The image contains a series of biology questions. We will focus on questions VII, VIII, 1, 2, and 3. VII. Given that each DNA molecule in human cells has $6 \times 10^9$ nucleotides, a. What is the total length of all DNA molecules in millimeters? b. Why can these lengths shrink and reside in the cell nucleus, which has a diameter of only 2-3 thousandths of a millimeter? VIII. One single strand of a DNA molecule has $3 \times 10^4$ nucleotides. 1. When the DNA molecule replicates twice, how many free nucleotides are needed?

1. Problem Description

The image contains a series of biology questions. We will focus on questions VII, VIII, 1, 2, and

3. VII. Given that each DNA molecule in human cells has $6 \times 10^9$ nucleotides,

a. What is the total length of all DNA molecules in millimeters?

b. Why can these lengths shrink and reside in the cell nucleus, which has a diameter of only 2-3 thousandths of a millimeter?

VIII. One single strand of a DNA molecule has $3 \times 10^4$ nucleotides.

1. When the DNA molecule replicates twice, how many free nucleotides are needed?

2. In a DNA molecule, the number of cytosine nucleotides is equal to 14% of the total number of nucleotides. Calculate the number of each type of nucleotide in the DNA molecule.

3. If this DNA molecule is used as a template, how many mRNA molecules are needed if each mRNA molecule has an average of $10^4$ nucleotides?

2. Solution Steps

VII. a. To determine the total length of all DNA molecules, we need additional information about the distance between two nucleotides along the DNA molecule. Let's assume that the distance between two nucleotides is 0.34 nm (a standard value). Then the total length would be

$6 \times 10^9 \text{ nucleotides} \times 0.34 \text{ nm/nucleotide} = 2.04 \times 10^9 \text{ nm}$. Converting this to millimeters ($1 \text{ mm} = 10^6 \text{ nm}$), we get:

$2.04 \times 10^9 \text{ nm} \times \frac{1 \text{ mm}}{10^6 \text{ nm}} = 2040 \text{ mm}$.

VII. b. The DNA is compacted through supercoiling and packaging around histone proteins to form chromatin. This allows the very long DNA molecule to fit within the small confines of the cell nucleus.

VIII.

1. The DNA molecule has two strands, so the complete DNA molecule has $2 \times 3 \times 10^4 = 6 \times 10^4$ nucleotides.

If the DNA replicates twice, the amount of new DNA is three times of the original DNA. Therefore, the number of free nucleotides needed is $3 \times 6 \times 10^4 = 18 \times 10^4 = 1.8 \times 10^5$ nucleotides.

VIII.

2. Total nucleotides: $6 \times 10^4$.

Cytosine (C): 14% of $6 \times 10^4 = 0.14 \times 6 \times 10^4 = 0.84 \times 10^4 = 8400$.

Guanine (G) = Cytosine (C) =

8

4

0

0. Adenine (A) + Thymine (T) = $6 \times 10^4 - 2 \times 8400 = 60000 - 16800 = 43200$.

Adenine (A) = Thymine (T) = $43200 / 2 = 21600$.

Thus, Cytosine: 8400, Guanine: 8400, Adenine: 21600, Thymine:

2

1

6

0

0.

VIII.

3. The DNA molecule contains $6 \times 10^4$ nucleotides. Each mRNA molecule has $10^4$ nucleotides. Therefore, the number of mRNA molecules that can be created is

$\frac{6 \times 10^4}{10^4} = 6$.

3. Final Answer

VII. a. 2040 mm

VII. b. DNA is compacted by supercoiling and histone packaging.

VIII.

1. $1.8 \times 10^5$

VIII.

2. Cytosine: 8400, Guanine: 8400, Adenine: 21600, Thymine:

2

1

6

0

0. VIII.

3. 6