A previously unknown protein has been isolated in your laboratory. Others in your lab have determined that the protein sequence contains 172 amino acids. They have also determined that this protein has no tryptophan and no phenylalanine. You have been asked to determine the possible tyrosine content of this protein. You know from your study of this chapter that there is a relatively easy way to do this. You prepare a pure 4900 μM solution of the protein, and you place it in a sample cell with a 1 cm path length, and you measure the absorbance of this sample at 280 nm in a UV-visible spectrophotometer. The absorbance of the solution is 0.511. How many (if any) tyrosines are there in this protein? You will need to use Beer's Law to solve this problem. The molar absorption coefficient of tyrosine is 1490 M-1cm-1. Number of tyrosines in this protein =
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511 Concentration (c) = 4900 μM = 4900 * 10^-6 M Path length (L) = 1 cm Molar absorption coefficient (ε) = unknown Rearranging the formula to solve for ε: ε = E / (c * L) ε = 0.511 / (4900 * 10^-6 * 1) ε = 104.2857 M^-1cm^-1 Show more…
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A previously unknown protein has been isolated in your laboratory. Others in your lab have determined that the protein sequence contains 172 amino acids. They have also determined that this protein has no tryptophan and no phenylalanine. You have been asked to determine the possible tyrosine content of this protein. You know from your study of this chapter that there is a relatively easy way to do this. You prepare a pure $50 \mathrm{mM}$ solution of the protein, and you place it in a sample cell with a $1-\mathrm{cm}$ path length, and you measure the absorbance of this sample at $280 \mathrm{nm}$ in a UV-visible spectrophotometer. The absorbance of the solution is $0.372 .$ Are there tyrosines in this protein? How many? (Hint: You will need to use Beer's law, which is described in any good general chemistry or physical chemistry textbook. You will also find it useful to know that the units of molar absorptivity are $\left.M^{-1} \mathrm{cm}^{-1} .\right)$
Dominador T.
Phosphate in urine can be determined by spectrophotometry. After removing protein from the sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate can be measured at $650 \mathrm{nm}$ and is directly related to the urine phosphate concentration. A 24 -hour urine sample was collected from a patient; the volume of urine was 1122 mL. The phosphate in a 1.00 mL portion of the urine sample was converted to the blue polymolybdate and diluted to 50.00 mL. A calibration curve was prepared using phosphate-containing solutions. (Concentrations are reported in grams of phosphorus (P) per liter of solution.) $$\begin{array}{|c|c|}\hline \text { Solution (mass P/L) } & \begin{array}{c}\text { Absorbance at } 650 \mathrm{nm} \\\text { in a } 1.0-\mathrm{cm} \text { cell }\end{array} \\\hline 1.00 \times 10^{-6} \mathrm{g} & 0.230 \\\hline 2.00 \times 10^{-6} \mathrm{g} & 0.436 \\\hline 3.00 \times 10^{-6} \mathrm{g} & 0.638 \\\hline 4.00 \times 10^{-6} \mathrm{g} & 0.848 \\ \hline \text { Urine sample } & 0.518 \\\hline\end{array}$$ (a) What are the slope and intercept of the calibration curve? (b) What is the mass of phosphorus per liter of urine? (c) What mass of phosphate did the patient excrete in the one-day period?
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