Data = Average of A595 nm of three wells ÷ 3 Example: A1+A2+A3 ÷ 3 In this case‚ 0.365+0.374+0.453 ÷ 3 = .397 Corrected Data =( (A595 nm well) – BackGround) or (Raw Data – Background) Background = negative control = Bradford Reagent + No Protein Background of this standard curve = A595 nm of well ‘A’ = .397 We used well ‘A’ as our negative control. DATA RAW DATA CORRECTED DATA BSA (µg) | A595 nm | 0
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allows it to be selective about what passes through it. It allows nutrients and appropriate amounts of ions to enter the cell and keeps out undesirable substances. For that reason‚ we say the plasma membrane is selectively permeable. Valuable cell proteins and other substances are kept within the cell‚ and metabolic wastes pass to the exterior. Transport through the plasma membrane occurs in two basic ways: either passively or actively. In passive processes‚ the transport process is driven by concentration
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The nurse is assessing a client with chronic hepatitis B who is receiving Lamivudine (Epivir). What information is most important to communicate to the physician? 1. The client’s daily record indicates a 3 kg weight gain over 2 days. 2. The client is complaining of nausea. 3. The client has a temperature of 99° F orally. 4. The client has fatigue. The nurse is assessing a client with hepatitis and notices that the AST and ALT lab values have increased. Which of the following statements
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antibody interaction. The orientations of the band will provide more information about the interaction of antibody and antigen. Hypothesis: For this experiment‚ antibodies will be placed in wells on an agar plate‚ surrounded by wells containing serum (proteins‚ antigens etc.). If the antibody interacts with the antigen of surrounding wells‚ then the presence
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of an Unknown Protein Solution with the use of the Bradford Method Paul Alcala‚ Andrea Basa‚ Melissa Caagbay‚ Frances Cayomba Department of Chemistry‚ University of Santo Tomas‚ Manila‚ Philippines Abstract The Bradford method used to determine the protein content of a certain solution (Menguito‚ 2010) and involves the acidic Coomassie Brilliant Blue G-250 as a coloring reagent. [1] The dye is originally pinkish-brown in color when it is in its acidic state. When protein is bound to the
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The liver is the largest organ in the entire‚ normal human body. It weighs anywhere from 2.5 to 3.3 pounds. With its large size it is also a very resilient organ. Up to 3/4 of its cells can be removed before is ceases to function. It is red-brown organ roughly shaped like a cone. The liver is located in the upper right abdominal cavity immediately beneath the diaphragm. Without the liver‚ we could not survive. It serves as the body’s chemical factory and it regulates the levels of most of the main
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diffuse glucose and albumin through the 200 MWCO membrane. How well did the results compare with your prediction? a. Glucose was able to diffuse through the 200MWCO membrane at a 0.0040 rate. However‚ albumin was not able to diffuse through the membrane as predicted. Albumin is a large protein that was not able to diffuse through the pore size of a 200MWCO membrane. 4. Put the following in order from smallest to largest molecular weight: glucose‚ sodium chloride‚ albumin‚ and urea. a
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Biological Molecules: Carbohydrates‚ Lipids‚ Proteins By the end of this lab‚ the student should be able to: Identify the functional groups for each of the biomolecules that react in the following biochemical tests: Benedict’s test‚ Iodine test‚ Brown Bag test‚ Sudan III/IV test‚ and the Biuret’s Test. Describe the mechanism of reaction for: Benedict’s test‚ Iodine test‚ Sudan III/IV test‚ and the Biuret’s Test. Interpret the results when presented with data for each of the biochemical
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This agreed with my predictions because after all the urea wasn’t able to diffuse in the 20 MWCO substance. 3. Describe the results of the attempts to diffuse glucose and albumin through the 200 MWCO membrane. How well did the results compare with your prediction? Glucose was diffused in the 200 MWCO membrane‚ but albumin wasn’t‚ this is because the weight difference in this two molecules. This agreed with my predictions because glucose was the only one that diffused in the 200 MWCO. ACTIVITY
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has many possible causes. People with either type of uncontrolled diabetes mellitus are at higher risk. A clinical indicator of early glomerulosclerosis is a change in renal function and is measured by the amount of albumin present in the urine. Microalbuminuria‚ or urinary albumin levels are checked and monitored in diabetic patient as a forecaster of possible future diabetic nephropathies. It is essential to educate
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