Revision of Fisher's analysis of Mendel's garden pea experiments.

R. A. Fisher (1936) made the assertion that the that Mendel would have counted those, correctly, as data in Mendel’s experiments with garden peas heterozygotes. Second, in those cases in which there (Mendel 1866) were too close to expectation. One of were 9 or fewer plants, all with dominant traits, Mendel the most striking examples seemed to be the six experiwould not have the 10 he specified as the number he ments with plant characters designed to test the theoreti“cultivated,” and because he would be less certain that cal 2:1 ratio of heterozygotes (Aa) to homozygotes (AA) the selfed F2 were indeed homozygous, it is highly plausiamong F2 plants exhibiting the dominant trait. In each ble that Mendel would have redone those sets of 10 (or of these experiments, 100 F2 plants showing only the used extra sets of 10 planted in anticipation of inevitable dominant trait were selfed, and 10 seeds from each were losses). planted. When a mixture of dominant and recessive The result of Novitski’s proposal is that there are two traits was observed among the 10 resulting plants, Meneffects on the expected ratio: the undercounting of del classified the F2 as heterozygous, and when the 10 heterozygotes in sets of 10 dominant-trait plants and plants all had dominant traits, Mendel classified the F2 the undercounting of homozygotes when sets of 9 or as homozygous. Mendel observed a 1.99:1 ratio overall fewer dominant-trait plants are discarded. The expected in these data, and he concluded that the six experiments quotient, R, of those counted as heterozygotes divided agreed with a 2:1 ratio. Fisher pointed out that occasionby those counted as homozygotes is calculated as follows. ally (5.6% of the time) a heterozygous F2 would have Those counted as heterozygotes are the sum of the 10 dominant-trait offspring in a row by chance and that following products: the fraction of F2 plants that are therefore the expected experimental ratio by Mendel’s actually heterozygous (two-thirds), the probability of methods should be 1.7:1, and not 2:1. Thus, Fisher consets of a certain number of surviving plants (based on cluded that some sort of bias must have entered into a failure rate, q), and the probability that that certain the execution of these experiments or the presentation number of plants includes at least one with the recessive of the data. trait. Those counted as homozygotes are the product This key conclusion of Fisher has been challenged by of the probability that all 10 plants will survive, and the E. Novitski (2004, accompanying article in this issue). sum of the fraction of the F2 plants that are actually First, it is highly unlikely that Mendel could plant 6000 homozygous plus the fraction of F2 plants that are actuplants (six experiments 100 F2 plants 10 seeds ally heterozygous times the probability of a heterozygous planted) with no losses. Novitski points out that Mendel F2 plant giving rise to all 10 dominant-trait offspring. does not give the rate of failure data for his 2:1 ratio This simplifies to the formula for R, the ratio of those experiments, but in a subsequent experiment of Mencounted as heterozygotes (Aa) to those counted as hodel’s, he mentions a 2% (11 of 556) failure of seeds to mozygotes (AA), germinate and survive. Then, if some sets of 10 were observed to have 9 or fewer surviving plants, what would R 10 i 1P(i){1 (3/4)i } P(10){(1/2) (3/4)10} , Mendel have done? It would be perfectly clear to Mendel that a set of 9 or fewer plants that had a mixture of where P(i) is the binomial distribution, dominant and recessive traits must have come from a selfed heterozygote, and Novitski persuasively argues P(i) k !