Time ÷ duration of the refractory period
1 ÷ duration of the refractory period
Figure 1 shows the changes in the permeability of a section of an axon membrane to two ions that are involved in the production of an action potential. Use the information in Figure 1 to calculate the maximum frequency of action potentials per second along the axon. Show your working.
Step 1: Determine the duration of the refractory period
From the graph: 2.75 milliseconds
Step 2: Convert this to seconds
0.00275 seconds
Step 3: Insert relevant figures into the equation
Time ÷ duration of the refractory period
1 ÷ 0.00275 = 363.63
Answer = 364 action potentials sec-1
It is very important to check the units given in the question. For example, the duration of the refractory period may be given in milliseconds but the answer must be in action potentials sec-1. 1000 milliseconds = 1 seconds!
转载自savemyexams
以上就是关于【AQA A Level Biology复习笔记6.2.6 Maths Skill: Calculating Maximum Impulse Frequency】的解答,如需了解学校/赛事/课程动态,可至翰林教育官网获取更多信息。
往期文章阅读推荐:
全网破防!ALevel CIE数学M1疑似错题?经济P2难度飙升?5月6日大考考情分析必看!
A-Level CIE就大规模泄题发布最严处罚!哪些考生必须重考?你的成绩怎么办?

© 2026. All Rights Reserved. 沪ICP备2023009024号-1