Answer:
CHK2, p53, p21, cell cycle progression.
Explanation:
In a normal cell with no LFS mutation, during the G1 stage of cell cycle progression, the CHK2 activates if there is damage in the DNA. CHK2 activates p53, which is a tumor suppressor protein that will hold the cell cycle in G1/S until the DNA is repaired. The p53 protein activates p21, a protein that binds to CDK2 and stops the cell cycle. The cell cycle will continue once the damage is repaired.
Genetic information.The main job of DNA is to carry the code for making protein.A gene is a stretch of DNA that can be read by proteins called ribosomes, and copied I to type of nucleic acid called massenger RNA.
Explanation:
O type blood can be a donor to Suzanne
Answer:
Science can not solve all of our problems. While scientific understanding can help battle things like disease, hunger, and poverty when applied properly, it does not do so completely and automatically. Furthermore, there are many areas of life where science can have little impact.
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.
