510 University of California Publications in Zoology [Vol. 13 



Table XXII shows that whether the parent with extra In-istles be 

 male or female the extra liristles are inlierited in practically the same 

 number by both sons and daughters, thus showing that the distribution 

 is simihir in the reciprocal crosses. This is sufficient to indicate that 

 there is no sex linkage involved, a conclusion reached also by Mac- 

 Dowell (1915). 



In almost all cases we were dealing with only one extra bristle and 

 the occasional two bristles were found in both males and females. 

 Neither did there seem to be any difference in the number of bristles 

 inherited by male and female; thus differing from MacDowell, who 

 found a larger percentage of extra-bristled flies among the females. 



Summary of Experiments on Mendelian Inheritance 



1. From the occurrence of the ratios approximating 3 to 1, 15 to 1. 

 and 63 to 1, we seem to be dealing with a Mendelian inheritance in- 

 volving, in the various crosses, 1, 2, and 3 equivalent factors, concerned 

 in the development of the extra bristles, or rather in their failure to 

 appear. 



2. There is also some evidence in favor of regarding this as a 

 fluctuating variation. 



3. Thickened hairs give evidence of a partial inheritance. 



4. There is no sex linkage involved and extra bristles appear with 

 the same regularity as to number and frequency in male and female. 



Selection 

 Selection of normals seemed to result in a partial weeding out 

 of extras in the normal cultures, but really this may have been a 

 fluctuation since extras appeared in all cultures. On the other hand. 

 a high-grade race was not obtained by the selection of extras. In 

 crossing a five-bristled fly with a seven-bristled fly, a six-bristled fly 

 with a six-bristled fly, and a four-bristled fly with a six-bristled fly, 

 the same results were obtained. 



TABLE XXIII 

 Comparing Number of Extra Bristles in P^vrents and Offspring 



Number of bristles in parents Normal 



.5X7 = 6 extra-bristled flies, 5 flies with 5 bristles, 1 fly with 6 bristles .55 

 6X6^8 extra-bristled flies, 8 flies with 5 bristles 46 



4X6 = 4 extra-bristled flies, 4 flies with 5 bristles 83 



6X5 = 5 extra-bristled flies, 5 flies with 5 bristles 70 



