A student performs a monohybrid cross and counts 240 round and 60 wrinkled peas (n = 300). A 3:1 ratio predicts 225 round and 75 wrinkled. χ² = (240−225)²/225 + (60−75)²/75 = 1.00 + 3.00 = 4.00. With 1 degree of freedom, the critical value at p = 0.05 is 3.84. What should the student conclude?
AThe data fit the 3:1 ratio because the observed counts are visually close to expected
BThe data statistically deviate from the 3:1 ratio, suggesting the simple Mendelian model may be insufficient
CA χ² of 4.00 is too small to be meaningful; far more data are needed before any conclusion
DThe hypothesis is disproven because expected values differ from observed values
The χ² value (4.00) exceeds the critical value (3.84) at p = 0.05 with 1 df, so we reject the null hypothesis that the data fit a 3:1 ratio. Option A is the classic intuitive error: 'close enough by eye' is not a statistical standard. Option D is too strong — a significant chi-square doesn't disprove the hypothesis, it means the deviations exceed what random sampling variation explains at this significance level, warranting further investigation. Option C misunderstands how chi-square works — the value is always evaluated against the critical value, not in absolute terms.
Question 2 Multiple Choice
After failing to reject a 9:3:3:1 ratio in a chi-square test, a student writes: 'The chi-square test proves that these two genes assort independently.' What is wrong with this conclusion?
AThe student should have used a t-test for genetic ratio data
BFailing to reject the null only shows the data are consistent with the model — it does not prove the model is correct
CChi-square cannot be applied to dihybrid crosses with four phenotypic classes
DThe conclusion is correct — a non-significant result means the hypothesis is confirmed
A non-significant chi-square means the observed deviations are within the range expected from random sampling variation if the model were true. Many alternative models might also fit the same data — this is especially true with small samples where statistical power is low. The correct phrasing is: 'The data are consistent with independent assortment' or 'We failed to find evidence against the hypothesis of independent assortment.' Failing to falsify ≠ proving. This is the distinction between failing to reject H₀ and confirming H₀.
Question 3 True / False
In a chi-square goodness-of-fit test for a dihybrid cross producing four phenotypic classes, the correct degrees of freedom is 3.
TTrue
FFalse
Answer: True
Degrees of freedom = number of phenotypic classes − 1 = 4 − 1 = 3. The subtraction of 1 accounts for the constraint that expected frequencies must sum to the observed total N (one degree of freedom is 'used up' fixing the total). With 3 df, the critical value at p = 0.05 is 7.82 — substantially higher than the 3.84 for a monohybrid (1 df) test. More phenotypic classes allow more ways for data to deviate from expectations, so a higher threshold is needed to distinguish real deviations from chance variation across more categories.
Question 4 True / False
A statistically significant chi-square result in a genetics experiment identifies which specific alternative mechanism — epistasis, linkage, or differential viability — is responsible for the deviation from expected ratios.
TTrue
FFalse
Answer: False
Chi-square is a goodness-of-fit test: it tells you only whether observed and expected frequencies differ more than sampling chance explains. It does not identify *why* they differ. A significant deviation from 9:3:3:1 could reflect epistasis (giving modified ratios like 12:3:1 or 9:7), linkage reducing recombinant class frequencies, differential viability of certain genotypes, or even systematic data collection errors. The chi-square flags the problem and tells you your simple model is wrong; identifying the correct alternative requires additional crosses, different experimental designs, and biological reasoning.
Question 5 Short Answer
Why does increasing sample size generally improve the usefulness of chi-square analysis in genetics? What statistical property does a larger sample improve?
Think about your answer, then reveal below.
Model answer: Larger sample sizes increase statistical power — the probability of detecting a real deviation from the expected ratio when one actually exists. With small samples, random sampling variation is large relative to expected counts, so even substantial proportional deviations may not produce a χ² value above the critical threshold. The same proportional deviation produces a larger χ² with a larger sample because the expected values (and therefore the denominators in each (O−E)²/E term) scale with sample size while the proportional deviations remain constant, causing the total χ² to increase. Equivalently: small samples can both fail to detect genuine non-Mendelian patterns (false negatives) and fail to exclude chance deviations (low precision). This is also why Mendel's published data — which fit expected ratios almost perfectly with few apparent sampling deviations — have been questioned statistically: with his sample sizes, some random deviation was expected, and suspiciously good fits may indicate selective reporting.