Biochemistry Lab I Final Review PDF

Biochemistry Lab I Final Review I. Introduction to the Pipettor a. Introduction i. The pipettor is designed to measure and transfer very small volumes of liquid with accuracy and precision. Accuracy refers to the margin of error bet...

Biochemistry Lab I Final Review I. Introduction to the Pipettor a. Introduction i. The pipettor is designed to measure and transfer very small volumes of liquid with accuracy and precision. Accuracy refers to the margin of error between the dialed volume and the volume delivered by the instrument. Precision refers to the ability of the instrument to deliver a given dialed volume repeatedly. b. Identifying the Parts of the Pipettor i. The pipettor is held in the hand with the fingers curled around the plastic body and the thumb placed on the plunger button. A plastic tip is placed on the bottom of the shaft to prevent cross-contamination between samples and to prevent liquid from being pulled into the shaft. After use, the tip is ejected with the metal ejector rod by depressing the tip ejector button with the thumb. Volume is adjusted by turning the volume adjustment knob or dial. The volume setting is viewed in the volume indicator window. ii. Dialing the pipettor ABOVE the maximum volume range of the pipettor distorts the spring, leading to inaccurate measurements. c. Reading the Scales on the Pipettor i. For the P10 and the P20, the last digit in the window is a decimal. The digits representing decimal places are often red, rather than black to help remind you that these numbers represent decimal places. ii. The P200 has no decimal places. iii. The P1000 is perhaps the most confusing of the scales to read. On the P1000, the “ones” place or the final digit in the number 1000 is not represented on the scale. This is because the P1000 is not accurate in the range of single µl volumes anyways d. Choosing the Correct Pipettor for the Task i. The general rule of thumb to keep in mind is that any given pipettor is accurate down to 10% of the total volume it can transfer. e. Operating the Pipettor i. Set the volume ii. Attach a disposable tip 1  iii. Load the sample: If you slowly depress the plunger you will reach a point where you feel resistance. This is the first stop. If you depress the plunger further, you will hit the second stop. When loading the sample, depress the plunger ONLY TO THE FIRST STOP. iv. Expelling the sample from the tip II. Lab Math and Making Solutions a. The X Dilution Concentration i. If you have 1L of 10X stock solution of TBE, you can make 10L of 1X TBE running buffer (1L of 10X stock + 9L water). b. The Percent (%) Solution i. The key to understanding the meaning of a % solution is to remember the number refers to the # of grams (weight to volume measurement) or mLs (volume to volume measurement) in a total volume of 100 mL ii. Example: 1. 80 grams of sucrose dissolved in a total volume of 100 ml = 80% w/v solution. 2. 10 ml glycerol (a liquid) in 90 ml water = 10% glycerol c. Molarity i. Molarity (M or mol/L) is a CONCENTRATION. Moles represent how much of something you have. ii. Example: 1. You need to prepare 1L of a 1M glycine (F.W. 75.07) solution. How would you accomplish this? a. (75.07 g/mol) (1M) (1L) = 75.07 g glycine plus water UP TO 1 L. 2. You wish to make 10 mL 100 mM glucose solution (F.W. 180.2). How would you prepare this? a. (180.2 g/mol)(100 mM)(1M/1000 mM)(10 ml)(1L/1000 mL) = 0.18 g glucose plus water up to 10 mL. iii. A mole refers to the quantity of the substance you are working with that has the same number of particles as 12g of carbon-12. This number is Avogadro’s number or 6.02 x 1023. d. Unit conversion i. Conversion table 2  ii. ii. Example 1. How many microliters (µl) are in 0.82 milliliters? a. (0.82 ml)(1000 µl/ml) = 820 µl e. Dilution of a solution (C1V1 = C2V2) i. Example: 1. You have a 1 M stock solution of phosphate buffer. You need a working concentration of 150 mM. Describe how you would make 200 ml of 150 mM buffer using the 1 M stock solution. a. (1M stock)(x) = (200 ml)(150 mM)(M/1000mM)  x = 30 ml stock + 170 ml water (total is 200 ml) III. Plasmid Purification Using Alkaline Lysis a. Introduction i. Plasmids are extrachromosomal pieces of DNA that carry genetic information useful to the bacterial cell. They are most commonly found in bacteria as small circular (not all), double-stranded DNA molecules. Some can transfer themselves from cell to cell (within a population or between bacterial cells of 2 different species). Plasmid reproduction coordinated with replication of the host chromosome. ii. Homologous expression is when a gene is expressed in its native organism. Heterologous expression is when a gene is expressed in a non- native organism b. Centrifugation i. When loading a centrifuge, be sure to balance the sample tubes across the rotor. The rotor MUST BE BALANCED. The volume of sample in tubes directly across from one another in the rotor must be the same. The tubes must have equal weight. ii. A “blank” tube can be used if enough sample tubes are not available to balance the rotor. The “blank” should be equal in weight to the tube across from it in the rotor. c. The science behind the plasmid purification method – Steps of Plasmid Prep i. Plasmid DNA can be separated from chromosomal DNA using differential centrifugation. The principle is because the chromosomal DNA is much larger and heavier than the plasmid DNA and can be separated by centrifugation. The chromosomal DNA will pellet out leaving the plasmids in solution. ii. Solution I will be used to resuspend the plasmid after removal of the LB broth. Solution I contains Tris (pH8) as a buffering agent. The slightly 3  alkaline pH helps reduce electrostatic interactions between the DNA and DNA scaffolding proteins once the cells are lysed. iii. It also contains either dextrose or glucose. This helps prevent the cells from lysing from osmotic shock as they are resuspended. Ethylene- diamine-tetra-acetic acid (EDTA) is included as a chelator. EDTA binds up any free metal ions in solution. This is important because once the cells lyse, not only is the DNA free, so are DNases which can destroy the product you are trying to purify. iv. Solution II is the lysis buffer. It contains sodium hydroxide (NaOH) and sodium dodecyl sulfate (SDS). The NaOH and SDS help lyse the cells. SDS is a detergent. It helps break down the cell wall. Detergents are amphipathic in nature. In other words, they have a hydrophilic and hydrophobic side which allows them to interact with the phospholipids (also amphipathic) in membrane bilayers. v. Steps: 1. Pellet overnight culture. Resuspend cells in glucose/EDTA solution 2. Lyse cells 3. Add sodium acetate/acetic acid a. To neutralize the lysis buffer b. Precipitates large proteins and lipids. c. Chromosomal DNA partially renatures; plasmid DNA renatures. 4. Centrifuge to separate plasmid (in solution) from chromosomal DNA (in pellet) 5. Wash with ethanol to remove salts. 6. Dry! Ethanol will interfere with enzyme activity in downstream reactions if it is not all removed. a. Pellet will be clear if DNA is pure. 7. Resuspend in TE/Rnase. TE=Tris/EDTA a. EDTA will continue to inhibit activity of any contaminating enzymes. d. Agarose Gel Electrophoresis i. The higher the % of agarose in a gel, the less porous the gel and the higher the resolving power of the gel. ii. DNA can be found in linear, supercoiled circular, or nicked circular forms. These three species migrate differently on a gel, even if they contain the same number of nucleotides. For this reason, size estimates of DNA cannot be made unless the DNA is in its linear form. iii. High molecular weight species in a plasmid prep are genomic DNA. iv. If you have a diffuse fuzzy band of low molecular weight, it is most likely RNA that was not completely degraded by the Rnase. 4  e. Visualizing DNA on the Gel i. Ethidium bromide is used to visualize DNA on the gel. When exposed to UV light, ethidium bromide glows orange, allowing one to “see” the DNA. Ethidium bromide carries a positive charge at neutral pH. This causes it to migrate in the opposite direction of the DNA on the gel. ii. In this lab, the gel can be soaked in running buffer to remove background staining and allow visualization of the DNA bands. GelGreen will be used. GelGreen is also a DNA intercalator, consisting of two acridinium orange subunits connected by a spacer. GelGreen can also be excited to fluoresce under blue light, making it a safer option than dyes which require excitation by UV light. f. DNA Ladders i. DNA ladders are useful tools to determine if a gel ran as expected and to estimate the size of DNA bands in other lanes of a gel. ii. In this lab we used λ-HindIII digest marker. IV. Restriction Digest of pUC18 DNA a. Restriction Enzymes i. Restriction enzymes can be thought of as a bacterium’s defense system. Restriction enzymes are molecular scissors. They cleave double-stranded DNA at very specific recognition sites. ii. Bacteria prevent their own DNA from being cleaved by methylating the recognition sites in its own DNA, thus preventing cleavage. Recognition sites are typically 4-8 base pairs in length. iii. When restriction enzymes cleave DNA, they leave either “sticky” ends or blunt cuts. Sticky ends leave single-stranded overhangs whereas blunt cuts leave no overhang. Sticky ends are ligated (pasted) together more efficiently than blunt ends in downstream cloning applications. The other advantage of sticky ends is that the direction the DNA fragment can be inserted into a plasmid is controlled. iv. Restriction Endonucleases are enzymes that recognize and cut a specific sequence of DNA. Endonucleases cleave within a DNA strand. Exonucleases cleave DNA progressively from the ends. v. Isoschizomers are restriction endonucleases that recognize the same sequence and cleave at the same site. Neoschizomers are restriction 5  endonucleases that recognize the same sequence but cleave at different sites within the recognition sequence. b. What might you want to clone into a Vector? i. DNA that you need a lot of copies of (making a probe, binding assays) ii. Gene for a protein product you want to express in large quantities (medical applications, crystallization studies, kinetics assays) iii. A mutant version of a gene so you can create mutant protein to study c. “Formula” for a Restriction Digest i. For a 50 µl reaction, 10 units of enzyme are typically added. A unit of restriction enzyme is the amount of enzyme needed to digest to completion 1 μg of λ DNA in a 50 μl reaction in 1 hour at 37oC. ii. The enzyme is suspended in glycerol. The final volume of enzyme in any restriction digest should not be greater than 10% of the reaction volume, because the glycerol can inhibit the function of the restriction digest. iii. The final concentration of the buffer should be 1X. The buffer will help maintain the proper pH, salt, and ion concentrations needed for the enzyme. The final concentration of BSA in a reaction is 1X. Enough time must be allowed for the enzyme to digest the DNA. d. Bovine Serum Albumin (BSA) i. BSA is sometimes required in restriction digests. It stabilizes certain restriction enzymes in solution. BSA coats the plastic pipet tip and microcentrifuge tube which helps reduce loss of the restriction enzyme due to the restriction enzyme sticking to the plastic. V. Polyacrylamide Gel Electrophoresis of Serum Proteins (PAGE) a. Introduction i. Polyacrylamide gels are used when resolving proteins or DNA fragments of nearly similar size. Polyacrylamide has a much higher resolving power than agarose gels. ii. Unpolymerized acrylamide is a neurotoxin and should be handled with care. Once polymerized polyacrylamide gel has formed, it is no longer harmful. b. Native Polyacrylamide Gels versus SDS-Polyacrylamide Gels i. In a native gel, proteins migrate through the gel in their native conformation. Because the proteins vary in terms of size, shape, and charge, the molecular weight cannot be determined from this type of gel as size is not the only factor affecting the migration rate. ii. SDS stands for Sodium Dodecyl Sulfate. In SDS gels, the proteins are denatured. They are not in their native-folded conformation. SDS disrupts the secondary, tertiary, and quaternary structure of proteins, leaving them in their linear, primary form. In addition, SDS also coats the unfolded proteins in the sample with a uniform negative charge. The amount of SDS bound to the protein is proportional to the molecular weight of the linear polypeptide and is independent of the actual amino acid sequence. 6  iii. In this case, the proteins migrate through the gel based on their size and thus molecular weight estimates of the protein sample can be made if an appropriate molecular weight ladder has been run on the gel as well. Glycosylated proteins, however, are an exception to this rule. The sugar side-chains linked to the polypeptide backbone cause the glycosylated protein to migrate differently than would be expected for the actual molecular weight of the protein. c. Stacking gels versus resolving gels i. Polyacrylamide gels consist of two gels stacked edge to edge vertically. The top gel is called the stacking gel and the bottom gel is called the resolving gel. Proteins migrate toward the positive electrode. ii. The stacking gel is where the wells, formed by the presence of a plastic comb positioned before the polyacrylamide has set into a gel, allows the sample to be loaded. iii. The resolving gel is denser and therefore provides greater separation of the proteins. The pore size decreases as the ratio of bis-acrylamide to acrylamide increases and as the overall concentration of acrylamide in solution increases. d. Coomassie Brilliant Blue Staining of Proteins i. Coomassie Brilliant Blue dye stained wool, it would likely stain proteins well too. e. Separation of sera proteins i. The most abundant protein in the blood sera of mammals is albumin. Albumin can bind up to 7 fatty acids and deliver them to where they are needed. f. Western Blotting of Proteins i. Western blotting is a method to identify a specific protein from a background of many proteins. It is used to solve the problem: There could be several proteins in the sample that are approximately the same size. Even when purifying protein from a sample, a method is needed to demonstrate that the protein purified is truly the protein one intended to purify. ii. To complete a Western blot, the sample of interest is run on either a native or an SDS-polyacrylamide gel. If the protein must be folded to be recognized by the antibody, a native system must be used. The antibody recognition site is called the epitope. iii. After the polyacrylamide gel has been run, the protein is transferred to a PDVF or nitrocellulose membrane by electrophoretic transfer. The membrane is then incubated in a prehybridization buffer for two hours up to overnight. Blocking the membrane before hybridizing with antibodies prevents the antibody from binding non-specifically to the membrane. After prehybridization, the primary antibody (or dry milk), specific to the protein of interest, is added and can hybridize to the target protein on the membrane. 7  iv. The membrane is washed to remove excess primary antibody that has not bound tightly to the target protein. The membrane is then incubated with a secondary antibody specific to the primary antibody. Secondary antibodies are typically labeled to allow an indirect way to detect the presence of the bound primary antibody. g. Molecular Weight Markers i. Prestaining proteins alters their actual molecular weights, depending on the dyes that are used. Therefore, the apparent molecular weights will vary slightly between the Kaleidoscope and original pre-stained standards (Kaleidoscope weight marker was used in the lab). VI. Fold-it Protein a. Protein structures i. Primary: refers to the linear string of amino acids (a) (amino acid sequence) ii. Secondary: refers to the folding of the linear polypeptide into alpha helices (b) and beta sheets (c). Secondary structure results from hydrophobic interactions, hydrogen bonding, and ionic bond formation. Disulfide bridges also contribute to secondary structure. 1. Alpha helices a. Right-handed in nature b. 3.6 residues/turn with a pitch of 5.4A 2. Beta structures a. Are either parallel or anti-parallel b. Extended parallel b sheets: (-Gly-Ser-Gly-Ala-Gly-Ala-)n iii. Tertiary: the folding of the sheets and helices into a globular form (d). 1. All the bonds that contribute to secondary structures apply again to tertiary structures (with the addition of metal ion coordination complexes). iv. Quaternary: the association of multiple, separate subunits to form a single, functional protein product (e). b. Protein Folding i. Rules governing protein folding: 1. Peptide bonds are planar and rigid 2. No overlap between amino acid side chains 3. Two side groups (R) with the same charge will be as far away from each other as possible. ii. Folding has 4 Tendencies (sub-rules): 1. Amino acids with polar (charged) side chains tend to be on the surface of the protein. 2. Non-polar (uncharged) amino acid side chains tend to be on the inside of the protein. 8  3. Hydrogen bonds tend to form between adjacent peptide linkages, particularly between a carbonyl oxygen and a hydrogen/nitrogen of the next peptide bond 4. Sulfhydryl groups of cysteine react together to form covalent disulfide linkages. VII. Crystallization of Lysozyme a. Introduction i. To accurately predict the tertiary structure of a protein from its linear primary amino acid, either crystallography or nuclear magnetic resonance (NMR) was used. ii. NMR has been used to solve structures up to 25 kD in size. X-ray diffraction, however, has traditionally been used to resolve structures of larger proteins. 87% of the structures which were entered into the database were resolved based on crystallography data. iii. Bragg’s Law: 2dsinθ = nλ (Where d is the distance between layers of the lattice, θ is the angle between the between the incident ray and the lattice plane, λ is the wavelength of the x-ray and n is an integer describing the order of the reflection) b. Determining the structure i. Follow these steps to create a model of a 3d structure: 1. A suitable crystal of the purified protein of interest must be grown a. To resolve the structure of a protein using X-ray diffraction, the protein of interest must be allowed to form crystals. Ordered lattices of the protein provide organized repeats of the same information over and over. b. Protein crystals must be kept in solvent always. If the solvent is allowed to dry completely, the lattice loses its periodicity and the diffraction data cannot be resolved. 2. The crystal must be subjected to an x-ray beam and the diffraction pattern captured on x-ray film a. The diffraction angle and intensity of the resulting spots on the X-ray film as the X-rays pass through the crystal are recorded. The data is converted to an electron density map. b. A minimum of a 2 angstroms resolution (minimum separation needed for two chemical groups) is needed to accurately determine the position of the amino acid side chains. 3. The diffraction data must be “transformed” to yield the model of the protein ii. Hanging-drops method 1. The concentrated protein is suspended as a hanging drop from a slide coverslip over a well containing a buffer of higher ionic strength than the buffer in which the protein is suspended. 9  2. The cover slip is typically sealed to the reservoir using vacuum grease or silicon oil to prevent the solutions from evaporating into the ambient air. 3. Within the sealed well, the solution in the drop will equilibrate with the solution in the reservoir. 4. Since the reservoir is at a higher ionic strength than the drop, liquid from the drop will move to the reservoir. This crowds the protein and increases the likelihood of protein molecule interactions and the formation of a lattice c. Lysozyme (Lysozyme crystals are commonly tetragonal) i. Lysozyme is one of the easiest proteins to crystallize. It is present in tears and saliva and in abundant quantity in chicken egg whites. ii. Lysozyme serves to weaken bacterial cell walls by cleaving β-1,4-linkages between N-acetylmuramic acid and N-acetyl-D-glucosamine. This property helps protect developing chick embryos from bacterial invasion. It also helps prevent bacterial infections of the nasal passages and conjunctiva around the eye. iii. Just because crystals form in a drop, does not mean you have protein crystals. Sometimes what forms are salt crystals since salt is a common ingredient in the reservoir buffer (salting out effect). 1. Verify that the crystals are actually protein: a simple method to do this is to add a dye of very small molecular size to the drop in which the crystals have formed. The small dye penetrates the space between the lattice of protein molecules as it diffuses through the solvent. 2. Salt crystals are so tightly packed, they exclude the dye. d. The experiment i. When looking for the right combination of conditions to allow crystal formation of purified protein, a screen is usually set up. A variety of conditions are tested. Salt conditions, pH, and precipitating agents such as ammonium sulfate or polyethylene glycol (PEG) concentrations are varied. ii. PEG serves as a crowding agent to aid the assembly of the protein into the lattice. NaCl helps keep the protein soluble, preventing it from 10  precipitating from the solution. Temperature may also be varied. Crystals may form at room temperature or in at 4°C. VIII. How to Write a Journal Manuscript - A Practical Guide to Scientific Writing a. Content i. Title 1. State nature of the work 2. States the name of the organism under study 3. Be concise (avoid unnecessary word or phrasing) ii. Abstract 1. The summary of the experiment – should NOT be long and detailed. a. The purpose of the study b. BRIEF overview of methods c. The key results d. Main conclusions iii. Introduction 1. Should NOT be too long. 2. Include a. Purpose of the experiment b. Summary of techniques (no details) c. Describes theory behind procedures d. Describes past work in the field e. A hypothesis as to what results were expected and WHY f. Appropriate citations of references iv. Materials and Methods 1. Describe how the experiment was conducted. 2. Must include enough information for another researcher to repeat the experiment. 3. Not a shopping list and does not include “common sense” items. v. Results 1. Report of the data. NO interpretation of the data here. 2. Include tables, graphs, and figures or drawings as appropriate for the experiment. Label appropriately. Each should have a description beneath it, not merely a reference in the text. Graphs should have titles. 3. Include text to describe your data verbally. Refer to your well labeled tables, graphs, etc. vi. Discussion 1. Explain your results and how they relate to the scientific principle explored by this experiment. 2. Explain whether the results match your hypothesis. 3. Summarize the main points of the experiment. 4. Explain the importance of the experiment and/or technique and possible current or future applications of this knowledge. 11  5. Propose a few questions that arise from the study or propose further experiments that would be interesting to do. vii. Reference 1. Bibliography and in-text citations must be in APA format b. Writing Proficiency and Format i. No plagiarism allowed. ii. Using past tense (not future because the experiment was done already) iii. Using passive voice instead of active voice. iv. Direct Quotes: Only used in science if quoting something in historical context v. Avoid “qualifying” words such as very, higher, about, etc. Use precise language (use exact numbers). IX. Size Exclusion Chromatography a. Introduction i. In size exclusion chromatography, also called gel filtration chromatography, molecules are separated by their relative size and molecular weight. Molecules move through the resin based on their Stokes radii. b. Stokes Radius i. This is the effective radius of a molecule moving through the medium based on a perfect sphere (or the effective radius of a molecule as it tumbles through a column). ii. If you have 2 proteins of the same molecular weight but one is a sphere and one is a rod, the rod will work its way through the column much more quickly because its Stokes radius is much larger. c. Size exclusion (gel filtration) column i. The resin contains porous beads. As the buffer flows through the column, the smaller molecules enter the beads while the larger ones are excluded. This causes the smaller molecules to mover more slowly through the column than the larger molecules. ii. The size of the pores is dependent on the degree of crosslinking or concentration of the agarose or polyacrylamide (composed the resin). The larger the pore size, the greater the risk of the resin collapsing under pressure. 12 d. The Kav equation i. The partition coefficient (Kav) is a measure of the degree of interaction of a molecule with a column resin. (V e−V o) (V e−V o) ii. (V t−V o) = Vi 1. Vo = void volume = the volume outside the beads a. Can be determined by eluting a molecule too large to interact at all with the resin from the column. b. The volume of buffer that flows through the column before the colored tracking molecule begins to elute is the void volume. c. Blue dextran is usually used. 2. Vi = space within the beads [Vi = (Vt-Vo)] 3. Ve = elution volume for the sample 4. Vt = the bed volume a. Can be determined using a very small tracking molecule. b. Potassium chromate is often used. It is very small and easily flows into the pores in the beads. c. The volume at which the potassium chromate is eluted represents the volume inside and outside the beads and is equivalent to Vt. 13  e. Plot of Kav i. A plot of Kav versus log of the molecular weight of proteins. The molecular weight of a protein can be estimated based on the Kav using a chart of known standards ii. The Kav values are plotted on the y-axis and the log of the molecular weights are plotted on the x-axis. X. Kinetics of a Turnip Peroxidase a. Introduction i. Enzymes exhibit substrate specificity. They will not catalyze all reactions. They will only catalyze ones where the substrate can bind to the enzyme active site. Enzymes (names are typically based on the reactions they catalyze and usually end in –ase), like chemical catalysts, are not changed by the reaction and return to their original state once product is formed. 1. Enzyme + substrate → Enzyme-substrate complex → Enzyme + Product ii. The turnip peroxidase catalyzes a reaction in which hydrogen peroxide (H2O2) is converted to water 1. DH2 + H2O2  D + 2 H2O by peroxidase a. D stands for a donor which provides the two extra hydrogen atoms needed to produce two molecules of water in the presence of peroxidase. b. Peroxidase “neutralizes” the hydrogen peroxide to prevent damage to the cells. c. The substrate is hydrogen peroxide iii. Thermodynamics is the study of whether a reaction will occur. It answers the question of whether the reaction is energetically favorable. 14  1. If the reaction can occur, but occurs at a very slow rate, an enzyme catalyst can be added to speed up the process. 2. Enzymes help bring together molecules into the same small space which greatly enhances the chances of the reaction occurring iv. Kinetics provides information on the reaction mechanism and the reaction pathway. Kinetic data is immensely useful in determining the mechanisms of naturally occurring biochemical processes. 1. By changing the conditions of the reaction, the rate of the reaction changes. 2. Parameters such as enzyme concentration, substrate concentration, temperature, pH, and the presence of inhibitors change the kinetics of the reaction. v. In order to actually determine the rate of a reaction, we need an assay that allows us to follow the reaction. We call this an activity assay. Colorimetric assays allow the researcher to follow the reaction based on a change in color that can be recorded using spectrophotometry readings b. The effect of pH on Enzyme Activity i. The pH of the environment influences the rate of an enzyme catalyzed reaction ii. Each ionizable side groups of an amino acid has a pK value, the point where half the ionizable side chains available are protonated and half are unprotonated. At too high or too low a pH, the enzyme may denature and thus be inactive. c. The Michaelis-Menten Equation i. The interaction between an enzyme (E) and its’ substrate (S) results in the formation of an intermediate enzyme-substrate complex (ES) which ultimately yields product (P) and regenerates the enzyme (E) ii. The Michaelis-Menten equation describes the rate of the reaction as a function of substrate concentration. The Vmax is the maximum velocity of a reaction where the substrate concentration is saturating. The Km is the substrate concentration when the velocity of the reaction is 50% of the Vmax and can be found by plotting the reaction velocity at several concentrations. d. Lineweaver-Burk plot 15  i. This linearizes the data and can be used to construct a double-reciprocal plot. 1. Slope = KM/Vmax 2. Y-intercept is equal to 1/Vmax 3. The extrapolated x intercept is equal to -1/KM 4. Alternately, the KM can also be determined by calculating the slope of the line. e. Inhibitors of Enzyme Activity i. A competitive inhibitor directly competes with the substrate for the active site of the enzyme. The enzyme can bind the inhibitor, but it cannot process it the same way it can with the substrate, so the enzyme’s activity is reduced because some of the enzyme binds the substrate while some binds the inhibitor ii. An uncompetitive inhibitor binds to the enzyme-substrate complex (ES) but not to the free enzyme. In this case the binding of the inhibitor interferes with the catalytic function of the enzyme, most likely by altering the shape of the protein and active site, but does not actually bind the active site. iii. In mixed inhibition, the inhibitor binds both the enzyme and the enzyme- substrate (ES) complex 16  iv. A special case of mixed inhibition is noncompetitive inhibition. In noncompetitive inhibition, the inhibitor binds the enzyme and the enzyme- substrate (ES) complex with equal affinity. In other words, the binding of the inhibitor to the enzyme does not change the enzyme’s affinity for the substrate and the binding of the substrate does not change the binding affinity of the inhibitor. v. Hydroxylamine (an inhibitor) binds to the active site in heme-containing enzymes. vi. Wavelength in the spec is set to 470 nm. f. Beer’s Law: A = Elc  c = A/El 1. Where A = absorbance 2. E = absorption coefficient 3. c = concentration of solution 4. l = cell path length XI. Chloroplasts and the Hill Reaction a. Introduction i. Photosynthesis: light energy is converted into chemical energy. Two processes involve in photosynthesis: one reaction depends on light and one reaction occur in the absence of light. 1. nH2O + nCO2  (CH2O)n + nO2 + H2O 2. Light reaction: a. Occur in the thylakoids of the chloroplast b. Produce ATP and NADPH 3. Dark reaction: a. Occur in the chloroplast stroma b. Synthesis of sugar 17  ii. Hill Reaction: H2O is the electron donor and A is the electron acceptor. Water is the only electron donor required. 1. H2O + Aox  AH2red + O2 2. The role of NADP as a Hill Reagent is summarized as: Water serves as the source of electrons. NADP+ serves as the final electron acceptor which is reduced to NADPH. a. H2O + NADP  NADPH + H+ + O2 iii. The Photosystems operate sequentially 1. Photosystem I: A system of photoreactions that absorbs maximally far-red light (>680 nm), oxidizes plasto-cyanin, and reduces ferredoxin. 2. Photosystem I: A system of photoreactions that absorbs maximally red light (680 nm), oxidizes water and reduces plastoquinone. Operates very poorly under far-red light iv. Isolated chloroplasts could form ATP from ADP and Pi in the presence of light, a process known as photosynthetic phosphorylation. v. A photoinduced electron transport system which results in the production of NADPH and ATP. These two provide the energy for the “dark reactions” in which CO2 and H2O are converted to carbohydrates. b. The experiment i. Functions of DCPIP and DCMU 1. DCPIP: was used as the terminal electron acceptor, replacing NADP which was not available due to the extraction of the chloroplasts from the plant cellular environment 2. DCMU: It functions to block both electron transport and phosphorylation by interrupting electron flow at the beginning of the major electron transport chain. An herbicide that blocks electron flow to plastoquinone. 18