Base ratio of a DNA is `0.03` and the amount of `A-TN_(2)` base contents is 30000, then what is the amount of `G-C,N_(2)` base contents in this DNA?

Options

A. 1000
B. 10000
C. 100000
D. None

Answer

To solve the problem step by step, we will use the information given about the base ratio and the amount of adenine and thymine in the DNA.

### Step-by-Step Solution:

1. **Understand the Base Ratio**:
The base ratio of DNA is given as 0.03. This means that the ratio of the total amount of adenine (A) and thymine (T) to the total amount of guanine (G) and cytosine (C) is 0.03. We can express this mathematically as:
\[ \frac{A + T}{G + C} = 0.03 \]

2. **Identify the Amount of A + T**:
We are given that the amount of adenine and thymine combined (A + T) is 30,000.

3. **Set Up the Equation**:
Let \( G + C \) represent the total amount of guanine and cytosine. We can substitute the known values into our equation:
\[ \frac{30,000}{G + C} = 0.03 \]

4. **Cross-Multiply to Solve for G + C**:
To eliminate the fraction, we can cross-multiply:
\[ 30,000 = 0.03 \times (G + C) \]

5. **Isolate G + C**:
Now, we can isolate \( G + C \) by dividing both sides by 0.03:
\[ G + C = \frac{30,000}{0.03} \]

6. **Calculate G + C**:
Performing the division:
\[ G + C = 1,000,000 \]

7. **Conclusion**:
The amount of guanine and cytosine (G + C) in this DNA is 1,000,000.

### Final Answer:
The amount of G-C base contents in this DNA is **1,000,000**.

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