Answer:
<h2>Upper epidermis.</h2>
<em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>h</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u>s</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
I belive the correct answer is D
Hope this helps :)
Answer:
A) A decrease in Na+ permeability, and an increase in K+ permeability.
Explanation:
When the potential action reaches a peak of about +40 mv Na+ channels are open and a high number of Na+ ions are entering the inside of the cell. Shortly after this happens the K+ channels will start to open their gates increasing the cell K+ permeability while the Na+ channels will start to close their gates, so the Na+ permeability will decrease. This happens in order to valance the positive charge on the inside of the cell. Normally the inside has a negative charge while the outside has a positive one, as the inside is more positive due to the increase in Na+ permeability at the beginning of the potential action, K+ cations (which are in abundance in the cell) will have to go out through the K+ channels so as to restore the charge valance, that means that there is an increase in K+ permeability.