0.50 | - 13.00 |
The end mass of the potato disc placed in 0.30 mol dm-3 sucrose solution was 2.44g. Which of the following is the correct percentage change in mass for this disc?
2.4%
-2.4%
0.9%
-0.9%
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1 markEnzymes extracted from fruits have many potential uses. The time taken for different concentrations of papain, a protease enzyme extracted from papayas, to break down the protein gelatine was investigated
The data from the experiment is shown below:
Concentration of papain (%) | Rate of protein digestion (s-1) | Standard deviation |
0.010 | 0.0056 | 0.00026 |
0.025 | 0.0083 | 0.00094 |
0.050 | 0.0126 | 0.00390 |
0.075 | 0.0173 |
|
0.100 | 0.0197 | 0.00416 |
The following results were used to calculate the mean rate of protein digestion at 0.075% concentration of papain:
0.0173, 0.0161, 0.0157, 0.0185, 0.0189
Calculate the standard deviation of these data and use the result to answer the following question.
Which of the following statements explains what the standard deviation shows in the table?
In the first four concentrations, as the concentration increases, the standard deviation decreases.
The 0.100% concentration deviated the most.
The difference between the results for 0.025% and 0.050% concentrations of papain is statistically significant.
As the concentration increases, the repeatability of the results decreases.
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1 markThe graph below shows data collected from an investigation into the effect of different concentrations of ethanol on cell membrane permeability in samples of beetroot tissue.
Which of the following statements provide an explanation for the data in the graph?
1 | Ethanol has increased the pH across the cell surface membrane, increasing its permeability. |
2 | Ethanol has dissolved cholesterol in the cell surface membrane, increasing its permeability. |
3 | Ethanol has denatured the proteins in the cell surface membrane, increasing its permeability. |
4 | Ethanol has dissolved phospholipids in the cell surface membrane, increasing its permeability. |
Statements 1 and 2.
Statements 2 and 4.
Statements 3 and 4.
Statements 2, 3 and 4.
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1 markThe optimum pH for pepsin is stated in a data table as 1.8. A student carries out an experiment to find out if this value is correct. Their data is determined to have a high level of precision and accuracy.
Which row in the table below shows the correct definition of both precision and accuracy for this data?
| Precision | Accuracy |
A | Results are recorded to a high number of decimal places | Mean result is close to 1.8 |
B | Results are recorded to a high number of decimal places | Repeated readings are close together |
C | Results have a small standard deviation | Mean result is close to 1.8 |
D | Results have a small standard deviation | Repeated readings are close together |
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1 markAn investigation was completed into the effect of the medication glyburide on the secretion of insulin from pancreas cells in culture.
The results are shown below:
Condition | Mean insulin secreted (%) |
Without glyburide | 6.9 ± 0.72 |
With glyburide | 10.2 ± 0.13 |
A statistical test was used by the researchers to analyse their data. It gave a value of p < 0.001. Which of the following give an accurate analysis of the data?
The results for mean insulin secreted with glyburide are more valid than those without glyburide.
There is a less than 1 in 1000 probability that the results are due to chance so the null hypothesis should be accepted.
The results are significant as there is a 99.9% probability that the results are not due to chance.
The mean insulin secreted without glyburide is a more precise set of data compared to with glyburide.
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