Answer: The flask A will have the most number of the <em>E.coli</em> cells. B. The flask C will have the least number of the <em>E.coli</em> ce
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Explanation:
<em>Escherichia coli</em> is a bacterium member of the enterobacteriaceae family and is part of the microbiota of the gastrointestinal tract of homeothermic animals, such as humans. It is a gram-negative and facultatively anaerobic bacillus, whose preferred growth temperature is 37 °C (called mesophilic).
<u>Facultatively anaerobic means it can develop in both the presence and absence of oxygen</u>. Generally, they can develop a respiratory metabolism, using the oxygen present; or fermentative, in the absence of oxygen. This means oxygen is not toxic to them.
<em>E. coli</em> is the most abundant facultative commensal anaerobic bacterium in the microbiota of the gastrointestinal tract where, together with other microorganisms, it is essential for the correct functioning of the digestive process. It also ferments glucose and lactose as a source of carbon, with the production of acid and gas after 24 hours, and it reduces nitrates to nitrites with energy production that is conserved as a driving proton force that is then used for ATP synthesis.
Flask A contains glucose and nitrate (which is used as a source of carbon) and oxygen.
Flask B contains glucose and nitrate and no oxygen.
Flask C contains only glucose and no oxygen.
<u>The most optimal growth conditions are found in flask A </u>which contains glucose and nitrate, and the oxygen is used to perform cellular respiration which is a process that provides more energy than fermentation. Besides, the presence of nitrate also provides more energy used to grow.
<u>Flask C will have the least number of cells</u> because there is no oxygen there and the bacterium has to perform fermentation instead of cellular respiration.
