Chi-Square Tests for Mendelian Ratios - Genetics

A researcher performs a dihybrid cross expected to yield a 9:3:3:1 phenotypic ratio. From a total of 320 progeny, the number of individuals in the double-recessive class is 15. What is the contribution of this phenotypic class to the total chi-square ( $\chi^2$ ) value?
Answer: 1.25 The chi-square contribution for a single class is calculated as $\frac{(O-E)^2}{E}$ , where O is the observed value and E is the expected value. First, calculate the expected number for the double-recessive class. Based on a 9:3:3:1 ratio, this class represents 1/16 of the total progeny. Expected (E) = (1/16) * 320 = 20. The observed value (O) is given as 15. The contribution is $\frac{(15 - 20)^2}{20} = \frac{(-5)^2}{20} = \frac{25}{20} = 1.25$ .
Two experiments test the same null hypothesis of a 3:1 Mendelian ratio. Experiment A has 100 progeny and a calculated $\chi^2$ value of 4.0. Experiment B has 1000 progeny, and the observed phenotypic proportions are identical to Experiment A. What is the $\chi^2$ value for Experiment B?
Answer: 40.0 The chi-square statistic is calculated as $\sum \frac{(O-E)^2}{E}$ . If the proportions stay the same but the sample size increases by a factor of $k$ (in this case, 10), then the observed (O) and expected (E) counts also increase by $k$ . The difference $(O-E)$ also scales by $k$ , so $(O-E)^2$ scales by $k^2$ . The formula becomes $\frac{(k(O-E))^2}{kE} = \frac{k^2(O-E)^2}{kE} = k \frac{(O-E)^2}{E}$ . Therefore, the new chi-square value is $k$ times the original value. Here, $k=10$ , so the new $\chi^2$ value is $10 \times 4.0 = 40.0$ .
A cross involving a recessive lethal allele is expected to produce offspring in a 2:1 phenotypic ratio of wild-type to mutant. A researcher observes 140 wild-type and 60 mutant individuals. What are the correct degrees of freedom (df) and null hypothesis for a chi-square test of this result?
Answer: df = 1; the observed results do not differ significantly from a 2:1 ratio. The null hypothesis must state the expected outcome, which in this case is a 2:1 ratio due to the lethal allele. The degrees of freedom are calculated as the number of phenotypic categories minus one. Since there are two observable phenotypes (wild-type and mutant), the degrees of freedom are $2 - 1 = 1$ . Therefore, the correct setup is df = 1 and a null hypothesis based on the 2:1 ratio.
A student crosses two heterozygous individuals and obtains 400 progeny. They want to test the null hypothesis that the results fit a 3:1 Mendelian ratio. What are the correct expected counts for the dominant and recessive phenotypes needed to perform the chi-square test?
Answer: 300 dominant, 100 recessive When you encounter chi-square problems in genetics, you're testing whether observed data fits an expected theoretical ratio. The key insight is that expected values depend only on your hypothesis and total sample size, never on what you actually observed. For a cross between two heterozygotes (Aa × Aa), Mendel's laws predict a 3:1 phenotypic ratio of dominant to recessive traits. With 400 total progeny, you calculate expected counts by applying this ratio: $\frac{3}{4} \times 400 = 300$ dominant and $\frac{1}{4} \times 400 = 100$ recessive. This makes answer A correct. Answer B (200:200) represents a 1:1 ratio, which you'd expect from a testcross (Aa × aa), not a cross between two heterozygotes. Answer C contains a fundamental misconception—expected values in hypothesis testing are determined by your null hypothesis and sample size, completely independent of observed results. If expected values depended on observed counts, you couldn't perform meaningful statistical tests. Answer D (250:150) doesn't correspond to any standard Mendelian ratio and appears to be a distractor. Remember this pattern: for chi-square tests in genetics, always start with the theoretical ratio from your null hypothesis, then multiply each fraction by your total sample size to get expected counts. The observed data only comes into play when calculating the chi-square statistic itself, never when determining expected values.
Which of the following would increase the calculated chi-square value for a given experiment, assuming the null hypothesis and total number of progeny remain the same?
Answer: A larger deviation between the observed and expected phenotypic counts. Chi-square analysis tests whether observed data significantly deviates from expected patterns under a null hypothesis. The chi-square statistic is calculated as $\chi^2 = \sum \frac{(observed - expected)^2}{expected}$ for each category. Looking at this formula reveals what increases the chi-square value. The numerator contains the squared difference between observed and expected counts. When this difference gets larger, the chi-square value increases proportionally. Choice A correctly identifies that larger deviations between observed and expected phenotypic counts will increase the calculated chi-square value. Choice B incorrectly suggests that changing the p-value threshold affects the calculated chi-square. The p-value threshold (0.05 vs 0.10) only determines how we interpret the chi-square value for statistical significance—it doesn't change the actual calculated number. Choice C is wrong because adding more phenotypic categories doesn't automatically increase chi-square; it depends on how well each category fits expectations. More categories could actually decrease chi-square if the additional data fits the expected pattern well. Choice D describes the opposite scenario—closer agreement between observed and expected counts means smaller deviations, which would decrease the chi-square value. Remember that chi-square measures "goodness of fit"—how well your data matches expectations. Larger deviations from expected patterns always increase the chi-square statistic, while better fits decrease it. Focus on the mathematical relationship in the formula rather than getting distracted by significance thresholds or experimental design elements.
Three separate chi-square tests are performed on crosses, each with 2 degrees of freedom. The resulting $\chi^2$ values are: Test X = 1.2, Test Y = 5.8, Test Z = 9.5. Which statement correctly ranks the p-values associated with these tests?
Answer: p-value X > p-value Y > p-value Z There is an inverse relationship between the chi-square ( $\chi^2$ ) value and the p-value. A larger $\chi^2$ value indicates a greater deviation between observed and expected results, making it less likely that the deviation is due to random chance. This corresponds to a smaller p-value. Conversely, a smaller $\chi^2$ value indicates a good fit to the null hypothesis and corresponds to a larger p-value. Given the $\chi^2$ values X < Y < Z, the corresponding p-values will be in the reverse order: p-value for X > p-value for Y > p-value for Z.
In a monohybrid cross, a student observes 78 dominant and 22 recessive phenotypes. They calculate a $\chi^2$ value of 0.48. A second student repeats the cross and observes 156 dominant and 44 recessive phenotypes. What is the $\chi^2$ value for the second student's experiment?
Answer: 0.96 The second student's sample size is double the first (200 vs 100), but the proportions are identical (156/200 = 78/100 = 0.78). The chi-square statistic scales linearly with sample size when proportions are constant. If the sample size is multiplied by a factor $k$ , the $\chi^2$ value is also multiplied by $k$ . In this case, $k = 200/100 = 2$ . Therefore, the new $\chi^2$ value is $2 \times 0.48 = 0.96$ .
A chi-square analysis of a dihybrid cross yields a $\chi^2$ value that corresponds to a p-value between 0.50 and 0.75. What is the most accurate interpretation of this result?
Answer: The observed deviation from the expected ratio is consistent with random sampling error. A p-value in the range of 0.50 to 0.75 is much greater than the standard significance level of 0.05. This indicates that we fail to reject the null hypothesis. A high p-value means that if the null hypothesis were true, observing a deviation as large as (or larger than) the one in the data is very likely due to random chance alone. Therefore, the data are consistent with the null hypothesis, and the deviation is attributable to random sampling error.
A researcher analyzes a dihybrid cross by first performing a chi-square test for a 9:3:3:1 ratio (Test 1). Finding a significant deviation, they decide to analyze each trait separately. They pool the data for the first trait and perform a second chi-square test for a 3:1 ratio (Test 2). How do the degrees of freedom (df) for Test 2 compare to Test 1?
Answer: The df for Test 2 are less than for Test 1. Degrees of freedom are calculated as the number of phenotypic categories minus one (n-1). In Test 1 (dihybrid cross), there are four phenotypic categories (e.g., round/yellow, round/green, wrinkled/yellow, wrinkled/green), so df = 4 - 1 = 3. In Test 2, the data are pooled to analyze a single trait (e.g., round vs. wrinkled), resulting in two phenotypic categories. Therefore, df = 2 - 1 = 1. The degrees of freedom for Test 2 (df=1) are less than for Test 1 (df=3).
Which statement provides the best evaluation of the geneticist's conclusion?
Answer: The conclusion is flawed; statistical tests can support, but not definitively prove, a scientific hypothesis. A high p-value (typically > 0.05) means that we fail to reject the null hypothesis. It indicates that the observed data is consistent with the expected ratio and that any deviations can be reasonably attributed to random chance. However, failing to reject the null hypothesis does not prove it is true. Science proceeds by falsifying hypotheses, not proving them. There might be other models that also fit the data. Therefore, the conclusion is an overstatement of what can be inferred from the statistical test.
A chi-square test for a 3:1 ratio yields a p-value of 0.04. Based on the conventional significance level of $\alpha = 0.05$ , what is the most valid conclusion?
Answer: The deviation from the 3:1 ratio is statistically significant, suggesting it is not due to random chance alone. The p-value (0.04) is less than the alpha level (0.05), which means the result is statistically significant. This leads us to reject the null hypothesis. The correct interpretation is that the observed deviation from the expected 3:1 ratio is unlikely to have occurred simply by random chance. While this deviation might be due to a biological phenomenon like linkage or experimental error, the chi-square test itself only indicates that the deviation is significant; it does not identify the cause.
A researcher hypothesizes that a mutation is semi-dominant, and a cross between two heterozygotes should yield a 1:2:1 phenotypic ratio. From 200 offspring, they observe 45 of phenotype 1, 110 of phenotype 2, and 45 of phenotype 3. They perform a chi-square test. What is the calculated $\chi^2$ value?
Answer: 2.0 First, determine the expected counts (E) based on the 1:2:1 ratio and 200 total offspring. E(phenotype 1) = (1/4) × 200 = 50. E(phenotype 2) = (2/4) × 200 = 100. E(phenotype 3) = (1/4) × 200 = 50. The observed counts (O) are 45, 110, and 45. Now, apply the chi-square formula: $\chi^2 = \sum \frac{(O-E)^2}{E}$ . $\chi^2 = \frac{(45-50)^2}{50} + \frac{(110-100)^2}{100} + \frac{(45-50)^2}{50} = \frac{25}{50} + \frac{100}{100} + \frac{25}{50} = 0.5 + 1.0 + 0.5 = 2.0$ .
A test cross between a heterozygous individual (AaBb) and a homozygous recessive individual (aabb) is performed. The resulting offspring show a significant excess of parental phenotypes (AaBb and aabb) and a deficit of recombinant phenotypes (Aabb and aaBb). Without performing a calculation, what is the most likely outcome of a chi-square test against the null hypothesis of independent assortment (1:1:1:1 ratio)?
Answer: A high $\chi^2$ value and a low p-value (< 0.05), leading to rejection of the null hypothesis. When you encounter test cross data showing an excess of parental types and deficit of recombinants, you're looking at evidence of genetic linkage rather than independent assortment. This setup tests your understanding of chi-square analysis and what the statistical results mean biologically. Independent assortment predicts a 1:1:1:1 ratio in test cross offspring. However, the observed data shows significant deviation from this expected ratio, with parental combinations (AaBb and aabb) appearing more frequently than recombinants (Aabb and aaBb). This pattern indicates the genes are linked on the same chromosome. In chi-square testing, large deviations between observed and expected values produce high $\chi^2$ values. High $\chi^2$ values correspond to low p-values, and when p < 0.05, we reject the null hypothesis. Since the data strongly deviates from the 1:1:1:1 ratio expected under independent assortment, we'd reject that null hypothesis. This makes D correct. Option A incorrectly suggests low deviation (low $\chi^2$ ) when we actually have high deviation. Option B makes the fundamental error of thinking high $\chi^2$ values give high p-values - it's the opposite relationship. Option C incorrectly pairs low $\chi^2$ with low p-values and rejection, but low $\chi^2$ means small deviations that wouldn't lead to rejection. Remember: significant deviations from expected ratios in genetics problems signal high $\chi^2$ values and low p-values, typically leading to rejection of simple Mendelian predictions like independent assortment.
A chi-square test on data from a genetic cross results in a $\chi^2$ value of 7.5 with 3 degrees of freedom. Using the provided table of critical values, which range contains the p-value for this result?
Answer: p > 0.05 To find the p-value range, locate the row for the given degrees of freedom (df=3) in the chi-square table. Then, find where the calculated $\chi^2$ value of 7.5 falls among the critical values in that row. The values for df=3 are 6.251 (for p=0.10) and 7.815 (for p=0.05). Since our calculated value of 7.5 is between 6.251 and 7.815, the corresponding p-value must be between 0.10 and 0.05. Therefore, the p-value is greater than 0.05.

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