Respuesta :
Answer:
a) In this protein, there are 3 Tryptophan residues
b) c = 3.128 x 10^-5 M
Explanation:
Molar extinction coefficient (ε) tells us how strongly some chemical substance can absorb light at particular wavelength. In proteins, aromatic amino acids such as Tyrosine and Tryptophan are capable of absorbing the light at 280 nm.
a) ε (protein) = 17900 M-1cm-1
We know that protein has one Tyr, which ε equals 1400 M-1cm-1
If we deduct this value from total ε of protein we wil get ε that comes from all Trp in protein.
ε (Trp) = ε (protein) - ε (Tyr)
ε (Trp) = 17900 M-1cm-1 - 1400 M-1cm-1
ε (Trp) = 16500 M-1cm-1
Then, to get the exact numbers od Trp in protein, we divide this ε with known ε of Trp which is 5500 M-1cm-1
number of Trp: 16500 / 5500 = 3
In this protein, there are 3 Tryptophan residues.
b) The equation which connects all this values is:
A = εlc
A is the amount of light absorbed by the sample for a particular wavelength
ε is the molar extinction coefficient
l is the distance that the light travels through the solution
c is the concentration of the absorbing species per unit volume
A = 0.56
ε = 17900 M-1cm-1
l = 1cm
c = ?
c = A/(εxl)
c = 0.56/(17900 M-1cm-1 x 1cm)
c = 3.128 x 10^-5 M
The amino acid structures that build up the polypeptides chains and help in the body function, signal and repair are called proteins. The molar extinction (ε) is the value that tells about the amount of light a substance can absorb at a specific wavelength.
In the given protein there are a. three tryptophan residues and b. [tex]\rm c = 3.128 \times 10^{-5}\;\rm M[/tex].
The number of Trp can be estimated as:
(a) In proteins the amino acids like Tryptophan and tyrosine can absorb light at 280 nm.
Given,
- Molar coefficient of the Trp = [tex] 5500 \rm M^{-1}cm^{-1}[/tex]
- ε (protein) = [tex] 17900 \rm \;M^{-1}cm^{-1}[/tex]
- The given protein has one Tyr (ε ) = [tex]1400 \;\rm M^{-1}cm^{-1}[/tex]
The molar coefficient of all Trp can be calculated by subtracting the ε Tyr from the ε total .
[tex]\begin{aligned}\epsilon (\rm Trp) &= \epsilon (\rm protein) - \epsilon (\rm Tyr)\\ \\ \epsilon (\rm Trp) &= (17900 \;\rm M^{-1}cm^{-1}) - (1400\; M^{-1}cm^{-1})\\ \\ \epsilon (\rm Trp) &= 16500\;\rm M^{-1}cm^{-1}\end{aligned} [/tex]
The number of Trp can be estimated as:
[tex]\begin{aligned}\text{Number of Trp} &= \dfrac{16500 }{5500} \\ \\ &= 3\end{aligned}[/tex]
Thus, there are 3 Trp in the protein sample.
Concentration of the protein can be calculated by:
(b) The equation that will be used is:
[tex]\text{A} = \epsilon \text{lc}[/tex]
Where A = amount of light absorbed, ε = molar extinction coefficient, l = distance traveled by the light and c = concentration
Given,
- A = 0.56
- ε = [tex]17900\;\rm M^{-1}cm^{-1}[/tex]
- l = 1 cm
- c = ?
Formula will be:
[tex]\rm c = A}{(\epsilon\times l)}[/tex]
Putting values in the formula:
[tex]\begin{aligned}\rm c &= \dfrac{0.56}{(17900 \;\rm M^{-1}cm^{-1} \times 1 \rm cm)}\\ \\ c &= 3.128 \times 10^{-5}\;\rm M\end{aligned} [/tex]
Thus, the concentration of the protein in the solution is [tex]3.128 \times 10^{-5}\;\rm M[/tex].
Therefore, a. there are three residues of Trp and b. the concentration of the proteins is [tex]3.128 \times 10^{-5}\;\rm M[/tex].
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