Values obtained in lab: reference values: Tyrosine e = 1280 M-1C-1, Tryptophan e

Question

Values obtained in lab:

reference values: Tyrosine e = 1280 M-1C-1, Tryptophan e= 5690 M-1C-1

I believe for this problem we use the refence values, but im really confused so I'm not sure. Please show steps.

Concentration (M) Absorption of Tyrosine Absorption of Tryptophan 0 0 0 1.9 x 10^-5 -5.9 x 10^-2 0.068 3.9 x 10^-5 0.057 0.11712 7.9 x 10^-5 0.11771 0.28056 1.5 x 10^-4 0.18507 0.62605 3.125 x 10^-4 0.31571 1.31400 6.25 x 10^-4 0.71224 1.25 x 10^-3 1.34390 Use your knowledge: find protein concentration e protein p53. Known as the "guardian angel" of the human genome, p53 is critical for maintaining the genetic stability of SSVPSQKTYQGSYGFRLGFLHSGTAKSVTCTYSPALNKMFCOLAKTCPVQLWVDSTPPPGTRVRAMAIYKOSQHMT e sequence below is the the cell and preventing cancer by inducing apoptosis in damaged cells. EVVRRCPHHERCSDSDGLAPPOHLIRVEGNLRVEYLDDRNTFRHSVVVPYEPPEVGSDCTTIHYNYMCNSSCMGGM NRRPILTIITLEDSSGNLLGRNSFEVRVCACPGRDRRTEEENLRKKGEPHHELPPGSTKRALPNNT Estimate the molar extinction coefficient of the p53 core domain by adding the molar extinction coefficients of all tyrosine and tryptophan residues. (Use the values for tyrosine and tryptophan above for your calculation.) Report the value in the discussion section of your lab report. If a pure sample of p53 core domain had an A28o value of 0.88, using a standard 1 cm cuvette, what is the protein's concentration in molar AND micromolar? To get the answer, rearrange the Beer's law equation and solve for concentration. Report the value in your Discussion section (include calculations in your Methods or Results section).

Explanation / Answer

total number of tyrosine residue - 8

Total number of tryptophan residue- 1

Molar extinction coefficient = 8*1280 + 1*5690

= 10240+ 5690 = 15930

So Molar extinction coefficient of p53 is 15930 M-1cm-1

B- We need to find out the concentration of protein whose OD is 0.88

According to Beer-Lambert law

A= ECl

Here E= molar extinction coefficient

C= concentration in Molar

l = path legnhr = 1cm

0.88 = 15930 *C *1

C = 0.88/15930

5.52 * 10-5 Molar

or 5.52 *10-2 mM

or 55.2 microMolar

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.