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Câu hỏi: Cho biết mỗi gen quy định một tính trạng, các alen trội là trội hoàn toàn và không xảy ra đột biến, các gen liên kết hoàn toàn. Theo lý thuyết, phép lai nào sau đây cho đời con có tỷ lệ phân li kiểu gen giống tỷ lệ phân li kiểu hình?
A. $\dfrac{AB}{ab}Dd\times \dfrac{AB}{ab}Dd$
B. $\dfrac{Ab}{ab}Dd\times \dfrac{aB}{ab}dd$
C. $\dfrac{Ab}{aB}{{X}^{D}}{{X}^{d}}\times \dfrac{aB}{ab}{{X}^{D}}Y$
D. $\dfrac{AB}{ab}{{X}^{D}}Y\times \dfrac{aB}{ab}{{X}^{d}}{{X}^{d}}$
A. $\dfrac{AB}{ab}Dd\times \dfrac{AB}{ab}Dd$
B. $\dfrac{Ab}{ab}Dd\times \dfrac{aB}{ab}dd$
C. $\dfrac{Ab}{aB}{{X}^{D}}{{X}^{d}}\times \dfrac{aB}{ab}{{X}^{D}}Y$
D. $\dfrac{AB}{ab}{{X}^{D}}Y\times \dfrac{aB}{ab}{{X}^{d}}{{X}^{d}}$
A: $\dfrac{AB}{ab}Dd\times \dfrac{AB}{ab}Dd\to \left( 1\dfrac{AB}{AB}:2\dfrac{AB}{ab}:1\dfrac{ab}{ab} \right)\left( 1DD:2Dd:1dd \right)$
$\to KG:\left( 1:2:1 \right)\left( 1:2:1 \right),KH:\left( 3:1 \right)\left( 3:1 \right)$
$B:\dfrac{Ab}{ab}Dd\times \dfrac{aB}{ab}dd\to \left( 1\dfrac{Ab}{aB}:1\dfrac{Ab}{ab}:1\dfrac{aB}{ab}:1\dfrac{ab}{ab} \right)\left( 1Dd:1dd \right)$
$\to KG,KH:\left( 1:1:1:1 \right)\left( 1:1 \right)$
$C:\dfrac{Ab}{aB}{{X}^{D}}{{X}^{d}}\times \dfrac{aB}{ab}{{X}^{D}}Y\to \left( 1\dfrac{Ab}{aB}:1\dfrac{aB}{aB}:1\dfrac{Ab}{ab}:1\dfrac{aB}{ab} \right)\left( 1{{X}^{D}}{{X}^{D}}:1{{X}^{D}}{{X}^{d}}:1{{X}^{D}}Y:{{X}^{d}}Y \right)$
$\to KG:\left( 1:1:1:1 \right)\left( 1:1:1:1 \right),KH:\left( 1:2:1 \right)\left( 3:1 \right)$
$D:\dfrac{AB}{ab}{{X}^{D}}Y\times \dfrac{aB}{ab}{{X}^{d}}{{X}^{d}}\to \left( 1\dfrac{AB}{aB}:1\dfrac{AB}{ab}:1\dfrac{aB}{ab}:1\dfrac{ab}{ab} \right)\left( 1{{X}^{D}}{{X}^{d}}:1{{X}^{d}}Y \right)$
$\to KG:\left( 1:1:1:1 \right)\left( 1:1 \right);KH:\left( 1:2:1 \right)\left( 1:1 \right)$
$\to KG:\left( 1:2:1 \right)\left( 1:2:1 \right),KH:\left( 3:1 \right)\left( 3:1 \right)$
$B:\dfrac{Ab}{ab}Dd\times \dfrac{aB}{ab}dd\to \left( 1\dfrac{Ab}{aB}:1\dfrac{Ab}{ab}:1\dfrac{aB}{ab}:1\dfrac{ab}{ab} \right)\left( 1Dd:1dd \right)$
$\to KG,KH:\left( 1:1:1:1 \right)\left( 1:1 \right)$
$C:\dfrac{Ab}{aB}{{X}^{D}}{{X}^{d}}\times \dfrac{aB}{ab}{{X}^{D}}Y\to \left( 1\dfrac{Ab}{aB}:1\dfrac{aB}{aB}:1\dfrac{Ab}{ab}:1\dfrac{aB}{ab} \right)\left( 1{{X}^{D}}{{X}^{D}}:1{{X}^{D}}{{X}^{d}}:1{{X}^{D}}Y:{{X}^{d}}Y \right)$
$\to KG:\left( 1:1:1:1 \right)\left( 1:1:1:1 \right),KH:\left( 1:2:1 \right)\left( 3:1 \right)$
$D:\dfrac{AB}{ab}{{X}^{D}}Y\times \dfrac{aB}{ab}{{X}^{d}}{{X}^{d}}\to \left( 1\dfrac{AB}{aB}:1\dfrac{AB}{ab}:1\dfrac{aB}{ab}:1\dfrac{ab}{ab} \right)\left( 1{{X}^{D}}{{X}^{d}}:1{{X}^{d}}Y \right)$
$\to KG:\left( 1:1:1:1 \right)\left( 1:1 \right);KH:\left( 1:2:1 \right)\left( 1:1 \right)$
Đáp án B.