NESA Biology Analysing Data and Information
15 sample questions with marking guides and sample answers · Avg. score: 65.1%
A computer simulation was used to observe genotypic changes in the gene pool of 20 randomly selected rabbits. The simulation was set with these parameters:
- each rabbit's coat colour was either black or white
- black alleles were dominant; white alleles were recessive
- the number of rabbits was constant in each generation and breeding was random throughout the population
- an environmental factor was chosen in the simulation to provide selection pressure.
The table shows the results of the simulation at the start and after 20 generations.
| Initial population genotypes | Population genotypes after 20 generations |
|---|---|
| BB BB BB BB BB BB | BB BB |
| Bb Bb Bb Bb Bb Bb Bb Bb Bb Bb | Bb Bb Bb Bb Bb Bb Bb Bb |
| bb bb bb bb | bb bb bb bb bb bb bb bb bb bb |
Contrast the initial allele frequency with the allele frequency after 20 generations to draw a conclusion about the effect of the selection pressure on the rabbit population.
Reveal Answer
Initial allelic frequencies were B 0.55 and b 0.45.
Allelic frequencies after 20 generations were B 0.3 and b 0.7.
B decreased (from 0.55 to 0.3) and b increased (from 0.45 to 0.7).
This selection pressure was in favour of white rabbits as both genotype and allelic frequencies shifted toward the white phenotype and the white allele.
| Descriptor | Marks |
|---|---|
Provides the correct initial and final allele frequencies | 1 |
Identifies consequentially correct change in allele frequency | 1 |
States a consequentially valid conclusion | 1 |
Use the following information to answer the question.
Students investigated the proteins found in several different types of human cells.
Students completed their investigation and analysed their results. They suggested their results were affected by systematic errors.
Systematic errors
result in a spread of readings.
affect the precision of a measurement.
are easy to identify and eliminate.
cause readings to differ from the true value by a consistent amount each time.
Reveal Answer
result in a spread of readings.
Incorrect. Random errors cause a spread or scatter of readings, whereas systematic errors shift all measurements in the same direction.
affect the precision of a measurement.
Incorrect. Systematic errors affect the accuracy of a measurement, while random errors affect its precision.
are easy to identify and eliminate.
Incorrect. Systematic errors are often difficult to identify because the measurements may still appear highly precise despite being consistently inaccurate.
cause readings to differ from the true value by a consistent amount each time.
Correct. By definition, a systematic error consistently shifts all measurements by the same amount and in the same direction from the true value.
Use the following information to answer the question.
Dengue is a disease primarily transported by mosquitoes. A group of students wish to investigate whether mosquitoes can be effectively controlled by spraying insecticide. The students will investigate the effect of spraying five different insecticides on mosquitoes.
The students want to check the reproducibility of their results.
To do this, they should ensure that the investigation
uses only recently calibrated equipment.
has both a positive and a negative control.
is conducted by only one student, in the same laboratory.
is conducted on different days by different students.
Reveal Answer
uses only recently calibrated equipment.
Using recently calibrated equipment improves the accuracy of the measurements, but it does not test whether the experimental results can be reproduced.
has both a positive and a negative control.
Controls are essential for the validity of an experiment to ensure the independent variable is causing the observed effect, but they do not check reproducibility.
is conducted by only one student, in the same laboratory.
Having the same student perform the experiment in the same lab tests repeatability (consistency under identical conditions), not reproducibility.
is conducted on different days by different students.
Reproducibility is the ability of an experiment to be duplicated by different researchers under different conditions (such as on different days) and still yield the same results.
A survey of grasshopper species was conducted across four eucalypt communities. Counts were conducted and the average abundance per 400 m recorded.
Grasshopper species (A–F)
| Community | A | B | C | D | E | F | Total |
|---|---|---|---|---|---|---|---|
| I | 32 | 18 | 1 | 3 | 0 | 46 | 100 |
| II | 3 | 2 | 0 | 1 | 3 | 12 | 21 |
| III | 3 | 2 | 28 | 3 | 18 | 51 | 105 |
| IV | 18 | 13 | 12 | 14 | 16 | 15 | 88 |
Which community has both the highest species richness and highest evenness for grasshoppers?
I
II
III
IV
Reveal Answer
I
Community I has a species richness of 5 (species E is absent), which is lower than communities III and IV. Additionally, it has low evenness because the population is dominated by species F and A.
II
Community II has a species richness of 5 (species C is absent), which is lower than the maximum richness observed in other communities. It also shows low evenness due to the dominance of species F.
III
Although Community III has the highest species richness (6 species), it has low evenness. The abundance is skewed heavily toward species F (51) and C (28), while other species have very low counts.
IV
Community IV has the highest species richness (all 6 species are present) and the highest evenness. The abundance counts for all species are very similar, ranging tightly between 12 and 18.
Use the following information to answer the question.
A Year 12 student conducts an investigation into the effects of different amounts of water on the growth of lettuce plants. The following table outlines some aspects of the investigation.
| Lettuce plant | Amount of water (mL/day) | Temperature (°C) | Light wavelength (nm) | Percentage increase in mass of lettuce (%) |
|---|---|---|---|---|
| 1 | 150 | 10 | 660 | 5 |
| 2 | 200 | 15 | 660 | 10 |
| 3 | 250 | 20 | 660 | 15 |
| 4 | 300 | 25 | 660 | 13 |
The investigation is not considered valid because
it contains more than one dependent variable.
only lettuce is used in the investigation.
the light wavelength is set at 660 nm.
the temperature is different for each plant.
Reveal Answer
it contains more than one dependent variable.
The investigation only has one dependent variable, which is the percentage increase in mass of the lettuce.
only lettuce is used in the investigation.
Using only one type of plant is necessary to control variables and ensure a fair test, so this does not invalidate the experiment.
the light wavelength is set at 660 nm.
Keeping the light wavelength constant at 660 nm is a controlled variable, which actually supports the validity of the experiment rather than undermining it.
the temperature is different for each plant.
For an experiment to be valid, only the independent variable (amount of water) should change. Because the temperature also changes, it acts as a confounding variable, making it impossible to know whether the water or the temperature caused the difference in growth.
Three factors that affect the rate of photosynthesis in plants are stated below:
- Factor 1 – light intensity
- Factor 2 – carbon dioxide concentration
- Factor 3 – temperature
Scientific data consistently shows specific trends for each factor when other variables are controlled.
Graphs were plotted showing the rate of photosynthesis against an increasing change in each factor.
If plotted over a large range, which of the following would show graphs with the same trend?
factors 1 and 2
factors 1 and 3
factors 2 and 3
factors 1, 2 and 3
Reveal Answer
factors 1 and 2
Both light intensity and carbon dioxide concentration show the same trend: the rate of photosynthesis increases initially and then plateaus when another factor becomes limiting.
factors 1 and 3
Temperature shows a different trend than light intensity; it increases to an optimum point and then rapidly decreases as enzymes denature, whereas light intensity plateaus.
factors 2 and 3
Temperature increases to an optimum and then decreases due to enzyme denaturation, which is a different trend from carbon dioxide concentration, which simply plateaus.
factors 1, 2 and 3
Temperature has a distinct peak and decline curve due to enzyme denaturation, which differs from the plateauing curves of light intensity and carbon dioxide concentration.
Students designed a controlled experiment. After they had performed the experiment, another group of students gave them feedback suggesting that they should modify the experiment to improve the accuracy of their results.
A change that the first group of students could make to improve the accuracy of their results could include
ignoring outlying results.
repeating the experiment many times.
carefully calibrating the equipment used.
having many people take the measurements.
Reveal Answer
ignoring outlying results.
Ignoring outliers does not improve accuracy and can lead to biased results; outliers should be investigated to understand their cause rather than simply discarded.
repeating the experiment many times.
Repeating the experiment many times improves the reliability and precision of the results, but it does not correct systematic errors that affect accuracy.
carefully calibrating the equipment used.
Carefully calibrating equipment reduces systematic errors, ensuring that the measurements are as close to the true value as possible, which directly improves accuracy.
having many people take the measurements.
Having multiple people take measurements introduces human variability and potential error, which can decrease precision without improving accuracy.
In an experiment studying the population dynamics of the house fly, two adult house flies were placed in a fly cage with a limited food supply. Population data was collected, as summarised in the table.
| Generation | Number of eggs | Number of larvae | Number of pupae | Number of adults |
|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 2 |
| 2 | 120 | 110 | 95 | 88 |
| 3 | 250 | 225 | 213 | 210 |
| 4 | 500 | 475 | 462 | 12 |
| 5 | 20 | 2 | 0 | 0 |
Determine the population growth model exhibited by the house fly. Explain your reasoning.
Reveal Answer
Initially there is a rapid increase in growth.
This is followed by a sudden drop in population numbers.
This pattern is typical for J-curve population growth.
| Descriptor | Marks |
|---|---|
States initial rapid population increase | 1 |
States sudden population drop | 1 |
States J-curve | 1 |
The effect of an invasive species on plant biodiversity was investigated by collecting this data from an ecosystem.
| Percentage cover (invasive species) | |||||
|---|---|---|---|---|---|
| 0–20% | >20–40% | >40–60% | >60–80% | >80–100% | |
| Species richness | 7 | 7 | 7 | 4 | 2 |
| Simpson’s diversity index | 0.83 | 0.77 | 0.55 | 0.49 | 0.30 |
Contrast species richness in areas of low invasive species cover (0–20%) with areas of high invasive species cover (>80–100%).
Reveal Answer
Species richness is higher in areas with low invasive species cover (seven different species compared to two).
| Descriptor | Marks |
|---|---|
Identifies species richness is higher in areas with low invasive species cover | 1 |
Draw a conclusion about the effect of the invasive species on plant biodiversity in this ecosystem. Justify your response.
Reveal Answer
The presence of the invasive species has a negative effect on biodiversity. While the same number of species were observed in the area when invasive species cover was 0–60% (7, as indicated by species richness), the SDI decreased from 0.83 to 0.3 as percentage cover increased. This indicates that overall, biodiversity decreased as percentage cover increased.
| Descriptor | Marks |
|---|---|
Identifies a relevant trend in the data | 1 |
Draws a valid conclusion about the effect of the invasive species | 1 |
Explain why having data on species richness and Simpson’s diversity index is more informative than a single measure for the purpose of this investigation. Refer to the table to support your response.
Reveal Answer
Species richness indicates the number of species that can coexist in the area; however, it doesn’t account for the abundance of each species, so one or two species may be dominating. SDI considers the number of species and their relative abundance (evenness), but the number of species is not obvious from the index.
Having both values is more informative because it allows the researcher to identify the number of species in the area (7) and infer evenness from the SDI. For example, the same number of species was present for 0–60% coverage (7), but the evenness was not the same, as the SDI decreased as % coverage got higher.
| Descriptor | Marks |
|---|---|
Recognises species richness is the total number of species present | 1 |
Recognises SDI considers species number and abundance/evenness | 1 |
Uses data to explain why data on species richness and SDI is more informative than a single measure | 1 |
Evolutionary relationships were investigated by sequencing a section of protein from five different species. Each letter represents an amino acid.
| Species I | D | E | V | G | W | E | A | L | G | R | L | V |
| Species II | D | E | V | G | W | E | G | L | G | R | A | V |
| Species III | D | E | A | G | S | E | G | L | A | R | L | E |
| Species IV | D | E | V | G | S | E | G | L | G | R | L | E |
| Species V | D | E | V | G | W | E | A | L | A | R | L | V |
It can be inferred that Species I is most closely related to
Species II.
Species III.
Species IV.
Species V.
Reveal Answer
Species II.
Species I and Species II differ by 2 amino acids. Since Species V has fewer differences, Species II is not the most closely related.
Species III.
Species I and Species III differ by 5 amino acids. This high number of differences suggests they are more distantly related compared to the other options.
Species IV.
Species I and Species IV differ by 3 amino acids. Because there is another species with greater sequence similarity, this is not the correct answer.
Species V.
Species I and Species V differ by only 1 amino acid. High sequence similarity implies a more recent common ancestor, making them the most closely related pair.
To determine the reliability of the capture–recapture technique and the Lincoln index , two researchers collected data on a mouse population from the same site at the same time.
| Researcher | Number of mice in first capture | Number of mice in second capture | Number of marked mice in second capture |
|---|---|---|---|
| I | 160 | 140 | 80 |
| II | 100 | 60 | ? |
If the reliability of the technique is high, how many marked mice would be expected in the second capture for researcher II? Show your working.
Reveal Answer
| Descriptor | Marks |
|---|---|
calculates N for researcher I | 1 |
determines expected m for researcher II | 1 |
Species richness (S) can be determined using the Menhinick’s Index, where is the number of different species represented in a sample and is the total number of individual organisms in the sample.
Species richness (S) =
Sampling of two communities (Community A and Community B) were completed, with Community A having S = 0.5 for the six species identified. The results for Community B are shown in the table.
| Community B | |
|---|---|
| Species | Number of individuals |
| A | 12 |
| B | 15 |
| C | 19 |
| D | 22 |
| E | 25 |
| F | 7 |
Use S to compare Community A with Community B.
Reveal Answer
Community B has greater richness than Community A.
Community B has a same number of different species as Community A (i.e. 6); however, Community A must have been for a larger sample.
| Descriptor | Marks |
|---|---|
States that S = 0.6 | 1 |
States that Community B has greater richness than Community A | 1 |
States that Community A and Community B have same number of species | 1 |
Infers that Community B’s sample size is smaller than Community A | 1 |
A type of pea plant can have red or purple flowers. The table below shows the flower colour of parental plants and their offspring for five different crosses. Use this information to answer the question.
| Cross | Male Parent | Female Parent | Offspring |
|---|---|---|---|
| 1 | red | red | all red |
| 2 | red | purple | all purple |
| 3 | purple | red | all purple |
| 4 | purple | red | some with red and some with purple |
| 5 | ? | purple | some with red and some with purple |
Which of the following statements about Cross 5 is correct?
The female parent could be a homozygote for the dominant allele.
The male parent could be a heterozygote with red flowers.
The male parent could be a homozygote with purple flowers.
The male parent could be a homozygote with red flowers.
Reveal Answer
The female parent could be a homozygote for the dominant allele.
If the female parent were homozygous for the dominant allele, all offspring would express the dominant phenotype, which contradicts the observed results of Cross 5.
The male parent could be a heterozygote with red flowers.
Based on the cross results, the red flower trait is recessive. Therefore, a plant with red flowers must be homozygous, not heterozygous.
The male parent could be a homozygote with purple flowers.
If the male parent were homozygous for the dominant purple flower trait, all offspring would have purple flowers, which does not match the data for Cross 5.
The male parent could be a homozygote with red flowers.
Because the red flower phenotype is a recessive trait, a plant expressing red flowers must be homozygous for the recessive allele.
The figure below shows the numbers of families of two groups of marine invertebrates, trilobites and ammonites, over evolutionary time.
The change in the numbers of trilobite and ammonite families over evolutionary time is an example of
macroevolution.
microevolution.
diversifying selection.
directional selection.
Reveal Answer
macroevolution.
Macroevolution refers to large-scale evolutionary changes that occur above the species level, such as the origin, diversification, and extinction of entire taxonomic groups like families over geological time.
microevolution.
Microevolution describes changes in allele frequencies within a single population over a relatively short period of time, not the large-scale changes seen in entire families over millions of years.
diversifying selection.
Diversifying (or disruptive) selection is a mechanism of microevolution that favors extreme phenotypes over intermediate ones within a population, which does not describe the long-term changes in the number of taxonomic families.
directional selection.
Directional selection is a microevolutionary process where a single extreme phenotype is favored within a population, rather than a description of broad evolutionary trends across entire taxonomic families.
The table shows data from a transect study along a sand dune.
| Zone | ||||
|---|---|---|---|---|
| A | B | C | D | |
| Distance from sea (m) | 0–100 | >100–150 | >150–250 | >250–300 |
| Age of dune (years) | 0–50 | >50–100 | >100–125 | >125–150 |
| pH of soil | 8.4 | 7.4 | 6.9 | 6.0 |
| Organic matter in soil (%) | 1 | 2.5 | 5 | 30 |
| Number of grass species | 2 | 4 | 6 | 2 |
| Number of tree species | 0 | 1 | 3 | 8 |
Contrast species richness in zones A and D. Refer to data in your response.
Reveal Answer
Species richness is five times higher in zone D than zone A. Zone A has two different species, whereas zone D has 10.
| Descriptor | Marks |
|---|---|
Identifies species richness is higher in zone D | 1 |
Provides appropriate data | 1 |
Infer how organic matter affects the pH of soil. Justify your response.
Reveal Answer
As organic matter increased from 1% to 30%, the pH decreased from 8.4 to 6. This suggests that organic matter lowers soil pH.
| Descriptor | Marks |
|---|---|
Infers that organic matter lowers soil pH | 1 |
Justifies response using data | 1 |
Predict which zone would have the highest proportion of K-selected species. Explain your reasoning.
Reveal Answer
Zone D. Zone D is the oldest dune and has the highest species richness. This suggests it may be further along in succession. The number of K-selected species tends to increase as succession progresses.
| Descriptor | Marks |
|---|---|
Predicts zone D | 1 |
Provides appropriate reasoning | 1 |