Fastidious organisms are bacteria that difficult to isolate because it's required special type nutrients for growth. Multiplication and reproduction are common character for all type of bacteria. Fast growth is shown by bacteria when it get sufficient nutrients and it is also common character for all bacteria. Fastidious organisms are required special type of nutrients and it can not grow in normal medium. For this cause this is very difficult to isolate, because sometime special media composition are not known to isolate fastidious bacteria. A suitable example is Neisseria sp is a fastidious organisms which required blood hemoglobin (and also common requirement) to isolate.
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please right answer Question 6 fastidious bacteria is bacteria that is DIFFICULT TO ISOLATE REPRODUCING FAST...
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What signals termination of translation in bacteria? the stop amino acid is attached to the growing peptide chain a tRNA specific to the stop codon enters the ribosome's A site RF1 and RF2 ribozymes EF-Tu and EF-G
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At the end of the Gram stain, gram-positive bacteria will be seen as purple cells. True False
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9.) Which of the following equilibriums favors the right side? Br + – HI + Br t: HCIO + H2O H0* + CIO C. HNO2 + NOHNO, + NO d, H2Se+ HS - HS+Hise e. HS + CI HS+ HCI
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25) Bacteria which grow best at temperature of 50°C are called (1pts) psychrophiles O mesophiles thermophiles microaerophiles 26) Yeast extract is added to a peptone medium primarily to provide additional amounts of O B vitamins O carbohydrates gelatin inorganic salts Opeptidoglycans 27) Facultative anaerobic bacteria (1 pts) respire in the presence or absence of free oxygen reuqire an excess of carbon dioxide for growth Ogrow only in the presence of minute quantities of oxygen are obligate thermophiles...
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QUESTION 29 A. What is the benefit for the rhizobia bacteria and what is the benefit for the legume plant in the legume-nodule symbiosis (2pts)? B. How and why do these benefits impact the growth of the bacteria and the plant? (2pts) TTT Arial 3 (12pt) T.E.E. . Path:p
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Given the equation y = 4 sin 3 2- 19.) + 9T 2 +7 The amplitude is: Preview The period is: Preview The horizontal shift is: Preview units to th Select an answer Left The midline is: y = Preview Right
A bacteria has a doubling period of 6 days. If there are 3300 bacteria present now, how many will there be in 41 days? First we must find the daily growth rate (Don't round up, keep all the decimal places) . The growth rate is _____ Then we use this rate to answer the question. Please include work :) Thanks! There will be ______ bacteria.
The population of bacteria in a culture can be modeled by P left parenthesis t right parenthesis equals negative 0.01 t cubed plus 12.96 t plus 10, where t is the time in hours after the culture was started and P left parenthesis t right parenthesis is the population in thousands. Complete the table to determine the population of the bacteria for the given values of time, t. This is a "Fill in the blank" question so please give ONLY...
lake-up Micro Lab: Identifying Unknown bacteria: dentify the Unknown Bacteria Lab Experiment elow are results from the unknown lab experiment. The identity of this bacteria is not known. Your ask is to use the data below to form a hypothesis of the identity of this bacteria. Record your results on he chart (Table 1) below and make a prediction on the bacteria based on your results. There is a list of ossible bacteria below. You may need to look back...
Dont't copy please answer all parts of this question and explain everything please. thank you 7) You isolate a mutant strain of mice that grow at an unusually fast rate, perform a blood test on the mice, and find that they have elevated levels of insulin-like growth factor2. The phenotype is the result of a mutation in a region of the genome containing a gene encoding a DNA methylase. What kind of mutation is causing the rapid growth of these...