Gene Arrangement and Chiasmal a 



133 



crossovers in our example can be calculated 

 in the following way: since each region has 

 a 0.2 chance for a single chiasma, each has 

 a 0.1 chance for a single crossover, and the 

 expected chance for a double crossover is 

 0.1 X 0.1, or 0.01. If, as before, the actual 

 double chiasmata in our example occurred 

 with a frequency of .02, then, since only 4 of 

 the 16 meiotic products appear as double 

 crossovers (Figure 17-4), the observed fre- 



CHIASMATA 



quency of double crossovers would be 

 .02/4 = .005. The coefficient of coincidence 

 determined from double crossovers (ob- 

 served frequency /expected frequency) would 

 be .005/.01, or .5, which is the same as the 

 value previously obtained using chiasmata 

 frequencies. In practice, therefore, one can 

 determine the coefficient of coincidence by 

 dividing the observed frequency of double 

 crossovers by their expected frequency, the 



MEIOTIC PRODUCTS CROSSOVERS 



2 Doubles 



2 Noncrossovers 



b c 



c 1 Double 

 c 2 Singles 

 4. 1 Noncrossover 



3-STRAND 



^ 1 Double 

 + 2 Singles 

 c 1 Noncrossover 



FIGURE 17-4. 



Double chiasmata 

 types and their 

 genetic consequences. 



^+ + + 



4-STRAND 



a +_ 



a b 



4 Singles 



