Final Protein Concentration: 8 uM Final Gdn HCI Concentration (8 M stock) uL Gdn HCI stock uL HAstV-1 stock uL Storage Buffer Final Volume (uL) 50 0.25 50 0.5 50 1 50 1.5 50 2 2.5 25 50 50 3 3.5 4 4.5 5.25 5.5 5.75 6 50 50 50 50 50 50 50 50 50
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stock aliquot - 1mg/ml
use the sequence below to determine the molarity using ExPASy Translate and protProt Program
atgggcgaggagtataaagttgtcctcacatttggatccccaatgagccctaatgcaaataacaaacagacttgggttaataaacctcttgatgcgccttcgggccattacaatgtgaaaattgcaaaggatgttgaccactatctaaccatgcagggtttcacttctatagcatctgttgactggtacactatagattttcaaccatctgaggcgcctgccccgataaaaggcttgcaggtacttgtgaacatctcgaaaaaagctgatgtgtatgccgtcaaacaatttgtcacagcgcagaccaacaacaagcaccaggttacaagcctgttcctagtaaaagtaacaactggttttcaggtgaacaactacctgagttacttttacagggcgtctgctactggggatgccacaactaacctgttggttagaggagacacatacacagcagggataagttttacccagggtggatggtatttgttgacaaatacatctattgttgatggggctatgccacctggctgggtctggaataacgtggaacttaaaactaacacagcgtatcacatggacaaaggcttggtccatctaataatgcctttgcctgagtccacgcaaatgtgttatgagatgctgacatctattccagcggccgcagagctcgctctggtgccacgcggtagttccgctcatcaccaccatcatcaccatcaccaccact
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- Calculate the unknown concentration of the PROTEIN C with an absorbance value of A412 given the standard curve indicated in the table below. Write your final answer (NUMBER ONLY) in two decimal places rounded off. Protein concentration (µg/mL) 0 0.02 0.04 0.06 0.08 0.10 Your answer APIENT Absorbance 0.000 0.161 0.284 0.438 0.572 0.762Given the following information, calculate the total activity in the undiluted protein sample. Activity of 1 ml of diluted sample = 0.5 Total volume of sample = 5 ml Dilution factor = 10 25 50.5 250 2.5Using the data from the standard curve done previously, you record the absorbance of your extracted milk protein sample at 0.499 at 540nm. What is the mass of the protein in mg assuming you collected 10mL worth of the protein? Your standard curve should have a line of best fit with the equation y=0.0001x+0.0038.
- Protein Solubility 3.5 5.5 5 5 0.068 0.028 Absorbance 0.098 Conc.(mg/ml) 0.195 0.130 0.044 0.9% 2.786% 9.429% 11.3: % solubility 4.179% The above table indicates the concentration of protein in the diluted supernatant and the supernatant before dilution at different pH. Dilution Factor 30 22.5 15 PH 7.5 1.5 4.5 5 0.027 3 0.042 Protein Solubility Versus pH 6.5 6 4.5 pH Value The above figure shows a plot of protein solubility versus pH. 50 7.5 7.5 50 0.032 0.053 8.5 50 0.054 0.100 21.429% 9 Please provide a brief discussion and explanation of the results. (using isoelectric point and net charge to explain)What is the amount of dextrose in mg/kg/min provided by the following TPN for a 60 kg patient? Round to 1 decimal place, do not enter units Dextrose (70%) 320 g Amino acids (8.5%) 92 g Lipids (10%) 44 g sodium chloride 60 mEq sodium acetate 40 mEq Sodium phosphate 15 mmol potassium chloride 80 mEq potassium acetate 30 mEq calcium gluconate 18 mEq magnesium sulfate 8 mEq Multivitamins 10 mL Trace elements 1 mL Total volume: 2640 mL Rate: 110 mL/hrA liter of a TPN solution contains 500 milliliters of 50 percent dextrose solution and 500 milliliters of 5 percent amino acid solution. Determine the daily energy and protein intakes of a person who receives 2 liters per day of such a solution. Calculate the average daily energy intake if the person also receives 500 milliliters of a 20 percent fat emulsion three times a week.
- Does this calculation look correct? My goal isto have a target mass of 10 µg of my protein with a total volume of 30 µl. Protein was measured and found to have a concentration of at 308.35 µg/ml Recall, mass = concentration * volume 10 µg = 308.35 µg/ml * V V = 0.01 mg/ 0.30835 mg/mL = 0.032430679 mL = 32.43067942 µL Note: cannot measure this amount; two dilutions required. Dilute by 1/40: measure 2.5 µl of protein at 0.30835 µg/µl and add it to 97.5 diluent to prepare a solution of protein at 7.71 mg/ml. This diluted sample will be used further. Mass = concentration * volume 10 µg = 7.71 mg/ml * V V = 0.01 mg/ 7.71 mg/ml = 1.3 µL Therefore, will measure 1.3 µl of protein at 7.71 mg/ml and add it to 28.7 µl of diluent.Size Size Crude Anion Cation 2) Exclusion Exclusion Lysate Exchange Exchange 1 2 Total Protein Concentration 15.2 6.6 2 3.75 4.7 (mg/mL) Final Sample Volume 60 30 20 4 3 (mL) Enzyme Specific Activity 0.43 1.7 7.25 18.3 12.5 (units/mgprotein) Based on your protein purification sample data, in which of the purification steps in your protocol did you effectively purify your enzyme? Select all that apply. a) anion exchange correct b) cation exchange correct c) size exclusion 1 correct d) size exclusion 2CAR) All una Red stock soluton concentration: wis strue Somple #yelume dilution 1 ete total tock AR volume 50.00m 50. 00mL 50.00L AR COC conentrastion Absorbance Q dilutibn 2 dilution 3 10 O. 16.4 20 0. 237 turn dilution 4 25 So00 O. 392 shift unknarn 1 0.485 (Include a printed copy of the colibratlon curve) Colibrotion urve cayotion: R2 valve
- 3 i. Define Biological Value (BV) of Proteinsii. What are the advantages of this protein evaluation method when compared to Nbalance and chemical score?5iii. A bull consumes 7 kg DM of feed containing 60 g N/kg. The bull excretes 120 g N inthe faeces and 50 g N in the urine. The feacal and urinary N had 20 g metabolic feacalnitrogen (MFN) and 6 g endogenous urinary nitrogen (EUN) respectively.Calculate the BV of the protein in the diet.amino glycine pka is 2.4, 9.8 Calculate the most effective buffering range. if there is two pka, since there are 2 buffer region, is the buffering range 1.4~3.4 and 8.8~10.8? i would like to know the correct answer and detailed solution/reason.Abs undiluted 0.208 Diluted (1/10) 0.105 Diluted (1/20) 0.09 Use the chart above to back calculate the protein concentration of the LDH sample based on the Abs562 measurements of the undiluted and 2 diluted LDH samples. Show calculation for each.