Answer:
20 million years
Explanation:
If we have a neutral mutation rate of one mutation per 5 million years, then the total of eight mutation between the two different species would be 20 million years. This is because both species will have 4 mutations in those 20 million years, so combined, both by 4, will have 8 mutations between them. So few mutations on so much time will result in two species that are very similar to each other even after 20 million years of evolution, even making them hardly distinguishable, especially if it comes to defining fossil records from them both. A nice example of this are the members of the felidae (cat) family, which are all very closely related, and are almost identical, thus making it extremely hard to distinguish two species of the same or similar size by their fossils.
Answer:
Repair mechanism for base cleavage (BER)
Explanation:
Repair by base cleavage (BER)
The altered bases are specifically recognized by glycosylases and removed, generating an AP site. The hole is filled by a DNA polymerase that takes the healthy strand as a template. This system arises not only by exposure to external agents, but also by the cell's own activity.
In case of damage in more than one nucleotide, repair by nucleotide excision (NER) is performed.
Nucleotide excision repair (NER)
The damaged area is recognized by UvrA and B, then A and B separate and UvrC enters which forms a complex with endonuclease activity with B. This enzyme cuts the T-dimer and the gap is filled by a DNA polymerase. There is also the TC-NER system (transcription-coupled nucleotide repair system). The alteration of these mechanisms gives rise to diseases such as: Xeroderma pigmentosum, Trichotiodystrophy or Cockayne Syndrome
Answer:
The genotype of the F1 was wy+/w+y.
Explanation:
One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.
<u>We have two genes with two alleles each:</u>
Red eyes (w+) is dominant over white eyes (w).
Brown body (y+) is dominant over yellow body (y).
The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:
- wy+/ey (white-eye, brown body): 670
- w+y/wy (red-eye, yellow body): 650
- wy/wy (white-eye, yellow body): 38
- w+y+/wy (red-eye, brown body 56
If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).
Every individual in the offpsring has a <em>wy</em> chromosome, as this was the gamete inherited from the test cross individual.
In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.
Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.
Answer:
Thylakoids.
Explanation:
Thylakoids absorb light through light harvesting complexes.
Answer:
if you f her she will let you get straight A+
Explanation:
ik becuase ive done it
