Using G-25 Sephadex beads (Fractionation range 1000-5000) KD, a sample of Vitamin B12 (MW ~1500 KD) will be: Fully excluded Partially included Fully included the activity of lysozyme was determined by looking at the increase in light scattering per unit time increase in absorbance per unit time. decrease in absorbance per unit time. increase in concentration per unit time.
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Using G-25 Sephadex beads (Fractionation range 1000-5000) KD, a sample of Vitamin B12 (MW ~1500 KD) will be:
|
Fully excluded |
||
|
Partially included |
||
|
Fully included |
the activity of lysozyme was determined by looking at the
|
increase in light scattering per unit time |
||
|
increase in absorbance per unit time. |
||
|
decrease in absorbance per unit time. |
||
|
increase in concentration per unit time. |
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- Using G-25 Sephadex beads (Fractionation range 1000-5000) KD, a sample of Vitamin B12 (MW ~1500 KD) will be: Partially included Fully excluded Fully includedin food microbiology, how do you compute for concentration (M) and absorbance (A) of peroxidase activity on hydrogen peroxide at different pH values? (wavelength used = 415nm). can you please explain the calculations thank you if given: Temperature = 75 C Molar absorptivity coefficient = 10.5 Path length (cm) = 2 I = transmitted light = 0.45Table 1 - Comparison of the effect of catechol concentration on the amount of product formed. Absorbance Potato extract Absorbance 0 mins after 30mins (2nd reading) (mL) 1st reading 1 Tube # la blank 2a 3a 4a 1 1 1 dH₂O Catechol (mL) (mL) 7 5 3 1 0 2 4 6 0.00 0.060 0.033 0-05-2 Q4) Give 2 reasons for adding dH₂O to these tubes in Table 1? Time for reading: 3:21 -0.11 Absorbance: Time for reading: 3.36 Q5) Tube la serves as a control, but why is this control needed? Absorbance: 0.197 Time for reading: 3.37 Based on the data from Table 1 answer these questions: Q1) What is the name of the enzyme found in potato extract? Answer: catechol Q2) What is the substrate? Answer: THO Q3) Name of product of this enzyme catalyzed reaction? Answer: Absorbance: 0.152 Time for reading: 3:39 Absorbance: . 166 ness Catechol Benzoquinone Subtract 1st from 2nd reading -0.01 0-137 0.11.19 0.119 Q6) Notice that your 1st absorbance reading in tubes 2a-4a are quite similar but it then becomes very different…
- Given this, if you used 6g of vitamin Z powder to make 20 ml of solution, what is the % concentration of this solution? (I gave the image since I don't know if that info is needed to solve this question.)It also gives a follow-up, if you can help here too: You work in a lab as a summer student. One of your tasks is to make sure that there is enough cell culture medium containing antibiotics to grow bacteria. One day you realize that there is only 5 ml of 10% Antibiotic stock solution in the freezer. You decide to use it all to prepare the working culture medium with 0.01% antibiotic. In the lab there is plenty of growth medium without antibiotics. (Note: dilution in medium is like dilution in water). You remember the equation to make dilutions of stock solutions. You usually use this formula to calculate the required volume of a stock solution, but you realize it can apply here as well, even though the unknown is the final volume. So, you make that dilution. Given that each bacterial…The reults for the macroscopic part: 0.30M glycerin – solution was translucent (could see text behind the test tube) 0.15M NaCl – solution was opaque (could not see text behind the test tube) 0.30M NaCl – solution was opaque (could not see text behind the test tube) 0.15M glucose – solution was translucent (could see text behind the test tube) 0.30M glucose – solution was opaque (could not see text behind the test tube) 0.30M Urea – solution was translucent (could see text behind the test tube) Results for microscopic part: 0.30M glycerin – no cells present 0.15M NaCl – normal sized cells 0.30M NaCl – crenated (shrunken and star-shaped) cells 0.15M glucose – no cells present 0.30M glucose – normal sized cells 0.30M Urea – no cells present Determine the osmolarity (hypoosmotic, isosmotic, or hyperosmotic) and tonicity (hypotonic, isotonic, hypertonic) of the following solutions.In which solutions did the osmolarity NOT match the tonicity? For those solutions, why did the osmolarity…the 10 mg/mL sample of lysozyme shows a lower activity than 2mg/mL lysozyme sample (refer to the graph shown below). It is potentially due to Inefficiency of lysozyme at high concentrations. Highest concentrated lysozymes do have lowest activity. Loss of data due to user inefficiency. Loss of data due to spectrophotometer inefficiency.
- Standard curves You are setting up a standard curve for lysozyme. You have a stock vial containing 2 mg/ml of lysozyme and you have 2 ml in the vial. You will want to have a minimum of 1 ml of each concentration for the spectrophotometer. How would you proceed? (Saline is added to make the various concentrations). Concentration Am't stock lysozyme or am't of a previous dilution (note which) Amount saline Total volume wanted 1.00 mg/ml 0.75 mg/ml 0.50 mg/ml 0.375 mg/ml100ml of LB media with 25 μg/ml of Amp and 100 μg/ml of Kan final concentration. You have 100ml of LB provided and Amp and Kan stocks at 100 mg/ml and 50 mg/ml provided. Determine how much of each antibiotic stock solution you need to add to 100ml of LB to reach desired antibiotic concentration.The Bradford reagent gives a linear response only from 0.1 mg/mL to 1.4 mg/mL of protein concentration. Would we be able to use an "absorbance vs. concentration" line graph for samples outside of the range indicated? Justify your response.
- Infuse heparin at 1,200 units per hour from a solution containing 40,000 units of heparin in 500 mL D5W. How many mL/hr will deliver the ordered dosage?Many heavy metals are used in antiseptics. Explain the oligodynamic effect and provide a clinical example of the use of heavy metals.Example of a Protein Purification Scheme: Purification of the Enzyme Xanthine Dehydrogenase from a Fungus Volume Total Total Specific Percent Fraction (mL) Protein (mg) Activity Activity Recovery 1. Crude extract 2. Salt precipitate 3. Ion-exchange chromatography |4. Molecular-sieve chromatography 5. Immunoaffinity chromatography 3,800 22,800 2,460 0.108 100 165 2,800 1,190 0.425 48 65 100 720 7.2 29 40 14.5 23 1.8 275 152.108 11 Calculate the specific activity of step#4. Note that percent recovery=% Yield.