Sex-Linkage 



each dyad separate and go to opposite poles 

 at anaphase II. The result here will be two 

 daughter nuclei each containing two X's, one 

 from each dyad. Therefore, at the end of 

 meiosis, the failure of dyads to disjoin at 

 anaphase I will result ultimately in four 

 nuclei, 2 with no X and 2 with two X's. The 

 nucleus which becomes the gametic nucleus, 

 therefore, has a 50% chance of carrying no 

 X and a 50% chance of carrying two X's. 



The second manner in which normal sepa- 

 ration of X's may fail to occur during meiosis 

 is as follows. In this case (Figure 12-9C), 

 anaphase I is normal so that at telophase I 



two nuclei are formed each containing one X 

 dyad. In one of the two nuclei at metaphase 

 II meiosis is completed normally, producing 

 two telophase II nuclei, each containing one 

 X. In the other metaphase II nucleus, how- 

 ever, the members of the X dyad fail to sepa- 

 rate at anaphase II and go instead into the 

 same telophase II nucleus. As a result of this 

 failure to disjoin two nuclei are produced, 

 one containing no X, and the other containing 

 two X's. As a consequence of these events, 

 the gametic nucleus has a 25% chance of 

 carrying no X, a 25% chance of carrying two 

 X's, and a 50% chance of carrying one X. 



FIGURE 12-9. Consequences of normal segrega- 

 tion of X chromosomes {A) and of its failure to 

 occur {B and C). 



METAPHASE 

 I 



