12 Jun Ottawa BIO 2133 – Kernels of corn are separate individual
Question
Kernels of corn are separate individual progeny (each kernel is the result of a pollen grain that landed on one strand of corn “silk” and that grew all the way down the cob to fertilize an ovule). Therefore, a corn cob represents a large sample of progeny from a single maternal parent plant. These corn cobs display four phenotypes, as listed in column 1 below. We want to test the hypothesis that kernel color and surface quality are caused by two separate genes with two alleles each, with one allele completely dominant at each locus.
Phenotype Genotype (H1) Genotype – Observed counts
purple, smooth P- S- 1692
purple, wrinkled P- ss 559
yellow, smooth pp S- 642
yellow, wrinkled pp ss 242
Use a ??X^2-test to determine whether observed genotype frequencies differ significantly from what you’;d expect under this model of inheritance (in which case you’d reject this model) using only the observed kernel count data (column 4 above). To do this you need to calculate the allele frequencies, then using these the expected numbers of each genotype, then the X^2 and P-value (to get the latter use the chidist function in Excel or this site: http://www.socscistatistics.com/pvalues/chidistribution.aspx). Clearly state whether the results of your analysis lead you to reject this model of inheritance or whether they support it.
Note this this is NOT a test of Hardy-Weinburg. It is a test of a particular model of inheritance for these traits (there are others) and you have to assume HW to get the allele frequencies. This is not circular: you assume a particular model of inheritance AND HWE, then calculate the genotype frequencies you would expect if this model of inheritance was true. If the results indicate you should reject the null hypothesis, than this doesn’t mean you reject HW, rather you reject the model of inheritance. The test has one degree of freedom: 4 classes of data (i.e. genotype counts) – 2 allele frequencies you have to estimate (think about this, for both hypotheses you need to calculate two allele frequencies and from these you can get the others) -1 = 1.
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