D. Genetic Processes · Genetic ProcessesInvestigating Crosses: Real and Simulated Data, Probability and Chi-Square
D2.4 D2.3 A1.1 A1.5 A1.6 A1.8 A1.10 A1.13About 75 minutes
Learning goals- I can plan an investigation of a monohybrid or dihybrid cross, real or simulated.
- I can compare observed results with expected ratios, using percent deviation and the chi-square test.
- I can work backward from offspring ratios to the parents' genotypes.
- I can explain why sample size matters in genetics.
Words to know
expected (E)
The number predicted by a hypothesis, for example 3/4 of the total for a 3 : 1 ratio.
observed (O)
The number actually counted in an experiment.
percent deviation
|O − E| ÷ E × 100%: how far an observation is from its expected value, as a percent.
chi-square (χ²) test
A statistical test that measures how far observed counts are from expected counts: χ² = Σ (O − E)² ÷ E.
degrees of freedom (df)
For a chi-square test of a ratio: number of categories − 1.
critical value
The χ² value at which a deviation becomes too large to blame on chance (at the 0.05 level: 3.84 for df = 1, 5.99 for df = 2, 7.81 for df = 3).
null hypothesis
The hypothesis that there is no real difference: the observed counts differ from the expected ratio only by chance.
Is 381 : 119 "close enough" to 3 : 1?
A student counts kernels on a corn ear and gets 381 purple and 119 yellow. That is not exactly 3 : 1 (it is 3.20 : 1). Is the hypothesis wrong, or is it just chance? Geneticists face this question every time they count offspring. This lesson is about doing genetics like a scientist: planning a cross, collecting data, and deciding whether the data fit a hypothesis.
Planning a genetics investigation
Real crosses in school labs often use organisms that grow fast and produce many offspring:
- Wisconsin Fast Plants (a rapid-cycling Brassica rapa): a complete life cycle in about 5 to 6 weeks. Seedlings with purple stems (anthocyanin pigment) versus non-purple stems are easy to score.
- Fruit flies (Drosophila melanogaster): a generation in about 10 to 14 days at room temperature, many offspring, traits such as eye colour and wing shape.
- Genetic corn ears: ears grown from controlled crosses, where each kernel is one offspring.
An investigation should state a testable question and a hypothesis (for example, "Purple stem is dominant, so an F2 will show 3 purple : 1 non-purple"), identify variables (same soil, light and water for all plants), describe how traits will be scored, and plan a large sample. Simulations (coins, beads, or online virtual fly labs) let you test the same ideas in less time.
Step 1: Expected numbers
Multiply the total by each fraction in the hypothesis. For 500 kernels and 3 : 1: E(purple) = 500 × 3/4 = 375, E(yellow) = 500 × 1/4 = 125.
Step 2: Percent deviation (a quick check)
Percent deviation = |O − E| ÷ E × 100%. For purple: |381 − 375| ÷ 375 × 100% = 1.6%. Small deviations are expected by chance; bigger samples usually give smaller percent deviations.
Step 3: The chi-square (χ²) test
The chi-square test gives one number for all categories together:
χ² = Σ (O − E)² ÷ E
Then compare it with a critical value from a table, using degrees of freedom = number of categories − 1. Biologists usually use the 0.05 probability level: if χ² is less than the critical value, the deviation could easily be due to chance, and we do not reject the hypothesis. If χ² is greater, a deviation this large would happen by chance less than 5% of the time, so we reject the hypothesis and look for another explanation (linked genes, a different dominance pattern, errors, or offspring that did not survive).
| Degrees of freedom | 1 | 2 | 3 |
|---|
| Critical value (p = 0.05) | 3.84 | 5.99 | 7.81 |
Worked exampleExample 1: Testing the corn ear
381 purple and 119 yellow kernels. Hypothesis: 3 : 1. Use a chi-square test.
- Total = 500. E(purple) = 375, E(yellow) = 125.
- Purple: (381 − 375)² ÷ 375 = 36 ÷ 375 = 0.096.
- Yellow: (119 − 125)² ÷ 125 = 36 ÷ 125 = 0.288.
- χ² = 0.096 + 0.288 = 0.384. Two categories, so df = 1; critical value 3.84.
- 0.384 < 3.84, so the difference is small enough to be due to chance.
χ² ≈ 0.38. Do not reject the hypothesis: the data support a 3 : 1 ratio (two heterozygous parents).
Reading a χ² result for one degree of freedom. The corn ear's χ² (0.38) is far below the critical value, so a 3 : 1 ratio fits. A test cross giving 62 : 38 (from the practice questions) has χ² = 5.76, too big to blame on chance.Worked exampleExample 2: Mendel's dihybrid data
Mendel counted 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green (556 seeds). Test the 9 : 3 : 3 : 1 hypothesis.
- Expected: 556 × 9/16 = 312.75; 556 × 3/16 = 104.25 (twice); 556 × 1/16 = 34.75.
- Round yellow: (315 − 312.75)² ÷ 312.75 = 5.0625 ÷ 312.75 ≈ 0.016.
- Round green: (108 − 104.25)² ÷ 104.25 = 14.0625 ÷ 104.25 ≈ 0.135. Wrinkled yellow: (101 − 104.25)² ÷ 104.25 = 10.5625 ÷ 104.25 ≈ 0.101.
- Wrinkled green: (32 − 34.75)² ÷ 34.75 = 7.5625 ÷ 34.75 ≈ 0.218.
- χ² ≈ 0.016 + 0.135 + 0.101 + 0.218 = 0.47. df = 4 − 1 = 3; critical value 7.81.
0.47 < 7.81: the data fit 9 : 3 : 3 : 1 very well.
Keep expected values unrounded
Use the exact expected values (such as 312.75) in the χ² calculation, even though you can't have three-quarters of a seed. Round only the final χ² (two decimal places is plenty). Report counts as whole numbers.
Working backward to the parents
Often you know the offspring and need the parents. Find the ratio (divide every count by the smallest), match it to a known pattern, then check with a Punnett square or a χ² test.
Worked exampleExample 3: Fast Plants
Two purple-stemmed Fast Plants are crossed. Offspring: 74 purple, 26 non-purple. What are the parents' genotypes?
- Ratio: 74 ÷ 26 ≈ 2.85, close to 3 : 1.
- A 3 : 1 ratio means both parents are heterozygous: ***Pp* × Pp** (purple dominant).
- Check with χ²: E = 75 and 25. (74 − 75)²/75 + (26 − 25)²/25 = 0.013 + 0.04 = 0.053 < 3.84. ✓
Just for funA statistician and a geneticist flip a coin 10 times and get 7 heads.
The geneticist says, "Let's do a chi-square test." The statistician says, "Let's flip it 990 more times." (Both are right: sample size matters!)
Hands-on activityBead (or candy) simulation with chi-square
You need: a paper bag, 20 beans or beads of one colour and 20 of another (or two kinds of dried pasta), paper, pencil and a calculator.
- Each kind of bead is an allele: one kind = A, the other = a. Put 20 of each in the bag: this is the gamete pool from Aa parents.
- Without looking, draw one bead (the allele from parent 1), note it, put it back and shake; then draw a second bead (the allele from parent 2) and put it back. The two beads together are one offspring: record its genotype.
- Make 40 offspring. Count dominant phenotypes (AA + Aa) and recessive (aa).
- Calculate expected numbers for 3 : 1 (30 and 10), percent deviation and χ².
- Is your χ² below 3.84? Combine with a second set of 40 draws and recalculate.
Hypothesis to write first: "If both parents are Aa, then about 3/4 of offspring will show the dominant phenotype."
What to look for: nearly everyone's χ² will be below 3.84; occasionally (about 1 time in 20) a fair simulation gives a value above it just by chance. That is what "p = 0.05" means.
Your turn
1200 offspring from Aa × Aa. What are the expected numbers of dominant and recessive phenotypes?
200 × 3/4 = 150 dominant; 200 × 1/4 = 50 recessive.
2Observed: 160 dominant, 40 recessive (E = 150, 50). Calculate χ².
(160 − 150)²/150 + (40 − 50)²/50 = 100/150 + 100/50 = 0.667 + 2.000 = 2.67.
3Is the χ² from the last question below the critical value? What do you conclude?
df = 1, critical value 3.84. 2.67 < 3.84, so do not reject the 3 : 1 hypothesis; the deviation can be explained by chance.
4Offspring: 51 tall, 49 dwarf. What are the likely parent genotypes?
About 1 : 1: a test cross, ***Tt* × tt**.
5Why do geneticists prefer large samples?
Chance deviations have a smaller effect on large samples, so the observed ratio is more likely to be close to the true ratio, and conclusions are more reliable.
6Give two reasons Fast Plants or fruit flies are good for genetics investigations.
Any two: short generation time; many offspring; easy-to-see traits; small and easy to raise; can be crossed in a controlled way.
More practice: warm-up
1Write the chi-square formula.
2How many degrees of freedom for a test with 4 categories?
3What is the critical value for df = 1 at p = 0.05?
4Expected 9 : 3 : 3 : 1 out of 160: give the four expected numbers.
5If χ² is larger than the critical value, what do you do with the hypothesis?
Reject it (the deviation is unlikely to be due to chance alone).
More practice: on your own
1A test cross gives 62 dominant and 38 recessive offspring. Test the 1 : 1 hypothesis.
E = 50 and 50. χ² = (12²)/50 + (12²)/50 = 2.88 + 2.88 = 5.76. df = 1, critical 3.84. 5.76 > 3.84: reject 1 : 1. The deviation is too large to blame on chance; investigate (for example, are recessive offspring less likely to survive, or was a parent misidentified?).
2Calculate the percent deviation for 62 observed when 50 were expected.
|62 − 50| ÷ 50 × 100% = 24%.
3Snapdragon offspring: 23 red, 52 pink, 25 white. Test 1 : 2 : 1.
E = 25, 50, 25. χ² = 4/25 + 4/50 + 0/25 = 0.16 + 0.08 + 0 = 0.24. df = 2, critical 5.99. 0.24 < 5.99: data fit 1 : 2 : 1.
4Fruit-fly offspring: 296 normal wings, 104 vestigial. What were the parents' genotypes, and what is χ² for 3 : 1?
296 ÷ 104 ≈ 2.85: about 3 : 1, so ***Vv* × Vv. E = 300 and 100. χ² = 16/300 + 16/100 = 0.053 + 0.16 = 0.21** < 3.84. ✓
5A student's simulation gave χ² = 4.1 for a fair 3 : 1 setup. Does that prove the simulation was unfair? Explain.
No. At p = 0.05, about 1 fair experiment in 20 gives a χ² above the critical value just by chance. The student should repeat or enlarge the sample before concluding anything.
Challenge
1A dihybrid test cross (AaBb × aabb) gives 140 AB, 130 ab, 18 Ab and 12 aB phenotypes (300 total). Test 1 : 1 : 1 : 1 and explain the result.
E = 75 each. χ² = (65²)/75 + (55²)/75 + (57²)/75 + (63²)/75 = (4225 + 3025 + 3249 + 3969)/75 = 14 468/75 ≈ 192.9. Far above 7.81 (df = 3): reject independent assortment. The parental types are much more common, so the genes are linked; the 30 recombinants (10%) come from crossing over.
2Explain why the chi-square test must use counts, not percents.
χ² depends on sample size: the same percent deviation is more significant in a large sample than a small one. Using percents would hide the sample size and give misleading results.
3Plan a Fast Plants investigation to find out whether purple stem is dominant. Include the cross, the expected results and two controlled variables.
Cross a true-breeding purple plant with a true-breeding non-purple plant; grow the F1 (if all purple, purple is dominant). Let F1 plants cross with each other and grow at least 100 F2 seeds; expect 3 purple : 1 non-purple and test with χ². Controlled variables (any two): same light, same soil/growing medium, same water and fertilizer, same temperature, scoring all seedlings at the same age.
Quiz
1. Expected numbers for a 3 : 1 ratio out of 120 offspring are:
2. A chi-square test with 2 categories has how many degrees of freedom?
3. χ² = 1.2 for a 3 : 1 hypothesis (critical value 3.84). The best conclusion is:
4. Offspring 3 : 3 : 1 : 1 suggest which parents?
5. Why are large sample sizes important in genetics?
6. Two Fast Plants are crossed and give 48 purple : 52 non-purple. The parents are most likely:
Big idea: This lesson is the investigation skill of D2.4 (crosses, real or simulated, with probability to find parents' genotypes), and it uses scientific-investigation skills from strand A: hypotheses (A1.1), controlled inquiry (A1.5), organizing data (A1.6), analysing deviation and error (A1.8), conclusions (A1.10) and appropriate precision (A1.13). The chi-square test goes slightly beyond the minimum, but many Ontario classes use it and it builds directly toward SBI4U.
How to help: The bead activity is quick and needs nothing special. Ask your teen to write a hypothesis before drawing any beads, and to explain in words what χ² below 3.84 means ("the difference is small enough to be chance").
Common difficulties:
- Using percents instead of counts in χ².
- Getting degrees of freedom wrong (it is categories − 1, not total − 1).
- Saying χ² "proves" a hypothesis; it can only support or reject it.