4.2 Chi-squared Testing of Genetic Ratios

Genetic counts rarely match ratios exactly. Chi-squared asks whether deviations are plausible sampling variation under a stated model. The null hypothesis says there is no significant difference between observed and expected frequencies. Expected counts come from the model ratio multiplied by total sample size.

For a 9:3:3:1 ratio among 320, expected counts are 180, 60, 60 and 20. For a 1:1:1:1 test cross among 800, each is 200. Categories must be mutually exclusive and expected counts sufficiently large for the approximation.

For every category, subtract expected from observed, square, divide by expected and sum. Squaring removes sign and magnifies large deviations; division by expected prevents the same numerical difference from having equal weight in a rare and common class.

Degrees of freedom are categories minus one when the ratio is fixed. A two-class monohybrid comparison has df = 1; a four-class dihybrid has df = 3. At $p=0.05$, critical values are 3.841 and 7.815 respectively.

If calculated chi-squared exceeds the critical value, reject the null: a discrepancy at least this large would occur by chance less than 5% of the time under the model. Otherwise, do not reject it. This does not prove the model true; evidence is merely insufficient to reject it at that threshold.

Sample size changes sensitivity. The same proportional deviation produces larger chi-squared with more observations. A significant departure might indicate linkage, but selection, scoring error, unequal viability or a wrong phenotype rule are alternatives. Statistics identifies mismatch; biology explains it.

Organize a calculation in columns for phenotype, expected ratio, expected count, observed count, $O-E$, $(O-E)^2$ and $(O-E)^2/E$. The signs in $O-E$ should balance because observed and expected totals are equal, but the squared contributions are all positive. Keeping intermediate values avoids arithmetic errors and makes it possible to identify which phenotype contributes most strongly to the final statistic.

Suppose a 3:1 cross produces 232 dominant and 88 recessive offspring from 320. Expected counts are 240 and 80. The contributions are $64/240=0.267$ and $64/80=0.800$, giving $\chi^2\approx1.067$. With one degree of freedom, this is below 3.841, so the result does not justify rejecting the 3:1 model. The same absolute deviation contributes more in the smaller expected class.

For four classes expected in a 1:1:1:1 ratio, degrees of freedom equal three because once three category counts and the fixed total are known, the fourth is determined. Choosing degrees of freedom from sample size is wrong. The probability column of a critical-value table also needs careful reading: the 0.05 column sets the conventional threshold; it is not the probability that the null hypothesis is true.

Rejecting a 9:3:3:1 model does not by itself demonstrate linkage. Linked loci are one candidate explanation, especially when reciprocal parental classes are high and recombinant classes low. A lethal genotype may selectively remove one category, poor scoring may merge phenotypes, and fertilization or survival may be unequal. Examine the direction of deviation and design another cross before assigning mechanism.

Failure to reject can also have several meanings: the model may fit well, the sample may be too small to detect a real difference, or deviations may cancel across broad categories. Statistical power increases with sample size and effect size. A careful conclusion therefore states the model, threshold, statistic, degrees of freedom and evidence decision without claiming proof.