HERITABILITY AND RUST RESISTANCE IN WHITE PINE 405 



1. Heritability when individuals are selected on the basis of their 

 being healthy after exposure to the pathogen (stage C) , m = 1, is 



4 o 2 n 

 h 2 = 



o 2 m + o 2 _ + o 2 mn + = a 2 + (a 2 - = a 2 , ) 

 T C TC m b e m a 



4(8.50) 



1.81 + 8.50 + 5.75 + 821 + 217.07 

 34.00 



1062.13 



= 0.032 . 



2. Heritability when candidates are selected for general combining 

 ability on basis of their progeny performance and selected candidates are 

 mated together (stage B) is 



h 2 = °- , 



C + a V J + ° 2 TC /T + °\CR /K 



and again using the mid elevational zone data as an example: 



2 (8.50) 



8.50 + 9.81/4 + 5.75/4 + 316/10 ' 



17.00 



43.98 



= 0.386 



These heritabilities were used to determine the selection gains for 

 stages B and C (Table 6) . The gain for stage A was obtained by subtracting 

 the mean of the controls (33.56%) from the mean of the crosses for the 

 particular elevation. 



Table 6. Predicted selection gains, in percent, by stages and 

 elevations 





Selection basis 





Elevation 





Stage 



Low 



Mid 



High 



A 



Natural selection in forest 



23.1 



11.3 



20.7 



B 



Progeny tests of candidates 



5.4 



3.5 



8.4 



C 



Individuals artificially 











inoculated 



0.1 



0.1 



1.4 



A+B+C 





28.6 



14.9 



30.5 



