Assignment 1

docx

School

Athabasca University, Athabasca *

*We aren’t endorsed by this school

Course

325

Subject

Biology

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by ElderHawk3269

March 4 th , 2023 Assignment 1 (covers Units 1–5; Chapters 1–8 of the textbook) Due: After Unit 5 Weight: 10% of course final grade This assignment is out of 100 marks. The assignment mark is 10% of your final grade for the course. 1. What is the Theory of Spontaneous Generation? How did Louis Pasteur disprove this theory? What did his observations provide the basis for, and how has this impacted microbiology? ( 10 marks ) The Theory of Spontaneous Generation was a widely held belief from the nineteenth century that simple life, such as small organisms, could arise spontaneously from nonliving materials. For example, scientists commonly believed that maggots, the progeny of flies, could result from manure or decaying organisms. Scientists such as Redi, Needham, and Spallanzani contested with this theory until 1861, when the French scientist Louis Pasteur demonstrated that air does not in fact create life from nonlife but that microorganisms exist on nonliving matter such as on solids, in liquids, and in the air. To prove this, Pasteur boiled flasks filled with beef broth, then sealed one. After allowing them to cool, he discovered microbes only present within the flask that were not sealed. Following this, Pasteur used a swan neck flask, a flask with its neck in a unique S-shape, and again boiled and cooled beef broth. The S-shaped design allowed air to travel into the flask but trapped airborne microorganisms, thus, the broth was absent of any microbes. This confirmed Pasteur’s theory that microbes contaminate nonliving matter through the air, not the air itself, disproving the Theory of Spontaneous Generation. Rudolf Virchow’s Theory of Biogenesis, the hypothesis that living cells can only arise from preexisting living cells, became the next leading theory. Through Pasteur’s experiments, he formed the basis of aseptic techniques drastically impacting the world of microbiology. He demonstrated that not only can microbes be destroyed by heat but that preventative measures can be used to

control the access of microorganisms. These aseptic practices are now commonly pervasive in laboratory and medical settings allowing for safer and more controlled procedures. Today, aseptic techniques save thousands of lives by helping prevent infection in surgical wounds and allow researchers to manipulate populations of microorganisms in experiments. 2. List and explain one beneficial effect and one detrimental effect of bacteria. ( 5 marks ) Bacteria are not only present in our environment but also live symbiotically on and in the human body. The microbiome refers to the bacterial cells that humans and animals rely on to maintain good health. The bacteria in our gastrointestinal tract, such as Escherichia coli, help with the breakdown and digestion of some molecules, such as cellulose, an insoluble fibre present in fruit and vegetables. Some bacteria that live on and in us can also cause harm, these are pathogenic bacteria. Streptococcus mutans is a bacterium that lives in our mouths and is the main cause of tooth decay in humans. It lives in the pits and fissures on teeth and feeds on the sugar that we consume producing biofilms with enamel-eroding acids causing tooth cavities and oral infections. 3. Column A lists bacterial cell structures. Column B lists the function of these structures. For each item listed in Column A, indicate the correct description from Column B. ( 10 marks ) Column A Column B _i_ cell wall a. energy reserves in cell _e_ capsule b. filamentous appendage for movement _b_ flagella c. contains the cell’s DNA _g_ plasma membrane d. substance inside the cell that contains enzymes and cell structures _f_ ribosomes e. polysaccharide attached to the cell wall that resists phagocytosis _J_ fimbriae f. sites of protein synthesis _h_ LPS g. is composed of lipids and proteins, and serves as a selective barrier _a_ inclusion bodies h. is composed of lipid A, which is an endotoxin _c_ nucleoid i. maintains cell shape _d_ cytoplasm j. external protein structures for adherence 2

4. Describe each of the following stains, and indicate when it is appropriate to use each one: (5 marks) a. simple stain Single, basic, positively charged, aqueous or alcohol solution/dye that adheres to the cell’s surface (negatively charged). A mordant may be used to intensify the staining by increasing the affinity or coating a structure to make it more prominent to see with the dye. e.x., methylene blue, crystal violet, safranin Used to determine cellular shape, size, arrangement, and basic structures (fixing required). b. differential stain Use of dyes that react differently with different kinds of bacteria and often involves using more than one dye. Used to distinguish between different kinds of bacteria (i.e., gram-negative vs. gram-positive, acid-fast stain). c. negative stain Acidic dyes used to contrast the background by staining it and leaving the bacteria colourless (bacteria are negatively charged and repel the dye’s negative ions). Used primarily to distinguish capsules. e.x., eosin, acid fuchsin, nigrosin Used to distinguish overall cell shape, size, and capsules – helps determine the virulence of pathogen (do not need to fix cells to use). d. acid-fast stain Carbolfuchsin (red dye) is applied, heated and then cooled, washed with water, and treated with an acid-alcohol which decolourizes the bacteria then counterstained with methylene blue to stain the non-acid-fast cells. The bacteria with a cell wall that contains lipids will retain the carbolfuchsin as it is more soluble (fixing required). 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Helps detect pathogenic/acid-fast, bacteria: Mycobacterium tuberculosis (causes tuberculosis), Mycobacterium leprae (causes leprosy), and genus Nocardia e. flagella stain A mordant is added to increase the diameter of flagella and carbolfuchsin is used to stain to observe flagella under a light microscope. Used to determine the presence of flagella. 5. Give an example of an organism that can survive best in the following conditions: (5 marks) a. acidic Acidophiles are organisms that grow at an optimal pH below 4. e.x., Acidithiobacillus ferrooxidans – chemoautotrophic bacteria, found in coal mine drainage water, able to oxidize sulphur to sulfuric acid, can survive at a pH of 1 Fungi (yeasts and moulds) b. basic Alkaliphiles are organisms that grow at an optimal pH of around 10. e.x., Natronomonas pharaonis – present in highly saline soda lakes Cyanobacteria c. neutral pH Most organisms grow best between a pH of 6.5 and 7.5 and are referred to as neutrophiles. e.x., Escherichia coli – present within the large intestine which has close to neutral pH 6. Distinguish between respiration and fermentation. Which process is anaerobic? Which process is aerobic? What are the substrates and products of lactic acid fermentation and alcohol fermentation? (10 marks) 4

Cellular respiration and fermentation are two processes microorganisms use to produce energy from the catabolism of sugars. Both pathways begin with glycolysis, the oxidation of glucose, releasing NADH and 2 ATP, resulting in 2 Pyruvic acids. During Cellular respiration, electron carriers NAD + and FAD are reduced to supply electrons to the electron transport chain, producing ATP. It begins by converting Pyruvic acid into Acetyl Coenzyme A. The Kreb cycle (Tricarboxylic acid cycle) uses this Acetyl Coenzyme A to produce ATP through substrate-level phosphorylation releasing CO 2 and reducing the electron carriers to NADH and FADH 2 . These electron carriers then supply electrons to the electron transport chain to produce many ATP by oxidative phosphorylation releasing water. This process is aerobic as it involves the use of oxygen as the final electron acceptor. Overall, cellular respiration takes Glucose, 6 Oxygen molecules, 38 ADP, and 38 inorganic phosphates and produces 6 Carbon dioxide molecules, 6 water molecules, and 38 ATP (36 in eukaryotes). Fermentation, in contrast to cellular respiration, is an anaerobic pathway that does not require the use of oxygen, but instead uses an organic molecule, other than oxygen, synthesized in the cell as the final electron acceptor (i.e., Pseudomonas and Bacillus use nitrate ion, Desulfovibrio uses sulphate). During fermentation, the catabolism of Pyruvic acid allows for the reduction of NADH, needed for glycolysis, the only ATP-producing step. Pyruvic acid is converted into various end products depending on the type of organism. For example, Streptococcus, Lactobacillus, and Bacillus have an end product of Lactic acid, referred to as lactic acid fermentation. After glycolysis, the 2 molecules of pyruvic acid are reduced by 2 molecules of NADH to produce 2 molecules of lactic acid. The lactic acid stores most of the energy produced by the reaction yielding only the energy from glycolysis. Similarly, Saccharomyces (yeast) has an end product of Ethanol and CO 2 , called alcohol fermentation, during which the 2 molecules of pyruvic acid (product of glycolysis) are converted into 2 molecules of acetaldehyde, releasing 2 molecules of carbon dioxide, and then subsequently converted into 2 5

molecules of ethanol by reducing 2 NADH. Similar to lactic acid fermentation, in alcohol fermentation, the only ATP production comes from glycolysis. Lactic Acid Fermentation Glucose ↓ 2 Pyruvic acid ↓ 2 Lactic Acid Alcohol Fermentation Glucose ↓ 2 Pyruvic acid ↓ 2 Acetaldehyde ↓ 2 Ethanol 7. Define the following terms, and identify which of these four nutritional groups most of the medically important microbes would belong to. (5 marks) a. photoautotroph Organisms that use light as a source of energy and carbon dioxide as their main source of carbon. e.x., photosynthetic bacteria, algae, green plants b. photoheterotroph Organisms that use light as a source of energy and organic compounds (alcohols, fatty acids, organic acids, etc) as their main source of carbon (cannot convert carbon dioxide to sugar). 6 1. Glyc olysis 2. 2 NAD 3. 2 4. 2 5. 2 6. 2 7. 2 NAD 8. Glyc olysis 9. 2 NAD 10. 2 11. 2 12. 2 13. C O 14. 2 15. 2 NAD

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

e.x., nonsulfur bacteria (green and purple) c. chemoautotroph Organisms that use electrons from reduced inorganic compounds as a source of energy and carbon dioxide as their main source of carbon. e.x., sulphur-oxidizing bacteria, nitrogen-fixing bacteria d. chemoheterotroph Organisms that receive their energy and carbon from the same organic compound, such as glucose, use the electrons from hydrogen atoms as their energy source. Most bacteria and all fungi, protozoa, and animals are chemoheterotrophs. This means most medically important microorganisms are chemoautotrophs as infectious organisms rely on catabolizing substances from their host for nourishment. 8. A buffet restaurant uses heat lamps at a temperature of 50 o C to keep the food warm. If the restaurant opens for service at 4 pm and closes at 10 pm, some of the food items that are not refilled frequently may sit for 6 hours at this temperature. The following experiment was conducted to determine whether this temperature is suitable to maintain the food and prevent bacterial growth: Beef cubes were inoculated with 500,000 bacterial cells and incubated at 43-53 o C to establish temperature limits for bacterial growth. Table 1 contains the numbers of bacteria after incubation for 6 hours, and plating on nutrient agar given in the numbers of colonies for a particular dilution. Calculate the cfu/ml for each organism at each temperature using the formula cfu/ml = # colonies x dilution factor x plating factor (10) and draw a line or bar graph to indicate the amount of growth after 6 hours for each temperature. What holding temperature would you recommend to maintain the food with minimal contamination? Is 50 o C sufficient? Assuming that cooking kills bacteria in food, how could these bacteria contaminate the cooked foods? What disease does each organism cause? (10 marks) For Staphylococcus aureus 43 o C: 7

cfu/ml = # colonies x dilution factor x plating factor = 15 x (1/1/1000000) x 10 = 1.5 x 10 8 51 o C: = 8 x (1/1/10000) x 10 = 8.0 x 10 5 53 o C: = 6 x (1/1/100) x 10 = 6.0 x 10 3 For Salmonella typhimurium 43 o C: = 32 x (1/1/100000) x 10 = 3.2 x 10 7 51 o C: = 9 x (1/1/10000) x 10 = 9.0 x 10 5 53 o C: = 2 x (1/1/1000) x 10 = 2.0 x 10 4 For Clostridium perfingens 43 o C: = 17 x (1/1/100000) x 10 = 1.7 x 10 7 51 o C: = 12 x (1/1/1000) x 10 = 1.2 x 10 5 53 o C: = 4 x (1/1/100) x 10 = 4.0 x 10 3 TABLE 1. Numbers of colony-forming units per ml of bacteria in beef cubes after incubation at different temperatures. 8

Organism Temp. ( o C) # Colonies Dilution D. Factor cfu/ml Staphylococcu s aureus 43 15 1:100000 0 10 6 1.5 x 10 8 51 8 1:10000 10 4 8.0 x 10 5 53 6 1:100 10 2 6.0 x 10 3 Salmonella typhimurium 43 32 1:100000 10 5 3.2 x 10 7 51 9 1:10000 10 4 9.0 x 10 5 53 2 1:1000 10 3 2.0 x 10 4 Clostridium perfringens 43 17 1:100000 10 5 1.7 x 10 7 51 12 1:1000 10 3 1.2 x 10 5 53 4 1:100 10 2 4.0 x 10 3 Figure 1. Number of colony-forming units per ml of bacteria in beef cubes after a 6-hour incubation at different temperatures on a semi-logarithmic plot. Given the data in Table 1 containing the number of bacteria colonies present on beef cubes after being incubated at 43°C, 51°C, and 53°C, only temperatures at 53°C produce a negative microbial growth rate. With this dataset, I would contest that 50°C is 9

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

not sufficient and that a temperature of at least 53°C should be used to promote minimal bacterial contamination of the species Staphylococcus aureus, Salmonella typhimurium, and Clostridium perfingens . Though cooking can kill bacteria in food, microbes are still present in the air, on cooking utensils, and transferred from humans surrounding the food that will be sitting for 6 hours, giving plenty of time for it to be contaminated by foreign microorganisms. Additionally, endospores, an inactive form of bacteria, are resistant to heat and may stay present in food even after cooking. Staphylococcus aureus , which causes gastroenteritis (food poisoning/stomach flu) and toxic shock syndrome, is present in nasal secretions and on the skin. It can produce many toxins and is resistant to penicillin making it of particular concern. Salmonella typhimurium causes salmonellosis (salmonella infection/poisoning) resulting in fever and gastroenteritis. This comes from food that is not properly cooked. Clostridium perfingens causes gangrene and gastroenteritis (foodborne diarrhea) and typically infects food through endospores that are resistant to heat. 9. Table 2 summarizes results obtained from an experiment of microbial sampling onto media that can help detect Enterobacter spp . from a freshwater spring. Based on these results, answer the questions below. You may wish to consult the Photographic Atlas for supplemental information. (10 marks) a. Compare the numbers of colony-forming units between nutrient agar and EMB agar for each temperature. In both the nutrient agar and Eosin Methylene Blue agar (EMB) the colony- forming units of Enterobacter spp. follow the same growth pattern in response to temperature changes. At the low temperature of 4°C, there are the fewest colonies, 0 on the EMB agar and 3.0 x 10 1 on the nutrient agar. When the temperature increases to 20°C there is a significant increase in colony-forming units present, 20 on the EMB agar and 7.2 x 10 2 on the nutrient agar. The biggest increase in the colony-forming unit is at 37°C, the preferred temperature of most bacteria, with the EMB agar having 1.7 x 10 4 colonies and the nutrient having 2.1 x 10 6 . At a temperature of 42°C, the number of colony-forming units decreases significantly to 9 on the EMB and 1.5 x 10 2 on the nutrient agar. 10

During all temperatures, the nutrient agar samples produced a higher number of colony-forming units than did the EMB agar. b. Why is there a difference in the number of organisms between the two types of media? The difference in the number of colony-forming units between Eosin Methylene Blue agar and nutrient agar is due to their different nutrient composition. Nutrient agar contains partially hydrolyzed proteins, peptone, beef and yeast extract which provides vitamins, carbohydrates, nitrogen, salts, distilled water, and agar. Eosin Methylene Blue agar contains the digest of gelatin, lactose, and the dyes eosin Y and methylene blue. These dyes prohibit the growth of gram-positive organisms. As Enterobacter is gram-negative it is still able to grow but is limited by the altered availability of nutrients. c. Describe the purpose of EMB agar in this experiment, and describe how this purpose is achieved. Eosin Methylene Blue agar is used in the isolation of fecal coliforms by streaking and the inhibition of gram-positive organisms. The lactose in the EMB agar provides nutrients for the selective growth of Enterobacter spp. while inhibiting other gram-positive organisms in the freshwater spring sample. d. Does Enterobacter spp . cause disease? Enterobacteriales commonly inhabit the intestinal tract. Two species of Enterobacter, E. Cloacae and E. aerogenes, can cause urinary tract infections and healthcare-associated infections in immunocompromised hosts. e. What can you deduce about the temperature requirements of these organisms? Similar to other bacteria found in the human body, Enterobacter spp. thrive at 37°C during which they were able to grow the most in both agar plates despite nutrient differences. TABLE 2. Numbers of colony-forming units per ml of sample obtained from spring water subjected to different temperatures. Incubation temperature (  C) Nutrient agar Eosin Methylene Blue agar (EMB) 4 3.0  10 1 0 20 7.2  10 2 20 11

37 2.1  10 6 1.7  10 4 42 1.5  10 2 9 10. Fill in the table below. (10 marks) Chemical Agent Mechanism of Action Preferred Use Phenol Disrupts lipid-containing plasma membranes causing cellular leakage. Not commonly used due to its irritating qualities and pungent smell. Able to stay active and stable in the presence of organic substances and thus are suitable for disinfecting pus, saliva, and feces. Chlorhexadine Disrupts cell membrane (particularly gram-positive bacteria). Bactericidal. Used for surgical hand scrubs and preoperative skin disinfection. Iodine Forms complexes with amino acids and unsaturated fatty acids impairing protein synthesis and altering cell membranes. Very effective antiseptic (active against bacteria, many endospores, various fungi, and some viruses). Skin disinfection and wound treatment. Can be used in water treatment. Alcohol Denatures proteins and dissolves lipids. Bactericidal and fungicidal. Not effective against endospores and nonenveloped viruses. Used for mechanical debridement of dirt/microorganisms (degerming) and disruption of skin oils before an injection. Silver nitrate Denatures proteins and enzymes. Biocidal or antiseptic. Used in newborn’s eyes to guard against Opthalmia neonatorum (slowly being replaced by antibiotics). Soap Mechanical removal of microbes through scrubbing. Used as degerming and debris removing agents. Quaternary Ammonium Compounds Disrupt plasma membrane changing cell permeability. The chemical mode of action is unknown. Enzyme inhibition and protein denaturation. Antiseptic, bactericidal, bacteriostatic, fungicidal, virucidal (enveloped viruses). Used on medical instruments, utensils, and rubber. Nitrates Inhibits iron-containing enzymes preventing the growth of endospores. Added to meat to prevent Clostridium botulinum (botulism). Ethylene Oxide Kills microbes and endospores by inhibiting vital cellular function. Used in sterilization of materials that would be damaged by heat (e.x., hospital mattresses). 12

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Aldehydes Denatures proteins. Most effective antimicrobials. Bactericidal and virucidal. Used in disinfecting hospital equipment and by morticians for embalming. 11. Is the lac operon a repressible or an inducible operon? What happens in the absence of lactose? What happens in the presence of lactose? Is the trp operon a repressible or an inducible operon? What happens in the absence of tryptophan? What happens in the presence of tryptophan? (10 marks) The operon model of gene expression permits the regulation of protein synthesis through the induction and repression of gene transcription. This mechanism allows for the up-and-down regulation of specific proteins in response to the presence or absence of specific precursor molecules. The lac operon is an inducible operon in Escherichia coli used to control the metabolism of lactose. Lactose must be present for the structural genes of the operon to be expressed. When Lactose is present the repressor protein is inactive and can no longer bind to the operon. RNA Polymerase can then bind to the promotor and the genes that encode the proteins that help break down lactose are expressed. When lactose is absent, the repressor is active and is able to bind to the promotor, blocking the binding of RNA Polymerase. The cell is then able to limit the production of lactose-metabolizing proteins as a way to save energy and resources. The trp operon is a repressible operon in Escherichia coli used to control tryptophan synthesis. Tryptophan is needed constantly for the survival of E. coli as it is an amino acid used to make proteins, thus should only be limited when excess amounts are present in the cell. The regulatory gene in the trp operon is derepressed and codes for an inactive repressor that is unable to bind with the operon allowing for the constant production of enzymes that help make Tryptophan. When there is an excess, Tryptophan acts as a corepressor and binds to the repressor, changing it to its inactive shape so that it can now bind to the operon. This blocks RNA Polymerase from transcribing the trp operon genes. Through this, 13

the cell is able to save energy by limiting the amounts of Tryptophan production when high concentrations are present. 12. A researcher is interested in isolating an auxotrophic mutant in the laboratory that is unable to synthesize the amino acid histadine. Explain how this researcher would have to use negative selection to isolate and identify a histadine mutant. (5 marks) An auxotroph is a mutant microorganism that has some nutritional requirements deviant from parental generations. Negative (or indirect) selection is the process that selects for cells that have lost a specific function or gene. To isolate a histadine auxotrophic mutant a researcher would have to make use of replica plating techniques through negative selection. To isolate and identify a histidine mutant, researchers innoculate 100 bacterial cells onto an agar plate, the master plate, containing histidine. After 18 to 24 hours a pad of sterile material is placed on top of the newly formed colonies on the master plate allowing for some to transfer to the sterile material. This pad is then transferred to two new sterile medium plates, one containing and one lacking histidine, and incubated. As mutants are rare, many plates may need to be screened. Colonies that cannot grow on the medium lacking histidine will be histadine mutant auxotrophs. 13. Differentiate between vertical and horizontal gene transfer. List and describe the three types of horizontal gene transfer in bacteria. (5 marks) Vertical gene transfer is the movement of genes from a parental generation to progeny through sexual or asexual reproduction. In addition to vertical gene transfer, Bacteria are also able to share their genetic information through horizontal gene transfer. Horizontal gene transfer is the movement of a selective portion of DNA from a donor cell to a recipient cell. The donor cell then incorporates the DNA into its own and degrades the rest with cellular enzymes, the resultant DNA is now recombinant. 14

This process is of significant medical importance as it allows for the spread of antibiotic resistance, though it only occurs in less than 1% of a population. The three mechanisms of horizontal gene transfer in bacteria include transformation, transduction, and conjugation (the most frequent). Transformation in bacteria occurs when fragments of DNA, not associated with any other molecule, are transferred in a solution from one bacterium to another. Transduction is the bacterial transfer of DNA through a virus called a bacteriophage. Occasionally, during the lifecycle of a bacteriophage, it may acquire parts of a donor cell’s DNA. When the donor cell lyses due to the viral infection, the bacteriophage may go on to infect another recipient cell releasing the previous bacterium’s DNA, and creating recombinant DNA. Lastly, conjugation is the direct cell-to-cell transfer of a single- stranded copy of plasmid DNA via a sex pilus (gram-negative) or a mating bridge (gram-positive) between two bacteria of opposite mating types. An F factor (fertility factor), a plasmid/DNA sequence, uses the sex pilis or mating bridge to transfer from the donor cell (F + cell) to a recipient cell (F - cell). The F factor then integrates into the recipient's DNA to create a high-frequency recombination cell (Hfr cell). This cell then goes on to pass a portion of its DNA to a new recipient cell (recombinant F - cell) via conjugation. 15

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Assignment 1

Related Documents