Can you answer the B part of the 2nd question. However, after doing a short lite

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Can you answer the B part of the 2nd question.

However, after doing a short literature search, I find out that alanine is a non-competitive inhibitor of this enzyme (Ki 0.5 mM). Since alanine is also an amino acid, I n with the other 19 amino acids in my CFPS reactions to synthesize protein. If I provide 33 mM PEP in my CFPS reaction, what is the maximum amount of alanine I can add (ie, maximum alanine concentration) so that the reaction rate is at least at 80% of its maximum velocity? (Assume ADP is in excess) (6 pts) 2. I am in charge of formulating the growth medium for a pilot-scale E. coli fermentation (1000 liter working volume). My cells are composed of 0.2% iron, 0.5% (atomic) chlorine, and 14% nitrogen (on a dry weight basis). My growth target is 120 g/liter of dry cell weight. ) How much ferric chloride (FeCls, FW-162.2) do I need to add (g/liter) ifl want a 25% safety factor? (2 pts) B) If NH4OH is my main nitrogen source, and the culture does not make acid or base, how many kg of NH4OH will be fed to a 1000-liter fermentation to control the pH? (2 pts) There is a lot of interest and investment in "clean technology". One plausible option is the production of biodiesel from corn starch to replace gasoline. We want to develop a process using an engineered E. coli (CHi.sOos No.22) to make a fatty acid (CisH3602, MW 284 g/mol) from the starch hydrolysate. We have a 60,000 L fermentor available to use and wil be filling it with 50,000 liters of medium. After sterilizing the empty vessel, we will use the 3. continuous sterilizer to sterilize the medium as we pump it into the clean and sterile vessel tiuous sterilization temperature of 145°C. We estimate that our in 1000 chance of

Explanation / Answer

2.b.

growth rate is 120 gm /liter

weight of nitrogen/liter = 120*0.14 = 16.8gm

required nitrogen for 100 liter = 16.8 * 1000 gm

weight of NH4OH = 35

weight of nitrogen in 35 gm of NH4OH = 14 gm

hence, 16.8 * 1000 gm nitrogen will be obtained from = (35 *16.8 * 1000) / 14 = 42000 gm =42 kg

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