HS-LS3-2: Claims and Evidence for the Sources of Genetic Variation
In plain English: HS-LS3-2 is the NGSS high school performance expectation that asks students to make and defend an evidence-based claim that inheritable genetic variation comes from new combinations of chromosomes in meiosis, from viable errors when DNA is copied, and from mutations caused by environmental factors. It is usually taught in Biology.
Make and defend a claim based on evidence that inheritable genetic variations may result from: (1) new genetic combinations through meiosis, (2) viable errors occurring during replication, and/or (3) mutations caused by environmental factors.
Clarification Statement: Emphasis is on using data to support arguments for the way variation occurs.
Assessment Boundary: Assessment does not include the phases of meiosis or the biochemical mechanism of specific steps in the process.
Next Generation Science Standards · Disciplinary Core Idea: Heredity: Inheritance and Variation of Traits · Official standard
Students make and defend claims about where inheritable genetic variation comes from. They work with three sources named in the standard: new combinations of chromosomes and alleles when sperm and eggs form (independent assortment and crossing over), rare errors made when DNA is copied that the cell survives and passes on, and mutations caused by environmental factors such as UV light, X-rays and some chemicals. For each source, students analyze data, state a claim, cite evidence, explain their reasoning and answer a rebuttal. A thread through the lesson: a change is inheritable only if it is in the DNA of cells that make sperm or eggs.
The NGSS Clarification Statement says: "Emphasis is on using data to support arguments for the way variation occurs." The Assessment Boundary says: "Assessment does not include the phases of meiosis or the biochemical mechanism of specific steps in the process." So the lesson treats meiosis only as the process that gives each sperm or egg one chromosome from every pair, with pairs sometimes exchanging segments. It does not name or order the phases, and it does not describe how enzymes copy or repair DNA (beyond this standard). All data sets are invented for teaching, with values set in realistic ranges; mutation rates are approximate published estimates, stated as such.
Learning Objectives
By the end of this lesson, students will be able to:
Use chromosome-pair counts and cross data to support the claim that meiosis produces new genetic combinations
Use mutation-rate estimates and data to support the claim that viable errors in DNA copying add new inheritable variants
Use exposure data to support the claim that environmental factors such as UV light and X-rays increase mutations
Explain why a change in DNA is inheritable only when it is in cells that form sperm or eggs
Write a claim, evidence and reasoning argument and defend it against a rebuttal
Prior Knowledge Required
Students should already be comfortable with:
Genes are stretches of DNA on paired chromosomes, one member of each pair from each parent HS-LS3-1
Mutations can change proteins and traits and can be harmful, helpful or neutral MS-LS3-1
Sexual reproduction mixes genetic information from two parents MS-LS3-2
Show a photo of two siblings who look quite different (a stock image works). Ask students to answer on their own first.
Warm-Up Prompt
"These two sisters have the same mother and father. List every way their DNA could differ from each other. Then circle any way that would also give one of them an allele that neither parent has."
Collect answers and sort them into three unlabeled columns on the board: reshuffling of the parents' chromosomes, brand-new changes during copying, and changes caused by something in the environment. Tell students the lesson will give each column a name and the evidence for it.
Direct Instruction15-20 minutes
Three sources of inheritable variation. (1) Meiosis forms sperm and eggs that carry one chromosome from each pair. Which member of each pair goes in is a matter of chance (independent assortment), and before the pairs separate, the two members can exchange matching segments (crossing over). Both make new combinations of existing alleles. Use Diagram 1. (2) Replication errors. Cells copy their DNA before every division. The copying is very accurate but not perfect; an error that the cell survives and passes on (a viable error) becomes a new allele in every cell that descends from it. (3) Environmental mutagens such as UV light, X-rays and some chemicals in tobacco smoke damage DNA and raise the rate of mutations. Sources (2) and (3) can create alleles that neither parent had. Only changes in cells that form sperm or eggs can be inherited; a mutation in a skin cell stays in that person.
Making and defending a claim. A claim answers the question, evidence is the specific data, reasoning explains why the data support the claim, and a defense answers the strongest counterargument with more evidence.
Counting new combinations (independent assortment)
Human body cells have 23 chromosome pairs. Claim: two children of the same parents are very unlikely to inherit identical sets of chromosomes, even with no new mutations.
Result: One parent can make 223 = 8,388,608 kinds of sperm or eggs. Two parents give 223 × 223 = 246 ≈ 7.0 × 1013 chromosome combinations, before crossing over adds even more.
Cross data for crossing over
Invented testcross data: a fly carrying alleles AB on one chromosome and ab on its partner is crossed with an ab/ab fly. Of 1,000 offspring, 412 are AB and 406 ab (parental types), 93 Ab and 89 aB (new combinations).
Result: Recombinant fraction = (93 + 89) ÷ 1,000 = 18.2%. The Ab and aB offspring carry combinations that were on neither of the parent's chromosomes, which is evidence that partner chromosomes exchanged segments.
How many new mutations per child (replication errors)
An approximate published estimate for humans is about 1 new mutation per 100 million base pairs per generation, and a child's body cells carry about 6.4 billion base pairs.
Result: 6.4 × 109 × 1 × 10-8 = 64 new mutations per child. Claim: every child carries dozens of variants that neither parent had, though a large share fall in DNA that does not code for proteins.
UV exposure and mutations (environmental factor)
Invented data: bacteria were exposed to UV light for 0, 10, 20 and 40 seconds. Mutant colonies per 100 million surviving cells: 3, 15, 31 and 58.
Result: Mutant counts rise with exposure: about 1.4 more per second of UV between 0 and 40 s ((58 - 3) ÷ 40 ≈ 1.4). The unexposed plate still has 3, which shows that some mutations arise without UV, from copying errors.
Defending a claim against a rebuttal
Claim: "A mutation caused by a sunburn will not be passed to your children." Rebuttal: "But UV causes mutations, and mutations are inherited."
Result: Defense: the UV reaches skin cells, not the cells in the testes or ovaries that form sperm or eggs, and a child gets DNA only from a sperm and an egg. A mutation is inherited only if it is in those cells.
After each example, ask students to label the claim, the evidence and the reasoning, and to name which of the three sources it supports.
Guided Practice15 minutes
Pairs analyze the invented data below: the average number of new mutations found in children, grouped by the father's age at conception. Background: cells that produce sperm keep dividing throughout adult life, so sperm from an older man come from cells that have copied their DNA more times. Compare with Diagram 2.
Invented data: father's age and new mutations per child
Father's age (years)
20
25
30
35
40
45
New mutations per child (average)
46
56
65
76
85
95
Describe the pattern. About how many more new mutations are there for each extra year of the father's age? (The best-fit slope is about 2 per year.)
Write a claim that names which source of variation these data support.
Explain the reasoning: why would more rounds of DNA copying lead to more new mutations?
A classmate says, "These data prove that older fathers are exposed to more chemicals." Write a defense of your claim that uses the data, and say what extra data would settle the question.
Independent Practice10-15 minutes
Each student analyzes invented data modeled on classic fruit fly experiments. Male flies were exposed to X-rays and mated. For each dose, 2,000 offspring lines were checked for a new lethal mutation on the X chromosome passed on from the father.
Invented data: X-ray dose and new mutations in fly offspring
X-ray dose (Gy)
Lines checked
Lines with a new X-linked lethal mutation
Percent
0
2,000
4
0.2%
10
2,000
56
2.8%
20
2,000
110
5.5%
30
2,000
166
8.3%
Students write a claim about X-rays and inheritable variation, cite the percent at each dose, and explain why the mutations must have been in the father's sperm-forming cells. Then they answer the rebuttal: "The 0 Gy group also had mutations, so X-rays are not the cause." Safety note for the class: these are data only; X-ray sources are never used in a school lab.
Closure5 minutes
Exit ticket: "Return to your warm-up list. Label each item with source 1, 2 or 3, or with not inheritable. Then write one sentence explaining which sources can give a child an allele that neither parent has." Collect and use it to open the next lesson.
Differentiation Strategies
For Struggling Students
Give a three-column organizer (meiosis, copying errors, environmental factors) with a picture cue and one sentence frame per column: "The data show ___, so ___ causes variation because ___."
Use the paper-strip model with 2 pairs before 3 pairs, and list all 4 combinations together before counting with powers of 2
Provide the rebuttal and two possible defenses, and have students choose the stronger one and explain why
For Advanced Students
Ask students to use the fly X-ray data to estimate how many lines out of 2,000 would carry a new mutation at 15 Gy, and to state what assumption the estimate makes
Have students compare the number of chromosome combinations for a fruit fly (4 pairs) and a dog (39 pairs) and argue which species' siblings are more likely to share a set
Ask students to write an argument that crossing over adds more variation than independent assortment alone, and to propose data that would test it
Assessment Guidance
What to Look For
Check that every claim names one of the three sources and cites specific numbers from the data rather than general statements. Look for reasoning that explains why the evidence supports the claim, and for defenses that answer the rebuttal with data. Listen for students who think any mutation anywhere in the body is inherited, who think meiosis creates new alleles, or who treat every mutation as harmful.
02
Classroom Activities
3 Activities
1
Chromosome Shuffle and Swap
20 minPairs
Pairs model how meiosis makes new combinations. Each student has three chromosome pairs made from paper strips (blue for chromosomes from their mother, tan for their father). They flip coins to build sperm or eggs, tally the combinations for the class, and then cut and swap matching segments to model crossing over.
Procedure
Each student makes 6 strips: a long, a medium and a short strip in blue and the same three in tan. Write alleles on the long pair: D on blue, d on tan.
For each pair, flip a coin: heads puts the blue strip in the sperm or egg, tails puts the tan strip. Record the three colors. Repeat to make 4 sperm or eggs.
Record your 4 results on the class tally of the 8 possible color patterns.
Crossing over: tape a blue and a tan long strip side by side, cut both at the same point, and swap the end pieces. Write the new allele combination on each strip.
Sample Class Tally (invented)
Twenty-eight students made 112 sperm or eggs. The counts for the 8 color patterns were 15, 13, 16, 12, 14, 15, 13 and 14.
Discussion Questions
With 3 pairs there are 23 = 8 patterns, so each should appear in about 1/8 of 112, or 14 results. Were all 8 patterns found? What claim do the tallies support?
After the swap, the long strips carry pieces of both colors. What does this add that coin flipping alone cannot?
Does anything in this model create a new allele? Which sources of variation would?
Modification for Distance Learning
Students use a random number generator (1 = mother, 2 = father) for each pair and enter results in a shared spreadsheet that tallies the 8 patterns.
2
Copy Chain: Errors in Copying
20 minGroups of 5
Each group passes a 60-base sequence down a chain of copiers. Student 1 copies the original card, student 2 copies student 1's copy, and so on, with 30 seconds per copy. Groups count errors, decide which copies are still viable, and compare their error rate with a cell's.
Procedure
Student 1 receives the original 60-base card and copies it in 30 seconds. Students 2, 3 and 4 each copy only the copy before theirs. Student 5 checks every copy against the original, base by base.
Viability rule for this model: a copy whose first three bases (the start signal ATG) are wrong is not viable and is discarded; any other error is passed on.
Record the number of errors in each generation and whether each copy is viable.
Sample Results (invented)
Six groups each made 4 copies, 1,440 bases copied in all. They made 7 errors in total, and one copy had a wrong start signal. Error rate: 7 ÷ 1,440 ≈ 0.49%, or about 1 error per 200 bases.
Discussion Questions
An approximate estimate for human cells is about 1 error per 10 billion bases copied. How many times more accurate is the cell than the class? (About 5 × 107 times.)
Why did errors made by student 1 show up in every later copy?
In what kind of cell would a copying error need to happen to be inherited by a child?
Challenge Variation
Give one group a "UV lamp" rule: a timer beeps twice during each copy and the copier must look away for 3 seconds. Groups predict and then compare the error counts, and argue what the rule models.
3
Claim Court
25 minGroups of 4
Each group gets four case files. For each case, one pair writes a claim naming the source of the variation with evidence and reasoning; the other pair writes the strongest rebuttal; then the first pair defends. Groups switch roles for each case.
Case Files (invented)
Case 1: Two brothers with the same parents differ at 11 of 20 genes where both parents carry two different alleles. DNA tests find no alleles in either brother that the parents lack.
Case 2: A girl has an allele of a bone-growth gene that neither parent carries in any tested cell. Her father was 44 at her conception.
Case 3: A strain of yeast grown in the dark gives 2 mutant colonies per plate; the same strain after 15 minutes of midday sunlight gives 19, and after 15 minutes of sunlight under SPF 30 sunscreen film gives 4.
Case 4: A corn plant has kernels with a combination of color and texture alleles that sat on different partner chromosomes in its parent plant.
Answer Key
Case 1: meiosis (new combinations, no new alleles). Case 2: a viable copying error in the father's sperm-forming cells is the best-supported claim. Case 3: an environmental factor (UV in sunlight), with the dark plate showing background copying errors. Case 4: meiosis (crossing over).
Discussion Questions
In Case 3, the sunscreen plate had 4 mutant colonies and the dark plate had 2. What does this say about how well the film blocks the effect of UV?
In Case 2, what data would you need to rule out an environmental cause?
Which case was hardest to defend, and what extra evidence would have helped?
Modification for Distance Learning
Post one case per discussion thread. One student posts a claim, the next posts a rebuttal and the first student replies with a defense.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: New Combinations From Two Chromosome Pairs
A parent cell with two chromosome pairs (blue from the mother, navy from the father) can pass on one chromosome from each pair in 2 × 2 = 4 ways. With n pairs there are 2n kinds of sperm or eggs. The model shows independent assortment only; crossing over would add more combinations.
Diagram 2: Father's Age and New Mutations per Child
The invented Guided Practice data, drawn to scale with a least-squares line. The number of new mutations rises by about 2 for each year of the father's age, which supports the claim that more rounds of DNA copying in sperm-forming cells produce more viable copying errors.
04
Homework Assignment
~30 min
HS-LS3-2 Homework: Arguing From Evidence About Genetic Variation
Directions: All data sets are invented for practice; mutation rates are approximate. For every problem, show your calculation and write a claim, the evidence you used and your reasoning. When a rebuttal is given, defend your claim.
Part 1: New Combinations From Meiosis (Problems 1-2)
Dogs have 39 chromosome pairs. (a) How many kinds of sperm can one dog make from independent assortment alone? Write the answer as a power of 2 and in scientific notation. (b) Write a claim about how likely two puppies in a litter are to inherit the same chromosome set from their father, and support it with your number.
Invented tomato testcross: a plant carrying alleles R (red) and S (smooth) on one chromosome and r (yellow) and s (fuzzy) on its partner is crossed with an rs/rs plant. The 1,000 offspring are 356 red smooth, 344 yellow fuzzy, 148 red fuzzy and 152 yellow smooth. (a) Which offspring show new combinations? (b) What percent are they? (c) Write a claim about crossing over and defend it against the rebuttal "these new plants must be the result of new mutations."
Part 2: Copying Errors and Environmental Factors (Problems 3-4)
A flask holds 2 × 109 bacteria, each produced by a cell division. For one particular gene, an error that changes it happens in about 1 of every 10 million divisions. (a) About how many bacteria in the flask carry a new change in that gene? (b) A student says, "Copying errors are too rare to matter." Use your answer to make a claim that answers the student.
Invented data: plates of the same bacteria were exposed to 30 seconds of UV light either uncovered or under a UV-blocking film, and a third set stayed in the dark. Mutant colonies per plate (average of 5 plates): uncovered 48, under film 6, dark 4. (a) Compare each exposed group with the dark group. (b) Write a claim about UV as a cause of mutations. (c) Explain why the dark plates are needed for the argument.
Part 3: Building and Defending Arguments (Problems 5-6)
A classmate argues: "All genetic variation comes from mutations, so meiosis does not really matter." Invented data: three pairs of siblings were tested at 30 genes where both parents carry two different alleles. The pairs differed at 18, 20 and 19 of the 30 genes, and none of the siblings had an allele that their parents lacked. (a) Find the mean number of genes at which the siblings differ. (b) Write a rebuttal to the classmate that uses the data.
For each observation, name the best-supported source of variation (meiosis, copying error or environmental factor) and whether the change can be inherited, with one sentence of reasoning: (a) a pilot has more mutations in skin cells than people who work indoors; (b) a calf has an allele found in neither parent, and the parents lived in a normal barn with no unusual exposures; (c) two kittens in one litter have different coat patterns, and both patterns appear in the parents' families.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Claim
Names the correct source of variation and whether it can be inherited
Names a source but misses whether it is inheritable, or is vague
Evidence and Calculation
Uses specific numbers from the data, with correct calculations
Cites the data in general terms or has one calculation error
Reasoning
Explains why the evidence supports the claim, including where in the body the change happened
Restates the evidence without linking it to the claim
Defense
Answers the rebuttal directly with data
Answers the rebuttal with an opinion or without data
05
Quiz: 20 Questions
Interactive, with answers
Instructions
All data sets are invented and mutation rates are approximate. 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
Pea plants have 7 chromosome pairs. From independent assortment alone, how many kinds of pollen sperm cells can one pea plant make?
Answer: C
Each of the 7 pairs gives 2 choices, so there are 27 = 128 kinds. Choice A doubles the number of pairs instead of multiplying 2 by itself 7 times. Choice B computes 72, swapping the base and the exponent.
Question 2 of 20 · Multiple Choice
Which source of variation creates new combinations of alleles the parents already had, but no new alleles?
Answer: D
Meiosis reshuffles existing chromosomes and swaps matching segments, so it makes new combinations but not new alleles. Choices A, B and C all change the DNA sequence itself and can create alleles that neither parent had.
Question 3 of 20 · Multiple Choice
Which new mutation in a man could be inherited by his future children?
Answer: A
Children get DNA only from a sperm and an egg, so only a mutation in the sperm-producing cells can be passed on. Choices B, C and D are body cells: a mutation there stays in that person and the cells that descend from it.
Question 4 of 20 · Multiple Choice
Invented testcross data: of 1,200 offspring, 510 and 486 show the parental combinations and 108 and 96 show new combinations. What percent of the offspring show new combinations from crossing over?
Answer: B
New combinations: 108 + 96 = 204, and 204 ÷ 1,200 = 17%. Choice D is the percent of parental types (996 ÷ 1,200). Choice A uses only one of the two recombinant classes (96 ÷ 1,200 = 8%).
Question 5 of 20 · Multiple Choice
What does "viable error occurring during replication" mean in HS-LS3-2?
Answer: B
Viable means the cell lives with the change and copies it into its descendants, so it can become a new allele. Choice A is the opposite of viable. Choice C describes independent assortment, which is not an error. Choice D describes an error that would leave no variation.
Question 6 of 20 · Multiple Choice
Which invented data set best supports the claim that a chemical causes mutations in yeast?
Answer: C
Choice C compares treated with untreated plates under the same conditions and shows many more mutants with the chemical. Choice A has no control group, so there is nothing to compare. Choice B shows no effect, and choice D shows fewer mutants with the chemical.
Question 7 of 20 · Multiple Choice
A plant's egg carries 500 million base pairs of DNA. If new mutations arise at about 2 per 100 million base pairs per generation, about how many new mutations does one egg carry?
Answer: D
5 × 108 × 2 × 10-8 = 10. Choice A uses 2 per billion base pairs, ten times too low (5 × 108 × 2 × 10-9 = 1). Choice B uses 2 per 10 million, ten times too high, and choice C divides 500 million by 2 million.
Question 8 of 20 · Multiple Choice
A student wants to test the claim that the heat from a warm incubator (37 °C) increases mutations in a yeast strain usually grown at 30 °C. Which plan gives the best evidence?
Answer: B
Choice B changes only the temperature, keeps every other condition the same and uses several plates, so a difference in mutant colonies can be tied to temperature. Choice A has no comparison group. Choices C and D change a second variable (sugar or time), so any difference could come from that instead.
Question 9 of 20 · Multiple Choice
A student claims, "Siblings differ only because each child has new mutations." Which evidence best defends the claim that meiosis also makes siblings different?
Answer: D
Choice D shows that differences come from different combinations of the parents' existing alleles, which only meiosis explains, while new mutations are too few to account for them. Choice C is a difference, but on its own it does not show where it came from. Choices A and B are not genetic.
Question 10 of 20 · Multiple Choice
What is the result of crossing over between the two members of a chromosome pair?
Answer: A
In crossing over the partners exchange matching segments, so one chromosome ends up with alleles from both of the organism's parents. Choice B describes mutation. Choices C and D confuse crossing over with errors in chromosome number.
Question 11 of 20 · Multiple Choice
A culture has 4 × 109 bacteria. A change in one particular gene arises in about 1 of every 100 million divisions. About how many bacteria in the culture carry a new change in that gene?
Answer: C
4 × 109 ÷ 108 = 40. Choice A divides by 107 instead of 108, and choice B divides by 109. The point for an argument: rare errors still produce many new variants in a large population.
Question 12 of 20 · Multiple Choice
Invented data: bacteria exposed to 0, 1, 2 and 4 units of a chemical from tobacco smoke gave 5, 12, 21 and 39 mutant colonies per plate. Which claim do the data best support?
Answer: A
Every increase in dose goes with more mutant colonies, from 5 to 39. Choice B is contradicted by the 5 mutant colonies at dose 0, which come from copying errors. Choices C and D ignore the upward trend.
Question 13 of 20 · Multiple Choice
Most copying errors in a person's DNA are never passed on to children. Which reason is correct?
Answer: C
Only viable changes in the germ line reach a child, and most dividing cells are body cells. Choice B is false: errors do change the sequence. Choice D is wrong because many errors are neutral.
Question 14 of 20 · Multiple Choice
A child has an allele that DNA tests find in neither parent. Which source of variation does this evidence rule out as the only explanation?
Answer: D
Meiosis only reshuffles alleles the parents already carry, so it cannot produce an allele found in neither parent. Choices A, B and C all create new alleles in the cells that form sperm or eggs, so each remains possible.
Question 15 of 20 · Short Answer
A model organism has 5 chromosome pairs. (a) How many kinds of sperm or eggs can it make from independent assortment alone? (b) How many different chromosome combinations can two parents produce in an offspring? (c) Why is the real number of combinations even larger?
(a) 25 = 32. (b) 32 × 32 = 1,024. (c) Crossing over exchanges segments between partner chromosomes, making chromosomes with new combinations of alleles, so there are far more possibilities than 1,024.
Question 16 of 20 · Short Answer
Invented data: UV-sensitive yeast plates were left in the dark, in shade or in direct sunlight for 10 minutes. Mutant colonies per plate: dark 3, shade 7, sun 26. Write a claim, cite the evidence and explain the reasoning.
Claim: UV light in sunlight increases mutations, an environmental source of variation. Evidence: the sun plates have 26 mutant colonies, about 8.7 times the 3 on the dark plates, and shade (less UV) gives an in-between 7. Reasoning: the only difference among the groups is light exposure, and more exposure goes with more mutants. The dark plates show the background rate from copying errors.
Question 17 of 20 · Short Answer
Rebuttal: "UV gave me freckles and a skin mutation, so my children will inherit it." Defend the claim that this mutation will not be inherited, and say what kind of cell a mutation must be in to be passed on.
UV reaches the skin, so the mutation is in skin cells, which are body cells. Children receive DNA only through a sperm and an egg, so only a mutation in the cells that form sperm or eggs can be inherited. The skin mutation stays in that person. (Freckles themselves depend on inherited alleles and sun exposure, which is a separate question.)
Question 18 of 20 · Short Answer
A mouse sperm or egg carries about 2.7 × 109 base pairs. With about 5 new mutations per billion base pairs per generation, about how many new mutations does a mouse pup carry (it gets one sperm and one egg)? What claim does this support?
Per sperm or egg: 2.7 × 109 × 5 × 10-9 = 13.5. Two per pup: about 27 new mutations. Claim: every pup carries new variants that neither parent had, from viable copying errors (and any environmental mutations) in the parents' sperm- and egg-forming cells.
Question 19 of 20 · Short Answer
Two parents both have genotype AaBb, and genes A and B are on different chromosome pairs. (a) What is the probability that a child is aabb? (b) Which source of variation produces this combination, which neither parent has?
(a) Each parent passes a and b together with probability 1/2 × 1/2 = 1/4, so P(aabb) = 1/4 × 1/4 = 1/16. (b) Meiosis (independent assortment): the combination is new, but both alleles already existed in the parents.
Question 20 of 20 · Short Answer
A student argues: "DNA copying is so accurate, about 1 error per 10 billion bases, that copying errors cannot be a real source of variation." A human cell holds about 6.4 billion base pairs, and sperm-forming cells divide hundreds of times over a man's life. Use numbers to defend the opposite claim.
Per division: 6.4 × 109 × 10-10 ≈ 0.64 errors, so about one new change every one or two divisions. Over hundreds of divisions in sperm-forming cells, that adds up to dozens of changes, so each child can inherit new variants. Accurate per base is not the same as rare per genome.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-LS3-2 mean?
HS-LS3-2 asks students to make and defend an evidence-based claim about where inheritable genetic variation comes from: new combinations formed in meiosis, viable errors in DNA copying, and mutations caused by environmental factors. The emphasis is on using data in an argument, not on memorizing the steps of meiosis. The code means High School, Life Science, core idea 3 (Heredity), performance expectation 2.
Do students need to know the phases of meiosis for HS-LS3-2?
No. The Assessment Boundary excludes the phases of meiosis and the biochemical mechanism of the steps. Students need to know that each sperm or egg receives one chromosome from every pair, chosen by chance, and that partner chromosomes can swap segments. Many teachers still teach the phases in the same unit, but they are beyond this standard.
What are the three sources of genetic variation in HS-LS3-2?
They are new genetic combinations through meiosis, viable errors during DNA replication, and mutations caused by environmental factors. The first reshuffles alleles that already exist; the other two change the DNA sequence and can create new alleles.
What is a viable error during replication?
It is a mistake in copying DNA that the cell survives and passes on to its descendants. Cells copy DNA very accurately and fix many mistakes, so the errors that remain are rare per base, but a genome has billions of bases, so some survive in every generation.
Are all mutations inherited?
No. Only mutations in cells that form sperm or eggs can be passed to offspring. Mutations in skin, lung or other body cells stay in the person, although they can still affect health, for example by leading to cancer.
Are mutations always harmful?
No. Many mutations have no effect on traits, some are harmful and a few are helpful in a given environment. A large share of new mutations fall in DNA that does not code for proteins. Treating every mutation as harmful is a frequent student misconception.
How do I teach students to defend a claim in biology?
Use claim, evidence and reasoning, and add a rebuttal round. Students state a claim, cite specific numbers, explain why the numbers support the claim, and then answer the strongest counterargument with more evidence. The Claim Court activity on this page builds this routine.
Is crossing over part of HS-LS3-2?
Yes, as one way that meiosis creates new genetic combinations. Students use cross data where some offspring show combinations that were not on either parental chromosome. The molecular steps of how the exchange happens are beyond this standard.
What lab activities are safe for teaching mutations?
Paper models, copying games, simulations and data analysis are safe and effective. If a class tests UV, use UV-sensitive yeast in natural sunlight with sunscreen or film as a treatment, and never use a UV lamp without eye protection and teacher supervision. X-rays and chemical mutagens appear as data only.
How does HS-LS3-2 connect to evolution?
Variation is the raw material of natural selection. HS-LS4-2 and HS-LS4-4 ask students to explain how differences in traits lead to differences in survival and reproduction, and HS-LS3-2 explains where those differences come from. HS-LS3-3 then describes how traits are distributed in a population.
07
Related Standards
6 standards
These standards connect to HS-LS3-2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-LS3-1Prerequisite
Model why gene mutations on chromosomes may change proteins and traits
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.
