Question 1
在孟德尔的豌豆杂交实验中,一对相对性状的杂合子 自交产生 。若 的表现型比例为 ,则该比例反映了减数分裂过程中的哪一遗传规律?
- A基因的自由组合定律
- B基因的分离定律
- C基因的连锁与交换定律
- D孟德尔的自由独立定律
Answer: B. 基因的分离定律
根据分离定律,在形成配子时,等位基因会彼此分离,使配子携带不同的等位基因,从而导致杂合子自交产生的后代出现 的表现型比例。
UEC Senior 3 · Biology
高三Biology · 遗传学 (Genetics)
遗传学 (Genetics) is part of the UEC Senior 3 Biology syllabus. Practice the MCQs here to sharpen your exam technique — each question comes with a worked explanation.
Free sample questions
Question 1
Answer: B. 基因的分离定律
Question 2
Answer: A. $1:1:1:1$
Question 3
Answer: A. $0$
Question 4
Answer: A. DNA分子是由两条反向平行的多核苷酸链盘旋而成的双螺旋结构
Question 5
Answer: D. 蛋白质 $ ightarrow$ DNA
Other Biology chapters
You can try the questions without an account. Sign up free to save your progress, see what you got wrong, and pick up where you left off.
Create free account →