Answer:
The cell interior would experience higher than normal Na+ concentrations and lower than normal K+ concentrations.
Explanation:
Na+/K+ ATPase exists in two forms: Its phosphorylated form has a high affinity for K+ and low affinity for Na+. ATP hydrolysis and phosphorylation of the Na+/K+ pump favor the release of Na+ outside the cell and binding of K+ ions from the outside of the cell. Dephosphorylation of the pump increases its affinity for Na+ and reduces that for K+ ions resulting in the release of K+ ions inside the cells and binding to the Na+ from the cells.
The presence of ATP analog would not allow the pump to obtain its phosphorylated form. Therefore, Na+ ions would not be released outside the cells. This would increase the Na+ concentration inside the cell above the normal. Similarly, the pump would not be able to pick the K+ from the outside of the cell resulting in reduced cellular K+ concentration below the normal range.
The answer would be ( D. formation of cell walls)
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Hope this helps.
