240 Dark to 121 Pale: A Ratio That Refuses to Be 3 : 1
In a school genetics practical, dark-bodied beetles from the same laboratory stock are crossed with one another and the offspring counted. In the largest single brood, 240 beetles are dark and 121 are pale. The rule the class has learned predicts a 3 : 1 ratio for a cross between two heterozygotes, which would give about 271 dark and 90 pale, so a student concludes that the counting was careless or that the parents were not both heterozygous. The teacher then shows the breeding records. In a repeat cross, 36 eggs were laid; 9 failed to hatch, and of the 27 that hatched, 18 were dark and 9 pale. She adds that in twenty years no dark beetle has ever bred true: every dark beetle test-crossed with a pale one has produced dark and pale offspring in equal numbers.
1. Test the 3 : 1 and the 2 : 1 hypotheses against the brood of 240 dark and 121 pale (total 361), using χ² with a critical value of 3.84 at one degree of freedom, and state which model survives.
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3 : 1 hypothesis. Expected dark = 361 × 3/4 = 270.75; expected pale = 90.25.
χ² = (240 − 270.75)²/270.75 + (121 − 90.25)²/90.25 = 945.6/270.75 + 945.6/90.25 = 3.49 + 10.48 = 13.97
This far exceeds 3.84, so the 3 : 1 hypothesis is rejected: the deviation is too large to be sampling noise, and looking for counting errors will not rescue it.
2 : 1 hypothesis. Expected dark = 361 × 2/3 = 240.67; expected pale = 120.33.
χ² = (0.67)²/240.67 + (0.67)²/120.33 = 0.002 + 0.004 = 0.006
This is far below 3.84, so the data are entirely consistent with 2 : 1. The student's conclusion that something went wrong should be replaced by the conclusion that the model was wrong.
2. Propose a genetic model that yields 2 : 1, and show how the egg records support it.
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Model: dark (D) is dominant to pale (d) for body colour, but DD is lethal - the homozygote dies before hatching, so it never appears among the counted adults. A cross Dd × Dd gives 1 DD : 2 Dd : 1 dd, and when the DD class is removed only Dd (dark) and dd (pale) remain, in the ratio 2 : 1.
The egg record fits this exactly. Of 36 eggs, 9 fail to hatch and 18 dark : 9 pale hatch. In a 1 : 2 : 1 segregation, 9 of 36 is precisely the expected DD quarter, and the surviving 27 split 18 : 9, i.e. 2 : 1. The unhatched eggs are not an accident of incubation; they are the missing genotype, and they explain the deficit of dark beetles in the adult count. The twenty-year breeding record confirms it: no dark beetle breeds true, because the only surviving dark genotype is Dd.
3. Predict the outcome of a test cross between a dark beetle and a pale beetle under your model, and explain what this prediction rules out that a simple dominance model does not.
Read the worked answer
Under the model, every dark beetle is Dd, so Dd × dd gives 1 Dd : 1 dd - half dark, half pale. That prediction is identical to what simple dominance predicts when the dark parent happens to be heterozygous, so the test cross alone does not separate the two models.
What separates them is what the test cross can never produce. Under simple dominance, a pure-breeding DD dark beetle exists, and crossing it with a pale beetle would give 100 % dark offspring. Under the lethal model no such parent can exist, so a test cross that yields all dark offspring is impossible, and any dark × dark cross must always give 2 : 1 rather than 3 : 1. The discriminating evidence is therefore the combination: the 2 : 1 ratio, the 25 % failure to hatch, and the total absence of a true-breeding dark line - no one of these alone is decisive, but together they exclude simple dominance.