Answer:
Population studies look at factors for one individual species.
Explanation:
In a garden, there are multiple species of plants growing. You could study different factors of each population, which is just all the individuals of one species in an area. For example, you could study how adding fertilizer affects the growth of tomato plants. To do this, you would have some tomato plants in the garden that have fertilizer added to the soil and some that don't. At the end of a certain period of time, maybe a month, measure the plants, find the average height and compare their growth. This is just one example, but there are lots of different things you could study.
Answer:
Explanation:
This question is incomplete. However, what is described in this question is a method of DNA damage repair mechanism. The method described is called base excision repair. Generally, glycosylases play important roles in base excision repair (as they detect and remove these damaged bases).
One thing that can be considered while identifying the bond to be broken is the codon that will be formed after the removal of base. If this codon (a three base sequence) will still produce the same protein as when a normal cytosine was present.
NOTE: During this base excision repair, a glycosylase will detect and remove the damaged cytosine. Once removed, the remaining nucleotide backbone (of the removed cytosine) is also removed and the gap is filled and sealed by other enzymes in this repair mechanism.
*The key function of each of the two photosystems is to absorb light and convert the energy of the absorbed light into redox energy, which drives electron transport.
In PS II (the first photosystem in the sequence), P680 is oxidized (which in turn oxidizes water), and the PS II primary electron acceptor is reduced (which in turn reduces the electron transport chain between the photosystems).
In PS I, the PS I primary electron acceptor is reduced (which in turn reduces other compounds that ultimately reduce NADP+ to NADPH), and P700 is oxidized (which in turn oxidizes the electron transport chain between the photosystems).
Phat1 = 9/142
Phat 2 = 5/268
Phat 3 14/410
Test statistics = (Phat 1 - Phat 2) / (Square root of (phat × (1- phat))
times the square root of (1/n1 + 1/n2))
Test statistics = 2.373
P- value = 2× (1- normal cdf (2.373, 0, 1))
= 0.0176
At the level of significance, there is difference between the proportion of drug-resistant wihtin the two states.
Because meiosis only has half of the genes.
In mitosis, the cell produced will have an exact number of chromosome like the parent cells. But in meiosis, only half of the chromosome will be given. That will make many variations since the chromosome is split in random. The process to split the gene doesn't include a process to making it move in the same manner so the gene position can be changed in the different case.
