1) Homeostasis is the condition in which the body maintains a. changing state‚ within an unlimited range. b. an equilibrium point that can change over a narrow range that is compatible with maintaining life c. an equilibrium change that balances external and internal environment such that values of each are equated 2) Anabolism is the a. breakdown of matter. b. expulsion of matter. c. synthesis of matter. d. All of the answers are correct. 3) The sum
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T2:BIOCHEMISTRY AND ANALYSIS:DETECTION OF FATS‚PROTEINS AND CARBOHYDRATES OBJECTIVE The main purpose of the experiment is to understand some general tests that detect fats‚proteins and carbohydrates in foods. INTRODUCTION Carbohydrates are also known as sacharides. There are 4 main groups of carbohydrates‚which are monosaccharides‚ disaccharides‚ oligosaccharides and polysaccharides. Carbohydrates play an important role in living organism as it is the energy storage‚ and it also plays
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With an improved procedure the question could have been answered with this methodology‚ but the aim was to find out the affect of carbohydrates on blood glucose levels‚ but the methodology provided an answer for the question without going into depth of the biochemistry and carbohydrate structure. This could be solved by amending the aim or by constructing some sort of pre-experiment with the bread that would be used to compare the structure of Danskt Rågbröd
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namely carbohydrates and lipids. Carbohydrates‚ which have the empirical formula (CH2O)n‚ acts as the predominant source of metabolic energy in muscle cells during exercise. They are stored in the liver and muscle cells as glycogen which is a branched polymer. Lipids mainly exist as triglycerides‚ which are made up of three fatty acids joined to one glycerol molecule‚ and are stored in adipose tissue. However‚ lipids exist as ketone bodies when in excess. Before a
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In this lab we employed various assays utilizing a biuret reagent‚ coomassie brilliant blue reagent‚ and ultraviolet light in order to determine the identity of six unknown solutions and the concentration of a bovine serum albumin sample. We were given three samples that lacked protein‚ and three samples containing proteins‚ and using a spectrophotometer we assessed the amount of light absorbed versus the light transmitted‚ based on the principles of the Beer-Lambert Law. The three proteins used included
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Bita Heydari Lab report 3 The Effects of Differentiation on Enzymatic Activity Introduction HL-60 cells are capable of undergoing differentiation to induce different cell types. HL-60 cells can undergo morphological changes‚ changes in gene expression‚ and changes in protein synthesis. In the past weeks‚ we were able to conclude that HL-60 cells treated with DMSO and HL-60 cells treated with PMA will differentiate into granulocytes and monocytes upon treatment (1). We were also able to observe
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To purify the protein in the cell lysate from lab 1 through nickel affinity chromatography. Protein purification should result in only one type of protein ideally‚ which is the protein of interest‚ wt-DHFR and mut-DHFR in this case. A pure protein allows for further analysis on the protein to be conducted‚ such as its concentration (Bradford assay)‚ its molecular weight‚ and its biological activity. 2. Overview of experiments Buffer Preparation Add the liquid sodium phosphate‚ solid sodium chloride
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If pH > pI‚ then the protein will have a negative charge and if pH < pI‚ the protein will have a positive charge. Buffer I has a pH >5‚ meaning both proteins carry a negative charge and bind to the DEAE (a positively charged resin). (b) pH = pKa + log10(Base/Acid) [Base = mM of sodium acetate; Acid = mM of acetic acid] = 4.7 + log10 (40/40) = 4.7 In order for the catalase to elute from the column‚ it must have lost its negative charge and stopped binding to the DEAE. Lowering the pH
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Task 1 • Describe the structure of an enzyme as a protein‚ in terms of tertiary/ quaternary structures. 1) Primary Structure This is in reference to the order of way that amino acids are connected to form a protein. These are built up from 20 amino acids‚ and follow these structures o A carbon (the alpha carbon) bonded to the four groups below: o A hydrogen atom (H) o A Carboxyl group (-COOH) o An Amino group (-NH2) o A "variable" group or "R" group 2) Secondary Structure This is in reference
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see looking at the Total Protein column on Table 3‚ the most effective step with regard to the percent of remaining protein removed was affinity chromatography because it was able to remove 98.6% of the remaining proteins. In comparison to 81.93% removed during the 65% ammonium sulfate precipitation and 81.3% during the size exclusion. This means that the affinity chromatography removed a big percentage of contaminating proteins. However‚ removing this huge amount of protein left us with a small amount
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