GENETICS AND EUGENICS 



833 



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 dominant gene 



Fig. 434. — In the cross between a color-blind male and a normal female, tlio 

 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 offspring, while 

 the sons are entirely free from it. 



Fig. 435. — Results (N, N), (N, cb), (N), (cb). In a cross between a normal 

 male and a female who is heterozygous for color blindness, 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. 



for normal. Therefore, the daughters will be heterozygous for the 

 trait and might be spoken of as carriers. Fig, 435 shows a cross be- 

 tween such a daughter and a normal male. 



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 



