(1)两对相对性状在遗传上遵循
(2)让F1与纯合品种乙杂交,理论上子代的基因型为
(3)让F2中所有不耐镉耐盐植株自交,发现有的植株后代不发生性状分离,有的植株后代性状分离比为3:1,还有的植株后代性状分离比为9:3:3:1。其中后代性状分离比为3:1的F2植株的基因型有
(4)若用较高浓度的盐水浇灌F2植株幼苗,让剩余植株在开花前自交,则理论上F3中表现耐盐植株的比例为

同类型试题

y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2


y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2

