Chi-square analysis of crosses Knowledge-Based Study

Study notes

1. Testing Genetic Ratios

   In genetic crosses, observed offspring counts rarely match expected theoretical ratios perfectly. Random sampling error and chance fluctuations occur naturally during fertilization and gamete formation. Geneticists use the Chi-square ($\chi^2$) goodness-of-fit test to determine whether observed deviations from theoretical expectations are due to chance alone or whether another biological mechanism is operating.

   Formulating Hypotheses

   Every Chi-square analysis begins with two formal hypotheses:

   • Null Hypothesis ($H_0$): There is no significant difference between the observed and expected phenotypic counts. Any observed deviation is caused by random chance, meaning the data fit the specified inheritance model.
   • Alternative Hypothesis ($H_A$): The observed counts differ significantly from the expected values. The deviation is too large to be attributed to chance, meaning the data do not support the hypothesized inheritance model.

   Calculating Expected Counts

   Before applying the Chi-square formula, expected values ($E$) must be calculated as counts rather than percentages or proportions. To find the expected number for each phenotypic class, multiply the theoretical fraction expected from the genetic cross by the total number of individuals observed ($N$):

   $E = \text{fraction} \times N$

   The Chi-Square Formula

   The test statistic is calculated by summing the squared deviations between observed ($O$) and expected ($E$) counts divided by the expected count for each phenotypic category:

   $$\chi^2 = \sum \frac{(O - E)^2}{E}$$

   • $O$: The observed count in each phenotypic class.
   • $E$: The expected count in each phenotypic class based on the theoretical model.
   • $\Sigma$: The summation across all phenotypic classes.

   Degrees of Freedom ($df$)

   The distribution of the Chi-square statistic depends on the degrees of freedom ($df$), which reflects the number of independent values that can vary. In goodness-of-fit tests for phenotypic crosses, degrees of freedom are calculated as:

   $df = k - 1$

   Here, k represents the total number of distinct phenotypic classes observed.

   The Critical Value and P-Value

   In biological sciences, the standard significance threshold (alpha level, $\alpha$) is conventionally set at 0.05. By locating the degrees of freedom on a standard Chi-square critical values distribution table:

   • Fail to Reject $H_0$ ($\chi^2 \le \text{critical value}$, $p \ge 0.05$): The difference between observed and expected results is not statistically significant. The inheritance model is supported.
   • Reject $H_0$ ($\chi^2 > \text{critical value}$, $p < 0.05$): The observed deviations are statistically significant. The proposed genetic hypothesis is rejected, suggesting phenomena such as gene linkage, differential viability, or epistasis.

   Evaluating an Incomplete Dominance Cross

   Consider a cross between two pink-flowered four o'clock plants (Mirabilis jalapa). Under incomplete dominance, the theoretical phenotypic ratio is 1 Red : 2 Pink : 1 White (a 1:2:1 ratio). A cross produces 200 total offspring with the following counts:

   • Red: Observed = 42 (Expected = 50)
   • Pink: Observed = 112 (Expected = 100)
   • White: Observed = 46 (Expected = 50)

   Step-by-Step Calculation

   1. Red: $\frac{(42 - 50)^2}{50} = \frac{(-8)^2}{50} = \frac{64}{50} = 1.28$
   2. Pink: $\frac{(112 - 100)^2}{100} = \frac{12^2}{100} = \frac{144}{100} = 1.44$
   3. White: $\frac{(46 - 50)^2}{50} = \frac{(-4)^2}{50} = \frac{16}{50} = 0.32$

   Summing these components gives:

   $$\chi^2 = 1.28 + 1.44 + 0.32 = 3.04$$

   Interpretation

   With 3 phenotypic classes, $df = 3 - 1 = 2$. At $p = 0.05$, the critical value for $df = 2$ is 5.991. Because our calculated value (3.04) is less than 5.991, we fail to reject the null hypothesis. The observed data are consistent with the hypothesis of incomplete dominance.