Question
What is the expected absorbance of a solution of DNA that has a concentration of 34.00 ng DNA/μL?
Report your answer to two decimal places.
Explain how you got the answer.
Include equation used to solve the problem, what numbers you substituted for which variables, the math necessary to solve this answer, and the appropriate units with this answer
Report your answer to two decimal places.
Explain how you got the answer.
Include equation used to solve the problem, what numbers you substituted for which variables, the math necessary to solve this answer, and the appropriate units with this answer
Answers
GPT-4o mini
To calculate the expected absorbance of a solution of DNA, we can use the Beer-Lambert Law, which is given by the equation:
\[ A = \varepsilon \cdot c \cdot l \]
Where:
- \( A \) is the absorbance (unitless)
- \( \varepsilon \) is the molar absorptivity (extinction coefficient) in L/(mol·cm)
- \( c \) is the concentration in mol/L
- \( l \) is the path length of the cuvette in cm (typically 1 cm)
For nucleic acids, the average molar absorptivity (\( \varepsilon \)) at a wavelength of 260 nm is approximately **0.020** L/(ng·μL) when the concentration is reported in ng/μL. When using this value, we can convert DNA concentration directly into absorbance without needing to convert to molarity.
Given:
- **Concentration (\( c \))** = 34.00 ng/μL
- **Path length (\( l \))** = 1 cm (standard path length for a spectrophotometer)
- **Molar absorptivity (\( \varepsilon \))** = 0.020 L/(ng·μL)
Now we can substitute these values into the equation:
\[
A = \varepsilon \cdot c \cdot l
\]
Substituting the values:
\[
A = 0.020 \, \text{L/(ng·μL)} \cdot 34.00 \, \text{ng/μL} \cdot 1 \, \text{cm}
\]
Calculating:
\[
A = 0.020 \cdot 34.00 \cdot 1 = 0.68
\]
So the expected absorbance of the solution is **0.68**.
### Final Answer:
**Absorbance: 0.68**
### Summary of the Calculation:
1. Used the formula: \( A = \varepsilon \cdot c \cdot l \)
2. Substituted \( \varepsilon = 0.020 \, \text{L/(ng·μL)} \), \( c = 34.00 \, \text{ng/μL} \), and \( l = 1 \, \text{cm} \) into the equation.
3. Performed the multiplication: \( 0.020 \cdot 34.00 = 0.68 \).
4. Reported the absorbance as 0.68, unitless.
\[ A = \varepsilon \cdot c \cdot l \]
Where:
- \( A \) is the absorbance (unitless)
- \( \varepsilon \) is the molar absorptivity (extinction coefficient) in L/(mol·cm)
- \( c \) is the concentration in mol/L
- \( l \) is the path length of the cuvette in cm (typically 1 cm)
For nucleic acids, the average molar absorptivity (\( \varepsilon \)) at a wavelength of 260 nm is approximately **0.020** L/(ng·μL) when the concentration is reported in ng/μL. When using this value, we can convert DNA concentration directly into absorbance without needing to convert to molarity.
Given:
- **Concentration (\( c \))** = 34.00 ng/μL
- **Path length (\( l \))** = 1 cm (standard path length for a spectrophotometer)
- **Molar absorptivity (\( \varepsilon \))** = 0.020 L/(ng·μL)
Now we can substitute these values into the equation:
\[
A = \varepsilon \cdot c \cdot l
\]
Substituting the values:
\[
A = 0.020 \, \text{L/(ng·μL)} \cdot 34.00 \, \text{ng/μL} \cdot 1 \, \text{cm}
\]
Calculating:
\[
A = 0.020 \cdot 34.00 \cdot 1 = 0.68
\]
So the expected absorbance of the solution is **0.68**.
### Final Answer:
**Absorbance: 0.68**
### Summary of the Calculation:
1. Used the formula: \( A = \varepsilon \cdot c \cdot l \)
2. Substituted \( \varepsilon = 0.020 \, \text{L/(ng·μL)} \), \( c = 34.00 \, \text{ng/μL} \), and \( l = 1 \, \text{cm} \) into the equation.
3. Performed the multiplication: \( 0.020 \cdot 34.00 = 0.68 \).
4. Reported the absorbance as 0.68, unitless.