Answer:
The solution would be HYPOTONIC. The concentration of water is higher inside the cell
Explanation:
Hypertonic: cell shrink
Isotonic: cell will stay the same
Hypotonic: cell will swell
Answer:
Among others, two adaptations might be
- Avoiding corporal heat loss
- Increasing oxygen absorption
Explanation:
Up in the mountains, there is low oxygen, food is scarce, and adverse meteorological conditions. Animals and plants need to develop different strategies to survive. These adaptations involve not only physical and physiological changes but also behavioral changes. To mention a few adaptations, we can name:
- Avoiding heat loss. Temperature tends to be very low at highs, so, to <u>avoid heat loss,</u> animals develop shorter legs, tails, and ears. By doing this they reduce the area or surface of heat loss and also avoid getting frozen. In mammals, the coat is also very important. A thick coat helps them maintain a constant body temperature and keep warm. Some amphibians might also develop a thicker skin as they can not regulate their temperature, and it also helps them not to dehydrate.
- Camouflage: Coat is also helpful in camouflaging. Mammals´ hair color depends on their environment. Some animals, such as hares, can also change their fur color depending on the season. During snow seasons they turn white, and during the warmer season, they turn yellow or brown.
- Size and metabolism: Small mammals lose heat very fast, so they need to keep active and feeding most of the time. They have an elevated metabolism to keep warm. On the contrary, big animals, such as bears, need to hibernate to reduce their metabolism and get to survive, otherwise, they would need many reserves to cover their energetic requirements.
- Oxygen absorption: Some animals have adapted to the lack of oxygen by increasing their heart and lungs capacity as well as their capability to absorbing more oxygen from the blood.
Since it has no nucleus it's most likely a<u><em> prokaryote</em></u>
Answer:
When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0
Explanation:
According to Levin's model (1969):
<em>dP/dt = c - e</em>
where P represents the proportion of occupied patches.
<em>c</em><em> </em>and <em>e </em>are the local immigration and extinction probabilities per patch.
Thus, the rate of change of P, written as dP/dt, tells you whether P will increase, decrease or stay the same:
- if dP/dt >0, then P is increasing with time
- if dP/dt <0, then P is decreasing with time
- if dP/dt = 0, then P is remaining the same with time.
The rate dP/dt is calculated by the difference between colonization or occupancy rate (<em>c</em>) and extinction rate (<em>e</em>).
c is then calculated as the number of successful colonizations of unoccupied patches as a proportion of all available patches, while e is the proportion of patches becoming empty. Notice that P can range between 0 and 1.
As a result, if the patch occupancy rate (c) equals the patch extinction rate (e), then patch occupancy P equals to 0.
