micb questions answers

micb questions answers - is still a probability that a can...

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1. Invaginations increases surface area of plasma membranes, allowing more Electron  Chain Transport proteins/enzymes to be embedded.   Since there are more  proteins/enzymes more reactions can occur, therefore more ATP are produced.  This  makes up for the low yield of energy from aerobic oxidation of inorganic N-compounds 2. Since the bacteria causes food bourne infection, that means that exotoxin is produced in  the intestine and not in the food, therefore sterilization is a possible method to reduce  food poisoning.  From the sterilization equation, it is not possible to reduce the number  of endospore to zero, regardless for how long we sterilize.  Therefore, at best we reduce  the number of endospore the longer we sterilize.  That is for any length of time of  sterilization,  we reduce the possibility of a viable endospore existing in a can, but there 
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Unformatted text preview: is still a probability that a can may contain a viable endospore causing food poisoning. 3. A) Limestone CaCO3 is a base, therefore neutralizing the acid in the mine which is required by the bacteria to oxidize Fe2+. B) This blocks the supply of CO2, O2 and N2. CO2 is required as a carbon source, O2 is required as a TEA. N2 may be required for nitrogen fixation, we don’t know for sure 4. A) From the plant roots B) Urease produces NH3 which is basic (alkaline) which exists as NH4+ C) The dye is used as an electron acceptor by the bacteria, which performs redox D) By sealing the container, this prevents the bacteria from using O2 as a TEA, therefore reducing the dye instead. Most likely the bacteria is a facultative anaerobe....
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This note was uploaded on 01/25/2010 for the course MICB MICB201 taught by Professor Kion during the Spring '10 term at UBC.

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