Labs 17-24, 26-27, 29-31

Question 1

In what scenario do you need to use selective and differential media?

Answer 1

Selective and differential media are used when isolating bacteria from a source where it is suspected that there are more than one organism present/Isolate desired microbes from mixed cultures/Selective agents: Prohibit growth of certain organisms

Question 2

What is a selective medium? 2 examples?

Answer 2

Prohibit the growth of some organisms while encouraging the growth of others.
Uses a selective agent or agents added to the medium.

The selective medium allows the organism of interest to grow, while inhibiting the growth of others. MAC and MSA plates are both examples of selective media.

Question 3

How does MacConkey’s agar work?

Answer 3

it contains crystal violet and bile salts that inhibit the growth of gram-positive organisms and don’t inhibit the gram-negative bacteria. The bile salts also aid in the isolation of bacteria that are found in the intestinal tract
(enteric bacteria)

Question 4

How does Mannitol salt agar work?

Answer 4

MSA selects for the growth of Gram positive bacteria and inhibits gram-negative bacteria. This is because it contains high levels of salt and gram-negative bacteria cannot survive in high salt concentrations because they have thin cell walls/peptidoglycan layer. On the other hand, staphylococci and gram positives can tolerate high levels of salt present in MSA.

Question 5

What is differential media? How is it different from selective media? 2 examples?

Answer 5

Its different from selective in that the growth is not inhibited. Differential media contains substances that allows certain groups of bacteria to appear differently than others when grown on the media. It therefore differentiates between two groups of bacteria. This type of media allows it to be determined if a certain species of bacteria is growing on the media just by observing the colonies.

Differential media contain one or more agents that react with different substances produced by the bacteria (end products) to yield a color change in the medium and/or the colonies growing on it. Often, the differential agent is a pH indicator that can distinguish between acidic or basic end products of metabolism. The indicator will change the color of the medium and sometimes the colonies.Differential media are used to differentiate closely related organisms or groups of organisms

-MSA and MAC

Question 6

How are MSA and MAC both selective and different?

Answer 6

MSA and McConkey’s are both considered selective media
● McConkey’s contains crystal violet and bile salts, prohibiting gram+ growth
● MSA contains high levels of salt, prohibiting the growth of gram-

MSA and McConkey’s are also both considered differential media
● MSA contains mannitol and phenol red - When mannitol is fermented, the pH drops, and the medium around the colony turns yellow. The medium around non-mannitol fermenters remains pink.

● McConkey’s contains lactose and neutral red - if the colony ferments lactose, pH drops and
the colony will turn red, otherwise it will stay transparent

if an organism can utilize lactose, it produces acidic end products. These end products lower the pH pf the media

Question 7

general procedure for selective and differential media

Answer 7

-Take a plate of MSA/MAC and divide the bottom into four
-Label each quarter with the four organisms
-Streak each organisms in its quarter
-Incubate at 37 C for 24 hours

Question 8

What is the Kligler Iron Agar (KIA)

Answer 8

KIA is a differential medium that serves as a determiner of both carbohydrate fermentation and gas production. Its useful for determining whether a Gram-negative rod is a member of the family Enterobacteriaceae. In other words, it identifies Enterics

Question 9

What are Enterics?

Answer 9

They are gram-negative facultative bacilli that ferment the carbohydrate glucose to acidic end products. They are very similar morphologically. Many are pathogenic.

Facultative: runs “normal” respiration in the presence of oxygen, ferments without it.

Question 10

How is the KIA medium prepared?

Answer 10

It is prepared as an agar slant 4 cm deep. It is inoculated with an innoculating needle NOT A LOOP. It is inoculated by first stabbing the agar butt and then streaking the slant.

Question 11

What is KIA formulated with?

Answer 11

KIA contains 2% polypeptone, 1% lactose, and 0.1% glucose, phenol red, ferrous sulfate

Question 12

Carbohydrate (glucose or lactose) fermentation will leave? Where does utilization occur?

Answer 12

acidic by products, making the agar yellow due to phenol red. Which is yellow at a pH lower then 6.8 and red above it. utilization can occur on the slant (aerobically) or in the butt (anaerobically)

Fermentation reactions are read on the slant, and in the butt, indicated by a color change from red (alkaline) to yello (acid)

Question 13

Organisms can ferment what in KIA?

Answer 13

Organisms can ferment both glucose and lactose, others only ferment glucose, some cannot ferment either

Question 14

What does the KIA slant tell us if its color changes?

Answer 14

How to interpret your results (always read your tube “ slant/ deep”)
1. Yellow slant/vellow deep = (A/A) glucose and lactose are fermented; if the butt has bubbles in it or is displaced, the tube is gas positive.
2. Red slant/yellow deep = (K/A) glucose, but not lactose is fermented.
3.
Red slant/black deep = (K/A) glucose but not lactose is fermented, and HS is produced. The deep is acid (yellow).
4. Red slant/red deep = (K/NC) neither carbohydrate is fermented. Peptone is degraded aerobically

Question 15

When does gas production take place? What are the gases? How do they appear in the tube?

Answer 15

CO2 and H2 occur when sugar utilization take place. Gas produced in the kIA tubes appear as a single gas bubble or many gas bubbles trapped in the agar. It may just split the slant completely

Question 16

What occurs when an organism does not ferment glucose or lactose?

Answer 16

It will utilize the peptone and turn the medium red. only on the slant if the organism is aerobic, and in the deep if it is anaerobic. Both will turn red if it can catabolize the peptone both anaerobically and aerobically.

Question 17

By product of carbohydrate metabolism for KIA?

Answer 17

Hydrogen sulfide (H2S)
This occurs by the reduction of thiosulfate in the medium or by the breakdown of cysteine in the peptone. Its colorless so have to use indicator, ferrous sulfate, forming a black precipitate in the butt of the tube.

will create a black
precipitate at the bottom of the tube. If this obscures the color of the agar, it will be counted as yellow, since the acidic conditions are required to make it

Question 18

Overview of KIA procedure?

Answer 18

Label each tube w/ the organism being inoculated
● Inoculate a tube for each organism with an inoculating needle
○ Don’t stab all the way to the bottom or into the sides of the tube - this needs to be kept anaerobic
● Carefully withdraw the needle and streak the slant
● Incubate tubes at 37 C for 24 hours

Question 19

Compare and contrast Streptococcus and Staphylococcus:

Answer 19

● Both Gram positive, non-spore forming and spherical shaped
● Staphylococcus sp. grow in clusters
● Streptococcus sp. grow in chains
● Staphylococcus sp. contain catalase Streptococcus sp. do not

Question 20

Why is Gram-staining the organism the first step in distinguishing Streptococcus and Staphylococcus?

Answer 20

When grown in thioglycolate broth and then Gram-stained, it is possible to visualize the groupings of the cells.

Question 21

Explain the catalase test

Answer 21

A biochemical test can then be performed to distinguish between Staphylococcus and Streptococcus.. Catalase breaks down hydrogen peroxide (H202) to water (HO) and oxygen gas (O2). Catalase protects the cell from the damaging effects of H2O2 to enzyme systems. When 3% H2O2 is dropped onto a Staphylococcus colony, visible bubbles are formed from the release of oxygen gas. Streptococcus colonies do not form bubbles when H2O2 is added because they do not contain catalase.
Instead, the cells die.

When the determination of the Staphylococcus genus has been made, it is important to determine whether the organism is S. aureus using coagulase test.

Question 22

Explain the coagulase test.

Answer 22

If the determination of the Staphylococcus genus has been made, it is important to determine whether the organism is S. aureus using coagulase test.
Coagulase is an enzyme that causes plasma to coagulate (clot) at 37°C. Produced by many species.

S. aureus can be distinguished from other species in the genus because it sometimes grows as yellow-pigmented colonies, produces hemolysins, ferments mannitol, and produces coagulase.

Question 23

Why is S. aureus a significant microorganism?

Answer 23

Causes more diseases in the US than any other bacteria. It causes abscesses, boils, conjunctivitis. food poisoning and toxic shock syndrome.

Question 24

Describe the identification of staphylococci procedure.

Answer 24

1. take a piece of para film, remove backing and dimple.
2. add H202 into para film
   3.Using loop transfer isolated colony of bacteria into hydrogen peroxide
3. Look for vigorous production of oxygen aka bubbles

Coagulase test
1. Demonstration: inject .1 mL of rabbit plasma into Durham tube, transfer
colony of S. epidermidis and incubate at 37 ̊C for 24 hours.

Question 25

Purpose of IMViC tests? What are the tests?

Answer 25

to identify enteric bacteria. There are four different tests. ○ Indole production- tryptophanase breaks down tryptophan into pyruvic acid, ammonia, and indole. Indole medium contains peptone and is enriched with tryptophan. After incubation for 48 hours, a small amount of Kovac's reagent is added, and turns rose colored if indole is present. ○ Methyl red and Voges-Proskauer - both tests are done in the same tube. The media contains peptone, glucose, and d-potassium phosphate. Glucose utilization with produce different results when methyl red is added. ■ Voges-Proskauer - made of KOH and alpha-naphthol. Tests for 2,3-butanediol by looking for acetoin, its precursor. The chemicals that make up the VP broth react with acetoin and produce a pink or red color ○ Citrate utilization test - Tests for citrate utilization by citrase, containing only sodium citrate as a carbon source. Bromothymol blue is added and turns blue if citrate is utilized.

Question 26

explain IMViC test procedure.

Answer 26

● Label tubes of tryptone (indole test), MRVP, and citrate with the organisms you will inoculate ● Inoculate tryptone broth, MRVP broth, and citrate slant (do not stab slant) ● Repeat for each organism ● Incubate all tubes at 37 C for 48 hours For indole: ● Dip sterile cotton swab in grown culture ● Add one or two drops of reagent to swab tip ● Repeat with a control (uninoculated tryptone) ● Kovac's reagent will turn red in the presence of indole For MR: ● Pipette one mL from MRVP tube into an empty tube, labeled with MR and the organism name ● Add approx. 20 drops of methyl red ● Red color is positive, orange to yellow is negative For VP: ● Use the remaining MRVP broth for this ● Add about 12 drops Barritt's solution A and shake from side to side ● Add 3 drops Barritt's solution B and shake vigorously, and reshake in 2-3 minute intervals ● Positive result is an intense rose color For Citrate: ● Examine citrate slants for growth and compare to uninoculated control for color change ● Change from green to royal blue is a positive test

Question 27

Describe urea hydrolysis.

Answer 27

Some Enterobacteriaceae contain the enzyme Urease. Urease catalyzes the hydrolysis of urea into ammonia and carbon dioxide. The ammonia is used for production of nucleotides and amino acids Members of the genus Proteus are distinguished from other enteric bacteria due to their fast urease activity

Question 28

Similarity of Genus proteus and other enteric pathogens

Answer 28

cannot utilize lactose just like most enteric pathogens. Must Distinguish the genus Proteus from enteric pathogens.

Question 29

The medium used to test for urease activity?

Answer 29

Stuart's urea broth tests for urease activity. ○ The broth contains urea, phenol red as a pH indicator, and is highly buffered ○ Only the genus Proteus can perform urea hydrolysis in high enough amounts to overcome the buffers. produces enough ammonia to overcome high buffer ○ When the solution reaches a pH of 8.4, it turns from orange to pink ○ Along with the negative lactose result, this indicates the presence of Proteus

Question 30

Overview of urea hydrolysis procedure.

Answer 30

Label 3 tube of Stuart's urea broth "Blank," "M. luteus," and "E. coli." ● Inoculate one tube with M, luteus and the other with E. coli, leaving the third uninoculated ● Incubate at 37 C for 48 hours Observe for hot pink/bright red color for positive result

Question 31

Explain streptococci in the mouth. The ability of an organism to produce what are associated with virulence? where are these tested?

Answer 31

At birth, the mucous membrane of the mouth and throat are sterile, yet streptococci of the viridian's establish themselves within 4 to 12 hours. These bacteria which include S. mitis and S. salivarne are a part of the normal microbiota. Some pathogenic bacteria may occur as permanent or short lived members of the human throat. These include: S. pnenmoniae, S. Progenes, Neiseria meningitis, and Hasmophilas influenza and can cause a variety of diseases. The ability of an organism to produce hemolysins is associated with virulence, and many of these species produce them. Hemolysins are enzymes which lyse red blood cells (RBC), and therefore the production of these enzymes can be tested in blood agar plates. There are three categories of hemolysins: alpha, beta and gamma.

Question 32

What are alpha, beta, and gamma Hemolysins?

Answer 32

○ Beta hemolysins - completely lyse RBCs, and then the bacteria further degrade the hemoglobin ■ Shown in blood agar as a clear round region around the colony - the zone of hemolysis ○ Alpha hemolysis - RBCs are partially disrupted, and hemoglobin is methylated, which give it a green color. ■ Shown in blood agar as a green zone around the colony ○ Gamma hemolysis - no destruction of RBCs ■ No change in the blood agar

Question 33

What does the blood agar do?

Answer 33

As a differential medium, blood agar differentiates between hemolytic and non-hemolytic bacteria as well as the types of hemolysins produced.

Question 34

What do Bacitracin and Optochin do? What agar?

Answer 34

- antibiotics, inhibit the growth of S. pyogenes and S. pneumoniae, respectively to test for the 2 using Mitis-Salivarius agar

Question 35

What is the purpose of the bile esculin test? What is a positive result?

Answer 35

The search for the identification of group D Streptococci. This group includes pathogens such as Enterococcus faecalis and S. faecium (opportunistic urinary tract infections and endocarditis). In some people E. faecalis can be part of the normal microbiota. The bile- esculin test is based on the ability of E. faecalis to propagate in the presence of 4% bile, and to hydrolyze esculin. Esculin hydrolysis results in a dark brown color in the bile-esculin medium. Only this organism grows and blackens bile-esculin agar.

Question 36

What organism is sensitive to the Bacitracin disc?

Answer 36

S. pyogenes

Question 37

What organism is sensitive to the Optochin disc?

Answer 37

S. pneumoniae

Question 38

Appearance of S. mutant, S. salivarías, and S.mitis on the Mitis-Salivarius plate?

Answer 38

-S. mutans forms light blue colonies approximately 3-4 mm in diameter. These colonies have a "Ground glass" or "frosted" appearance. -S. salivatius forms fairly large (1-5 mm in diameter), heaped, blue "gum drop" colonies. -S. mitis is another member of the oral microbiota that you may isolate on your Mitis-Salivarius agar plate. This organism forms small or minute navy blue to black colonies.

Question 39

questions

Answer 39

4.Why is optochin more effective on S. pneumoniae than S. pyogenes? And why is bacitracin more effective than S. pyogenes than S. pneumonia? Optochin is more effective on S. pneumoniae than S. pyogenes because S. pneumoniae is more sensitive to it and has a high affinity uptake to the antibiotic while S. pyogenes does not have this high-affinity uptake system making it resistant to optochin. Bacitracin is more effective on S. pyogenes than S. pneumoniae because S. pyogenes is also more sensitive to the antibiotic. Bacitracin alters the peptidoglycan and S. pyogenes is affected by this function of bacitracin due to its cell wall structure. S. pneumoniae has a different cell wall structure so it's less susceptible to bacitracin. 5.Compare your results to the back of the book (Table of Biochemical Tests). Do any of your tests differ from the expected result? If your results differ, please explain what happened? The Bile esculin experiment results do align with the results in the book. 6.Certain gum products have a sugar alcohol called Xylitol. Does Xylitol enable bacteria to grow or kill them? Unlike regular sugar, bacteria cannot metabolize xylitol in order to grow or reproduce. Because of this, the bacteria starve to death when xylitol gum is chewed. So Xylitol does not directly kill bacteria upon impact, but it causes the bacteria to starve and die off, keeping it from having a negative impact on the teeth. 3.How are our oral microbiota involved in tooth decay? Our oral microorganisms are responsible for the formation of acid in dental plaque which through buildup can lead to tooth decay. This is why it is important to maintain a healthy oral microbiome.

Question 40

Procedures for streptococci in mouth

Answer 40

Activity 1 - Throat Swabs ● Label streak plate ● Rotate a sterile swab over your tonsillar of you mouth without touching your tongue or any other area (a tongue depressor may be used to facilitate this) ● Streak the 0 quadrant with the swab and use a loop for the rest of the plate ● Incubate at 37 C until next lab Activity 2 - Normal Oral Microbiota ● Label 1⁄2 of your Mitis-Salivarius agar plate "tongue" and the other "gum line" ● Using a sterile swab, rub your gum line and streak the "gum line" half of the plate ● Repeat the process with your other swab, taking a sample from your entire tongue ● Incubate the plate at 37 C until the next lab Activity 3 - Snyder Test ● Chew a rubber band to create saliva and spit into a test tube ● Pipette 0.2 mL saliva into Snyder test agar and lightly mix ● Incubate at 37 C Activity 4 - Bile Esculin Agar Slant ● Streak the surface of a bile esculin slant with E. faecalis ○ Don't stab the slant ● Incubate at 37 C for 48 hours

Question 41

Overview of nitrate reaction exercise.

Answer 41

Facultative organisms contain enzymes that reduce nitrate and in some cases nitrite. Nitrate can be reduced to several compounds depending on the ability of the bacteria to produce two enzymes: Nitrate reductase and Nitrite reductase Nitrate reductase converts nitrate to nitrite. Nitrite reductase converts nitrite to ammonia or gaseous nitrogen. Most enteric bacteria are nitrate reducers such as e.coli and salmonella typhi

Question 42

What does reduction and facultative microbes mean?

Answer 42

reduction:The addition of hydrogen and or electrons to a molecule with the removal of oxygen form the molecule. Oxygen is used as the hydrogen acceptor in anaerobic respiration. Facultative microbes: microbes that can utilize both anaerobic and aerobic metabolism depending on oxygen availability. These microbes can produce ATP through aerobic metabolism, however, can switch to ATP production through fermentation during anaerobic conditions

Question 43

What media is used for detection of nitrate reduction?

Answer 43

Tht type of media used is nitrate broth which is undefined meaning it contains several nutrients but only KNO3 as a nitrogen source.

Question 44

When an organism is incubated in nitrate broth, only a couple things can occur.

Answer 44

-Nitrate is unaltered, meaning no enzymes were present. -Nitrate is converted to nitrite, meaning only nitrate reductase is present -Nitrate is converted to nitrite then converted to ammonia or gaseous nitrogen, meaning both enzymes are present.

Question 45

What must be done after incubation during nitrate reduction test?

Answer 45

After incubation, Reagents containing sulfuric acid and alpha-naphthylamine are added to detect the presence of nitrite. If nitrite is present, it'll produce an intense red color indicating the reduction of nitrate to nitrite. Positive test. If no color change occurs it means either the organism is unable to reduce nitrate, or it reduces nitrate to some compound other than nitrite such as nitrogen gas. In order to find that out, zinc dust is added to the medium. Zinc directly reduces nitrate to nitrite. So when added the zinc will reduce the original nitrate of the broth, and the red color is observed. This means that a red color upon addition of zinc is a negative result for nitrate reduction, neither enzyme is present. If no color change appears this means that the nitrate has been reeduces all the way to ammonium or nitrogen gas and is a positive test.

Question 46

describe nitrate reduction procedure.

Answer 46

Need: 3 tubes of nitrate broth Label each tube: Blank E. coli P. stutzeri Inoculate each tube with appropriate bacteria Incubate at 37*C for 48 hr. 2.

Question 47

Describe carbohydrate fermentation exercise?

Answer 47

The carbohydrate fermentation test is used to determine whether or not bacteria can ferment a specific carbohydrate. ad convert to glucose. Carbohydrate fermentation patterns are useful in differentiating among bacterial groups or species. It tests for the end product presence of acid and/or gas produced from carbohydrate fermentation use media with different carbohydrates

Question 48

Media for carbohydrate fermentation exercise? also list colors, pH, what to make sure to do when reading the results after incubation

Answer 48

Carbohydrate Broth- includes peptone, pH indicator phenol red, and specific carbohydrate being tested. Phenol red willturn yellow at pH < 6.8 and red to fuchsia pH > 7.4. When acid is produced the broth will turn yellow. Peptone metabolized will produce ammonia thus raising the pH turning the broth red. Durham tube with bubble indicates gas production. Read test no later than 48 hours to avoid missing ability to produce acid because they'll exhaust the carbohydrate quickly.

Question 49

Describe carbohydrate fermentation procedure and what the results mean

Answer 49

-Inoclate carbohydrate tubes with bacteria at 37C -Appearance of a yellow-colored broth indicates the fermentation of that particular carbohydrate. Acid has been produced. This is a positive (A) carbohydrate result. -Presence of a bubble in the inverted Durham tube (G) indicates the production of gas during metabolism. Appearance of a pink colored broth indicates the utilization of peptones in the broth. This is a negative carbohydrate result. A no change condition in the broth indicates that no fermentation of that particular carbohydrate occurred. This is a negative (-) carbohydrate result. Note that the organism is also unable to breakdown the peptone). Note. Sometimes the medium turns into a color, which is somewhat intermediate between the original color and yellow. This is interpreted as a weak (W) positive.

Question 50

Antibiotics experiment purpose?

Answer 50

to determine the susceptiblitly of bacteria to various antibiotics. Antibiotics are Used to treat various bacterial infections and ailments. NOT effective on viruses. used in clinical settings to identify microbe causing infection or to determine effective antibiotic in controlling the microbe

Question 51

How do antibiotics work differently on bacteria?

Answer 51

* Bacteriostatic effect – Prohibits further growth of bacteria * Bactericidal effect – Kills the bacteria outright

Question 52

How is the antibiotic experiment done? how can you tell if it is effective or not?

Answer 52

This is done by growing pure colonies of the diagnosed infectious agent on agar plates and placing antibiotic-soaked paper discs on top of the agar. As the plate incubates, the antibiotics diffuse from the discs into the culture medium. This diffusion causes the antibiotics to become more diluted as they move away from the disc. Bacterial growth will occur up to the disc if the antibiotic is not effective (the organism is resistant to it). However, if the antibiotic inhibits bacterral growth, it will create a clearing around the disk where the organism Cannot grow because the antibiotic is too concentrated. This ring is called "zone of inhibition"

Question 53

“zone of inhibition”, and its size depends on what various factors: How is it measured?

Answer 53

1) the sensitivity of the organism to the specific antibiotic; (2) the antibiotic’s ability to diffuse from the disc to the agar (how solid it is in water); (3) the concentration of the antibiotic in the disc measure the diameters of the zones of inhibition around each antibiotic disc on the plate

Question 54

Difference between disinfectants and antiseptics

Answer 54

*Disinfectants- bacteriostatic chemicals used on inanimate objects such as counter surfaces and non-living objects. The most common is phenol (carbolic acid). *Antiseptics- are applied to living tissue with the intent to rid tissue of microbes. Think of it as disinfectants for living tissue. such as throat sprays and skin creams *Both are measured by zones of inhibition.

Question 55

Who is Joseph lister?

Answer 55

Joseph Lister discovered washing surgical tools with carbolic acid lowered the risk of patient infection/post-surgery infection in 1887 * Because of this, phenol is the standard to which all disinfectants and antiseptics are held

Question 56

Which, gram pos or neg is more resistant to the chemicals?

Answer 56

The Gram-negative bacteria were found to be more resistant to the chemicals than Gram-positive bacteria. This is probably because Gram-negative bacteria have a more complex cell wall that cannot be easily penetrated. Gram-negative bacteria that have an outer membrane rich in lipopolysaccharide

Question 57

What are strict aerobes, strict anaerobes, facultative anaerobes, aerotolerant bacteria, and microaerophiles? What are fungi classified as?

Answer 57

Strict aerobes require O2 for growth. Strict Anaerobes cannot grow in aerobic conditions and may die in the presence of O2. Between these two groups of microbes lie the facultative anaerobes, aerotolerant, and microaerophiles. Facultative anaerobes grow in the presence or absence of O2, but exhibit best growth in oxygenic environments. Aerotolerant bacteria are not harmed by O2, but they cannot utilize it. Microaerophiles require decreased O2 and sometimes increased CO2 concentrations. All fungi are strict aerobes

Question 58

What determines whether or not oxygen is a requirement ?

Answer 58

The enzyme systems of bacteria. This depends on whether the enzymes are present or absent from glycolysis, the CAC and/or the electron transport chain.

Question 59

What do bacteria utilize as an electron acceptor?

Answer 59

Bacteria utilize O2 as a terminal electron acceptor in the electron transport chain but nitrate, sulfate, chlorate, etc may be used instead.

Question 60

How do you create conditions for microaerophilic growth?

Answer 60

The use of a candle jar and a large container such as a mayo jar that holds several petri dishes and can be tightly sealed. Place a candle on top of the inverted plates and light it. Now tightly seal the jar. The burning candle will consume some of the O2 and replace it with CO2 leaving a 10% CO2 atmosphere.

Question 61

How do you create conditions for anaerobic growth?

Answer 61

The use of a Brewer-type anaerobic jar in solid medium cultures. Plates of bacterial cultures are stacked at the bottom of the jar. A gas generator is opened and placed next to the plates in the jar. Water is added to the gas generator packet and the lid on the jar is sealed. In the gas generator packet, sodium borohydride and sodium bicarbonate begin reacting with water to form H2 and CO2 gasses then H2 and O2 react to form H2O in the presence of palladium. Next, add a methylene blue strip to the jar, Methylene blue is blue when oxidized and colorless when reduced. Shortly after the conversion of O2 to H2O, the methylene blue strip should become colorless as O2 is removed from the environment.

Question 62

Another method to culture anaerobic organisms is the use of thioglycollate broth. Describe it

Answer 62

It contains sodium thioglycolate, which reduces O2 to H2O to create a more anaerobic environment in the broth. Again, an oxygen indicator such as methylene blue or resazurin may be included to indicate where oxygen remains in the medium. Resazurin dye is pink if oxidized and colorless if reduced. In addition to allowing growth of anaerobes, thioglycolate may also be used to determine aerotolerance by observing the pattern of growth of the organisms. Strict aerobes will grow on the surface region of the tubes, while obligate anaerobes will grow towards the bottom of the tube. Facultative and aerotolerant anaerobes grow throughout the medium. Finally, microaerophiles will grow somewhere between the oxygenated surface and the anaerobic deep of the tube (they find the appropriate oxygen tension).

Question 63

Why is temperature one of the most principal factors affecting bacterial growth?

Answer 63

This is because bacterial cells lack homeostatic mechanisms to maintain a constant temperature. Bacteria cells are sensitive to fluctuations in temperature. These fluctuations cause changes in the enzyme systems of bacteria.

Question 64

Enzymes function at an optimum temperature range, depending on the type of bacteria. What are they? If the temperature falls below or above this optimum range, what happens?

Answer 64

Different types of organisms have different enzyme systems and therefore have different temperature requirements. Psychrophilic organisms function best between 0-20°C. Mesophilic organisms function best between 20-40°C. Thermophilic organisms thrive best at 40-80°C. the enzymes become inactive.

Question 65

Which is more damaging, lower or higher temperatures?

Answer 65

Lower temperatures are less damaging than high temperatures. Denaturation of enzymes occurs at high temperatures causing irreversible damage.

Question 66

What are the adaptions needed for surviving the high temperatures?

Answer 66

Between 60 and 100°C, bacterial cells are usually killed, but endospores are not affected. specialized enzymes and more dense cell membrane

Question 67

Between 60 and 100°C, bacterial cells are usually killed, but endospores are not affected. Although this temperature range is lethal for most vegetated bacterial cells, WHAT is also important for killing

Answer 67

the length of time

Question 68

Since lethal effects of temperature are time-dependent, two methods of determining lethal limits of time and temperature are commonly employed:

Answer 68

The thermal death point (TDP), which is the temperature at which an organism is killed in 10 minutes of exposure. The thermal death time (T'DT), which is the time required to kill a suspension of cells or spores at a given temperature.

Question 69

What are the Effects of UV Radiation on Bacteria?

Answer 69

No detectable change, and a mutation induced by the UV radiation was fixed by the cells DNA repair system. Obvious phenotype changes such as loss of pigment, or inability to utilize certain nutrients inna growth medium. It also may die due to a mutation. Radiation is commonly used to kill bacteria or inhibit their growth. Types of radiation include, UV, X-rays, beta-rays, and gamma rays.

Question 70

What is a mutation and mutant?

Answer 70

Mutation is Any change in the DNA sequence of a cell. A mutant is an organism that has different genetic characteristics from its parents.

Question 71

Why is UV radiation used in operating rooms and clean rooms where pharmaceutical products are housed?

Answer 71

This helps to reduce the number of bacteria on surfaces

Question 72

Why is UV light an inefficient mode of sterilization?

Answer 72

1.UV sterilizers work effectively in its direct light path. It becomes ineffective when light is blocked by objects. Hence it has to be ensured that objects should be placed directly in the path of UV light. 2. The killing efficiency related to both the distance between the organism and the UV light as well as the time of exposure./Many studies claim that the efficacy of UV device depends on the duration of irradiation and the exposure time. The microorganism killing efficacy of UV light was greater when the radiated distance was shorter, and the exposure time was longer. 3. The effectiveness of the light source is dependent on the wavelength it produces. Short UV wavelengths (about 265 nm) are more effective than long wavelengths (about 365 nm).