One turn of the helix in a `beta`-form of DNA is approximately

Options

A. 20 nm
B. 0.34 nm
C. 3.4 nm
D. 2 nm

Answer

To determine the length of one turn of the helix in the beta form of DNA, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of DNA**: The beta form of DNA, also known as B-DNA, is a right-handed helix. Each complete turn of the helix consists of a specific number of base pairs.

2. **Identify the Number of Base Pairs in One Turn**: In B-DNA, one complete turn of the helix contains approximately 10 base pairs.

3. **Determine the Distance Between Base Pairs**: The distance between each base pair in B-DNA is approximately 3.4 angstroms.

4. **Calculate the Total Length of One Turn**: Since there are 10 base pairs in one turn, we can calculate the total length by multiplying the number of base pairs by the distance between them:
\[ \text{Total Length} = \text{Number of Base Pairs} \times \text{Distance Between Base Pairs} \]
\[ \text{Total Length} = 10 \times 3.4 \text{ angstroms} = 34 \text{ angstroms} \]

5. **Convert Angstroms to Nanometers**: To express the length in nanometers, we convert angstroms to nanometers. Since 1 nanometer is equal to 10 angstroms:
\[ 34 \text{ angstroms} = \frac{34}{10} \text{ nanometers} = 3.4 \text{ nanometers} \]

6. **Final Answer**: Therefore, the length of one turn of the helix in the beta form of DNA is approximately **3.4 nanometers**.