Results These are the images taken of the gel after electrophoresis. Gel 1 Gel 2 Lane 7. This is Maddie’s (MCB) sample. Gel 2 Lane 6. This is Madi’s (MN) sample. From our sample of the gel electrophoresis‚ both Madi and me are homozygous positive (+/+) for the Alu gene. This can be determined by looking at the ladder and comparing our sample to it‚ to find out if we are homozygous or heterozygous. Discussion For this lab‚ DNA from our cheek cells were separated through PCR
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and Function 16 October 2014 Gel Filtration and Electrophoresis Objective The essential goal of the experiment was to separate proteins in a solution based on size in different fractions. The relative protein content for each one fraction was found through the utilization of an amido black-based protein assay. Later in the trial polyacrylamide gel electrophoresis was utilized to separate BSA from hemoglobin. Methods I. Gel Filtration and Protein Assay: 1. A slurry of Bio-Gel P-100 beads in water
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Gel Electrophoresis Adventure Intro The final goal of this lab was to successfully measure the size of different samples of DNA by placing each sample into a well in agarose gel and running a current through a charged chamber. The DNA samples will move through the gel towards the positive charge. Ideally‚ the DNA will move and create and sequence of smallest to largest. This lab exposes us to DNA technology. Backround Gel electrophoresis is used to separate macromolecules like DNA or RNA by size or
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gel and perform gel electrophoresis in six different dyes. Also‚ to extract DNA from wheat germ. INTRODUCTION: Agarose gel is a substance that is used in science for gel electrophoresis and size exclusion chromatography. These processes use agarose gel to separate and analyze proteins and DNA. The medium is composed of a purified agarose powder that has been boiled in a buffer solution and then cooled into a gel. Agarose gel is most commonly associated with gel electrophoresis. In this procedure
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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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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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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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DATE: 11/10/12 COURSE CODE: BIOL 2365 Comparative Biochemistry TITLE: Proteins and Amino Acids RESULTS: Table 1: The results of experiment 1; the Lowry Test Volume of Standard Protein/ Unknown (mL) Absorbance at 750 nm 0 0.000 0.1 0.017 0.3 0.135 0.3 0.155 0.5 0.230 0.7 0.323 0.7 0.310 1.0 0.457 1.0 Unknown 1a 0.463 1.0 Unknown 1b 0.433 1.0 Unknown 2a 0.237 1.0 Unknown 2b 0.159 Table 2: The results of Experiment 2; Ninhydrin Test Amino acid Color X
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Gel Electrophoresis Lab SBI4U1 May 13th‚ 2013 Gel Electrophoresis Lab Purpose: The purpose of this lab is to learn how restriction enzymes cut DNA molecules at specific sequences‚ thus producing DNA fragments of various lengths. Students learn how fragments form unique patterns‚ which help to distinguish the base for DNA identification. This lab answers the question “whose DNA was left behind?”. Materials: * Transfer pipets * Agarose Gel * Dyed DNA samples *
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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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