Biology 101 Lab Manual Seattle Community College south XanEdu Metric Lab Name Length With all measurements in this class you need to measure to the appropriate number of significant digits. The metersticks have marks for each millimeter so you can use these marks to measure to the nearest millimeter in addition you need to estimate the next digit or in the nearest tenth of a millimeter. The last digit will always be a estimate it allows us to be more accurate when measuring. Use the correct number of digits for answers. Using the meter stick draw a line in the margin of this page that is 10cm long. Measure the length of this line in inches ____________________________________________ How many inches is 100 cm? _________________________________________ How tall are you in meters? ___________________________________________ Area and Volume Procedure Measure the length and width of the room in metric. Questions 1. What is the width of the room? ______________________________________ 2. What is the length of the room? _____________________________________ 3. Calculate the area of the room the area is equal to 1W and is in m2). ______ 1. What is the height of the room? 2. Calculate the volume of the room area *H in m3. 3. Do you think this volume is equal to the volume of air in the room? Why or why not? page 13 Weight In this class we will be using triple beam balances. Before you weight anything you first need to zero the balance. You do this by turning the knob on the left side just underneath the platform until the two lines on the right side meet exactly. Make sure the platform is clean and all the weights are on zero. Weight of Object A _____ Weight of Object B _____ When weighing dry chemicals use a piece of weigh paper on the balance. How much does the weigh paper weigh? _____ Weigh 2.7 g of table salt and put that amount aside. This is the amount of salt that is recommended in your diet daily. Weigh 11.6 g of table salt compare this to the other pile. This is average amount of salt consumed in the US daily. Which pile do you think is closer to the amount you eat every day? Find the nutrition label on the soda how much sugar is found in this can? Weigh this out and estimate the number of tablespoons this is? Volume How many pints, cups, tablespoons and or teaspoons of water does it take to fill a 1 liter graduated cylinder exactly to the 1 liter mark? How many cups are in a quart of water? Which is larger a liter of water or 1 quart? circle your answer. page 14. Special Relationships Pennies made before 1982 were made using a large percentage of copper it became too expensive to continue putting as much copper into each penny. Pennies made after 1982 have less copper in them and correspondingly they weigh less. Try it yourself. 20 pre 1982 pennies weigh? 20 post 1982 pennies weigh? Why do they weigh less? they are less dense or they are smaller. Write a hypothesis to answer this question it should be a sentence that answers this question and is something you can test in our class. Design and perform a experiment to test your hypothesis. Compare groups of at least 10 pennies to measure the difference in weight or volume. Volume can be calculated either in mm3 or in milliliters. Hypothesis Experimental Design Results Conclusions page 15 Navigating on internet How would you modify your search based on these results? 3. Now try 3 different search engines using your modified terms. altavista.com hotbot.com infoseek.go.com excite.com webcrawler.com google.com lycos.com opentext.com keep track of sites you visited. 4. How do the results from the different pages compare? Which seemed to be the easiest to use? Google was easiest. Which gave the best results? Were the results similar for each site? 4. Find a page with lots of links to other biology sites? What is the url of this page? page 107 Other Biology Projects microspheres viewed through a microscope. Forming a coacervate you need test tube dropper gradulated cylinder glass microscope slides coverslips microscope 1% gelatin 1% gum arabic solution 0.1 NHC1 NHC1 is a acid. pour into test tube. Predator and Prey Interactions observations about the organisms. species. life, frogs, obtain food or escape predators. Some members of a species are more successful than others in building nests or attracting mates or flowering plants attracting pollinators. materials needed beans white, red, speckled, black. foraging devices forceps, forks, spoons, knives. paper cups calculator. Predator prey relations between populations of two species. bean species of different colors. page 79 Questions 1.Which prey appears to be best adapted to this environment? Why? Which prey appears to be least adapted to this environment? In what environment might this prey type have had a better chance for survival? 2. Which predator appears to be most successful at capturing prey? 3. Did any predator go extinct? 4. What do the phrases natural selection and differential reproduction mean? 5. How is variation within a species the raw material of natural selection? 6. How can changing environments change population characteristics? page 85. Population studies Patterns of survivorship page 87 Name ___ Populations increase in size when the number of individuals entering through birth and immigration exceeds the number leaving through death and emigration. Materials required For each lab table Bucket containing 50 dice Soap bubble solution and wand stopwatch or digital watch 15 cm ruler Overhead transparency of Figures 1 and 2. One set of red, blue, black, green pens. Overhead projector. Exercise 1 Dice Survivorship Procedure First population Empty the bucket of 50 dice onto the floor. Assume that all individuals that come up as ones die of heart disease. All other survive. Pick up all the ones set them aside in the cemetery and count the number of individuals who have survived. Record the number of survivors in this generation. Return the survivors to the bucket. Dump the survivors onto the floor again and remove the deaths ones that occurred during this second generation. Count and record the number of survivors. Continue this process until all the dice have died from heart disease. Population two Start again with full bucket of 50 dice. Assume that ones die of heart disease and twos die of cancer. Proceed as described for population 1 recording the numbers in table 1 untill all dice are dead. Determine the percentage of survivors for each generation with the following formula table 1 is percentage of survivors = number surviving x 100% 50 Exercise II Soap bubble survivorship blow bubbles Three different populations of soap bubbles will be formed. Population A Once the bubble leaves the wand group members wave blow or fan in a effort to keep the bubble in the air and prevent it from breaking. dying. Population B Group members do nothing to interfere with the bubbles or keep them up in the air. Population C This group uses a wand mounted on a wooden frame. The group member blowing bubbles tries to blow the bubbles through the opening in the frame. Procedure Practice blowing bubbles for a few minutes until they can be generated with the single end of the wand. Once the bubble is free of the wand the timer should start the watch. When the bubble bursts the timer notes the time. Obtain data on 50 bubbles. Summarize your data as follows Count the number of checks the number of bubbles dying at each age. Record the number in the column marked total number dying at this age. By subtracting the number dying at each age from 50 determine and record the surviving at each age. For example if 5 bubbles broke died at age 1 second then 50 -5 = 45 survived at least one second. Calculate the percentage surviving at each age. the formula to use is percentage surviving = number surviving x 100% to this age 50 Plot the percentage surviving on the graph. Table 1 Dice population data population 1 population 2 heart disease only cancer & heart disease number surviving percentage surviving Generation 0 50 100% 50 100% 1 2 3 4 5 interpretation of the survivorship curves Examine first the plots from the dice populations. In the first population heart disease only a constant 1/6 or 17% of the population dies at each age. In the second population 2/6 or 33% dies at each age. on the arithmetic plot these data form a smooth curve while on the logarithmic plot they form a straight line. What types of curve would you expect for each of the three experiments? make a arithmetic plot for survivorship of the dice and soap bubble populations. Use open circles to plot percentage of dice surviving at each age. closed circles for soap bubbles. 30 25 20 15 10 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Logarithmic 1 9 8 7 6 5 4 3 2 1 5 10 15 20 Observations in the field page 95 the opportunity to observe the diversity of organisms. All organisms need a source of energy to power themselves and a source of matter nutrients to build and repair themselves. Characteristics of different food chain trophic levels Primary producers provide the basic source of organic molecules in communities. green pigments green plants Primary consumers are the organisms that eat plants rabbits or deer eat plants. Secondary consumers ladybugs and spiders are avid insect eaters. small birds eat earthworms coyotes eat rabbits mountain lions eat deer bears eat people. snakes eat frogs and toads. marine organisms found in puget sound brown algae red algae green algae BIO 100 Biological Principles immersion oil for seeing smaller things in under microscope 100 is 1000x. KASALA http://waol.blackboard.com/webapps/login Login: Username: guest Password: password *Login is the same for everyone. Washington DNS Servers: Primary: 64.40.40.51 Secondary: 209.102.96.10 Northwest Access Numbers Seattle: (206) 424-1111 Seattle: (206) 337-5050 Seattle: (206) 337-5252 http://www.nocharge.com/connect.htm Login: Username: guest Password: password Western Washington Free Dial-up Service DNS Servers: Primary: 64.40.40.51 Secondary: 209.102.96.10 netzero hispeed to work correctly connections tab in the dial up settings box hightlight netzero and click the settings button. 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Box 3268 Lafayette, LA. 70502-2458 Techniques * Flora of the skin* ~ Aseptic technique ~ - used almost always when transferring organisms from and to different media The technique is simple !! 1) Flame sterilize inoculating loop by use of a bunsen burner 2) Flame sterilize mouth of tube that culture will be transferred to, and sterilize mouth of tube that culture is taken from 3) Dip sterile loop into culture, then transfer to media 4) Flame sterilize inoculating loop and mouth of both tubes by bunsen burner once transfer is complete *Note- only steps 1 and 4 are required for flora of the skin experiments. pictures from : http://www.mc.uky.edu/oaa/curriculum/iid98/manual/00labtechniques.htm ~ Swabbing technique ~ - media used would be a Nutrient agar plate and a Malt extract agar plate supplemented with 0.1% erythromycin and 0.1% olive oil The technique : 1) Sterilize cotton swab by dipping into sterile water or saline 2) Swab area on skin 3) Inoculate plate with one streak from cotton swab near the top of plate 4) Then using inoculating loop, streak away from main streak Image taken from Applications in General Microbiology, A Laboratory Manual; Kerr/McHale; 6th edition;copyright 2003; pg.333 ~ Additional techniques ~ - Gram Staining - TSI agar slant and motility stab *Flora of the throat and nasal passage* - Different media are used to differentiate between pathogenic and non-pathogenic organisms Media and Techniques used: ~ Blood agar plates~ -aseptic and swab techniques are used the same way as in flora of the skin except Blood Agar plate is used -shows different types of hemolysis * alpha hemolysis- partially lysed; greenish or brownish zone around colony beta hemolysis- completely lysed; clear zone around colony * gamma hemolysis- no lyse; no change around colony pictures from: http://gold.aecom.yu.edu/id/micro/hemolysis.htm -shows difference in colony morphology * Streptococcus colonies appear small and translucent on Blood agar plates * Staphylococcus colonies appear large and opaque *Note---> incubation should be in a candle jar or a carbon dioxide incubator -Steps to follow to differentiate S.pneumoniae and S. mitis: 1) Take a throat and nasal swab and inoculate on Blood agar plate. 2) Pick and transfer an alpha hemolytic organism into a tube of saline solution 3) Pipet 0.5 ml of of this solution into tube with Bile solution, and another 0.5 ml into a tube of saline 4) If Bile tube has lost turbidity and becomes clear, then organism is bile soluble ~ Staphylococcus Medium No. 110~ -differentiates Staphylococcus aureus from other Gram-positive cocci - Steps to follow: 1) Take a throat and nasal swab and inoculate on Blood agar plate 2) Plate Staphylococcus colonies (yellow colonies with beta hemolysis) on Staphylococcus medium No. 110, and incubate 48 hours in 37C 3) For detecting coagulase activity---> using aseptic technique remove yellow or orange colonies from Staphylococcus medium and mix in tube of BHI (brain heart infusion) Broth. Then with a Pasteur pipet add 2 drops of BHI broth to a tube with rabbit coagulase plasma. Any clotting within 3 hours is positive for coagulase. 4) Add Brom cresol purple indicator to Staph No.110 plate where the colony was picked. Any change in color means there is fermentation of mannitol. 5) Flood plate with saturated ammonium sulfate and incubate. Clear zone around colonies picked--->gelatinase positive. Image taken from Applications in General Microbiology, A Laboratory Manual; Kerr/McHale; 6th edition;copyright 2003; pg.334 *References* Applications in General Microbiology, A Laboratory Manual; Kerr/McHale; 6th edition;copyright 2003; pg. 327-335 created by: Alicia Turner adoptuskids.org www.microscope-depot.com Crystal Violet Stain Iodine Methylene Blue Stain, Gram's Iodine Solution, 1 oz Item No S-08020 $4.95. postage $7.95 The Microscope Depot 392 W. Larch Rd. Bldg. 28 Tracy, CA 95304-1643 Kacey inc. 60 McCormick Place Asheville, NC 28801-4255 Microbiology Water 1. Water is the most abundant substance in cells. 2. Because water is a polar molecule it is a excellent solvent. 3. Water is a reactant in many of the decomposition reactions of digestion. 4. Water is a excellent temperature buffer. page 50 Introduction to microbiology in a microbiology class you have a opportunity to explore a extremely small biological world that exists unseen in our own ordinary world. The microscope can be used to see extremely small organisms. A few of these many and varied organisms are pathogens capable of causing disease. isolate and identify. Photosynthetic bacteria are a important source of the Earth's supply of oxygen. Microorganisms also make major contributions in the fields of antibiotic production food and beverage production, food preservation, and recombinant DNA technology. This course microbiology is a introduction to the microbial world. pathogenic organisms. Microbiology Experiments Fourth Edition Exercise 1 Laboratory Report Ubiquity of microorganisms Results Room Temperature about 25C Plate. plate quadrant plate numbers 1 2 3 4 Source Colony appearance 37c plate 1 2 3 4 Source Colony appearance Questions 1. Give three reasons why all the organisms you placed on the Ts agar plates might not grow? temperature 2. Why were some agar plates incubated at 37c and others at room temperature? 3. Why do you invert agar plates when placing them in the incubator? 4. Name on place that might be free of microorganisms? Exercise 2 Bright field light microscopy Microbiology is the study of living organisms too small to be seen with the naked eye. A optical instrument the microscope allows you to magnify microbial cells for visualization. living organisms called animalcules known as bacteria in scrapings from teeth, and larger animalcules known as protozoa and algae in droplets of pond water and hay infusions. Achromatic objective A microscope objective lens in which the light emerging from the lens forms images practically free from prismatic colors. Bright field light microscope A form of microscopy in which the field is bright and the specimen appears opaque. Compound microscope A microscope with more than one lens. Magnification The ability of a microscope to increase specimen size. Parfocal Having a set of objectives so mounted on the microscope that they can be interchanged without having to appreciably vary the focus. Simple microscope A microscope with only one lens. Wet mount A microscope slide preparation in which the specimen is immersed in a drop of liquid and covered with a coverslip. objective how to use the microscope. materials Cake of baker's yeast Tube containing 10 ml distilled water Plastic dropper Prepared stained slides of various bacterial forms coccus rod spiral. Procedure 1. Place the microscope on a clear space on your desk and identify the different parts. 2. read information turn microscope on. 3. Sample preparation wet mount. Prepare a yeast cell suspension by adding to water in a test tube just enough yeast to cause visible clouding approximately 1 loopful per 10 ml of water. Remove a small amount of the suspension with plastic dropper and carefully place a drop on the surface of a clean slide. Cover the drop with a clean coverslip. Focus with the oil immersion objective. Results 1. Draw a few yeast cells from each magnification. Include any interesting structural changes evident at the three magnifications. Magnification _____ _____ _____ Objective _____ _____ _____ 2. Examination of prepared bacteria slides. Examine with the oil immersion objective and draw a few cells of each morphological form. Coccus Rod Spiral 3. Answer the following questions about your microscope. a. What is the magnification and numerical aperture NA stamped on each objective of your microscope? Objective magnification Numerical Aperture _______________________ __________________ _______________________ __________________ _______________________ __________________ _______________________ __________________ 4. What is the magnification stamped on the oculars? 5. Calculate the total magnification of the objective/ocular combination with The lowest power objective The highest power objective Questions 1. Discuss the advantages of a modern compound microscope over a early microscope. 2. Why must the distance from slide to objective increase rather than decrease when coarse focusing with the high dry and oil immersion objectives? 3. How does increasing the magnification affect the resolving power? 4. How does lens immersion oil help to increase the resolving power of the oil immersion objective? 5. How can you determine that the ocular and objective lenses are free of sweat, oil, and dust contaminants? 6. What are the functions of the substage condenser? 7. What is meant by the term parfocal? Does it apply to your microscope? True False Questions 1. Van Leeuwenhoek's microscope was corrected for spherical not chromatic aberrations. 2. Spherical lens aberrations are easier to correct than chromatic lens aberrations. 3. The objective NA is more important than the condenser NA for increasing resolving power. 4. The working distance is the distance from the tip of the objective to the tip of the condenser lens. 5. Excess oil on the oil immersion objective can safely be removed with lens paper containing a drop of solvent. Exercise 3 Depression slide a microscope slide with a circular depression in its center used for viewing liquid like liquid water. Hanging drop slide A microscope specimen observation technique in which the specimen hangs suspended from a inverted coverslip mounted on a depression slide. Wet mount slide a microscopic specimen observation technique in which a drop containing the specimen is placed on the surface of a clean slide followed by the addition of a coverslip over the drop. objective To become familiar with the advantages and limitations of wet mount and hanging drop preparations for observing living cell material. Materials Cultures 12-18 hour nutrient broth cultures of Staphylococcus epidermidis and Spirillum volutans showing visible clouding. 12-18 hour nutrient broth cultures of Bacillus cereus and Pseudomonas aeruginosa showing visible clouding. A yeast suspension previously prepared by suspending sufficient baker's yeast in a tube of glucose yeast fermentation broth to produce visible clouding followed by 6-8 hour incubation at 25C. A hanging drop depression slide. Vaseline and toothpicks. pipets. acidified methylene blue A star diaphragm for dark field microscopy. Procedure Wet Mounts for study of Bacterial form and motility. 1. Prepare six clean microscope slides and seven clean coverslips. 2. Suspend your broth culture of S. epidermidis by gentle tapping on the outside of the culture tube. mix contents. 3. Remove the test tube cover and with a pipet 0.1 ml of the broth culture. 4. Transfer a drop of this suspension to the surface of a slide. 5. Grasp a clean coverslip and place on droplet. 6. Insert the wet mount on the stage of your microscope and examine for cell motility and form with the oil immersion objective. S. volutans B. cereus P. aeruginosa. use of hanging drop slides for study of bacterial form and motility. 1. Prepare a clean depression drop slide and coverslip. 2. With a toothpick spread a thin ring of Vaseline 1/4 inch outside the depression slide concavity. 3. Using your suspended B. cereus broth culture and a wire loop transfer 2 loopfuls to the central surface of a coverslip. If done correctly the droplet will remain suspended and will not come in contact with the well bottom. Use of a vital stain methylene blue to determine yeast viability. 1. transfer a small drop of methylene blue to the surface of a clean slide. 2. With a pipet add a small drop of the baker's yeast suspension. Carefully place a clean coverslip over the surface of the droplet. 3. Observe the wet mount with bright field microscopy using the low and high dry microscope objectives. 4. Prepare drawings of representative cells and show calculations for determing the percent of viable yeast cells. dead cells, viable cells remain colorless. Exercise 3 Laboratory report Determination of cell motility, form, viability using wet mount and hanging drop preparations. Results 1. Wet mounts for study of bacterial form and motility. Drawings of representative cells showing their relative sizes, shapes and arrangements. Record magnification x and motility + or -. S. epidermidis S. volutans B. cereus P. aeruginosa ___ x ___ x ___ x ___ x ___ motility ___ motility ___ motility ___ motility 2. Hanging drop slide. B. cereus Make observations similar to those above and indicate any difference from the B. cereus wet mount observations. Differences B. cereus ___ x ___ motility 3. Dark field microscopy of baker's yeast. Drawings of cells showing their size, shape, and arrangement and visual appearance of living and dead cells. Record magnification used. ___ x ___ x show calculations for determining the percent of viable cells. 4. Bright field microscopy of baker's yeast stained with methylene blue. Record magnifications used. ___ x ___ x Discuss your yeast cell visibility results by the two methods shown above. If a wide viability variance > 10% exists between the two methods what other method might you use to prove which method is more accurate? Questions 1. What advantages are there in determining cell motility microscopically rather than with a stab culture? 2. What advantages does a hanging drop preparation have over a wet mount preparation? Disadvantages? 3. How did you obtain optimal results with dark field microscopy? 4. Why is it difficult to employ the oil immersion objective for dark field microscopy? 5. What might be a reason for employing a actively multiplying culture when examining viability microscopically? 6. In addition to determining cell viability what other useful morphological determination can sometimes be made with dark field microscopy? 7. What difficulties might there be in attempting to determine the viability of bacterial cells with stains such as methylene blue? prokaryotic makeup of bacteria. Yeasts are eukaryotic cells. Exercise 4 Simple Stains Positive and Negative Stains Negative stain is a simple stain in which the organisms appear clear against a dark background. Simple stain is a procedure for staining bacteria consisting of a simple stain. Smear is a dried mixture of bacteria and water or broth on a glass slide in preparation for staining. Objective to learn to prepare and stain a bacterial smear using s dimple stain. Observe stained organisms under the oil immersion lens. Materials Cultures Bacillus subtilis or Bacillus cereus Staphylococcus epidermidis Enterococcus faecalis Micrococcus luteus Staining bottles with crystal violet methylene blue safranin glass slides pen tap water in small dropper bottle Inoculating loop. Alcohol sand bottle is a small screw cap bottle half full of sand and about three quarters full of 95% alcohol. Procedure Simple stain 1. Clean a glass slide. 2. You write or label the glass slide. 3. Add a drop of water to the slide on top of each circle. Use a loop to transfer or dropper to transfer water. 4. Sterilize a loop by holding it at a angle in the flame of the Bunsen burner. 5. Hold the loop a few seconds to cool it then remove a small amount of a bacterial culture and suspend it in one of the drops of water on the slide. water become turbid cloudy. 6. Heat the loop red hot. 7. Permit the slide to dry. 5-10 mintues. 8. heat fix the organisms to the slide by passing the slide through the bunsen burner flame two or three times. 9. Place the slide on a staining loop over a sink or pan. 10. pour off the stain in 20 seconds. 11. Carefully blot the smear dry with a paper towel. 12. Observe the slide under the microscope. Use oil immersion lens to see bacteria. cover the smear with immersion oil. 13. Record your results. The Negative stain observe capsules Materials Culture Some cultures used for simple stain Bottle of India ink. procedure 1. Place a drop of water on a clean slide and add organisms with a loop until the drop is cloudy. 2. Mix a loopful of India ink into the drop and spread the mixture out into a thin film. 3. Let dry and examine under the microscope. Bacteria can be seen as clear areas on a black background. 4. Record your results. Results 1. Simple stain Staphylococcus Bacillus Micrococcus Enterococcus Draw shape and arrangment 2 Negative stain Questions 1. What are the advantages of a simple stain over a wet mount? 2. Do you need more or less light when viewing a stained preparation compared to a wet mount? 3. What information can you observe in a wet mount that cannot be seen in a stained preparation? 4. How does the negative stain compare to the simple stain? 5. How many um are in a millimeter mm? How many um are in a meter m? Exercise 5 Multiple and Differential Stains The Gram stain is especially useful as one of the first procedures in identifying organisms because it reveals not only the morphology and the arrangement of the cells, also information about the cells. Appearance of the cells after each procedure Gram + Gram - Crystal violet Purple Purple Iodine Purple Purple Alcohol Purple Colorless Safranin Purple Pink Objective is to learn the Gram stain procedure. Gram positive organisms and Gram negative organisms. Materials Staining bottles of the following crystal violet iodine acetone/alcohol or 95% alcohol safranin Clothespin or forceps Staining bars overnight cultures growing on TS agar slants Escherichia coli Bacillus subtillis Staphylococcus epidermidis Enterococcus faecalis Micrococcus luteus Procedre for Gram Stain 1. Put two drops of water on a clean slide. Gram positive gram negative 2. Place the slide on a staining bar across a sink or can. 3. Flood the slide with crystal violet. 6 - 30 seconds. 4. Flood the slide with Gram's iodine. 12 - 60 seconds. Acid fast stain materials Cultures Mycobacterium smegmatis Carbolfuchsin acid in staining bottles Methylene blue in staining bottles Acid Alcohol in staining bottles Beaker Metal or glass staining bars. procedure 1. Prepare a smear of the material and heat fix. 2. Cover the smear with carbolfuchsin and stain for 3 5 minutes. 3. Rinse with water. 4. Decolorize with acid alcohol for 10-30 seconds. 5. Rinse with water. 6. Counterstain with methylene blue for 20-30 seconds. 7. Rinse with water. 8. Blot dry carefully and examine under the oil immersion lens. 9. Record resuls Endospore Stain Materials Culture Bacillus cereus on nutrient agar slant after three or four days incubation at 30C. Malachite green in staining bottles Safranin in staining bottles Metal or glass staining bars. beaker or can. Procedure for endospore stain 1. Prepare a smear on a clean slide and heat fix. 2. Add about a inch of water to a beaker and bring it to a boil. 3. Place two short staining bars over the beaker and place a slide on them. 4. Tear a piece of paper towel a little smaller than the slide and lay on top of the smear. The paper prevents the dye from running off the slide. 5. Flood the slide with malachite green and steam for 5 minutes. Continue to add stain to prevent the dye from drying on the slide. 6. Decolorize with water for 30 seconds by flooding with water. The vegetative cells dividing cells will lose the dye the endospores will retain the dye. 7. Counterstain with safranin for 30 seconds. Then wash with water for 30 seconds. 8. Observe with the oil immersion lens. The endospores will appear green. The vegetative cells will appear pink. Results Gram reaction arrangement E. coli B. subtilis S. epidermidis E. faecalis M. luteus optional stains Organism appearance Acid Fast Capsule Endospore Storage granules Flagella Questions 1. What is the function of each one of the Gram stain reagents? 2. Give two reasons Gram positive organisms sometimes appear Gram negative. 3. What is the purpose of using a control in the Gram stain? 4. What is a capsule? 5. What are storage granules and why are they important to the cell? 6. How does a endospore appear draw and indicate color? a. when Gram stained? b. When spore stained? 7. What is another way you can determine whether a organism was motile besides observing a flagellar stain? 8. Why can't you Gram stain a acid fast organism? Introduction to environment and microbial growth A organism cannot grow and divide unless it is in a favorable environment. Environmental factors include temperature, availability of nutrients, moisture, oxygen. Exercise 6 Pure culture and aseptic technique Aseptic Free of contamination. Incubate is store cultures under conditions suitable for growth often in a incubator. Broth to broth transfer with a wire loop. Materials 2 Tubes of TS broth. Inoculating loop. use inoculating loop to place culture of bacteria on agar. Transferring broth with a pipet. Materials TS broth tubes from previous procedure. 1 ml pipet. bulb to fit on end of pipet. Streak plate technique materials 2 Trypticase soy agar TSA plate petri dishes. Cultures Broth culture containing a mixture of two organisms Micrococcus and Staphylococcus. Protect agar surface from contamination. mark petri dish with a pen. streak with a loop of bacteria. flame loop and cool. streak second section on petri dish. flame loop and cool. streak last section. clear turbid tube 1 tube 2 Questions 1. What is the definition of a pure culture? 2. Why is sterile technique important? 3. What is the purpose of a strak plate? 4. Why is it important to avoid digging into the agar with the loop? 5. Is there anything you can do to improve your streak plate technique? Exercise 7 Defined, Undefined, Selective and Differential media. Defined medium A synthetic medium composed of inorganic salts and usually a carbonsuch as glucose. Differential medium Medium permitting certain organisms to be distinguished from others by the appearance of their colonies. Selective medium Medium formulated to permit the growth of certain bacteria. Undefined medium A complex medium in which the exact amounts of components and their composition are unknown because it is made of extracts or enzymatic digests of meat, plants or yeast. Materials Cultures growing in trypticase soy broth. Escherichia coli Staphylococcus epidermidis Pseudomonas aeruginosa Enterobacter aerogenes Trypticase soy TS agar plate Glucose mineral salts agar plate EMB eosin methylene blue agar plate. 37 C for 48 hours. observe growth. Results E. coli Staphylococcus Pseudomonas Enterobacter Questions 1. Which organisms could not grow on the glucose salts medium? Which organisms could grow on it? 2. Which organisms do not require any growth factors? 3. did some organisms grow better on TS agar then glucose salts? 4. Which organisms could grow on the EMB agar? 5. Which organisms could ferment lactose? 6. Could you differentiate E. coli from other organisms growing on EMB? 7. In general EMB selects for what kind of organisms? 8. What kinds of organisms does EMB differentiate? Exercise 8 Quantification of microorganisms important to count bacteria. bacteria colonies. turbid is cloudy. viable bacteria growing and dividing. Materials A turbid suspension 1 99 ml water blank. 2 9.9 ml water blanks. 3 9.0 ml water blanks 2 1 ml pipets. 4 TS agar deeps 4 sterile petri dishes Suspension A A overnight TS broth culture without shaking of E. coli diluted 1:1 with TS broth. Results 1. Which plate dilution had between 30 and 300 colonies? 2. How many colonies did you count? 3. How many organisms ml were in the original suspension? Questions 1. How many organisms ml can be in a cloudy broth? 2. how many organisms ml can be in a clear broth? 3. What are two sources of error in procedure? 4. If you serially dilute a sample with three 1:10 dilutions what is the final dilution of the last tube? 5. If you add 1.0 ml to 99 ml of water what is the dilution of the sample? 6. If you had a solution containing 6,000 organisms ml how would you dilute and plate a sample so you had a countable plate? exercise 8 Understanding Dilutions 1. To make a dilution use the following formula sample/(diluent + sample) = the dilution. Example 1 How much is a sample diluted if 1 ml is added to 9.0 ml of water the water is sometimes called a diluent? 1/(1 + 9) = 1/10 also expressed 10 -1 Example 2 How much is a sample diluted if 0.1 ml is added to 9.9 ml of water? 0.1/(0.1 + 9.9) = 0.1/10 = 1:100 or 10 -2 To calculate the nuber of organisms in the original suspension use the formula The number of organisms /ml in the original sample = number of colonies on plate x 1/volume of sample x 1/dilution. Example 3 Suppose you counted 120 organisms on a plate diluted 10 -2. The sample size was 0.1 ml. Solution 120 number of organisms on plate x 1/0.1 x 1/10 -2 = 120 x 10 x 100 =120 x 10 3 or 1.2 x 10 5 organisms/ml. Example 4 Suppose you counted 73 colonies on the plate marked 10 -6. If the sample size is 1.0 ml then 73 x 1/1.0 x 1/10 -6 = 73 x 1 x 10 6 = 73 x 10 6 or 7.3 x 10 7. organisms/ml in the original suspension. it is important to label the answer per ml. tube 1 9ml 10 -1 (1/10) tube b 9ml 10 -2 (1/100) Exercise 9 Aerobic and Anaerobic growth Aerobic in the presence of air. Air contains about 20% oxygen. Agar deep a test tube filled with agar almost to the top. Anaerobic in the absence of air. Materials TS broth cultures labeled A B and C. Escherichia coli Micrococcus Clostridium 3 TS + 0.5% glucose agar deeps 3 TS agar slants session Boiling agar deeps for 10 minutes to drive out dissolved oxygen. Cool the agar in a 50C water bath for 10 minutes. Inoculate agar with a loop pf culture. 25c 48 hours incubation. observe the surface of the slants. Which cultures are aerobes and anaerobes. appearance Questions 1. Would you expect a obligate anaerobe to grow on a slant incubated aerobically? Why? 2. Which kind of organism would you expect to grow both in the agar deep and on the slant? 3. Which kind of organisms can grow aerobically on slants? 4. could grow through agar 5. Why did you boil the agar deeps longer than it took to melt the agar? 6. how were the obligate anaerobes grown in the broth? Reference from Nester et al. Microbiology A human perspective 4th ed. 2004 chapter 4 section 4.3. Exercise 10 The effect of incubation temperature on generation time. Every bacterial species has a optimal temperature for the fastest growth. Organisms associated with animals grow fastest at about 37C. Generation time is the time it takes for a population of cells to double. growth curve. Materials Cultures Escherichia coli TS broth cultures in log phase. 1 TS broth in a tube that can be read in a spectrophotometer or klett colorimeter. Prewarmed in a water bath. Water bath at 30C. Water bath at 37C. plot the data on semi log graph paper. Calculate the generation time for E. coli at each temperature. by selecting a point on the line and noting the OD. Find the point on the line where this number has doubled. The time between these two points is the generation time. Growth curve of cells growing in log phase at 37C and 30C. Temperature and Generation Time. 1.0 9 8 7 6 5 4 3 2 1-09 0 20 40 37 C temperature Time O.D. 0 .08 20 .126 40 .20 60 .33 80 .50 Generation time 30 minutes. 30 C temperature Time O.D. 0 .08 20 .110 40 .152 60 .210 80 .30 Generation time 40 minutes. Results Date your temperature _______ average temperature _____ Time Reading Time Reading 1 2 3 4 5 Generation Time E. coli at 37C. E. coli at 30C. Questions 1. What is the generation time of a organism? 2. Why is it important to keep the culture at the correct incubation temperature when measuring the generation time? 3. Why is it important to use cells in log phase? 4. If the growth of two cultures were plotted on semi log paper one slower than the other which would have the steeper slope? Exercise 11 Moist and Dry Heat sterilization Thermal death point and thermal death time Autoclaving is the most commonly used method of moist heat sterilization. pasteurization or boiling hot water. Typical sterilization times and temperatures are 2 hours at 165 C for dry heat. 15 minutes at 121 C for moist heat. Materials Cultures 24 hour 37C Escherichia coli cultures in 5 ml TS broth. Spore suspension in 5 ml of sterile distilled water of a 4-5 day 37C nutrient agar slant culture of Bacillus subtilis. 5 ml tube of TS broth. Two large beakers or cans one for use as a water bath and the other for use as a reservoir of boiling water. 1 hot plate. 1 thermometer. water bath. A vortex apparatus. using a hot plate. Results 1. Determination of thermal death point and thermal death time. Bacterial growth at Assigned temperatures and times Culture 40C 55C 80C 100C C 10 20 30 40 C 10 20 30 40 C 10 20 30 40 C 10 20 30 40 E. coli B.Subtilis C is control. 10 is in minutes a. Determine the thermal death time for each culture. Thermal death time minutes Escherichia coli __ Bacillus subtillus __ b. thermal death point 2. Evaluation of materials sterilized in laboratory with moist autoclave and dry heat hot air oven. a. List of materials sterilized with the autoclave. b. List of materials sterilized with the hot air oven. Questions 1. Discuss similarities and differences between determining thermal death point and thermal death time. 2. How would you set up a experiment to determine to the minute the TDT of E. coli? Begin with the data you have already collected. 3. A practical question related to thermal death time TDT relates to a serious outbreak of E. coli infection in early 1993 when people ate insufficiently grilled hamburgers. Assume that the thermal death point TDP is 67.2 C 157 F the temperature required required for cooking. What factors would you consider in setting up such a exoperiment? 4. What is the most expedient method for sterilizing a heat sensitive liquid that contains a spore forming bacterium? 5. List one or more materials that are best sterilized by following processes. a. membrane filtration b. Ultraviolet light. c. dry heat d. moist heat e. Tyndallization f. radiation 6. What are three advantages of using metal caps rather than cotton for test tube closures? Are there any disadvantages? 7. How would you sterilize a heat sensitive growth medium containing thermoduric bacteria? 8. Was the Bacillus subtillis culture sterilized after 40 minutes of boiling? If not what is necessary to assure sterility by boiling? Exercise 12 Control of microbial growth with ultraviolet light. Ultraviolet light is the component of sunlight that causes for sunburn. It can also kill microorganisms by acting on their DNA and causing mutations. It consists of very short nonionizing wavelengths of radiation 200 to 400 just below blue light 450 - 500 nm in the visible spectrum. Materials Cultures Suspension of Bacillus spores in sterile saline. E. coli in TS broth. Raw hamburger or soil mixed with sterile water. 3 TS agar plates. UV lamp with shielding. 18 to 36 inch fluorescent bulb. 3 sterile swabs. dark box for storing plates after UV exposure. Dip a sterile swab in Bacillus subtilis and rub on agar plate. Results 1. Record your observations for control and treated sides of petri dishes exposed to UV light at a distance of ___ cm for ___ minutes. Make a drawing of each plate. a. Plate containing Bacillus spores. b. Plate containing E. coli. c. Plate containing either a raw hamburger suspension or soil suspension. Indicate which one you used. 2. Which organisms were most resistant to UV? ___ Least resistant? ___ Questions 1. Why can't you use UV to sterilize microbiological media agar or broth? 2. How does UV cause mutations? 3. Give a possible reason some organisms in the soil or meat were able to grow after exposure to UV not others? 4. Frequently organisms isolated from the environment are pigmented while organisms isolated from the intestine or other protected places are not. Can you provide a explanation for this? 5. Mutations can lead to cancer in animals. Explain why persons living in the southern half of the United States have a higher incidence of skin cancer than those in the northern half. Exercise 13 Osmotic Pressure and its effect on the rate and amount of Microbial Growth. Osmosis is the process of flow or diffusion that takes place through a semipermeable membrane. When the solute concentration outside the cell is less than the solute concentration inside the cell a inward osmotic pressure occurs and water enters the cell. Osmotic pressure is the pressure exerted by water on a membrane as a result of a difference in the concentration of solute molecules on each side of the membrane. Materials Cultures Use TYEG salts agar slants for Escherichia coli Micrococcus luteus and Saccharomyces cerevisiae. Use American type culture collection ATCC medium 213 for the Preceptrol strain of Halobacterium salinarium. Incubate E. coli, M. luteus and S. cerevisiae cultures for 24 hours at 35C. Incubate H salinanium culture for 1 week at 35C. TYEG salts agar plates containing 0.5, 5, 10. and 20% NaCl 4 plates. TYEG salts agar plates containing 0, 10, 25, and 50% sucrose 4 plates. streak petri dish plates. incubate 8 plates at 30C. observe the plates for growth upto 1 week. Bacteria Colony Growth Colony color Colony size Colony texture sucrose % NaCl % 3. Microscopic examination of wet mounts of bacteria and yeast colonies showing marked changes in visual appearance from the controls. Questions 1. From your studies which organisms tolerate salt best? ___ Least? ___ 2. Which organisms tolerate sugar best? ___ Least? ___ 3. Compare bacteria and yeast with respect to salt tolerance. colonial and cellular appearance. 4. to sugar appearance. 5. What evidence did you find of a nutritional requirement for salt or sugar in the growth medium? 6. Matching 1. Halobacterium ___ osmosensitive 2. Saccharomyces ___ long generation time 3. escherichia coli ___ saccharophilic 4. Micrococcus ___ osmotolerant 7. Matching 1. Plasmolysis ___ isotonic solution 2. Plasmoptysis ___ hypotonic solution 3. Normal cell growth ___ hypertonic solution ___ swelling of cells Exercise 14 Antiseptics and Antibiotics moldy loaf of bread. a thin slice from the outside of the loaf was cut off mixed into a paste with water and applied to the wound. antibiotics are chemicals produced and secreted by microorganisms. bacteria, fungi, actinomycetes. Alcohol Penicillin materials Cultures Bacteria 24 hour 37 c TS broth cultures. Staphylococcus epidermis a gram positive coccus. vials of antibiotic discs pencillin ethanol hydrogen peroxide agar 6 plates Procedure antiseptics and disinfectants first session 1. divide the plate with a marking pen. 4 sectors. 2. Record codes for the 4 antisepticsand disinfectants underneath the plate. alcohol, ethanol, hydrogen peroxide. 3. Label the cover of 1 petri dish S. epidermis. and the cover of the other dish E. coli. 4. Suspend S. epidermis. streak swab on agar. 5. repeat for step 4 for E. coli. 6. Sterilize forceps by dipping them in 95% alcohol and then touch to the flame of the bunsen burner. Air cool. 7. Using forceps remove one of the filter paper discs from the container and dip it into solution 1: 70% ethanol. 8. Drain the disc. 9. Repeat steps 5 6 7 and place the disc in the center of quadrant 1 of the plate labeled E. Coli. 10. Repeat steps 5 - 8 for remaining 3 and use hydrogen peroxide. hexachlorophene. 11. Invert the petri dishes and incubate 37C for 48 hours. Antibiotics procedure first session. 1. Use broth culture. 2. divide agar sections into 6 pie shaped sections. 3. Record the codes of the seven antibiotic discs underneath dishes. 4. Label the cover for each plate. for each bacterium. 5. streak the first broth culture aseptic technique. 6. Heat sterilize forceps and remove a antibiotic disc from the container. place in pie sections on agar. 7. Continue placing the remaining six. 8. Repeat steps 5 -7 with remaining 3 cultures. 9. Invert and incubate plates at 37 C for 48 hours. Antiseptics and disinfectants second session. 1. Turn over S. epidermis plate and measure clear zone around each disc. 2. record your results. second session antibiotics. 1. Observe plates using the same method describe in step 1 of the second session for antiseptics and disinfectants. Note any large colony maybe resistant. 2. Record your findngs. 3. Compare all results, the susceptibility of your test cultures to the antibiotics as resistant R intermediate I or susceptible S. results 1. Fiter paper disc technique for antiseptics and disinfectants. Bacteriostatic activity of various antiseptics and disinfectants. Antiseptic or disinfectant zone of inhibition mm Staphylococcus epidermis Escherichia coli 70% ethanol e. 3% hydrogen peroxide HP Listerine L 0.3% hexachlorophene H others What general conclusions can you make from this study? What differences if any did you observe on your plates between antiseptic and disinfectant preparations. 2. Filter paper disc technique for antibiotics. Antibiotic susceptibility Test organism zone of inhibition mm Susceptibility Notes Chl Ery Pen Str Sul Tet Nys Chl Ery Pen Str Sul Tet Nys S. epidermidis E. coli P. aeruginosa M. smegmatis Questions 1. What relationship did you find if any between the Gram staining reaction of a microorganism and its susceptibility to antiseptics and disinfectants? 2. You may have noted that nystatin was not listed. The reason for its omission is that it is a antifungal antibiotic. Was it antibiotically active against any of the bacteria you studied? Is there a organism that could have been tested that might have been susceptible? 3. Name some other factors affecting the size of the zone of growth inhibition that were not included in your test. Why were they omitted? their importance? 4. To what general groups of organic compounds does hexachlorophene belong to? advantage disadvantage for scrubbing down cleaning? 5. Matching 1. 70% ethanol ___ antibiotic 2. 5% phenol ___ antiseptic 3. nystatin ___ coal tar dye 4. prontosil ___ disinfectant 5. sulfanilamide ___ drug Exercise 15 Selection of bacterial mutants resistant to Antibiotics. All the bacterial cells in a pure culture are derived from a single cell. Antibiotic is a substance produced by one organism usually a microorganism which kills or inhibits other organisms. Materials First session 2 flasks or bottles containing 50 ml TS broth. 2 TS agar deeps. 2 petri dishes. 2 1 ml pipets. Overnight broth cultures -18 hours of Escherichia coli K12 about 10 9 cells ml. Streptomycin solution at 30 mg ml. Second session 2 TS agar deeps. 2 petri dishes. 2 1-ml pipets. 2 tubes of 0.5 ml sterile water. 0.1 ml streptomycin. Procedure first session 1. Melt and place 2 TS agar deeps in a 50C water bath. 2. Label one petri plate and one flask with streptomycin. 3. Add 0.3 ml streptomycin to the flask labeled streptomycin and 0.1 ml to one of the melted cooled agar deeps. 4. inoculate the agar deep with 1 ml of the bacterial culture mix and pour in the plate labeled with streptomycin. 5. Add 1 ml of bacteria to the tube of melted cooled agar without streptomycin. 6. Add 1 ml of bacteria to each of the flasks. 7. Incubate the plates and flasks at 37C. Second session 1. Melt and cool two tubes of TS agar in 50C water bath. 2. Pour one tube of melted agar into a petri dish labeled without streptomycin and let harden. 3. Add 0.1 ml streptomycin to the other tube of melted agar pour into a petri dish labeled with streptomycin and let harden. 4. Examine the bottles and plates inoculated last period. 300 colonies would be to many to count. 5. Test the bacteria growing in the bottles and on the plates for sensitivity or resistance to streptomycin. 6. Dig a isolated colony out of the agar plate suspend in water and incubate at 37C. 7. which bacteria will be sensitive to streptomycin and which will be resistant. results after second session Source Growth no growth TS broth control TS broth plus streptomycin Source Number of colonies TS agar plate control TS agar plate plus streptomycin Results after third session Source Growth on TS Agar Plate Growth on TS Agar + Strp TS broth control TS broth plus streptomycin TS agar control TS agar plus streptomycin How many organisms ml were streptomycin resistant in the original overnight culture of sensitive E. coli? Questions 1. Two bottles of TS broth with and without streptomycin were inoculated in Session 1 with 1 ml of a overnight culture of E. coli. After incubation why was one population streptomycin sensitive and the other streptomycin resistant? 2. How were you able to estimate the number of streptomycin resistant organisms already present in the overnight culture of E. coli growing in the TS broth? 3. Why should antibiotics not be used unless they are necessary? 4. What is correct? a. A organism becomes resistant after it is exposed to a antibiotic. b. A antibiotic selects organisms that are already resistant. Exercise 16 Transformation a form of genetic recombination Transformation is used to transfer the genes of one bacterium to another. objective to understand the process of transformation and observe it in the laboratory. Materials 1 TSY Agar plate. 1 TSY Agar plate with streptomycin Broth culture of Acinetobacter Str r resistant to streptomycin Broth culture of Acinetobacter Str s sensitive to streptomycin Tube with 0.1 ml detergent SDS sodium dodecyl sulfate in 10 x saline citrate Solution of DNase 1 ml pipet. 60 C water bath with test tube rack. Procedure first session 1. Transfer 1.0 ml of Str R Acinetobacter broth culture into the tube of SDS. incubate at 60C water bath for 30 minutes. 2. divide the TSY agar plate into 5 sections. 3. Inoculate the plate by adding a loop of broth culture. 4. Incubate the plates at room temperature at 37C for 48 hours. second session 1. Observe the plate you prepared in the first session. There should be growth. 2. Divide the TSY. 3. Streak a loopful of cells from the first plate to the streptomycin plate. Cells growing. DNA. 4. Incubate at room temperature at 37C for 48 hours. third session 1. Observe the TSY + streptomycin agar plate inoculated last period and record results. Did you transform the cells sensitive to streptomycin to cells that were resistant and could now grow on streptomycin? Results Indicate growth + or no growth - in each sector on petri dishes. TSY Agar TSY Agar + streptomycin Yes or No Demonstrates observation Were Acinetobacter StrS cells sensitive to streptomycin? Were Acinetobacter StrR cells resistant to streptomycin? Was the DNA cell lysate free of viable cells? Did transformation take place? Did the DNase prevent transformation? Questions 1. What two components were mixed together to show transformation? 2. What is the action of DNase? 3. What control showed that transformation and not conjugation or transduction was responsible for the results? 4. If the STRs cells had grown on the TSY + streptomycin agar would you have been able to determine if transformation had taken place? 5. If you had used a DNA lysate containing viable cells would it have been possible to determine whether transformation had taken place? 6. How does transformation differ from conjugation and transduction? Exercise 17 Laboratory Report Bacterial conjugation Conjugation is one of the three mechanisms responsible for genetic transfer in bacteria. A Auxotroph is a organism. Conjugation is a method of transferring DNA between bacteria requiring cel to cell contact. F factor is F plasmid Genes giving the cell the ability to transfer DNA via conjugation. F- cells lacking the F factor. recipient bacteria. F+ cells containing the F factor. donor bacteria. Objective 1. To increase knowledge of concepts and techniques used in the study of genetics. 2. auxotrophic organisms Exercise 17 Materials TS broth culture in log phase of E. coli AF+ methionine - Hfr donor TS broth culture in log phase of E. coli BF- threonine- recipient 1 TS agar plate 1 Sterile test tube 5 Sterile 9 ml water blanks 4 Agar plates of mineral salts + 0.5% glucose mineral medium mm 6 Sterile 1.0 ml pipets 1 Sterile bent glass rod Procedure First session 1. Label plates and tubes. 2. Transfer 1 ml of culture A into the water blank labeled A 10-1. 3. Transfer 1 ml of culture B into the water blank labeled B 10-1. using a pipet transfer a second 1 ml of culture A into the tube labeled A + B. 4. Gently mix the A + B tube and incubate it wothout shaking for 30 minutes at room temperature. 5. Incubate mm plate labeled a and half of the TS agar plate labeled a with loopfuls of culture. 6. Incubate the agar plates at 37C for 2-3 days. Results Indicate where growth occurred. Culture A Culture B TS Agar MM Agar Culture a auxotroph? Plate counts A + B 10-1 A + B 10-2 A + B 10-3 Number of colonies 1. How many recombinant organisms resulted from the mating? 2. If there were 2 x 10 8 organisms in the A + B mixture what percent of the original mixture resulted in recombinants? 1 x 10 8 cells/ml 3. If the back mutation rate for threonine is 1 in 10 8 cells how many back mutants would you expect to be present per ml in the A + B mixture? Questions 1. Why can the same pipet and bent glass rod be used to inoculate plates when starting with the most dilute mixture? 2. Why is a auxotrophic organism not able to grow on MM agar? 3. If either A or B could grow on the MM agar how would that change the resuts? 4. Using the answers you calculated in part 2 in the results section how did the number of recombinants resulting from conjugation compare with the number you expect from back mutation? 5. Compare the effects of mutation and conjugation to produce organisms with new genetic capability? Exercise 18 Gene regulation Induction and catabolite repression A bacterial cell has all the genetic information to produce and operate a new cell. enzymes for obtain energy. Materials Mineral salts + 0.2 % glucose broth. 5 ml/tube Mineral salts + 0.2% lactose broth. 5 ml/tube Mineral salts + 0.2% glycerol broth. 5 ml/tube. 3 ml of ONPG ortho nitrophenyl B. galactopyranoside. Overnight Ts broth cultures of Escherichia coli cultures of Enterobacter and Klebsiella. Procedure first session 1. Inoculate each tube with a drop of E. coli. 2. Incubate at 37C for 48 hours. second session 1. Examine the tubes for growth. 2. Add 1 ml of ONPG to each tube. 3. Incubate at room temperature for 30 minutes. 4. Examine the tubes to determine if the broth has turned yellow. 5. record the results Catabolite repression materials Cultures Bacillus growing on TS agar slants. Nutrient + starch agar plate. Nutrient + starch + glucose agar plate. Gram's iodine. Procedure first session 1. Label each plate. 2. Inoculate the middle of each plate with the Bacillus in a area a few mm square. 3. Incubate at 30C for 1 or 2 days. second session 1. Flood the agar plate with Gram's iodine. The starch will turn purple. 2. record the results. Results of induction Glucose Lactose Glycerol Color B galactosidase present? Results of catabolite repression Starch Starch + Glucose Zone of clearing present Presence of amylase Questions 1. Which substrate induced B galactosidase? 2. What reaction produced the yellow color? 3. What results would you expect if B galactosidase were a constitutive enzyme? 4. In the catabolite repression exercise did the Bacillus have the capacity to synthesize amylase amylase + ? How did you determine? 5. Did you observe catabolite repression when glucose was added to the starch? How did you determine? 6. What results would you expect if amylase were a constitutive enzyme? Exercise 19 page 157 Fungi materials culture Sabouraud's dextrose broth cultures 48 hr 25 C of Saccharomyces cerevisiae and Candida albicans. pathogen. dextrose agar slant cultures 48 hr 25 c candida albicans. covered slide cultures of the three aabove filamentous fungi on Sabouraud's dextrose agar 3-5 days 25C. Tubes of glucose, maltose, and lactose broth containing Durham tubes. 2 tubes of each Glucose acetate yeast sporulation agar 1 plate. cornmeal agar 1 plate. Test tube 12 by 75 mm containing either 0.5 ml of serum or raw nonsterile 1 egg white. 4 sterile droppers. tweezers. Dissecting microscopes. mm ruler. Dropping bottle containing methylene blue. Procedure first session Suspend the broth cultures of Saccharomyces cerevisia S.c. and Candida albicans C.a. second session 1. Yeast fermentation study. Examine tubes for a. Presence or absence + or - of cloudy broth growth. b. Presence or absence + or - of gas in the inverted Durham tube. c. Change in color of the pH indicator dye. A change to a yellow color is indicative of acid production. sabourauds agar plate 5 7 days observing for growth of yeast organisms. colony diameter mm, colony surface smooth or rough. flat or raised. color of colony 5. Why would the growth of a pellicle or film on the surface of a broth growth medium be advantageous to the physiology and viability of that yeast? 6. What are some ways in which you might be able to differentiate Rhizopus nigricans from Aspergillus niger simply by visually observing a petri dish culture? 7. How can you determine whether or not a green woolly looking colony is a Aspergillus or Penicillium? 8. What problems might you have in identifying a pathogenic fungus observed in a blood specimen? What might you do to correct such problems? 9. In what ways can we readily distinguish a. fungi from algae? b. fungi from bacteria? c. fungi from actinomycetes? 10. Define a opportunistic fungus. Are all medically important fungi opportunistic? 11. Name three pathogenic fungi that exhibit dimorphism. Describe the type of dimorphism each exhibits and the laboratory conditions necessary to elicit it. Exercise 19 Laboratory Report Microscopic identification of Fungi Results nonfilamentous fungi 1. Fermentation study examine tubes and record results + or -. C. albicans and S. cerevisiae fermentation activity in tubes of broth containing different carbohydrate sugars. Yeast strain Glucose Maltose Lactose Cloudy Gas Acid Cloudy Gas Acid Cloudy Gas Acid C. albicans S. cerevisiae 2. Yeast colonial and Vegetative Cell morphology study. a. Colony characteristics C. albicans and S. cerevisiae colonial characteristics on Sabouraud's dextrose agar plates. yeast strain colonial morphology Colony color Consistency Diameter mm Surface appearance edge appearance Candida albicans Saccharomyces cerevisiae b. Vegetative cell morphology and enter results below. Candida albicans Saccharomyces cerevisiae 3. Sexual Sporulation study S. cerevisiae. Drawings of asci and ascospores. 4. Chlamydospore formation study. C. albicans. Drawings of chlamydospores. 5. Germ tube formation study C. albicans. Drawings of germ tubes. 6. Colonial characteristics of Rhizopus, Aspergillus, and Penicillium when grown on Sabouraud's dextrose agar. Colonial characteristics of three filamentous Fungi cultured for ___ days on Sabouraud's dextrose agar. Rhizopus Aspergillus Penicillium Colony color Colony diameter mm Colony texture Colony convolutions Colony margin Soluble pigments in agar 7. Drawings of their asexual reproductive structures. Rhizopus Aspergillus Penicillium 8. Drawings of their asexual spores. Rhizopus Aspergillus Pencillium Questions 1. List four ways of differentiating Candida albicans from Saccharomyces cerevisiae. 2. What are two ways in which you were able to differentiate pathogenic from nonpathogenic Candida species? 3. Explain the physiological differences between yeast fermentation and yeast assimilation of glucose. 4. Why is a loop rather than a pipet used to inoculate the sugar fermentation tubes? Fungi 1. Yeast fermentation study. Inoculate each of the carbohydrate fermentation tubes glucose, lactose, maltose with a loopful of S.c. repeat using fresh tubes with C.a. incubate for 48 hours 25c to 30c. again repeat for sabouraud dextrose agar plates. Caution Never smell fungus cultures spore inhalation may cause infection. spore dispersal. Exercise 20 page 169 Parasitology Protozoa and Helminths parasitic life As a introduction you will have a opportunity to observe the movements and structure of some living nonparasitic protozoans and worms often found in pond water. parasites tapeworms Materials Cultures Living cultures of a Paramecium species Amoeba proteus and a Dugesia species fresh sample of stagnant pond water Planaslo solution 1 or more dropping bottles Depression slides following commercially prepared slides Subkingdom Protozoa Phylum Sarcodina 1. Examination of free living cultures. Pond water examination. oil immersion and view pond water on a depression slide. examine a free living flatworm and amoeba proteus. Name __ Results 1. Examination of free living cultures a Pond water examination Descriptions of movements and drawings of any protzoans found in pond water. b. Examination of fresh samples of a free living amoeba paramecium and flatworm. description of movements and drawings with labels. 2. Examination of stained slides for trophozoites and cysts Prepare drawings of the trophozoite and cyst stage of either Entamoeba histolytica or Giardia lamblia label. 3.Examination of Protozoans Present in stained blood slides. a. Examine blood smears of Plasmodium vivax. b. Blood smear of Trypanosoma gambiense. 4. Comparison of a free living worm with its parasitic relative. a. Comparison of Dugesia species free living with Clonorchis sinensis parasitic. b. Study of a parasitic tapeworm Taenia species. 5. Life cycle of Schistosoma mansoni and possible methods of control. a. For each space in this life cycle the appropriate stage using the prepared microscope slides. in the human host in the water free living larva in the snail host sporocyst stage miracidium stage b. Propose a plan for public health control of schistosomiasis. various strategies interupt cycle. Question 1. Which form the trophozoite of the cyst is most infective when found in a feces sample? 2. In what ways are free living and parasitic worms similar they can be identified as closely related? 3. In what ways do the parasitic species differ from the free living planaria? Planaria Fluke Tapeworm Outside covering Organs of attachment Sensory organs digestive system reproduction 4. Estimate the length and width of a trypanosome. 5. How is the Echinococcus tapeworm transmitted to humans? Does it cause a serious disease? What are two ways in which its transmission to humans can be prevented? Exercise 20 page 175 materials Plasmodium vivax ring amoeboid schizont stages Subkingdom helminths worms Phylum platyhelminths flatworms class turbellaria free living Dugesia species class trematoda flukes schistosoma mansoni adult male adult female ovum egg ciliated miracidium infective ciliate cercaria sporocyst stage in snail liver tissue clonorchis sinesis class cestoda tapeworms taenia solium trophozoite Procedure 1. Pond water examination on a slide. 2. oil immersion amoeba proteus view. 3. Examination of protozoans present in stained blood slides. 4. Plasmodium vivax and located blood cells containing the parasites. merozoites form into parasites after a mosquito bite. they penetrate into the the blood and red blood cells. 5. Comaprison viewing of a free living worm. observe a prepared slide of a free living worm. Exercise 21 Prokaryotic viruses bacteriophage isolation and titering Bacteriophage phage are viruses that infect bacteria. Phage are too small 200 nm to be seen in a light microscope. can be detected if grown on a bacterial lawn. Phage and host cells are mixed in a small tube of soft agar and then poured on top of a agar base plate. bacterial lawn the confluent growth of bacteria on a agar plate. attempt to isolate phage that infect Escherichia coli. Materials first session raw sewage filtered through 0.45 um membrane filter. Host bacterial Escherichia coli K12 in a late log phase OD - 0.4. Tubes of 4 ml soft tryptone overlay agar 2 Tryptone agar base plates 2 Bacteriophage isolation and culture filtered sewage. filtered sewage may contain harmful animal viruses. handle with extreme care. First session 1. Label plates 1 ml and 0.1 ml. 2. Melt overlay agar in boiling water and place in 50 C water bath for at least 5 minuts. 3. Add 1 ml filtered sewage to one tube overlay. 4. Add o.1 ml filtered sewage to other tube of soft overlay. 5. add several drops of E. coli to each tube. Mix tubes and pour onto labeled tryptone agar base plates. 6. Permit to harden for 5 minutes. 7. Incubate inverted at 37 C overnight. Second Session 1. Examine the plates for plaque formation. Notice any different types of plaques and their relative sizes due to different kinds of phages. Count the number of each kind. Third session 1. Examine plates for plaques that have increased insize. 2. If you do suspect Bdellovibrio stab the plaque with a loop and prepare a wet mount. Look for very small motile bacteria. 3. Record results. Bdellovibrio can only reproduce in actively diving cells. Titering a Phage suspension materials Host bacteria for phage in late log phase Escherichia coli phage T4 suspension 9 ml tryptone blanks 4 4 ml overlay agar tubes 4 Tryptone agar base plates 4 Sterile 1 ml pipets 5 Procedure First session 1. Label four 9 ml tryptone blanks 10-1 10-2 10-3 10-4 2. Transfer 1 ml of the bacteriophage to the tube labeled 10-1 with a sterile 1 ml pipet. 3. Mix and transfer 1 ml of the 10-1 dilution to the 10-2 tube and discard pipet. 4. mix and transfer 1 ml to the 10-3 tube and siscard pipet. 5. Mix and transfer 1 ml to the 10-4 tube and discard pipet. 6. Label four tryptone hard agar petri plates. 10-1 10-2 10-3 10-4 7. Melt four tubes of soft overlay agar and place in a 50 C water bath let cool for 10 minutes. 8. Add 0.1 ml or several drops of E. coli broth to each tube of melted overlay agar. 9. Starting with the most diluted phage tube 10-4 add 1 ml to the overlay agar and pour on the tryptone agar base plated labeled 10-4. 10. Using the same pipet add 1 ml of the 10-3 dilution to a tube of overlay agar and pour into the plate labeled 10-3. 11. Repeat for the 10-2 and 10-1 phage dilution. 12. Incubate the plates inverted at 37C after the agar has hardened. Second session 1. Examine the plates. Select a plate containing between 30 and 300 plaques. 2. Estimate the numbers on the other plates. 10. 3. To determine the titer use formula No. of plaques x 1 dilution x 1 ml of sample = plaque forming units ml. Example If 76 plaques were counted on the 10-4 dilution then 76 x 1/10-4 x 1/1 = 76 x 10 4 pfu/ml. 4. Record results. exercise 21 Results 1. Isolation and Culture from filtered sewage. a. How many different types of plaques observed? ___ type 1 appearance ___ number ___ type 2 appearance ___ number ___ type 3 appearance ___ number ___ b. Did any plaques increase in size after reincubating? ___ If yes were small, very motile bacteria from the edge of the plaque observed? 2. Titering a phage suspension Dilution Control 10-1 10-2 10-3 10-4 numbers of plaques a. Which dilution resulted in a countable plate? b. Did the number of plaques decrease 10-fold with each dilution? c. How many phage ml were in the original suspension? Questions 1. Why was the sewage sample filtered? 2. How can you distinguish a lytic phage from a temperate phage when observing plaques from the filtered sewage sample? 3. Why can a plaque be considered similar to a bacterial colony? appearance, size, shape 4. Why do plaques formed by Bdellovibrio continue to increase in size after 24 hours not plaques formed by phage? Exercise 22 Normal skin flora The organisms growing on the surfaces and in the orifices of the body are called normal flora. Normal flora prevent harmful organisms from colonizing the skin. Staphylococcus epidermidis is a gram positive coccus is found on the skin. Objectives To learn to identify organisms making up the normal skin flora. To understand the importance of skin flora. To learn about the anaerobe jar. materials First Session TSY agar plates or TSY contact plates 2 inches in diameter 2 GasPak anaerobe jar or other anaerobic system. Sterile swab. Sterile saline. 70% ethanol cotton balls. Second Session Tubes of sterile water 0.5 ml tube 6. TSY + glucose agar deeps yeast extract and glucose are added to TS agar to encourage the growth of Propionbacterium 4. TSY + glucose + bromcresol purple agar slants 2. Magnifying glass is optional helpful Third Session Plasma Caution Staphylococcus aureus part of normal flora is a pathogen. First Session 1. Saturate a cotton ball with 70% ethanol and rub the forehead for 20 seconds. This will remove any transient organisms you might have on your skin, which are not part of your normal flora. 2. Let the forehead dry for 20 minutes. 3. Moisten a sterile swab with saline and rub it on a area of your forehead the size of a quarter for 15 seconds. 4. Swab the first third of a TsY agar plate discard the swab and finish the streak plate with a loop. 5. Repeat the procedure swabbing a second TSY agar plate. 6. Incubate one of the TSY agar plates aerobically at 37 C. Incubate the second TSY agar plate anaerobically in a GasPak or other anaerobe jar at 37 C. 7. After 48 hours of incubation should store the aerobic plate at room temperature to prevent the plates from drying out. Staphylococcus and Micrococcus can be observed after 48 hours. Propionibacterium must be incubated five days before colonies can be seen. Second Session 5 days later 1. Examine the aerobic TSY plate and circle two different colony types with a marking pen on the bottom of the plate. Make a gram stain of each circled colony. 2. If the colonies are gram positive cocci suspend the remainder of the colony used for the gram stain in 0.5 ml sterile saline. Use this suspension to inoculate. a. a glucose + bromcresol purple TS agar slant. b. a cooled melted agar deep. Incubate at 37C. 3. Examine the plate incubated in the anaerobe jar. Some colonies will be visible. Third session 5 days later 1. Observe the glucose + bromcresol slants. If the organism is able to ferment glucose the acid produced will turn purple agar yellow. 2. Observe the agar deeps. Obligate aerobes are only able to grow on the top while facultative anaerobes will grow throughout the entire tubes. coagulase test Place a drop of water on a slide and make a very thick suspension of cells from a yellow colony. Place a drop of plasma next to it and mix the two drops together. Look for clumping. clumped cells indicate a coagulase positive result. Drop the slide in boiling water and boil for a few minutes to kill the organisms before cleaning the slide 3. Identify your isolates. yellow colonies clusters. Staphylococcus are Gram positive cocci in clusters. S. aureus yellow Micrococcus are gram positive cocci. Staphylococcus epidermidis gram stain + cocci colony color whte gray Staphylococcus aureus yellow Micrococcus yellow Propionibacterium + rods white buff. Exercise 22 Laboratory report normal skin flora Results isolate 1 isolate 2 isolate 3 isolate 4 gram stain colony appearance glucose fermentation agar deep possible identity what organism seemed most numerous in amount Questions 1. How could normal skin flora be helpful to the host? 2. Why did you wipe your forehead with ethanol before sampling it? remove other 3. Why was Staphylococcus the only organism that could grow on both plates? 4. How can you distinguish Staphylococcus from Propionibacterium in a gram stain? 5. Why does Staphylococcus cause more contamination than Propionibacterium? Are most agar plates incubated aerobically or anaerobically? Exercise 23 Exercise 25 Gram stain of original specimen _____ describe cell shape, arrangement, and gram reaction. Gram stains of TS agar subcultures _____ describe cell shape, arrangement, and gram reaction. Test organism 1 organism 2 Colony description trypticase soy agar or blood agar gram stain ___ colony appearance macconkey or EMB ___ special stains ___ ___ capsule ___ ___ endospore ___ ___ lactose fermentation glucose fermentation sucrose fermentation mannitol fermentation Indole production methyl red voges proskauer citrate utilization urea hydrolysis motility catalase test coagulase oxidase final report 1. What is the identification of your organisms? Discuss the process of identification reason for choosing specific tests, and other comments. Organism 1. Organism 2. Exercise 25 Clinical unknown identification Your unknown specimen represents either a urine infection or a wound infection. Materials first session Unknown mixture labeled with hypothetical source. Blood agar plate or TS agar plate. MacConkey agar plate or EMB agar plate. Mannitol salt agar plate. second and third sessions TS agar plates Nutrient agar slants. Citrate agar slants. Urea slant. Glucose + bromcresol purple agar slants. Fermentation broths of glucose lactose sucrose. MR VP broth for the Voges Proskauer and methyl red test. Tryptone broth for the indole test. Kovacs reagent. Voges Proskauer reagents A and B. Methyl red for methyl red test. Plasma for coagulase test. Staining material for endospores and capsules. H2O2. first session 1. Make a gram stain of the broth culture. observe microorganisms. 2. Inoculate a complete medium agar plate TS agar or blood agar. MacConkey agar can be used for negative rod. Streak the plates for isolated colonies. 3. Incubate at 37 C. Second session 1. Examine the streak plates after incubation and identify organisms. Using stains. 2. Restreak each organism on a complete medium for isolation. Third session Observe the plates after incubation. 2. Try to identify which colonies are Staphylococcus Micrococcus. Bacillus gram positive rod Exercise 26 Differential White Blood Cell Stains the cellular forms of the immune system. Differential blood stains are important in disease diagnosis. objective to prepare 2 stained blood slides. Materials outdated blood bank whole blood human blood smears whole blood 3 new microscope slides. plastic droppers for dispensing blood. Wright's stain dropping bottle. Phosphate buffer pH 6.8 dropping bottle. Wash bottle containing distilled water. A Coplin jar with 95% ethanol. staining rack. Procedure. 1. Prepare three clean microscope slides. clean with 95% alcohol. 2. Place a drop of blood on one end of a clean slide. 3. Spread the drop of blood on the slide. 4. Flood or add 15 drops of Wright's stain to each blood smear. 5. add a equal volume of phosphate buffer. 6. Let stand until a green metallic scum forms on the surface of the slide. 7. Wash off the stain with water. 8. Examine blood stain smears. oil immersion lens WBC basophils eosinophils . a differential white blood cell count. Eexercise 28 Use of microbial antigens for diagnosis of infectious diseases. Materials Use of Proteus Antigens to Detect Rickettsial antibodies test. For use with rapid slide and tube test. Proteus OX-2 or OX-19 antigen and antiserum. Suitable light source for observing agglutination gooseneck fluorescent or fiber optic lamp A rubber bulb for pipetting serum and antiserum. Isotonic saline 0.85g NaCl 100 ml distilled water. For use with rapid slide test. 3 clean microscope slides. toothpicks. dropper delivering 0.03 ml pipets 0.2 ml For use wth tube test. test tubes 10 x 100 mm 10 Test tube rack for holding test tubes. sterile 5 ml pipets cotton plugged. Water bath 37C. Use of a nonmicrobial antigen USR antigen and test control serum. negative control serum. Hypodermic needle 18 gauge 1/45 ml per drop. syringe luer type 1-2 ml. alcohol and acetone for rinsing syringe with needle. 2 clean microscope slides. pipets 0.2 ml cotton plug 3. ruler mm. Procedure The rapid slide test 1. mark three clean microscope slides with two 16 mm 5/8" circles per slide. 2. Using a 0.2 ml pipet amounts of proteus antiserum into first five circles. 0.08 0.04 0.02 0.01 0.005 ml. sixth circle add 0.08 ml of 0.85% saline. 3. add one drop of proteus antigen to each slide with a pipet. 4. Mix each antiserum antigenwith toothpick. 5. Hold the slide and gently rotate 15-20 times. 6. Observe for macroscopic agglutination clumping. 7. Record the amount of clumping. 1 + approximately 25% cell clumping. 2 + approximately 50% cell clumping. 3 + approximately 75% cell clumping. 4 + complete agglutination. 4. The rapid slide agglutination test is defined as a qualitative method. Why then are a series of doubling dilutions evaluated for use with this method? 5. Discuss ways in which the rapid slide agglutination test differs from the USR agglutination test. Consider variables test antigens and observation techniques. 6. Discuss the pros and cons of using nontreponemal antigens cardiolipin rather than treponemal antigens for syphilis diagnosis. questions 1. What titers did you find in the bacterial agglutination test? __ rapid slide __ tube dilution. 2. Why is a positive titer with the Weil felix test not necessarily conclusive the patient has a rickettsial infection? 3. What other tests would be necessary to confirm a rickettsial infection? Difco manual for information. Exercise 29 2. What are some reasons for including positive and negative controls? 3. What advantages are there of the latex slide test over the capillary precipitin test for Lancefield grouping of pathogenic streptococci? 4. Streptococcus pneumoniae bacteria often possess cell wall surface antigens that react with Lancefield group c antiserum Slifkin and Pouchet Melvin 1980. In this event how would you determine if the positive agglutination test is due to a group c Streptococcus or to S. pneumoniae? Exercise 26 Laboratory report differential white blood cell stains Results 1. Color drawings of RBCs and various WBCs found in blood smears stained with Wright's stain. RBCs Neutrophils Eosinophils Basophils Monocytes Lymphocytes 2. Differential WBC count. record the kinds of leukocytes found as you examine each microscopic field.After counting 100 WBCs calculate their percentages from the totals found for each type. Also record the number of microscopic fields examined to find 100 WBCs. neutrophils Eosinophils Basophils Lymphocytes Monocytes Total Percent Questions 1. What problems if any did you find in preparing and staining your blood smears? Indicate any differences noted between thin and thick smears? 2. Were your blood stains satisfactorily? Did the stain cells resemble those in figure 26.1 and color plate 23? Were they better? 3. Did your differential white blood cell count percentages compare with the percentages in normal blood table 26.1? If not give a explanation. 4. Were there any WBC types that you did not find in your blood smear? If so which ones? Why did you not find them? 5. Matching 1. Neutrophils ___ Involved in antibody production 2. Basophils ___ A minor phagocytic cell. c. Monocytes ___ Increased number in parasitic infections. d. Eosinophils ___ Largest WBC. e. Lymphocytes ___ Inflammatory WBC. Exercise 27 Materials Nutrient broth cultures 37C 24 hours of Staphylococcus epidermis and E. coli. Sabouraud's dextrose broth culture 25 c 24 hour of saccharomyces cerevisiae. Melted nutrient agar 2 deeps. sabourad's dextrose aga deep held in a 48C water bath. 4 sterile petri dishes. petri dish containing 9 sterile filter paper discs 1/2" diameter. petri dish containing 1-2 ml of aseptically prepared raw egg white. petri dish containing 1-2 ml of lysozyme chloride sigma cat# L-2879 with activity of 60,000 units per mg of protein. Diluted 1:10 with sterile distilled water. Test tube containing 9 ml sterile distilled water. Raw onion. knife. mortar and pestle. tweezers. Sterile 1 ml 4 pipets. ruler mm. 4. Procedurally what additional important step would be necessary to evaluate the lysozyme activity of nasal secretions? Why? 5. Why are most gram negative bacteria not lysed by lysozyme yet they have a peptidoglycan cell wall structure similar to that of the gram positive bacteria? Exercise 29 grouping of pathogenic Streptococci materials Fresh 24 hr 37C unknown cultures of S. pyogenes group A and S. pneumoniae on blood agar plates labeled 1 and 2. Fresh TS broth cultures of unknown 1 and 2. diluted cell wall extraction enzymes 0.3 ml. aliquots contained in 3 serological test tubes. One vial with dropper of a 1% latex bead suspension coated with strep Gr A antibodies prepared in a glycine buffer. A vial coated with strep Gr B antibodies. A vial of polyvalent positive control antigen a extract of strep Gr A B C F and G. Sterile physiological saline 0.85% NaCI. Calibrated 1 ml pipets 2. toothpicks. plastic droppers. 37 C water bath. vortex for mixing tubes. high intensity incandescent light source. mechanical rotator for slides. mm ruler. 1. Prepare gram stains of both the broth and agar unknown cultures. use oil immersion when examining. 2. If the gram stains are indicative of streptococci gram positive cocci in pairs or chains. Laboratory report 1. Drawings of unknown gram stained bacteria seen with the oil immersion objective. unknown #1 blood agar and TS broth. unknown #2 blood agar and TS broth 2. Describe the type of hemolysis found on blood agar. Unknown #1 Unknown #2 3. Record the latext agglutination reactions + or -. Test antigen Lancefield group a group b 1. Unknown #1 2. Unknown #2 3. Negative control 4. Polyvalent + control 4. From the three studies morphology, hemolysis, latex agglutination which unknown did you identify as S. pyogenes _______. and S pneumoniae _______. 5. Were all of your findings consistent with the literature? If not describe any inconsistencies observed and if possible provide a explanation. Questions 1. Why are group d streptococci not included in the polyvalent positive control? Exercise 30 use of enzyme linked immunosorbent assay. Test for Coccidioides immitis identification The ELISA test is used for identification of plant and animal pathogens. including viruses. Materials A Meridian diagnostics inc. premier Coccidioides EIA package insert contains Antigen coated microwells 96. breakaway plastic microwells each coated with a mixture of TP tube precipitin and CF complement fixation antigens. positive control 2.7 ml prediluted positive human serum with a preservative. do not dilute further. Exercise 30 Questions 1. Discuss the test results and their significance Visual observation of processed samples sample yellow color IgM IgM control IgG IgG control Sample Absorbance at 450 nm with a plate reader Sample Absorbance Negative control IgM IgG 2. Discuss laboratory safety considerations related to handling of a. Antigen coated microwells b. Positive serum control c. Immunoglobulin enzyme conjugates d. Urea peroxide e. Tetramethylbenzidine 3. What is the importance of rinsing when conducting the ELISA test? 4. Why is the ELISA test for coccidioidomycosis a more definitive test than the Ouchterlony immunodiffusion test? Is the latter test still of value as a diagnostic tool? 5. Discuss the pros and cons of using the enzyme linked immunosorbent assay ELISA as used here and the double antibody sandwich ELISA assay. Exercise 31 page 261 A ouchterlony double immunodiffusion test for Coccidioides immitis identification. Precipitin reaction tests such as the Ouchterlony test are widely used for serodiagnosis of fungal diseases. Double diffusion precipitin reactions observed in agar gel plates. Materials A Meridian Diagnostics Inc. test kit. 603096 which contains 0.5 ml Coccidioides ID antigen Diluted anti coccodioides ID control serum Immunodiffusion agar plates Capillary pipets with bulb Moist chamber a dish with a tight fitting cover containing moist paper toweling is satisfactory provided the ID plates remain level and hydrated during the incubation period. 60 watt desk lamp distiled water Hold the petri plate close to desk lamp or reading light. 1. Using a capillary pipet with a attached bulb fill the pipet 1/4 full with Coccidioides antigen. 2. Next fill the center well with Coccidioides antigen. 3. Using a fresh capillary pipet repeat steps 1 and 2 with positive control antiserum. 4. If negative control sera or positive control sera from patients known to have a Coccidioides infection are available they can be added to other external wells of the ID agar plate 5. Identify the contents of each well. antiserum ws diluted and how much? 6. Place your name and date on the ID plate cover. Incubate it in the moist chamber at room temperature for 24 to 48 hours. 7. After 24 to 48 hours incubation read and record the ID bands. A light source is preferred for observing the nature of the bands. orientation of the bands bent bands. Bands on agar. Introduction to the prevention and control of communicable diseases. Communicable, or infectious diseases are transmitted from one person to another. Transmission is either by direct contact with a previously infected person example by sneezing, or by indirect contact with a previously infected person who has contaminated the surrounding environment. A classic example of indirect contact transmission is a epidemic of cholera that occurred in 1854 in London. During a 10 day period more than 500 people became ill with cholera and subsequently died. Anthrax from bacillus anthracis killed 5 people in U.S.A. in 2001. Epidemic The occurrence in a community or region of a group of illnesses of similar nature clearly in excess of normal expectancy. Name ___ Exercise 31 Laboratory report A ouchterlony double immunodiffusion test for Coccidioides immitis identification Results Date ID plate incubated __ Date observed __ 1. Make drawings of all precipitin lines. 1 2 3 4 5 6 7 Test well pattern of the immunodiffusion agar plate. 2. Ouchterlony Fungal immunodiffusion analysis form Well No. Well description Reading observed 1 Psitive control serum 2 3 4 5 6 7 Coccidioides Indicate under well description in which wells if any the serum was diluted. Test is invalid if positive control is negative after 24 hour incubation. 3. Discuss your results and their significance. 4. What future studies might be suggested from your results? Exercise 32 Epidemiology A Staphylococcus Carrier Study Materials TS broth 24hr 37 C reference culture of S. aureus with its antibiogram. Mannitol salt agar plates with phenol red indicator 7. 9 sterile swabs each in a test tube. 2 tongue depressors. 2 tubes of sterile water. 2 blood agar plates. 4 tubes of TS broth. tubes containing 0.5 ml of coagulase plasma 2 3 sterile pasteur pipets. mueller hinton agar 3 plates. a dropper bottle containing fresh 3% hydrogen peroxide. 2 forceps. mm ruler. antibiotic discs 3 of each of the following. penicillin 10 ug erythromycin 15 ug. streptomyin 10 ug tetracycline 30 ug. sulfanilamide 300 ug chloramphenicol 30 ug. S. aureus can cause wound infections. food poisoning, toxic shock syndrome. First session 1. Assemble 3 mannitol salt agar plates nose throat skin label. 2. label 3 test tubes. 3. moisten the swab from tube labeled nose. 4. repeat with throat. 5. Repeat step 3 with the swab labeled skin rub it over the skin surface located between your fingers and finger tips return the swab to the empty tube. 6. Streak each swab over 1/3 of the surface of the appropriately labeled mannitol salt agar plate. Continue to streak with a loop to obtain isolated colonies. 7. Repeat steps 3 through 6. 8. Streak a loopful of the S. aureus refrence culture on the surface of a mannitol salt agar plate. 9. Invert plates and incubate at 37 C until next session. second session 1. Examine the reference strain mannitol salt agar plate and note the appearance of typical S. aureus colonies large opaque colonies. yellow hallows. 2. Select several typical appearing staph colonies and test for catalase production. A positive catalase test is only suggestive of the presence of S. aureus. 3. Twenty four hours before the next laboratory session. subculture a colony from your body isolate mannitol salt agar plate. subculture the S. aureus reference strain. They should be subcultured in two tubes of TS broth and on a plate of blood agar. Both your body culture and the reference strain can be streaked on one blood agar plate by dividing the plate into halves. Label the tubes and plates and incubate at 37C for 24 hours. 4. Determine the coagulase activity of another catalase and mannitol positive body plate colony and the reference colony culture. Use the method outlined in exercise 22. Third Session 1. Record the presence or absence of hemolysis on the blood agar plates in table 32.1. Compare it for similarity with the reference strain of S. aureus. 2. If the coagulase test is negative report your culture as negative for S. aureus. turn in filled out sheet to instructor. 3. If the coagulase test is positive set up a antibiotic susceptibility test. The six antibiotic discs to be used are listed in the materials section. The inocula to be used are the three TS broth cultures prepared in the second session. The procedure is as follows. a. with a permanent marking pen divide the bottom of a petri dish containing Mueller Hinton agar into six pie shaped sections. Repeat the marking procedure with the remaining two dishes of Mueller Hinton agar. Label one dish Reference culture. The remaining two dishes represent body cultures. b. Inoculate the reference plate by moistening a sterile swab with the TS broth reference culture and spreading it uniformly over the plate surface by moving the swab back and forth in three directions. Repeat the inoculation procedure with the remaining two plates using your two TS broth cultures as inocula. c. Assemble the discs and record the code of each in part 3 of the laboratory report. d. Heat sterilize forceps by dipping in 95% alcohol and flaming. Air cool. e. Remove one disc aseptically from container. Place gently in the center of one pie shaped section of the reference plate culture. Tap disc gently with forceps to fix it in position on the agar surface. f. Continue placing the remaining five discs in the same way. Make certain that you sterilize the forceps after placing each disc since there is a possibilty of contaminating stock vials with resistant organisms or even occasionally with drug dependent bacteria. g. Repeat the procedure with your two body culture plates. h. Invert and incubate the plates at 35 C for 48 hours. Fourth session 1. Using a mm ruler measure the diameter of the zone of inhibition around each antibiotic disc and record their diameters. Consult table 14.2 and determine from the information if the cultures are susceptible S or resistant R to the antibiotic in question. Record s or R in the appropriate square of table 32.1. The reference culture is expected to be susceptible to all six antibiotics. 2. Fill out the information sheet of the laboratory report. Turn in to instructor for tabulation of data to be inserted in table 32.2. Name ___ Date ___ Session ___ instructor information sheet for tabulation of Staphylococcus Carrier Study Which sources showed growth and fermentation on mannitol salt agar? Throat __ Nose __ Skin __ Do you work in a clinical setting this quarter? Yes __ No __ If so where? ___ Have you taken antibiotics this quarter? Yes ___ No ___ If so which? ___ Isolate Mannitol Beta Antibiotic susceptibility S or R Source Fermentation hemolysis Coagulase Peni- Strepto- Tetra- Chloram Erythro- Sulfanil- Tested + or - + or - + or - cillin mycin cycline phenicol mycin amide Your strain reference strain Deviation from reference strain Results 1. Record the initial results you obtained from your three body cultures on mannitol salt agar. Note At least one colony should show good colony growth with a yellow color change indicating a positive test for mannitol fermentation. Record each culture as mannitol positive + or negative -. Nose __ Throat __ Skin __ 2. Catalase test results + or - Nose __ Throat __ Skin __ 3. Record the code of each antibiotic disc. Penicillin ___ Erythromycin ___ Streptomycin ___ Tetracycline ___ Sulfanilamide ___ Chloramphenicol ___ 4. Record results obtained with the one strain you chose to study in table 32.1. indicate deviation of any of the results obtained for your strain from the refrence strain. Table 32.1 Test results obtained with reference S. aureus culture and Mannitol Salt agar positive body culture. Antibiotic Susceptibility S or R Isolate Mannitol Beta source Fermentation hemolysis Coagulase Peni Strepto- Tetra Chloram- Erythro- Sulfanil tested + or - + or - + or - cillin mycin cycline phenicol mycin amide Yor strain Reference strain Deviation from Reference strain. page 281 5. Based on the information for the class given to you by your instructor complete table 32.2. Table 32.2 Classroom Summary of S. Aureus Epidemological Study Health professionals General population Total class results students working in students not workng in clinical setting in clinical setting Number % of sample Number % of sample Number % of sample S. aureus carriers positive cultures Noncarriers negative cultures Total Carriers of susceptible strains Carriers of resistant strains Total Carriers of strainlike reference strain Carriers of deviant strains total Different types of deviant strains Total number and % of mannitol possitive strains from Throat ___ Nose ___ Skin ___ Total ___ Questions 1. What proportion of students are carriers of the potential pathogen Staphylococcs aureus? 2. What parts of the body harbor this pathogen and which parts carry it most commonly? 3. To what extent has S. aureus acquired resistance to antibiotics to which it was originally susceptible? 4. How many different strains can be isolated from the student population that deviate from the typical S. aureus? 5. Do working health professionals have a higher carrier rate than the general population? 6. Do the strains of S. aureus carried by health professionals have a higher proportion of resistant strains than those isolated from the general population? 7. Is there a greater number of different strains of S. Aureus among health professional carriers than among the general population? Name ___ Date ___ Section ___ Exercise 33 Laboratory Report Bacteriological examination of water. Multiple tube fermentation and membrane filter techniques. Results 1. Multiple tube fermentation technique a. Record the results of the presumptive test. Presumptive test for the presence or absence of gas and turbidity in multiple tube fermentation media. Water sample size ml Tube #1 Tube #2 Tube #3 Tube #4 Tube#5 10 1 0.1 Use a + sign to indicate gas and a circle O around the plus sign to indicate turbidity. 2. Membrane filter technique Number of Coliform Colonies present in various dilutions of the water sample. Undiluted Sample 10-1 Dilution 10-2 Dilution a. Calculate the number of coliform colonies ml present in the original water sample. b. Determine the MPN below. Test readout ___ MPN ___ 95% Confidence limits ___. c. Confirmed test results gas production in brilliant green lactose bile 2% broth. Sample Number Gas + or - 24 hours 48 hours d. Appearance of colonies on LES Endo agar. e. Completed test results. Sample Number Gas + or - Gram Stain Reaction Undiluted Sample 10-1 Dilution 10-2 Dilution. a. Calculate the number of coliform colonies ml present in the original water samplr. Questions 1. What nutritional means might be used to speed up the growth of the coliform organisms using the membrane filter technique? 2. Describe two other applications of the membrane filter technique? 3. Why not test for pathogens such as Salmonella directly rather than use a indicator organism such as the coliform bacteria? 4. Why does a positive presumptive test not necessarily indicate the water is unsafe for drinking? 5. List three organisms that are apt to give a positive presumptive test. 6. Describe the purpose of lactose and Endo agar in these tests? 7. What are some limitations of the membrane filter technique? 8. Define the term coliform? 9. Briefly explain what is meant by presumptive, confirmed, and completed tests in water analysis? 10. See Note 3 MPN determination on page 291 for the question. individual project 35 page Individual exercise 34 Identifying DNA with restriction enzymes. DNA from one organism can be distinguished from the DNA from another organism by the use of a type of enzyme called restriction enzymes. DNA from blood stains. EcoRI is the first restriction enzyme from Escherichia coli strain R. size of the fragments? running the DNA on a electrophoresis agarose gel. definitions Buffer solution is a salt solution formulated to maintain a particular pH. Electrophoresis is a procedure used to separate components by electrical charge. Restriction enzymes are enzymes isolated from bacteria that are used to cut DNA at specific sequences. objective 1. To understand the use of restriction enzymes to cut DNA into segments. 2. To learn how to separate different lengths of DNA using gel electrophoresis. 3. To understand the use of DNA fingerprinting to identify DNA. Materials Lambda DNA 0x 174 DNA unknown phage either lambda or 0X 174 DNA size standard restriction enzyme Dra I microfuge tubes or Eppendorf tubes micropipettors gel box agarose TBE buffer for electrophoresis TRIS EDTA Borate. TE buffer Ethidium bromide stop mix includes tracking dye. water bath. microfuge UV transilluminator. goggles spatula. disposable gloves. procedure 1. Pour a agarose gel. The amount of agar will depend on the size of the gel box. The usual concentration is 0.7% agar and it is dissolved in the buffer TBE. Add ethidium bromide at 15 ul/200 ml agarose gel. 2. Add sterile distilled water to a sterile Eppendorf tube and then 1u1 10X TE buffer. Buffer is added before the enzyme so that the conditions are immediately optimal for the enzyme. 3. Add lambda DNA to the tube. This is usually 0.5-1.0 ug/uml. 4. Add the restriction enzyme Dra I. 5. Repeat for 0X 174 DNA and the unknown virus DNA. 6. Each tube contains Distilled water to bring total volume to 10 ul including enzyme. 10X TE buffer DNA Enzyme. 7. Mix and incubate for 30 minutes in a 37 degrees water bath. 8. Add stop mix 1 ul of 10X. this stops the reactions sample sink to bottom of well. 9. Place gel in a gel box. Cover the agarose gel with running buffer and load the samples into the wells. 10. Attach the electrodes to the gel. 11. Transfer the gel to a illuminator with a spatula while waearing gloves. 12. View on a UV transilluminator. 13. Photograph the gel. Compare the number of bands that result from cutting of the phage DNA with Dra I. Exercise 34 Laboratory Report Identifying DNA with Restriction Enzymes Results 1. Record your results. Results of Phage DNA Electrophoresis Lambda oX174 Unknown Phage X Number of fragments About what size is the largest piece? About what size is the smallest piece? 2. What is the probable identity of phage X? Questions 1. How were you able to estimate the size of the DNA fragments? 2. What is the purpose of adding ethidium bromide to the agarose gel? 3. What is the purpose of adding a tracking dye to the DNA before adding to the well? 4. Would you have the same number of fragments from each phage if you used a different restriction enzyme? Why or why not? Exercise 35 page 307 Identification of Bacteria using the Ribosomal Data Project Identifying and classifying bacteria has always been more difficult than identifying and classifying plants and animals. Bacteria have very little differences in their structure. This exercise will give you a chance to send the DNA sequence of the 16S ribosomal RNA of a organism to the Ribosomal Database Project and determine the identification of the organism. Procedure 1. Open Netscape or Explorer on the computer. 2. Type in the url http://rdp.cme.msu.edu/ 3. Click on Online analysis. 4. Find sequence match in purple column. Click on gray arrow in run column for the sequence match. 5. In the box cut and paste a sequence from your machine at the bottom of the page. enter the sequence in the box. Try typing in the first 150 bases as it may be sufficient. 6. Click on submit sequence a few screens down. use standard. 7. The iditification of the genus and species appears. The organisms were studied in 32 22 23 13. tctctgatgt tagcgggcgga Name ___ Date ___ Section ___ Exercise 35 Laboratory report Identification of Bacteria using the Ribosomal data project. Results Organism Bacteria or Archaea 1. _____ ___________________ 2. _____ ___________________ 3. _____ ___________________ 4. _____ ___________________ 5. _____ ___________________ Questions 1. What is a advantage of identifying a organism by using the Ribosomal data project? 2. What is the disadvantage? 3. It is generally thought by microbiologists that you cannot randomly create a sequence that the database will identify as a organism. Can you prove them wrong, organism 2 taacacgtgg ataacctacc individual project 36 page 315 Hydrocarbon degrading Bacteria, Cleaning up after oil spills. The biosphere contains a great variety of organisms. The hydrocarbons in oil are natural compounds found almost anywhere in nature. Enrichment The first step in these kinds of isolations is enrichment as a way of increasing the numbers of the desired bacteria. 1. For the enrichment prepare mineral salts medium broth MSM. Add 200 ml to a 500 ml flask and cover with a foil cap. Prepare several flasks to maximize your chances of successfully isolating the organisms. Minimal salts medium modified from E. Rosenberg. NaCl 2.5g 28.4g NaCl for marine organisms K2HPO4 4.74 g. KH2PO4 0.56 g MgSO4 7H20 0.50g CaCl2 H20 0.1g NH4NO3 2.5g Tap water 1 liter Agar for plates 20 grams 15 grams Difco Agar pH 7.1 2. Add 0.1% vol/vol of the hydrocarbon of the carbon source 0.2 ml hydrocarbon 200 ml borth. %vol/vol = ml/100 ml. 3. Add a gram of soil to the broth. 4. Incubate the flask for two weeks at room temperature obligate aerobe organisms. The second step is selection. After enriching for the organism in a broth culture. Streak the broth on mineral salts agar plates and add the hydrocarbon. 1. Pour mineral salts agar plates. 2. Add the hydrocarbon carbon source such as fuel oil. add 0.5 ml of the hydrocarbon to the filter paper and replace the inverted top of the agar plate. The organisms will be able to grow on the fumes. 3. Streak the MSM agar plates with a loopful of inoculum from your enrichment flasks to obtain isolated colonies. 4. Prepare a control by adding water to the filter paper in another plate. control organisms degrade agar. 5. Incubate the plates at romm temperature for 1 week. Isolation 1. Examine your plates carefully. Is there more than one type of colony? 2. Purify your isolates by restreaking the organisms on a similar plate using the same hydrocarbon. Repeat until you can be sure you have at least one pure culture. 3. Make a Gram stain for initial identification of the organism or organisms growing on the plates. If a gram negative rod is present detemine if it is oxidase positive. Pseudomonas is oxidase positive and well known for the ability to degrade unusual compounds. 4. You might try different hydrocarbons to determine whether your organism can degrade other petroleum products. Does it grow on standard laboratory media? Can you think of other experiments using your isolated organism? individual project 37 page 317 Luminescent Bacteria Bacteria that produce light. Some marine bacteria have the ability to emit light a process called bioluminescence. Most of these organisms live in relationship with tropical fishes. Many marine dinoflagellates luminesce when the salt water is warm. Procedure for isolating Luminescent Bacteria 1. Prepare several plates of sea water complete agar. Materials sea water 750 ml. glycerol 3.0 ml peptone 5.0 grams yeast extract 0.5 grams agar 15.0 grams water 250 ml pH 7.5 2. Obtain a whole saltwater fish or purchase a squid at a seafood store. 3. Dip a swab in the intestinal content of the fish and swab a third of the agar plate Try swab the inside organs inside of the squid. 4. Incubate plates at room temperature 12-48 hours. 5. Observe for luminescent colonies. Take the plate in a dark room and permit your eyes to adjust for a few minutes. 6. Restreak a glowing colony onto another plate to obtain a pure culture. The organism is probably either Photobacterium or Vibrio. individual project 38 page 319 Methylotrophs Organisms that grow on one carbon compounds. a reliable place to look for methylotrophs was on the underside of green leaves. Plants produce a one carbon compound. If you press a leaf on a agar plate and then remove it, the pattern of the leaf can be seen in the resulting bacterial growth. Procedure 1. Prepare mineral salts + methanol + cyclohexamide agar plates. Prepare mineral salts agar without a carbon source. Autoclave the agar cool to 50C and then add the methanol the source of carbon and the cyclohexamide before pouring plates. Cyclohexamide is a antifungal antibiotic. Add the cyclohexamide to the cooled agar plates. Prepare a control plate without the methanol. 2. Press the underside of a leaf on the agar and immediately remove. Try different leaves and agar plates. 3. Incubate at room temperature for 3-7 days and look for pigmented colonies. Are the colonies growing in the pattern of the leaf? Many methylotrophs are pink. Gram stain. 4. Restreak your organism on the same medium to obtain a pure culture. 5. After you have isolated your organism try to determine what other carbon sources it can utilize in addition to methanol. 6. If your organism is in the second group report it as a methanotroph. If it is in the first group it could be many different organisms including Bacillus, Pseudomonas, or Vibrio. Materials Mineral Salts medium NH4 2SO4 1 gram K2HPO4 7 grams KH2PO4 3 grams MgSO4 7H2O 0.1 grams Trace elements if available 1 ml Tap water 1 liter Agar Difco 15 grams pH 7.0 Autoclave at 120C for 20 minutes. Agar can be stored until ready to use. Before pouring plates melt agar heating in a container of boiling water 30 to 45 minutes. Cool to 50C. add the methanol and cyclohexamide and then pour the plates. Pour the plates a day before you plan to use them so the surface will be dry. After autoclaving and before pouring plates add the following 20 ug/ml cyclohexamide or 1.0 ml of a 100 x stock solution 100 ml agar. 0.1% methanol 0.1 ml methanol 100 ml autoclaved cooled agar before pouring plates. Stock solutions Antibiotics are required in very small amounts. make up a concentrated stock solution and then add it to the melted and cooled agar before pouring plates. Cyclohexamide Stock solution A 100 x solution would be 100 x 20 ug/ml or 2,000 ug.ml. Since 2,000 ug is 2 milligrams a stock solution would be 2 mg/ml. If you prepared 100 ml of a stock solution you would add 200 mg 0.2 grams to 100 ml of water. individual project 39 page 321 Deinococcus, Bacteria with out of the world capabilities. Deinococcus are aerobic gram positive cocci that for their ability to repair damage to their DNA. are organisms. Enrichment procedure In a enrichment procedure the organism you are trying to isolate is encouraged to grow while other organisms are discouraged. Deinococcus will be present to be seen on the irradiated plate. 1. Obtain some hair from the back or tail of a cow or any other source you think has possibilities. 2. Incubate the hair in tryptone yeast extract glucose TYEG broth at least three days at 30 celsius. The organism seems to require growth factors in yeast extract. It grows slowly so incubate at least three days . The optional temperature for growth is 30 C. materials required Tryptone 5.0 grams Yeast extract 3.0 grams. Glucose 1.0 gram. Tap water 1 liter Agar 15 grams. 1. Place 0.1 ml of the enrichment culture on a TYEG medium agar plate and spread with a swab or a bent glass rod. This will result in a lawn of bacterial growth. 2. Irradiate the plates 1-3 hours under a UV lamp. 3. Incubate several days at 30C. The Bacillus species plates can be incubated at 30 C. Bacillus spores are resistant. colonies of Deinococcus are frequently pigmented usuaaly yellow, orange or red. A gram stain of the colonies should show gram positive cocci arranged in packets of four or eight. How would a gram stain help you distinguish Deinococcus from Bacillus? Restreak your Deinococcus isolate on TYEG to obtain a pure culture. Once you have isolated Deinococcus perhaps you can think of other experiments you might try with it. Regain control of your computer with Do you feel like surfing the net is no longer what it used to be? Well you're not alone! Today's websites bombard visitors with popups, banners, cookies and all sorts of information. As if that was not enough, your computer keeps track of all your site visits, passwords & credit card details (just to mention a few)! And then came ! Key Features Popup Stopper automatically kills most popups, pop-unders and banner ads (click here to see why most popups get eliminated, but not all). 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Download a free, fully-enabled 7-day trial version here: Register you SmoothSurfer here The expression of color in homing pigeons is an exception to Mendel's Laws of inheritance follows a simple pattern of dominance follows a pattern of codominance a and c only answer is Karyotype analysis is a means of detecting and reducing mutagenic agents. is a surgical technique that separates chromosomes that have failed to segregate properly during meiosis II. is used to detect chromosomal mutation and metabolic disorders in embryos. substitutes defective alleles with normal ones. answer is Recombinant DNA has occurred in sexually reproducing forms can be produced with new biological techniques has produced changes that resulted in evolution all of the above answer is can be produced with new biological techniques. The appropriate adjective to describe DNA replication is nondisruptive semiconservative progressive natural answer is semiconservative Replication of DNA REPLICATION OF DNA How is cellular DNA copied? DNA replication begins with a partial unwinding of the double helix at an area known as the replication fork Deoxyribonucleic acid (DNA) was first identified in 1868 by Friedrich Miescher, a Swiss biologist, in the nuclei of pus cells obtained from discarded surgical bandages. The substance he found contained an acidic part, nucleic acid, and a basic (alkaline) part, which we now know to be histone proteins, which bind to the nucleic acid. DNA Replication DNA replicates semiconservatively. Replication starts by opening of the DNA helix at a particular sequence called an origin of replication (ori). Bacteria, even during logarithmic growth, have ONE map position where DNA can originate replication. Eukaryotes have MANY origins of replication, all of which run concurrently. In either case, each origin of replication runs bidirectionally, with TWO replicating forks. Experiment to show Semiconservative Replication Bidirectional semiconservative replication can be demonstrated by observing DNA from cells replicating in the presence of radiolabeled nucleotides. Replication of DNA is mediated by enzymes and binding proteins. http://biology.kenyon.edu/courses/biol114/Chap01/Chapter_01b.html rDNA stands for recombinant DNA. Before we get to the "r" part, we need to understand DNA. Those of you with a background in biology probably know about DNA, but a lot of ChemE's haven't seen DNA since high school biology. DNA is the keeper of the all the information needed to recreate an organism. All DNA is made up of a base consisting of sugar, phosphate and one nitrogen base. There are four nitrogen bases, adenine (A), thymine (T), guanine (G), and cytosine (C). The nitrogen bases are found in pairs, with A & T and G & C paired together. The sequence of the nitrogen bases can be arranged in an infinite ways, and their structure is known as the famous "double helix" which is shown in the image below. The sugar used in DNA is deoxyribose. The four nitrogen bases are the same for all organisms. The sequence and number of bases is what creates diversity. DNA does not actually make the organism, it only makes proteins. The DNA is transcribed into mRNA and mRNA is translated into protein, and the protein then forms the organism. By changing the DNA sequence, the way in which the protein is formed changes. This leads to either a different protein, or an inactive protein. Recombinant DNA is the general name for taking a piece of one DNA, and and combining it with another strand of DNA. There are three different methods by which Recombinant DNA is made. They are Transformation, Phage Introduction, and Non-Bacterial Transformation. Each are described separately below. Transformation The first step in transformation is to select a piece of DNA to be inserted into a vector. The second step is to cut that piece of DNA with a restriction enzyme and then ligate the DNA insert into the vector with DNA Ligase. The insert contains a selectable marker which allows for identification of recombinant molecules. An antibiotic marker is often used so a host cell without a vector dies when exposed to a certain antibiotic, and the host with the vector will live because it is resistant. The vector is inserted into a host cell, in a process called transformation. One example of a possible host cell is E. Coli. The host cells must be specially prepared to take up the foreign DNA. Selectable markers can be for antibiotic resistance, color changes, or any other characteristic which can distinguish transformed hosts from untransformed hosts. Different vectors have different properties to make them suitable to different applications. Some properties can include symmetrical cloning sites, size, and high copy number. Non-Bacterial Transformation This is a process very similar to Transformation, which was described above. The only difference between the two is non-bacterial does not use bacteria such as E. Coli for the host. In microinjection, the DNA is injected directly into the nucleus of the cell being transformed. In biolistics, the host cells are bombarded with high velocity microprojectiles, such as particles of gold or tungsten that have been coated with DNA. Phage Introduction Phage introduction is the process of transfection, which is equivalent to transformation, except a phage is used instead of bacteria. In vitro packagings of a vector is used. This uses lambda or MI3 phages to produce phage plaques which contain recombinants. The recombinants that are created can be identified by differences in the recombinants and non-recombinants using various selection methods. How does rDNA work? Recombinant DNA works when the host cell expresses protein from the recombinant genes. A significant amount of recombinant protein will not be produced by the host unless expression factors are added. Protein expression depends upon the gene being surrounded by a collection of signals which provide instructions for the transcription and translation of the gene by the cell. These signals include the promoter, the ribosome binding site, and the terminator. Expression vectors, in which the foreign DNA is inserted, contain these signals. Signals are species specific. In the case of E. Coli, these signals must be E. Coli signals as E. Coli is unlikely to understand the signals of human promoters and terminators. Problems are encountered if the gene contains introns or contains signals which act as terminators to a bacterial host. This results in premature termination, and the recombinant protein may not be processed correctly, be folded correctly, or may even be degraded. Production of recombinant proteins in eukaryotic systems generally takes place in yeast and filamentous fungi. The use of animal cells is difficult due to the fact that many need a solid support surface, unlike bacteria, and have complex growth needs. However, some proteins are too complex to be produced in bacterium, so eukaryotic cells must be used. Why is rDNA important? Recombinant DNA has been gaining in importance over the last few years, and recombinant DNA will only become more important in the 21st century as genetic diseases become more prevelant and agricultural area is reduced. Below are some of the areas where Recombinant DNA will have an impact. Better Crops (drought & heat resistance) Recombinant Vaccines (ie. Hepatitis B) Prevention and cure of sickle cell anemia Prevention and cure of cystic fibrosis Production of clotting factors Production of insulin Production of recombinant pharmaceuticals Plants that produce their own insecticides Germ line and somatic gene therapy Item 11006-026 CNA Agar w/ 5% Sheep Blood 10/sleve $12.95 postage $7.75 $33.65 IPM Scientific P.O. Box 1576 Eldersburg, MD 21784-1576 Registration Office South Seattle Community College 6000 16 Ave SW Seattle, WA 98106-1499 764-7938 AdvisorSouth@sccd.ctc.edu CHAPTER 1 - INTRODUCTION TO BIOLOGY HOME CHAPTER 2 - CHEMISTRY OF LIFE Isotopes - Atoms of the same element can have different numbers of neutrons; the different possible versions of each element are called isotopes. For example, the most common isotope of hydrogen has no neutrons at all; there's also a hydrogen isotope called deuterium, with one neutron, and another, tritium, with two neutrons. There are "preferred" combinations of neutrons and protons. Light elements tend to have about as many neutrons as protons; heavy elements apparently need more neutrons than protons in order to stick together. Atoms with a few too many neutrons, or not quite enough, can sometimes exist for a while, but they're unstable. Chemical properties- of an atom are due to the number of valence electrons Oxidation/Reduction Reactions Any reaction where electrons are transferred between reactants is called an reduction-oxidation reaction or redox reaction. These are reactions where one substance wants an electron so badly that it takes it away from another substance. Whether or not it succeeds depends on who it meets. For example, FeCl2 + CeCl 4 -->FeCl3 + CeCl 3 What happened here? If we remove the spectator ions and write the net ionic equation we find: Fe2+(aq) + Ce4 +(aq) -->Fe3+(aq) + Ce 3+(aq) Ce4+ took an electron from Fe2 +! This is an oxidation/reduction reaction. In this example, Fe2+ is oxidized and Ce4+ is reduced. The charge of Fe went from +2 to +3, that is, it lost an electron. This process is called oxidation. Oxidation: The loss of an electron by a substance. Likewise, the charge of Ce went from +4 to +3, that is, it gained an electron. This process is called reduction. Reduction: The gain of an electron by a substance. Generally, metals "lose" electrons and are said to be "oxidized". Nonmetals gain electrons and are said to be reduced BIOLOGICAL EXAMPLES: photosynthesis, nitrogen fixation, and metabolism (below) - C6H12O6 + 6 O2 ----------> 6 CO2 + 6 H2O + Energy In the above example, the carbon atoms in glucose are oxidized, undergoing an increase in oxidation state (each carbon loses 2 electrons) as they are converted to carbon dioxide. At the same time, each oxygen atom is reduced by gaining 2 electrons when it is converted to water. Part of the energy is released as heat and the remainder is stored in the chemical bonds of "energetic" compounds such as adenosine triphosphate (ATP) and nicotinamide adenine dinucleotide (NADH) discussed in chapter 5. (list from small to large)- electrons, atoms, inorganic compounds such as water (usually), organic compounds such as sugar BIOCHEMISTRY: CARBOHYDRATES Monosaccharides The 5th carbon bonds to the first to form a hectagon shape These are the simplest carbohydrate molecules. The most commonly occurring monosaccharides in food are glucose, fructose and galactose. The formula for glucose is C6H1206. Sugars are called simple carbohydrates because your body digests them quickly and easily. Simple carbs are usually sweet tasting, like cookies, candy, soda, and other sugary foods. And some foods from nature - like many fruits - are sources of simple carbohydrates. Disaccharides sucrose is a disaccharide. These sugars are formed when two monosaccharide molecules join together with the elimination of one molecule of water (dehydration synthesis). They have the general formula C12H22011. Examples of disaccharides are sucrose (glucose and fructose), lactose (glucose and galactose) and maltose (2 molecules of glucose). When these bonds break to again form simple sugars, the opposite reaction occurs and is called a hydrolysis reaction. Polysaccharides (also known as complex carbohydrates) Polysaccharides are made up of many monosaccharide molecules, (usually glucose), joined together. They have the general formula (C6H1005)n where `n' is a large number. Examples of polysaccharides are starch, glycogen, cellulose, beta glucan and pectin. Starchy carbohydrates or complex carbohydrates take longer to be digested than simple carbohydrates do. Complex carbs are found in foods like bread, noodles, and rice, and When you eat foods with carbohydrates in them, your body breaks them down into two different types of fuel. For energy that you'll use right away, your body takes those carbs and turns them into glucose Glucose is carried in your blood to all the cells in your body, and gives you energy. It powers every part of your body. But your cells can only use so much glucose at one time. So when there is glucose left over that can't be used right away, your cells save in in your liver and muscles, and it's called glycogen. Glycogen is released for quick energy when you're exercising. If you're sprinting or doing another quick exercise, your body calls on glycogen for energy. But if you are exercising for a long time, your body turns to its "reserve tank" of fuel for energy: fat. BIOCHEMISTRY LIPIDS (FATS) It is also possible for fat cells to take up glycogen and amino acids, which have been absorbed into the bloodstream after a meal, and convert those into fat molecules. The conversion of carbohydrates or protein into fat is 10 times less efficient than simply storing fat in a fat cell, but the body can do it. Given a choice, a fat cell will grab the fat and store it rather than the carbohydrates because fat is so much easier to store. FATS TRIGLYCERIDE. The "R" denotes long carbon-hydrogen chains. The blue is the glycerol molecule (if not bonded to other molecules, each carbon is normallly bonded to an OH or hydroxyl group) Fat insulates our bodies from the cold and provides some cushioning for our organs. Fat gives our bodies energy. Some fats help to make up important hormones that we need to keep our bodies at the right temperature or keep our blood pressure at the right level. Fat helps you have healthy skin and hair. It helps vitamins A, D, E, and K hang out and get transported through your bloodstream when your body needs them! Even though our bodies need some fat to work properly, they don't need as much as most people eat. It's a good idea to avoid eating a lot of fat, because it can contribute to obesity (when a person weighs much too much for his or her height) and other illnesses that can occur when you're older, like heart disease or adult-onset diabetes. Why do athletes need to keep hydrated? Drinking fluids to keep you well hydrated will enhance your performance. Water helps maintain body temperature. When you become dehydrated (low body water), heat builds up in the muscles causing them to fatigue early and impair your performance. Which nutrients provide the body with energy?Your body uses a mixture of carbohydrates and fats to fuel your muscles.When you sprint or perform an explosive exercise such as jumping or weight lifting, your body primarily uses carbohydrates (about 70 to 85%of total fuel). Performing an endurance exercise taps into both fat and carbo-hydrates in about equal amounts.Therefore, you need to replace these fuels after exercise. Carbohydrate stores deplete quicker than more abundant and calorie-dense fat stores. Which nutrient provides the body with repair tools? Eating protein helps repair body tissues. Muscles tear during exercise and repair during resting and sleeping. If you under eat on carbohydrates and fats, your body will steal protein for repairing muscles and use it as a fuel source. This takes protein away from the repair process causing loss of muscle mass. BIOCHEMISTRY PROTEINS Some types of amino acids are made by you, right inside your body,. These are called nonessential amino acids, and there are 11 of them. And they are necessary, but they are not essential as part of the food you eat. The essential amino acids - all nine of them - must come from food. That's where eating foods with protein comes in, to give your body the amino acids it needs. Peptide bonds - The amino acids are linked covalently by peptide bonds. The images show three typical amino acids. NOTE: the red is the same in all amino acids. This includes the amine group, a carboxyl group and a CH. The part in black is often designated as an "R" to show there is a variable chain to distinguish one amino acid from another. valine glutamine phenylalanine The diagram below shows how three amino acids linked by peptide bonds into a tripeptide. This is an example of a dehydration synthesis chemical reaction. Protein's biggest job is to build up, keep up, and replace the tissues in your body. Your muscles, your organs, even some of your hormones are made up mostly of protein. Protein helps your body in other ways, too. It makes hemoglobin, the part of red blood cells that carries oxygen to every part of your body. It even makes antibodies, the cells that fight off infection and disease. It's what helps make your cuts and scrapes heal! Primary Proteins - have a "primary" structure (a specific peptide chain composed of a specific sequence and number of amino acids such as some hormones) Secondary Proteins - have both a primary and secondary structure (pleated sheet such as collagen) Tertiary Proteins - have primary, secondary and tertiary structure (spiral which as been folded into a complicated spherical or globular shape such as the muscle protein myoglobin) Globular Proteins - The tertiary structure of globular proteins reflects their interaction with water. A globular protein consists of a hydrophobic core surrounded by a hydrophilic external surface which interacts with water. The tertiary fold of the polypeptide is such that the apolar side chains are buried in the center, while the polar sides remain exposed. This principle is held by many to be the dominant driving force behind the folding of the polypeptide chain into the compact globular form: the aggregation and burial of the hydrophobic surface reduces the number of unfavourable interactions of these groups with water; thus the hydrophobic effect. Quaternary proteins - have a "quaternary structure (a complex of two or more globular (tertiary) proteins such as hemoglobin) BIOCHEMISTRY ENZYMES (usually proteins) Enzyme sequence - substrate + enzyme-->intermediate compound(s) + enzyme (may be repeated)-->product + enzyme Enzymes include RNA& DNA polymerase (proteins) and catalyst of peptide bonds (not a protein, but a nucleic acid) and many others Enzymes work best under optimal conditions of pH and temperature. If these environmental factors are well above or below the optimum, the structure of the enzyme begins to change, thereby altering their activities HOME -------------------------------------------------------------------------------- CHAPTER 3 - THE CELL THE CELL: AN INDIVIDUAL UNIT OF LIFE The first sightings of the internal action of the cell were made by Robert Brown (described the nucleus as an integral part of the cell.) Theodore Schwann created the term "cell theory" and declared that plants consisted of cells. This declaration was made after that of Matthias Schlieden's (1804 - 1881) that animals are composed of cells. German pathologist Rudolf Virchow (1821 - 1902) altered the thought of cellular biology with his statement that "every cell comes from a cell". Not even twenty years after this statement, processes of cell reproduction were being described--Virchow had completed the thought behind the basic cell theory. Major Organelles: Mitochondrial DNA - Mitochondria are the powerhouses of the cell. They take in fuel molecules derived from sugars and fats, harvest the energy in their chemical bonds with the aid of oxygen, and spit out ATP, the universal energy carrier needed throughout the cell to fuel the energy-hungry reactions of life. This cutaway graphic shows the mitochondrion's two membrane layers: a smooth outer membrane and a folded inner membrane where enzymes (depicted as balls)help with the synthesis of ATP. Human mitochondria DNA is less than 1/300,000th of the total length of DNA molecules in the nucleus of a human cell. It codes for rRNAs and tRNAs used in the mitochondrion, and contains 13 recognizable genes that code for polypeptides. It is efficient - nearly every part of the molecule is transcribed. Each gene codes for a different polypeptide that makes up part of the electron transport chain in the inner mitochondrial membrane. Cell Membranes Most organelles are limited by one or more membranes. To perform the function of the organelle, the membrane is specialized in that it contains specific proteins and lipid components that enable it to perform its unique roles for that cell or organelle. In essence membranes are essential for the integrity and function of the cell. Membrane components may: be protective regulate transport in and out of cell or subcellular domain allow selective receptivity and signal transduction by providing transmembrane receptors that bind signaling molecules allow cell recognition provide anchoring sites for cytoskeletal filaments or components of the extracellular matrix. This allows the cell to maintain its shape and perhaps move to distant sites. help compartmentalize subcellular domains or microdomains provide a stable site for the binding and catalysis of enzymes. regulate the fusion of the membrane with other membranes in the cell via specialized junctions ) provide a passageway across the membrane for certain molecules, such as in gap junctions. allow directed cell or organelle motility HOME -------------------------------------------------------------------------------- CHAPTER 4 - CELL MEMBRANE Humans are are much better protected against water loss, because our skin cells are covered with a protein called keratin. However, osmosis is still a life-and-death matter for us in other ways. For example, the functioning of our kidneys relies heavily on osmosis. Osmosis can be harmful to humans as well; excessive salt in the diet is thought to be one factor promoting high blood pressure, because the salt goes into the fluids surrounding your cells and draws water out by osmosis. Is this osmosis or diffusion? How do you know? Drug Addictions (a physiological response caused by repeated use of a drug that alters the normal functing of cell membranes) People can get addicted to all sorts of substances including alcohol, nicotine (in cigarettes), cannabis, opiates, cocaine, antidepressants even chocolate! Scientists think that addictive substances stimulate what can be thought of as a 'pleasure pathway' in the brain. Making certain activities pleasurable - like eating, drinking and sex - ensures that we continue these essential body functions for our own survival and that of the species. Addictive drugs increase the levels of dopamine (naturally produced in body) in the neurons and synapeses of nerve cells. As a chemical messenger, dopamine is similar to adrenaline. Dopamine affects brain processes that control movement, emotional response, and ability to experience pleasure and pain. Neurons containing dopamine are clustered in the midbrain . Drugs can stimulate or fail to stimulate dopamine receptors in the cell membranes. Some drugs bind to dopamine receptors in the membrane in place of dopamine and directly stimulate those receptors. In this case, the cell produces more receptors (proteins) In contrast some drugs bind but don't stimulate dopamine receptors keeping dopamine from attaching to receptors. Here, the cell produces more dopamine. Cocaine and other drugs of abuse can alter dopamine function. Such drugs may have very different actions. The specific action depends on which dopamine receptors the drugs stimulate or block, and how well they mimic dopamine. Some drugs increase dopamine by preventing dopamine reuptake, leaving more dopamine in the synapse. An example is the widely abused stimulant drugs, amphetamine and cocaine . Both drugs increase the amount of dopamine in the synapse. However, cocaine achieves this action by preventing dopamine reuptake, while amphetamine helps to release more dopamine. So, these drugs with similar effects produce their actions through entirely different processes. HOME -------------------------------------------------------------------------------- CHAPTER 5 - PHOTOSYNTHESIS & CELLULAR RESPIRATION Joseph Priestly (1772) demonstrated plants provide oxygen for animals. Jan Ingenhausz (1779) demonstrated that plants need sunlight for photosynthesis Stages one and two of photosynthesis are commonly called the light dependent reactions Stage three (Calvin Cycle) is commonly called the light independent reactions of photosynthesis. The Krebs Cycle is frequently called the citric acid cycle Summary of the changes that occur as a result of the citric acid cycle: We put in: One acetylSCoA Four oxidized coenzymes (3 NAD+, 1 FAD) One ADP and a phosphate. We get out: Four reduced coenzymes (3 NADH, 1 FADH2) One ATP (energy rich) Two CO2 molecules from an acetyl group One coenzyme A The two carbons of the acetyl group have been oxidized to CO2 and energy from that has been deposited in ATP, NADH and FADH2. The rest of the story is about the oxidation of NADH and FADH and the transfer of the energy stored there to ATP. The body must manage metabolic energy through the breakdown of complex substances to simpler ones (catabolism) and the reverse of that process, the synthesis of complex substances from simpler ones (anabolism). Catabolic processes tend to release energy, whereas anabolic processes require input of energy. Catabolic pathways include glycolysis, the citric acid cycle Anabolic pathways include photosynthesis Oxidation reactions release energy. Catabolic pathways involve oxidation. Catabolic processes can ultimately collect the energy released in the form of ATP. Reductions require energy. Anabolic pathways involve reductions. Electron carriers are important in Redox reaction. NAD, FAD, and NADP are molecules that can accept electrons. After they accept electrons, they become NADH, FADH2, and NADPH, respectively. NADH, FADH2, and NADPH serve as electron donors for anabolic reactions. NAD, FAD, and NADP serve as electron acceptors for catabolic reactions. Futile cycles result when cells have catabolic and anabolic reactions going on at the same time. To avoid futile cycles, cells must exert control over when and where metabolic reactions occur. There are four main methods of control of enzymatic reactions. Two of these are the control of enzyme levels andthe control of enzyme activity . The other two are compartmentation and hormonal regulation which you will more next year. The key to ATP's usefulness is its three phosphate groups, which are shown in red in the cartoon. The bond between the second and third phosphate groups can be broken to release a small amount of energy, just the right amount to power activities inside a cell--such as those involved in flexing a muscle. In this way, ATP ends up providing all the immediate energy needs in a cell. After a cell has exerted itself, its ATP is depleted. But the molecule that is left over (called ADP, because it is a "diphosphate," with only two phosphate groups) can be recharged again to produce more ATP. To recharge your ATP, all you have to do is provide an energy source, such as the carbohydrates, fats, and proteins in a cheeseburger. So the next time you need to "recharge your batteries," remember that you are doing just that--although the "batteries" are actually ATP. ANAEROBIC CELLULAR RESPIRATION: It has been generally accepted for decades that the production of lactic acid causes muscle fatigue. A recent study published in the "Journal of Science" has strong evidence to the contrary. Contracting (exercising) muscles without lactic acid loos electrical activity. Acidity (excess of H+ ions) keeps electrical activity going which keeps tired muscles going. Lactic acid is responsible for the "burn" in tired muscles during intense exercise. HOME -------------------------------------------------------------------------------- CHAPTER 6 - MITOSIS -------------------------------------------------------------------------------- CHAPTER 7 - MEIOSIS Crossing over: Crossing-over is part of a complicated process which can occur during cell division. In meiosis, the precursor cells of the sperm or ova must multiply and at the same time reduce the number of chromosomes to one full set. During the early stages of cell division in meiosis, two chromosomes of a homologous pair may exchange segments in the manner shown above, producing genetic variations in germ cells. SISTER CHROMATIDS DUPLICATE CROSS OVER DURING PROPHASE i RECOMBINANT CHROMOSOMES Prior to prophase I the cell completed a typical interphase stage where the cell grew and DNA was synthesized. During prophase I, the same events that occurred in mitosis also occur here. The chromatin condenses into visible chromosomes, the nuclear membrane disintegrates, and centrioles migrate toward opposite poles. The unique event that occurs during prophase I is synapsis where homologous chromosomes line up and fuse together. This fusion allows the exchange of genetic material between two chromosomes in a process called crossing over. After the chromosomes separate, they are parceled out into individual sex cells. Each chromosome moves independently of all the others - a phenomenon called independent assortment. So, for example, the copy of chromosome 1 that an egg cell receives in no way influences which of the two possible copies of chromosome 5 it gets. Assortment takes place for each of the 23 pairs of human chromosomes. So, any single human egg receives one of two possible chromosomes 23 times, and the total number of different possible chromosome combinations is over 8 million (2 raised to the 23rd power). And that's just for the eggs. The same random assortment goes on as each sperm cell is made. Thus, when a sperm fertilizes an egg, the resulting zygote contains a combination of genes arranged in an order that has never occurred before and will never occur again. Mitochondrial Inheritance - When sperm and egg fuse to form a diploid zygote, the new individual gets half of its nuclear genetic information, 23 chromosomes, from each parent. That 50/50 split is the basis of Mendelian inheritance. However, due to the sheer size of the egg cell, all (or nearly all) of the mitochondria in the embryo come from the mother. In other words, mitochondrial inheritance is maternal, and that's why some diseases that are not sex-linked are passed only from the mother to her offspring. CHAPTER 8 - MENDEL & HEREDITY Mendel: First to base genetic theories on experimentation. Conclusions: Two inherited factors govern each trait studied One factor comes from each parent Traits come in several different forme (alleles) In heterozygotes, one trait (dominant) may mask the other (Law of Dominance) Only one of two possible alleles is passed from each parent to the offspring (Law of Segregation) Alleles sort independently Modern Genetics: Chromosomes come in homologus pairs that share the same length, centromere position, and gene loci (position on chromosomes where a gene for a particular trait is located. Evidence supports that hundreds to thousands of genes are arranged in linear order on each chromosome. Genes that occupy the same chromosome but different loci are "linked" genes. Linked genes must move together (do not follow the Law of Segregation) unless mutations or crossing over occurs. Mitochondrial DNA - A few years ago, Doug Wallace, a researcher at Emory University in Atlanta, was puzzled by an unusual inherited disorder known as Leber's Hereditary Optic Neuropathy ("Leber's" for short). Leber's results in a rapid loss of vision that usually begins in adolescence and can result in total blindness. Leber's runs in families, so it had been suspected as a genetic disorder, but there were two puzzling aspects to the disorder. First, the disorder was highly variable, causing complete blindness in some people and only minor loss of vision in others. Second, and most puzzling, was the fact that only women seemed to be able to pass the disorder along to their male AND FEMALE children. The children of men with Leber's never inherited the disorder, but the children of women with Leber's very often did. This did not mean that Leber's was sex-linked. These two characteristics -- variable effect and maternal inheritance -- seemed to violate Mendel's principles of genetics. Wallace, and everyone else who had worked on Leber's, was puzzled. What could account for this strange pattern of inheritance? Wallace discovered that Leber's patients had a point mutation -- a single DNA base change -- in their mitochondrial tDNA that normal patients did not. The base change changed a single codon in the gene for a protein in the electron transport pathway. When the slightly altered mRNA from this gene is translated, a single amino acid in the protein is changed. The protein still works, still transfers electrons, but it's just a bit less efficient. This reduced efficiency of electron transport, which slightly lowers the rate at which ATP can be made. CHAPTER 9 - DNA After Watson and Crick determined the structure of DNA, there were three hypothesis as to how DNA could pass on genetic information. 1. Conservative Replication: the two parent strands of DNA serve as templates for the new double-stranded DNA, leaving the parent molecule intact. 2. Semi-conservative Replication: the two partner strands separate, each acting as a template in the formation of a new strand of DNA 3. Dispersive Replication: the two parent strands are broken up into smaller double-stranded segments that were somehow used as templates for the new double-stranded DNA molecule. The amount of DNA in the cells of an organism is consistent, while the amount of protein varies from cell to cell within the same organism. CHAPTER 10 & 11 - PROTEIN SYNTHESIS & GENETIC ENGINEERING DNA SYNTHESIS The carbon atoms in the sugar ribose are numbered. The 5' carbon is bonded to the phosphate group and is the "leading" strand during DNA synthesis. The 3' carbon is called the "lagging" strand. The strand that has the 5' carbon exposed at the replication fork replicates with the aid of DNA polymerase in sequence down the strand. This replication happens quickly and is called the leading strand. The opposite side has the 5' carbon facing away from the replication fork and therefore replicates in reverse order. This process takes longer and is therefore called the lagging strand. Genetic Engineering: The manipulation of genes in an organism is genetic engineering. This process includes (1) gene isolation (2) gene synthesis (3) coupling (combining synthesized gene with bacterial or viral gene (4) transplantation of newly constructed gene into a cell to undergo cell replication. Rings of DNA (called plasmids) can be extracted from bacteria, sliced into fragments and inserted into DNA from another source. The plasmid fragments and the "NEW" DNA can be reformed into another plasmid. The altered plasmid is known as recombinant DNA. Recombinant DNA can be introduced into another bacterial cell. The transformed cell can make a protein that it was u nable to make before. When the cells with the new plasmid reproduce new cells with the plasmid, the new cell is considered to be cloned. Application: (1) production of proteins that serve as medicines such as insulin and growth hormone (2) production of proteins not normally made in the cell. (3) production of cloned DNA fragments used to diagnose infections and genetic diseases. (4) alteration of genetic inheritance. HOME -------------------------------------------------------------------------------- CHAPTER 12, 13 - GEOLOGIC TIME & THEORY OF EVOLUTION Evolution in a nutshell Evolution is a theory that groups of organisms change morphologically and physiologically over the course of many generations. The result of evolution is that the descendants are different from their ancestors. Evolution involves a change in allele frequences in a population's gene pool over successive generations. Therefore, a population may evolve, but an individual can not. The geological time scale is divided into four eras: 1. Precambrian (prior to 600 million years ago) is characterized by prokaryotic life and the beginning of eukaryotes. 2. Paleozoic (225 to 500 million years ago) is characterized by marine invertebrates dominated in the beginning of this era, followed by fish and amphibians 3. Mesozoic era (65 to 225 million years ago) age of reptiles, dinosaurs were abundant 4. Cenozoic era (present to 65 million years ago) characterized by dominance of mammals, birds and insects. The lasst three eras are broken down into periods and epochs as follows: Paleozoic Era Permian Period Carboniferous Period Pennsylvanian Period Mississippian Period Devonian Period Silurian Period Ordovician Period Cambrian Period (oldest) Mesozoic Era Cretaceous Period Jurassic Period Triassic Period Cenozoic Era Quaternary Period (current period) Holocene Epoch Pleistocene Epoch Tertiary Period Pliocene Epoch Miocene Epoch Oligocene Epoch Eocene Epoch Paleocene Epoch Darwin in a nutshell All organisms overproduce gametes Not all gametes form offspring and of the offspring formed, not all survive The most competitive organisms will have a greater likelihood of survival Survival traits vary form individual to individual Survival traits are passed on to the next generation, therefore, the best adaptations for survival are maintained The environment determines which traits will be selected for or against and will change with time (a selected trait may later be disadvantageous) Darwin did not theorize on the cause of variations in traits. It is know known that variation may be due to genetic mutations gene flow due to migration genetic drift (especially in small populations) natural selection of genotypes (ability to survive and/or reproduce) Cline: a character gradient in space (a) A "smooth" cline. If environmental character changes gradually [for example, a north / south temperature gradient], the fitness of the "A" allele (red line) changes gradually: the frequency of the allele changes gradually. (b) A "stepped" cline. If the environment changes abruptly [for example, in an ecotone transition between forest & grassland], the fitness of the "A" allele also changes abruptly. Depending on the degree of gene flow, the cline will be more or less steep. Industrial Melanism clarified Our text is misleading on the cause of the peppered moths apparent change in color. The dark colored allele is dominant, while the light colored allele is recessive. As long as the barks of trees were light colored, it was harder for the moth's prey to spot them When coal-burning industry caused a layer of soot to cover many tree trunks, the dark color became environmentally advantaged. The homozygous recessive moths were less likely to survive and pass on the recessive gene. The moths did not mutate themselves. A Word on Isolation Isolation can be categorized as pre-mating (no mating will occur) or post-mating (mating occurs unsuccessfully) Pre-mating mechanisms are more common in nature and include geographical isolation (separation of species due to land and/or climate seasonal isolation (breeding seasons do not overlap) behavioral isolation (communication is species specific. Example: a bird of one species may not respond to the mating call from another) mechanical isolation (structures used in mating are incompatible due to size or shape) Post-mating mechanisms include hybrid sterility (offspring are produced, but not fertile. Example: a donkey & mare produce a sterile mule) Failure of gametes to fuse Abnormal development and early death of the hybrid after birth HOME -------------------------------------------------------------------------------- .CHAPTER 15 & 20 - CLASSIFICATION & KINGDOMS Classification of Man Kingdom Animalia includes birds, worms, horses, apes, clams, jellyfish, lancelets, humans, mosquitoes, upright prehistoric primates, and many more Phylum Chordata includes birds, horses, apes, lancelets, humans, upright prehistoric primates and many more Subphylum Vertebrata includes birds, horses, apes, humans, upright prehistoric primates and many more Class Mammalia includes horses, apes, humans, upright prehistoric primates and many more Order Primate includes apes, humans, upright prehistoric primates and more Family Hominidae humans and upright prehistoric primates Genus Homo humans and some more advanced upright prehistoric primates Species Homo sapiens includes humans Most Used Criteria for Classification Morphology - body structure and form Phylogeny - evolutionary history Check out your textbook for more information. Multicellularity The advent of multicellular organisms allowed cells to specialize in form and function. The process of becoming cells becoming specialized is call differentiation. This allows cells to form tissues; tissues to form organs;. organs to form organ systems HOME -------------------------------------------------------------------------------- CHAPTER 37 ANIMAL BEHAVIOR Innate Behaviors are inherited and instinctive programmed by genes; (Ethology is the study of innate behaviors) highly stereotyped (similar each time in many individuals) Four Categories 1. Kinesis: "change the speed of random movement in response to environmental stimulus". Kinesis refers to random, undirected movement, 2. Taxis: a directed movement toward or away from a stimulus. Taxis is stimulus dependent: different stimuli produce different types of taxis. Some common types of taxis include phototaxis (a reaction to light), chemotaxis (reaction to chemicals), and magnetotaxis (reaction to magnetism). Movement toward a stimulus is called positive taxis and movement away from a stimulus is called negative taxis. Different taxes (plural of taxis) result in response to different types of stimuli. Each of these forms of taxis can be described by simply adding a prefix to the word taxis. The table below shows the most common forms of taxis. 3. Reflex: movement of a body part in response to stimulus. 4. Fixed Action Pattern (FAP): "stereotyped and often complex series of movements, e.g. a herring gull parent regurgitates food for its young Sign Stimulus (or releasor)isa specific environmental cue that triggers (releases) the fixed action pattern. being tapped on on its bill by a gull chick releases regurgitation Characteristics of Innate Behaviors - especially FAPs include: The behavior is performed without prior experience There is a stereotypic releaser stimulus breeding crosses produce hybrid behaviors the behavior is adaptive - signs that natural selection is at work Rhythmic behaviors Animals typically display particular behaviors at different times of the day or times of the year. The mechanisms underlying these rhythmic behaviors may be categorized as endogenous or exogenous (coming from within the animal and coming from the animal's environment, respectively) Typically the exogenous signal is light (i.e., the stuff that comes from the sun). This endogenous-exogenous system allows animals to innately display appropriate behaviors at nearly appropriate times independently of fallible exogenous signals, but still adapt to changes in exogenous signals, e.g., as day lengths change with the seasons Learned Behavior: Five Categories Imprinting Habituation - means by which animals adapt behaviorally to their environment (i.e., learn). Habituation is the learned ignoring of a stimulus. Conditioning - laboratory setting classical conditioning operant conditioning Trial and Error Learning - nature Insight, reasoning HOME -------------------------------------------------------------------------------- CHAPTER 16 - POPULATIONS A closed population has no input of individuals and therefore no new genetic material. These populations frequently die out. An open population has an input of individuals and therefore new genetic material. This allows for more variation and greater likelihood of the survival of the population Three phases in the growth of a typical population: lag phase - population is new to the environment. They may not have adjusted to existing predators/prey or some environmental factors. Therefore, growth rate is slow exponential phase - a period of rapid population growth. steady state - The average growth rate is zero. The population size still rises and falls somewhat. A horizontal line on the graph through the middle of the steady state is called the carrying capacity. The black line in the graph above is an example of a Malthusian population growth which predicts that a population will exceed the carrying capacity and then enter stages of decline and regrowth The logistic model is graphed as the letter "S" slightly slanted. There are no sudden fluctuations in the population growth rate. This is represented by the purple line in the graph above Limiting factors - a condition that sets a limit on how many organisms can live in an area, such as food water, light and living space. Density independent - usually operate independent of the population size/density. These include drought, frost, storms, volcanic activity, earthquakes and other natural weather/climate geological phenomena. Density dependent - usually operate only when a population is large and crowded. They do not affect small, widely scattered populations much. These include compeition for food, nesting materials, space, predation, parasitism and crowding & stress due to crowding HOME -------------------------------------------------------------------------------- CHAPTER 17 - ECOSYSTEMS Chemical elements cycle in four different spheres: the atmosphere, biosphere, geosphere, and hydrosphere. biosphere The region inhabited by living organisms on land , in the oceans and in the atmosphere. atmosphere A mixture of gases surrounding the Earth. Earth's atmosphere consists of 79% nitrogen (by volume), 20.9% oxygen, 0.03% carbon dioxide, and trace amounts of other gases. It can be divided into a number of layers according to thermal properties (temperature). The layer nearest Earth is the troposphere (up to about 10-15 km above the surface), next is the stratosphere (up to about 50 km) followed by the mesosphere (up to 80-90 km) and finally the thermosphere or ionosphere which extends into space. There is little mixing of gases between layers. geosphere The physical elements of the Earth's surface crust, and interior. hydrosphere The totality of water encompassing the Earth, comprising all the bodies of water, ice, and water vapor in the atmosphere BIOGEOCHEMICAL CYCLES: Study the four in your text. Here are two more The Law of Conservation of Mass (or Matter) : In a chemical reaction, matter is neither created nor destroyed. biomass The mass of living or organic material, usually expressed as dry weight per unit area. Since all organisms are made of roughly the same organic molecules in similar proportions, a measure of their dry weight is a rough measure of the energy they contain. biomass is measured in kilograms Analysis of various ecosystems indicates that those with biomass pyramids with one trophic level and the next averaging 10% or better difference are less likely to be disrupted by physical or biotic changes. Those averaging less than 10% are more likely to be easily disrupted. Energy pyramid (refer to text) are measured in kilocalories (or Joules) Hypothetical Energy Pyramid LAW of CONSERVATION of ENERGY Energy exists in different forms and can be transformed from one form into another. When these transformations occur, energy is not destroyed and neither is energy made. Transformation of energy just exchanges one form of energy for another. Number pyramid. Many grass plants are needed to feed fewer snails on which, in turn, even fewer chickens would be able to feed. This in turn requires only a few people to eat the chickens that ate the snails. The number pyramid shows the number of organisms in each trophic level and does not take into consideration the size of the organisms and over-emphasizes the importance of small organisms. In a pyramid of numbers the higher up one moves, so each consecutive layer or level contains fewer organisms than the level below it. HOME -------------------------------------------------------------------------------- CHAPTER 18 - COMMUNITIES SUMMARY OF INTERACTIONS RELATIONSHIP A B DESCRIPTION Neutralism 0 0 Populations A and B are independent and do not interact Mutualism (symbiotic) + + Both species benefit from the relatoinship Commensalism (symbiotic) + 0 One species benefits, the other neither benefits nor is harmed Competition - - Both species have a negative effect on the other. They compete for the same resources Predation + - The predator benefits. The prey is harmed Parasitism + - The parasite benefits. The host is harmed TYPES OF BIOMES Deserts - limiting factors include water, temperature, radiation hot - lack of moisture in the air allows extreme fluctuations in temperatures. Excessively hot during the day, but extremely cold at night. Primary producers make less than 200 kilocalories of energy per year per square meter of hot desert cold - Found on peaks of high mountains and near the poles. The snow never melts, so the annual amount of precipitation is relatively insignificant. Cold air can not hold much moisture. Primary producers produce about 8,000 kilocalories of energy in the spring in one square meter of space. This amount fluctuates greatly throughout the year. All the primary producers are in the ocean Coniferous Forests - due to short growing season, the forest floor producers about 3,500 kilocalories of food per year per square meter. For every 100 pound deer in these forests, they need approximately 150 square meters to graze. Grasslands - divided into short, medium, and tall grasslands. Much of Oklahoma is grassland Deciduous Forests - Primary producers provide approximately 6000 kilocalories per year per square meter Rain Forests tropical temperate - western edges of North and South America - Primary producers provide about 9,000 kilocalores per square meter per year Tundra - Primary producers make about 600 kilocalories per year per square meter. Aquatic - Temperatures change slowly without extremes. Light does not penetrate far into water intertidal zone - affected by high & low tides (wet & dry). Most organisms have exoskeletons to protect from predators & dehydration estuary - fresh & salt water mix. high in nutrients. animal nursery oceanic shallow mud flats - gentle tides. burrowing animals (clams & worms) oceanic shallow water rocks - strong waves. animals/plants that can cling to rocks. shallow sandy - sand shifts. few plants. burrowing crabs coral reefs - warm, clear, deep water. great biodiversity kelp forests - deep, but light penetrates. Huge kelp (algae) protects many kinds of animals abyssal zone - deep ocean bottom. cold. dark. no plants. little life, but highly varied hydrothermal vents - underwater volcanic eruptions. high in mineral content. organims include bacteria, worms, clams streams and rivers - fresh water. varied plant & animal life lakes - fresh water. varied plant & animal life HOME
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