Answer:
1. The cells in our bodies are surrounded by these types of solutions. → Isotonic solution.
3. When an animal cell is places in this solution, it will burst (get layer) → Hypotonic solution.
4. When an animal cell is placed in this solution, it will shrivel or shrink (get smaller) → Hypertonic solution.
5. This is a solution with more solute than the cell. Hypertonic solution.
Explanation:
The cells in the body are in a balance of substances —concentration of solutes— between their cytoplasm and the extracellular space. This balance is dynamic in living beings, due to the constant exchange of ions and substances between the intracellular and extracellular space. For this reason, the extracellular medium is isotonic with the cytoplasm.
<u>A cell can lose or gain water depending on the amount of solutes that a medium has in which it is found</u>, with respect to the cytoplasm. This difference in solutes concentrations produces an osmotic gradient that drags water from the least concentrated solution to the most concentrated, through the process of osmosis, which seeks to achieve an equilibrium of concentrations.
- <em>When a animal cell is exposed to a </em><em>hypertonic solution</em><em> </em>—<em>with a higher concentration of solutes</em>— <em>it loses water and tends to </em><em>dehydrate and become smaller</em><em>.</em>
- <em>An animal cell in a </em><em>hypotonic solution</em><em> receives water, so it can </em><em>expand and even burst</em><em>.</em>
In practice, the concentrations of intracellular and extracellular solutes depend not only on the osmotic gradient, but also on the concentration gradient of substances.
Answer:
1.5 micro liter.
Explanation:
Antibodies may be defined as the substance hat are produced by the immune cells of the body against the particular antigen. The antibodies concentration is helpful to determine the infection stages.
The amount of primary antibody can be calculated by teh following formula:
N1V1 = N2V2
Here, N1 = amount of primary antibody = 1, N2 is dilution factor 1/3333 and V2 is 5 ml = 5000 micro liter.
Then V1 can be find as follows;
V1 × 1 = 1/3333 x 5000
V1 = 1.5 micro liter.
Thus, the answer is 1.5 micro liter.
Their exoskeleton keeps arthropods from drying out.
In order for the local residents to lower the frequency of landslides in the region that experiences frequent landslides from heavy rains, they should do the following:
A. Start contour plowing on the slopes where landslides happen.
B. Plant vegetation on the slopes that experience landslides.
Answer:
The answer to this question is given below in this explanation section.
Explanation:
"advantages of interconnecting air sacs for birds"
The air sacs permit a unidirectional flow of air through the lungs.Unidirectional flow means that air moving through birds lungs is largely fresh air and has a higher oxygen content.In contrast air flow is bidirectional in mammals moving back and forth into and out of the lungs.As a result air coming into a mammals lung is mixed with old air has less oxygen. so in bird lungs,more oxygen is available to diffuse into the blood(avian respiratory system).
- The air sacs are thin membrane structures connected to the primary or secondary bronchi via ostia and they comprise most of the volume of the respiratory system.
- Air sacs are poorly vascularized by the systemic circulation and do not directly participate in significant gas exchange but act as a bellows to ventilate the lungs.In most species,there are nine air sacs which can be considered in cranial and caudal functional groups.
- The cranial group consists of the paired cervical air sacs,the unpaired clavicular air sacs and the paired cranial thoracic air sacs.
