Grow Mendel’s pea plants, build your own Punnett squares, crack a test-cross mystery and watch the 3 : 1 and 9 : 3 : 3 : 1 ratios appear in real data.
Every offspring gets one allele of each gene from each parent. A Punnett square lays out the parents’ gametes and fills in every possible combination, so you can read off the odds of each genotype and phenotype. In the 3D model and the 2D calculator you can choose the parents, switch between a one-gene monohybrid and a two-gene dihybrid cross, and watch the classic 3:1 and 9:3:3:1 ratios appear on the live chart. Seven missions guide you from Mendel’s garden to two-trait crosses, with a quiz, a certificate, a printable worksheet and answer key, and an AP chi-square test.
In the 1850s and 1860s Gregor Mendel crossed pure-breeding pea plants and counted thousands of seeds. Repeat his experiment: predict each generation, grow it, then plant small and big samples of F2 seeds and watch the ratio settle near 3 : 1.
Each F2 seed has a 3/4 chance of the dominant trait. Plant at least 3 samples, one of them 1,000 or more (or Mendel’s sample).
| Run | Seeds (N) | Round | Wrinkled | Ratio |
|---|
♀ Parent 1 (down the side)
♂ Parent 2 (across the top)
⬇ Next: build your own square in Mission 2. The 2D Punnett square calculator with phenotype charts is further down, after Mission 7.
Build the Punnett square yourself. Parent 1’s gametes go down the side, parent 2’s gametes go across the top. Each box joins the allele from its row with the allele from its column. Solve 3 different crosses.
Tap each gray header to choose the allele that gamete carries.
Crosses solved: 0 / 3 · right on the first check: 0
Tap a card, then tap the group it belongs to (keyboard: Tab to a card, Enter, then choose). Only your first placement counts. Get 16 of 20 right to complete the mission.
Purple (P) is dominant over white (p), so a purple-flowered pea can be PP or Pp, and you cannot tell by looking. Pick a partner, breed offspring and decide. Solve 3 cases.
No offspring yet.
Mendel also crossed plants that differ in two traits: seed shape (R round > r wrinkled) and seed color (Y yellow > y green). Both parents are RrYy. Each makes 4 kinds of gametes, so the square has 16 boxes.
| Seed type | Mendel 1866 | Expected | Your simulation |
|---|
Five word problems with peas, guinea pigs, dogs, rabbits, cats, cattle, tomatoes and corn. Draw a quick Punnett square on paper if it helps. Get 4 of 5 right; every answer comes with a worked solution.
10 questions from a bank of 20. Score 7 or more to complete the mission. Each answer is explained after you check.
Each parent is heterozygous (Aa) for one gene. Both make two kinds of gamete, A and a, so the 2×2 square gives a 1:2:1 genotype ratio and, because A is dominant, a 3:1 phenotype ratio.
| Cross | Genotypes | Phenotypes |
|---|---|---|
| RR × rr | all Rr | all round |
| Rr × Rr | 1 RR : 2 Rr : 1 rr | 3 round : 1 wrinkled |
| Rr × rr (test cross) | 1 Rr : 1 rr | 1 round : 1 wrinkled |
| RR × Rr | 1 RR : 1 Rr | all round |
| RrYy × RrYy | 9 genotypes in 16 boxes | 9 : 3 : 3 : 1 |
| Trait | Dominant : recessive | Ratio |
|---|---|---|
| Seed shape | 5,474 round : 1,850 wrinkled | 2.96 : 1 |
| Seed color | 6,022 yellow : 2,001 green | 3.01 : 1 |
| Flower color | 705 purple : 224 white | 3.15 : 1 |
| Two traits | 315 : 108 : 101 : 32 (556 seeds) | ≈ 9 : 3 : 3 : 1 |
Do Mendel’s 556 seeds really fit 9 : 3 : 3 : 1, or are the differences too big to be chance? The chi-square (χ²) test answers that. Expected counts: 556 × 9/16 = 312.75, 556 × 3/16 = 104.25 (twice) and 556 × 1/16 = 34.75.
| Seed type | O | E | (O − E)² | (O − E)² ÷ E |
|---|---|---|---|---|
| round yellow | 315 | 312.75 | 5.0625 | 0.016 |
| round green | 108 | 104.25 | 14.0625 | 0.135 |
| wrinkled yellow | 101 | 104.25 | 10.5625 | 0.101 |
| wrinkled green | 32 | 34.75 | 7.5625 | 0.218 |
| Total | 556 | 556 | χ² ≈ 0.47 |
Degrees of freedom = 4 classes − 1 = 3. The critical value at p = 0.05 is 7.81. Because 0.47 < 7.81, we do not reject the 9 : 3 : 3 : 1 hypothesis: the small differences are easily explained by chance.
Mendel had never seen a chromosome. Today we know the two alleles of a gene sit on a pair of homologous chromosomes, and the pair separates in anaphase I of meiosis. That is the law of segregation. Different chromosome pairs line up independently at metaphase I: that is independent assortment. See it in the Meiosis simulation.
Cross a pure-breeding round-seeded pea with a pure-breeding wrinkled-seeded one and the whole first generation is round — the wrinkled trait seems to vanish. Let those round offspring breed with each other and wrinkled seeds reappear, in about one of every four seeds. That reliable reappearance is the puzzle Gregor Mendel solved in the 1860s: traits are carried by discrete “factors” (we now call them genes) that come in versions called alleles, and they do not blend — they hide and resurface.
Seed shape is one gene with two alleles: $R$ (round, dominant) and $r$ (wrinkled, recessive). Each plant carries two alleles — its genotype is $RR$, $Rr$, or $rr$ — but shows only one phenotype, round unless it is $rr$. By the law of segregation, the two alleles separate when gametes form, so an $Rr$ plant makes two kinds of gamete, half $R$ and half $r$. Those gamete probabilities are exactly what the Punnett square lists along its edges.
Each box of the square is one equally likely fertilization, so the offspring odds are just the fraction of boxes of each kind. For $Rr \times Rr$ the four boxes give genotypes $1\,RR : 2\,Rr : 1\,rr$, and since $R$ is dominant the phenotype ratio is $3:1$ — in probability terms, $P(\text{round}) = \tfrac{3}{4}$. Add a second, independent gene (seed color, $Y$ yellow dominant, $y$ green) and the law of independent assortment lets each gene run its own $3:1$. Multiply them: $\tfrac{3}{4}\times\tfrac{3}{4}=\tfrac{9}{16}$ show both dominants, and the full dihybrid ratio is $9:3:3:1$.
Start on 1 gene with the Aa × Aa preset and confirm the $3:1$ phenotype, $1:2:1$ genotype split. Switch Parent 2 to $aa$ for a test cross and watch it become $1:1$ — that is how you expose a hidden recessive. Set one parent to $AA$ and see every box turn dominant. Finally press 2 genes (dihybrid) with both parents $RrYy$: the square grows to $16$ boxes and the bar chart snaps to the $9:3:3:1$ pattern. Toggle Label boxes with phenotype to see how four genotype groups collapse into the colored classes.
| Step | What happens |
|---|---|
| 1. Genotypes | Each parent has two alleles per gene, e.g. $Rr$. You set these with the dropdowns. |
| 2. Segregation | The two alleles separate into gametes. $Rr$ makes $\tfrac12 R$ and $\tfrac12 r$; $RR$ makes only $R$. These become the row and column headers. |
| 3. Assortment | For two genes, each gamete gets one allele of each gene independently, so $RrYy$ makes four gamete types: $RY,\,Ry,\,rY,\,ry$. |
| 4. Fertilization | Every box unites one gamete from each parent into an offspring genotype, sorted dominant-first ($Rr$, not $rR$). |
| 5. Counting | Each box is equally likely, so the fraction of boxes of each genotype is its probability; grouping by dominance gives the phenotype ratio. |
A monohybrid cross follows a single gene. The signature result is the $Rr \times Rr$ cross: a $2\times2$ square of four boxes holding $RR$, $Rr$, $Rr$, $rr$. The genotype ratio is $1:2:1$, but because the dominant allele masks the recessive one in three of the four boxes, the phenotype ratio is $3:1$. Change the parents and the pattern changes predictably: $RR \times rr$ gives all $Rr$ (all dominant); $Rr \times rr$ — the test cross — gives $1:1$, the tell-tale sign that the dominant parent was carrying a hidden recessive allele.
A dihybrid cross tracks two genes that assort independently. With both parents $RrYy$, each makes four equally likely gametes, so the square is $4\times4 = 16$ boxes. Counting phenotypes gives $9$ round-yellow, $3$ round-green, $3$ wrinkled-yellow, and $1$ wrinkled-green — the famous $9:3:3:1$. It is simply two independent $3:1$ ratios multiplied together, which is why the product rule of probability ($P(A \text{ and } B) = P(A)\times P(B)$) is the fastest way to predict any single class without drawing the whole grid.
Mendel chose pea traits that happen to be controlled by single genes with clean dominance, which is why his ratios are so crisp. The same logic predicts single-gene human traits and genetic conditions such as cystic fibrosis (recessive) or Huntington’s disease (dominant). But many traits add complications — incomplete dominance (a blended heterozygote, so the phenotype ratio equals the $1:2:1$ genotype ratio), codominance (both alleles show, as in human AB blood), polygenic traits like height, sex linkage, and environmental effects. Mendel’s laws are the foundation everything else builds on.
Everything runs in the browser on classroom laptops, tablets and phones: no login and no personal data. The page uses Google Analytics for anonymous usage counts only; student answers, mission progress and the name on the certificate stay in the browser on the student’s device and are not sent anywhere. Simulated seed counts use random numbers, so every student’s data are a little different (a good discussion point).
Seven missions in one lesson is tight; for younger classes use two periods (Missions 1–4, then 5–7 and the quiz).
| Time | Phase | Activity |
|---|---|---|
| 0–5 min | Engage | Mendel’s puzzle: “Pure round × pure wrinkled peas gave only round seeds. Where did wrinkled go, and why did it come back?” Quick vote: blended away, hidden, or a new mutation. |
| 5–15 min | Explore | Mission 1 Mendel’s Garden: predict F1 and F2, plant 10, 100, 1,000 and 7,324 seeds; fill the data table on the worksheet. Mission 2 Punnett Builder. |
| 15–25 min | Explain | Mission 3 sort (genotype, phenotype, homozygous, heterozygous), then Mission 4 Test-Cross Detective in pairs. Discuss: why must the partner be white (pp)? |
| 25–33 min | Elaborate | Mission 5 two traits: predict 9 : 3 : 3 : 1 and compare with Mendel’s 556 seeds. AP classes: the χ² test in Going further. |
| 33–45 min | Evaluate | Missions 6–7: genetics problems and the final quiz (or homework). Exit ticket on the worksheet. |
Two printable pages per student: Mission 1 data table, two blank Punnett squares, test-cross reasoning, the two-trait cross, four problems and an exit ticket. The key has every answer, plus an example data set from the page’s own random-number model.
NGSS: MS-LS3-2 (model of how offspring get one allele of each gene from each parent, using Punnett squares), HS-LS3-3 (statistics and probability to explain the distribution of expressed traits). AP Biology: Unit 5 Heredity, topic 5.3 Mendelian Genetics (including the chi-square test). Practices: developing and using models, analyzing and interpreting data, using mathematics and computational thinking.
A Punnett square is a grid that predicts the possible genotypes of offspring from a cross. You write one parent’s gametes (reproductive cells: a sperm cell (carried by pollen) or an egg cell, each carrying one allele per gene) across the top and the other parent’s gametes down the side, then fill each inner box by combining the allele from its column with the one from its row. Because every box is equally likely, counting the boxes of each genotype or phenotype gives the expected ratio. A heterozygous cross, Aa × Aa, yields one AA, two Aa, and one aa — a 1:2:1 genotype ratio and a 3:1 phenotype ratio. It is named after Reginald Punnett.
Key takeaway: list each parent’s gametes on the two sides, fill every box with the combined alleles, and count the boxes to get the probability of each genotype and phenotype.The genotype is the set of alleles an organism carries, written as AA, Aa, or aa. The phenotype is the observable trait, such as round seeds or purple flowers. They differ because of dominance: a dominant allele masks a recessive one, so AA and Aa look identical and only aa shows the recessive trait. That is exactly why a heterozygous cross gives a 1:2:1 genotype ratio but a 3:1 phenotype ratio — the genotype is the instruction, the phenotype is the result after dominance acts.
Key takeaway: genotype is the allele combination; phenotype is the visible trait, and because dominance hides recessives two genotypes can share one phenotype.It states that each organism carries two alleles per gene, one from each parent, and that these two alleles separate during gamete formation so each gamete gets only one. At fertilization two gametes fuse and the offspring again has two alleles. This is why an Aa parent makes two equally common gametes, A and a, and why alleles line up one at a time along the edges of a Punnett square. Mendel deduced it from pea crosses in the 1860s; we now know it reflects the separation of homologous chromosomes during meiosis.
Key takeaway: the two alleles of a gene separate during gamete formation so each gamete carries just one, which is why heterozygotes make two equally common gametes.Independent assortment says the alleles of different genes go into gametes independently, so which allele a gamete gets for one gene does not affect what it gets for another (true for genes on different chromosomes). A doubly heterozygous RrYy parent therefore makes four equally likely gametes — RY, Ry, rY, ry — and crossing two of them fills a 16-box square. Counting phenotypes gives nine both-dominant, three and three for the mixed classes, and one both-recessive: 9:3:3:1. It is just two separate 3:1 ratios multiplied, since ¾ × ¾ = 9/16.
Key takeaway: different genes are inherited independently, so a dihybrid cross combines two 3:1 ratios into 9:3:3:1.A test cross finds the unknown genotype of an individual showing a dominant trait, which could be homozygous (AA) or heterozygous (Aa). You cross it with a homozygous recessive (aa) partner, whose genotype is certain. If the unknown is AA, every offspring inherits a dominant allele and all look dominant. If it is Aa, about half the offspring are aa and show the recessive trait, giving a 1:1 ratio. So any recessive offspring reveals a hidden recessive allele in the unknown parent.
Key takeaway: crossing a dominant-looking individual with a homozygous recessive partner tells you, from whether recessive offspring appear, if it was homozygous or heterozygous.Count the boxes and remember dominance. Each Aa parent makes A and a gametes equally, so the 2×2 square holds AA, Aa, Aa, aa — a 1:2:1 genotype ratio. But A is dominant, so AA and both Aa boxes (three of four) all show the dominant phenotype, and only the single aa box shows the recessive one: 3:1. A 1:1 ratio comes from a different cross, the test cross Aa × aa, where half the offspring are Aa and half aa.
Key takeaway: the 1:2:1 genotypes collapse into 3:1 phenotypes because the dominant allele masks the recessive in three of the four boxes.The ratios are probabilities, not guarantees — expected proportions over many offspring, like a coin landing heads half the time. A small family can stray from 3:1 by chance; the ratio shows up clearly only with large numbers. Many traits also break simple dominance: incomplete dominance gives an intermediate heterozygote (pink from red and white) and a 1:2:1 phenotype ratio; codominance shows both alleles (AB blood); polygenic traits like height involve many genes; and the environment, sex linkage, and gene interactions add further layers. Mendel’s laws are the foundation, not the whole story.
Key takeaway: Punnett ratios are expected probabilities seen clearly only with many offspring, and real inheritance often adds incomplete dominance, codominance, polygenic effects, and environment.In his 1866 paper Mendel reported the F2 counts for seven pea traits. For seed shape he counted 5,474 round and 1,850 wrinkled seeds (2.96 : 1); for seed color 6,022 yellow and 2,001 green (3.01 : 1); for flower color 705 purple and 224 white plants (3.15 : 1). In a two-trait cross he counted 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green seeds, close to 9 : 3 : 3 : 1. None is exactly 3 : 1, because every seed is a separate chance event; large samples land close to the expected ratio.
Key takeaway: Mendel’s big samples gave 2.96 : 1, 3.01 : 1 and 3.15 : 1, close to but never exactly 3 : 1.Treat each gene as its own one-gene cross and multiply the probabilities (the product rule). For RrYy × RrYy, P(round) = 3/4 and P(yellow) = 3/4, so P(round and yellow) = 3/4 × 3/4 = 9/16. In the same way round green = 3/4 × 1/4 = 3/16, wrinkled yellow = 1/4 × 3/4 = 3/16 and wrinkled green = 1/4 × 1/4 = 1/16. This works because genes on different chromosomes assort independently.
Key takeaway: split the cross into one-gene crosses and multiply: 3/4 × 3/4 = 9/16.Compare observed counts O with the counts E expected from the ratio. For Mendel’s 556 two-trait seeds, E = 312.75, 104.25, 104.25 and 34.75 for 9 : 3 : 3 : 1. Add up (O − E)² ÷ E for the four classes: 0.016 + 0.135 + 0.101 + 0.218 ≈ 0.47. With 4 classes there are 3 degrees of freedom, and the critical value at p = 0.05 is 7.81. Because 0.47 is much smaller than 7.81, the differences are easily explained by chance and the data fit 9 : 3 : 3 : 1.
Key takeaway: χ² = Σ(O − E)²/E ≈ 0.47 for Mendel’s data, far below 7.81 (df 3), so the data fit 9 : 3 : 3 : 1.