HS-LS2-2: Using Math to Explain Changes in Biodiversity and Populations
In plain English: HS-LS2-2 is the NGSS high school performance expectation that asks students to use averages, trends and graphs of several data sets to support explanations about what changes biodiversity and population sizes in ecosystems, and to revise those explanations when new evidence arrives. It is usually taught in Biology or Environmental Science.
Use mathematical representations to support and revise explanations based on evidence about factors affecting biodiversity and populations in ecosystems of different scales.
Clarification Statement: Examples of mathematical representations include finding the average, determining trends, and using graphical comparisons of multiple sets of data.
Assessment Boundary: Assessment is limited to provided data.
Next Generation Science Standards · Disciplinary Core Idea: Ecosystems: Interactions, Energy, and Dynamics · Official standard
Biologists rarely see a population or a community change directly: they see numbers. In this lesson students use simple math to explain what changes biodiversity (how many kinds of organisms live in a place and how evenly they are spread) and population size. They average counts from several samples, find trends over time, compare species richness and relative abundance, and graph several data sets side by side. Then they do the harder half of the standard: when a new data set does not fit, they revise the explanation instead of ignoring the data.
The NGSS Clarification Statement says: "Examples of mathematical representations include finding the average, determining trends, and using graphical comparisons of multiple sets of data." The Assessment Boundary says: "Assessment is limited to provided data." So every quiz and homework item on this page gives students the data they need; no item depends on outside facts about a real ecosystem. All data sets are invented for teaching and labeled as invented. The ecosystems range from a single rock to a region of forest fragments.
Learning Objectives
By the end of this lesson, students will be able to:
Find the average of counts from several samples and use it to compare populations at two or more sites
Describe a trend in population or biodiversity data as a rate of change per year, and judge whether the trend is steady
Compare the biodiversity of two communities using species richness, relative abundance and a given diversity index
Use graphs of several data sets to support an explanation of a factor that affects biodiversity or a population
Revise an explanation when new data do not fit it, and state what evidence changed the explanation
Prior Knowledge Required
Students should already be comfortable with:
How changes to physical or biological parts of an ecosystem affect populations MS-LS2-4
Finding the mean and describing a data set 6.SP.B.5
Percents and rates 7.RP.A.2
The idea of carrying capacity and limiting factors HS-LS2-1
Show two photos side by side: a mowed sports field and an overgrown field edge. Ask students to decide which has more biodiversity and what they would count to prove it.
Warm-Up Prompt
"Field 1 has 5 kinds of plants, but 95 of every 100 plants are grass. Field 2 has 5 kinds of plants in roughly equal numbers. Which field has more biodiversity? What numbers would you use to convince someone who disagrees?"
Take a quick vote. Many students say the fields are equal because both have 5 kinds. Use the disagreement to introduce two measures: species richness (how many kinds) and relative abundance (what share each kind makes up).
Direct Instruction15-20 minutes
Math as evidence. A single sample can mislead, so biologists take several and average them. A trend is the change per year, found from the first and last values or from a best-fit line. Species richness is the number of kinds in a sample; relative abundance is the count of one kind divided by the total. One common diversity index combines both: D = 1 - Σp², where p is the share of each kind. D is near 0 when one kind dominates and closer to 1 when many kinds are evenly spread. An explanation is supported when the numbers fit it, and it must be revised when new numbers do not. Use Diagram 1 with the fifth example and Diagram 2 with the third.
Averaging samples to compare populations
Mussels are counted in five 0.25 m² quadrats at two rocky sites (invented). Site A: 34, 41, 29, 38, 33. Site B, beside a storm-drain outlet: 12, 18, 15, 9, 16.
Result: Site A mean = 175 ÷ 5 = 35 per quadrat, or 140 per m². Site B mean = 70 ÷ 5 = 14 per quadrat, or 56 per m². The data support the explanation that runoff lowers the mussel population, but the sites could differ in other ways too.
Richness, relative abundance and a diversity index
Result: Richness: 5 kinds against 4. Midges make up 30% of the forest sample but 72% of the farm sample. D = 1 - (0.24² + 0.16² + 0.20² + 0.30² + 0.10²) ≈ 0.78 for the forest and 1 - (0.72² + 0.18² + 0.06² + 0.04²) ≈ 0.44 for the farm: lower biodiversity downstream of the farm.
Determining a trend
Pikas (small mountain mammals) counted on one rocky slope every 2 years, 2014-2024 (invented): 62, 58, 51, 47, 40, 36. Average July temperature at the slope rose over the same years.
Result: Change = (36 - 62) ÷ (2024 - 2014) = -2.6 pikas per year. The drop is steady, 4-7 pikas every 2 years. The data fit an explanation based on rising temperature, but a trend alone does not prove the cause.
Revising an explanation with new data
Frog egg masses in ponds (invented). Three ponds stocked with trout: 12, 9, 6. Three ponds without fish: 30, 27, 33. Explanation 1: "Trout eat tadpoles, so stocked ponds have fewer frogs." A later survey adds four fishless ponds that dry up by July: 5, 3, 4, 4.
Result: Means: stocked 9, fishless 30, fishless-but-drying 4. Drying ponds have no trout yet few frogs, so Explanation 1 is incomplete. Revised: frogs need ponds that are free of fish and hold water long enough for tadpoles to change into frogs.
Graphical comparison of several data sets
Bird species found in forest fragments of four sizes in two regions (invented). Region A, isolated fragments: 1 ha 8, 5 ha 14, 20 ha 21, 100 ha 30. Region B, fragments linked by strips of forest: 1 ha 11, 5 ha 18, 20 ha 25, 100 ha 33.
Result: In both regions bigger fragments hold more species. At every size the linked fragments hold 3 or 4 more species (3.5 on average), which supports the explanation that corridors let birds reach and use fragments.
After each example, ask: "What would a data set look like that would make us change this explanation?" This question prepares students for the revision tasks. Point out that the data here are all provided, which is also how the standard is assessed.
Guided Practice15 minutes
Pairs analyze invented data from eight kelp forest sites. Sea urchins graze on kelp, and sea otters eat sea urchins. Between 2015 and 2023 the water warmed by about the same amount at all eight sites.
Guided practice data: kelp forest sites (invented)
Site group
Urchins per m², 2015
Urchins per m², 2023
Kelp stalks per m², 2015
Kelp stalks per m², 2023
Sites 1-4, no sea otters (means)
5
40
8
1
Sites 5-8, sea otters present (means)
4
5
9
8
A diver proposes Explanation 1: "Warmer water caused the urchin increase and the kelp loss." Using only the rows for sites 1-4, do the numbers support it?
Now use sites 5-8. The water warmed there too. Find the change in urchins and kelp at those sites. Does Explanation 1 still work?
Write a revised explanation that fits all eight sites, and name the data that forced the revision.
Describe one more data set that could test the revised explanation, for example urchin counts after otters return to one of sites 1-4.
Independent Practice10-15 minutes
Each student analyzes invented data from grassland plots grazed by cattle at three intensities. The table gives the number of plant species in each 1 m² plot.
Independent practice data: plant species per plot (invented)
Grazing
Plot 1
Plot 2
Plot 3
Plot 4
None
9
11
10
10
Light
15
17
16
16
Heavy
6
8
7
7
Students find the three means (10, 16 and 7 species), graph them, and test the explanation "grazing lowers plant diversity." The data do not fit it everywhere: light grazing has the most species. Students revise the explanation (for example, light grazing keeps a few tall grasses from shading out smaller plants, while heavy grazing removes plants that cannot recover) and say what evidence led to the change.
Closure5 minutes
Exit ticket: "Three traps at a pond caught 14, 11 and 8 dragonflies. Find the mean. A student explains the low count with pesticide spraying nearby. Describe one new data set that would make you revise that explanation." (Mean 11. Example: counts that were already this low in the years before the spraying began.) Return to the warm-up and have students name which measure settled the vote.
Differentiation Strategies
For Struggling Students
Give data tables with an empty "mean" column and a worked first row, and let students use a calculator for every average
Provide an explanation frame: "Our first explanation was ___. The new data show ___. So we now think ___ because ___."
Compute relative abundance as "out of 100" before introducing the diversity index
For Advanced Students
Have students compute the diversity index for the Activity 1 data in a spreadsheet and test how much it changes when one rare species is removed
Ask students to add a best-fit line to the pika data and compare its slope with the first-to-last rate
Ask students to write two competing explanations for one data set and design the data that would tell them apart
Assessment Guidance
What to Look For
Check that students average several samples before comparing sites, and that trends are given as a rate with units (species per year, individuals per year). Look for explanations that cite a specific number and name a factor. The key sign of the practice is revision: when new data do not fit, students should change the explanation and say which data forced the change, not explain the data away. Watch for students who treat richness alone as biodiversity and ignore how evenly the species are spread.
02
Classroom Activities
3 Activities
1
Schoolyard Quadrat Survey: Lawn Against Field Edge
30 minGroups of 3-4
Groups place square quadrats in two parts of the school grounds, a mowed lawn and an unmowed edge or garden bed. They count the kinds of plants and the number of each kind, then use averages, relative abundance and the diversity index to compare the two areas.
Procedure
Each group places four 0.5 m × 0.5 m quadrats in the lawn and four in the edge, spaced a few meters apart. Toss a beanbag gently underhand to choose each spot so the choice is random.
In each quadrat, record the number of kinds of plants. Give a kind you cannot name a label such as "fuzzy leaf A" and take a photo.
For the whole group, count the individuals of each kind in the lawn quadrats and in the edge quadrats.
Safety: the teacher walks the area first and rules out any spot with poison ivy or poison oak, stinging insect nests, broken glass or animal waste. Students wear gloves, do not pick or taste plants, stay within sight of the teacher and away from roads and parking lots, and wash hands afterward. Students with a bee-sting allergy tell the teacher and keep their medication with them; bring water and sunscreen on hot days.
Sample Data (invented)
Kinds of plants per quadrat: lawn 3, 4, 3, 2; edge 7, 9, 8, 8. Individuals in the lawn quadrats: grass 180, clover 12, dandelion 6, plantain 2. Individuals in the edge quadrats: grass 60, clover 30, goldenrod 25, aster 20, yarrow 20, vetch 20, plantain 15, dandelion 7, violet 3.
Discussion Questions
The edge averages 8 kinds per quadrat and the lawn 3. How much of the difference could come from where you placed the quadrats?
Grass makes up 90% of the lawn plants but 30% of the edge plants. Use the diversity index (about 0.19 for the lawn and 0.83 for the edge) to explain why richness alone undersells the difference.
Which factor, mowing, foot traffic or soil, best explains the difference? What data would you collect to tell them apart?
Modification for Distance Learning
Students survey one quadrat in a yard or park and one on a lawn, with an adult present, or analyze the sample data above, and post their calculations and explanation.
2
Stream Evidence Rounds: Support, Then Revise
25 minGroups of 3-4
Groups receive three rounds of invented evidence cards about a stream where insect diversity dropped below a road crossing. After each round they write or revise their explanation on a whiteboard and record what changed their thinking.
Evidence Cards
Round 1: Kinds of stream insects in three samples: upstream of the road 13, 15, 14; downstream 7, 9, 8. Stoneflies per sample: upstream 12, downstream 1.
Round 2: Water chemistry: salt content (conductivity) 310 µS/cm upstream and 320 µS/cm downstream; dissolved oxygen 9.1 mg/L upstream and 9.0 mg/L downstream.
Round 3: Fine silt covers 10% of the streambed upstream and 55% downstream. A construction site beside the road has bare soil that washes into the stream when it rains.
Procedure
After Round 1, compute the mean number of kinds at each location and write a first explanation.
After each later round, keep, change or replace the explanation, and write one sentence naming the evidence that led to the decision.
This activity uses data cards only. If you extend it to real stream sampling, go only with trained adults, sample in shallow, slow water, wear gloves and closed shoes, and never sample after heavy rain.
Discussion Questions
Round 1 shows 6 fewer kinds downstream on average. Many groups first blame polluted road runoff. How do the Round 2 numbers (a difference of only 10 µS/cm and 0.1 mg/L) weaken that explanation?
How does silt covering 55% of the streambed explain the loss of stoneflies, which cling to clean rocks?
Which additional measurement would best confirm the revised explanation?
Modification for Distance Learning
Release the rounds as three slides at timed intervals in a video call. Each group edits a shared document and keeps every version so the revisions are visible.
3
Habitat Fragmentation Grid Simulation
20 minPairs
Pairs model a forest as a 10 × 10 paper grid of 100 cells. Eight forest species each need a block of connected forest of a certain size: 1, 2, 4, 9, 16, 25, 36 and 64 cells. A species survives if at least one block is large enough. Pairs shade in roads, count the largest block and record how many species survive.
Procedure
Scenario 1: no roads, one block of 100 cells.
Scenario 2: a road fills column 6 (10 cells), leaving blocks of 50 and 40 cells.
Scenario 3: roads fill row 5 and column 5 (19 cells), leaving four blocks of 16, 20, 20 and 25 cells.
Scenario 4: roads fill rows 4 and 8 and columns 4 and 8, leaving nine small blocks, the largest of 9 cells.
Scenario 5: go back to Scenario 3 and add wildlife bridges that let animals cross the roads, so the four blocks act as one 81-cell block.
Record the largest block and the number of surviving species for each scenario, then graph them.
Expected Results
Surviving species: 8, 7, 6, 4 and 8 in Scenarios 1-5.
Discussion Questions
Scenario 3 loses only 19 of 100 cells but 2 of 8 species. Why is the loss of species larger than the loss of area?
Scenario 5 restores all 8 species without adding any forest. What does this suggest about corridors?
Name one way this model is simpler than a real forest, and how that could change your explanation.
Challenge Variation
Pairs design a road layout that uses 20 road cells and keeps as many species as possible, then compare layouts across the class.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Bird Species in Isolated and Linked Forest Fragments
Invented bird species counts for forest fragments of four sizes in two regions, drawn to scale. Species rise with fragment size in both regions, and at every size the fragments linked by corridors hold 3 or 4 more species than isolated fragments of the same size.
Diagram 2: Trend in Pika Counts on One Slope
Invented pika counts, one survey every 2 years, drawn to scale with a best-fit line. The slope of the line, about -2.7 pikas per year, is close to the first-to-last rate of -2.6 per year. A steady trend fits an explanation but does not by itself prove its cause.
04
Homework Assignment
~30 min
HS-LS2-2 Homework: Explaining Biodiversity and Population Data
Directions: All data sets are invented for practice, and you need only the data given. Show each calculation. Where the problem asks for an explanation, cite specific numbers, and where it asks you to revise, say which data changed your mind.
Part 1: Averages and Trends (Problems 1-2)
Salamanders were counted under five cover boards in a forest before and after nearby logging. Before: 7, 9, 6, 8, 10. After: 3, 5, 2, 4, 6. (a) Find both means. (b) Find the percent change. (c) Write an explanation that the data support, and name one other factor that could also explain the change.
Bumblebee species found in a meadow survey: 2012: 18, 2015: 17, 2018: 15, 2021: 14, 2024: 12. The percent of the meadow covered by flowers in the same years: 40, 37, 31, 28, 22. (a) Find the average change in bee species per year from 2012 to 2024. (b) Is the decline steady? Use the change between surveys. (c) Do the flower data support an explanation for the decline? What else would you want to know?
Part 2: Comparing Biodiversity (Problems 3-4)
Samples of 50 pond animals. Pond A: beetles 12, dragonfly nymphs 9, snails 11, water boatmen 8, mayfly nymphs 10. Pond B, which receives fertilizer runoff: snails 38, beetles 7, water boatmen 5. (a) Find the species richness of each pond. (b) Find the share of snails in each pond. (c) Compute D = 1 - Σp² for each pond. (d) Explain the difference.
Butterfly species counted in three years along a gradient: city center 6, 5, 7; suburbs 11, 12, 10; farmland edge 18, 17, 19. The percent of ground covered by pavement and buildings is 85% in the city center, 45% in the suburbs and 5% at the farmland edge. (a) Find the mean for each location. (b) Sketch a graph that compares the three data sets. (c) Write an explanation of the pattern.
Part 3: Revising Explanations (Problems 5-6)
Young yellow perch caught per net in a lake: 2016-2019: 44, 38, 41, 37; 2020-2023: 18, 15, 16, 15. Anglers explain the drop by overfishing of adults. New data: adult perch per net were 22, 21, 23, 22 in 2016-2019 and 21, 22, 23, 22 in 2020-2023. A small invasive fish that eats zooplankton was first found in 2019, and zooplankton fell from 31, 29, 30, 30 to 13, 11, 12, 12 per liter. (a) Find the mean of each data set before and after. (b) Do the data support the overfishing explanation? (c) Write a revised explanation.
Fish species counted on reefs: four protected reefs: 42, 38, 45, 39; four fished reefs: 30, 27, 33, 26. (a) Find both means and write an explanation. (b) New data: the protected reefs are 12-20 km from river mouths, the fished reefs 1-4 km, and rivers carry mud that clouds the water. Two more fished reefs far from any river have 37 and 35 species. Revise your explanation and estimate how much of the difference each factor could account for. (c) Describe a comparison that would separate the two factors more clearly.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Calculations
Means, rates, shares and the diversity index are correct, with units where needed
Correct method with one calculation error
Supported Explanation
Explanation names a factor and cites specific numbers from the data
Explanation names a factor but cites no numbers
Revision
Revised explanation fits all the data and names the evidence that forced the change
Explanation is changed but the evidence for the change is not named
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All data sets are invented, and each question gives you the data you need. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Oysters are counted in five quadrats on a reef (invented data): 18, 24, 21, 15 and 22. What is the mean count per quadrat?
Answer: C
The mean is (18 + 24 + 21 + 15 + 22) ÷ 5 = 100 ÷ 5 = 20. Choice A is the sum, not divided by the number of quadrats. Choice B is the range (24 - 15). Choice D is the median, the middle value, which is a different measure.
Question 2 of 20 · Multiple Choice
A bird survey records (invented data): robins 12, sparrows 30, jays 3, sparrows 5 (a second flock), finches 8, robins 4. What is the species richness of the sample?
Answer: A
Richness counts kinds, not lines or birds: robins, sparrows, jays and finches make 4 species. Choice B counts the six lines of the record, so it counts robins and sparrows twice. Choice C is the number of birds. Choice D leaves out one species.
Question 3 of 20 · Multiple Choice
A sample from a meadow contains 250 insects, and 90 of them are ants (invented data). What is the relative abundance of ants?
Answer: D
Relative abundance is 90 ÷ 250 = 0.36, or 36%. Choice A treats the count as a percent. Choice C is the share of insects that are not ants (160 ÷ 250). Choice B divides the ants by the other insects (90 ÷ 160 ≈ 0.56) instead of by the whole sample.
Question 4 of 20 · Multiple Choice
A herd was counted every 2 years (invented data): 2010: 480, 2012: 450, 2014: 420, 2016: 390. What is the trend?
Answer: B
The change is (390 - 480) ÷ (2016 - 2010) = -90 ÷ 6 = -15 per year. Choice A is the change per 2-year survey, not per year. Choice C is the total change over 6 years. Choice D has the wrong sign: the herd is shrinking.
Question 5 of 20 · Multiple Choice
Two meadows each have 4 plant species and 100 plants (invented data). Meadow A: 25, 25, 25, 25. Meadow B: 85, 5, 5, 5. Using D = 1 - Σp², which meadow has higher biodiversity?
Answer: A
A: 1 - 4(0.25²) = 1 - 0.25 = 0.75. B: 1 - (0.85² + 3 × 0.05²) = 1 - 0.73 = 0.27. The evenly spread meadow has the higher index. Choice C uses richness alone and ignores evenness. Choice D reports Σp² instead of 1 - Σp², which reverses the answer. Choice B confuses one dominant species with diversity.
Question 6 of 20 · Multiple Choice
Live coral covered 40% of a reef in one survey and 26% five years later (invented data). By what percent did coral cover decrease?
Answer: B
The drop is 40 - 26 = 14 percentage points, and 14 ÷ 40 = 0.35, a 35% decrease. Choice A gives the drop in percentage points, not the percent change. Choice C divides by the new value (14 ÷ 26). Choice D is the cover that remains as a share of the start (26 ÷ 40), not the decrease.
Question 7 of 20 · Multiple Choice
A town explains a decline in songbirds by a new cell tower (invented case). Which new data would most clearly require the town to revise that explanation?
Answer: D
If birds far from the tower declined just as much, the tower cannot explain the pattern, so the explanation must change. Choices B and C are consistent with the tower explanation, and C is not measured data. Choice A says nothing about the decline.
Question 8 of 20 · Multiple Choice
Trout were counted in three streams for five years (invented data). Stream A: 50, 48, 51, 49, 52. Stream B: 60, 52, 45, 38, 30. Stream C: 20, 26, 31, 37, 42. Which statement do the data support?
Answer: C
Stream C went from 20 to 42 (2.1 times the start) and Stream B from 60 to 30 (half). Choice A is wrong because A is roughly steady and C rises. Choice B is wrong: A changed by only 2 trout. Choice D is wrong because C was below A every year.
Question 9 of 20 · Multiple Choice
A sample has three species with 6, 3 and 1 individuals. What is D = 1 - Σp²?
Answer: A
The shares are 0.6, 0.3 and 0.1, so Σp² = 0.36 + 0.09 + 0.01 = 0.46 and D = 1 - 0.46 = 0.54. Choice B stops at Σp². Choice C is the richness. Choice D assumes three equal shares: 1 - 3 × (1/3)² ≈ 0.67.
Question 10 of 20 · Multiple Choice
Neighborhoods with more street lights have fewer moth species (invented data). Which additional data would best test the explanation that the lights cause the decline?
Answer: D
A before-and-after comparison with dark control sites separates the effect of the lights from other differences between neighborhoods. Choice A counts only lights, not moths. Choice B relies on impressions rather than measurements. Choice C repeats the same kind of comparison that produced the pattern.
Question 11 of 20 · Multiple Choice
Students caught 36 beetles in a 4 m² plot and 50 beetles in a 10 m² plot (invented data). Which statement is correct?
Answer: B
Density is count divided by area: 36 ÷ 4 = 9 per m² and 50 ÷ 10 = 5 per m². Choice A compares totals instead of densities. Choice D is wrong: dividing by area is exactly how plots of different size are compared.
Question 12 of 20 · Multiple Choice
A native mussel and an invasive mussel were counted in a lake (invented data, per m²). Invasive: 0, 50, 400, 1,500. Native: 20, 18, 9, 2 in the same four years. Which statement is best supported by these data?
Answer: C
The native count fell from 20 to 2 as the invasive count rose from 0 to 1,500, so the data fit the explanation. The pattern alone cannot rule out another factor that changed at the same time, which is why choice A goes too far. Choice B misreads the native counts. Choice D ignores a clear opposite trend.
Question 13 of 20 · Multiple Choice
Wetland area in a county fell from 5,000 ha to 3,000 ha over 30 years, and the number of duck species that breed there fell from 12 to 8 (invented data). Which statement is supported?
Answer: D
Area: 2,000 ÷ 5,000 = 40% lost. Species: 4 ÷ 12 ≈ 33% lost. So both fell and species fell by a smaller percent. Choice B reverses the comparison. Choice C ignores the difference. Choice A treats the raw drops as percents.
Question 14 of 20 · Multiple Choice
An explanation says a beetle population dropped because of a pesticide sprayed on Farm X. Counts per trap at Farm X fell from 40 to 12, and at Farm Y, which never sprays, from 38 to 13 over the same years (invented data). What should a scientist do?
Answer: B
Farm Y fell by 25 per trap without any pesticide, almost the same as the 28 at Farm X, so something affecting both farms is a better explanation. Choice A ignores the new evidence. Choice C throws out data that do not fit, which is the opposite of revising. Choice D does not follow: both farms declined.
Question 15 of 20 · Short Answer
Pitfall traps caught ground beetles in a burned forest (4, 7, 5, 8) and in a nearby unburned forest (12, 15, 11, 14) (invented data). Find both means and the percent difference, and write an explanation that the data support.
Burned mean = 24 ÷ 4 = 6; unburned mean = 52 ÷ 4 = 13. The burned forest has (13 - 6) ÷ 13 ≈ 54% fewer beetles. Explanation: the fire removed the leaf litter and logs that ground beetles shelter and hunt in, so the population is smaller. The explanation is supported, but the two forests could differ in other ways, such as soil moisture.
Question 16 of 20 · Short Answer
A town blames a new factory, opened in 2019, for fewer mayflies in its river. Mayflies per sample upstream of the factory: 2016: 60, 2018: 54, 2020: 48, 2022: 42. Downstream: 2016: 58, 2018: 52, 2020: 46, 2022: 40 (invented data). Find the trend at each location and revise the explanation if needed.
Both locations fall by 18 over 6 years, or -3 mayflies per year, and the decline began before 2019. The upstream site, which the factory cannot affect, declines at the same rate, so the factory explanation does not fit. Revised explanation: something that affects the whole river, such as warming water or land use upstream of both sites, is causing the decline. A next step is to measure temperature and sediment at both sites.
Question 17 of 20 · Short Answer
Two gardens each have 5 species and 40 plants (invented data). Garden A: 8, 8, 8, 8, 8. Garden B: 24, 4, 4, 4, 4. Compute D = 1 - Σp² for each and explain what the difference shows.
A: 1 - 5(0.2²) = 0.80. B: shares 0.6, 0.1, 0.1, 0.1, 0.1, so Σp² = 0.36 + 4(0.01) = 0.40 and D = 0.60. The gardens have the same richness, but in B one species makes up most of the plants, so B is less diverse. The index shows evenness that a species count misses.
Question 18 of 20 · Short Answer
Rats were removed from Island X in 2016; Island Y, nearby, was left alone. Seabird species nesting on each island (invented data): Island X: 2016: 6, 2018: 7, 2020: 9, 2022: 11. Island Y: 6, 6, 5, 6. Compare the two data sets and write an explanation.
Island X gained 5 species in 6 years, about +0.8 species per year, while Island Y stayed near 6. Because the islands started equal and only X changed, the data support the explanation that rats were keeping seabirds from nesting (rats eat eggs and chicks). Island Y acts as a comparison that rules out a change affecting the whole area, such as weather.
Question 19 of 20 · Short Answer
Wildflower species were counted in three fenced plots that keep deer out (13, 15, 14) and three unfenced plots (8, 10, 9) (invented data). Find the means and the difference, and write an explanation. Then state one limit of these data.
Fenced mean = 14; unfenced mean = 9; the fenced plots have 5 more species on average. Explanation: deer eat some wildflowers before they can flower and spread, so browsing lowers plant diversity. Limit: there are only three plots of each kind and one year of data, so the difference could partly reflect where the plots were placed.
Question 20 of 20 · Short Answer
Beekeepers in a region explain winter colony losses by harsh winters. Colonies lost per 100 (invented data): three mild winters: 28, 30, 26; three harsh winters: 31, 29, 33. Beekeepers who treated their hives for mites every year lost 12, 14 and 13 per 100. Use the data to revise the explanation.
Means: mild winters 28, harsh winters 31, mite-treated hives 13 lost per 100. Harsh winters add only about 3 losses per 100, while treating for mites cuts losses by about 15 to 18 per 100. Revised explanation: winter weather has a small effect, but mites are a much larger factor in colony loss in this region.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-LS2-2 mean?
HS-LS2-2 asks students to use math, such as averages, trends and graphs, to support explanations of what changes biodiversity and population sizes, and to revise those explanations based on evidence. The code means High School, Life Science, core idea 2 (Ecosystems: Interactions, Energy, and Dynamics), performance expectation 2.
What is the assessment boundary for HS-LS2-2?
The official Assessment Boundary says that assessment is limited to provided data. Students are not expected to recall facts about particular real ecosystems or to collect their own data for a test: every item gives the numbers, and students reason from them. Field work like a quadrat survey is still valuable in class.
Is HS-LS2-2 biology or environmental science?
It is usually taught in the ecology unit of high school Biology, and it fits Environmental Science just as well. The math is middle school level: means, percents, rates of change and graphs.
What is the difference between biodiversity and population size?
Population size counts the individuals of one species. Biodiversity describes the variety of species in a community: how many kinds there are (richness) and how evenly the individuals are spread among them. A place can have a large population of one species and still have low biodiversity.
How can students measure biodiversity in class?
Count the kinds of organisms in several samples (species richness) and the individuals of each kind. Relative abundance is each count divided by the total. A diversity index such as D = 1 - Σp² combines richness and evenness in one number between 0 and 1. Students can work from quadrats on the school grounds, stream samples collected by trained adults, or provided data sets.
What does "support and revise an explanation" mean in HS-LS2-2?
To support an explanation, students show that the numbers fit it, for example that the site with more runoff has a lower mean count. To revise it, they change the explanation when a new data set does not fit, and say which data forced the change. Revising is a normal part of science, not a sign that the first explanation was a failure.
What factors affect biodiversity and populations?
Common factors include habitat loss and fragmentation, pollution and runoff, invasive species, predators and their removal, disease, harvesting, fire and grazing, and climate. Data often show several factors acting at once, which is why a second data set can change an explanation.
Why do biologists average several samples?
One sample can land on an unusually rich or poor spot. The mean of several samples is a steadier estimate, and the spread between samples shows how much a difference between sites could be due to chance. Comparing two sites on single samples can easily lead to the wrong explanation.
How do you find a trend in population data?
Divide the change in the count by the change in time to get a rate, such as 3 fewer animals per year. For data with ups and downs, a best-fit line gives a steadier rate. Check whether the change between surveys is steady, and remember that a trend shows what changed, not why.
How does HS-LS2-2 connect to other standards?
It builds on MS-LS2-4, which asks for arguments about how changes to an ecosystem affect populations, and it pairs with HS-LS2-1 on carrying capacity. HS-LS2-6 asks students to evaluate claims about ecosystem stability, and HS-LS2-7 asks them to design solutions that reduce human impacts on biodiversity. The statistics standards on comparing data sets use the same math.
07
Related Standards
5 standards
These standards connect to HS-LS2-2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-LS2-4Prerequisite
Argue from evidence that changes to ecosystem components affect populations
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