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:
what color light is least useful to a plant...
ans - Green Light.
A crystals shape is determined by the arrangement of the atoms or molecules inside the crystal.
Answer:
The order must be K2→K1, since the permanently active K1 allele (K1a) is able to propagate the signal onward even when its upstream activator K2 is inactive (K2i). The reverse order would have resulted in a failure to signal (K1a→K2i), since the permanently active K1a kinase would be attempting to activate a dead K2i kinase.
Explanation:
- You characterize a double mutant cell that contains K2 with type I mutation and K1 with type II
mutation.
- You observe that the response is seen even when no extracellular signal is provided.
- In the normal pathway, i f K1 activat es K2, we expect t his combinat ion of two m utants to show no response with or without ext racell ular signal. This is because no matt er how active K1 i s, it would be unable to act ivate a mutant K2 that i s an activit y defi cient. If we reverse the order, K2 activating K1, the above observati on is valid. Therefore, in the normal signaling pathway, K2 activates K1.
Answer: The correct answer is- to see what new traits might appear next.
As per the information in the question, pigeon breeder selects the offspring of pigeons with spotted bodies.
Spotted body is a trait that would be depicted by two different alleles ( such as one for black and other for white color).
As both the alleles for the color are represented in the pigeon offspring, therefore, it is a case of co-dominance ( let us take its genotype as Bb, where 'B' is allele for black and 'b' for white color).
When pigeons with spotted bodies are cross bred, they can produce pigeon with spotted and non spotted ( such as black 'BB', white 'bb', or spotted 'Bb' ) body depending upon the allele that is passed to the offspring.
Thus, pigeon breeder selects the spotted bodies pigeons and then breed those offspring to see what new traits might appear next.
