解:两对等位基因共同控制生物性状时,F2中出现的表现型异常比例分析:(1)12:3:1即(9A_B_+3A_bb):3aaB_:1aabb或(9A_B_+3aaB_):3A_bb:1aabb(2)9:6:1即9A_B_:(3A_bb+3aaB_):1aabb(3)9:3:4即9A_B_:3A_bb:(3aaB_+1aabb)或9A_B_:3aaB_:(3A_bb+1aabb)(4)13:3即(9A_B_+3A_bb+1aabb):3aaB_或(9A_B_+3aaB_+1aabb):3A_bb(5)15:1即(9A_B_+3A_bb+3aaB_):1aabb(6)9:7即9A_B_:(3A_bb+3aaB_+1aabb)根据题意分析:显性纯合子(AABB)和隐性纯合子(aabb)杂交得F1,再让F1测交,测交后代的基因型为AaBb、Aabb、aaBb、aabb四种,表现型比例为1:3,有三种可能:(AaBb、Aabb、aaBb):aabb,(AaBb、Aabb、aabb):aaBb或(AaBb、aaBb、aabb):Aabb,AaBb:(Aabb、aaBb、aabb).因此,让F1自交,F2代可能出现的是:①15:1即(9A_B_+3A_bb+3aaB_):1aabb;②9:7即9A_B_:(3A_bb+3aaB_+1aabb);③13:3即(9A_B_+3A_bb+1aabb):3aaB_或(9A_B_+3aaB_+1aabb):3A_bb共三种情况.故选:B.