# Formulas chi square e e o 2 2 − σ = χ o =

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Unformatted text preview: Formulas: Chi Square e e o 2 2 ) ( − Σ = χ o = observed individuals with observed genotype e = expected individuals with observed genotype Degrees of freedom equals the number of distinct possible outcomes minus one Degrees of Freedom p 1 2 3 4 5 6 7 8 0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 0.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09 Example problem: Wisconsin Fast Plants have two very distinctive visible traits (stems and leaves). Each plant will either have a purple (P) or green (p) stem and also have either have green (G) or yellow (g) leaves. Suppose that we cross a dihybrid heterozygous plant with another plant that is homozygous purple stem and heterozygous for the leaf trait. Make a Punnett square to figure out the expected ratios for the phenotypes of the offspring. PpGg x PPGg PG Pg PG Pg PG PPGG PPGg PPGG PPGg Pg PPGg PPgg PPGg PPgg pG PpGG PpGg PpGG PpGg pg PpGg Ppgg PpGg Ppgg Purple stem/Green leaves 12 Purple stem/Yellow leaves 4 so the expected ratio of the phenotypes is 12:4 or 3:1. Suppose a class observed that there were 234 plants that were purple stem/green leaves and 42 that were purple stem/yellow leaves. Does this provide good evidence against the predicted phenotype ratio? Find expected values. 234 + 42 = 276 total. 207 4 3 * 276 = 69 4 1 * 276 = df= n – 1 = 1 o e o – e e e o 2 ) ( − 234 207 27 3.52 42 69 -27 10.57 so X 2 = 3.52 + 10.57 = 14.09 Because 14.09 > 3.84, P-value < 0.01 The data did not occur purely by chance alone. Reject the null hypothesis Using your understanding of genetics, what might be one reason why the class got these results? One of the conditions must have been violated, probably the independent assortment condition. This could lead you to a test cross for linkage, of which you could check map units. Sample size could also be the culprit. The test cross for detection of linkage is somewhat different from a regular test cross. In the former case, the genotypes of both of the parents are known. The parent with the dominant phenotype is known to be a heterozygote (for example, AaBb) and the tester parent is known to be completely homozygous recessive (aabb). The heterozygous parent with the dominant phenotype, if the genes show independent assortment, there should be a 50% recombination frequency. In other words 50% of the children have one of the two parental genotypes and 50% of the children are recombinants (An offspring whose phenotype differs from that of the parents). If the recombination frequency is less than 50% the genes linked and thus are not assorting independently. The smaller the recombination frequency, the closer the two genes are to each other on the chromosome. Hardy-Weinberg Review Formulas: p 2 + 2pq + q 2 = 1 p = frequency of the dominant allele in a population p + q = 1 q = frequency of the recessive allele in a population Example problem: For people, being right handed (R) is the dominant trait over being left handed (r). Suppose there is a sample of 20 people that reveals the following genotypes:...
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Formulas Chi Square e e o 2 2 − Σ = χ o = observed...

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