Abstract
Evidences indicate that a man, doubly hemizygous for the abnormal genes of both series of color-blindness, manifests anomaly either of the protan- or the deutan-type. A method for estimating recombination value between the loci for the two color-blindness genes has been devised by modifying Smith's method for linkage test based upon Bayes' theorem, and z(θ) scores for various values of family size and recombination fraction are presented.
Families usable for such study should have (1) a phenotypically normal mother, (2) three or more boys examined for color vision of whom (3) at least one shows the protan-type anomaly and another the deutan-type anomaly.
Calculation based on 6 suitable pedigrees has yielded that the recombination value, at which the probability obtaining such a sample is highest, is about 0.06 under the assumption that the deutan type anomaly is epistatic over the protan type anomaly and about 0.15 if it is assumed that the protan type is epistatic over the deutan type.
