HS-LS4-6: Simulating Solutions That Protect Biodiversity
In plain English: HS-LS4-6 is the NGSS high school performance expectation that asks students to create or revise a simulation that tests a solution to a human impact on biodiversity, such as bycatch, road deaths or isolated habitat. Students set the rules, run the model with and without the solution, and judge the results for population size and genetic variation. It is usually taught in Biology or Environmental Science.
Create or revise a simulation to test a solution to mitigate adverse impacts of human activity on biodiversity.
Clarification Statement: Emphasis is on designing solutions for a proposed problem related to threatened or endangered species, or to genetic variation of organisms for multiple species.
Next Generation Science Standards · Disciplinary Core Idea: Biological Evolution: Unity and Diversity · Official standard
Students build and revise simple simulations of populations that people are harming, then use them to test a solution before anyone spends money in the real world. A simulation here is a set of rules (a growth factor, a loss from a human activity, a random event) applied year after year or generation after generation, by hand, with beads or dice, or in a spreadsheet. Students run each model with and without a solution, compare the outputs with a stated goal, and name what the model leaves out.
The NGSS Clarification Statement says: "Emphasis is on designing solutions for a proposed problem related to threatened or endangered species, or to genetic variation of organisms for multiple species." HS-LS4-6 has no official Assessment Boundary. The lesson therefore works with two kinds of output: population size for threatened species (sea turtles, salamanders, seabirds, manatees) and genetic variation, measured as heterozygosity, for several species at once. Every data set and every model parameter on this page is invented for teaching and labeled as such. Students do the practice of using a simulation: they set up the rules, run them, revise them when they miss something real, and argue from the outputs.
Learning Objectives
By the end of this lesson, students will be able to:
Build a simulation of a population with a growth factor and a loss caused by a human activity, and run it for several time steps
Add a solution to a simulation as a change in one rule, and compare outputs with and without the solution against a stated goal
Use the rule Hnext = H × (1 - 1/(2N)) to simulate the loss of genetic variation in small populations, and test a solution such as a corridor, translocation or seed bank
Revise a simulation that leaves out a real factor (a human impact, a carrying capacity, random variation or genetic variation) and explain how the revision changes the result
Write a claim, evidence and reasoning argument for a solution that cites simulation output and names one limit of the model
Prior Knowledge Required
Students should already be comfortable with:
Comparing design solutions for maintaining biodiversity MS-LS2-5
How genetic variation helps some individuals survive in a changing environment MS-LS4-4
Percent change and multiplying by a growth factor
Basic probability: the chance of an event as a fraction of equally likely outcomes
Show a photo of a two-lane road between a forest and a breeding pond. Tell students that salamanders cross it every spring, and many are hit by cars.
Warm-Up Prompt
"A town can afford one solution: tunnels under the road with fences that guide the salamanders to them. Before building anything, how could the town find out whether tunnels would save this population? List three numbers your test would need."
Collect answers on the board. Students usually name the number of salamanders, how many die on the road and how many young are born. Point out that they have just listed the parts of a simulation: a starting value, a rule for growth, a rule for the human impact and a rule for the solution.
Direct Instruction15-20 minutes
What a simulation needs. A population simulation applies the same rules each time step. A growth factor λ multiplies the population to account for births and natural deaths (λ > 1 means growth without people). A human impact is either a number removed each step (bycatch, poaching) or a fraction removed (road deaths). A solution changes one of those rules. For genetic variation, a population of N breeding adults keeps, on average, the fraction 1 - 1/(2N) of its heterozygosity H each generation, so small isolated populations lose variation fastest. This rule is an idealized expectation: it assumes random mating, no migration, mutation or selection, and an equal chance for every adult to pass on its genes. Real populations with uneven sex ratios or uneven breeding success usually lose variation faster than the rule predicts. To create a simulation, students choose the rules; to revise one, they add a factor it missed and run it again. Use Diagram 1 with Example 3 and Diagram 2 with Example 2.
Threatened species: bycatch and a device that frees turtles
Invented model of a sea turtle population: 2,000 adults, growth factor 1.02 per year without fishing, and 60 adults drowned each year in shrimp nets. A net device is tested in the model that lets 90% of trapped turtles escape.
Result: Without the device: 2,000 × 1.02 - 60 = 1,980, a loss of 20. With it, bycatch is 6: 2,000 × 1.02 - 6 = 2,034. The population adds 40 turtles a year from growth, so any solution that cuts bycatch to 40 or fewer (a cut of at least one third) stops the decline.
Genetic variation: a corridor between two populations
Invented model: a mountain population has 20 breeding adults and heterozygosity H = 0.60. A proposed wildlife corridor would join it to a neighboring population, giving 50 breeding adults that act as one population.
Result: Isolated: H = 0.60 × (1 - 1/40)10 ≈ 0.47 after 10 generations. With the corridor: H = 0.60 × (1 - 1/100)10 ≈ 0.54. The corridor slows the loss of variation; it does not stop it.
Revising a simulation: adding road deaths, then tunnels
A first model of 800 salamanders used a growth factor of 1.10 and predicted steady growth. Field counts show that 15% of the adults are killed on the road each spring. Tunnels with guide fences would cut road deaths to 3% (invented values).
Result: Revised factor: 1.10 × 0.85 = 0.935, so the population shrinks to about 572 in 5 years. With tunnels: 1.10 × 0.97 = 1.067, and the population grows to about 1,106. The revision reversed the prediction, which is why the missing factor mattered.
Comparing two solutions for an endangered bird
Invented model: 100 birds. Option A releases 10 captive-raised birds each year while the growth factor stays at 0.95. Option B protects nesting habitat, which raises the growth factor to 1.03, with no releases.
Result: After 10 years: Option A gives about 140 birds and Option B about 134. But Option A levels off at 10 ÷ 0.05 = 200 birds and needs releases forever; Option B keeps growing (until the habitat is full) because it removes the cause of the decline.
Genetic variation for many species: sampling for a seed bank
A seed bank wants to store a rare allele that has frequency 0.05 in a wild plant population. Each plant carries two copies of the gene, so sampling n plants samples 2n gene copies.
Result: The chance of catching at least one copy is 1 - 0.952n: about 0.64 for 10 plants and about 0.95 for 30 plants. The same rule, run for each species with its own allele frequency, sets how many plants to sample per species.
After each example, ask: "Which rule stands for the human impact, which rule stands for the solution, and what output tells you whether the solution works?" Stress that a simulation shows what follows from its assumptions: its answer is only as good as its rules and numbers.
Guided Practice15 minutes
Pairs run a simulation by hand for an invented reef fish (a grouper) that is being overfished. The reef can support about 10,000 adults (the carrying capacity K), the population grows at r = 0.20 when it is small, and fishing removes a fraction h of the fish each year. The rule is Nnext = N + 0.20 × N × (1 - N/10,000) - h × N. The proposed solution is a no-take reserve on part of the reef, which cuts the yearly harvest fraction from 15% to 5%. Pairs fill in the blank rows, rounding to the nearest fish.
Guided practice simulation: grouper starting at 3,000 adults (invented)
Fill in years 2-4. Answers: 2,942, 2,916, 2,892 with no reserve and 3,547, 3,827, 4,108 with the reserve.
Find the population where the model stops changing: growth equals harvest when 0.20 × (1 - N/10,000) = h. That gives 2,500 fish with no reserve and 7,500 with it.
Write one sentence that states whether the reserve works and cites two numbers from the simulation, then name one thing the model leaves out (for example, fish moving out of the reserve into fished water).
Independent Practice10-15 minutes
Each student revises a simulation for an invented seabird colony on an island where people accidentally brought rats. The first model used only adult survival. Students revise it so that each year's growth factor is λ = s + f × p, where s = 0.85 is adult survival, f = 0.40 is the number of chicks that would fledge per adult if every nest survived, and p is the fraction of nests that survive rat predation.
Independent practice data (invented)
Situation
Nest survival p
Growth factor λ
Rats present
0.25
?
After rat removal
0.60
?
Students compute both growth factors (0.95 and 1.09), state which one shows a growing colony, and find the smallest nest survival that stops the decline (p = 0.375). They finish with a claim-evidence-reasoning paragraph and one revision they would make next, such as a rule for rats returning on supply boats.
Closure5 minutes
Exit ticket: "A lizard population of 600 has a growth factor of 1.05 without people, and pet collectors remove 45 lizards a year. Does the population grow or shrink in year 1? What yearly collection limit would stop the decline?" (585 lizards, so it shrinks; the limit is 30 or fewer.) Return to the warm-up list and ask students to label each number as a starting value, a growth rule, a human impact or a solution.
Differentiation Strategies
For Struggling Students
Give a partly filled simulation table with the first two rows done and a column for each step of the rule (multiply, then subtract)
Start with a number removed each year (bycatch) before a fraction removed each year (road deaths), and use whole-number populations
Provide a sentence frame: "With the solution, the model predicts ___ after ___ years, compared with ___ without it, so the solution ___ the goal of ___."
For Advanced Students
Build the grouper model in a spreadsheet and find the largest harvest fraction that keeps the long-run population above half the carrying capacity
Add random good and bad years to a population model with a random-number function, run it 50 times and report the fraction of runs that fall below 100 animals
Combine a population rule and the heterozygosity rule in one model, so that a corridor changes both the number of animals and the loss of variation
Assessment Guidance
What to Look For
Check that students apply the rules in the right order each step and carry the new value into the next step, rather than applying the growth factor to the starting value every time. Look for arguments that compare runs with and without the solution against a stated goal, not just one run. Listen for students who treat a simulation result as a certain prediction; they should say what it assumes and what it leaves out. For random simulations, check that students run them several times before drawing a conclusion.
02
Classroom Activities
3 Activities
1
Bead Drift Simulation: Can a Corridor Keep Genetic Variation?
25 minGroups of 3-4
Groups simulate a tiny isolated population of 5 animals (10 gene copies) with beads of two colors, then revise the simulation to add a corridor that brings in one migrant gene copy per generation. They record how often one color disappears, which is a loss of genetic variation.
Procedure
Put 5 red and 5 blue beads in a cup. This is generation 0.
Make the next generation: draw one bead without looking, record its color, put it back and shake. Repeat 10 times, then fill a fresh cup with beads matching the 10 colors you recorded.
Repeat for 10 generations or until one color is gone. That is one run of the isolated simulation.
Revise the simulation for the corridor: after each new generation, remove one bead at random and replace it with a bead drawn from a paper bag holding 30 red and 30 blue beads (a large neighboring population). Do one run.
Combine class results so there are at least 8 runs of each version.
Sample Data (invented)
Isolated version, 8 runs: 4 runs had lost one color by generation 10. Corridor version, 8 runs: 1 run was missing a color at generation 10.
Discussion Questions
In the sample data, half of the isolated runs lost a color, and only one corridor run did. What does the lost color stand for in a real population?
Why would one run of each version not be enough to judge the corridor?
How would you revise the simulation to test a corridor that lets a migrant through only every other generation? Predict the result.
Modification for Distance Learning
Students run the same rules in a spreadsheet: each gene copy in the next generation is red if a random number between 0 and 1 is below the current fraction of red copies. They copy the formula down 10 generations, press recalculate 20 times per version and post the counts.
2
Sea Turtle Nest Simulation: Screens or Lighting Rules?
20 minPairs
Pairs turn invented beach data into the rules of a dice simulation, then use it to test two solutions for a threatened sea turtle: wire screens over nests that keep raccoons out, and a town rule that turns off beachfront lights that lead hatchlings away from the sea.
Beach Data and Rules (invented)
Last season, of 100 nests, 35 were dug up by raccoons, 15 failed because most hatchlings crawled toward lights, and 50 sent hatchlings to the sea. Pairs build the rules for a 20-sided die: 1-7 raccoons, 8-10 lights, 11-20 success. In the model, screens cut raccoon losses to a roll of 1 only, and the lighting rule cuts light losses to a single number.
Procedure
Write the rule table for each version: no solution, screens only, lighting only, both.
Roll the die 20 times per version (one roll per nest) and record the number of successful nests.
Compare your counts with the expected counts from the rules: 10, 16, 12 and 18 successful nests out of 20.
Discussion Questions
If the town can pay for only one solution this year, which one gives more successful nests in the expected values, and did your 20 rolls agree?
Why can a run of 20 nests differ from the expected count? How many runs would you want before advising the town?
The model counts each failed nest as a total loss. How would you revise it if lights led only part of each nest's hatchlings astray?
Challenge Variation
Revise the simulation to add a storm rule: before rolling for each nest, flip a coin, and on heads roll a second die where 1-3 means the nest is washed out. Decide whether screens or lighting still gives the better result, and explain why the storm rule affects both solutions the same way.
3
Simulation Critique Gallery: Find the Flaw, Revise the Model
20 minGroups of 3-4
Three posters each show an invented simulation that a student team used to argue for a conservation solution. Each has one flaw. Groups rotate every 5 minutes, name the flaw, write a revised rule, and then fully revise one simulation of their choice.
Simulation Cards
Card A, river otters after a pollution cleanup: 60 otters, Nnext = 1.15 × N for 25 years. The team predicts about 1,975 otters on a 40 km stretch of river and claims the cleanup is a complete success.
Card B, frog ponds and a pesticide ban: the team rolled a die once for each of 10 ponds using the survival rule with the ban in place, got 7 surviving ponds, and claims the ban saves 7 ponds.
Card C, island foxes and predator control: 15 breeding foxes, stable in number for 20 generations. The team claims the population is safe without new foxes, but their model tracks only the number of foxes.
Discussion Questions
Card A predicts about 49 otters per kilometer of river. Why is that not believable, and what rule is missing?
Which runs are missing from the Card B test, and why can its result not show an effect of the ban?
Using Hnext = H × (1 - 1/(2N)), Card C's foxes keep only about 0.51 of their heterozygosity after 20 generations. What does that mean for the team's claim, and what solution could they add to the model?
Modification for Distance Learning
Post the three cards on a shared slide deck. Groups work in breakout rooms, add a comment with the flaw and a revised rule to each card, and upload one fully revised simulation as a spreadsheet.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Revised Salamander Simulation With and Without Road Tunnels
Invented simulation from Example 3, drawn to scale. Both runs start with 800 salamanders and a growth factor of 1.10 without the road. With 15% of adults killed on the road each spring (dashed), the revised factor is 0.935 and the population falls to about 467 after 8 years. With tunnels and fencing that cut road deaths to 3% (solid), the factor is 1.067 and the population reaches about 1,344.
Diagram 2: Simulated Loss of Genetic Variation, Isolated Versus Joined by a Corridor
Invented simulation from Example 2, drawn to scale, using Hnext = H × (1 - 1/(2N)) with a starting heterozygosity of 0.60. The isolated population of 20 breeding adults (dashed) falls to about 0.22 after 40 generations. Joined by a corridor to form 50 breeding adults (solid), it keeps about 0.40. Dots mark every 10 generations.
04
Homework Assignment
~30 min
HS-LS4-6 Homework: Build, Run and Revise Conservation Simulations
Directions: All populations and numbers are invented for practice. Show each step of every simulation in a table, round populations to the nearest whole animal, and end every problem with a one- or two-sentence claim that cites the simulation output.
Part 1: Running a Simulation (Problems 1-2)
A regional manatee population of 1,200 has a growth factor of 1.04 per year without boats, and boat strikes kill 70 manatees a year. A proposed slow-speed zone would cut strikes in half. (a) Simulate year 1 with and without the zone. (b) How many fewer strikes per year would be needed just to stop the decline? (c) Does the zone meet the goal of a growing population?
An island fox population has 40 breeding adults and a heterozygosity of 0.50. (a) Use Hnext = H × (1 - 1/(2N)) to predict H after 20 generations. (b) A captive-breeding plan could raise the number of breeding adults. Find the smallest number of breeding adults that keeps at least 90% of the starting heterozygosity after 20 generations. (c) Name one other species for which the same rule could test a solution.
Part 2: Revising a Simulation (Problems 3-4)
A first model of 5,000 bats near a new wind farm used a growth factor of 1.03 and predicted growth. Surveys show the turbines kill 4% of the bats each year. (a) Revise the model: find the new yearly growth factor and the population after 10 years. (b) Turning the blades off on low-wind nights would cut turbine deaths by 60%. Revise again and find the population after 10 years. (c) Does the solution meet the goal of no decline?
A student models a butterfly population of 10,000 with a growth factor of 1.025 every year, the average of a good-year factor of 1.30 and a bad-year factor of 0.75 caused by pesticide drift. (a) What does the student's model predict after two years? (b) Revise the model so that one good year and one bad year happen, in either order. What is the population after two years? (c) Planting milkweed field borders would raise the bad-year factor to 0.85. Run the revised model again and explain why using the average factor was misleading.
Part 3: Designing and Defending a Solution (Problems 5-6)
A seed bank plans to sample 15 plants each of two wildflower species. A rare allele has frequency 0.10 in species A and 0.02 in species B. (a) Use 1 - (1 - p)2n to find the chance of storing at least one copy of each rare allele. (b) How many plants of species B must be sampled for at least a 95% chance? (c) Write a sampling plan for both species and justify it with your numbers.
Groundwater pumping threatens an endangered desert pupfish. A simulation run 1,000 times gives a 0.20 chance that a single large refuge pond is lost within 50 years, and a 0.30 chance for each of three smaller spring pools. (a) If the pools are lost independently, what is the chance that all three are lost? (b) Which option does the simulation favor? (c) Pumping lowers the water table under all three springs at once. Explain how that breaks the model's assumption and how you would revise the simulation before advising the managers.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Running the Simulation
Every step shown in a table, rules applied in order, values carried forward correctly
Correct rules with one arithmetic or carrying error
Revising the Model
Names the missing factor, writes it as a rule and shows how it changes the output
Names the missing factor but does not change the rule or rerun the model
Testing the Solution
Compares runs with and without the solution against a stated goal
Runs the solution but does not compare or state the goal
Claim, Evidence, Reasoning
Claim cites specific output and names one limit of the simulation
Claim stated but evidence is general or the limit is missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All populations and model numbers are invented. Hnext = H × (1 - 1/(2N)) is the average loss of heterozygosity per generation for N breeding adults. 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
Why would a conservation team run a simulation before building a solution such as a wildlife tunnel?
Answer: B
A simulation applies stated rules to show what follows from them, so it lets the team compare runs with and without the solution before spending money. Choice A overstates it: the output is only as good as the rules. Choice C is backwards, because the rules come from field data.
Question 2 of 20 · Multiple Choice
A rhino population of 500 has a growth factor of 1.06 per year, and poachers kill 40 rhinos a year. What does the simulation give for year 1?
Answer: C
500 × 1.06 = 530, then 530 - 40 = 490, so the population shrinks. Choice A leaves out the poaching. Choice B subtracts the poaching but leaves out the growth, and choice D assumes growth and poaching cancel.
Question 3 of 20 · Multiple Choice
A shark population of 3,000 has a growth factor of 1.05 per year, and 200 sharks a year die as bycatch. By what percent must bycatch be cut so the population does not decline in year 1?
Answer: A
Growth adds 3,000 × 0.05 = 150 sharks, so bycatch must fall from 200 to 150 or less, a cut of 50 ÷ 200 = 25%. Choice B cuts 100 sharks, more than needed. Choice D divides 50 by 150 instead of by the original 200.
Question 4 of 20 · Multiple Choice
A population of 10 breeding adults has heterozygosity 0.50. What does the simulation rule predict after one generation?
Answer: D
1 - 1/(2 × 10) = 0.95, and 0.50 × 0.95 = 0.475. Choice A uses 1 - 1/N and forgets the 2 (each adult carries two gene copies). Choice C uses 1 - 1/100, and choice B halves H.
Question 5 of 20 · Multiple Choice
Which population does the heterozygosity rule predict will lose genetic variation fastest?
Answer: A
The fraction lost each generation is 1/(2N): 1/24 for 12 adults, 1/120 for 60 and 1/600 for 300. The isolated group of 12 loses the most. Choice C has the same small group, but the corridor makes it part of a much larger population and brings in new gene copies.
Question 6 of 20 · Multiple Choice
A team uses dice to simulate nest success. Their single run with a solution gave 14 successful nests, and a single run without it gave 13. What should they do before making a claim?
Answer: C
Random simulations give different results each run, so one run of each cannot separate the solution's effect from chance. Choice A treats one run as proof. Choice B changes the model to fit a wanted answer. Choice D is wrong: repeated runs make random models useful.
Question 7 of 20 · Multiple Choice
A student's simulation of a restored wetland predicts that a frog population will keep growing by 20% every year forever. Which revision makes the model more realistic?
Answer: B
Real habitats support a limited number of animals, so a rule such as N + rN(1 - N/K) slows growth as N nears the carrying capacity K. Choices A, C and D keep the unlimited growth rule, so they keep the flaw.
Question 8 of 20 · Multiple Choice
A toad population has a growth factor of 1.08 without roads. A new highway kills 10% of the adults each year. What is the revised yearly growth factor?
Answer: D
Survivors are 90% of the population, so the factor is 1.08 × 0.90 = 0.972, and the population declines. Choice A subtracts 0.10 from 1.08. Choice B adds the loss instead of removing it, and choice C multiplies 1.08 by 0.10, the fraction killed.
Question 9 of 20 · Multiple Choice
A rare allele has frequency 0.05 in a wild grass. A seed bank samples 6 plants (12 gene copies). About what is the chance of storing at least one copy?
Answer: A
The chance of missing it in all 12 copies is 0.9512 ≈ 0.54, so the chance of catching at least one is about 1 - 0.54 = 0.46. Choice C is the chance of missing it. Choice B multiplies 0.05 by 6 plants, which ignores that each plant has two copies and that chances do not simply add. Choice D is the allele frequency itself.
Question 10 of 20 · Multiple Choice
Which simulation result would best support the claim that a corridor protects genetic variation in a small population?
Answer: C
A test compares the solution with no solution, uses the output that matches the claim (heterozygosity) and uses many runs. Choice A is one run and measures numbers, not variation. Choice B has nothing to compare with, and choice D is not simulation output.
Question 11 of 20 · Multiple Choice
A 20-sided die simulates plover nests: 1-6 means a predator takes the nest, 7-8 means a high tide floods it, and 9-20 means it hatches. Fenced cages would change the predator rule to 1-2 only. How many of 40 nests are expected to hatch with the cages?
Answer: B
With cages, 2 numbers are predators and 2 are floods, so 16 of 20 numbers mean hatching: 16/20 × 40 = 32. Choice A is the expected count without cages (12/20 × 40). Choice C forgets the flooding rule.
Question 12 of 20 · Multiple Choice
Which change to a model of a forest bird population represents a solution rather than a human impact?
Answer: D
Replanting restores habitat, so it raises the carrying capacity: that is a mitigation. Choices A, B and C are the harms the solution would need to offset.
Question 13 of 20 · Multiple Choice
A simulation of a desert tortoise population uses the same growth factor every year, but rainfall in the real habitat varies a lot from year to year. Which revision best represents this?
Answer: C
Drawing the factor at random from real wet-year and dry-year values keeps the variation, and many runs show the range of outcomes. Choices A and B throw away the dry years that matter most for survival, and choice D does not change the rule at all.
Question 14 of 20 · Multiple Choice
A seed bank samples 10 plants of each of two species. Species X has a rare allele at frequency 0.20 and species Y has one at frequency 0.01. Which statement does the sampling rule 1 - (1 - p)2n support?
Answer: B
For X: 1 - 0.8020 ≈ 0.99. For Y: 1 - 0.9920 ≈ 0.18. The rarer allele needs far more plants. Choice C reverses the relationship, and choice D ignores that 2n is in the exponent.
Question 15 of 20 · Short Answer
A whooping crane flock of 400 has a growth factor of 1.03 per year, and collisions with power lines kill 20 cranes a year. Marking the lines with bird diverters would cut collisions to 6 a year. Simulate year 1 with and without the diverters and say whether each population grows.
400 × 1.03 = 412. Without diverters: 412 - 20 = 392, so the flock shrinks. With diverters: 412 - 6 = 406, so it grows. Growth adds 12 cranes a year, so any solution must keep collisions below 12.
Question 16 of 20 · Short Answer
A population of 25 breeding adults has heterozygosity 0.70. Simulate H after 5 generations. Then repeat for a translocation plan that raises the population to 100 breeding adults, and compare.
N = 25: H = 0.70 × (1 - 1/50)5 = 0.70 × 0.985 ≈ 0.63. N = 100: H = 0.70 × 0.9955 ≈ 0.68. The larger population loses about a quarter as much variation per generation, so the plan keeps more of it.
Question 17 of 20 · Short Answer
A model of 200 wild bee colonies uses a growth factor of 1.05 but ignores pesticides. Field data show pesticide exposure kills 12% of colonies each year. Pesticide-free buffer strips would cut this loss to 4%. Revise the model, give both new growth factors and the colonies after one year, and say whether the buffers work.
Revised factor with pesticides: 1.05 × 0.88 = 0.924, about 185 colonies after one year (decline). With buffers: 1.05 × 0.96 = 1.008, about 202 colonies (slow growth). The buffers turn a decline into slight growth, so they meet a no-decline goal, though only just.
Question 18 of 20 · Short Answer
Plan a simulation test of a wildlife overpass for a deer herd split by a highway. Name the rule you would change, what you would keep the same, the output you would measure and how many runs you would do.
A strong answer: change the road-death rule (for example, the fraction killed each year drops when the overpass is used) and, if genetic variation is included, treat the two halves as one larger population. Keep the starting herd, growth factor and number of years the same in both versions. Measure herd size (and heterozygosity) after a set number of years. If the model has random events, run each version many times, for example 50, and compare averages and the fraction of runs below a danger level.
Question 19 of 20 · Short Answer
A fishery model uses Nnext = N + 0.30N(1 - N/6,000) - hN. Find the population where the model stops changing for a harvest fraction h = 0.10 and for h = 0.20, and say which rule better protects the fish.
The population stops changing when 0.30(1 - N/6,000) = h. For h = 0.10: N = 6,000 × (1 - 1/3) = 4,000. For h = 0.20: N = 6,000 × (1 - 2/3) = 2,000. The lower harvest keeps twice as many fish, so it better protects the population.
Question 20 of 20 · Short Answer
Feral cats threaten a lizard population of 200 on an island. Option A removes the cats, raising the growth factor from 0.92 to 1.04. Option B leaves the cats and releases 30 captive-bred lizards a year. Simulate 3 years for each option, then give one reason from the model to prefer Option A anyway.
Option A: 200 × 1.043 ≈ 225. Option B: 200 → 214 → 226.88 → about 239. Option B is ahead after 3 years, but it levels off at 30 ÷ 0.08 = 375 and only while releases continue forever. Option A removes the cause, so the population keeps growing until the habitat is full, without depending on captive breeding (which can also narrow genetic variation).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-LS4-6 mean?
HS-LS4-6 asks students to create or revise a simulation that tests a solution to harm people cause to biodiversity. Students set up rules for a population (growth, the human impact, the solution), run them, and judge whether the solution meets a goal. The code means High School, Life Science, core idea 4 (Biological Evolution: Unity and Diversity), performance expectation 6.
Is HS-LS4-6 taught in biology or environmental science?
It is usually taught in high school Biology, often in the evolution or ecology unit, and it also fits Environmental Science. The math is multiplying by a growth factor, percent change and simple probability, so Algebra I is enough.
Does HS-LS4-6 have an assessment boundary?
No. HS-LS4-6 has no official Assessment Boundary. The Clarification Statement puts the emphasis on designing solutions for problems with threatened or endangered species, or with genetic variation of organisms for multiple species, and this lesson keeps its tasks there.
What counts as a simulation for this standard?
Any model that applies rules step by step to show what happens over time. That includes a table worked by hand, a bead or dice activity with random draws, a spreadsheet, or a ready-made computer model. What matters is that students can change a rule to represent a solution and compare the outputs.
Do students need to know how to code?
No. Hand tables, beads, dice and spreadsheets all meet the standard. Coding is a good option for students who already know it, but the science practice is choosing the rules, running them and judging the result.
What is the difference between creating and revising a simulation?
Creating means choosing the rules and numbers yourself from data, as in turning beach nest data into dice rules. Revising means adding something a model left out, such as a road-death rule, a carrying capacity, random good and bad years or genetic variation, and running it again to see whether the conclusion changes.
Why does genetic variation matter in HS-LS4-6?
A small population can hold steady in number while it loses genetic variation, which leaves it less able to adapt to disease or a changing climate. The rule Hnext = H × (1 - 1/(2N)) lets students simulate that loss and test solutions such as corridors, moving animals between populations, or seed banks that store variation for many plant species.
What are common mistakes students make with population simulations?
A common one is applying the growth factor to the starting value every year instead of to last year's result. Others are adding a percent loss instead of multiplying by the survival fraction, trusting a single run of a random model, and treating the output as a certain prediction instead of a consequence of the model's assumptions.
How can teachers assess HS-LS4-6?
Give students a simulation with a flaw or a missing solution and ask them to revise it, run it, and write a claim-evidence-reasoning argument from the output. Score whether they compare runs with and without the solution against a goal and whether they name a limit of the model. The rubric on this page follows that approach.
How does HS-LS4-6 connect to other NGSS standards?
It builds on MS-LS2-5, where students evaluate competing solutions for biodiversity. It sits alongside HS-LS2-7, which designs and refines solutions for human impacts, and HS-ETS1-4, which uses computer simulations to test solutions. HS-ESS3-3 extends the idea to simulations that link resource management, human populations and biodiversity.
07
Related Standards
6 standards
These standards connect to HS-LS4-6: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-LS2-5Prerequisite
Evaluate competing design solutions for maintaining biodiversity and ecosystem services
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