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.
<span>Actually baby Aiden will surely develop a strong secretory Iga antibodies which is passive immunity to most of the types of bacteria and viruses, ie, as a protection against infections, which inturn is a response to the baby to its immune system in the form active immunity.</span>
The sulphur would lable the capsule and the phosphorous the nucleic acid.
<h3><u>Explanation</u>:</h3>
Hershey and Chase experiment included growing of the pages in two batches, one in presence of 35S and other in presence of 32P. They then infected bacterial cells with these phages, cleaned them and then centrifuge the cells to isolate the marked elements in bacterial cells.
This was done to isolate which part of the phage is actually infective. Sulphur being a part of the proteins will mark the capsule whereas DNA having the phosphate bridges will be marked by 32P.
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