
Escherichia coli have lac operon produces the enzyme β-galactosidase which can break lactose to galactose and glucose. But the gene for β-galactosidase is generally switched off in the absence of Lactose. In this procedure, a sample of E. coli is treated with lactose, and then the β-galactosidase activity of this sample and glucose treated sample are compared. ONPG (ortho-nitrophenyl-β-D-galactoside) is used as a substrate for the enzyme action which produces galactose and a compound which is yellow in alkaline conditions. The intensity (or optical density) of the yellow colour produced is a qualitative indicator or quantitative measure of the β-galactosidase activity.
In Lab activity 2, when we look into the activity of reaction 1 and reaction 3 that is the culture conditions in LB with Lactose and Glucose respectively, it is evident that the Glucose containing LB culture possess more activity when compared to the Lactose containing culture. A similar response is seen when we look into reaction 2 and 4. Thus in the presence of glucose, lactose is least preferred and thus less ONPG is converted to product, that is optical density is low and absorbance is high, and hence the activity is higher when glucose is present in the culture. The procedure indicates that the gene which produces β-galactosidase in E. coli is induced switched on by the presence of lactose. Hence sugars do interfere the assay. Thus the presence of sugar affects the assay reading.
Lab Activity 2: Table 3 CultureRan ulUnused Lysis E. ColiZbuffer Vol. ONPG Conditions | # | 20% |...
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Lab 3 Using the blood glucose calibration curve and your knowledge of what a diabetics glucose levels are, your job will now be to determine whether or not insulin is needed. You are going to imagine that you are a person with diabetes, but you've broken your meter. Luckily you had all of the reagents needed to test your blood glucose tests on your own. Please determine what your blood glucose level is at each time interval and determine...
C E F G В Activity 1 $20 D Activity 3 $30 Activity 2 $40 Unit Profit NM 4 Resourceſ Resource Used per Unit Produced 1 Totals 400 100 200 Available 400 100 200 | Activity 3 Activity 1 50 Activity 2 50 | 10 Units Produced Total Cost $3,000 0 Where are the changing cells located? Multiple Choice 0 E5:E7 £5:07 0 B10:D10 0 G10 610 0 B2:02 B2:02 0 B2:D2, B4:07, and G5:67
Bio 121
I need to make (yeast fermentation) lab
report.
This is the lab experiment and results:
This is a guide to making the lab report:
General Biology BIO121 Yeast Fermentation Lab Introduction Organisms stay alive by the utilization of energy through metabolism. The energy acquiring pathways in photosynthesis convert radiant energy from the sun into the chemical bond energy of carbohydrates. This photosynthetic process is limited to the producers or autotrophs, which include plants, photosynthetic bacteria and some protists....
he he tolenwine table presents the aetivitiles of s small grojest and thetr tme The precedence relationships among the activities are also p Duration Activity Predecessor 2 4 C, D, E Draw the network of the project showing the activities and their time durations Note that this should be a AON network). a) Using the time durations for each activity, determine the critical path for the project network. b) decide whether on. She 52. The local operations manager for the...
Graph the rate of cellular respiration in
germinating (Table 7-3) vs. non-germinating (Table 7-2) lentils by
plotting levels of CO2 detected in the
BioChamber 250 over ten minutes, collected at 1-minute
intervals. Plot CO2 results of the
germinating lentils and the non-germinating lentils on the same
graph and use a key to distinguish between the two lines. The
independent variable, time elapsed in minutes, should be plotted on
the x-axis, and the dependent variable, CO2 levels in
ppt, should be plotted...
LEILU.23.27 Eugenia Bueso-Alas: Attempt 2 UUTUUU-Dye SD Incorrectly Low Question 3 (20 points) An unknown food-dye is analyzed by the method used in this experiment. The absorbance of 5 standard solutions is plotted vs. their concentration (molarity) and a straight line is obtained with the formula: Abs = 4.350x +0.06 An unknown food-dye soluton is diluted and it's absorbance is found to be 0.65. What is the molarity of the diluted unknown, to two significant figures? Do not type units...
lab question 1. What is the basis of the different purification methods? 2. What are some of the factors the might have interfered with your results? 3. How might you improve the process to increase the yield and purity? lab process E. coli BL21 (DE3) cells were transformed with the pET Topo-1521 vector containing a reading frame encoding the green fluorescent protein (GFP). Cells were cultured in M9ZB media at 37°C until the absorbance at 600 nm reached 0.7, at...
Case Study for Care Plan Assignment: fill each attached column with its appropriate answer, use the below scenario. A retired 69-year-old man "Mr. Casey" with a 5-year history of type 2 diabetes. Although he was diagnosed 5 years ago he had symptoms indicating hyperglycemia for 2 years before diagnosis. His fasting blood glucose values of 118–127 mg/dl, which was explained to him as “borderline diabetes.” He also states he has had past episodes of nocturia with large pasta meals and...