Introduction To Microbial Genetics
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Published: Mon, 01 May 2017
In this experiment, handling bacteria was learned and phenotypes of bacteria in microbial genetics are investigated. For the phenotypes, E. coli and its mutants are used. The second part was done to observe plate assays to see whether or not the β-galactosidase was produced or not. The last part was to assay plaque forming units in E. coli medium infected with a sample of B. subtilus phage SPO1.
Auxotrophic mutants require specific nutrient or compound to grow and express its phenotypes (Madoka). Bacterial growth involves lac operon. Therefore, it is crucial to understand the functionality of lac operon. There are three genes of lac operon which needs for lactose metabolism: lac Z, lac Y, and lac A. lac Z encodes β-galactosidase enzyme, lac Y encodes β-galactosidase permease and lac A encodes lactose transacetylase. (2)(1)
This experiment introduces a few ways to test lac phenotypes. One way is to use minimal lactose medium. In this medium, Lac- will not grow whereas Lac+ will grow. Second way is to use MacConkey medium. MacConkey medium is a pH sensitive medium which would show red in the acidic medium. In this medium, both Lac+ and Lac- will grow on the plate but Lac+ would be the only one that would turn red. The reason is because the fermentation of lactose by Lac+ produces acidic metabolite that lowers pH of the media. Third way is to use X-gal plate where the inducer is also introduced along with bacteria. X-gal cleaves the β-galactosidase enzyme then it is shown as blue. (1)(2)
There are four ways to test lac phenotype we perform in this experiment. One is to use minimal lactose medium, and Lac- will not grow on the plate while Lac+ will. Second is to use MacConkey medium. Both Lac+ and Lac- can grow on the plate but only Lac+ will be pink on the plate because the fermentation of lactose by Lac+ produces acidic metabolites which lower the pH of the media and the pH indicator turns the plate pink. Third is to use X-gal plate which has IPTG as an inducer in the plate. Both Lac+ and Lac- can grow on the plate but only Lac+ will be blue. The last method is to use ONPG which turns yellow with β-galactosidase, and units of enzyme activities are calculated. (3)(1)
Material and Methods:
All procedures are performed according to the BIOL 368 lab manual (Concordia Biology Department 2013) except for the following modifications: After adding ONPG, forgot to verify the colors.
Part A. Growth of E.coli
Cultures of E. coli were grown in flask containing liquid medium and containing solid medium with agar. E. coli in the plate that contains liquid medium was yellowish. Those were circular in shape and flat. The surface was smooth and opaque. When the bacteria was introduced to liquid medium, bacterial lawn was observed. These were small and rough. Some were not opaque, more like flat.
Part B. Phenotypes of WT and mutant strains
I. Auxotrophic mutants
To examine auxotrophic mutants, JF1754 was used. This strain requires methionine, histidine and leucine. Therefore, we predicted that it will not grow on the media where all three amino acids are not contained. As predicted, JF1754 only grew on the medium with all three M, H and L amino acids. Grown JF175 was seen as white. This is because auxotrophic mutant requires all of the required nutrients to carry out the biosynthetic pathway. On the other hand, CAG12033 requires only a minimal media to grow which means in the minimal media, it is able to carry out the biosynthetic pathway.
II. Antibiotic resistance
In this part, strain EC5827 was used. Along with CAG12033, it was introduced to LB medium and LB medium containing streptomycin. As a result, EC5827 grew in both of the plates whereas CAG12033 did not grew in the plate with streptomycin. This is due to the fact that the strain EC5827 is resistant to streptomycin whereas CAG12033 is not. CAG12033 does not have the resistance, and therefore, it eventually did not grow on plate with streptomycin.
III. The many colours of lac
In min+ lactose:
Strain NK6042 growth was not observed. This is because the lac operon was deleted from the chromosome of strain NK6042 and therefore, it cannot metabolize the sugar which prevents it to grow. On the other hand the wild type grew eventually.
Strain NK6042 and wild type grew on the plate because this medium contains complex carbon source. The wild type is able to metabolize lactose and peptone and NK6042 can metabolize peptone only. However, they differed in color. NK6042 was white and the wild type was pink. This is because MacConkey medium contains pH indicator that turns pink under acidic condition. In addition, Lac+ acidifies the medium by fermenting lactose and by excreting the mixture of organic acids whereas Lac- does not. The result makes sense because wild type contains lac operon to produce Lac+ cells and NK6042 does not have lac operon.
Strain NK6042 does not appear as blue, it is just transparent whereas the wild type is totally blue. This is due to what X-gal does. X-gal cleaves the β-galactosidase enzyme. Then, it turns blue. Therefore, it indicates whether or not the β-galactosidase enzyme is produced. NK6042 is not able to produce β-galactosidase since it has its lac operon removed. However, the wild type is still able to produce β-galactosidase in the minimal medium, and as a result, its color turns into blue.
Part C. β-galactosidase assay
The objective of the lab was to practice handling bacteria and learn methods that investigate phenotypes of bacteria. Table 3 shows the anticipated activity and respective volume of culture and z-buffer. There are three strains involved, CAG12033, MH321 and ML 308. Each of the strains are examined with and without IPTG. IPTG is an inducer where it binds the repressor that prevents the transcription. Then, repressor is no more able to repress. Therefore, inducer stimulates the transcription, and allows the production of -galactosidase. Low activity is expected without IPTG and high activity is expected with IPTG. In addition, cAMP level is expected to be low in presence of glucose so the CAP protein would bind to the promoter minimally. ONPG is used for quantitative -galactosidase activity in the sense that it breaks it into o-nitrophenol and galactose. When it turns yellow, it comes from o-nitrophenol. Looking at table 2, MH321 always has no activity because, lac Z gene is mutated, so it cannot produce β-galactosidase no matter what medium. IPTG only binds repressor and prevents it from repressing, so the inducer has no effect. On the other hand, ML 308 has a mutation in repressor. Most probably, it is ID and it works constitutively. Also, IPTG will not be able to bind this repressor, but this repressor does not prevent the transcription. Therefore, high activity predicted in any cases.
Table 5 shows that the prediction is eventually correct. We have highest value for the low activity with CAG12033 without IPTG: 20.38 min-1 ml-1 Au-1. This is considered as low activity compared to high activity values we have and any other low activity predicted values have lower values than this. However, we earned low activity for MH321 where we predicted none. This is due to experimental error and this value is low enough to be considered as none. Therefore, our prediction for the low activity was precise. For the high activity, the highest value was found with the CAG12033 strain: 1506.4 min-1 ml-1 Au-1. With ML308, we have 570.4 and 767.4 min-1 ml-1 Au-1. These values are much bigger than the ones we have as low activities. They are enough high to be considered as high activity.
For MH321, we got low activity because of the following reasons: mutation and contamination. One of the reasons that the mutant strains still have detectable activity is that the silent mutation might be occurred. That is, the mutation of a base does not change the amino acid sequence. Also, MH321 is the strain that has a mutation in lac Z and there may be some that still has lac Z activity. The other reason is the contamination, which is the most common source of error. The container might not have been washed enough, or microorganisms in the air may penetrated in the solution containing MH321. Speaking of the expected range, we expected the activity range from 1000 to 3000. However, our values are smaller: ML308 had 570.4 and 767.4 min-1 ml-1 Au-1.This is due to the fact that we did not have enough time for it to turn yellow, so, the values of the absorbance or optical density was lower than it is supposed to be.
For part D, we have various mediums and various strains are used to predict which unknown is which strain. To figure out which is which, we made a table predicting in which medium the strain will grow or not. Looking at unknown 2, it only grows on Min M. We have the same with MH142 where it only grows with methionine plate. Also, both gives in MacConkey, white. Therefore, unknown 2 is supposed to be MH142. Because MH142 requires methionine to grow as an auxotrophic strain, it only grows in Min M. For unknown 3, the growth pattern matches MH807 and therefore, it should be MH807. With MacConkey plates, MH142 are lac- due to the white color presence of the plates. With the same principle, we figured that the unknown 5 is CAG8209 and 6 is CAG 18475. CAG8209 is tetracycline resistant and requires leucine and CAG 18475 has tetracycline resistance and requires methionine to grow. However, unknown 4 and 1 cannot be determined. They both have same growth pattern and also, both gives red color in MacConkey plate. They should be CAG12204 or D10. We need further test to identify which unknown is which strain. CAG12204 has kanamycin resistance whereas D10 does not. Thus, we can introduce this unknown to kanamycin+ minimal medium. If it grows, it is CAG12204 and if not, it should be D10. Finally, the result of unknown 7 strain matches none of the theoretically predicted result. It is due to the experimental error. We may have introduced some other unknown or solution in the medium that leaded to this error. Fortunately, it is the last unknown, and the last strain left is BW6165. So, it should be BW6165. BW6165 is only expected to grown in tetracycline and min R since it has tetracycline resistance and requires arginine to grow. However, it did not grow in our plate, therefore, if may have put something other than BW6165 by mistake such as CAG8209.
In part E, SPO1 phage was introduced to E. coli and different concentration of B. subtilis cells. When B. subtilis was 10-folded and 100-folded, bacterial lawn was observed, so there were a lot of small and adjacent colonies. When it was diluted to 10-3 and 10-4, isolated colonies were found, 174 and 16 colonies respectively. We had 1.74*106 pfu titer. The section average was 1.75*106 and therefore, our value is very close to the section average. They are almost identical, and we can say that we successfully diluted and obtained proper colonies. In addition, when there was no phage and when 10-folded E. coli was plated in each plate, there was not a single colony found. This is because E. coli does not have the specific receptors for infection to occur by SPO1 whereas B. Subtilis does.
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