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A comprehensive set of exam questions and answers for the portage learning biod 171 microbiology course. The questions cover a wide range of topics related to microbiology, including cell structure and function, metabolism, genetics, and microbial diseases. The answers provided are 100% verified and correct, ensuring that students can use this resource to prepare for their exams with confidence. Updated regularly, with the latest version released in 2024, guaranteeing that the information is current and relevant. Whether you are a university student enrolled in the biod 171 course or a lifelong learner interested in microbiology, this document can serve as a valuable study aid and reference material to help you succeed in your academic or personal pursuits.
Typology: Exams
1 / 46
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[Date] 1
as such are not considered microorganisms, they can, however, be classified as microbes, a more general term that includes microorganisms and viruses.
smallest biological unit of life is the cell.
Lipids, Nucleic acids and Polysaccharide
____ known forms. Amino acids, 20
major types of nucleic acids: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Nucleic acids are chemical molecules that carry genetic information within the cell. DNA contains a vast amount of hereditary information and is responsible for the inheritable characteristics of living organisms. RNA is
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[Date] 2 responsible for deciphering the hereditary information in DNA and using it to synthesize proteins.
for each complementary pair: 3’ GGUCAUCG 5’ 5’ CC AGC 3’ 3’ GGUCAUCG 5’ 5’ CCAGUAGC 3’ There are 2 bonds formed between A and U, and 3 bonds between G and C.
A. Only restricts movement of materials into the cell B. Is often a bilayer comprised of lipids? C. Cannot prevent essential nutrients from escaping D. Contains hydrophobic tails pointing inward
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[Date] 3
macromolecule? Give an example. Polysaccharides. Glucose, sucrose and cellulose are all acceptable answers. True or False: The genetic material within a prokaryotic cell is contained within a membrane enclosed region. False. Only eukaryotic cells contain its genetic material within a nucleus True or False: Prokaryotic cells can be subdivided into Bacteria and Archaea. True Describe the 4 basic bacterial morphologies. Coccus (round/spherical), bacillus (rod), vibrio (curved rod) or spirillum (spiral/corkscrew). True or False: Archaea is noted for its ability to survive under harsh conditions. True. Archaea can often be found in harsh conditions such as high salt levels, high acid conditions, high temperatures and even oxygen-poor conditions.
organism A. Bacteria B. Eukarya C. Archaea D. Virus
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[Date] 4
False—they are heterotrophic.
converting light energy into chemical energy. Plantae
Which of following also contain chitin? Select all that apply. ACD A. Mushrooms B. Bacteria C. Yeast D. Molds
form tissue layers. True.
A. Plants B. Fungi C. Bacteria
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[Date] 5 D. Mammalian cells E. Algae
A. Lipid synthesis B. Protein synthesis C. To produce energy (ATP) D. Protein modification and distribution E. Waste disposal via hydrolytic enzymes
occur in living organisms in order to maintain life. True.
reactions. False. Enzymes are proteins that catalyze chemical reactions.
reaction? Cofactors are usually metal ions and assist enzyme during the catalysis reaction.
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[Date] 6
of macromolecules from small molecular units into much larger complexes.
would be active? Catabolism would be active as proteins are made up of amino acids. Therefore, the process described (proteins into amino acids) is the breakdown, or catabolism of protein.
the energy (phosphate group) to donate while ADP can accept energy in the form of a phosphate group. Thus, ATP can be reduced (ATP →ADP + Pi) while ADP can be built into ATP (ADP + Pi →ATP).
called a _________? Phototrophic microorganism. An organism that derives its energy by removing electrons from elemental sulfur would be classified as a __________? Lithotroph
A. Photophosphorylation B. Substrate-level phosphorylation
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[Date] 7 C. Oxidative phosphorylation
transitions? The breakdown of glucose goes through (1) Glycolysis then (2) Fermentation or Respiration and finally through (3) the electron transport chain (ETC).
electron transport chain yields 34 ATP while both glycolysis and fermentation (or respiration) each yield only 2 ATP.
Glucose + 2NAD+ → 2 NADH + 2 Pyruvate + 2 ATP The reactants are to the left of the arrow: Glucose and 2NAD+
start? Glucose-6phosphate (G6P)
Fermentation only reduces NADH back to NAD+
A. NAD+
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[Date] 8
C. NADH D. FADH 2
electron transport system. False. The products of the TCA cycle (reduced electron carriers) enter and drive the production of ATP via the electron transport system.
energy sources? Select all that apply. A, C, D can all be used as alternative energy sources A. Lactose B. Nucleic acids. Carbohydrates D. Lipids.
used? Select all that apply and D. Proteases breakdown proteins while lipases breakdown lipids.
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[Date] 9 A. Ligases B. Proteases C. Transferases D. Lipases
lipids. True
Chloroplasts are specific to algae and plants only.
select all that apply. A, C, D, and E. Carbon fixation uses the ATP/NADPH produced during light reactions to convert CO2 and H2O into useful sources of energy (carbohydrates). A. ATP B. Glyceraldehyde- 3 - phosphate C. CO 2 D. NADPH
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[Date]
Similar to the electron transport chain, one of its main functions is to generate a proton concentration gradient to generate ATP.
True. The term ‘dark reactions’ (also known as the Calvin Cycle) simply denotes the second stage in photosynthesis—dark reactions do not actually require darkness in order to occur.
one molecule of glucose? Six.
indicated. __CO 2 + __ATP + __NADPH + __H 2 0 → __C 6 H 12 O 6 + __ADP + __NADP+ 6 CO 2 + 18 ATP + 12 NADPH + 12 H 2 0 → C 6 H 12 O 6 + 18 ADP + 12 NADP+
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[Date]
Ab + C → A + Cb A- Lyases B- Transferases C- Oxioreductaces D- Hydrolases E- Ligases F- Isomerases 1E 1. A + B → A-B Ligases
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[Date] 2A 2. A-B → A + B Lyases 3C 3. A—^ + B → A + B—^ Oxioreductaces 4B 4. Ab + C → A + Cb Transferases
A. 10
B. 10 -^6 C.10-^9 D. 10-^12 B. A micrometer is one-millionth of a meter.
False. A nanometer is 1,000 times smaller than a micrometer.
an object. Explain each.
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[Date] Resolution refers to the distance between two objects at which the objects still can be seen as separate. Poor or low resolution means two (or more) objects may appear as one. Contrast is the difference in light absorbance between two objects. Poor contrast gives a high background and makes the visualization of multiple objects difficult. For instance, trying to identify 2 dark colored objects at night (low light = low contrast) versus the same 2 objects in the middle of a sunny afternoon (bright light against 2 dark objects = high contrast).
the microscope you would adjust to limit the amount of light entering the microscope. Select all that apply. A. Objective B. Condenser C. Iris diaphragm D. Eye piece
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[Date] C. The iris controls the amount of light that passes through the sample and into the objective lens. Thus, it can be adjusted (opened or closed) to alter the amount of light.
60x objective and a 10x eyepiece? Show your math. 60 x 10 = 600x magnification
using bright field microscopy. False. Adherent, flat cells are almost invisible due to the limits on both resolution and contrast.
apply. A. Bacteria with diameter of 24 μm B. Protozoa with diameter of 150 μm C. Virus with a diameter of 0.2 μm D. Skin cell with diameter of 1500 μm
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[Date] B and D. The unaided eye can, on average, clearly resolve objects > 100 μm
answer: Phase-Contrast Dark Field Fluorescence Confocal This type of microscope is best suited for visualizing GFP, RFP and YFP proteins. Fluorescence
staining. Phase-Contrast
samples and background by reflecting light off of the specimen. Dark Field
rendering a specimen in 3D Confocal
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[Date]
layer in the cell wall. Purple; Thick
peptidoglycan layer in the cell wall. Pink; Thin
the presence of LPS in the outer membrane. True.
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[Date]
heat fix, but you can chemically fix the specimen. False. Both heat and chemical fixation strategies will kill the cell, making motility observations impossible.
staining technique that you could perform? Name at least one dye you would use during this process. Simple stain. You could use any of the following: methylene blue, crystal violet, safranin or fuschin.
sent off to the lab for an acid-fast stain. If the patient were infected with TB, describe what you would expect to see on the stained slide. You would expect to see red cells (TB+) on a blue background (TB negative).
pathogenic bacteria. True. Pathogenic bacteria would appear pink while nonpathogenic bacteria would appear purple.
two similar species of bacteria. True. Growth media does not contain restrictive factors, while selective media is best used to encourage the growth of one microbe while simultaneously discouraging the growth of the other. Since two similar species of microbes are being studied they must be differentiated under similar but just slightly different conditions (differential media).
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[Date]
meningitides. Knowing Neisseria meningitides is slow growing and other foreign microbes may also be present in the culture, which type of media would be best suited A. Growth media B. Differential media C. Selective media D. Selective and Differential media
False. LB agar is the most basic type of agar and like LB media supports the growth of virtually all microbes without restriction.
smooth surface on which microbes can grow.
hemolysis- Incomplete hemolytic activity. Beta- Complete, Gamma- No hemolytic
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[Date] A. Trypticase Soy Agar B. MacConkey Agar C. Blood agar D. EMB agar
enriched media. False. Chocolate (cocoa) is never added to the media. The name is derived simply based on the color that actually comes from the presence of ‘cooked’ (lysed) red blood cells in the media.
MacConkey plate. However, the researcher can’t remember if E. coli is Gram- positive or Gram-Negative. Would a Gram stain be necessary to confirm? Why or why not? No. A Gram stain would not be necessary; as only Gram-negative microbes will grow on MacConkey agar. Thus, E. coli is a Gram-Negative microbe.
O157:H7, a researcher spread a culture onto a MacConkey agar with failed results. What type of agar should they (correctly) try next? Why? The microbe should be plated on SMAC (Sorbitol-MacConkey agar) as it is specifically formulated to detect O157:H7. Pathogenic E. coli (O157:H7) cannot ferment
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[Date] sorbitol while non-pathogenic E. coli can ferment both sorbitol and lactose. Therefore, colonies that ferment (acidic conditions; non-pathogenic) can be differentiated from non-fermenters (neutral to basic conditions; pathogenic).
Eosin Methylene Blue (EMB) agar plates? Gram-Negative. EMB plates specifically restrict the growth of Gram-positive bacteria.
strain Staphylococcus aureus? Yellow. Pathogenic Staph aureus will turn the agar from red to yellow.
in a liquid media or on a solid (agar) media? Why? Solid (agar) media. The primary advantage is that cells are held into place. When grown in a nutrient broth, bacterial cells can multiply but are free to move around in solution. When grown on agar within a petri dish the fixed in such a way as to form colonies.
cells that have often multiplied a million times over. True. To form a bacterial
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[Date] colony, the initial cell must have multiplied many times over, often greater than a million, in order for the naked eye to resolve it.
colonies of a bacterial population. True. The purpose of the quadrant streak is to generate individual colonies such that a single (pure) bacterial sample can be isolated.
strategy? A quadrant (aka phase-dilution) streak. The resulting gradient should always contain within it the growth of individual colonies.
concentration of bacteria, P2 or P4? P2 (Phase 2) would contain a higher concentration of bacteria than Phase 4 (P4). The phases rank (from highest to lowest), P1 > P2 > P3 > P4.
must be used for each phase. True. Failure to do so would prevent the establishment of a dilution gradient, as the same bacterial concentration would be spread across both phase regions.
researcher do during a phase dilution streak? A researcher may either (1) opt to
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[Date] perform only a 3-phase dilution streak or (2) pass the loop through the previous phase multiple times (as opposed to only once).
might decrease an incubator from 37°C to 25°C. True. Pathogenic strains of bacteria tend to grow faster than nonpathogenic strains at 37°C, so researchers may set incubators at 25°C to restrict its growth. When given an unknown bacterial sample the first step is to expand the current bacterial population. Which form of media best suites this need? Why? A. MSA agar B. LB media containing ampicillin and neomycin C. MacConkey agar D. Blood agar D. Blood agar. All other options (A, B and C) are all forms of selective media, meaning they may potentially inhibit the growth of the unknown sample. Although blood agar is considered a differential media, it is, most importantly, a non-selective media. Given the alternatives, this is the best option.
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[Date]
any and all samples, whether known or unknown, are to be treated as potentially hazardous (or pathogenic) materials.
assessing an unknown microbial sample A lab researcher would be certain to note:1. Size and shape2. Any observable motility3. Gram status (positive or negative)4. The presence of any chemical reactions5. Changes in color localized to the organism or the surrounding media6. Capture (or draw) images of any of the characteristics described above
determine the following observations: (1) the presence of any motility and (2) its Gram status using the same sample—the liquid sample cannot be divided. Which would you determine first and why? You must determine motility before determining the Gram status. Motility requires a wet mount, while Gram staining requires heat fixing the sample. If one were to begin with the Gram stain the heat fixation process would kill the organism, making any observations regarding motility impossible. The correct approach would be to place the liquid culture on a glass slide and determine its motility status. Next, the same liquid culture can be heat fixed and Gram stained.
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[Date]
conditions? A facultative anaerobe is capable of growth under aerobic (with oxygen) and anaerobic (without oxygen) conditions.
peroxides? Why? Streptococcus would not survive in the presence of peroxides—it is unable to breakdown peroxides (catalase negative). Left unchecked, peroxides would damage the cellular integrity of Strep causing lysis/cell death.
A. Chocolate agar B. EMB agar C. Blood agar D. Spirit Blue agar 3.True or False. The Lancefield groups are used to subdivide antigenic groups of gamma-hemolytic Streptococcus. False. The Lancefield groupings are used to subdivide beta-hemolytic Strep.
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[Date] A. Group B B. Group D C. Group A D. Group C
hemolytic activity A. Strep. Pharyngitis; beta B. Strep. Septicemia; gamma C. Rheumatic fever; beta D. Strep. Pharyngitis; alpha
of human population where it remains pathogenic. False. While staphylococcus can be found in ~30% of the human population, it remains non-symptomatic.