An enzyme was purified from a yeast homogenate using column chromatography. Enzyme activity of the homogenate and column eluate was measured and found to be 0.74 nkat and 0.52 nkat, respectively. What percentage of enzyme activity was lost during the purification process? % of enzyme activity was lost during the purification process. Round your answer to the nearest integer where necessary. You are performing a colourimetric assay in order to determine protein concentration of a cellular lysate. You set up the assay for several standards of known concentrations and the lysate that has been diluted 5-fold and measure absorbance using a spectrophotometer. You then plot the absorbance values against the concentration of standards and obtain a line of best fit with the following equation: A = 0.027 + 0.737C where A is absorbance and C is protein concentration in mg/mL. Your diluted lysate gave absorbance 0.244. What is the protein concentration of the undiluted lysate? The concentration of the undiluted lysate is mg/mL. Write your answer in decimal form rounded to 2 decimal places where necessary. Column chromatography was performed to obtain pure protein. Fractions with 3 mL volume were collected and their protein concentration was determined (see table below). What is the total amount of protein obtained from the column, i.e. contained in the three fractions? Fraction | Concentration (mg/mL) 1 | 2.01 2 | 4.15 3 | 1.07 The total amount of protein obtained from the column is mg. Write your answer in decimal form without rounding.
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First, we need to find the percentage of enzyme activity lost during the purification process. We can calculate this using the formula: Percentage loss = $\frac{Activity_{initial} - Activity_{final}}{Activity_{initial}} \times 100$ Percentage loss = $\frac{0.74 Show moreā¦
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Adi S.
A protein assay requires the reaction of the protein solution with a protein-binding dye for one minute, followed by measuring the absorbance of the protein-dye complex at 505 nm. The absorbance data for four standard protein solutions and the blank is provided in the table. The absorbance of each solution was measured three times. Construct a calibration curve using the data in the table and find the slope (m), intercept (b), and their respective standard deviations using the method of least squares. Note that the calibration curve should be created from absorbance values corrected for the absorbance of the blank, and not the raw data in the table. Obtain the corrected data by subtracting the average absorbance of the three blank values from each absorbance value in the table, including the individual blank values. When creating the calibration curve, do not use the average absorbance at each protein concentration. Instead, each data point should be used separately, meaning there will be three data points at each protein concentration, for a total of 15 points, including each individual blank value replicate 1, replicate 2, and replicate 3. blank: 0.0332, 0.0331, 0.0335 0.10 g/dL: 0.0985, 0.0982, 0.0981 0.20 g/dL: 0.230, 0.230, 0.230 0.50 g/dL: 0.523, 0.525, 0.524 1.00 g/dL: 1.015, 1.016, 1.014 Calculate the concentration and uncertainty of a protein solution that produced an average absorption of 0.800 when measured three times.
Sri K.
An enzyme is extracted and purified from a sample of E. coli cells. The enzyme's catalytic activity is measured, and its absorbance spectrum is recorded. Then this same batch of enzyme is put through a series of modifications as outlined in the table below. The first modification step involves adding a large concentration of urea to the solution containing the enzyme molecules. Then the urea is removed using a procedure known as dialysis, which involves placing the enzyme solution inside a semipermeable membrane that allows only the urea to diffuse out. Following this, the enzyme is then treated again with 6 M urea followed by a chemical modification step. During chemical modification, the six lysine residues of the enzyme react with a chemical reagent in the presence of urea. As a result of the reaction, a new methyl group becomes covalently bonded to the end of each of the six lysine residues in the enzyme. In the final step, the chemically modified enzyme is dialyzed to remove urea. Can you compare the results after each of the two dialysis steps? And explain why the results were different?
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