90695_Jaylah_Holmes_Holmes_Complete_Lab_Report_1815204_810398667

Chitinase is a protein that breaks down chitin, a primary component of the cell wall in fungi, scales in fish and exoskeletons of arthropods. The activity of chitinase extracted from a plant was shown to be optimum at pH 5. You were tasked to prepare 300 mL of 150 mM buffer solution for further analysis of the extracted chitinase. REAGENTS Ka 2.5M Acetic acid Solid NaOAc•3H2O [136.08g/mol] 1.76 x 10-5 2.5M NH3 Solid NH4Cl [53.49g/mol] 5.6 x 10-10 2.5M Lactic acid Solid sodium lactate [112.06g/mol]  4.0 x 10-5 5 M HCl 5M NaOH   Pls show sol'ns 1. Given the following reagents, give the moles of each component (acid & base).2.  What are the mass/volume of the components needed to prepare the buffer? 3. What will the pH of the buffer be if 1mL of 5 M NaOH was added?

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1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…

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Result nad Discussion    Lead Acetate Reaction: Samples: lysine, cysteine, methionine Reagents: 10% Sodium Hydroxide (NaOH) and Lead Acetate Pb(CH3COO)2 -To 1 ml of the amino acid solution taken in a test tube, add few drops of sodium hydroxide (40%) and boil the contents for 5-10 mins over a bunsen burner. Cool the contents and add few drops of 10% Lead acetate solution and observe.

+ 4G 4G ← 0.00 12:23 KB/s S▸ OGLE, O ( 30 Input title July 25 2:23 PM Thu What is the role of acid in our stomach & Û

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Two-dimensional gel electrophoresis of proteins in a cell extract provides a qualitative way to compare proteins with respect to intracellular abundance. Describe a quantitative approach to determine the number of molecules of an enzyme per cell.

correct answer should be 20.6 but how?

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2c which is altered by a catalyst such as an enzyme, the net free energy change of the reaction or the activation energy of the reaction?

During successful purification of every enzyme, the following may be expected: Select ALL that apply. 1. Solubility in NaCl increases 2. The activity increases 3. The specificity increases 4. The number of subunits increases 5. the epitope number increases

During successufl purification of every enzyme, the following may be expected: 1. Solubility in NaCl increases 2. The activity increases 3. The specificity increases 4. The number of subunits increases 5. the epitope number increases

A purified protease enzyme from the fungus Aspergillus sp. Is tested in a laboratory. This enzyme was lyophilized as a white powder. When reconstituting with phosphate buffer pH 7.2 the active enzyme is obtained. To check its purity, an electrophoresis is performed where a single band of approximately 70,000 molecular weight is observed. The solution with enzymatic activity was stored at 4ºC for later use. A few days later it was found that the enzymatic activity had been lost and in the electrophoretic analysis, instead of a single band there were three bands of weights 40,000, 20,000 and 10,000. Come up with a reasoned explanation of what might have happened to the enzyme

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solve the question. And draw an elution parameter graph and explain why.

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Typical activity curves of enzymes that are analyzed in the test tube, like the one presented below, flatten out with time. Provide two possible reasons why this is observed. Must be less than 50 words total

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Please help me find the Michaelis-Menten and the Lineweaver-Burk plots for this. I am stuck.

9026238page=9 La-donya 021 Exam Containers Undergraduate Exam Only Containers | S1, 2020/21 EXAM CHEM0901 | S1 2020/21 mistry) ISTRY A If the enthalpy of standard formation enthalpies of CO and CO, are -110 kJ mol and -393.5 kJ mol, respectively, what is the enthalpy change for the reaction? co+1/20 CO. (e) Select one O a 283.5 kJ mo Ob-283.5 kJ mol O C -503.5 kJ mol Od 141 8 k mol 223 pm 10/12/2020 A ENG irsert prt sc delete home 6 8 Y enter H. K pause ctrl alt

If protein X produces a higher absorbance reading at pH 7 than it does at pH 9, what can you conclude?    Question 1 options:   A lesser amount of protein X was used in the reaction at pH 7 than at pH 9.   A greater amount of protein X was used in the reaction at pH 9 than at pH 7.   Protein X is a more efficient catalyst at pH 7 than it is at pH 9.   Protein X is a more efficient catalyst at pH 9 than it is at pH 7.

Q1. What is the theoretical yield (product mass) of Aspirin if you started 2.50 g of Salicylic Acid? Q2. From the experiment, the actual yield is 2.51 g, calculate the percent yield and use the theoretical yield? Q3. What is the reaction name of Aspirin synthesis? Answer/ Q4. The chemical name of Aspirin is: Answer/

Question: Estimate the pl of the tripeptide ? Step 1: Determine what can be protonated/deprotonated Abundance Step 2: Estimate where the net charge of the protein would be = 0 Step 3: Use the pka's that bracket the neutral form. +H₂N COOH تیار کیا ہے CH₂ H CC H +1 H₂N H COOH CH3 CH3 0 1 2 cook CH₂ coof H pkg= 2.19 O H H₂N No charge H H соон CH3 CH3 3 4 pka= 3.86 Coo COO™ CH₂ H H H coof CH3 CH3 pka= 4.25 COO - 2 5 6 6 Coo J. CH₂ 7 pH O H H 8 H coo- CH3 CH3 COO" 9 + Н. H3N pka= 9.82 10 pka= 9.82 pka= 3.86 CH₂ H₂N 11 H H H coo CH₂ p - 3 O=C H H H 12 H pka= 4.25 O CH3 1 CH3 coo- CH3 CH3 COO 13 COO pka= 2.19 coo 14

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