Name : Student number : . Midterm exam #1 of BIO3124: General...

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1 ( Name : Student number : . Midterm exam #1 of BIO3124: General microbiology This exam is closed book and is composed of two parts. The first part worth 30 points consists of 31 multiple choice questions. Make sure to indicate your answers to these questions on both this copy of the exam as well as on the scantron sheet provided. DO NOT forget to write your name and student number on BOTH this exam paper and the scantron sheet. The second part of the exam worth 15 points consists of 4 short answer questions. You must answer ONLY 3 of the 4 questions, or else only the first three questions answered will be corrected. Cellular phones, unauthorized electronic devices or course notes (unless an open-book exam) are not allowed during this exam. Phones and devices must be turned off and put away in your bag. Do not keep them in your possession, such as in your pockets. If caught with such a device or document, the following may occur: you will be asked to leave immediately the exam, academic fraud allegations will be filed which may result in you obtaining a 0 (zero) for the exam. By signing below, you acknowledge that you have ensured that you are complying with the above statement. Formulas : = = / = / − = )

Transcript of Name : Student number : . Midterm exam #1 of BIO3124: General...

Page 1: Name : Student number : . Midterm exam #1 of BIO3124: General microbiologymysite.science.uottawa.ca/jbasso/micro/exams/micro e… ·  · 2017-07-26Name : Student number : . Midterm

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Name : Student number : .

Midterm exam #1 of BIO3124: General microbiology

This exam is closed book and is composed of two parts. The first part worth 30 points consists

of 31 multiple choice questions. Make sure to indicate your answers to these questions on both

this copy of the exam as well as on the scantron sheet provided. DO NOT forget to write your

name and student number on BOTH this exam paper and the scantron sheet.

The second part of the exam worth 15 points consists of 4 short answer questions. You must

answer ONLY 3 of the 4 questions, or else only the first three questions answered will be

corrected.

Cellular phones, unauthorized electronic devices or course notes (unless an open-book

exam) are not allowed during this exam. Phones and devices must be turned off and put

away in your bag. Do not keep them in your possession, such as in your pockets. If

caught with such a device or document, the following may occur: you will be asked to

leave immediately the exam, academic fraud allegations will be filed which may result in

you obtaining a 0 (zero) for the exam.

By signing below, you acknowledge that you have ensured that you are complying with

the above statement.

Formulas :

𝑵 = 𝑵𝟎𝟐𝒏

𝑵 = 𝒕/𝒈

𝝁 = 𝒍𝒏𝟐/𝒈

−𝒌𝒕 = 𝐥𝐧 𝑵𝟎

) 𝑵

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Part 1: Multiple choice questions. Provide your answers on this exam paper and the scantron sheet provided.

1. A researcher inoculated each of three sealed flasks (A, B, and C) containing a sterile growth medium with one of three different bacteria (Chlamydia psittaci, Bacillus thuringiensis, or Shigella sonnei). After a 24 hour incubation period, he observed growth in flasks A & B. The experiment was then repeated, but the inoculated flasks were boiled for 10 minutes before incubating them. This time, growth was only observed in flask A. According to these observations, which flasks were inoculated with which bacteria?

A. A: Chlamydia psittaci; B: Bacillus thuringiensis; C: Shigella sonnei B. A: Bacillus thuringiensis; B: Chlamydia psittaci; C: Shigella sonnei C. A: Shigella sonnei; B: Bacillus thuringiensis; C: Chlamydia psittaci D. A: Bacillus thuringiensis; B: Shigella sonnei; C: Chlamydia psittaci

2. The growth of different bacteria was examined in different media. The results obtained and a

brief description of the media compositions is indicated below.

Media composition A: Complex blood based medium B: Defined medium with 50 ingredients including vitamin B12 C: Same medium as B, but lacking vitamin B12

According to these results which bacteria most probably belong to different biotypes?

A. W & X B. X & Y C. Y & Z D. W & Z

3. According to the results presented in the previous question, which bacterium most probably has the lowest biosynthetic capacity?

A. W B. X C. Y D. Z

4. A higher percentage of hybrids would be expected between the ribosomal RNA genes from which of the following pairs of organisms?

A. A Clostridium and a Bacillus B. A bacteroid and a protist C. A cyanobacteria and a green algae D. An archaebacteria and a proteobacteria

Media Bacteria

W X Y Z

A + + + -

B - + + -

C - - + -

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5. Two different views of dust mites are shown. Dust mites typically ingest organic detritus, such as flakes of human skin, and measure approximately 0.2-0.3 mm. To which kingdom would these organisms belong to?

A. Animalia B. Protista C. Eukarya D. Protozoa

6. The list below indicates some of the characteristics of the organism shown in the previous question. Why would this organism not be considered a microorganism?

a) It is multicellular b) It possesses different cell types/tissues c) It is 0.2-0.3 mm d) It feeds by ingestion e) It reproduces sexually

A. a, d & e B. b & c C. a, b, c & e D. Only c

7. Which of the following hypotheses were shared by Aristotle, Jan Batista Van Helmont, and Francesco Redi?

A. All life can be spontaneously generated from the transformation of nonliving materials B. Living organisms can reproduce to generate other organisms such as themselves C. Life invisible to the naked eye can be generated from nonliving matter D. Air is essential to the generation of all living things

8. Bacterial cells of the genus Mycoplasma with a cytoplasm containing 0.9% solutes

dissolved in water are placed in a glass of pure water. What will happen to these cells?

A. Nothing much. B. The cells will shrink and get smaller C. The cells will become turgid D. The cells will swell and explode

9. Which of the following is mismatched?

A. Catabolic reaction – digestion of lipids B. Anabolic reaction – transpeptidation

C. Reduction – metabolism of glucose to CO2 D. Lithotroph – oxidation of inorganic compounds

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10. A Gram stain was performed on a smear of the illustrated bacteria by using the following reagents in the order indicated: Gram’s iodine, crystal violet, ethanol, and safranin. After each step, the bacterial smear was washed with water to remove excess reagent. What color would these bacteria be?

A. Purple B. Red C. Blue D. Colorless

11. Which of the following bacterium is not classified amongst the Gram positive bacteria?

A. Spirillum volutans B. Staphylococcus aureus C. Clostridium botulinum D. Mycobacterium tuberculosis

12. What was the most significant difference between Pasteur’s and Needhams’/Spallanzani’s experiments that allowed to definitely prove that spontaneous generation did not occur?

A. Pasteur’s flask allowed air to access the medium B. Pasteur’s flask allowed the medium to have access to air but not to particulates in the air C. Pasteur used a selective medium

D. Pasteur did not seal the flask

13. A newly developed antimicrobial agent that binds with sterols is marketed by a pharmaceutical company. Such a drug would affect which type of microorganisms?

A. Archaebacteria B. Gram negative bacteria C. Gram positive bacteria D. Fungi

14. It is generally accepted that the use of antibiotics in animal feed leads to antibiotic-

resistant bacteria. Which of the following reasons best explains this phenomena?

A. Antibiotics cause mutations in the bacteria from the host animal, which results in resistance to antibiotics

B. Resistant bacteria from other animals replace those killed by the antibiotics in the original host

C. Antibiotics kill sensitive bacteria, but the few that are spontaneously resistant survive and repopulate the host animal

D. The host animal develops an immunity against the antibiotics allowing the bacteria to survive

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15. Which of the following is the BEST definition of fermentation?

A. Complete catabolism of glucose to carbon dioxide and water B. The reduction of organic molecules through the oxidation of a carbohydrate C. Production of ethanol from glucose D. Reduction of glucose to pyruvic acid

16. Which of the following groups of organisms is most likely to carry out the following

reaction? 6CO2 + 12H2S + light → C6H12O6 + 12S0

A. Cyanobacteria B. Green algae C. Purple sulfur bacteria D. Purple nonsulfur bacteria

17. Which is NOT a function of the bacterial plasma membrane?

A. It is a selectively permeable barrier B. It is a site of enzymatic reactions C. It is the site of aerobic respiration in bacteria D. It prevents cell lysis in hypotonic solutions

18. Which of these is NOT a function of the bacterial glycocalyx?

A. Adhesion to environmental surfaces B. Establishment of multispecies communities C. Nutrient storage D. Protection from phagocytosis

19. In which of the following ways does aerobic respiration differ from anaerobic respiration?

A. Aerobic respiration results in more electrons being extracted from glucose as compared to anaerobic respiration resulting in a higher ATP yield

B. Aerobic respiration is the complete oxidation of glucose while anaerobic respiration is the fermentation of glucose

C. Aerobic respiration uses an electron transport chain while anaerobic respiration does not D. The final electron acceptor is oxygen for aerobic respiration and some other inorganic

molecule for anaerobic respiration

20. Anabaena uses carbon dioxide as its carbon source and obtains its energy from light by the oxidation of water. This organism would be classified as a…?

A. A microaerophilic chemoautotroph B. A strictly anaerobic lithoautotroph C. An aerotolerant photoautotroph D. An aerobic photoautotroph

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21. What is the most likely reason why mycobacteria are classified as Gram positive?

A. Because they stain purple following a Gram stain B. Because the structure of their cell wall is the same as that of Gram positive bacteria C. Because their cell wall is impermeable D. Because just like Gram positive bacteria, their cell wall traps the primary dye

22. A fluorescently conjugated antibody which binds teichoic acid was used on different

bacteria. Which bacteria or bacterium would be stained by this antibody?

a) Mycoplasma pneumoniae b) Mycobacterium tuberculosis c) Moraxella catarrhalis d) Bacillus megaterium e) Staphylococcus aureus

A. a, b and c B. d and e C. c, d and e D. Only e

23. The schematic depicts compartments separated by a lipid bilayer. The

black spheres are an impermeable solute. Among the different transport mechanisms listed, which would allow movement of the solute in the direction indicated until its concentration in compartment “A” is zero?

a) Passive diffusion c) Active transport b) Facilitated diffusion d) Group translocation

A. b and c B. Only c C. a, b and c D. c and d

24. Illustrated are oxidation reduction reactions

carried out by Thiobacillus. What is the order of

the Eo values of each of the compounds; from the most negative value to the most positive value?

A. NADPH, H2S, Cyt1 and O2 B. H2S, NADPH, Cyt1, and O2 C. H2S, Cyt1, O2, and NADPH

D. Cyt1, NADPH, O2, and H2S

25. Oxygen, a product of oxygenic photosynthesis, is generated by the oxidation of…?

A. Carbon dioxide B. Glucose C. Hydrogen sulfide D. Water

A B

Cyt1red + O2 + ADP → Cyt1ox + H2O + ATP

H2S + Cyt1ox → So + Cyt1red + E

H2S + NADP + E → So + NADPH

Cytred + NADP+ + E → Cyt1ox + NADPH

Ox : oxidized; Red : reduced; E : energy

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26. How does penicillin act on bacteria of the genus Streptococcus?

A. It inhibits linkage between NAG and NAM B. It cleaves the glycine bridge between the peptidoglycan polymers C. It prevents the linkage between the lysine and alanine residues of peptidoglycan units D. Penicillin does not act on bacteria of the genus Streptococcus

27. Which of the following represents growth of a psychrophile?

A. Mold growing on bread on a kitchen counter B. Mold growing on fruit in the refrigerator C. Bacteria growing in the throat of a child D. Bacteria growing in salt cured meats such as salami

28. Compare the following statements:

I. The percentage of 108 Gram negative cells with a D100 value of 1 minute killed after 1 minute.

II. The percentage of 106 bacterial spores with a D120 value of 5 minutes killed after 5 minutes.

A. I is greater than II B. I is less than II C. I is exactly equal to II D. This cannot be determined from the information given

29. Which treatment would be ineffective to inactivate hydrophilic (naked) viruses?

A. Autoclave B. A halogen based disinfectant C. Gluteraldehyde D. Filtration

30. You inoculate a medium with a bacterium with a growth rate constant (k) of 2. At 4 PM the culture has an optical density of 2.0. At what time would the optical density of the culture have been 0.5?

A. At 2:00 PM B. At 2:30 PM C. At 3:00 PM D. At 3:30 PM

31. What is the target of the chemical shown here?

A. Lipid bilayers B. The cell wall C. Proteins D. DNA

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Part 2. Answer ONLY 3 of the 4 questions or else only the first 3 questions will be corrected! (5 points/question)

1. Indicate whether each of the following statements is true (T) or false (F). If a statement is false, replace ONE of the underlined words to make the statement true. You may only change the underlined words! Indicate in the space provided the original word and the replacement word.

Ex. F The professor who teaches this class prefers being referred to as Dr. Basso. Dr. Basso to John

If lysozyme is added to a culture of Micrococcus luteus growing in an isotonic

medium it would die as a result of cell lysis.

The percentage of G/C within one strand of a bacterial genome is 42%. Consequently the percentage of G/C within the complementary strand of this genome must be 58%.

The primary function of the bacterial cell wall is to protect the cell from the excess

entry of water.

Lipopolysaccarides are the most abundant structural component of the cytoplasmic membrane of Gram negative bacteria.

The cell membrane provides the general shape of the bacterium.

The acid-fast stain is a differential staining technique that allows the identification of the bacterial pathogen responsible for leprosy based upon its cell wall composition.

Eukaryotic flagella are composed of a single protein, tubulin, and their structure is an extension of the cell membrane.

A medium that contains lauryl sulfate, which inhibits the growth of Gram-positive bacteria, is a differential medium.

The cell membranes of growing bacteria are permeable to DAPI.

A positively charged stain could not be used for the negative staining of bacteria

fixed on a glass slide.

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2. Bacterial cultures were grown in a defined medium (Control) containing radioactive

glucose (14C), radioactive methionine (35S), glutamate, and ammonium phosphate. The experiment was also performed in the same medium either in the presence of an ATP synthesis inhibitor (AI) or lacking glutamate (-Glu). At regular intervals the ratio between the external and internal concentrations of each of the isotopes was assessed ([External]/[Internal]). The table shows the results of this investigation.

Time (min)

Isotope ratios : [External]/[Internal]

Control +AI -Glu

14C 35S 14C 35S

0 7.5 10 7.5 10

15 5.5 5.0 7.3 9.0

30 3.5 2.0 7.0 8.0

60 1.5 1.0 6.8 7.0

120 0.5 1.0 6.6 6.0

240 0.2 1.0 6.4 5.0

a) Indicate all the possible transport mechanisms for glucose in the presence and in the absence of the ATP synthesis inhibitor.

b) Indicate all the possible transport mechanisms of methionine in the presence and in the absence of glutamate.

c) Indicate all possible types of ammonium sulfate transporters which would be in agreement with the results presented. Indicate none if no transporter is involved.

d) Name two cellular structures seen in class in which you would expect 35S to be incorporated.

e) At time zero the cultures contained 1 X 104 cells. At the end of the experiment (time =

240 minutes) the cultures reached a density of 1 X 107 cells and 1 X 105 cells in the control and –Glu media respectively. The growth rate (µ) was approximately how many times faster in the control medium?

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3.

a. At 60 minutes the optical density of culture “C” was 0.25. What would be the predicted

optical density of this culture at 180 minutes?

b. After 180 minutes, culture “A” reached a density of 5 X 1010 cells/ml. How much time

(total) would be required for culture “B” to reach the same density observed for culture

“A” at 180 minutes? Please give your answer in hours.

c. A sample of culture “B”, taken at 80 minutes, was diluted 10 fold and applied to a

hemocytometer slide. How many bacteria would be seen in one of the squares with the

following dimensions: 0.5mm X 0.2mm X 0.01mm?

d. A viable count was done on a sample of culture “B” taken at 80 minutes. The sample

was diluted by a factor of 105 of which 0.1ml was plated. If the bacterium was diplococci,

how many CFUs would you expect?

e. Cultures “A” and “C” represent the growth of a bacterium of the genus Streptococcus.

Culture “A” is in an isotonic medium, whereas culture “C” is in a hypotonic medium.

Samples of both cultures were taken at 80 minutes, treated with lysozyme, and then

Gram stained. What color would bacteria from each of the cultures be?

A

B

C

Time (Minutes)

Nu

mb

er

of

cell

s/m

l X

10

8

Lo

ga

rith

mic

sca

le

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4.

1000

100

10

1

90 100 110 120 130 140 150

Temperature oC

a. What is the mortality constant (k) at 170oC?

b. A can of beans contains 104 spores of Clostridium. Based on the above profile what

would be the thermal death time (TDT) at 120oC?

c. What is the approximate thermal death point (TDP) of the can of beans described in the previous question?

d. How much time would be required at 100oC to sterilize the can of beans containing 104

spores?

e. What is the inactivation factor achieved in 10 seconds at 130oC?

D V

alu

es (

Seco

nd

s)

Lo

gari

thm

ic s

cale

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Q. Ans. Q. Ans.

1 D 16 C

2 B 17 D

3 D 18 C

4 A 19 D

5 A 20 C

6 B 21 D

7 A 22 B

8 D 23 D

9 C 24 A

10 B 25 D

11 A 26 C

12 B 27 B

13 D 28 C

14 C 29 D

15 B 30 C

31 A Short answer: 1. F (isotonic to hypotonic), F(Either G/C to A/T or 58% to 42%), T, F Lipopolysaccharides to

phospholipids), F(membrane to wall), T, F(Cell membrane to cytoskeleton), F(Differential to selective), T, T

2. a) Control: active transport; +AI: Passive diffusion b) Control: facilitated diffusion; -Glu: Passive diffusion c) symporter d) Any 2 of the following: Flagella, membrane, transporters, gas vesicles, glycocalyx, fimbriae e) approx.. 3X

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3. a) g = 60minutes; n = (180-60)/60 = 2; N=No(2n); N=0.25(22) =1.0 b) N=No(2n); 5 X 1010 = 1 X 108(2n); 2n = 500; n =9 ; g=40 min. ; 9 X 40 min= 3600 min c) (4 X 108)/10 X 1 X 10-6 = 40 d) 4 X 108/105 = 4000; 4000 X0.1 = 400; 400/2 =200 e) A: red, C: no cells or no color

4. a) -23 cells/sec. b) 120 sec. c) Approx. 105oC d) Approx. 4800 sec. or 80 min. or 1.3 hours

e) 10