The anatomy of the inner ear is dominated by large fluid-filled spaces. The inner ear consists of a complex series of tubes, running through the temporal bone of the skull. The bony tubes (sometimes called the bony labyrinth) are filled with a fluid called perilymph. Within this bony labyrinth is a second series of tubes made out of delicate cellular structures (called the membranous labyrinth). The fluid inside these membranous structures is called endolymph, The different spaces of both the perilymphatic and endolymphatic compartments are interconnected by a series of ducts.
Answer:
Viruses and Bacteria
Explanation:
Bacteria are single-celled, prokaryotic microorganisms that exist in abundance in both living hosts and in all areas of the planet. By their nature, they can be either "good" or "bad" for the health of plants, humans, and other animals that come into contact with them. A virus is acellular (has no cell structure) and requires a living host to survive; it causes illness in its host, which causes an immune response. Bacteria are alive, while scientists are not yet sure if viruses are living or nonliving; in general, they are considered to be nonliving.
Infections caused by harmful bacteria can almost always be cured with antibiotics. While some viruses can be vaccinated against, most, such as HIV and the viruses which cause the common cold, are incurable, even if their symptoms can be treated, meaning the living host must have a strong enough immune system to survive the infection.
Answer:
Memory cells remain ready to respond to a subsequent encounter with a pathogen quickly and efficiently. This so-called secondary response is always stronger than the primary infectious response
-i hope this helps sorry if its wrong
Explanation:
<em>Membrane proteins that aid in the passive transport of substances do so without the use of ATP. During active transport, ATP is required to move a substance across a membrane, often with the help of protein carriers, and usually against its concentration gradient.</em>
*To figure the length of one cell, divide the number of cells that cross the diameter of the field of view into the diameter of the field of view. For example, if the diameter of the field is 5 mm and you estimate that 50 cells laid end to end would cross the diameter, then 5 mm/50 cells = 0.1mm/cell.
