Linkage and Crossing Over Between Genes 



110 



In another species, the sweet pea, the 

 trait purple flowers is due to a single gene 

 ( + ) whose recessive allele (r) produces 

 red flowers. Long pollen ( + ) is dominant 

 to round pollen (ro). Assume two pairs 

 of genes are involved in a cross between a 

 pure line of purple long (-)- -+-/+ +) and 

 red round (r ro rro). The F] produces all 

 purple long (-| — \-/rro) and self-fertiliza- 

 tion of the F, produces in F 2 too many P, 

 phenotypes and too few new recombina- 

 tional types (purple round and red long) 

 for independent segregation. Therefore, 

 these genes must be linked. But, as before, 

 linkage is incomplete. 



In this case, the crossovers obtained can 

 be accounted for if the Pj(Fj) dihybrid 

 forms gametes in the relative proportions 

 10 + -f:10rro:l + ro:l r + . This fre- 

 quency of crossovers is obtained no matter 

 how the genes enter the dihybrid. Notice, 

 however, that the constant frequency (1/11) 

 in the sweet pea differs from the frequency 

 (1/64) observed previously in the garden 

 pea. 



Consider also, the following cases: 



1 . In Drosophila you recall, the mutant 

 gene (w) for white eye is X-linked. So 

 also is another (presumably nonallelic) mu- 

 tant gene which produces miniature wings 

 (m). Using pure lines, a white-eyed long- 

 winged fly is crossed to a dull-red-eyed min- 

 iature-winged fly. The Fi female carries 

 two X's and is, presumably, w +/+ m - 

 This female is then mated, and the sons 

 are scored phenotypically. (Any male can 

 be used as parent since it will usually trans- 

 mit to each son a Y chromosome lacking 

 alleles of the genes under consideration. In 

 fact, the Y is found to lack alleles of almost 

 all of the genes known to be present on the 

 X except the gene for bobbed bristles, bb. 

 Moreover, the Y contains several genes for 

 male fertility that have no alleles on the X.) 

 Since sons normally receive their single X 



from their mother, their phenotypes directly 

 indicate which hemizygous X-linked alleles 

 each has received. Among the sons of this 

 mating, about one crossover type appears 

 for every two that are noncrossovers. 



2. In man. color-blindness (c) and he- 

 mophilia type A (h) are recessive X-linked 

 mutant genes absent on the Y chromosome. 

 Though rare, some women have the geno- 

 type + h/c H with one of these mutants 



on each X. Available data indicate that 

 crossover (c h or + +) and noncrossover 

 ( + h or c + ) sons occur in the approxi- 

 mate ratio of 1:9. 



These examples show that when linkage 

 between nonalleles is incomplete, the per- 

 centage of progeny carrying crossovers is 

 constant for a given case but can be quite 

 different in different organisms. 



The possibility that the strength of link- 

 age varies in the same organism can be 

 tested using two mutants, b (black body 

 color) and vg (vestigial wings), of Drosoph- 

 ila melanogaster. 



A P, cross between vg -f- vg -f- (vestig- 

 ial) 1 females and + b /+ b (black) males 

 produces all normal F, (vg + /+ b). As 

 shown in Figure 9-4A, a test cross of the Fj 

 female {vg + + b 9 by vg b/vg b $ ) pro- 

 duces in Fj only 20% with recombinant 

 chromosomes. (All Fj carry vg b from the 

 father, their maternal chromosome 40% of 

 the time is vg +; 40%, + b; 10%, + +; 

 10%, vgb.) Since these results are inde- 

 pendent of sex, we conclude that b and vg 

 are linked autosomally. Recall that the X- 

 1 inked genes m and w showed 33% cross- 

 overs; therefore, the linkages between differ- 

 ent pairs of nonalleles on different pairs of 

 homologs can have different strengths. 



When the reciprocal cross ( vg -f-/+ b 6 



1 The convention used here, and usually hereafter, 

 is to describe the phenotypes of individuals only 

 with respect to the appearance of mutant traits, 

 all traits not mentioned being of the normal type. 



