BIO161 Lab Report #11 PDF

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Lab Report #11
Gene Action: Synthesis of β-galactosidase in E. coli
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Lab Report #11 Gene Action: Synthesis of β-galactosidase in E. coli

Overview: Biological Significance: Escherichia coli  (E. coli) can produce the enzyme β-galactosidase which breaks lactose into galactose and glucose. However, the gene for β-galactosidase is normally switched off, except in the presence of lactose. The lac operon is an operon required for the transport and metabolism of lactose in E. coli  and some other bacteria. It consists of three adjacent structural genes, a promoter, a terminator, and an operator. The lac operon is regulated by several factors including the availability of glucose and of lactose. The way the operon works is more important than what it does. Gene regulation of the lac operon was the first genetic regulatory mechanism/transcription regulation to be studied. Hypothesis: The Lac Operon regulates β-galactosidase expression with a repressor molecule, which binds to DNA in presence of glucose or absence of lactose. Therefore, E.Coli exposed to glucose will have lower rates of β-galactosidase enzymatic activity than E. Coli exposed only to lactose. Methods: In this experiment, glucose and lactose metabolism is investigated using three different E. coli  strains. The strains being a wild type (WT, or normally acting, and two different mutant strains, Mutant X and Mutant Y. The goal is to determine which mutations in Mutant X and Mutant Y, may have that explains the sugar metabolism results. The purpose of the wild type is for it to serve as a positive control for the response to lactose or lactose+glucose. How each culture responds to different sugar conditions is observed, over a 40 minute incubation period, in increments of 20 minutes while starting at 0 minutes. First of all, three cuvettes labeled “A,” “B,” and “C” are obtained for each time point. Then a flask with 27 ml of of E. coli  is prepared according to the culture that is assigned by lab benches (ultimately, each E. coli  strain is tested.) In a 15 ml tube, 3 ml of the appropriate sugar, either 10%  lactose only or 10% lactose + glucose, for the culture is added. Then 6 plastics tubes are labeled accordingly: “B-0,” “C-0,” “B-20,” “C-20,” “B-40” and “C-40”. 0, 20 and 40 being in minutes and the time points in which the culture is observed. Into the three “B” tubes, 200 μl  of H2O and 10 μl of chloroform are added. Additionally, in the C tubes, 200 μl of ONPG solution and 10 μl of chloroform are added. Once everyone in the lab bench is ready, at 0 minutes the sugar is poured into the E. coli  flask, swirled immediately. Using a pipet, 1 ml of E. coli  solution is transferred to the cuvette labeled “A” and 1 ml to tubes B-0 and C-0. The B and C tubes are then inverted several times to mix and placed in a heat block, while being checked every minute to see if a reaction has occurred (tube C solution turning pale yellow, this indicates that the enzyme β-galactosidase is present.) The culture flask is placed in a shaking incubator set at 37°C. In between monitoring the tubes in the heat block, record the absorbance of cuvette A at 600 nm. Once there is reaction, 400 μl of Stop Solution (1M Na2CO3) to both tube B and tube C is added. If after 15 minutes no pale yellowness has appeared for tube C solution proceed to adding the Stop Solution. Once the Stop Solution has

been added, in order to record the absorbance, 1 ml from the solution in tube B is pipetted to a clean cuvette and 1 ml from the solution in tube C is pipetted to another clean cuvette. Have the spectrometer set to 420 nm and use the B cuvette as a blank for the C cuvette. Read and record the absorbance. Then, obtain the flask from the incubator and repeat the same procedure for the time points at 20 minutes and 40 minutes.

Results: - see report booklet

Discussion/Conclusions: β-galactosidase activity Discuss expression of β-galactosidase in the wild type and Lac  mutant strains (mutant X and mutant Y) when grown under the different conditions. What specific gene or sequence from the Lac operon is altered in mutant X and mutant Y? The wild-type E. Coli had expected amounts of β-galactosidase with and without glucose. When glucose was present, no β-galactosidase was expressed because when glucose bounds to the galactosidase permease, it does not allow lactose to enter the cell. Without glucose, β-galactosidase will be expressed slowly at first and will increase more over time after the E. Coli has the ability to mix with the sugar. Mutant X had fast reaction times and high levels of β-galactosidase enzyme. That implies that Mutant X did not have galactosidase permease binding to glucose where glucose would normally stop lactose from entering the cell. Also, it is likely that the repressor protein was mutated or not present due to the very quick reaction. RNA polymerase would have nothing inhibiting or stopping it from continuously translating the Lac Z gene and producing high levels of β-galactosidase. Mutant X may also have an enzyme that speeds up it’s reaction of lactose. It was very hard to distinguish what exact mutation occurred from this experiment. Mutant Y had extremely low levels of β-galactosidase and long reaction times. It is most likely that Mutant Y’s repressor molecule could not bind to lactose, which creates a physical barrier for RNA polymerase from reaching Lac Z, therefore β-galactosidase would not be expressed. It is also possible that the promoter molecule, Lac P, was deleted, so RNA polymerase would not be able to bind with anything in order to begin translation. Then RNA polymerase would never reach Lac Z and β-galactosidase would not be expressed. There is no definite conclusion for which of these two mutations occurred in Mutant Y.

Growth of E. coli Did all E. coli cultures grow at the same rate? Discuss growth profiles of all 6 cultures and justify these patterns based upon what you know of the lac  operon and the role of β-galactosidase activity in cellular metabolism. Some of the E. coli culture had some similarities in the growth rates, however not one of the cultures was an exact match for another. The wild-type E. coli with just lactose did not have the growth that was expected it had since it only had lactose the growth rate should have been a slow incline. The number of cells actually decreased from time 0 to 20 minutes then an increase from 20 minutes to 40. The β-galactosidase activity also had a massive decrease from time 0 to 20 minutes. This can be due to the fact that some cells used up more of the lactose in the beginning of the experiment and had a shortage sugar cause some cells to die off and a lower β-galactosidase activity. The 20 minutes to 40 minutes can be explained by there being a more stable number of cells for the sugar left in the flask . This is also supported from the drop of β-galactosidase activity. The wild-type with both lactose and glucose had cell growth from time point 0 to 20 minutes and a decrease of β-galactosidase activity. This goes along with what is know about the lac operon in the E. coli, the cells would rather use glucose than use spend its resources making β-galactosidase. However, the growth does not decrease but increase since it still had a energy source. Then for the time 20 to 40 minutes the growth rate decreased and the β-galactosidase activity increased, this could be due to the fact that the cells depleted their glucose supply and need to start to make β-galactosidase to start using the lactose as an energy source. The mutant X with only lactose had a decrease in the the cells during the time 0 to 20 minutes and increase of cell growth during 20 to 40 minutes. The mutant β-galactosidase activity dropped from 0 to 20 minutes now this does not show much but with the data of the mutant X with both sugars shows the same drop in β-galactosidase activity unlike that of the wild-type E. coli. The wild-type shows a drastic change in β-galactosidase activity when glucose was adding into the solution. The mutant X also did not show a that much of change in the growth rate with and without the glucose unlike that of wild-type. This shows that there is a mutation in the lac operon where glucose does not affect the production of β-galactosidase. Mutant Y had very little to no growth what so ever when it it was only introduced to lactose and showed a low and declining β-galactosidase activity throughout the experiment. Also mutant Y with glucose had more of a growth in the beginning of the experiment and after while, most likely after the glucose was depleted the cells went back to a using lactose with very low β-galactosidase activity. This can be caused by a mutation in the lac operon that make lactose binding more difficult or impossible, or damage to the promoter sequence that does not allow the RNA polymerase to bind to the DNA....