Genetics Quiz: Chi Square Tests For Mendelian Ratios

What this quiz covers

This quiz focuses on Chi Square Tests For Mendelian Ratios, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 0.24
2. 0.48
3. 0.96 (correct answer)
4. 1.92

Explanation: 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 $2 \times 0.48 = 0.96$.

Question 2

1. 0.25
2. 1.25 (correct answer)
3. 2.50
4. 5.00

Explanation: 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 $\frac{(15 - 20)^2}{20} = \frac{(-5)^2}{20} = \frac{25}{20} = 1.25$.

Question 3

1. 1.0
2. 1.5
3. 2.0 (correct answer)
4. 2.5

Explanation: 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: $\chi^2 = \sum \frac{(O-E)^2}{E}$. $\chi^2 = \frac{(45-50)^2}{50} + \frac{(110-100)^2}{100} + \frac{(45-50)^2}{50} = \frac{25}{50} + \frac{100}{100} + \frac{25}{50} = 0.5 + 1.0 + 0.5 = 2.0$.

Question 4

1. df = 1; the observed results do not differ significantly from a 3:1 ratio.
2. df = 1; the observed results do not differ significantly from a 2:1 ratio. (correct answer)
3. df = 2; the observed results do not differ significantly from a 2:1 ratio.
4. df = 2; the observed results do not differ significantly from a 1:2:1 ratio.

Explanation: 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.

Question 5

1. 0.4
2. 4.0
3. 40.0 (correct answer)
4. 400.0

Explanation: 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 $10 \times 4.0 = 40.0$.

Question 6

1. The df for Test 2 are greater than for Test 1.
2. The df for Test 2 are less than for Test 1. (correct answer)
3. The df are the same for both tests because the total number of individuals is unchanged.
4. The df are the same for both tests because the underlying cross is the same.

Explanation: 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).

Question 7

1. 300 dominant, 100 recessive (correct answer)
2. 200 dominant, 200 recessive
3. Depends on the observed counts of the progeny
4. 250 dominant, 150 recessive

Explanation: 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.

Question 8

1. The conclusion is correct; a high p-value confirms the null hypothesis is true.
2. The conclusion is flawed; statistical tests can support, but not definitively prove, a scientific hypothesis. (correct answer)
3. The conclusion is incorrect; a p-value of 0.85 is statistically significant and the null hypothesis should be rejected.
4. The conclusion is questionable; such a high p-value suggests the data may be fabricated or the experiment was biased.

Explanation: 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.

Question 9

1. A low $\chi^2$ value and a high p-value (> 0.05), failing to reject the null hypothesis.
2. A high $\chi^2$ value and a high p-value (> 0.05), supporting the null hypothesis.
3. A low $\chi^2$ value and a low p-value (< 0.05), leading to rejection of the null hypothesis.
4. A high $\chi^2$ value and a low p-value (< 0.05), leading to rejection of the null hypothesis. (correct answer)

Explanation: 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.

Question 10

1. p-value X > p-value Y > p-value Z (correct answer)
2. p-value Z > p-value Y > p-value X
3. p-value X = p-value Y = p-value Z
4. p-value Y > p-value X > p-value Z

Explanation: 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.

Question 11

1. A low $\chi^2$ value and a high p-value (> 0.05), failing to reject the null hypothesis.
2. A high $\chi^2$ value and a high p-value (> 0.05), supporting the null hypothesis.
3. A low $\chi^2$ value and a low p-value (< 0.05), leading to rejection of the null hypothesis.
4. A high $\chi^2$ value and a low p-value (< 0.05), leading to rejection of the null hypothesis. (correct answer)

Explanation: 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.

Question 12

1. 300 dominant, 100 recessive (correct answer)
2. 200 dominant, 200 recessive
3. Depends on the observed counts of the progeny
4. 250 dominant, 150 recessive

Explanation: 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.

Question 13

1. 0.24
2. 0.48
3. 0.96 (correct answer)
4. 1.92

Explanation: 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 $2 \times 0.48 = 0.96$.

Question 14

1. 1.0
2. 1.5
3. 2.0 (correct answer)
4. 2.5

Explanation: 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: $\chi^2 = \sum \frac{(O-E)^2}{E}$. $\chi^2 = \frac{(45-50)^2}{50} + \frac{(110-100)^2}{100} + \frac{(45-50)^2}{50} = \frac{25}{50} + \frac{100}{100} + \frac{25}{50} = 0.5 + 1.0 + 0.5 = 2.0$.

Question 15

1. 0.4
2. 4.0
3. 40.0 (correct answer)
4. 400.0

Explanation: 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 $10 \times 4.0 = 40.0$.

Question 16

1. The df for Test 2 are greater than for Test 1.
2. The df for Test 2 are less than for Test 1. (correct answer)
3. The df are the same for both tests because the total number of individuals is unchanged.
4. The df are the same for both tests because the underlying cross is the same.

Explanation: 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).

Question 17

1. The data prove that the two genes assort independently as predicted by the null hypothesis.
2. The null hypothesis is rejected, but with very low confidence in the decision.
3. The fit between the observed and expected data is poor, warranting further investigation.
4. The observed deviation from the expected ratio is consistent with random sampling error. (correct answer)

Explanation: 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.

Question 18

1. 0.25
2. 1.25 (correct answer)
3. 2.50
4. 5.00

Explanation: 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 $\frac{(15 - 20)^2}{20} = \frac{(-5)^2}{20} = \frac{25}{20} = 1.25$.

Question 19

1. The conclusion is correct; a high p-value confirms the null hypothesis is true.
2. The conclusion is flawed; statistical tests can support, but not definitively prove, a scientific hypothesis. (correct answer)
3. The conclusion is incorrect; a p-value of 0.85 is statistically significant and the null hypothesis should be rejected.
4. The conclusion is questionable; such a high p-value suggests the data may be fabricated or the experiment was biased.

Explanation: 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.

Question 20

1. df = 1; the observed results do not differ significantly from a 3:1 ratio.
2. df = 1; the observed results do not differ significantly from a 2:1 ratio. (correct answer)
3. df = 2; the observed results do not differ significantly from a 2:1 ratio.
4. df = 2; the observed results do not differ significantly from a 1:2:1 ratio.

Explanation: 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.