Answer: for me the best option is D.
Explanation: lets explain this.
Cellular respiration begins with a process that divides the glucose within the cells making it readily available as a source of energy. This process can occur without oxygen (anaerobic respiration) or in the presence of oxygen (aerobic respiration). Anaerobic respiration generates more excess waste (lactate) than aerobic . Besides, high levels of lactate build within the muscle cells. Excess lactate slows the cellular respiratory process and is experienced as a burning sensation in the muscles if exercise continues.
The nurse should carry a plan of care in the client as the
hair of the client is thinning and the skin on the client’s forehead is getting
irritated because of the client’s disorder. The highest priority of plan of
care to the client is being able to set limits on the behavior of the client, In
order to prevent further harm from occurring
to the client as well as towards other people.
Answer:
In order to find average speed during each interval, we need to divide the distance during those intervals with the period of time. So, for the first interval (day 0 to day 2) hawksbill started from 0 and reached 10 kilometers by the end of the second day. That means that it crossed 10 kilometers in 2 days, so the average speed is 10/2 which is 5 km/day. Similarly, we can calculate speed for other intervals:
• day 2 - day 3: it went from 10 to 12 km in one day, which means it crossed 2 km in one day, so the average speed is 2/1 = 2 km/day
• day 3 - day 4: at the end of the third day it reached 12 km and at the end of the day 4 it remained at 12 km. That means the hawksbill wasn't moving in that interval so the speed was 0
• day 4 - day 5: it went from 12 km to 18 km, which means it crossed 18-12=6 km in one day, so the average speed is 6/1=6 km/day
• day 5 - day 6: it went from 18 to 24 km, which means it crossed 24-18=6 km in one day, so the speed was 6/1=6 km/day
So, to summarize, during the first interval turtle was moving with average speed of 5 km/day, then 2 km/day, in the third interval it wasn't moving and in the last two intervals, it moved in average speed of 6 km/day.
Answer:
A) They have low-maintenance and are easy to keep track of for mutations.
B) The deduction can be "Single Gene Mutation"
Explanation:
After examining the example given in the question on Neurospora crassa and the details about how they reproduce, the following points can be made regarding the questions;
A) It is stated that they form a colony in time and that they are asexual spores and the first reason to choose them would be because they contain somatic cells (which refer to the cells other than reproductive cells) and non-motile gonidia which can multiply by dividing themselves and these properties make the colony's maintenance easy. And since they multiply by division, it is easier to keep track of the occuring mutations.
B) Given the information in the question that the mating is between an albino strain and a wild type, and then between two albino strains which have the same genotype. The results indicate that the strains have gone through single gene mutation during the process.
I hope this answer helps.
