Because in F 1, high pole: short pole =3: 1, disease resistance: susceptibility =3: 1.

So parents: TtRr×TtRr

The Tt probability in F 1 is 1/3, and TT probability is 2/3; The Rr probability is 1/3 and the RR probability is 2/3 (because only the high reactance type is selected).

Because F 1 and double hidden hybridization.

So Tt probability in F2 is 0, TT probability1/3 *1+2/3 *1/2 = 2/3, TT probability 1-2/3= 1/3.

Similarly, Rr 2/3, rr 1/3.

Therefore, the phenotypic high antibody (Tt binding Rr) is 2/3*2/3=4/9, the high sensitivity (Tt and RR) is 2/3 * 1/3 = 2/9, and the short antibody (tt and RR) is1/3 * 2/3 = 2.

To put it bluntly, it is to use mathematical step-by-step probability calculation.