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BIOL 130L Lab Study Notes.docx

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University of Waterloo
Dragana Miskovic

BIOL 130L Lab Study Notes LAB 1: Identification of Some Macromolecules Theory Part I: Treatment Controls Used in Lab • Treatment controls were used in this lab o These controls are specific samples that are used to check if the experimental conditions were set properly and if the procedure was followed correctly  Ensures that any changes in the experiment were as a result to the actual experimental results o These controls can be positive or negative  A positive control = designed/prepared in such a way that the sample should give a positive result  A negative control = designed/prepared in such a way that the sample should give a negative result o WATER was used a treatment control and it served as a negative control for all tests o 1% Starch solution was a positive control for starch in the Iodine test o 1% Glycogen solution was a positive control for glycogen in the Iodine test Theory Part II: Introductory Theory behind the Lab • The most abundant elements in living organisms are “CHON+PS” o Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, and Sulfur o These elements combine in a variety of ways to form biological molecules • Biological molecules (i.e. macromolecules) generally fall under 4 categories: o Carbohydrates (monosaccharides, disaccharides, and polysaccharides) o Lipids o Proteins o Nucleic Acids Theory Part III: The Tests Used to Identify Macromolecules in Various Substances • There were 3 tests used in this experiment: o Iodine Test for Starch or Glycogen o Benedict’s Test for Reducing Sugars o Biuret Test for Protein • Note: Nucleic acids were not identified in this first lab; only carbs (starch and glycogen), lipids, and proteins were! • The Iodine Test for Starch and Glycogen: o The iodine solution used in this test is a pale yellow o Positive result: Blue-black for starch; reddish-brown for glycogen  Starch is a mixture of 2 different polymers (amylose and amylopectin) • Amylose is an unbranched helical molecules with alpha 1,4 linkages • Amylopectin is a branched straight molecule with alpha 1,6 linkages • Iodine solution reacts with the amylose to form the intense blue- black colour  Glycogen is similar to amylopectin but is more branched • Iodine solution reacts with glycogen’s multi-branched structure to form the less intense reddish-brown colour o Negative result: substance remains yellow:  When applied to a substances that has neither starch nor glycogen, then a negative result should simply be that the substances remains yellow because the yellow iodine solution did not react with any starch or glycogen present • The Benedict’s Test for Reducing Sugars: o Benedict’s test solution contains blue cupric (copper) ions Cu++ o A simple sugar will have a free aldehyde group, which characterizes it as an aldose sugar  The cupric ions of the Benedict’s test will react with the free aldehyde groups at the end of the straight-chain sugar to reduce the ions into Cu+  Also bear in mind, substances like fructose with free ketone group can also reduce Benedict’s solution o Positive result:  Reducing sugars with FREE aldehyde or ketone groups in their straight- chains will be able to reduce Cu++ ions into Cu+ causing the blue solution, when added into the substance, and boiled, to develop into a precipitate ranging from yellow, green, red, brown, or a combination o Negative result:  If no reducing sugars are present OR if the sugar does not have a free aldehyde/ketone group, then the substance should remain blue, which is the same colour as the added Benedict’s solution • The Biuret Test for Proteins: o Proteins are formed by peptide bonds between each of their monomer amino acid components  The solution used in the Biuret test has Cu++ ions with alkali that reacts with peptide bonds in proteins o Positive result:  Presence of peptide bonds within proteins will cause the Cu++ ions and alkali to react with the bonds to create a PURPLE/VIOLET colour o Negative result:  The original colour of the Biuret Test is a light blue hue  Therefore if the substance being tested does not have proteins, which means by extension no peptide bonds, then the substance should just remain a light blue colour See expected results for experiment #1 on the next page! Expected Results for Lab #1: Qualitative Result  Qualitative Result  Qualitative Result  for Iodine Test for  for Benedict’s Test  for Biuret Test for  Contents of Beaker  Starch or Glycogen for Reducing  Proteins Sugars  Beaker # (15 mL each) (Pos. or Neg.)   (Pos. or Neg.)   (Pos. or Neg.)   1% Glucose  Solution 1 Negative Positive Negative  0.3% Glucose­1­ phosphate 2 Negative Negative Negative 1% Maltose solution 3 Negative Positive Negative 4 Honey solution  Negative Positive Negative 1% Sucrose solution 5 Negative Negative Negative 6 1% Lactose solution Negative Positive Negative 7 1% Glycogen  Positive  Negative Negative solution (for Glycogen) 8 1% Starch solution Positive  Negative Negative (for Starch) 9 Protein Negative Negative Positive  10 Beer Negative  Positive Negative  11 Distilled water Negative Negative  Negative  LAB 2: Isolation of Some Macromolecules Theory Part I: Isolating Just the Proteins and Nucleic Acids from Yeast Cells Using TCA • One package of yeast and a few teaspoons of sand were placed in a mortar and grounded for a few minutes o Yeast cells are held together by glucan, a starchy binding material similar to cellulose, that makes up yeast cells walls o Glycogen, proteins, and nucleic acids are present in the cytoplasm o The grinding process is done to rupture the cell walls and cell membranes • Trichloroacetic acid (TCA) was then added! o Polysaccharides like starch and glycogen are soluble in TCA and will go into solution o Nucleic acids and proteins are insoluble in TCA and will remain in suspension • Decant the suspension into 4 clean tubes equally in terms of volume o The nucleic acids and proteins should be in suspension o The carbohydrates should be in solution • Centrifugation now occurs on the contents of the 4 tubes: o This is a fractionation technique where centrifugal force is applied by controlling speed and time in order to separate particles of different sizes in a solution o Causes the sediments (i.e. the precipitate/pellet) at the bottom to be separated from the liquid (i.e. the solution/supernatant) • After centrifugation is complete, the supernatant is dumped out, leaving only the pellets in each tube o Again, the supernatant contained polysaccharides o The pellet contains proteins and nucleic acids Theory Part II: Separating Proteins from the Nucleic Acids Using NaCl • NaCl was then added to the 4 tubes containing the pellets of protein and nucleic acids o Nucleic acids are soluble in NaCl and will go into solution o Proteins are insoluble in NaCl and will go into suspension • The tubes are then boiled will helps coagulate the suspended proteins • After boiling, the contents are transferred into new tubes and centrifuged again o Centrifugation helps to pack the coagulated proteins into a pellet at the bottom o Nucleic acids will remain in solution as the supernatant • The supernatants are then decanted into a beaker and the pellets remain in the tubes o Again, the supernatant in the beaker contains nucleic acids o The tubes contain the pellets which are comprised of proteins Theory Part III: Nucleic Acid Fraction • Chilled ethanol is added to the beaker containing the supernatant with nucleic acids in it o Nucleic acids are insoluble in ethanol, so the addition of chilled ethanol along with placing the beaker in ice forces the nucleic acids out of suspension o The presence of a WHITE suspension should appear whilst stirring • Pour the liquid and its white suspension evenly in 4 tubes and centrifuge once more o Centrifugation causes the white suspension to form into a pellet at the bottom, thus separating it with the solution • H 2O i4 added into one of the tubes containing a white pellet of nucleic acids to re- suspend it o There is an annoying pouring step in the procedure that is unimportant; the point is that all 4 white pellets will be re-suspended in hydro sulphuric acid because nucleic acids are soluble in H2SO 4 • The liquid of nucleic acids is poured into 2 large tubes o One of them is going to be boiled  Boiling is a hydrolyzing process that breaks up nucleic acids into its nitrogenous base, sugar, and phosphoric acid subunits  After boiling, the contents are then poured into a new tube and labelled “hydrolyzed nucleic acids” o The other tube will not be boiled  It will simply be labelled “unhydrolyzed” nucleic acid • The contents of both tubes are neutralized with barium hydroxide Ba(OH) 2 o At the end of the titration white precipitate should be at the bottom of the top  This is barium sulfate salt that has formed as a result of the acid-base reaction o The reason why barium hydroxide and sulfuric acid was used, as opposed to say NaOH and HCl, is because the reaction between B.H and S.A forms an insoluble salt that can be removed  If the salt was not insoluble and could not come out of solution to be removed, then it would interfere with chromatography later on • At the end both tubes (unhydrolyzed and hydrolyzed nucleic acids will receive a yellow cap to be refrigerated Theory Part IV: Protein Fraction • There is an annoying pouring step in the procedure that is unimportant; but what is important is that you will eventually get 2 tubes: o One tube is going to have a re-suspended pellet of protein with pancreatic enzyme and phosphate buffer  “Hydrolyzed Protein” stoppered with a red cap  Red cap = incubate for 37 degrees for 24 hours then refrigerate o The other tube will have a re-suspended pellet of protein with phosphate buffer, but no enzyme  “Unhydrolyzed protein” stoppered with a blue cap  Blue cap = refrigerate immediately • For obvious reasons, the pancreatic enzyme is used to breakdown the protein to hydrolyzed it into its individual amino acid monomers o Pancreatic enzyme is used in order to mimic the naturally occurring hydrolytic process in living cells using an actual enzyme • Thymol crystals are also put into both tubes in order to prevent or minimize bacterial growth LAB 3: Characterization of Some Macromolecules Theory Part I: Introduction to Chromatography • Chromatography is a technique that separates mixtures into their individual components o The degree of separation of the components depends on the structure of each component  The one that adheres most to the matrix of the chromatography will be slowed down the most; therefore, the difference of the rate of movement between components results in the separation • The chromatographic system has a number of critical components o Stationary phase:  A solid, liquid, or gel that is immobilized and held by a supporting matrix; these substances must be inert o Chromatographic bed:  The environment where the stationary phase will be packed in (e.g. a glass or metal column) o Mobile phase:  a liquid or gas that acts as a solvent to carry the sample through the stationary phase o Delivery system:  a means to pass the mobile phase through the chromatographic bed o Detection system:  a means to monitor the test substance • How does chromatography work? o A test mixture is applied to a spot on the matrix of the stationary phase o The matrix is then partially immersed in the mobile phase (solvent) in such a way that the application is ABOVE the solvent level o Thus, when the solvent begins to migrate along the matrix, it will dissolve the mixture to allow the components of the mixture to migrate as well  Choice of solvent is VERY important  If the test mixture is completely soluble to the solvent, then they would move so rapidly that no separation occurs while migrating  But if the test mixture is completely insoluble to the solvent then no migration will occur • Factors that affect overall mobility of substances during chromatography: o The nature of the mixtures relative to the solvent o Molecular weight of the test mixture o Polarity of the test mixture Theory Part II: The R falue • Chromatography is terminated when the solvent has almost reached the opposite end of the matrix, where the leading edge is marked • The distance of each substance travelled from the origin is then measured and compared to the distance travelled by the solvent itself o The ratio of the 2 differences is a constant (i.e. with no units), known as fhe R value (relative mobility factor): R = Distance(¿origin)travelled bysubstance(cm) f Distance(¿origin)travelledbysolvent(cm) Theory Part III: Separation & Identification of Proteins • Mostly procedural stuff… • Each test substance was applied 6 times on the same spot along the chromatography paper; each application was put on only after the previous one was dried • When stapling the chromatography paper to make a cylinder shape, stapling should be done in a way that does not make the ends of the paper overlap • The solvent used was formic acid, which is toxic and thus the chromatography was done under a fume hood within a jar Theory Part IV: Separation & Identification of Nucleic Acids • Same procedural stuff as proteins section • The only difference is that the nucleic acids can only be seen afterward by shining UV light since nitrogenous bases can absorb light and appear as a dark spot on a pale background • Few notes: o RNA in yeast cells is much more abundant than DNA, therefore all of the nucleic acid samples are derived from RNA o Also, guanine is insoluble and is hardest to detect by paper chromatography Expected Results: • Hydrolyzed Protein – 100% successful fraction will yield individual amino acid subunits o Hydrolyzed protein also travelled the furthest up the matrix • Unhydrolyzed Protein – Should still contain full proteins/polypeptides • Hydrolyzed Nucleic Acid – 100% successful fraction will yield nitrogenous bases, sugar, and phosphoric subunits o Hydrolyzed nucleic acid travelled the furthest up the matrix • Unhydrolyzed Nucleic Acid – Should still contain chains of nucleotides LAB 4: Spectroscopy Theory Part I: The Spectrophotometer • The human eye is sensitive to wavelengths between 400nm (violet) to 750nm (red), which is the visible light spectrum • The spectrophotometer works in the following way: o There is a white light source that is focused on a prism o The prism serves to separate the white light into its component portions in the spectrum o Each distinct energy band can be focused through a narrow slit onto a sample specimen o That specimen is dissolved in a suitable solvent in a cuvette o The focused incident beam this the specimen and can either be  Absorbed, transmitted, or reflected o The transmitted remnant of the incident beam can then strike an photoelectric cell  The photoelectric cell generates a current that is proportional to the intensity of the beam o That electricity is measured by a galvanometer with a graduated scale either by:  % transmittance – arithmetic from 0 to 100%  Optical density – logarithmic scale with unequal divisions 0 to 2.0 Theory Part II: Beer’s Law • First, the substance needs to be dissolved in a suitable solvent • Second, a cuvette containing only the solvent (called the blank) is inserted into the machine o The scale is zeroed by manually setting it to 100% T or 0.0 absorbance • The blank is taken out and the specimen is then put in o Now, any reading less than 100% (i.e. more than 0.0 absorbance) is solely because the specimen absorbed some of the incident light • Beer’s Law: Beer’s Law: A = εl[C] o Absorbance of light passing through a solution is proportional to solute concentration o Absorbance of light pass through a solution is also proportional to the length of the solute Theory Part III: Analysis Using a Spectrophotometer • First step in analyzing a substance is to plot the substance’s absorption spectrum o Read the absorbance of that substance at many different wavelengths o Absorbance on y-axis; wavelength on x-axis o Once the absorption spectrum is plotted, the peaks of the graph represent the wavelengths of maximum absorbance • Now a concentration curve can be constructed o Set the spectrophotometer at the max absorbance wavelength o Then measure the absorbance values for 2-3 different known concentrations of the substance o Plot the results:  Absorbance on y-axis; concentration on x-axis o The points can be extrapolated via a straight line b/c of Beer’s Law • The concentration of an unknown can easily be determined: o The unknown’s max absorbance reading is taken, located on the curve, and a perpendicular line is dropped down to the x-axis, which will identify its concentration. Expected Results: • The concentration of the unknown fast green solution was found to be approximately 0.0081 mg/mL • For chlorophyll A, the max absorption was found at 665nm • For chlorophyll B, there were 2 peaks o Max absorption @ around 455nm o Max absorption @ around 645nm • Conclusions: o Most leaves of plants are green because they are unable to absorb wavelengths that relate to the colour green on the spectrum. o Therefore those wavelengths are reflected or transmitted and we simply register it as being green using our eyes and brain. LAB 5: Enzymes Theory Part I: Introduction to Enzymes • Enzymes are biological catalysts that combine with a substrate (reactant) to form a substrate-enzyme complex, which then converts the substrate into a product all while not altering itself o Substrate + enzyme  Substrate-enzyme complex  Product + Enzyme (unaltered) • Things that affect enzyme activity: o Temperature o pH o Enzyme concentration o Substrate concentration o Product concentration o Activation energy • Theoretically, all enzyme catalyzed reactions are reversible but the direction of the reaction depends on the conditions under which the reaction takes place in Theory Part II: Salivary Amylase • Digestive enzyme found in saliva • Acts on starch by breaking off maltose from the ends of the starch chain • Each time it breaks off a maltose, one water molecule is consumed o Therefore, this is a hydrolytic reaction o The process is called “hydrolysis” Expected Results for Salivary Amylase Portion of Lab: Contents of Beaker Results for Iodine Test Results of Benedict’s Test (For starch) (For reducing sugars) 10% salivary amylase - + 5% salivary amylase - + 2% salivary amylase - - 1% salivary amylase - - 1% starch solution + - • Any solution with salivary amylase in it did not show signs of having starch after the iodine test was done o Likely b/c the salivary amylase broke down all the starch • Only the 10% and 5% salivary amylase showed signs of a reducing sugar based on the Benedict’s test o This makes sense because salivary amylase works by taking off maltose from the ends of starch o However, 1% and 2% salivary amylase showed no signs of reducing sugar, despite having salivary amylase in it  Likely due to either contamination…  Or perhaps the amount of maltose removed was so little that the Benedict’s test did not pick it up Concentration of Salivary Amylase Time Taken for Starch to be Broken Down 10% 65 seconds 5% 75 seconds 2% 300 seconds 1% 420 seconds • This part of the experiment illustrates that the more enzymes there are the faster starch can be broken down Theory Part III: Phosphorylase • Phosphorylase is an enzyme the takes off glucose molecules by rupturing the glucose- glucose bonds in starch • Every time phosphorylase takes off a glucose, a phosphate ion is consumed o This process is called phosphorylysis o Unlike with salivary amylase, water’s role is simply a medium in this process  It is neither a product nor a reactant in the reaction o There needs to be phosphoric acid present for phosphorylysis of glucose to occur since the acid is the source of phosphate ions • Also keep
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