All questions and explanatory responses
1. Calculate the expected phenotype counts under the stated null model.
Aa × Aa predicts AA, Aa, Aa and aa with equal probabilities in this simple model. Complete dominance therefore gives expected proportions 3/4 dominant and 1/4 recessive. Expected counts are 60 and 20 out of 80.
- Use a dominant:recessive probability ratio of 3:1.
- Multiply 80 by 3/4 to obtain 60 dominant.
- Multiply 80 by 1/4 to obtain 20 recessive.
Common confusion: The observed 64:16 ratio is the result to test, not the source of the null-model expected ratio.
2. Calculate χ² using Σ[(O − E)²/E]. Show the two contributions and round the total to three significant figures.
χ² = 16/60 + 16/20 = 0.2666… + 0.8 = 1.0666…, giving 1.07 to three significant figures. Both expected counts exceed five. Sum unrounded contributions before rounding the final result.
- Calculate (64 − 60)²/60 = 16/60.
- Calculate (16 − 20)²/20 = 16/20.
- Sum the unrounded terms to obtain 1.07 to three significant figures.
Common confusion: Dividing both terms by 80 or adding signed differences does not calculate this statistic.
3. Determine the degrees of freedom for this fully specified two-category goodness-of-fit model.
There are two categories with a fixed total, and no parameters have been fitted from these observations. Degrees of freedom = 2 − 1 = 1 for the stated model. A model with fitted parameters or a different test design would require reconsideration.
- Count two phenotype categories.
- Use categories minus one with no parameters fitted from the data.
- Give one degree of freedom.
Common confusion: Degrees of freedom are not the number of offspring minus one in this goodness-of-fit question.
4. Use the supplied critical value to assess whether these results reject the null model at the 5% level. State what the conclusion does not establish.
1.07 is below 3.841, so the null model is not rejected at the stated 5% level. The observed deviation is not large enough to cross this test’s rejection threshold. This is not proof of the inheritance model, and other models or limited power may also explain compatible data.
- Compare 1.07 with the supplied 3.841 critical value.
- State that the statistic is below the threshold, so the null is not rejected at 5%.
- Explain that this does not prove the model or show that no other model could fit.
Common confusion: Do not call the probability exactly 5% or claim that the null hypothesis has been proved true.
5. Explain why phenotype-dependent survival before counting could weaken this test of segregation.
The counted distribution reflects both which offspring were produced and which survived to observation. If survival depends on phenotype, an initially 3:1 distribution could change before counting. The stated expected counts assume equal survival, so this confounding process would weaken an inference about segregation alone.
- Distinguish genotype or phenotype production from later survival.
- Explain that unequal survival can alter counted phenotype proportions.
- Relate this to a violated assumption of the stated null model.
Common confusion: Observed phenotype counts can be influenced by events after fertilisation as well as by gamete probabilities.
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