Interpreting a Genetic Cross with Unexpected Ratios
Two fruit flies with normal wings are crossed. In the offspring, 76 have normal wings and 24 have curled wings. A student expects a 3:1 ratio. A second cross between a normal-winged fly and a curled-winged fly gives 41 normal and 39 curled. The student must decide whether the data fit simple Mendelian inheritance and what the most likely genetic model is.
1. Test whether the first cross fits a 3:1 ratio and interpret the result.
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Total offspring = 76 + 24 = 100. Expected under 3:1: 75 normal, 25 curled. Chi-square = (76−75)²/75 + (24−25)²/25 = 1/75 + 1/25 = 0.0133 + 0.040 = 0.053. With 1 degree of freedom, the critical value at p = 0.05 is 3.84; 0.053 is far below it, so the data are entirely consistent with 3:1. Interpretation: both parents are heterozygous (Nn × Nn), normal (N) is dominant to curled (n), and the curled offspring are nn. The judgement is not just 'it fits' but identifying the parental genotypes from the ratio.
2. Interpret the second cross and state the parental genotypes.
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The second cross gives approximately 1:1 (41:39). A 1:1 ratio arises from a cross between a heterozygote and a homozygous recessive: Nn × nn → 1 Nn : 1 nn. So the normal-winged parent is heterozygous and the curled parent is nn. Chi-square for 1:1 with 80 offspring (expected 40 each): (41−40)²/40 + (39−40)²/40 = 0.025 + 0.025 = 0.05, well below 3.84, so the fit is good. The combined evidence from both crosses supports a single-gene model with complete dominance, and rules out the normal parent being homozygous NN (which would give all normal offspring).
3. A third cross of two curled-winged flies gives all curled offspring. What does this confirm, and what caveat applies?
Read the worked answer
Curled × curled giving all curled offspring is consistent with curled being homozygous recessive (nn × nn → all nn). This confirms the recessive assignment and shows curled flies breed true. Caveat: a small sample could hide rare events, and 'all curled' could also occur if curled were dominant and both parents homozygous - but that is excluded by the earlier 3:1 and 1:1 results. The judgement is to use the third cross as corroboration, not as independent proof, and to note that sample size limits confidence in 'all'.