HOW TO SOLVE BREEDING-DATA PROBLEMS
(courtesy of Dr. Miller)
(P1 = Purebred Stock) (F1 = First Generation) (F2 = Second Generation) (TC = Test Cross)
I. If wild type (normal, standard) occurs in starting stocks
A. Find crosses with it giving wild type F1 (only);
1. The other parent stock is homozygous recessive mutant (i.e. any and all mutants are recessive).
a. If the F2 is given, the irreducible ratio or frequency indicates the minimum number of mutants assorting. For example, 16ths implies two mutants; and similar reasoning is used for testcross data, or for mixtures of F2 and testcross data. Epistasis and mimics may reduce the number of phenotypic classes and allow fraction reduction, e.g. 4/16 to ¼.
b. Give symbols to each mutant (e.g. aa for albino).
c. If feasible, write down physiological action of each mutant (e.g. aa controls absence of all melanin pigment)-—helpful, but not mandatory.
B. Find crosses with wild type giving non-wild type F1 (only).
1. If the F1 look like the mutant parental type, the latter is homozygous for at least one dominant mutant.
a. If the F2 or backcross is given, the frequency will indicate how many mutants (as in 1.a. above).
b. Give the symbols to each mutant (e.g. M for multiple spurs).
c. Write down the physiological action of each mutant (e.g. controls growth of more than normal spurs).
2. If the F1 is intermediate (only), the mutant parent type is homozygous for a partial-dominant (incomplete or co-dominant) mutant or mutants.
a. If the F2 or backcross is given, the frequency will indicate how many mutants (as in 1.a above).
b. Give symbols (e.g. FF for extreme frizzle, FF+ for mild frizzle).
c. Write down physiological action of mutant (e.g. F causes feathers to curl).
C. Find any families in which the F1 show "crisscross" (reciprocal crosses not equal) indicating sex linkage.
II. If wild type (normal, standard) is not among starting stocks
A. Find crosses giving normal in the F1.
1. These indicate that each parent type is homozygous for one or more recessive mutants not possessed by the other.
a. F2 or backcross frequencies imply the number of mutants (as in I.1.a. above).
b. Give symbols.
c. Write down the physiological actions.
B. Are any starting purebred stocks unfixable? If so, intermediate dominance is implied for at least one gene pair. (Purebred does not always imply homozygous).
1. The phenotypic frequencies among the progeny show the number of mutants segregating or assorting and whether or not one is lethal homozygous (e.g. 3rds = 1 mutant, lethal homozygous missing).
2. If wild type constitutes part of the progeny from an unfixable type, the heterozygote can be symbolized e.g. AA+.
a. Write down the physiological action.
C. If the cross between two pure abnormal stocks gives all abnormal F1, but some wild type in the F2, one or both starting stocks must be homozygous for a dominant or partial-dominant mutant.
1. The F2 frequency, number of classes, etc. should indicate the number of mutants involved in the cross.
D. If wild type does not appear in P1, F1, or F2, the starting stocks concerned probably have at least one homozygous mutant in common, or there may be multiple mutant alleles involved. If any colors are the result of interaction of two (2) or more mutants, these frequencies will be "off" and higher numbers of young will be necessary to demonstrate it. Four (4) or more types within any one family is sufficient proof of more than one mutant being involved.
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