Đáp án: $C$

Giải thích:

$\text{P: $\dfrac{\text{Ab}}{\text{aB}}$X$_{\text{E}}^{\text{D}}$Y × $\dfrac{\text{Ab}}{\text{aB}}$X$_{\text{e}}^{\text{D}}$X$_{\text{e}}^{\text{d}}$}$

Xét riêng từng phép lai:

$\text{P: $\dfrac{\text{Ab}}{\text{aB}}$ × $\dfrac{\text{Ab}}{\text{aB}}$ → F$_1$: 1$\dfrac{\text{Ab}}{\text{Ab}}$:2$\dfrac{\text{Ab}}{\text{aB}}$:1$\dfrac{\text{aB}}{\text{aB}}$}$

$\text{P: X$_{\text{E}}^{\text{D}}$Y × X$_{\text{e}}^{\text{D}}$X$_{\text{e}}^{\text{d}}$ → F$_1$: 1X$_{\text{E}}^{\text{D}}$X$_{\text{e}}^{\text{D}}$:1X$_{\text{E}}^{\text{D}}$X$_{\text{e}}^{\text{d}}$:1X$_{\text{e}}^{\text{D}}$Y:1X$_{\text{e}}^{\text{d}}$Y}$

Xét đáp án A:

$\text{$\dfrac{\text{Ab}}{\text{Ab}}$X$_{\text{e}}^{\text{D}}$Y+$\dfrac{\text{aB}}{\text{aB}}$X$_{\text{e}}^{\text{D}}$Y+$\dfrac{\text{Ab}}{\text{aB}}$X$_{\text{e}}^{\text{d}}$Y=($\dfrac{1}{4}$×$\dfrac{1}{4}$)+($\dfrac{1}{4}$×$\dfrac{1}{4}$)+($\dfrac{1}{4}$×$\dfrac{2}{4}$)=$\dfrac{1}{4}$=25%}$

`->` Sai

Xét đáp án B:

Ở F$_1$ có số kiểu hình là:

$\text{3×3=9 (Kiểu Hình)}$

`->` Sai

Xét đáp án C:

$\text{$\dfrac{\text{Ab}}{\text{aB}}$X$_{\text{E}}^{\text{D}}$X$_{\text{e}}^{\text{D}}$+$\dfrac{\text{Ab}}{\text{aB}}$X$_{\text{E}}^{\text{D}}$X$_{\text{e}}^{\text{d}}$=($\dfrac{2}{4}$×$\dfrac{1}{4}$)+($\dfrac{2}{4}$×$\dfrac{1}{4}$)=$\dfrac{1}{4}$=25%}$

`->` Đúng

Xét đáp án D:

Đời có F$_1$ có tối đa số kiểu gen là:

$\text{3×4=12 (Kiểu Gen)}$

`->` Sai