This document contains biology questions written in Khmer. We will focus on solving the mathematical problems present in sections VII and VIII. Section VII asks us to calculate the total length of $6 \times 10^9$ nucleotide base pairs of ADN in millimeters, and question VIII consists of three sub-problems about ADN.

1. Problem Description

This document contains biology questions written in Khmer. We will focus on solving the mathematical problems present in sections VII and VIII.

Section VII asks us to calculate the total length of $6 \times 10^9$ nucleotide base pairs of ADN in millimeters, and question VIII consists of three sub-problems about ADN.

2. Solution Steps

Section VII:

a. The prompt asks for the total length of $6 \times 10^9$ nucleotide base pairs of ADN in millimeters, and gives the information that one nucleotide base pair has a length of 0.34 nanometers (not visible but implied in biological context).

First we must convert nanometers to millimeters:

$1 nm = 10^{-6} mm$

So, 0.34 nm = $0.34 \times 10^{-6} mm$

Next, we will multiply this length by the number of base pairs to calculate the total length:

$6 \times 10^9 \times 0.34 \times 10^{-6} mm = 2.04 \times 10^3 mm = 2040 mm$

Section VIII:

1. We have a single strand of ADN molecule with $3 \times 10^4$ nucleotides. After two duplications, we need to find the number of free nucleotides needed.

After the first duplication, we have $2 \times 3 \times 10^4$ nucleotides. The original strand serves as a template, so we need $3 \times 10^4$ free nucleotides to make the complimentary strand.

After the second duplication, we have $4 \times 3 \times 10^4$ nucleotides. We need to duplicate each double stranded DNA from the first duplication one more time. Therefore, we need another $2 \times 3 \times 10^4$ free nucleotides to make 2 new complimentary strand. In all, we needed $3 \times 10^4+ 2 \times 3 \times 10^4 = 3 \times 3 \times 10^4= 9 \times 10^4=90000$ free nucleotides.

2. In a ADN molecule, Cytosine (C) accounts for 14% of the total nucleotides. We need to calculate the number of each type of nucleotide. Since A=T and C=G, and the total percentage adds to 100%, we can say:

C=G=14%, then A=T=(100-14-14)/2= 72/2 = 36%.

Since there are $3 \times 10^4$ total nucleotides,

C = 0.14 * $3 \times 10^4 = 4200$

G = 0.14 * $3 \times 10^4 = 4200$

A = 0.36 * $3 \times 10^4 = 10800$

T = 0.36 * $3 \times 10^4 = 10800$

3. The ADN molecule can be used as a template. Each ARN molecule has an average of $10^4$ nucleotides. How many ARN molecules are needed?

Total ADN nucleotides = $3 \times 10^4$

Nucleotides per ARN = $10^4$

Number of ARN molecules = ($3 \times 10^4$)/$10^4$ = 3

3. Final Answer

VII. a. 2040 mm

VIII.

1. 90000

VIII.

2. C = 4200, G = 4200, A = 10800, T = 10800

VIII.

3. 3