There are many different mechanisms involved in controlling the regulation of gene expression in prokaryotes. In order for the cells to be energy efficient, they should not produce proteins and other cell components which are not necessary for the cell.
Amongst the bacteria, the bacterium E. coli has an efficient mechanism for metabolising lactose. Three proteins that are important in lactose metabolism are all encoded in a single expressible unit of DNA, called the lac operon which consists of three structural genes, a promoter, a terminator, regulator and an operator (Marks DB et al, 1996). The structural genes in lac operon encode three types of proteins which are involved in the metabolism of lactose in the E. coli. These include: β -galactosidase; which converts lactose molecules into glucose and galactose, β -galactoside permease; responsible for transporting lactose into the cell and finally β -galactoside transacetylase which its function is not yet known (Marks DB et al, 1996).
Get Help With Your Essay
If you need assistance with writing your essay, our professional essay writing service is here to help!
The bacterium does not waste energy expressing these proteins if lactose is not present in the growth medium. It only makes these proteins when lactose is available to be metabolized. In an E. coli cell growing in the absence of lactose, a repressor binds to the operator, preventing RNA polymerase II from transcribing the lac operon’s genes (Marks DB et al, 1996). When an E. coli cell is growing in the presence of lactose the inducer, allolactose; which is the product of lactose binds to the repressor and prevents it from binding to the operator region. As long as there is no repressor to bind to the operator, RNA polymerase can recognize the promoter and start transcribing the structural genes in the lac operon (Marks DB et al, 1996).
So, this experiment is aiming to investigate the expression of lac operon under different condition such as presence of an inducer such as IPTG, alternative source of energy to lactose such as glucose as well as presence of bactericidal antibiotics which prevents protein synthesis in E. coli.
The following experiment was carried out to examine the relative levels of B-galactosidase in growing E. coli K12 cells using B-isopropylthiogalactoside (IPTG) as the inducer and compares this to the situation with no induction as well as determining the induction time of the lac operon in E. coli . Moreover, this experiment will investigate the effects of glucose, chloramphenichol, rifampicin and streptomycin addition to the system as together with a comparison of lactose as the inducer rather than IPTG.
In order to carry out this experiment, the materials and experimental procedures described in BIOC2201 manual (2010) were followed carefully without introducing any changes to it. These include:
The experimental procedure for investigating the course of induction of β-galactosidase by IPTG experimental procedure, p 50-51.
Experimental methods for the assay of β-galactosidase activity, p 52-55.
The experimental procedure for investigating the characteristic of the induction of β-galactosidase, p 56-58.
Worked example of calculation:
The number of Units of β-galactosidase enzyme in each tube was calculated as following given that one unit of β-galactosidase is the amount of enzyme that catalyse hydrolysis of 1micro mole of ONPG to o-nitrophenol per minute, and the ϵ414 of o-NP is 21,300 M-1cm-1, the path length is 0.9 cm and the assay volume is 0.8ml.
The number of Units of β-galactosidase enzyme in tube 1 (induction time 1).
Assay volume = 0.8ml = 0.0008L
A = ϵCl
C = A/ϵl
C = (0.003)/(21,300M-1cm-1 x 0.9 cm) = 1.56 x 10-6 = 1.56 µM
Now using n=cv, the number of moles of o-NP in 0.0008L of assay is:
N = cv
N = 1.56µM x 0.0008L
N = 1.25 x 10-2µmol
As the induction time was five minutes, the number of Units of β-galactosidase can be calculated as:
Units of β-galactosidase = 1.25 x 10-3µmol/5min
= 2.50 x 10-4µmol/min
As this is the Units of β-galactosidase in 200µL of bacterial culture, now, the Units of β-galactosidase per one ml of bacterial culture would be five times greater, i.e.
2.50 x 10-4µmol/min x 5 = 1.25 x 10-3µmol/min
The same method was used to calculate the number of Units of β-galactosidase in the other tubes.
Worked example of calculation:
The number of Units of β-galactosidase enzyme in each tube was calculated as following given that one unit of β-galactosidase is the amount of enzyme that catalyse hydrolysis of 1micro mole of ONPG to o-nitrophenol per minute, and the ϵ414 of o-NP is 21,300 M-1cm-1, the path length is 0.9 cm and the assay volume is 0.8ml.
The number of Units of β-galactosidase enzyme in tube induced only by IPTG and labelled as 5 (induction time 5) can be found as following:
V = 0.8ml = 0.0008L
A = 0.088
A = ϵCl
C = A/ϵl
C = (0.088)/(21,300M-1cm-1 x 0.9 cm) = 4.59 x 10-6 M = 4.59 µM
Now using n=cv, the number of moles of o-NP in 0.0008Lof assay is:
N = cv = 4.59µM x 0.0008L = 3.67 x 10-3µmol
As the induction time was five minutes, the number of Units of β-galactosidase can be calculated as:
Units of β-galactosidase = 3.67 x 10-3µmol/5min
= 7.34 x 10-4µmol/min
As this is the Units of β-galactosidase in 200µL of bacterial culture, now, the Units of β-galactosidase per one ml of bacterial culture would be five times greater, i.e. 7.34 x 10-4µmol/min x 5
= 3.67 x 10-3µmol/min
The same method was used to calculate the number of Units of β-galactosidase in the other tubes.
GRAPH B (The graph of Units β-galactosidase per ml of bacterial culture against the time of induction of lac operon in the presence of IPTG, IPTG + Chloramphenichol, Lactose, IPTG “5 minutes” + glucose, IPTG “10 minutes” + glucose, IPTG + rifampicin or IPTG + streptomycin)
The amount of β-galactosidase production in each tube is the indictor of expression of lac operon in that tube. Hence, quantifying the amount of β-galactosidase in tubes, each under different condition is a very useful tool to understand the role of inducers and repressors in controlling the lac operon expression and in general, the control of gene expression in prokaryotes.
At the given time sets, CTAB was added to the tubes to kill the E. coli cells and lyses the cells to release its contents including galactosidase enzyme. (Lab Manual, 2010).
As it is difficult to directly quantify the amount of β-galactosidase produced in the tubes, but we can measure id indirectly. Β-galactosidase can convert ONPG to galactose and o-nitrophenol which has a yellow colour with an absorbance maximum at 414nm (Lab Manual, 2010). The absorbance reading in each tube indicates the amount of ONPG converted to galactose and nitrophenol by the enzyme β-galactosidase. This in turn indicates the amount of β-galactosidase enzyme produced in each tube, i.e. the greater the absorbance reading, the greater the amount of β-galactosidase in each tube. Using this method, the rate of production of β-galactosidase, hence the rate of expression of lac-operon in E. coli in the presence of inducer, effectors and repressors were investigated.
Find Out How UKEssays.com Can Help You!
Our academic experts are ready and waiting to assist with any writing project you may have. From simple essay plans, through to full dissertations, you can guarantee we have a service perfectly matched to your needs.
View our academic writing services
Graph A shows the amount of β-galactosidase produced in E. coli after adding the IPTG; the inducer in different period of times. At first glance, it can be seen that the graph does not cut the X axis at time 0 when the IPTG and CTAB were added at the same time. This is because the IPTG did not have enough time to induce the lac operon. The graph cuts the X axis at approximately 2-minutes after the addition of inducer (IPTG). This is when there was significant amount of β-galactosidase produced, i.e. the lac operon was first induced. This suggests that adding an inducer to the E. coli cell does not lead to an immediate production of β-galactosidase; rather it takes a while for the lac operon to pass all the expression stages and produce the final product; β-galactosidase and other enzymes. Looking at the control, there seems to be no β-galactosidase made in the process. This is because there was no IPTG added to the tubes and the repressor was not inhibited from binding to the operator, hence the RNA Polymerase II did not bind to the operator to initiate the transcription process.
Looking at the graph B, (IPTG only) there is a positive linear relationship between the time of induction with IPTG and the amount of β-galactosidase production in the tubes. IPTG is an inducer, binding to the repressor protein and inhibit its binding to the operator region in the lac Operon and allows the lac-Operon itself to be expressed (Lab Manual, 2010).
Moreover, in the tubes where the lactose was added, the amount of production of β-galactosidase was quite low compared to that of where there was no lactose present, i.e. IPTG only. When there was lactose added to the tube, it was converted to products; allolactose and glucose by the enzyme β-galactosidase. Allolactose which has a similar action to IPTG, act as inducer, binding to the repressor and inhibits it’s binding to the operator region in the lac operon and allows it to be expressed. Allolactose unlike IPTG can be easily hydrolysed so its inducing action lasts for a shorter period of time compared to that of IPTG (Lab Manual, 2010). As a result, its graph looks flatter.
When the lac operon expression was investigated in the presence of glucose, the graph initially increased but once the glucose was added, the graph started to flatten out. In E. coli cell, lactose is not always the preferred source of energy (Marks DB et al, 1996). Beside lactose, if there is glucose present in the cell, the bacteria prefer to use glucose before starting the turning the lac operon on. This suggests that there must be another level of gene expression control exists in order to prevent the lactose from being metabolised. In the lac operon, the promoter has two distinct binding sites; one for RNA polymerase enzyme binding and the other site is where catabolite activator protein (CAP) along with cyclic AMP (cAMP) binds to (Marks DB et al, 1996). For the lac operon to be transcribed, it is crucial for the CAP-cAMP complex to bind to the promoter site. On the other hand, wether this complex is present in the cell or not, is associated with the availability of glucose inside the bacterial cell. When there is high amount of glucose present n the cell, the amount of cAMP decreases and vice versa (Marks DB et al, 1996). Once the level of cAMP was declined, this lead to the inactivation of CAP+RNA Polymerase II complex. This inactivation in tern inactivates the promotor; hence the lac operon is getting turned off. Hence, after the time 5 minutes, the lac operon was repressed. In the tube where the glucose was added at the time 10 minutes, the graph started to flatten out right after adding the glucose. Here, the E. coli had more time to express the lac operon and make more of β-galactosidase enzyme. These two stages of the experiment show that glucose is of a major important in controlling the lac operon expression in E. Coli.
On the other hand, looking at the graph B, where there was chloramphenichol added to the tube at the time 10mins, the rate of production of β-galactosidase remained almost constant after the time of addition. This is because chloramphenichol is an antibiotic, inhibiting the peptidal transferase activity of the bacterial ribosome (Todar. K, 2007), hence blocking elongation of polypeptide chain, i.e. β-galactosidase enzyme.
Similarly, rifampicin inhibits DNA-dependent RNA polymerase in bacterial cells by binding to its beta-subunit (Todar. K, 2007), thus preventing DNA transcription and hence protein syntheses. Streptomycin, another antibiotic is also protein synthesis inhibitor. It binds to the 30S subunit of the bacterial ribosome, interfering with the binding of tRNA to the 30S subunit (Todar. K, 2007). This action blocks translation of lac operon mRNA in E. coli and prevents the β-galactosidase enzyme to be made. These types of actions of antibiotics block the expression of lac operon and keep the amount of β-galactosidase enzyme to the level which was present in the tubes before the antibiotics addition. This is clearly shown in the graph B. This inhibition of lac operon by antibiotics indicates that the level of production of β-galactosidase is regulated by the expression of lac operon.
Finally, from this experiment it can be concluded that the expression of Lac operon in E. coli is being controlled regularly depending on the availability of lactose in the cell, presence of inducers and the availability of alternative source of energy for metabolism instead of lactose. When there is glucose present in the cell, the lac operon is being repressed. Likewise, in the presence of lactose and the absence of glucose, the lac operon is being expressed to metabolise lactose molecules in the cell. Overall, these finding suggests that the bacterial cells are very efficient in terms of energy consumption; avoiding spending energy to produce unnecessary proteins.
Cite This Work
To export a reference to this article please select a referencing style below:
