AllBestEssays.com - All Best Essays, Term Papers and Book Report
Search

Phenylethyl Alcohol Agar

Essay by   •  November 12, 2016  •  Research Paper  •  15,067 Words (61 Pages)  •  1,619 Views

Essay Preview: Phenylethyl Alcohol Agar

Report this essay
Page 1 of 61

Exercise 4-1: Phenylethyl Alcohol Agar

Introduction: Selective media is media that encourage growth of some organisms and discourages growth of others. Phenylethyl Alcohol Agar is an undefined, selective medium that encourages growth of Gram-positive organisms and inhibits growth of most Gram-negative organisms. Sodium chloride is added to the medium to provide a stable osmotic environment and digests of casein and soybean are added for nutrition. PEA is selective against Gram-negative organisms because it breaks down their membrane barrier, which than allows substances that are usually blocked to leak into the cell while also causing leakage of a large amount of cellular potassium. Because of this, DNA synthesis is halted and ultimately disrupted. PEA is important because it helps isolate staphylococci and streptococci specimens, including enterococci and lactococci, when given a mixture of bacteria. In the end, the PEA plate is compared with the nutrient agar plate to determine how well specific micro-organisms grew on both plates.  

Results:   Refer To Data Sheet.

Discussion: From my results I obtained that Escherichia coli had poor growth in the PEA plate and because the PEA plate is a selective medium that encourages growth of Gram-positive bacteria and discourages growth of most Gram-negative bacteria I concluded that Escherichia coli was a gram-negative bacteria that was inhibited by the phenylethyl alcohol because it had better growth in the NA plate and very poor growth in the PEA plate. Using the website, http://www.niaid.nih.gov/topics/antimicrobialresistance/ examples/gramnegative/Pages/default.aspx I found that Escherichia coli is indeed Gram-negative, and therefore my results are correct.  Next, I found that Enterococcus gallinarum is a Gram-positive bacteria because it showed better growth in the PEA plate than the NA plate, thus showing that it is not inhibited by the phenylethyl alcohol, and only Gram-positive bacteria are not inhibited by the phenylethyl alcohol. Using the website, http://emedicine.medscape.com/article/216993-overview I confirmed that Enterococcus gallinarum  is indeed a Gram-positive bacterium. Last, I also found that Staphyloccous saprophyticus showed equally good growth in both the PEA plate and the NA plate, and therefore this organism was not inhibited by the phenylethyl alcohol and therefore is a Gram-positive bacteria, and my results are confirmed using http://textbookofbacteriology.net/staph.html which states that Staphyloccous saprophyticus is a Gram-positive bacteria. 

Conclusion: In conclusion, I confirmed that Enterococcus gallinarum and Staphyloccous saprophyticus were both Gram-positive bacteria because they grew equally well in the PEA plate, and only organisms that are not inhibited by phenylethyl alcohol grow well in this plate. Also, because they grew well in this plate, I also confirmed that they were Gram-positive bacteria because the PEA plate is a selective media that encourages growth of Gram-positive organisms. Also, I confirmed that Escherichia coli was a Gram-negative bacteria because it did not grow at all in the PEA plate when compared to its’ growth in the nutrient agar plate. Because phenylethyl alcohol agar is a selective medium that discourages growth of most Gram-negative organisms, I concluded that Escherichia coli was gram-negative since there was no growth prevalent, meaning it was inhibited by the phenylethyl alcohol.

References

http://www.niaid.nih.gov/topics/antimicrobialresistance/ examples/gramnegative/Pages/default.aspx

http://emedicine.medscape.com/article/216993-overview

http://textbookofbacteriology.net/staph.html

Exercise 4-3: Bile Esculin Test

Introduction:   Bile Esculin Agar is an undefined, selective and differential medium meaning that it inhibits growth of some organisms while encouraging growth of others while also exposing differences between organisms. In this agar, beef extract and gelatin provide energy and nutrients, bile acts as the selective agent to separate Streptococcus bovis and enterococci from other streptococci, and ferric citrate is added as an indicator of a positive test of Bile Esculin positive. This test is usually used to indentify enterocci and members of the Streptococcus bovis group, which are all positive, because only these two groups can tolerate bile and hydrolyze esculin. When esculin molecules are split the esculetin reacts with Fe3+ from the ferric citrate that is added and forms a dark brown precipitate that darkens the medium surrounding the growth, and illustrates that the organism is Bile-Esculin positive. If no dark precipitate is seen, the organism is Bile Esculin negative and therefore cannot tolerate bile or hydrolyze esculin.

Results: Refer to data sheet.

Discussion: The Bile Esculin Test is most commonly used for identification of enterococci and members of the Streptococcus bovis group, including S. equines, S. gallolyticus, S. infantarius, and S. alactolyticus.  As the website http://www.studymode.com/essays/Hydrolysis-Report-429546.html states, bile salts, the selective agent, can allow only Enterococcus and group D Streptococcus to hydrolyze esculin in the presence of bile salts, and inhibit growth of other gram positive bacteria, and therefore is the inhibitor while ferric citrate acts an indicator of a positive test. When esculin molecules are split esculetin rections with Fe3+ from the ferric citrate and forms a dark brown precipitate which can be interpreted as the organism being Bile Esculin positive, meaning it can tolerate bile and hydrolyze esculin.

From my results, I found that Lactococcus lactis had color darkening and therefore was positive for the Bile Esculin Test because dark precipitate was formed. The Bergey’s manual states that Lactococcus lactis is negative for esculin hydrolysis and therefore I know my result was not correct.  

Next, I concluded that Providencia alcalifaciens was Bile Esculin negative because it did not show a black precipitate, but rather stayed a light yellow.  The website http://www. ncbi.nlm. nih.gov /pmc/ articles /PMC275140/ states that Providencia alcalifaciens  is negative for this test. As the lab manual suggests, the esculetin did not react with the Fe3+ from the ferric citrate and therefore no dark precipitate formed, showing that is it bile-esculin negative.

...

...

Download as:   txt (101.3 Kb)   pdf (399.8 Kb)   docx (72.1 Kb)  
Continue for 60 more pages »
Only available on AllBestEssays.com