30 QUESTION 30Below is a standard curve generated by taking absorbances of standard solutions at concentration 0.2 mg/mL, 0.4 mg/mL,0.6 mg/mL and 0.8 mg/mL. Keeping in mind that the axes follow the normal standard curve rules, determine theconcentration of an unknown protein sample with an absorbance of 0.4. The slope of the line is shown on the graph.1.21y=1.25x.0.80.60.40.20.10.20.30.40.50.60.70.80.9O Not enough information0.5 mg/mLO 0.55 mg/mLO 0.32 mg/mL

Proteins are the biomolecules which are composed of amino acids. The concentration of proteins can…

Answered: QUESTION 30 Below is a standard curve…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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QUESTION 2 You obtained the following raw data when setting up a Biuret standard curve: Absorbancy BSA (mg/ml) 540nm 0.158 1 0.210 2 0.260 0.305 4 0.360 0.410 0.455 7 0.510 8 0.530 9. 0.550 10 0.554 After blanking against a biuret-dH20 sample, the protein concentration of an unknown sample was determined using the same method and an absorbancy of 0.324 was obtained. curve, excluding outliers (experimental and statistical) and determine the protein concentration in the unknown sample in mg / Set up a standard ml (up to 3 significant figures). 5
The results observed in a patient admitted to an emergency department with confusion and oliguria were reviewed (reference ranges are given in brackets): Sodium Potassium Urea Creatinine Alkaline phosphatase Alanine aminotransferase Albumin Bilirubin Calcium 146 mmol/L 7.2 mmol/L 18.5 mmol/L 256 μmol/L 319 IU/L 32 IU/L 42 g/L 13 μmol/L 2.47 mmol/L (135-145) (3.5-5.0) (3.5-6.6) (70-150) (95-320) (a) What is the electrolyte abnormality in this patient? (b) What do the data suggest? (c) What action should you take? (5-42) (35-50) (<17) (2.12-2.62) The sample was not visibly haemolysed and was centrifuged and analysed in the labora- tory within one hour of being collected.
Tirofiban (C22H36N₂O5S MWt = 440.6) is present as tirofiban HCI monohydrate (C22H36N2O5S.HCI.H₂O MWt=495.1) at 0.281 mg/mL in a concentrated solution. A solution for infusion is prepared by extracting 50 mL from a 250 mL bag of 5% glucose solution and adding 50 mL of concentrated solution. Sofia who weighs weighs 87 kg requires a tirofiban dose of 0.1 mcg/kg/min for 12 to 24 hours. What would the infusion rate be (mL/h)? (Answer to 1 decimal place.) Answer:
The chief pharmacist assigned you to prepare the Furosemide papertabs. The available preparation of Furosemide in the hospital pharmacy is a 40 mg tablet. How many papertabs can be made from a Furosemide 40 mg tablet based on the doctor’s order? Show your computation.You weighed one (1) tablet of Furosemide 40 mg using an analytical balance and the weight of one (1) tablet of Furosemide is 0.2598 grams. What is the weight of each papertab to provide the ordered dose?
Tirofiban (C22H36N2O5S MWt = 440.6) is present as tirofiban HCI monohydrate MWt=495.1) at 0.281 mg/mL in a concentrated solution. (C22H36N2O5S.HCI.H₂O A solution for infusion is prepared by extracting 50 mL from a 250 mL bag of 5% glucose solution and adding 50 mL of concentrated solution. Sofia who weighs weighs 87 kg requires a tirofiban dose of 0.1 mcg/kg/min for 12 to 24 hours. What would the infusion rate be (mL/h)? (Answer to 1 decimal place.)
A 5% dextrose in 1/2 normal saline (D5 1/2 NS) solution is commonly administered to patients needing post-operative IV fluids. How could you prepare 500 mL of this solution? Dextrose is 5% (m/v) and the NaCl is 0.45% (m/v).
After three minutes, the concentration of drug Zip in the red blood cells is 10 mmoles l-1. What is the average rate of entry of drug Zip into the red blood cells in units of moles min-1 red blood cell-1 during the first three minutes after placing the red blood cells into the bathing solution? Assume that each red blood cell occupies about 1 x 10-13 liters.
The standard curve provided below will allow you to answer the remaining questions. Generic Standard Curve 0.7 y = 0.9425x- 0.0034 0.6 0.5 0.4 0.3 02 0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Concentration (mg/ml) - Please describe two ways in which you could use this standard curve to determine the concentration of a pigment in an unknown sample.| #1: #2: -What is your rough estimation of the concentration if you determine the absorbance of a sample is 0.150? Absorbance
The protein concentration of a known standard is 100mg/mL If you prepared a serial dilution, mixing 10μL of protein with 40μL of water what would be concentrations of the first 3 dilutions?
QUESTION 1 Which protein standard curve would you use to read off the protein concentration of an unknown sample with an absorbancy value of 1.105 1.75 1.50 1.25 1.00 0.75 0.50- O- BSA + BGG 0.25 - 0.00 500 1,000 1,500 2,000 Protein Concentration (pg/ml) Download image. O BSA standard curve O BGG standard curve O Either BSA or BGG standard curves O Neither BSA or BGG standard curves Net Absorbance (595 nm)
QUESTION 2 You obtained the following raw data when setting up a Bradford standard curve: Absorbancy BSA (mg/ml) 595nm 0.225 0.310 2 0.420 0.510 4 0.610 0.720 0.810 0.915 0.950 9 0.980 10 0.990 After blanking against a bradford-dH20 sample, the protein concentration of an unknown sample was determined using the same method and an absorbancy of 0.583 was obtained. Set up a standard curve, excluding outliers (experimental and statistical) and determine the protein concentration in the unknown sample in mg / ml (up to 3 significant figures).
Tirofiban (C22H36N₂O5S MWt = 440.6) is present as tirofiban HCI monohydrate MWt=495.1) at 0.281 mg/mL in a concentrated solution. (C22H36N2O5S.HCI.H₂O A solution for infusion is prepared by extracting 50 mL from a 250 mL bag of 5% glucose s and adding 50 mL of concentrated solution. Sonya who weighs weighs 60 kg requires a tirofiban dose of 0.4 mcg/kg/min for 30 minute- What volume would remain after 30 minutes? Answer: 94.53 mL X The correct answer is: 236 mL

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