Answer:
a. 3′→5′ exonuclease activity = No proof reading
b. 5′→3′ exonuclease activity = No primer removal
c. 5′→3′ polymerase activity = No gap filling
Explanation:
<u>a) 3′→5′ exonuclease activity = No proof reading</u>
3′→5′ exonuclease activity of DNA polymerase I is responsible for proof reading so that if a wrong nucleotide is inserted during replication, it could be removed. So if a mutation will occur in the domain which is responsible for 3′→5′ exonuclease activity error prone DNA replication will occur.
<u>b. 5′→3′ exonuclease activity = No primer removal</u>
5′→3′ exonuclease activity is useful in removing primers so that appropriate nucleotides could be inserted in the lagging strand. But if mutation will occur in this part of DNA polymerase I then primer will not be removed and replication of lagging strand will be at halt.
<u>c. 5′→3′ polymerase activity = No gap filling</u>
5′→3′ polymerase activity of DNA polymerase I, joins the nucleotides in the lagging strand where the gaps have been created after removal of primer. So the gaps created after removal of primer in the lagging strand will not be filled if there will be a mutation in this segment of DNA polymerase I.
This would be global ecology.
This is because "Global ecology"<span> is the study of the interactions among the Earth's ecosystems, land, atmosphere and oceans. </span>Global ecology<span> is very important because it is used to understand large scale interactions and how they influence the behavior of the entire planet, including the earth's responses to future changes.
Hope this helps. :)</span>
Answer:
The percentage of the population that is heterozygous for this trait is 48%
Explanation:
They are two alleles, the phenylthiocarbamide tasters (PTC) and the non phenylthiocarbamide tasters (non PTC). PTC testers are dominant and non PTC tasters are recessive.
let the frequency of the dominant allele(A) be p
and the frequency of the recessive allele(a) be q
We are told that 64 percent of people living in a remote, isolated mountain village can taste phenylthiocarbamide (PTC) and must, therefore, have at least one copy of the dominant PTC taster allele (that is AA and Aa)
Frequency of AA = p², Frequency of Aa = 2pq and Frequency of aa = q²
Therefore p² + 2pq = 64% = 0.64
According to Hardy–Weinberg:
p² + 2pq + q² = 1 and
p + q = 1
Since p² + 2pq = 0.64
∴ 0.64 + q² = 1
q² = 1 - 0.64 = 0.36
q = √0.36 = 0.6
Since p + q = 1
p = 1 - q = 1 - 0.6 = 0.4
The frequency of heterozygous = 2pq = 2 × 0.4 × 0.6 = 0.48
Therefore the percentage of the population that is heterozygous for this trait is 48%
hellosphere. This is not an actual part of the "spheres" on Earth
Hope this helps
