[R-sig-ME] all pair comparisons in proportion analysis

Atsuko Sato atsuko.sato at zoo.ox.ac.uk
Fri Oct 5 17:45:42 CEST 2012

Dear all,

I was wondering how we could compare thermal tolerance between genotypes using mixed effect model. Experimental data tell us that two closely related species of marine invertebrate have different sensitivities to temperature, specifically one is more resistant to higher water temperatures. The project is seeking to compare thermal tolerance between genotypes as follows:

*        We generate crosses of defined genotype. The two species are known as A and B, and offspring are: AA, BB, AB, BA (the last two are hybrids, AB =A mother and B father, BA the reverse). These have been replicated using different parents (batches) collected from from the field at least 7 times.

*        For each replicate the fertilised eggs (many 100s) are split in two. Half are maintained at constant temperature as a control. Half are heat shocked. Both cultures then allowed to develop to the same point.  Both the proportion of embryos developing normally in the heat treated sample and that in control are calculated.

By analysing the data using mixt effect model, batches as random effect, the test showed that interaction model of genotype and treatment is significantly better fit than non-interaction model.

Questions here is then which genotype makes the difference and where are the differences are. Since proportion of normal larvae is normally not 100%, thermal tolerance should be a proportion of normal larvae in heat treated section normalised by a proportion of normal larvae in control section. How can we conduct all pair comparison of such complex proportion (proportion/proportion) data? I have used
but it seems incorrect because the p-value does not matches with the plot I will get. In the plot of thermal tolerance (proportion of normal larvae in heat treated section devided by proportion of normal larvae in control section) AA (=1) and BB (=4) shows a large difference, for example, but the p=0.981.

Details of how I did these tests (using R) and the dataset and the plot of thermal tolerance are attached.
Treatment: 1, control; 2, heat shock
Genotype: AA=1, AB=2, BA=3, BB=4

Thank you very much.

Yours sincerely
Atsuko Sato

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