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 best answer is D, birds.
For more information and details on the way a bird's respiratory system works:
https://asknature.org/strategy/air-flow-patterns-facilitate-efficient-gas-exchange/
Answer:
water
Explanation:
Water puts out fire by creating a barrier between the fuel source and the oxygen source (it also has a cooling effect which has to do with the energy required to convert liquid water into water vapor). It does this because it is a completely, 100% oxidized material. It simply cannot oxidize any further so it will not “burn”. This smothers the fire. The same thing would happen if you used the ashes that remained after a completely spent fire. Or, as I mentioned before, CO2.
Answer:
The age of the cell
Explanation:
Assuming your question was meant to be "Which of the following is NOT a difference between Prokaryotic and Eukaryotic cells?", everything except the age is different.
Prokaryotes are simple cells that have no nucleus and are generally small. Eukaryotes are complex cells that have a nucleus, organelles, and are much bigger than prokaryotes.
DNA polymerase III has a subunit for proofreading during DNA replication. As replication continues and
the polymerase III enzyme detects a mismatched base pair, through a deformity in the double helix structure, the
polymerase backs up, nicks the mismatched base and replaces it with the correct
base before continuing replication.
