GENETICS AND EUGENICS 



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which, have no X-chromosome and therefore, have no such gene. All 

 female gametes have one X-chromosome bearing a gene for normality. 

 The various possible combinations of the two kinds of sperm with the 

 single type of egg will produce two types of offspring; namely, sons 

 who will be strictly normal in regard to the trait in question and 

 daughters who will inherit the gene, but who will not show the trait 

 because its effect will be overcome by the presence of a gene for nor- 

 mal. Therefore, the daughters will be heterozygous for the trait and 

 might be spoken of as carriers. Fig. 81 shows a cross between such a 

 daughter and a normal male. 



Fig. 80. — In the cross between a color-blind male (at left) and a normal female 

 (at right) the results (N, cb), (N, cb), (N), (N) indicate that all daughters, being 

 heterozygous, will not show the defect though they might transmit it to their off- 

 spring, while the sons are entirely free from it. iV2V=normal female individual ; 

 cb = Color-blind individual ; and N, c&=normal function but carrier of the color- 

 blind gene. 





Fig. 81. — Results (N, N), (N, cb), (N), (cb). In a cross between a normal 

 male (at left) and a female who is heterozygous for color blindness (at right), 

 half of the sons will be normal and half will be color-blind. Half the daughters will 

 be homozygous for normal and half will be heterozygous. 



Inspection of this figure will show that such a cross will produce 

 four types of offspring; namely, normal daughters, ''carrier" daugh- 

 ters, normal sons, and color-blind sons. Thus, we see that half the 

 grandsons of a color-blind man will receive the trait through their 

 mothers, and that all daughters and half the granddaughters of the 

 color-blind male will be carriers of the trait, but will have normal 



