Normal No Spacing Exercise b. Controi of Growtn by UV Lignt Figure 1 Side A Side

Question

Normal No Spacing Exercise b. Controi of Growtn by UV Lignt Figure 1 Side A Side B Side A Side B B. cereus E. coli Figure 1 shows the results of UV light exposure to E. coli and B. cereus. One side of each plate was exposed to 30 seconds of UV light. The other side was not exposed to any UV light (0 seconds exposure). 1. Which side of each plate (A or B) was exposed to 30 seconds of UV-light? Which side was not exposed to any UV light? Explain how you know this to be true. side A of each plate was exposed to UV light, I know this to be true because UV light is used to control microbial growth, UV light causes irreparable damage to bacterial cell DNA. It uses a form of non-ionizing radiation that creates thymine-thymine dimers which interrupts DNA replication, this mutation is lethal to bacteria. 2. Based on figure 1, which species was more resistant to UV light? B. cereus was more resistant UV light than E. coli. 3. Explain why this strain was more resistant to UV light (hint: think about cell structure difference between the two...use your textbook for help as needed) B. cereus was more resistant UV light than E. coli because it can form endospores and has the endospores hard shell protecting it from UV light 4. 5. Looking at figure 1-what is the % reduction in growth between Oand30seconds UV light exposure? Percent reduction in growth between 0 and 30 seconds of UV light exposure is Using figure 1 as a guide, what do you expect the % reduction of growth to be for B. cereus when comparing 0 seconds to 60 seconds of UV light exposure? Explain. 6. States)

Explanation / Answer

1-ANSWER)

Side B was exposed to UV light in both plates. Side A was not exposed to UV Light for 30 seconds. Short wavelength UV radiation is used to control microbial growth. UV radiation causes change in structure of DNA by inducing formation of Pyrimidine dimers. This causes a mutation in DNA and replication is halted. The bacteria cannot grow further and dies.

2- ANSWER) B.

cerus was more resistant to UV radiation as some colonies can be observed forming in the UV radiated portion of plate.

3- ANSWER)

B.cersus is known to form endospores. These endospores can survive UV radiation because they have tightly bound DNA with a protective coating.

4-ANSWER)

Percent reduction is estimated by counting the colony forming units (CFU). It can be said that there is about 95% reduction in the CFUs in both plates.

5-ANSWER)

We can estimate that there will be 99 % reduction in growth. As the death of bacteria is directly proportional to exposure time. The longer it is exposed to UV the sooner it dies.

PART- 2

1-ANSWER)

We use a carbapenem disk and a susceptible strain for modified hogde test (MHT). The production of carbapenemase allows the growth of susceptible strain giving a characteristic cloverleaf- like indentation. We use Mueller Hinton broth (MHB) or 0.85% physiological saline, Mueller Hinton agar (MHA), meropenem or ertapenem susceptibility disk and E. coli ATCC 25922 strain. After 24 hrs check, IF Cloverleaf like indentation present result is positive for carbapenemase production. Negative result doesn’t show any growth.

2-ANSWER)

LIMITATIONS of MHT

1)Carbapenemase class cannot be determined

2) False positive show growth without carbapenemase production.

3)Lack of specificity

4) Delay in obtaining result.

3-ANSWER)

Main objective of study undertaken by Gilrich et al was to determine the value of the Modified Hodge Test for Detection of Emerging Carbapenemases in Enterobacteriaceae. This was done by studying the sensitivity of assay across different bacterial strain and media taken for growth.

4-ANSWER)

In image where the colonies are produced it would indicate that the test for SIDE A was positive for carbapenemase production since they are able to survive in media, while side B was negative with almost no colonies.

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