Bio II Lab Test 2 Labs PDF

BZE Autumn 2024 Lab 5 Alcohol Fermentation & Aerobic Respiration OVERVIEW (There are Prelab questions on last page) Cells obtain energy from food either by cellular respiration or by fermentation. These are two fundamental processes of life. These processes involve energy release by the breaking of...

BZE Autumn 2024 Lab 5 Alcohol Fermentation & Aerobic Respiration OVERVIEW (There are Prelab questions on last page) Cells obtain energy from food either by cellular respiration or by fermentation. These are two fundamental processes of life. These processes involve energy release by the breaking of chemical bonds in food molecules sugars, fats, proteins & the trapping of some of this energy under the form of ATP (some energy is invariably released as heat). Aerobic respiration requires oxygen gas, whereas fermentation does not. The respiration of sugar may be represented by the following chemical equation: C6H12O6 + 6 O2 ──────> 6 CO2 + 6 H2O + ENERGY (36 or 38 ATP) + Heat The fermentation of food substances takes place by various pathways, depending upon the organism. The most common fermentation pathways may be represented as follows: C6H12O6 ──────> 2 C2H5OH + 2 CO2 + ENERGY (2 ATP) + Heat (occurs for e.g. in yeast cells) Hexose ethanol C6H12O6 ──────> 2 C3H6O3 + ENERGY (2 ATP) + Heat (occurs for e.g. in human muscle cells) Hexose lactic acid OBJECTIVES o Understand the basic differences between the processes of cellular respiration & fermentation, including their overall rates; o Explain similarities or differences between common carbohydrates as fermentation substrates in yeasts; o Discuss the relevance of the R.Q. ratio (respiratory quotient) in terms of types of substrates being used (carbohydrates or fats) in cellular respiration; o Utilize a computer equipped with appropriate sensors, lab interface & software to collect & analyze gas pressure data obtained during fermentation in living organisms. INTRODUCTION For the yeast fermentation rates, 6 stations with laptops will be set-up for this experiment, you will use yeast (a unicellular fungus) as the experimental organism; to observe alcohol fermentation & measure the increase in pressure due to CO2 given off. Glucose, along with 3 disaccharides, sucrose, maltose, and lactose, will be tested as “food” sources to see whether yeast can ferment all equally well. A control with yeast extract and water will also be done. Yeast can metabolize common monosaccharides like glucose or fructose immediately. Cane sugar (sucrose) is a disaccharide. Most species of yeast have the enzyme sucrase. Sucrase catalyzes the breakdown of sucrose into the two monosaccharides glucose & fructose. Yeast can also hydrolyze maltose into its glucose monomers, though the metabolism of maltose by yeast is a slow process. In the absence of glucose, it’s hypothesized that the few molecules of maltose that enter the cell do so by non-specific transport mechanisms. However, once they do enter the cell, their presence stimulates the synthesis – through transcription & translation – of the enzymes maltose permease & maltase. This then allows the more rapid entry of maltose into the cell. The lag period for the appearance of these enzymes is yeast strain-dependent. In the end, maltose is hydrolyzed into two glucose molecules, which are then fermented into ethanol. In this experiment we will carry two runs with maltose, one immediately after adding maltose, the other allowing a 15 minute lag for the synthesis of transport proteins and enzymes Lactose is also a disaccharide, composed of glucose & galactose. It most often found in milk products (it has however a beta glycosidic linkage between the two monosaccharides). Yeast does not normally produce enzymes to metabolize a sugar that is not available in its environment. The yeast used in this experiment has been raised on sucrose & glucose. We predict, therefore, that the rate at which lactose is metabolized will be slower than that of the other two sugars (if it is metabolized at all). In all these trials, you will measure a change in gas volume. In this set of experiments, you will use a gas pressure sensor to measure rates of fermentation; this involves the use of a pressure tympanum (membrane) whose output is digitised for computer display – read below. For the respiration experiment, germinating pea seedlings can be used as the experimental organism, & oxygen uptake is measured. Since carbohydrate (starch) is the main food supply in these seedlings, we assume from the equation on page 1 that the carbon dioxide output will be the same as the oxygen uptake. The Gas Pressure Sensor The Gas Pressure Sensor is a probe consisting of a small tympanum which creates an electrical signal when it is distorted. This signal is digitized by an interface box & sent to the computer where, after referring to a calibration table, it is displayed in appropriate pressure units graphically & in a table form. The Gas Pressure Sensor has a sensitive tympanum which flexes as absolute pressure changes. The sensor produces an output voltage which varies with deformation of this membrane, hence measuring pressure due to the accumulation of gases (or the creation of a vacuum). The pressure range of the probe is  C6H12O6 ──────> 2 C2H5OH + 2 CO2 + ENERGY (Ethanol) Set-up: Add 10 ml of carbohydrate (or water, for controls) & collect the data for 5 minutes. Notice that the white stopper that fits the neck of the Erlenmeyer flask has two openings. The valve, when open, allows pressure to equalize between the environment & the air space. Leave it open & fit the stopper snuggly to the Erlenmeyer flask. Once it is firmly in place, do not press on it again so as not to damage the sensor’s membrane. Connect the tube to the sensor (already done for you), & close the valve. 2. Collect Data.  Be sure that the Leuer-locks on the sensor are closed so that air building up in the Erlenmeyer flask can build up pressure on the sensor’s tympanum. Press on the Collect button & watch the data being collected in the Graph & Table windows. You must gently swirl the solution during the 300 seconds you collect data. When the 300 seconds are finished, the “Collect” button should reappear. Pressing on the “Stop” button will abort the experiment at any other time. From the Experiment menu select Store latest Run. This stores data so each run can be displayed as data is collected for the next run. Label graph using Text Annotation from the Insert menu. Each tray (with materials for 4 students) will have a working laptop already connected to the Pressure Sensor next to it. There will be 6 trays in total in the lab room. In groups of four, run the fermentation assays in the following order: A. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml of the water solution to the Erlenmeyer flask. Store & analyze the results (control). This will provide a baseline for the other runs. B. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml of the glucose solution to the Erlenmeyer flask. Store & analyze the results (control). C. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml of the maltose solution to the Erlenmeyer flask. Store & analyze the results (control). D. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml of the maltose solution to the Erlenmeyer flask. Wait 15 minutes before gathering data. Store & analyze the results (maltose 15) E. Repeat control A. (water) F. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml of the sucrose solution to the Erlenmeyer flask. Store & analyze the results (control). G. Rinse the 125 ml Erlenmeyer flask & place 25 ml of stirred yeast (0.05 g/ml) solution in it. Add 10 ml  of the lactose solution to the Erlenmeyer flask. Store & analyze the results (control). Collect all your data by printing (or with screen shots), and by writing the rates down in the report section also. Once completed, proceed to do the respiration measurements (with pea seedlings). See page below. RATE OF AEROBIC RESPIRATION IN PEA SEEDLINGS PROCEDURE: 1. 2. An experimental test tube is filled it with germinating pea seedlings, leaving the top of the tube empty (about 20 g peas). 3. A thin layer of cotton is put over the seedlings & a layer of about one inch of soda lime over the cotton. This absorbs any CO2 gas produced by the seedlings during respiration & allows the measurement of only oxygen uptake. 4. Note that since starch is likely the main respiratory substrate in peas, the volume of O2 taken up should be the same as the volume of CO2 released. 5. The test tube is wrapped with aluminium foil. The aluminium foil is used to prevent chlorophyll from carrying out photosynthesis, which could interfere with measurements of respiration. 6. The tube is connected to the sensor, & the valve is closed. 7. The data is measured for 900s (or 15 minutes) & collected in the Graph/Table windows.  41.5/44 SamanthaSchwartz and Mia Zhang 10 A Lactic Acid Fermentation Alcohol Fermentation  1 25 0.005480 0.26304 0.008668 0.41606 158 0.003867 0.18562 70.6 0.005346 0.25661 97.6 0.004163 0.19982 76.0 0.003171 0.15221 58.0 sucrose Glucose Maltose 15 mins lactose Maltose Watefesstfegmentation 5th fermentation of sucrose into glucose if.tkntiiieriEhataieiiieafmpea YEtes59ohade.bg Fes ym.ee tosbetiifo'tiesdei as the yeast has notbeenexposed to lactose and is not -0.5 I'd metfsetshg.lifdtosselower It's still at 76% fermentation rate though. which is higher than maltose (immediate). How would you account for this relatively high rate? Lactic Acid Fermentation ion are t.is itn isieresi enine is ai 2 pyruvate 2 Lactate -0.5  Pyruvate is reduced NADT to lactic acid by NADH. NADT Ee ifsinitiates tatiigsya.ie ngtsb iaainTmtaic fanfeastation Yeast fermentation is exothermic soheat is producedand the temperature rises sucrose 191 lactose Water Maltose Glucose -1  It would be 1 because there is a 1 1 ratio of 02 to CO2 in thebalancedreaction ATP Heat bffetractfeeddemeansressetestohteracedould be equalto the orconsumedand no change inpressurewould why? There are more C-C and C-H bonds in fatty acids to break than glucose. This generates more electron carriers for the ETC which requires more oxygen. 1/2 -1  22/22 Samantha Schwartz and Mia Zhang True net gain of 2 ATP 2 invested 4gained False it consumes NADt which is reduced to formNADH False glycolysis can take place in aerobic or anaerobiccondions False all organisms undergoglycolysis beforefermentation or beforepyruvateoxidation False microorganisms without membranebound organelles raise.int ie ii nisiiiiniiiiiit substratelevel phosphorylation Falseethanolfermentation doesnot produceadditionalATPmolecules beyond thosegenerated in glycolysis It regenerates NAD from NADH to keep glycolysis runningbutdoesnot produceATP wantto sustain glycolysis this is necessary esasttofffr.to Truebacteriaandyeastpossess thenecessary enzymes tocarryout the fermentationprocess Falseethanolfermentation is an anaerobicprocess it can occur in the absence of oxygen Truethisreduction is achievedby transferring electrons from NADH to acetaldehyde forming ethanol  -10% late 54.5 - 5.5 (late) = 49/55 Stomatal Pore Guard cells Stoma Sponay mesophull Palisade mesophyll Xylem Pholem Vascular bundle Lower epidermis Upper epidermis Cuticle  active transport H + ions > - primary proton pump + H out = basic in so k" added + voltage k ions + K ions facilitated diffusion > secondary - Ho iOns sucrose increases water water turgid Blue light acts as an indicator of daylight , when photosynthesis activity is highest This blue. light triggers receptors in Guard cells to increase in transport and water influx This causes the. guard cells to swell and open the stumata to optimize (O2 uptake Light affects stomatal.. movement Chloroplasts are essential to light detection for Stomata regulation. Chloroplacts also produce ATP used in stomatal movement (open/close). The light detection also help detect changes in light intensity , leading to the adjustment of the stomatal opening , helping to balance CO2 intake. Mesophute = used as reference (normal There is no need to prevent Absence of a cuticle water loss since they live lor very thin layer in water Cow/no-risk. desiccation. The space allows for better It is left with a lot more buoyancy and facilitate gas space between (intercellular exchanges letting it float space) , and access light. Since water is below , there is no need for the stoma at The stomata are at the top the bottom when it can make gas exchanges directly with the air above. prevent excessive influx of water. Larger leaves mean larger areas to capture more sunlight especially , when light penetration is limited. This also allows for more efficient floating to keep the leaf on the water's surface for better light exposure and gas exchanges. Since they are supported by water , there is no need for long thick roots , only small feathery for Oxygen absorption and roots. They use these roots better buoyancy , anchorage. The lesser weight helps keep things afloat and they don't need roots (sometimes) to absory water.  L stomatal hair , epidermis hair It prevents water loss thanks to the barrier Thicker cuticle. helping to retain water. The many layers provide extra protection against Multiple layers of epidermis water and reflect sunlight to reduce heat absorption and evaporation. It reduces the Stomata's exposure Stomata in a sunken -0.5 to air/light and creates a micro-environment by trapping humic Dit moisture. It reduces water loss only on bottom epidermis and evaporation when open. The widespread reach of the roots allow for maximum water absorption through a larger area , even during extreme ory climates. The high intracellular osmolarity allows more efficient water absorption from the soil , even when soil moisture levels are low by maintaining the strong osmotic gradient. The leaf should be small and thick to reduce surface area exposed to the sun to limit water loss through evaporation. It also allows better conservation of water , minimizing transpiration. -0.5  carotene Xanthophyll aldenude more hydrophilic/polar cause carbonyl group ~ has methyl group ( carot = orange middle isn - yellow Xanthophyl  yellowish-green moderate 0 10. Chlorophyll B blueish-green nigh 0 23. Chlorophull A very pale yellow very low 0 43. Xanthophull very pale yellow very low 0 60. Xanthophyll yellowish-orange low 1. 0 Carotene than carotene * Xanthophyll more polar = 2 extra OH groups Since the solvent used is nonpolar and going by the concept of likes attracts like , the faster rate or higher Re is attributed to a nonpolar molecule as it has high solubility in a nonpolar solvent. The highest If traveled the farthest is for carotene, the least polar molecule (no polar groups) that also doesn't interact with the stationary polar molecules so it moves easily through the nonpolar solvent. Xanthophyll has an intermediate R I and travelled farthest after carotene, ove to slight polarity from higher hydroxul groups. The Chlorophyll A travelled slightly farther and had slightly Rf , because of its lower polarity since a CHO group (Chlorophyll B) is more polar than a A methyl Group (Chlorophyll A) but has more groups with O than Xanthophull Chlorophyll ,. and B are polar (don't like nonpolar) and Xanthophyll has 20 groups making it more polar than carotene who only has methyl groups (nonpolar). 0 31. 0 142. 0 071. 0. 398 * 0 166. 0 082. A 0 123. 0. 208 0 082 A. 0. 049 0 223. A 0 006. 0 037. 0 059. 0 036. 0. 043 0 043. 0. 024 0 045. 0 04/. 0 02/. 0 059. 0 046. 0. 019 0 071. 0 053. 0. 018 0. 088 0 04 8. 0. 017 0 163. 0. 11 A 0 016. 0 18. A 0 051. 0 015. 0. 028 0 019. 0 014. 425/675 Violet/red green green and 475/650 blue/red yellow orange and 425 - 450 Violet-blue reg It allows the plants to "last longer" and conserve more nutrients for winter , thanks to the boarder range of light they each can absorb. Chlorophyll, absorb blue and red light during warmer seasons , until they get denatured at lower temperatures , when carotenoids take over to absorb blue and green light to prolong photocunthesis and to store more nutrients. carotenoids are also able to absorb and dissipate excess light to prevent damage to the Chlorophyll pigments. This also prevents the formation of free radicals which could harm the cell. LABORATORY VII Bacterial Transformation Expression of GFP in Bacteria OVERVIEW This lab exercise will be carried out over 2 lab periods. The goal of this experiment is to add the gene for green fluorescent protein (GFP) from a bioluminescent jellyfish species to the genome of a bacterial species using bacterial transformation and the pGLO system. We will also directly view expression of this gene by observing a new trait: green fluorescence when the bacterial cells are exposed to UV light. The lab exercise will also introduce students to the use of restriction enzymes as tools in recombinant DNA technology such as the building of the pGLO plasmid. LEARNING OBJECTIVES 1. Describe transgenesis and briefly comment on its use in biological research, biotechnology, agriculture, and medicine using examples. 2. Describe the procedure used in a CaCl2 bacterial transformation and explain the function of each step. 3. Describe the general features of a plasmid vector and the specific elements of pGLO. 4. List the control treatments used and explain the purpose of each of these trials. 5. Analyze the results (actual or hypothetical) of a pGLO transformation experiment and state your conclusions. 6. Describe a restriction enzyme (endonuclease) and describe their role in the organisms that produce them 7. Describe how restriction enzymes can be used in recombinant DNA technology. 8. Use a restriction digest map of a plasmid to calculate the size of the fragments generated. 9. Describe electrophoresis and how it can be used as a tool in research. Identify DNA fragments by size in agarose gel electrophoresis using a DNA ladder. INTRODUCTION A bacterial transformation is a type of transgenesis technique. In transgenesis, the goal is to add a foreign gene to an organism, express the gene product, and alter a phenotype (trait) in that organism. Once added to its genome, the organism will also be able to pass on this gene to its offspring. Transgenesis creates transgenic organisms and is considered a component of genetic engineering. The gene to be transferred is isolated from the genome of an organism, cloned (amplified), and modified before it is inserted into the genome of another organism. A bacterial transformation is a mainstream but rather simple technique that allows a scientist to add a gene to a bacterial cell population. Bacteria are excellent candidates for transgenesis because they are unicellular, easily cultured and manipulated in a lab, reproduce asexually, and divide rapidly. However, bacteria cannot be used for all applications of transgenesis. For many applications, transgenesis in multicellular organisms (e.g., animals, plants, fungi) is necessary. This feat is regularly conducted but requires more complicated techniques and also special considerations. General Biology II Page 1 Laboratory Exercises Fall 2024  Lab 7 – Bacterial Transformation Transgenesis and transgenic organisms are essential in biological research, biotechnology, agriculture, and medicine. Table 1 lists some examples of uses of transgenesis in these areas. Table 1. Uses of Transgenesis Area: Goal: Example: Biological Research Observe and explain nature. Study the expression, regulation, and function of genes. Biotechnology Use organisms to carry out tasks. Design organisms to produce important protein products or carry out a bioremediation application. Agriculture Cultivation of productive and Increase the resistance of plants to pathogens healthy plants. and pests. Medicine To prevent and treat illness. Deliver of gene to replace nonfunctional alleles that cause disease (gene therapy). Some but limited success. Bacterial Transformation Although bacteria reproduce asexually, these organisms often transfer DNA from one cell to another within a population (and sometimes between species) by a process called horizontal gene transfer (HGT). Along with mutation, HGT is a driving force behind the genetic variability and evolution of bacterial populations. HGT also has implications in medicine by producing superbugs, which are bacteria that are resistant to several commonly used antibiotics. Transformation happens naturally in bacterial populations and is one (of three) mechanism for HGT. Bacteria will take up fragments of DNA in their environment (exogenous DNA) and incorporate the genes into its genome. These bacteria are said to be naturally competent. In 1928, Frederick Griffith was the first to document bacterial transformation in a lab. He transformed a non-pathogenic strain of Streptococcus pneumoniae into a pathogenic strain by incubating the non- pathogenic strain with a dead preparation of broken pathogenic cells that contained the various components of the broken cells. What happened was that the non-pathogenic strain incorporated the gene for a virulence factor (a factor that aids in pathogenesis) from the pathogenic strain. Specifically, expression of the gene that added a capsule to the cell structure and helped the bacteria evade the immune system and cause disease. Although Griffith was unaware of what was happening at a molecular level, he set the stage for the search for his transforming principle and the discovery of DNA as the genetic material in all organisms. In this lab, you will produce transgenic bacteria that express a foreign gene and gives rise to a new phenotype. We will 1. use a CaCl2 transformation to make an E. coli population artificially competent, 2. add the gene for Green Fluorescent Protein (GFP) using the pGLO system. General Biology II Page 2 Laboratory Exercises Winter 2024  Nfarethatoteins inserted gentes Lab 7 – Bacterial Transformation 3 The pGLO System Types of Sequences In addition to chromosomal DNA, bacteria (also Transcription Unit of Gene archaea and yeast) contain small circular DNA Gene Promoter Origin of DNA molecules called plasmids. These Replication Other Sequence extrachromosomal DNA molecules contain one or Ei more genes for traits that are beneficial to cell f it ParaC survival in new environments (e.g. resource use, antibiotic resistance). These plasmids can be r nsie ParaBAD transferred by HGT to other bacterial cells, which will spread these beneficial traits and adapt a iiiiii population to a new environment. EE.in Modified plasmids are also very important tools in Pbla genetic engineering. They act as vectors Figure 1. Map of pGLO. Each r component is (vehicles) for the amplification and transfer of described in Table 2. Arrows indicate the direction of genes or other DNA sequences to bacteria and transcription of the genes. other organisms. Using recombinant DNA technology, a DNA fragment, the insert, can be added into a defined spot in the plasmid making a recombinant plasmid. pGLO is a recombinant plasmid that can be used to transfer the GFP gene and express the GFP protein in bacteria. The major features of the pGLO plasmid are illustrated as a plasmid map in Figure 1 and described in Table 2. Table 2. Components of the pGLO Plasmid. Plasmid Type of DNA Label Function in System Component Sequence* Recruits DNA replication machinery and directs ori Bacterial Origin of A replication of the plasmid before cell division of a DNA Replication bacterial cell by binary fission. This allows the plasmid to be passed on to daughter cells. GFP Gene Encodes the amino acid sequence of the GFP GFP (Gene of Interest) C protein. Expression produces the GFP protein. Encodes the amino acid sequence of the beta- Beta-Lactamase lactamase enzyme that confers resistance to Gene certain antibiotics. Expression of the enzyme acts bla (Selectable C as a selectable marker. A selectable marker allows Marker Gene) you to select for cells that took in the plasmid and are expressing its genes. Encodes the amino acid sequence of the Ara C Ara C Protein protein that acts as a gene expression regulatory Gene protein. The protein can repress expression from araC C (Regulatory the ParaBAD promoter in the absence of the sugar Protein Gene) arabinose but activate expression from the same promoter when arabinose is present. Recruits RNA polymerase and directs transcription Bacterial Promoter of a linked transcription unit in bacteria. ParaBAD is a P (Regulatory Element) B regulated promoter and can be controlled by the AraC protein. ParaC and Pbla are constitutive promoters in bacteria (always on). * Complete with either: A. Regulatory DNA Sequence for DNA Replication; B. Regulatory DNA Sequence for Gene Expression or C. Transcription Unit of Gene General Biology II Page 3 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation This constructed plasmid contains a bacterial DNA replication origin (ori) and a selectable marker gene (bla) that encodes the beta-lactamase enzyme. Beta-lactamase chemically modifies beta- lactam antibiotics (contains a beta-lactam ring structure). After the transformation, the cells are cultured on agar plates containing the beta-lactam antibiotic, ampicillin. This gene will allow us to select for the bacterial cells that take up the pGLO plasmid because only these cells will be able to survive on this medium. ParaBAD Figure 2. Regulation of GFP GFP Expression from the pGLO plasmid. + AraC + AraC When arabinose is absent, AraC (Regulatory Protein) (Regulatory Protein) represses transcription from Para and + Arabinose GFP is not expressed. When arabinose (Sugar) is present, AraC activates transcription from Para and GFP is expressed. No Transcription AraC RNA Pol GFP AraC AraC GFP ParaBAD ParaBAD Transcription mRNA Translation Polypeptide GFP Protein Transcription of the GFP gene is initiated from the bacterial promoter: ParaBAD. This plasmid was designed for regulated expression of GFP from this promoter. The plasmid contains the regulatory gene araC, which produces the AraC regulatory protein. In the absence of arabinose, this protein binds to DNA control elements and represses transcription of mRNA from the ParaBAD promoter and, therefore, GFP expression. This negative regulation can be eliminated in the presence of the sugar, arabinose. Arabinose binds to the AraC protein and changes its conformation. The AraC protein in this conformation can bind to DNA control elements and activate transcription of mRNA from the ParaBAD promoter by RNA polymerase. The GFP mRNA can be translated into GFP protein. This regulation scheme is outlined in Figure 2. The ParaBAD promoter and araC gene are naturally present in bacterial genomes as part of the arabinose operon. Operons are gene regulation systems found in bacteria that regulate a cluster of genes and coordinate their expression. The genes in an operon encode proteins that have various functions in the same process. For example, the arabinose operon contains the genes that produce the enzymes needed by bacteria to process arabinose as a food source. The genes of this operon are only expressed when arabinose is present and gene expression is silenced when arabinose is not in the environment. Operons and other gene expression regulation systems allow cells to respond to changing environments and to prevent wasteful overproduction of unneeded proteins. Elements of these gene regulation systems can be used as tools for regulated gene expression during bacterial transformation. GFP and Other Fluorescent Proteins Are Important Research Tools GFP is a protein from the bioluminescent jellyfish, Aequorea victoria first discovered in 1962. GFP fluoresces bright green when exposed to ultraviolet (UV) light. It has become an indispensable tool to view biology in action. Figure 3. Spectrum of Florescent Proteins. Osamu Shimomura, Martin Chalfie, and Roger These fluorescent proteins are important tools in General Biology II biological research. Page 4 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation Tsien shared the 2008 Nobel Prize in Chemistry for the discovery and development of GFP. Several variants have been engineered by mutation over the years to shine more strongly and also produce different colours, such as cyan, blue, and yellow. Since then, other fluorescent proteins from nature (e.g., DsRED, a red florescent protein from corals) have been discovered and developed giving biologist an array of colored proteins to work with (see Figure 3). Restriction Enzymes Restriction enzymes are important tools to analyze DNA and create recombinant DNA for genetic engineering. These enzymes are endonucleases found in bacteria as part of a primitive protective system called restriction modification that defends these organisms from infection by viruses called bacteriophages (Latin/Greek roots for bacteria eaters). Currently, several hundred restriction enzymes from various bacterial species have been identified and are available commercially to molecular biologists. Restriction enzymes can recognize specific nucleotide sequences in DNA called restriction sites and cut the sugar-phosphate backbone at two defined locations within this site, one cut in each strand. Restriction sites range from 4-8 base pairs in length, but many are 6 base pairs in length. In addition, these sequences are typically palindromes, sequence of bases within the site on each strand are identical when read in the same direction (e.g., 5’ to 3’). Table 3 lists some examples of commonly used restriction enzymes and their activities. Restriction enzymes are named according to the bacterial species and strain in which it was discovered and the order of its discovery in this strain. The first three italicized letters reflect the genus and species, the next letter reflects the strain, and the number reflects the order. For example, EcoRI was the first enzyme isolated from strain R of the species E. coli. Table 3. Common Restriction Enzymes Used as Tools in Molecular Biology Enzyme Restriction Site Ends Generated iiiiii 5’-GAATTC-3’ Sticky EcoRI 3’-CTTAAG-5’ 5’ Overhang mm 5’-GATATC-3’ Blunt EcoRV 3’-CTATAG-5’ it T.ie 5’-AAGCTT-3’ No Overhang Sticky Jena state HindIII 3’-TTCGAA-5’ 5’ Overhang I 5’-CTGCAG-3’ Sticky PstI 3’-GACGTC-5’ 3’ Overhang onceattach to plasmid andtry to find For si i i i h phosphate ERY.gg jatt General Biology II Page 5 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation Sticky Ends and Recombinant DNA The fragments generated by a restriction enzyme can be either sticky or blunt. Sticky 5¢ 3¢ GAATTC ends are created when the restriction enzyme DNA CTTAAG cuts each strand at different points within the 3¢ 5¢ restriction site. The ends created have 1 Restriction enzyme overhangs, a single stranded region. Some cuts sugar-phosphate enzymes cut both stands at the same point backbones. 3¢ 5¢ within the restriction site and generate blunt 5¢ AATTC 3¢ G ends with no overhangs. CTTAA G Ends with overhangs are considered “sticky” 5¢ 3¢ 3¢ 5¢ because the overhangs are complementary and can hybridize by base pairing. 5’ vs. 3’ 5¢ AATTC 3¢ overhang refers to whether the single- G G stranded region of a sticky end represents the CTTAA 2 DNA fragment added 3¢ 5’ end or 3’ end of the DNA molecule of that from another molecule 5¢ strand. As shown in Fig. 4, EcoRI generates cut by same enzyme. sticky ends with a 5’ overhang when it cuts its Complementary ends restrictions site. If a restriction enzyme hybridize. generates sticky ends, then any fragment 5¢ 3¢ 5¢ 3¢ 5¢ 3¢ created with this enzyme will have the same G AATT C G AATT C sticky ends. This property can be used to join C TTAA G C TTAA G DNA fragments together with the help of the 3¢ 5¢3¢ 5¢ 3¢ 5¢ enzyme ligase to create recombinant DNA 3 DNA ligase (Fig. 4). Recombinant plasmids, such as seals the backbone. pGLO, were made using restriction enzymes 5¢ 3¢ as tools to piece together the elements of the system. 3¢ 5¢ Recombinant DNA molecule Figure 4. Using Sticky Ends to Create Recombinant DNA. Methylation as another protective mechanism Bacterial restriction endonucleases target viral DNA, not the bacterial cells' own DNA, because bacteria modify their own DNA, using enzymes known as methylases that add methyl (CH3) groups to some Figure 5. Example of methylation of the of the nucleotides in the bacterial DNA (Fig. 5) nitrogenous base of a nucleotide. When nucleotides within a restriction endonuclease's recognition sequence have been methylated, the endonuclease cannot bind to that sequence. Consequently, the bacterial DNA is protected from being degraded at that site (Fig. 6). Viral DNA, on the other hand, has not been methylated and therefore is not protected from enzymatic cleavage. Figure 6. Methylated restriction sites are protected from cleavage. General Biology II Page 6 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation Gel Electrophoresis Gel electrophoresis can be used to separate the components in 1 Load samples of a mixture of biopolymers such as proteins and nucleic acids biopolymers into wells at the Samples (Fig. 7). The molecules are separated based on their charge, top of the gel. Electrodes size, and shape. The gel is placed in an electrophoretic Well chamber and immersed in an appropriate buffer solution. The mixture is loaded into a well (depression) of the gel and an electric field is applied. The gel forms microscopic pores and Gel acts as a molecular filter. The migration of a molecule in electrophoresis refers to the direction and rate at which the molecule moves through a gel in an electric field. The charge of 2 Apply an the molecule determines the direction of its migration (towards electric field. – which electrode). The size and shape of the molecule will determine the rate of migration (how fast it migrates) through the gel in an electric field. The specific physical properties of a molecule will result in a distinct migration pattern during + electrophoresis. These characteristic migrations patterns allow us to use electrophoresis to separate and identify molecule types in a 3 Stop after a desired time, mixture of biopolymers. Although proteins can be separated by stain the biopolymers to visualize. Higher-mass molecules – charge, size, and shape, nucleic acids can generally only be separated by size since all nucleic acids have a negative Each band is a group of charge and a consistent linear, double-helix shape. biopolymers with the same physical characteristics: charge, Several conditions can vary in gel electrophoresis such as the mass, shape. + type of polymer used to make the gel or the concentration of polymer in the gel. For many applications, the gel is made from Lower-mass molecules agarose or polyacrylamide. In addition, protein samples can be run in native (non-denaturing) or denaturing conditions. Figure 7. Electrophoresis can be used to separate mixtures of biopolymers Nucleic acids and (most) proteins also require a staining based on charge, mass, and shape. technique to be visualized in the gel. Various specific techniques exist for nucleic acids and proteins that involve Lane: 1. 2. 3. reagents that give of a colored or fluorescent signal to identify separated proteins as bands in the gel. The gel and Direction of migration towards Wells its bands can be photographed resulting in an the positive electrode electrophoretogram (example for DNA electrophoresis shown in Fig. 8). Various bands can be visualized in the lanes of the gel image. The bands in Figure 8 represent many DNA molecules that are traveling (migrating) at the same rate since they have the same length/size (not Bands necessarily the same sequence!). As the DNA molecules in each well are pulled into the gel by the electric field, they follow a straight path towards the positive electrode and create a separation lane (think cars on a highway!). Figure 8. DNA agarose gel electrophoresis electrophoretogram. DNA molecules of similar lengths travel together and are visualized as bands using a fluorescent signal. Each lane contains the separation products of a given sample loaded into a well. General Biology II Page 7 Fikri t Effing Laboratory Exercises Winter 2024 war  etia resisten Eigron Lab 7 – Bacterial Transformation PART 1: pGLO Restriction Digest PROCEDURE Overview: Each group of 4 students will prepare one set of restriction enzyme digests of the pGLO plasmid (different restriction enzymes alone or in combination) and an “uncut” control. A. Restriction Enzyme Digest Table 4 describes the various trials to be prepared by the team. Table 4. Restriction Digest Trials of pGLO Trial 2 Buffer (10X 3 pGLO (L) 1 Distilled Total concentrated; [0.5 g/L] 4EcoRV (L) [0.5 U*/L] 5 HindIII (L) [0.5 U*/L] Water (L) Volume L) (L) ontrol 1: UC 2 2 0 0 16 20 2: E 2 2 2 0 14 20 3: H 2 2 0 2 14 20 4: E + H 2 2 2 2 12 20 * U=activity unit. 1 U is enough activity to digest 1 g of DNA in 1 hour in the appropriate buffer. Some questions to think about: 1. Why do we add 2 L of the buffer? 2. How much DNA was added to each sample? 3. Why do we add 2 L of the restriction enzyme? 1. Label 4 micro test tubes: E, H, E + H, and UC. 2. Locate the aliquots of reagents in the ice bucket. Each student in the team will be responsible for one reaction in the set. Put together the reagents (described for your trial in Table 4) in the following order: 1. Water, 2. Buffer 3. pGLO plasmid 4. Restriction Enzyme(s). Keep enzyme aliquots on ice in between steps to avoid reducing enzyme activity.Laststep flick spin pit 3. Add your group’s set of 4 micro test tubes to a floating rack. a) Each sample is incubated at 37 oC for 1 hour and then b) at 65oC for 20 minutes to stop the digestion. There are water baths set-up for this purpose.last down label w to teacher step spin namesymbol group give Some questions to think about: 1. Why do we incubate these reactions at 37oC? 2. Why does an incubation at 65oC stop the reaction permanently? the verities Iptimal forenzyme B. Loading and Running the Agarose Gel 2 teams of 4 will prepare and share one 1% agarose gel. 1. Prepare the casting tray for the gel. Your instructor will provide guidelines to do this. 2. Add the pre-weighed agarose powder (1 g) to the Erlenmeyer flask provided. YdÉ 3. Add 100 mL of TAE electrophoresis buffer (dispenser on side bench) to the flask using a graduated 2 cylinder and swirl gently to mix. 4. Your instructor will use a microwave to bring the sample to a boil to dissolve the agarose and make a solution. 5. Using a heat glove, bring the solution back to your bench and add the thermometer provided. 6. When the solution cools to 50-55oC, pour the gel into the casting tray. 7. After about 30 min., the gel will solidify. Prepare the gel. Your instructor will give you instructions on how to do this. General Biology II Page 8 Laboratory Exercises Winter 2024  Firstcentrifuge everything DNAladderGx dye all 4 id Lab 7 – Bacterial Transformation gpg 8. Add loading dye/DNA stain to each sample. This reagent is 6X concentrated. To figure out how much of this reagent to add to your sample to make it 1X concentration (the right concentration), Enttake the volume of the sample (20 L) and divide it by the concentration factor of the reagent (6X) - p add 1 = 5. How much of the loading dye/DNA stain so that it is at the right concentration (i.e., 1X) why in our samples? Vi 1 24m11 V added 2colourwe The loading dye/DNA stain is a dense, 6 solution red that has a) The dye molecules help you visualize the liquid during loading. many functions: In potnadge b) The density of the solution causes your sample to fall to the bottom of the well during loading. c) The dye molecules migrate during electrophoresis and allow you to monitor the process. d) The dye solution also contains a DNA stain. The stain molecule is an intercalating agent that bind to double-stranded DNA and fluoresce when exposed to UV light. The stain allows us to visualize and image the DNA in the gel after electrophoresis to produce an electrophoretogram. 9. Load 20 L into a well in the following order (left to right): Table 5. Gel Loading Scheme for Two Teams iii Well Team Sample 1 1 pGLO (uncut) 2 1 pGLO / EcoRV DNA 3 1 pGLO / HindIII 4 1 pGLO / EcoRV + HindIII 5 1 kb DNA Ladder 1 kb DNA Ladder 6 2 pGLO (uncut) 7 2 pGLO / EcoRV 8 2 pGLO / HindIII 9 2 pGLO / EcoRV + HindIII 10 1 kb DNA Ladder 1 kb DNA Ladder 10. Your instructor will load the DNA ladder samples. The DNA ladder is a set of standard DNA fragments of known sizes. It can be used as a tool (type of positive control) to estimate the size of an unknown fragment in an experimental sample. 11. Run the gel at 80 volts for approximately 60 minutes. 12. Wait until your instructor calls your team. You will use the gel documentation system to image the gel under UV. This will allow us to view the DNA “bands” and record an image (electrophoretogram). Your instructor will provide instructions on how to use this equipment. C. Record Data: Number of Restriction Fragments and Estimated Sizes 1. Observe the data for each lane of your electrophoretogram. For each lane: a) Count the number of distinct bands b) Use the results from standards in the DNA ladder sample to estimate the approximate size range of the fragments. 2. Record your data in Table 6 of the Results section General Biology II Page 9 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation PART 2: pGLO Transformation PROCEDURE Overview: You will be working in teams of 4 students to complete the inoculation of 4 experimental agar plates in this lab. Each sub-team of 2 lab partners should work across from each other at the same bench station. N.B. Please distribute the work, so that all team members participate in the procedure! Safety and Sterile Technique With any type of microbiology procedure (such as working with and culturing bacteria), two important considerations are 1. your safety and 2. avoiding contamination. In order to avoid unwanted exposure to the bacterial culture, a lab coat, safety glasses, and gloves will be necessary at all times during this laboratory exercise. Being prepared and focused during the lab period will minimize the chance that you will be exposed to the bacteria we will be using. If you do make contact with the bacteria, alert your teacher and calmly make your way to the nearest sink to clean the exposed area. Any spills in your bench area that contain bacteria must be cleaned immediately and disinfected. All materials that come into contact with the bacterial culture (material used in the experiment or material used in a clean-up) must be disposed of in the appropriate container (i.e., biohazard bag). It is also important not to introduce contaminating bacteria into the experiment by using sterile (aseptic) technique. Because contaminating bacteria are ubiquitous and are found on fingertips, benchtops, etc., it is important not to contacting these contaminating surfaces with sterile instruments used in the experiment (e.g., pipettes, inoculation loops, lid of the agar plate). In addition, agar plates should be opened with caution, only for the minimum amount of time required for a procedure, and then closed immediately after the procedure is completed. Being prepared and focused during the lab period will also minimize the chance that you will contaminate your bacterial cultures and negatively affect the results of your experiment. A. Preparation of Cell Suspensions 1. Locate the two micro test tubes in the plastic rack. Label one test tube +pGLO and the other one -pGLO. Label both tubes with your team’s name. Return the tubes back to the rack. 2. Take your group’s starter culture plate and observe the colonies of E. coli. Document the morphology (color, shape, and texture) of the colonies and record your data in Table R1 of the Results section. N.B. Remember, the goal of any transgenic experiment is to add a gene and express a new phenotype in an organism. Before any change in the phenotype of an organism can be detected, a thorough examination of its usual (pre-transformation) phenotype must be made. It is possible that transformation or the procedure will affect colony morphology. 3. Remove a sterile transfer pipet from its packaging and open the tubes. Transfer 500 μl of CaCl2 transformation solution (on ice) into the +pGLO tube. Temporarily store the tip of the pipette in the empty -pGLO tube. We will use it again for a transfer into this tube. General Biology II Day 1 what is a plasmid it Ee En at into Page 10 genes eight Laboratory Exercises Winter 2024 what is a restriction Restriction sites  Lab 7 – Bacterial Transformation N.B. The transfer pipette has graduations that allow you to measure the volume of liquid that you aspirate. (See the figure to the right to correlate the graduations and the volume they indicate.) Read Before You Proceed: The bacteria must be actively reproducing to achieve high transformation efficiency. In the next step, to choose a bacterial population produced from one ancestor cell that is actively reproducing, it is necessary to: a) Take individual colonies (not a swab of bacteria from the dense portion of the plate). b) Select large colonies (i.e., 1–2 mm in diameter) that are uniformly circular with smooth edges. 4. Use a sterile loop to pick up 6 colonies of bacteria (see description above) from your starter plate. Pick up the +pGLO tube and immerse the loop into the transformation solution at the bottom of the tube. Spin the loop between your index finger and thumb until the cells in the bacterial film are dispersed in the transformation solution. You want a uniform bacterial cell suspension with no floating chunks. Discard the used loop into the bacterial waste container. 5. Take the transfer pipette (used in Step 3) from the –pGLO tube. Before transferring a portion of the cell suspension to the –pGLO tube gently aspirate and dispense the bacterial cell suspension in the +pGLO tube a few times to evenly disperse the cells in the solution. Transfer 250 l (half) of the cell suspension to the –pGLO tube. Discard the used transfer pipette into the bacterial waste container. N.B. The cells are less robust in the transformation solution. Being too rough with the cells while aspirating and dispensing liquid will cause them to break open and die. Read Before You Proceed: For the next step, make sure that you DO NOT add plasmid DNA to the -pGLO cell suspension tube. 6. Use a P20 micropipette to transfer 10 μl of pGLO DNA stock solution (in ice bucket) into the cell suspension of the +pGLO tube. Using the micropipette and the same tip, mix the cell suspension by gently aspirating liquid from the bottom of the suspension and dispensing it back into the top of the suspension (referred to as pipetting up and down). Pipet up and down 5 times to mix well. Close the +pGLO tube. Make sure that the caps for both tubes are securely in place. N.B. Again be gentle with the cells so that you do not damage them. 7. Place both the –pGLO tube and +pGLO tube into the foam rack. Transfer the rack to the ice bucket and incubate the tubes on ice for 10 min. Make sure to push the tubes all the way down in the rack so the bottoms of the tubes stick out and make contact with the ice. General Biology II Page 11 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation What’s Happening? It is postulated that the Ca2+ cations of the transformation solution (50 mM CaCl2, pH 6.1) neutralizes the repulsive negative charges of the phosphate backbone of the DNA and the phospholipids of the cell membrane to allow the DNA to enter the cells during the heat shock procedure coming up in the next part of the protocol (heat shock in procedure P2). This is described as making the bacterial cells artificially competent. 8. During this incubation, label your 4 LB nutrient agar plates on the bottom of the plate (not the lid!). Label each as follows: a) Label one LB/amp plate: +pGLO b) Label the LB/amp/ara plate: +pGLO c) Label the other LB/amp plate: -pGLO d) Label the LB plate: -pGLO Label each plate with your team’s name as well. Leave the plates (with lid facing up) in the tray on your bench. B. Heat Shock and Recovery Read Before You Proceed: The procedure used to increase the bacterial uptake of foreign DNA is called heat shock. The heat shock procedure must be done with care and accuracy to obtain maximum transformation efficiency. It is important that students follow the directions regarding time. Also important is the rapid temperature change and the duration of the heat shock. To accomplish this, it is important to: a) Bring a timer and ice bucket (with samples) to the water bath. b) Be accurate with respect to temperature and time of incubation in the water bath. c) Carry out transfers between ice and water bath and water bath and ice quickly. d) Make sure that the bottoms of the tubes in the rack are in contact with the water or ice depending on the step. 1. Bring your ice bucket (containing your rack) to the water bath on the side bench. Make sure to push the tubes all the way down in the rack so the bottom of the tubes sticks out and will contact the warm water. Quickly transfer the rack from the ice into the water bath (set at 42oC), for exactly 50 sec (use a timer to measure the incubation time). 2. When the 50 sec. are done, quickly transfer all tubes in the rack back on ice. Again, make sure to push the tubes all the way down in the rack so the bottoms of the tubes stick out and make contact with the ice. Bring your ice bucket back to your station and leave the tubes on ice for 2 min. What’s Happening? The heat shock procedure (ice-heat-ice) increased the permeability of the cell membrane to DNA. While the mechanism is not known, the duration of the heat shock is critical and has been optimized for the type of bacteria used and the transformation conditions employed. The pGLO plasmid DNA entered the permeable cell membrane which the help of the CaCl2 transformation solution. General Biology II Page 12 Laboratory Exercises Winter 2024  fourth growth 25th L colonies Lab 7 – Bacterial Transformation 3. control manner Remove the tubes from the rack on ice and place in the plastic rack. Open a tube and, using a sterile transfer pipet, add 250 μl of LB nutrient broth to the tube and reclose it. Repeat with a sterile transfer pipet for the other tube. Incubate the tubes for 30 min at room temperature. Resuspend the cells in the samples by inverting each tube midway through (after 15 min.). What’s Happening? The incubation period following the addition of LB nutrient broth allows the transformed cells to recover from the harsh heat shock but also to make the ampicillin resistance protein beta-lactamase so that the transformed cells survive on the ampicillin selection plates. C. Plating the Bacteria Read Before You Proceed: The agar in the plates is a semi-solid gel and you must use care in the following steps to avoid damaging the plate. Be careful not to pierce the agar with the transfer pipet or apply too much pressure with the loop when inoculating your plates. If you crack or break the agar, the plate can no longer be used. Your instructor will provide guidelines for the following steps so that you do not damage your agar plates or contaminate them with unwanted microorganisms. 1. Use a sterile transfer pipette to gently resuspend the cells in the –pGLO cell suspension. Using the same transfer pipette, add 100 μl of the –pGLO cell suspension to both the –pGLO LB agar plate and the –pGLO LB/amp agar plate. N.B. Dispense the cell suspension over the surface of the agar at the center of the plate (do not touch or pierce the agar). 2. Using a new transfer pipet, repeat the procedure for the +pGLO cell suspension and the +pGLO LB/amp and +pGLO LB/amp/ara plates. 3. Using a new loop for each of the four plates, spread the cells evenly around the surface of the agar by gently gliding the flat surface of the loop back and forth across the entire surface of the agar. Remember: Be gentle! You will crack and break the agar by applying too much pressure. Uncover one plate at a time and re-cover immediately after spreading the suspension of cells. This step will minimize contamination. 4. Once all of the plates have been inoculated, let them sit for 5 min. to allow the liquid to be taken into the agar. After 5 min., stack them together upside down (with lids facing down) and tape them together. Label the tape (on the side of the stack) with your team’s identifier. 5. The stack of plates will be incubated upside down in a 37°C incubator for 20 hrs by our technicians. The plates are inverted to prevent condensation on the lid, which may drip onto the culture and interfere with colony formation. General Biology II Page 13 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation D. Record Data: Characterizing Bacterial Growth on Experimental Agar Plates 3. Observe the 4 experimental plates on the side bench. For each plate: a) Count the number of colonies b) Measure the average colony size c) Describe the morphology of the colonies. 4. Record your data in Table 7 of the Results section. 5. Wait for your instructor to call your group over to the UV transilluminator. Observe colony fluorescence under UV light for each plate. Record your data in Table 7. General Biology II Page 14 Laboratory Exercises Winter 2024  49.5/51 Lab 7 – Bacterial Transformation Names:_____________ Mia Zhang Lab Section: ______ _____________ Samantha Schwartz 08 RESULTS PART 1: pGLO Restriction 10000 bp 8000 bp Figure 9. DNA Ladder with Digest 6000 bp sizes of fragments. The three 5000 bp most intense bands (labelled 4000 bp 1. Observe your data from the 3500 bp with yellow stars) are easiest electrophoretogram provided by your 3000 bp to find and can help you locate instructor and determine the number of 2500 bp the bands in the ladder of your bands (representing fragments) and electrophoretogram. 2000 bp estimate the size range for each fragment in a digest. 1500 bp The most intense (brightest) bands of the DNA ladder sample are easiest to Direction of migration find as landmarks and represent 1000 bp (towards positive electrode) fragments that are 6000 base pairs (bp), 3000 bp and 1000 bp. 750 bp Table 6. Fragments Observed in Restriction Digests of pGLO Number of Fragment Sizes (bp) Digest Use the DNA ladder to estimate the size range of fragments (e.g., between Bands* 500 and 600 bp) within the band Can’t use the DNA ladder to estimate size of circular pGLO (uncut) 2 DNA forms. Standards are linear DNA fragments. pGLO / EcoRV 1 5000 6000 pGLO / HindIII 2 1 3000 6000 2 below 1000 pGLO / EcoRV + HindIII 3 1 4000 2.2000 1000 3 below1000 * Bands are distinct florescent signals produced by many DNA molecules traveling the same distance because they have the same characteristics. 2. Determine how many times you find the restriction site for each restriction enzyme in the pGLO sequence. Hint: How many bands to you detect when pGLO is digested with each restriction enzyme alone. EcoRV restriction site: 1 5’-GATATC-3’ 3’-CTATAG-5’ HindIII restriction site: 2 5’-AAGCTT-3’ 3’-TTCGAA-5’ General Biology II Page 15 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation 3. Complete the restriction enzyme digest map Base pair 1 of pGLO (Fig. 10) by determining the (Arbitrary start point) Site 1 (386 bp) restriction enzyme (either EcoRV or HindIII) Base pair number that cuts at restriction sites 1, 2, 3. You have increases to right enough information from your analysis of the pGLO digests in your electrophoretogram to figure out this puzzle. Some notes: pGLO (5371 bp) a) Circular molecules of DNA such as plasmids Site 2 (1465 bp) have an arbitrary start point for numbering the base pairs (bp). b) The total size of pGLO is 5371 bp c) The number of base pairs in brackets beside a Site 3 (2114 bp) site indicates how far away the site is from the start point. Site 1 in map is cut by: __________________ ECORN Figure 10. Restriction Map of pGLO. Circular DNA molecules have an arbitrary base pair that Site 2 in map is cut by: __________________ represents the 1 first pair. Hindill Site 3 in map is cut by: __________________ Hind 11 4. The molecules of an uncut plasmid (still circular) exist in two forms: circular open (not supercoiled) and circular supercoiled (compact shape of a helix of the DNA helix). How does this explain the number of bands observed and their respective migration compared to each other and to the plasmid when it was linearilized (cut once) in gel electrophoresis? Hint: Which characteristics are used to separate molecules in electrophoresis and which is important here? Since circular supercoiled is verycompact it can travel faster and farther compared to circularopen which is much lesscompact and slower Linearinzed is the fastestand goesthe farthest of all since it is thinner and long with no resistance unlike the circular ones It canpacethrough the agar more easilyofthan the others So for uncut plasmid, separation supercoiled and non-supercoiled is based on their shape -0.5 (And for linearized DNA samples, the bands are separated by size) 5. Why was running this uncut pGLO an important control sample in this assay to build a restriction site map of pGLO? Plasmids are artificially constructed and this control is to ensure that the plasmidbeingused is circular Thiscontrolchecks forplasmid integrity Anuncut circularplasmid will give 2 bands Onewill bebiggerandcloserto the well andthe other smallerand slightlyfurtherfromthe well General Biology II Page 16 Laboratory Exercises Winter 2024 -0.5  Lab 7 – Bacterial Transformation PART 2: pGLO Transformation Comparing Bacterial Growth on Experimental Agar Plates Bacterial growth can be characterized and compared by observing or measuring: a) Colony Number: reflects the number of individual cells from your initial population that survived the procedure and divided repeatedly on the plate to produce a colony of clones that is visible. b) Colony Size: reflects the number of cells in a colony. If comparing colonies that were produced over the same length of time, the size of the colony also reflects the rate of cell division. c) Colony Morphology: is a set of characteristics (e.g., color, shape, texture) specific to a bacterial species. Colony morphology may be altered by transformation or some aspect of the procedure. d) Fluorescence (when exposed to UV light): reflects whether or not the cells are expressing GFP. Table 7. Comparison of Bacterial Colony Number and Characteristics on Each Experimental Culture Plate. Experimental Culture Plate Starter Culture -pGLO +pGLO Type of Agar* LB LB LB/amp LB/amp LB/amp/ara Number of Colonies Count the number of colonies, if a lawn (film) is present, indicate lawn lawn colonies 84colonies 92 colonies Size of Colonies (mm) Take the mean of 3 colonies NA NA 1.0 0.9 Color of Colonies yellowish yellowish yellowish under visible light white white NA yellowish white Shape of Colonies [circular or irregular] circular NA NA circular circular Colony Texture [glossy or mat] glossy glossy N A glossy glossy Green Fluorescence Observed when exposed to UV [yes or no] no NA no yes * LB: Nutrient rich broth; amp: ampicillin; ara: arabinose. General Biology II Page 17 Laboratory Exercises Winter 2024  Lab 7 – Bacterial Transformation 1. Which characteristics of bacterial cells (listed in the table) can reflect a new phenotype if the transformants are different from the starter culture for this characteristic? -0.5 inattas it nightie.tt es iran icteithianPiietoitnasmia andhave adopted the green fluorescence phenotype What else? changes to colony morphology 2. Which of the four (4) experimental cultures do you expect to most closely resemble the starter culture in these characteristics? There.tn oiIitietretteihob th.fialnE o ias ie rr bEnethesusstaintees ampicilinandarabinose 3. Answer the following questions concerning the number of colonies counted on each plate for the –pGLO (control) and the +pGLO (transformation) cultures. a) Did the number of colonies observed for the –pGLO LB plate match your prediction? What information about the experiment does this control give you? Yes a lawn wasexpected The bacteria consumes the nutrient rich broth to grow and it's growth is nothinderedby the ampicilin This tells us that the bacterial fr nentaYEiure a'sediti toaIethatoo b.it ai.thritie b) Did the number of colonies observed for the –pGLO LB/amp plate match your prediction? What information about the experiment does this control give you? EE i c) Why do you see less colonies for the +pGLO LB/amp plate compared to the –pGLO LB plate? Anybacteriathat did notgettransfectedwiththep620plasmidwould not grow in the presence of ampicilin d) Did you observe roughly the same number of colonies for the +pGLO LB/amp plate and the +pGLO LB/amp/ara plate? Is this what you would expect? Briefly explain your response. We observed a very similar amount of colonies on both of these plates This was expected because the arabinose doesnot effect the amount i stick site.tn Etesaethsi etie ntiiert et iecoi's e) Which 2 plates would you compare colony numbers to estimate the transformation i gy efficiency (number of transformants produced / number of total bacterial cells in the transformation)? YEE.IS Iantaheam tn.iIthe i terriftea oithnot.hn Iatiisucessfully the transfected of theLBamp polo more closely amount platewill resemble the number of donies on the positive colonieson controlplate LB pGL0 General Biology II Page 18 Laboratory Exercises Winter 2024 -0.5  Lab 7 – Bacterial Transformation 4. Answer the following questions concerning the fluorescence observed in colonies on each plate for the +pGLO cultures. a) Would you expect to see green fluorescence if you exposed the pGLO plasmid DNA stock solution to UV light? Briefly explain your response. -0.5 I would notexpect to see green fluorescence in the polo plasmidDNA stock farragylatedby the Parabad promoter thatonlypromotes It's lithepiece The ability to glow is due to GFprotein, not the DNA itself b) You likely do not see green fluorescence in the colonies on LB/amp but you do see green fluorescence in colonies on LB/amp/ara. How can you explain this result? Eesthiepbtdas.inse.thEf fabetri piece thebacteriaininteglow because no GFP was transcribed 5. Answer the following questions concerning the colony morphology observed for colonies on each plate for the –pGLO and +pGLO transformations. a) Does the morphology of the colonies for the +pGLO culture plated on LB/amp resemble the morphology of the colonies of the -pGLO culture plated on LB? [If yes: what information can be gained from this? If not: what was different in the way these bacterial cells were treated that could have resulted in this difference in morphology?] Bothcolonies have the same texture colour and reaction to UVlight They are present in differentamounts havedifferentshapeand differentsizes This tells us that the Ecolibacteria was present on both plates Thedifferences are due to the bacteria on the LBamp polo plate that werenot successfully transfected withthe plasmid andcouldnot polo b) Does the morphology of the in thepresence ofampicilin colonies for the +pGLO growculture plated on LB/amp/ara resemble the morphology of the colonies of the +pGLO culture plated on LB/amp? [If yes: what information can be gained from this? If not: what was different in the way these bacterial cells were treated that could have resulted in this difference in morphology?] There.eeetthe I iiithtIn niiiiis'were p inert.ie it i e e i c c) If the morphology of colonies between these plates were different (comparison in b), is it c not necessarily due to the expression of a new gene and the appearance of a new phenotype? No environmental factorssuch as food temperature and pHcould affect the colony morphology General Biology II Page 19 Laboratory Exercises Winter 2024 -0.5