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题目内容
(请给出正确答案)
细胞膜在静息情况下,对下列哪种离子的通透性最大
A.Na
B.K
C.Cl
D.Ca
E.Mg![](https://assets.51tk.com/images/ebab89799c40b995_img/2a291387c89142e3.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/8a8441ffb7eddd17.gif)
B.K
![](https://assets.51tk.com/images/ebab89799c40b995_img/1992daaaad1ad6b6.gif)
C.Cl
![](https://assets.51tk.com/images/ebab89799c40b995_img/8ce99077a3c1e0d7.gif)
D.Ca
![](https://assets.51tk.com/images/ebab89799c40b995_img/0481c4e7d1383eca.gif)
E.Mg
![](https://assets.51tk.com/images/ebab89799c40b995_img/2a291387c89142e3.gif)
参考答案
参考解析
解析:该考试题是理解判断型题目,考查考生对生物电形成原理的理解。细胞膜在静息情况下对Na
不通透,所以,备选答案A是错误的;此时,对CI
(备选答案C)可以自由通透而不致对膜两侧浓度差产生很大影响;此时,Ca
(备选答案D)和Mg
(备选答案E)不易通透。由于膜两侧电一化学梯度较大,且此时对K
易通透,所以,K
的通透性最大。这样,有利于形成K
的平衡电位,这正是静息电位产生的原因。所以,正确答案为B。
![](https://assets.51tk.com/images/ebab89799c40b995_img/b71e7ddf1a984cdf.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/18171fd37d064c78.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/7fa41cd9c514f286.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/8ead7568853d247e.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/458df49a145f51a1.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/4fb8c76fcee108f8.gif)
![](https://assets.51tk.com/images/ebab89799c40b995_img/aea17f44d2b5297b.gif)
更多 “细胞膜在静息情况下,对下列哪种离子的通透性最大A.Na B.K C.Cl D.Ca E.Mg” 相关考题
考题
单选题细胞膜在静息情况下,离子通透性最大的是( )。A
K+B
Na+C
Cl-D
Ca2+E
Mg2+
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