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BIOL 130L Exam Review.docx

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University of Waterloo
Dragana Miskovic

BIOL 130L Exam – November 30, 2012 1. Identification of Some Macromolecules - positive control: designed/prepared in such a way that it should give positive results if experimental conditions were set properly and followed - negative control: designed/prepared in such a way that it should give negative results if experimental condition were set properly and followed - time points: time as a variable, have to have a control for every time point in experiment; zero time point control is extremely important; samples must be measured immediately after treatment, at time zero - most abundant elements in living material: C, H, O, N, S, P - four major biological macromolecules: carbohydrates (monosaccharaides – fructose, galactose, glucose, polysaccharides), lipids, proteins, nucleic acids Iodine Test - blue-black in the presence of starch and a reddish-brown in presence of glycogen - in plants, starch is the storage polysaccharide composed of multiple glucose units linked together by glycosidic bonds - in animals, glycogen is the counterpart to starch and is generally a larger polymer - starch is a mixture of two different polymers:  Amylose: unbranched, helical molecule where glucose units are joined by (1- 4) linkages; reacts with iodine to give blue colour  Amylopectin: straight and branched sections from (1-6) linakges - glycogen is similar to amylopectin in over structure but more highly branches  Reddish-brown colour due to multibranched components Benedict’s Test - basic carbohydrate molecules are simple sugars - all sugars can exist as straight chains or ring forms  Straight form = terminal group (aldehyde group O=C-H) – aldose sugar; it is this part that reacts in the Benedict’s test and makes glucose a reducing sugar - if blue solution develops a coloured precipitate (yellow or green to red or brown), the test is positive and indicated presence of reducing sugar - Benedict’s solution contains cupric ions (Cu++) that when mixed with reducing sugars, the aldehyde group will reduce cupric ions to cuprous ions (Cu+); in turn, Cu+ will combine with oxygen to form a precipitate of cuprous oxide (Cu O) 2  4Cu + 2OH + 2e  2Cu O + 22 + 2e + - cuprous oxide is generally proportional to the concentration of free aldehyde groups in solution - ketose sugar (fructose) can reduce Benedict’s solution  Carbon 1 of ketose is not oxidized directly but basic conditions catalyze isomerization of 2-ketose to an aldose (positive test) - false positive reaction could happen due to presence of other substance that could be oxidized Biuret Test - treatment of peptide bonds within proteins with Cu ions and alkali give a violet ++ Cu peptide complex - added NaOH (sodium hydroxide) to each tube as well as CuSO (cop4er sulfate) 2. Isolation of Some Macromolecules - yeast contains yeast cells held together with a starchy binding material - glucan (a cellulose-like polysaccharide) is present in the year cell walls, and glycogen, proteins and nucleic acids are present in the cytoplasm - grinding cells will rupture cell walls and membrane - TCA (tricholoracetic acid): starch and glycogen are soluble in TCA while proteins and nucleic acids are insoluble in TCA and remain in suspension - centrifugation: technique in which centrifugal force sediments suspended particles at the bottom of the tube  Sediment is precipitate (or pellet) and liquid remaining above is know as the supernatant  Supernatants contain polysaccharides; the pellets contain proteins and nucleic acids - nucleic acids are soluble in strong NaCl while proteins are insoluble in NaCl (suspension) - nucleic acids are insoluble in ethanol and will precipitate out of solution to form a suspension - nucleic acids are soluble in sulphuric acid and will go into solution - boiling in acid is a hydrolyzing process; it will break up nucleic acid initially into its component nucleotide subunits and eventually into the base and sugar phosphoric acid subunits  Hydrolysis involves breaking the peptide bonds through addition of water forming two molecules  Denaturation does not affect the primary structure; unfolding of secondary, tertiary or quaternary structure - litmus paper:  If unchanged, test solution is acid  Slightly blue = reached end of titration with the addition of drops of base (slightly past pH=7) - end of titration should be a white precipitate formed (barium sulfate) as a result of the reaction between the base and acid  H 2O +4Ba(OH)  B2SO + 2H O4 2 - barium hydroxide was used as a titrating base (rather than NaOH) and sulfuric acid as the titrating acid (rather than HCl) - salt formed is insoluble and will precipitate out, leaving a salt-free solution of nucleic acid (if didn’t precipitate out, remained in nucleic acid solution as a contaminant and interfere with chromatographic separation of nucleic acid) - (un)hydrolyzed protein = pancreatic enzyme (phosphate buffer) and thymol  Pancreatic enzyme will hydrolyze the proteins into their amino acids; simulates naturally-occurring hydrolytic process and works best at pH=7  Hydrolysis could be achieved also through boiling in a strong acid/base  Thymol prevents (or minimizes) bacterial growth 3. Characterization of Some Macromolecules - chromatography: technique that separates mixtures into their individual components - stationary phase: in any chromatography is the matrix (generally an inert substance like cellulose – chromatography paper) - mobile phase: is the solvent - test mixture is applied on spot on the matrix and then immersed in the mobile phase so that the application is above the level of solvent - solvent absorbed by the matrix and allows the components of the mixture in the substance to migrate along the matrix - choice of solvent is crucial: if substance was completely soluble in mobile phase, no separation would occur during rapid migration; if completely insoluble, the substances would not migrate at all - factors that influence mobility of substance: molecular weight and overall polarity - Rf= distance (from origin) travelled by substance / distance (from origin) travelled by solvent - chromatography for proteins: ninhydrin-acetone reacts with amino acids and will cause them to turn purplish-pink - chromatography for nucleic acids: all nitrogenous bases or compounds containing nitrogenous bases absorb U.V light (any compound with a base will appear as a dark spot) - amount of RNA in yeast cells is greater than amount of DNA - most of the substances in nucleic acid sample will be derived from RNA not DNA - base guanine is insoluble and has been found difficult to detect by chromatograph - 100% successful:  Hydrolyzed protein (amino acids, broken peptide bonds)  Unhydrolyzed protein (protein molecules with peptide bonds)  Hydrolyzed NA (nitrogen bases, sugars, phosphates, broken phosphdiester/glycosidic bonds)  Unhyrolyzed NA (nucleic acid molecules with full bonds) - partially successful:  Hydrolyzed protein (part protein molecules, part amino acids with some broken bonds)  Hydrolyzed NA (part nucleic acids, part nitrogen bases, sugars, phosphates with some broken bonds) 4. Spectroscopy - light is a form of energy: photons, light particles with no mass, move through space as waves - human eye sensitive to wavelengths between 400nm (violet) and 750nm (red) - spectrophotometer: consists of white light source, which is focused on a prism that separates the white light into distinct narrow portions of spectrum bands of radiant energy - beam hits the sample specimen which is dissolved in a solvent housed in an optically selected tube called a cuvette - beam reaches sample, gets absorbed/transmitted/reflected - transmitted remnant or beam strikes a photoelectric cell and generated an electrical current that is proportional to the intensity of transmitted beam - electric current is measured by a galvanometer with a graduated scale  Arithmetic scale: percent transmittance (%T) – 0% to 100%  Logarithmic scale: absorbance or optical density with unequal divisions graduated from 0.0 – 2.0 - to be analyzed, a substance is first dissolved in a solvent - cuvette containing only solvent (called a blank) is inserted into machine and scale is zeroed for that particular solvent by manually setting needle at 100% T or 0.0 absorbance - blank is replaced with a cuvette that contains solvent plus substance – any reading on scale now is less than 100% transmittance or more than 0.0 absorbance due to fact that substance absorbed some of the incident light - energy of transmitted beam was less than the energy of the incident beam - Beer’s Law: concentration of a light-absorbing solute it directly proportional to the absorbance over a given range of concentration - absorption spectrum: absorbance vs. concentration graph (line)  Result offers the absorption spectrum of substance  Able to determine wavelength(s) of max. absorption by observing where the peak(s) occur - once wavelength of max. absorption, construct a concentration curve  Once plotted, able to extrapolate by connecting known points with a straight line (Beer’s Law)  Easily able to determine concentration of an unknown sample (perpendicular line dropped to x-axis and value identified) - calculation for concentrations (C1 1= C2 2) - numerous pigments in chloroplasts that have the ability to absorb part (or parts) of the visible spectrum - pigment appears deeply coloured because it’s unable to absorb wavelengths of that particular colour; therefore reflected or transmitted - major pigments in chloroplasts are chlorophylls (chlorophyll a and chlorophyll b) and the carotenoids (carotenes and xanthophyll) - chromatograph: thin orange band (carotene), 2 yellow bands (xanthophyll), 1 st nd green band (chlorophyll a), 2 green band (chlorophyll b) - chlorophyll a and b strongly absorb red and blue; their peaks are within the red and blue spectrum and lowest points are covering the green portion (reflect) 5. Enzymes - biological catalysts - thousand of enzymes required to mediate all reactions needed for normal cell function - all enzymes (excluding ribosomes) are proteins - their specif
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