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Tamarin: Principles of Genetics, Seventh Edition
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II. Mendelism and the Chromosomal Theory
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2. Mendel s Principles
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Genotypic Interactions
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Mendel determined whether a dominant phenotype in the F2 generation was a homozygote or a heterozygote by self-fertilizing it and examining ten offspring. In an F2 generation composed of 1AA:2Aa:1aa, he expected a 2:1 ratio of heterozygotes to homozygotes within the dominant phenotypic class. In fact, this ratio is not precisely correct because of the problem of misclassi cation of heterozygotes. It is probable that some heterozygotes will be classi ed as homozygotes because all their offspring will be of the dominant phenotype. The probability that one offspring from a selfed Aa individual has the dominant phenotype is 3/4, or 0.75: the probability that ten offspring will be of the dominant phenotype is (0.75)10 or 0.056. Thus, Mendel misclassi ed heterozygotes as dominant homozygotes 5.6% of the time. He should have expected a 1.89:1.11 ratio instead of a 2:1 ratio to demonstrate segregation. Mendel classi ed 600 plants this way in one cross and got a ratio of 201 homozygous to 399 heterozygous offspring.
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This is an almost perfect t to the presumed 2:1 ratio and thus a poorer t to the real 1.89:1.11 ratio. This bias is consistent and repeated in Mendel s trihybrid analysis. Fisher, believing in Mendel s basic honesty, suggested that Mendel s data do not represent an experiment but more of a hypothetical demonstration. In 1971, F. Weiling published a more convincing case in Mendel s defense. Pointing out that the data of Mendel s rediscoverers are also suspect for the same reason, he suggested that the problem lies with the process of pollen formation in plants, not with the experimenters. In an Aa heterozygote, two A and two a cells develop from a pollen mother cell. These cells tend to stay together on the anther. Thus, pollen cells do not fertilize in a strictly random fashion. A bee is more likely to take equal numbers of A and a pollen than chance alone would predict. The result is that the statistics Fisher used are not applicable. By using a different statistic, Weiling showed that, in fact, Mendel need not have manipu-
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lated any numbers (nor would have his rediscoverers) in order to get data that t the expected ratios well. By the same reasoning, very little misclassi cation of heterozygotes would have occurred. More recently, Weiling and others have made several additional points. First, for Mendel to be sure of ten offspring, he probably examined more than ten, and thus he probably kept his misclassi cation rate lower than 5.6%. Second, despite Fisher s brilliance as a statistician, several have made compelling arguments that Fisher s statistical analyses were incorrect. In other words, for subtle statistical reasons, many of his analyses involved methods and conclusions that were in error. We conclude that there is no compelling evidence to suggest that Mendel in any way manipulated his data to demonstrate his rules. In fact, taking into account what is known about him personally, it is much more logical to believe that he did not cheat.
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cross the hens and roosters of this heterozygous F1 group, we will get, in the F2 generation, walnut-, rose-, pea-, and single-combed fowl in a ratio of 9:3:3:1. Can you gure out the genotypes of this F2 population before reading further An immediate indication that two allelic pairs are involved is the fact that the 9:3:3:1 ratio appeared in the F2 generation. As we have seen, this ratio comes about
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when we cross dihybrids in which both genes have alleles that control traits with complete dominance. Figure 2.21 shows the analysis of this cross. When dominant alleles of both genes are present in an individual (R- P-), the walnut comb appears. (The dash indicates any second allele; thus, R- P- could be RRPP, RrPP, RRPp, or RrPp.) A dominant allele of the rose gene (R-) with
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Table 2.3 Multihybrid Self-Fertilization, Where n Equals Number of Genes Segregating Two Alleles Each
Monohybrid n 1 Number of F1 gametic genotypes Proportion of recessive homozygotes among the F2 individuals Number of different F2 phenotypes, given complete dominance Number of different genotypes (or phenotypes, if no dominance exists) 2 1/4 2 3 Dihybrid n 2 4 1/16 4 9 Trihybrid n 3 8 1/64 8 27 General Rule 2n 1/(2n)2 2n 3n
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