Download MCB 2010C Exam 2: Microbiology Questions and Answers and more Exams Microbiology in PDF only on Docsity!
MCB 2010C EXAM 2 FSW 2024|2025 Microbiology
EXAM @Question And Answer (CORRECT ANSWER)
Salt and sugars work to preserve foods by creating a Lower osmotic pressure A depletion of nutrients A hypotonic environment A hypertonic environment A hypertonic environment Salt and sugars work to preserve foods by creating a hypertonic environment, which dehydrates microbial cells and inhibits their growth. Water has an unequal distribution of charges and is called a(n) a)nonpolar molecule. b)polar molecule. c)hydrophobic molecule. d)organic molecule. b)polar molecule : think of the north and south pole, opposite partial charges within a water molecule (partial positive hydrogen and partial negative oxygen) wrong answers: nonpolar molecule= no partial charges because the electrons are evenly distributed hydrophobic molecule= water-fearing organic molecule contains carbon and hydrogen
In a dehydration synthesis reaction, a)a larger molecule is broken down into smaller parts. b)old bonds are broken and new bonds are formed. c)neither the reactants nor end products are stable. d)atoms, ions, or molecules are combined to form a larger molecule. d)atoms, ions, or molecules are combined to form a larger molecule. Dehydration synthesis is the process of building larger molecules (synthesizing) and in the process you dehydrate the molecules forming a chemical bond and water. A-OH + H-B = A-B + H2O The building of complex organic molecules from simpler ones is called a)catabolism. b)anabolism. c)photosynthesis. d)oxidation. b)anabolism. Anabolism is the process of using smaller molecules to build larger ones. In this process chemical bonds are formed through dehydration synthesis and energy is used. An example is photosynthesis. Catabolism is the process of breaking larger molecules down into smaller ones. In this process chemical bonds are broken through hydrolysis or decomposition releasing energy. An example is cellular respiration.
Saturation is when every enzyme's active site is reacting with a substrate. In a graph the line for enzyme activity will plateau or level off flat. Competitive and noncompetitive inhibition interfere with the enzyme at either the active or allosteric site, respectively. Denaturation is when the 3-D shape of an enzyme falls apart. Inhibitors that fill the enzyme's active site and compete with the normal substrate are a)noncompetitive. b)allosteric. c)competitive. d)conformational. c)competitive. Competitive inhibitors are called such because they compete with the enzyme’s substrate by interacting with and blocking the active site. Non-competitive or allosteric inhibitors bind to the alternate (allosteric) site which causes a change in shape (conformational change) of the active site so the enzyme cannot interact with its substrate. A molecule that undergoes reduction a)becomes phosphorylated. b)loses a hydrogen atom. c)gains one or more electrons. d)loses one or more electrons.
c)gains one or more electrons. Oxidation reduction reactions are due to the removal and subsequent gain of electrons from an atom or molecule. Remember OIL RIG (oxidation is losing reduction is gaining) Which of the following is CORRECTLY matched? a)Glycolysis – glucose is broken down into citric acid b)Glycolysis – can occur with or without the presence of oxygen c)Krebs/citric acid cycle – where the majority of the ATP is produced d)Krebs/citric acid cycle – oxygen is used as the final electron acceptor e)Electron transport chain – citric acid is produced b)Glycolysis – can occur with or without the presence of oxygen The three steps in aerobic respiration are: Glycolysis – glucose is broken into pyruvic acid, 2 ATP are made, NADH (electron carrier) molecules are made, can occur in the presence or absence of oxygen Krebs/citric acid cycle – citric acid is made, 2 ATP are made, NADH and FADH2 (electron carriers) are made, CO2 is produced
Which of the following is true regarding fermentation? a)Does not require oxygen b)Makes as much ATP as aerobic cellular respiration c)The pathways can make alcohol or citric acid d)Starts with glycolysis and the Krebs/citric acid cycle before going a different pathway a)Does not require oxygen Fermentation is a pathway that occurs without oxygen. It still starts with glycolysis and then goes a different pathway to produce alcohol or citric acid. It only makes 2 ATP from glycolysis as the energy is stored in the end products. Photosynthesis in plants, algae, and cyanobacteria uses light, CO2 and H2O to make glucose and O2. Some bacteria can use an alternative mechanism that uses __________. a)NH b)CH c)H2S d)H2O c)H2S H2O is replaced with H2S (hydrogen sulfide) and instead of producing O2, these bacteria produce S (sulfur). ________ use electrons from reduced inorganic or organic compounds as a source of energy and CO2 as a source of carbon. a)Photoautotrophs b)Photoheterotrophs
c)Chemoautotrophs d)Chemoheterotrophs c)Chemoautotrophs Energy sources (first part of the word):
- Photo = light
- Chemo = chemicals (compounds) Carbon sources (second part of the word):
- Auto = CO
- Hetero = carbon compounds (not CO2) The question stated H2 for energy and CO2 for carbon, so chemoautotrophs. Which of the following uses glucose for carbon and energy? a)Photoautotrophs b)Photoheterotrophs c)Chemoautotrophs d)Chemoheterotrophs d)Chemoheterotrophs Energy sources (first part of the word):
- Auto = CO
- Hetero = carbon compounds (not CO2) The sentence stated light for energy and CO2 for carbon, so photoautotrophs. _____ Lowers microbial counts to safe public health levels Sanitization ____ Kills microorganisms Germicide _____ Destruction of vegetative pathogens on an inanimate surface Disinfection _____ An object or area is free of pathogens Aseptic _____ Removal or destruction of all living microorganisms including endospores Sterilization
_____ Limited heat treatment to kill endospores of Clostridium botulinum, but retain the quality of the food commercial sterilization _____ Removing most of the microbes in a limited area, typically through mechanical removal Degerming _____ Stops or slows the growth of bacteria Bacteriostasis _____ Treatment directed at destroying pathogens on living tissue antisepsis _____ Heat-loving microbes Thermophiles _____ Bacteria that grow in acid Acidophiles _____ Microbes that require high salt concentration for growth Obligate halophiles
_____ Bacteria that can grow at extremely high salt concentrations extreme halophiles An experiment began with 4 cells in log phase and ended with 128 cells. How many generations did the cells go through? 4 6 64 5 32 Start with 4 cells and remember that each cell doubles during 1 generation in log phase: 4 x 2 = 8 cells after 1 generation 8 x 2 = 16 cells after 2 generations 16 x 2 = 32 cells after 3 generations 32 x 2 = 64 cells after 4 generations 64 x 2 = 128 cells after 5 generations 5GENERATIONS Three cells with generation times of 60 minutes are inoculated into a culture medium. Assuming no lag phase takes place, how many cells are there after 5 hours? 180 96 15 900 32 Start with 3 cells and remember that each cell doubles during 1 generation (1 hour in this case) in log phase: 3 x 2 = 6 cells after 1 hour
6 x 2 = 12 cells after 2 hours 12 x 2 = 24 cells after 3 hours 24 x 2 = 48 cells after 4 hours 48 x 2 = 96 cells after 5 hours 96 CELLS Eight cells with generation times of 60 minutes are inoculated into a culture medium. Assuming no lag phase takes place, how many cells are there after 3 hours? •8 x 2 = 16 after 1 hour •16 x 2 = 32 after 2 hours •32 x 2 = 64 after 3 hours Assume you inoculated 100 cells, with a generation time of 20 minutes, into 100 ml of nutrient broth. You then inoculated 100 cells of the same species into 200 ml of nutrient broth. After incubation for 3 hours, you can reasonably expect to have__________ more cells in the 200 ml. more cells in the 100 ml. the same number of cells in both. The answer cannot be determined based on the information provided. the same number of cells in both. Because you used the same species in each volume of nutrient broth, they will divide at the same rate. The amount of nutrients or different volumes will not impact the final outcome. Since each culture started with 100 cells, the generation time is the same since it is the same species, and the time of incubation is the same at 3 hours, the number of cells in each will be the same.
The answer cannot be calculated. 18,700, First - the proper colony range is from 30 to 300. If the number falls below 30 or above 300 the answer cannot be calculated. Second - add as many zeros as the number of tubes used Third - if the volume plated was 1 ml do not add any more zeros. If the volume plated was 0.1 ml, add 1 more zero. This is because it would take 10 of these to equal 1 ml and you are always calculating for CFU/ml. For this problem, take the 187 + five 0s from the number of tubes, so 18,700, A sample is diluted in a 1:10 series. 0.1 milliliter of the third tube is plated and 550 colonies grew. How many CFU/ml? 550 5, 55, 550, 5,500, The answer cannot be calculated. The answer cannot be calculated. First - the proper colony range is from 30 to 300. If the number falls below 30 or above 300 the answer cannot be calculated. A sample is diluted in a 1:10 series. 1 milliliter of the third tube is plated and 263 colonies grew. How many CFU/ml? 263 2,
The answer cannot be calculated. 263, First - the proper colony range is from 30 to 300. If the number falls below 30 or above 300 the answer cannot be calculated. Second - add as many zeros as the number of tubes used Third - if the volume plated was 1 ml do not add any more zeros. If the volume plated was 0.1 ml, add 1 more zero. This is because it would take 10 of these to equal 1 ml and you are always calculating for CFU/ml. For this problem, take the 263 + three 0s from the number of tubes, so 263, A sample is diluted in a 1:10 series. 0.1 milliliter of the sixth tube is plated and 65 colonies grew. How many CFU/ml? 65 650 6, 65, 650, 6,500, 65,000, 650,000, The answer cannot be calculated 650,000, First - the proper colony range is from 30 to 300. If the number falls below 30 or above 300 the answer cannot be calculated. Second - add as many zeros as the number of tubes used
The organism Clostridium botulinum produces endospores that could survive improper heat treatment and since it is anaerobic they will germinate in the can and the vegetative cells will produce a toxin that can be fatal to humans. Damage to which one of the following structures causes cellular contents to leak into the surrounding medium? a)Cell wall b)Plasma membrane c)Nucleoid d)Glycocalyx b)Plasma membrane The plasma membrane is the surrounding structure in all cells and is responsible for selective permeability. If this lipid bilayer is disrupted the cell can become “leaky”. To sterilize heat-sensitive solutions such as culture media, enzymes, and vaccines, one should use a)direct flaming. b)an autoclave. c)filtration. d)boiling. c)filtration.
Anything that is heat-sensitive cannot be exposed to heat. Direct flaming, autoclaving and boiling all rely on heat for sterilization. Filtration is passing the liquid through a filter with pore sizes smaller than the average bacterial cell so they are trapped on the filter and the liquid coming through is sterile. Which method kills microorganisms primarily by denaturation? a)Filtration b)Low temperatures c)Osmotic pressure d)Moist heat d)Moist heat – works by denaturing proteins Filtration – works by physically trapping microorganisms on the filter paper Low temperatures – bacteriostatic process that slows down chemical reactions Osmotic pressure – works by drawing out water leading to dehydration The lowest temperature at which all of the microorganisms in a particular liquid suspension will be killed in 10 minutes is called the a)thermal death point. b)thermal death time. c)decimal reduction time. d)decimal reduction point. a)thermal death point.
