Suppose a class observed that there were 234 plants

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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
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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.
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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: (RR) (Rr) (RR) (Rr) (rr) (Rr) (RR) (RR) (Rr) (RR) (Rr) (rr) (Rr) (Rr) (RR) (RR) (Rr) (RR) (rr) (Rr) a. What percentage of the people are right handed? Left handed? 17/20 are right handed so 85% 3/20 are left handed so 15% b. Find p and q and interpret each in the context of the problem. p = 25/40 = .625 62.5% of the alleles in the sample are dominant R. q = 15/40 = .375 37.5% of the alleles in the sample are recessive r. Note: p and q are NOT telling us the percent of people that are right and left handed. They refer the allele frequency Now suppose that we took another sample of 10 people. This time we only know their phenotypes. (Right) (Left) (Right) (Right) (Right) (Right) (Right) (Right) (Left) (Right) c. What percentage of the people are right handed? Left handed? Right handed = 8/10 = 80% Left handed = 2/10 = 20% c. Can you find p and q exactly? Why? No. We can’t see the genotypes of each person. For each right handed person, we do not know if they are homozygous (RR) or heterozygous (Rr).
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