(courtesy of Dr. Miller)
If wild type (=normal=standard type) does not occur in a useful part of the problem, then the analysis is limited to the evident segregation and assortment, and cannot imply particular alleles, nor how many mutants are present.
The P1 and F1 phenotypes will recur among the F2, if enough such progeny are produced.
The class with the highest frequency in the F2 will have the same phenotype as the F1, unless mimics combine enough class frequencies.
If both mutants in a dihybrid are dominant, or both are recessive, then the F2 data will have each single mutant type in the 3/16 frequency class. If one mutant is recessive and one dominant, then the single recessive mutant form shows as the 1/16 class, and the single dominant form shows as the 9/16 class. For polyhybrids note equivalent frequency classes.
The F2 or testcross frequencies are clues to the minimum number of assorting gene pairs. For example, in any F2 sixteenths imply at least a dihybrid (2 gene pairs) assortment; thirty-secondths might imply two gene pairs in any F2 equivalent plus one gene pair in a testcross, etc.
The number of classes also is a clue to the minimum number of assorting gene pairs. For example, if more than three phenotypes are present in the F2 or testcross data, one is dealing at least with a dihybrid. Epistatic gene interactions and mimics may reduce the number of phenotypic classes and allow fraction (frequency of occurrence) reductions.
If a purebred stock does not breed true, then it is based on a heterozygous phenotype of a codominant (Medial dominant) If its "F2" equivalent generation shows only two phenotypes among classified progeny with a peculiar ratio or frequency based on thirds, then at least one such codominant mutant is lethal before classification when homozygous.
If an F2 shows a phenotype limited to the hemizygous sex, sex-linkage is implied. If an F2 shows two phenotypes reversed in the sexes compared to the parental types (crisscross), sex-linkage is implied with the hemizygous or hetergametic sex possessing the dominant.
GENETIC RELATIONSHIP OF MUTANTS:
In crosses of purebred mutant stocks, if the F1 is normal, any and all mutants are recessive and non-allelic. If the F1 is mutant then one or any combination of three alternatives may exist:
One or both mutant stocks contain a dominant or medial dominant; OR
Two or more of the mutants are allelic; OR
Both stocks hold one or more mutants in common.
If alternative number 1 occurs, then a testcross of the F1, or else production of F2 is necessary to demonstrate recovery of normal in a di-or polyhybrid frequency.
If alternative number 2 or 3 occurs then the F2 or testcross data will not contain a normal phenotype. So to distinguish 2 from 3 and the particular mutants involved, a cross with wild type is necessary, usually through the F2 since the testcrosses will not yield homozygous codominant phenotypes.
Reduced penetrance or variable expressivity may obscure the analysis. Among possible alternatives choose the a priori most likely situation first.
William of Occam’s razor is a practical necessity in pedigree analysis.
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