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Natalija [7]
3 years ago
11

Base excision repair corrects DNA by A. removing a double-stranded fragment of damaged DNA. B. detecting, removing, and replacin

g damaged or incorrect nucleotides in a single strand of DNA. C. excising the incorrect base from a nucleotide D. removing extraneous groups such as methyl or oxygen added by mutagens. E. correcting A
Biology
1 answer:
Firdavs [7]3 years ago
5 0

Answer:

Option C

Explanation:

Base excision repair involves, the detection, removal and replacement of incorrect base in a nucleotide on one strand while using the other undamaged strand as template for the correction. It involves the DNA glycosylases that aids in detection. Repair can be done in two pathways: long patch panther

Short patch panther

It corrects base lesions that arise due to alkylation, depurination/depyrimidation, deamination etc

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A student run10m/s what does it means<br><br>​
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Which statements describes characteristics of earths geosphere? Check all that apply
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Describe the function of each of the important structures in a virus. (
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Answer:

The main function of the virion is to deliver its DNA or RNA genome into the host cell so that the genome can be expressed (transcribed and translated) by the host cell. The viral genome, often with associated basic proteins, is packaged inside a symmetric protein capsid.

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5 0
2 years ago
In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,2
Alisiya [41]

Answer:

Suppose that we have a given function f(x)

The average rate of change of the function between two values x₁ and x₂ is given by:

r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}

a) We want to find the average (rate) of change on the size of population from 1985 to 1995.

We have that:

f(1985) = 750

f(1995) = 1500

Then we have:

r = \frac{1500 - 750}{1995 - 1985}  = 750/10 = 75

This means that the population of trees increases, in average, at a rate of 75 trees per year.

b) What is the density of trees each year that they were counted?

This will be equal to the quotient between the number of trees and the area.

1985: number of trees = 750 pines

         area = 250 ha

Then the density is:

D(1985) = (750 pines)/(250 ha) = 3 pines/ha

So 1985, there were 3 pines per hectare.

1990: number of trees = 1250 pines

         area = 250 ha

Then the density is:

D(1990) = (1250 pines)/(250 ha) = 5 pines/ha

1995: number of trees = 1500 pines

         area = 250 ha

The density is:

D(1995) = (1500 pines)/(250 ha) = 6 pines/ha

3) now we want to get the average change between 1985 and 1995 in the density, this will be:

r = \frac{D(1995) - D(1885)}{1995 - 1985}  = \frac{6 pines/ha - 3pines/ha}{10}  = 0.3 pines/ha

So, on average, each year the number of pines per hectare increases by 0.3

7 0
3 years ago
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