HS-LS3-3: Probability and Statistics of Trait Variation in Populations
In plain English: HS-LS3-3 is the NGSS high school performance expectation that asks students to apply statistics and probability to explain the variation and distribution of expressed traits in a population. Students use Punnett squares and probability rules, summarize trait data with center, spread and histograms, and separate genetic from environmental effects. It is usually taught in Biology.
Apply concepts of statistics and probability to explain the variation and distribution of expressed traits in a population.
Clarification Statement: Emphasis is on the use of mathematics to describe the probability of traits as it relates to genetic and environmental factors in the expression of traits.
Assessment Boundary: Assessment does not include Hardy-Weinberg calculations.
Next Generation Science Standards · Disciplinary Core Idea: Heredity: Inheritance and Variation of Traits · Official standard
Students use probability and statistics to explain why individuals in a population differ in a trait and how the trait is spread across the population. They predict offspring ratios with Punnett squares and the product rule, compare predictions with observed counts, and see why small families rarely match the expected ratio. They then move from single genes to traits controlled by several genes, where the number of contributing alleles follows a bell-shaped distribution, and they use means, spread and dot plots to separate the effect of genes from the effect of the environment.
The NGSS Clarification Statement says: "Emphasis is on the use of mathematics to describe the probability of traits as it relates to genetic and environmental factors in the expression of traits." The Assessment Boundary says: "Assessment does not include Hardy-Weinberg calculations." So every probability on this page comes from crosses between known parents or from sample data, and allele-frequency equilibrium equations are beyond this standard. All data sets are invented for teaching, with values in realistic ranges.
Learning Objectives
By the end of this lesson, students will be able to:
Use Punnett squares and the product rule to find the probability of genotypes and phenotypes in offspring
Compare observed offspring counts with expected counts and explain differences in terms of chance and sample size
Explain why traits controlled by several genes show a continuous, bell-shaped distribution in a population
Use means, spread and dot plots or histograms to describe the variation of a trait
Use data to separate genetic and environmental effects on the expression of a trait
Prior Knowledge Required
Students should already be comfortable with:
Genes and alleles on paired chromosomes, one from each parent HS-LS3-1
Probability of compound events with lists, tables and tree diagrams 7.SP.C.8
Mean, median and measures of spread for a data set 6.SP.B.5
Dominant and recessive alleles and simple Punnett squares
Put two questions on the board and have students answer them on their own before discussing.
Warm-Up Prompt
"Two parents who both carry one allele for a recessive trait have four children. A classmate predicts that exactly one child will show the trait. Is the prediction guaranteed? Then: why are there many different human heights, not just 'tall' and 'short'?"
Collect a few answers. Many students expect exactly one in four; others mention nutrition or "many genes" for height. Tell students they will test both ideas with probability and data.
Direct Instruction15-20 minutes
Probability of traits from one gene. Each parent passes one of its two alleles, each with probability 1/2, so a Punnett square lists four equally likely outcomes. For two genes on different chromosome pairs, multiply the probabilities (the product rule). An expected ratio describes a large number of offspring; small groups vary by chance.
Distributions of traits. A trait controlled by one gene with complete dominance falls into two clear classes. A trait controlled by several genes, each adding a small amount, gives many classes that form a bell shape (Diagram 1). The environment (food, light, temperature) shifts and spreads the distribution, so genetically identical individuals still vary (Diagram 2). Summaries such as the mean, the range and the standard deviation describe the variation.
One gene, complete dominance
Two tall pea plants, both Tt, produce 400 seeds that grow into plants. Observed: 291 tall and 109 short.
Result: Punnett square: TT, Tt, Tt, tt, so P(tall) = 3/4 and P(short) = 1/4. Expected: 300 tall and 100 short. The observed count of short plants is 9 more than expected, a 9% difference from 100, the kind of gap chance can produce in 400 offspring.
One gene, incomplete dominance
In snapdragons, red (RR) × white (WW) gives pink (RW). Two pink plants are crossed and 160 seeds grow.
Result: P(RR) = 1/4, P(RW) = 1/2, P(WW) = 1/4, so the expected distribution is 40 red, 80 pink and 40 white: three classes instead of two.
Two genes, product rule
In guinea pigs, black coat (B) is dominant to brown (b) and short hair (S) is dominant to long (s). The genes are on different chromosome pairs. Two BbSs guinea pigs mate.
Result: P(black) = 3/4 and P(long) = 1/4, so P(black and long) = 3/4 × 1/4 = 3/16.
Several genes (polygenic trait)
Kernel color in a grain is set by three genes, and each dark allele (A, B or C) adds the same amount of pigment. Two AaBbCc plants are crossed.
Result: Each offspring gets 6 alleles, each dark with probability 1/2. P(k dark alleles) = C(6, k)/64: 1, 6, 15, 20, 15, 6, 1 out of 64. The middle class (3 dark) is 20/64 ≈ 31%, and each extreme is only 1/64 ≈ 1.6%.
Same genes, different environment
Invented data: 20 cuttings from one basil plant (genetically identical) grow for 6 weeks, 10 with fertilizer and 10 without (Diagram 2).
Result: Mean height: 42.0 cm with fertilizer and 30.5 cm without, a difference of 11.5 cm. The standard deviation is about 2.2 cm in each group. Because the plants share the same genes, all of this variation comes from the environment.
After each example, ask: "Is this a prediction (probability) or a description of data (statistics), and which factor, genes or environment, explains the variation?"
Guided Practice15 minutes
Pairs compare invented kernel-color data from 640 offspring of the AaBbCc × AaBbCc cross in the fourth example with the expected distribution in Diagram 1.
Invented kernel-color data, 640 offspring
Dark alleles (color class)
0
1
2
3
4
5
6
Observed
13
55
162
191
144
66
9
Expected
10
60
150
200
150
60
10
Check the expected row: multiply 640 by 1/64, 6/64, 15/64 and so on.
Sketch a histogram of the observed counts. Describe its shape, center and spread.
Find the largest difference between observed and expected. Is it large compared with the count itself?
Write two sentences: one that uses probability to explain the shape, and one that names a factor that could spread the colors further in a real field.
Independent Practice10-15 minutes
Each student analyzes invented data from a field trial of two corn lines: a tall line and a dwarf line, each grown in well-watered and dry plots. Values are mean plant heights at harvest.
Invented corn trial: mean height (cm)
Line
Well-watered
Dry
Tall line
210
160
Dwarf line
95
70
Students compute the genetic effect (difference between lines) in each plot, the environmental effect (difference between plots) for each line, and the average of each. Then they write a claim about which factor explains more of the height variation in this trial, and explain why the answer could change in a different environment.
Closure5 minutes
Exit ticket: "Return to the warm-up. Explain in two sentences why the four-child family may not have exactly one child with the trait, and why human height shows many values rather than two." Collect answers and read two aloud to start the next class.
Differentiation Strategies
For Struggling Students
Give blank Punnett squares and a probability ladder (1/4, 1/2, 3/4, 1) to place each result on
Start the polygenic idea with two genes (4 alleles, 5 classes) and have students list all 16 outcomes before using the counting pattern
Provide a data-summary frame: "The center is about ___, the values range from ___ to ___, so the trait varies by ___."
For Advanced Students
Have students compute a chi-square statistic for the kernel-color data and interpret it with a critical-value table (an extension beyond the minimum of this standard)
Ask students to simulate 100 families of four children in a spreadsheet and compare the share of families with exactly one affected child with the probability from the binomial formula
Have students design a study that separates genetic and environmental effects on a trait in plants, naming the controls and the summary statistics they would report
Assessment Guidance
What to Look For
Check that students multiply probabilities only for independent events and add them for outcomes that exclude each other. Look for comparisons of observed and expected counts that mention sample size, and for descriptions of distributions that give center and spread with units. Listen for students who treat a 3 : 1 ratio as a guarantee for each family, who think a bell-shaped trait must come from one gene, or who assume all variation is genetic.
02
Classroom Activities
3 Activities
1
Coin-Flip Offspring
20 minPairs
Each partner plays one Aa parent and flips a coin to choose the allele passed on (heads A, tails a). Pairs make 40 offspring, record genotypes, and then pool the class results to see how the distribution changes with sample size.
Procedure
Both partners flip at the same time; record the two letters as one offspring genotype.
Make 40 offspring. After every 10, record the running counts of AA, Aa and aa.
Add your pair's 40 results to the class table and compute the class percentages.
Sample Class Results (invented)
Fourteen pairs made 560 offspring: 136 AA, 285 Aa and 139 aa. One pair's first 10 offspring were 1 AA, 5 Aa and 4 aa.
Discussion Questions
The expected class counts are 140, 280 and 140. How close did the class come, in percent?
One pair had 4 aa in its first 10, which is 40% instead of 25%. Why is this not evidence against the Punnett square?
What fraction of the class offspring showed the dominant phenotype? Compare it with 3/4.
Modification for Distance Learning
Students use an online coin flipper or a spreadsheet random function and enter their 40 results in a shared class table.
2
Polygenic Bead Draw
20 minGroups of 3
Groups model a trait controlled by three genes. Six cups stand for the six alleles an offspring receives (three from each parent); each cup holds 1 dark and 1 light bead. Drawing one bead from each cup makes one offspring, and the number of dark beads is its color class. Groups then add an environment rule and see how the distribution changes.
Procedure
Without looking, draw one bead from each of the 6 cups. Record the number of dark beads (0-6), then return the beads.
Make 50 offspring and draw a histogram of the color classes.
Environment rule: for each offspring, roll a number cube. A 1 moves it one class lighter, a 6 one class darker (staying within 0-6), and 2-5 leaves it unchanged. Redraw the histogram.
Discussion Questions
Which class appeared most often, and why is it the one with 20 of the 64 possible outcomes?
How did the environment rule change the spread of the histogram?
If a trait had 10 genes instead of 3, how would the histogram look?
Challenge Variation
Groups predict how many of 50 offspring should land in class 0 or class 6 (together 2/64 of the outcomes, about 1.6) and explain why their own count might be 0 or 3.
3
Measure a Trait: Hand Span
25 minGroups of 4, then whole class
Students measure the hand span of their dominant hand (tip of thumb to tip of little finger, fingers spread) to the nearest 0.1 cm, pool the class data, compute summary statistics and build a histogram. They then discuss which factors, genetic and environmental, could explain the spread.
Mean 20.1 cm, median 20.0 cm, range 17.5-23.1 cm, sample standard deviation about 1.4 cm.
Discussion Questions
Make a histogram with 1 cm bins starting at 17 cm. Is it roughly symmetric?
How many of the 24 values lie within one standard deviation of the mean (between about 18.7 and 21.5 cm)? Compare with the two-thirds expected for a bell-shaped distribution.
List two genetic and two environmental factors that could contribute to the spread. Which data would help tell them apart?
Modification for Distance Learning
Students measure a family member or use the sample data, enter values in a shared spreadsheet, and use its functions for the mean and standard deviation.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Expected Distribution of a Three-Gene Trait
Expected share of offspring from AaBbCc × AaBbCc in each color class, drawn to scale. Each class holds C(6, k) of the 64 equally likely outcomes, so the distribution is symmetric and bell-shaped, with the middle class (3 dark alleles) the largest. With more genes and some environmental effect, the classes blend into a continuous range.
Diagram 2: Genetically Identical Plants in Two Environments
Invented heights of 20 basil cuttings from one plant, drawn to scale: 10 grown without fertilizer and 10 with fertilizer. The plants share the same genes, so both the shift in the mean (30.5 cm to 42.0 cm) and the spread within each group come from the environment.
04
Homework Assignment
~30 min
HS-LS3-3 Homework: Explaining Trait Variation With Probability and Data
Directions: All data sets are invented for practice. Show each Punnett square or calculation, give probabilities as fractions and decimals, and end each problem with one or two sentences that explain the variation in the trait.
Part 1: Probability of Traits (Problems 1-3)
Two parents are both carriers (Cc) of the recessive allele for cystic fibrosis. (a) What is the probability that a child has cystic fibrosis (cc)? (b) That a child is a carrier? (c) What is the probability that none of their 4 children has cystic fibrosis? (d) Explain why one family with 4 children could have no affected children and another could have 2.
In cattle, red (RR) and white (WW) coats are codominant, and RW cattle are roan. A roan bull is mated with white cows and 30 calves are born: 13 roan and 17 white. (a) What are the expected numbers? (b) Is the observed result reasonable? Explain with sample size.
In tomatoes, tall (D) is dominant to dwarf (d) and red fruit (R) is dominant to yellow (r), on different chromosome pairs. A DdRr plant is crossed with a ddrr plant and 240 seedlings grow. (a) List the four phenotypes and the probability of each. (b) How many dwarf, red-fruited plants are expected?
Part 2: Distributions, Genes and Environment (Problems 4-6)
Invented data: a flower-size trait is controlled by two genes with additive alleles. From AaBb × AaBb, 320 offspring fell into classes of 0, 1, 2, 3 and 4 large-size alleles with counts 17, 86, 114, 83 and 20. (a) List the 16 equally likely allele combinations, or use a counting pattern, to find the probability of each class, then compute the expected counts. (b) Describe the shape of the distribution. (c) Explain why a single gene could not produce five classes.
Invented data: Himalayan rabbits (genotype chch) were raised at two temperatures. Rabbits raised at 15 °C had dark fur on an average of 18% of the coat; rabbits raised at 25 °C averaged 7%. Rabbits with genotype CC had 100% dark fur at both temperatures. (a) How many times larger is the dark area at 15 °C? (b) Which factor explains the variation among chch rabbits, and which explains the difference between chch and CC rabbits?
In Labrador retrievers, consider only the B gene: black (B) is dominant to chocolate (b). A Bb dog is mated with a bb dog and has a litter of 6 puppies. (a) What is the probability that a puppy is chocolate? (b) What is the probability that exactly 3 of the 6 are chocolate? (c) What is the probability that none is chocolate? (d) What do your answers say about how often a litter matches the expected ratio exactly?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Probability
Correct Punnett squares, product rule and fractions for every cross
Correct method with one calculation error
Data Analysis
Expected counts, comparisons and summary statistics correct, with units
Comparisons made but expected counts or statistics partly wrong
Explanation of Variation
Explains the variation with chance, sample size, number of genes and environment as each applies
Explanation names a cause but does not connect it to the data
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All data sets are invented. Give probabilities as fractions unless a question asks for a decimal. 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
One parent is AA and the other is Aa. What is the probability that a child is AA?
Answer: B
The AA parent always passes A, and the Aa parent passes A with probability 1/2, so the Punnett square gives AA, AA, Aa, Aa and P(AA) = 1/2. Choice A applies the Aa × Aa result to a different cross. Choice D confuses the genotype AA with the dominant phenotype, which every child shows.
Question 2 of 20 · Multiple Choice
A parent who is Aa has children with a parent who is aa. What is the probability that a child shows the dominant phenotype?
Answer: D
The children are Aa or aa with equal chance, so P(dominant) = 1/2. Choice B applies the Aa × Aa result to a different cross, and choice C assumes the dominant allele always gets passed on.
Question 3 of 20 · Multiple Choice
In snapdragons, red (RR) × white (WW) plants give pink (RW) flowers. What share of the offspring of an RR plant and a WW plant will be pink?
Answer: C
Every offspring gets R from one parent and W from the other, so all are RW, which is pink. Choice B is the share of pink from RW × RW. Choice D assumes one color must dominate.
Question 4 of 20 · Multiple Choice
Two AaBb parents mate, and genes A and B are on different chromosome pairs. What is the probability that an offspring shows both dominant phenotypes?
Answer: A
P(A_) = 3/4 and P(B_) = 3/4, and the product rule gives 3/4 × 3/4 = 9/16. Choice B uses only one gene. Choice C is P(aabb), and choice D is the probability of one dominant and one recessive phenotype in a single order.
Question 5 of 20 · Multiple Choice
A trait is controlled by two genes, each with an allele that adds the same amount. From AaBb × AABB, what fraction of offspring have exactly 3 of the adding alleles?
Answer: C
The AABB parent always passes A and B, 2 adding alleles. The AaBb parent passes AB, Ab, aB or ab with equal chance, adding 2, 1, 1 or 0 more, so offspring have 4, 3, 3 or 2 adding alleles and P(exactly 3) = 2/4 = 1/2. Choice A counts only one of the two ways (Ab or aB). Choice B treats the three classes (2, 3 and 4 adding alleles) as equally likely. Choice D is the chance of at least 3 adding alleles.
Question 6 of 20 · Multiple Choice
Height in a population of plants shows a smooth, bell-shaped distribution with many values. Which explanation fits best?
Answer: B
Many additive genes produce many classes, and the environment blurs them into a smooth range. Choice A would give two clear classes. Choice C ignores that tall and short parents tend to have tall and short offspring.
Question 7 of 20 · Multiple Choice
Genetically identical cuttings from one plant are grown in one greenhouse, yet their heights after 5 weeks vary from 18 to 25 cm. What causes this variation?
Answer: C
The cuttings share the same DNA, so their differences come from the environment. Choices A and B need genetic differences, which clones do not have.
Question 8 of 20 · Multiple Choice
Invented data: five plants have heights of 40, 44, 46, 50 and 55 cm. What is the mean height?
Answer: B
The sum is 235 cm, and 235 ÷ 5 = 47 cm. Choice A is the median, not the mean. Choice C is the range (55 - 40), and choice D is the sum.
Question 9 of 20 · Multiple Choice
Invented data: Group A leaf lengths are 10, 11, 12, 13 and 14 cm; Group B are 6, 9, 12, 15 and 18 cm. Which statement is correct?
Answer: D
Both means are 12 cm, but Group B ranges over 12 cm and Group A over 4 cm, so Group B varies more (its standard deviation is 3 times as large). Choice A ignores that the sums are equal. Choice C reverses the spread.
Question 10 of 20 · Multiple Choice
A cross of Aa × Aa gives 8 offspring, and 4 of them show the recessive trait instead of the expected 2. What is the best conclusion?
Answer: B
With only 8 offspring, chance alone can easily give 4 recessive. Choices A and C reject the model on too little data, and choice D is ruled out because aa × aa parents could not have any offspring with the dominant trait, yet 4 of the 8 show it.
Question 11 of 20 · Multiple Choice
Two carrier parents (Aa × Aa) have two children. What is the probability that at least one child shows the recessive trait?
Answer: A
P(neither shows it) = 3/4 × 3/4 = 9/16, so P(at least one) = 1 - 9/16 = 7/16. Choice D is the probability that neither does. Choice C is the probability that both do, and choice B adds 1/4 + 1/4, which counts the case where both show it twice.
Question 12 of 20 · Multiple Choice
Hydrangea plants of the same variety have blue flowers in acidic soil and pink flowers in alkaline soil. What does this show?
Answer: D
The same variety, with the same genes, gives different colors depending on soil, so the environment affects expression. Choice B contradicts "same variety." Choice C is wrong because the color changes back when the soil changes, which mutations would not do.
Question 13 of 20 · Multiple Choice
Siamese cats have dark fur on the ears, paws and tail, the coolest parts of the body. Which explanation is best supported?
Answer: B
The allele codes for a pigment enzyme that works only in cooler skin, so temperature, an environmental factor, controls where the trait appears. Choice A is wrong because all body cells carry the same alleles. Choices C and D describe purposes or learning, which do not explain how the pattern is produced.
Question 14 of 20 · Multiple Choice
From Bb × Bb, 480 offspring are expected. About how many are expected to be bb?
Answer: C
P(bb) = 1/4, and 480 × 1/4 = 120. Choice A uses 1/2 (the Bb share), choice B uses 3/4 (the dominant phenotype), and choice D divides by 3, a sign of reading 3 : 1 as "one third."
Question 15 of 20 · Short Answer
Invented data: a cross of two heterozygous purple-flowered plants gave 152 purple and 48 white offspring. (a) Find the expected counts. (b) By what percent does the white count differ from the expected count? (c) Do the data support a 3 : 1 prediction?
(a) Expected: 200 × 3/4 = 150 purple and 200 × 1/4 = 50 white. (b) (50 - 48) ÷ 50 = 4%. (c) Yes: a difference of 2 plants out of 50 is small for a sample of 200 and can be explained by chance.
Question 16 of 20 · Short Answer
A trait is controlled by four genes, each with an allele that adds the same amount, so an offspring of AaBbCcDd × AaBbCcDd receives 8 alleles. (a) What is the probability that it gets all 8 adding alleles? (b) Exactly 4? (c) What do these numbers say about the shape of the distribution?
(a) (1/2)8 = 1/256. (b) C(8, 4)/256 = 70/256 ≈ 0.27. (c) The extremes are very rare and the middle class is common, so the distribution is bell-shaped with 9 classes, closer to a smooth, continuous trait than with fewer genes.
Question 17 of 20 · Short Answer
Invented data: mean grain yields (tonnes per hectare) for two wheat varieties with and without nitrogen fertilizer are variety A 6.8 and 4.9, variety B 5.6 and 4.3. Compute the average genetic effect and the average environmental effect, and say which is larger in this trial.
Genetic effect (A minus B): 1.2 with nitrogen and 0.6 without, average 0.9 t/ha. Environmental effect (with minus without): 1.9 for A and 1.3 for B, average 1.6 t/ha. In this trial the fertilizer (environment) explains more of the yield variation than the variety (genes).
Question 18 of 20 · Short Answer
Two carrier parents (Aa × Aa) have 3 children. What is the probability that exactly one child shows the recessive trait? Show the calculation.
Each child has P(aa) = 1/4 and P(not aa) = 3/4. The affected child can be the first, second or third: 3 × 1/4 × (3/4)2 = 3 × 9/64 = 27/64 ≈ 0.42.
Question 19 of 20 · Short Answer
Invented data: seed masses in a bean population were sorted into classes: 0.30-0.39 g (4 seeds), 0.40-0.49 g (15), 0.50-0.59 g (32), 0.60-0.69 g (17) and 0.70-0.79 g (5). (a) How many seeds were weighed? (b) Which class holds the median? (c) Describe the distribution and give one genetic and one environmental reason for the spread.
(a) 4 + 15 + 32 + 17 + 5 = 73 seeds. (b) The median is the 37th seed, which falls in the 0.50-0.59 g class (seeds 20-51). (c) The distribution is roughly symmetric and bell-shaped, centered on 0.50-0.59 g. Genetic reason: several genes affect seed size. Environmental reason: the position of the pod on the plant, water or nutrients.
Question 20 of 20 · Short Answer
Parents who are both Aa have 2 children. What is the probability that exactly one of the 2 shows the recessive trait, and what does the answer say about families that "match" the 1 : 3 expected ratio?
Exactly one: 2 × 1/4 × 3/4 = 6/16 = 3/8. So more than half of such families (5/8) do not have exactly one affected child out of two. Expected ratios describe many offspring, not each family.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-LS3-3 mean?
HS-LS3-3 asks students to use statistics and probability to explain why a trait varies among individuals and how it is distributed in a population. Students predict offspring with probability, describe trait data with center and spread, and explain variation with genes and environment together. The code means High School, Life Science, core idea 3 (Heredity), performance expectation 3.
Does HS-LS3-3 include Hardy-Weinberg?
No. The Assessment Boundary says that assessment does not include Hardy-Weinberg calculations. Students work with probabilities from crosses between known parents and with statistics of trait data. Allele-frequency equilibrium equations are usually taught later, in evolution units or advanced biology courses.
What math do students need for HS-LS3-3?
They need fractions and probability of compound events (7.SP.C.8), the product rule for independent events, and summary statistics such as mean, median, range and standard deviation, with histograms or dot plots (HSS.ID.A.1 and HSS.ID.A.2). A calculator or spreadsheet helps with larger data sets.
Why doesn't a real family match the Punnett square ratio?
A Punnett square gives the probability for each child, not a guarantee for a family. Each child is a separate chance event, so small families often differ from the ratio. The ratio shows up when many offspring are counted, which is why geneticists work with large samples.
What is a polygenic trait?
A polygenic trait is influenced by several genes, often with each contributing a small effect. Human height and skin color and kernel color in wheat are classic examples. The more genes involved, the more classes there are, and the closer the distribution comes to a smooth bell shape.
How do genes and the environment both affect a trait?
Genes set the range of possible outcomes and the environment affects where an individual falls in that range. Nutrition affects human height, and sun exposure affects skin color. Studies of genetically identical plants or animals in different conditions show the environmental part directly.
Should I teach chi-square with HS-LS3-3?
It is optional. The standard asks for statistics and probability, and comparing observed with expected counts is the core idea. Many biology teachers add the chi-square test as an extension for students ready for it; the advanced differentiation on this page includes it.
What are common student mistakes in genetics probability?
Frequent errors are adding probabilities that should be multiplied, reading 3 : 1 as a probability of 1/3, expecting each family to match the ratio, and forgetting that "at least one" is easiest to find as 1 minus the probability of none. Asking students to state which events are independent prevents many of these.
How does HS-LS3-3 connect to natural selection?
HS-LS4-3 asks students to use statistics and probability to show that organisms with an advantageous heritable trait tend to increase in proportion over generations. HS-LS3-3 builds the tools for that argument: describing how a trait is distributed and why individuals differ.
Is HS-LS3-3 taught in biology or in math class?
It is a science standard and is usually taught in Biology, in the genetics unit. It uses math from middle school and Algebra I statistics, so it is a natural place to connect with the statistics standards in math courses.
07
Related Standards
5 standards
These standards connect to HS-LS3-3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.SP.C.8Prerequisite
Find probabilities of compound events using lists, tables, trees and simulation
NGSS is a registered trademark of WestEd. Neither WestEd nor the lead states and partners that developed the Next Generation Science Standards were involved in the production of this product, and do not endorse it.
