Lab E-copy SP20-BSI-014 PDF

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Biochemistry Lab E-copy

Name: Attique-ur-Rehman Registration number: Sp20-BSI-014 Instructor: Dr. Bushra Ejaz

Department of Bioinformatics Comsats University Islamabad

Experiment 1: Molarity, Molality, Normality and Dilution Equation Molarity: Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution.

Molarity equation M = moles solute / liters solution Molality: Molality (m), or molal concentration, is the amount of a substance dissolved in a certain mass of solvent. It is defined as the moles of a solute per kilograms of a solvent.

Molality equation m = moles solute / kilograms solvent Normality: Normality (N) is defined as the number of mole equivalents per liter of solution.

Normality equation normality = number of mole equivalents/1 L of solution Dilution Equation: Dilution is a process in which the concentration of a solution is reduced. This is typically done by adding more solvent to the solution which decreases the number of moles per liter. M1.V1 = M2.V2 M1 = the molarity of the original solution V1 = the volume of the original solution M2 = the molarity of the diluted solution V2 = the volume of the diluted solution

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Question1: An aqueous solution of HCl is 38% by mass & its density is 1.19 gm/ml. Calculate the molality & molarity of solution. Solution: Mass of solute (HCI) by percent = 38 % We know that: 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Mass % = x 100% 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Solute (38 grams) + Solvent (100 – 38 grams) = solution (100 grams) Solute (38 grams) + Solvent (62 grams) = solution (100 grams) Mass of solute in grams = 38 grams Molar mass of solute (HCI) = 36.46 g / mol 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 Moles of solute (HCI) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 38 𝑔 Moles of solute (HCI) = 36.46 𝑔 / 𝑚𝑜𝑙

Moles of solute (HCI) = 1.04 moles Mass of solution = 100 g Density of solution = 1.19 gm/ml 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Liters of solution = x 0.001 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Liters of solution =

100 𝑔

1.19 gm/ml

x 0.001

Liters of solution = 0.084 liters Mass of solvent in grams = 62 grams Mass of solvent in kilograms = 62 x 0.001 = 0.062 kg 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 We know that: Molarity = 𝑙𝑖𝑡𝑟𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1.04 𝑚𝑜𝑙𝑒𝑠 0.084 𝑙𝑖𝑡𝑒𝑟𝑠

Molarity = Molarity = 12.38 M We know that: Molality =

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡

Molality =

1.04 𝑚𝑜𝑙𝑒𝑠 0.062 𝑘𝑔

Molality = 16.77 m

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Question 2: A mixture of C2H5OH & water contains 54% water by weight. Calculate the mole fraction of alcohol & Water in this mixture? Solution: Amount of water in mixture = 54% by weight As the fixed amount of mixture is not mentioned int the question, so we suppose that the amount of mixture is 100g. We have to find, Mole fraction of water = ? Mole fraction of ethanol = ? Mass of water = 54 grams Mass of ethanol = 100 – 54 = 46 grams Moles of water = mass in grams/molar mass = 54/18 = 3 moles of water Moles of ethanol = mass in grams/molar mass = 46/46 = 1 mole of ethanol Mole fraction of water = no of moles of water/(no of moles of water + no of moles of ethanol) Mole fraction of water = 3/(1+3) = 0.75 Mole fraction of ethanol = no of moles of ethanol/(no of moles of water + no of moles of ethanol) Mole fraction of ethanol = 1/(1+3) = 0.25 Question 3: 49 grams of H2SO4 are present in 100 ml aqueous solution. What is the molarity of H2SO4? Solution: 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 We know that Molarity = 𝑙𝑖𝑡𝑟𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Mass of solute (H2SO4) = 49 grams Milliliters of aqueous solution = 100ml 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 Moles of solute (H2SO4) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 Molar mass of H2SO4 = 98.079 g / mol 49 𝑔 Moles of solute (H2SO4) = 98.079 𝑔/𝑚𝑜𝑙

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Moles of solute (H2SO4) = 0.499 moles. Liters of aqueous solution = 100 ml x 0.001 Liters of aqueous solution = 0.1 liters 0.499 𝑚𝑜𝑙𝑒𝑠 Molarity = 0.1 𝑙𝑖𝑡𝑒𝑟𝑠 Molarity = 4.99 M. Question 4: 4 grams of NaOH is dissolved in 200 ml of aqueous solution. What will be the molarity of this solution? Solution: Mass of NaOH = 4 grams Molar mass of NaOH(solute) = 40 grams Moles of NaOH = mass in grams/molar mass = 4/40 = 0.1 moles Volume of aqueous solution in liters = 200ml x 0.001 = 0.2 L Molarity = moles of solute/ liters of solution Molarity = 0.1/0.2 = 0.5M Question 5: A solution contains 410.3 grams of H2SO4 per liter of Solution at 200 C. If the density is 1.243 gm/ml. What will be its Molarity & Molality? Solution: Mass of solute (H2SO4) = 410.3 grams Molar mass of solute (H2SO4) = 98.079 g / mol 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 Moles of solute (H2SO4) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 410.3 grams Moles of solute (H2SO4) = 98.079 g / mol

Moles of solute (H2SO4) = 4.18 moles Liters of solution = 1 liter 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 We know that Molarity = 𝑙𝑖𝑡𝑟𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 4.18 moles 1 liter

Molarity = Molarity = 4.18 M

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Density of solution = 1.243 gm/ml Liters of solution = 1 liter = 1000 ml Mass of solution in grams = 1.243 gm/ml x 1000 Mass of solution in grams = 1243 grams Solute (410.3 grams) + Solvent (1243 – 410.3 grams) = solution (1243 grams) Solute (410.3 grams) + Solvent (832.7 grams) = solution (1243 grams) Mass of solvent in grams = 832.7 grams Mass of solvent in kilograms = 832.7 x 0.001 = 0.8327 kg 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 We know that: Molality = 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡

Molality =

4.18 𝑚𝑜𝑙𝑒𝑠 0.8327 𝑘𝑔

Molality = 5.01 m Question 6: Calculate the normality of the solution containing 5-gram NaOH dissolved in 250 ml. aqueous solution Solution: Normality is obtained by multiplying Molarity by the number of hydroxide ions in the base. Moles of NaOH in 250 mL= 5g/ 40 g/mol= 0.125 moles Molarity of the solution is; 0.125moles are present in 250 mL X moles are present in 1000 mL X= (0.125moles×1000 ml) ÷ 250 mL X= 0.5M The number of hydroxide ions in the base is 1. The normality of the solution therefore is; 0.5×1= 0.5 N

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Question 7: The density of a 2.03 M solution of acetic acid (molecular weight = 60) in water is 1.017 gm/ml. Calculate molality of the solution. Solution: The formula for molality is = moles of solute / kilograms of solvent. Mass of the solute in a liter of solution is = Molarity × molar mass =2.03×60=121.8g Density of solution =1.017g/mL Mass of 1 liter of solution =1000mL×1.017g/mL= 1017g Mass of water =1017−121.8=895.2g = 895.2/1000 = 0.8952 Kg Molality of the solution is; 2.03 moles are present in 0.8952Kg of solvent X moles are present in 1Kg of solvent X= (2.03 moles×1Kg of solvent)/ 0.8952Kg of solvent X=2.2676 m Question 8: If you dilute 275 mL of a 1.6 M solution of NaCl to 1.0 L, determine the new concentration of the solution. Solution: We know that: M1(initial concentration) x V1(initial volume) = M2(final concentration) x V2(final volume) Equation (A) In this case: M1 = 1.6 M V1 = 275 ml M2 = ? V2 = 1L = 1000ml Putting values in Equation (A) (1.6 M)(275 ml) = (M2)(1000 ml) 440 = (M2)(1000 ml) 440 M2 = 1000 M2 = 0.44 M

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Question 9: You need to make 15.0 L of 1.5 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it? Solution: We know that M1VI =M2V2 M1=? VI=2.5 L M2= 1.5 M V2=15 L M1 =(M2V2)/ VI M2= (1.5×15)/ 2.5 =9.0 M Question 10: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl? Solution: We know that: M1V1 + M2V2 = M3V3 Equation (B) In this case: M1 = 3.55 M V1 = 250 ml M2 = 5.65 M V2 =? M3 = 4.50 M V3 = 250+V2 Putting values in Equation (B) (3.55 M)(250 ml) + (5.65 M)(V2) = (4.50 M)( 250+V2) 887.5 + 5.65(V2) = 1125 + 4.50(V2) 887.5 – 1125 = 4.50(V2) - 5.65(V2) -237.5 = -1.15(V2) −237.5 V2 = −1.15 V2 = 206.52 ml Now we can easily find V3 by simply putting value of V2 in 7 V3 = 250+V2

Putting value of V2 V3 = 250 ml + 206.52 ml V3 = 456.52 ml

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Experiment 2: Acid Base Titration Learning Objective To calculate the concentration of an unknown acid or base given its volume and the volume and concentration of the standardized titrant. Key Points 1.An acid-base titration is a quantitative analysis of acids and bases; through this process, an acid or base of known concentration neutralizes an acid or base of unknown concentration. 2.The titration progress can be monitored by visual indicators, pH electrodes, or both. 3.The reaction’s equivalence point is the point at which the titrant has exactly neutralized the acid or base in the unknown analyte; if you know the volume and concentration of the titrant at the equivalence point, you can calculate the concentration of a base or acid in the unknown solution. Terms 1.Acid-base titration: determines the concentration of an acid or base by exactly neutralizing it with an acid or base of known concentration 2. Equivalence point: the point at which an added titrant’s moles are stoichiometrically equal to the moles of acid/base in the sample; the smallest amount of titrant needed to fully neutralize or react with the analyte 3. Titrant: the standardized (known) solution (either an acid or a base) that is added during titration. 4.Analyte: the unknown solution whose concentration is being determined in the titration.

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Acid Base Titartion Assignment A student conducted the following experiment in the lab.50.00 mL of acetic acid (of unknown concentration) was pipetted into a 150 mL Erlenmeyer flask. Indicator was added and the pH of the solution was recorded. 0.1 M NaOH (a strong base) was added to the Erlenmeyer flask from a buret. The contents of the flask were continuously stirred and the pH was recorded every 2 mL using a pH meter.

Data Collected:

The chemicals involved in the titration are NaOH (a strong electrolyte) and CH3COOH (a weak electrolyte). Predict the shape of the titration curve that would be obtained from the data above. Also predict what the pH at the equivalence point will be.

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The shape of the titration curve that is obtained from the data above is S shaped. Initially the pH is very low, but as the concentration of NaOH increase, the pH increases. There is a sharp change near the equivalnce point. The pH of the equivilance point is 8.72 What is the volume of NaOH at the equivilance point? The volume of the NaOH at the equivilance point is 50ml. What is the pH at the equivilance point? The pH at the equivilance point is 8.72 How does the value of pH at the equivilance point tell you about the strength of the acid? The pH at the equivilance point tell us that the given acid is a weak acid. If the given acid is a strong acid then the pH at the equivilance point would be equal to 7. 5. Calculate the concentration of acetic acid solution. We Know that M1VI(Acetic acid) = M2V2(NaOH) M1 = (M2 x V2) / V1 M1 = (0.1 x 50) / 50 = 0.1M

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Experiment 3: Yeast Fermentation Objective: 1.To find the ability of yeast in fermenting sugar in different conditions 2.To determine what is produced in fermentation Principle: Quantifying gas production is a standard approach to measure yeast fermentation. Although its approach is rather traditional, the apparatus is commonly used due to its low cost. Using this system, students can study effects of a wide range of parameters on the fermentation process. Factors Affecting Enzyme Activity: Several factors affect the rate of enzymatic reaction -temperature, pH, enzyme concentration, substrate concentration, and the presence of inhibitors or activators. Other factors, such as ionic strength, can also affect the enzymatic reaction. Each of these physical and chemical parameters must be considered and optimized in order for an enzymatic reaction experiment to be accurate and reproducible. Procedure: Go to the following lab, click start and follow the instructions: http://www.bch.cuhk.edu.hk/vlab2/animation/fermentation/index.html

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Review Questions Which flask had inflated balloon and actually produce gas? • The flask 2 (contain yeast + water + sugar) had inflated balloon because CO 2 is produced during the process of fermentation as yeast utilized glucose. • The flask 5(contain yeast + water + flour) had also inflated balloon but less than flask 2 because yeast took more time to break flour (mixture of complex carbohydrates) down into glucose that being used for fermentation. What do you think the gas is? How do you verify it? The gas produced during fermentation is CO2. The formation of bubbles in the flasks clearly indicate that the gas produced is CO2.

What is the principle used in this experiment? Yeasts contain the enzyme zymase which catalyze the breakdown of glucose to ethanol and carbon dioxide.

When we dissolve CO2 in water, carbonic acid id formed which is a weak acid.

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pH(acidic/basic)

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Experiment 4: Protein sequencing Objective: To determine the protein sequence by Edman degradation method Background information: Proteins Are Built from a Repertoire of 20 Amino Acids. Amino acids are the building block of proteins. An amino acid consists of a central carbon atom, called α-carbon, linked to an amino group, a carboxylic acid group, a hydrogen atom, and a distinctive R group. The R group is often referred as the side chain. Mechanism of protein sequencing: Amino Acid Sequence Can Be Determined by

Automated Edman Degradation. The amino acid composition of a protein can be ascertained by hydrolyzing it into its constituent amino acids in 6 M HCl at 110°C. The amino acids can be separated by ion-exchange chromatography and quantitated by reacting them with ninhydrin or fluorescamine. Amino acid sequences can be determined by Edman degradation, which removes one amino acid at a time from the amino end of a peptide. Phenyl isothiocyanate reacts with the terminal amino group to form a phenylthiocarbamoyl derivative, which cyclizes under mildly acidic conditions to give a phenylthiohydantoin-amino acid and a peptide shortened by one residue. Automated repeated Edman degradation by a sequencer can analyze a peptide sequence of about 50 residues. Longer polypeptide chains are broken into shorter ones for analysis by specifically cleaving them with a reagent such as cyanogen bromide, which splits peptide bonds on the carboxyl side of methionine residues. Trypsin, which cleaves on the carboxyl side of lysine and arginine residues (if the next residue is not proline), is also useful in splitting proteins.

Principle: In 1950 Pehr Edman developed a method of protein sequencing. It involves sequential identification of amino acids from N to C terminal. Phenyl isothyocynate (PITC) reagent is used for the Edman degradation. The amino terminal of a protein can be identified by reacting the protein with PITC that forms a stable covalent linkage with the free amino group prior to hydrolysis with 6M HCl.

Procedures: 1.Go to the following website, click start and then follow the instructions:http://www.bch.cuhk.edu.hk/vlab2/animation/protein/index.html 2.Phenyl isothiocyanate (PITC, Edman reagent) reacts with the amino acid at the Nterminus of the peptide. 3.The PTH-amino acid is released from the peptide and identified by chromatography. 4.Cycles of labelling and release allow one to determine sequence of the peptide.

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Review Questions What was the sequence of the amino acids analyzed by the Edman degradation method? The sequence of the amino acids analyzed by the Edman degradation method is shown below:

Why does the Edman reagent bind only to the N-terminal of amino acid?

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Why ninhydrin or fluorescamine is used to quantitate the amino acids? How does the quantitation take place?

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Experiment 5: Bradford Protein Assay Background information: The Bradford assay is very fast and uses about the same amount of protein as the Lowry assay. It is fairly accurate and samples that are out of range can be retested within minutes. The Bradford is recommended for general use, especially for determining protein content of cell fractions and assessing protein concentrations for gel electrophoresis. The method described below is for a 100 µl sample volume using 5 ml color reagent. It is sensitive to about 5 to 200 micrograms protein, depending on the dye quality. In assays using 5 ml color reagent prepared in lab, the sensitive range is closer to 5 to 100 µg protein. Scale down the volume for the "micro assay procedure," which uses 1 ml cuvettes. Protocols, including use of microliter plates are described in the flyer that comes with the Bio-Rad kit. Principle: The assay is based on the observation that the absorbance maximum for an acidic solution of Coomassie Brilliant Blue G-250 shifts from 465 nm to 595 nm when binding to protein occurs. Both hydrophobic and ionic interactions stabilize the anionic form of the dye, causing a visible color change. The assay is useful since the extinction coefficient of a dyealbumin complex solution is constant over a 10-fold concentration range. Equipment: In addition to standard liquid handling supplies a visible light spectrophotometer is needed, with maximum transmission in the region of 595 nm, on the border of the visible spectrum (no special lamp or filter usually needed). Glass or polystyrene (cheap) cuvettes may be used; however, the color reagent stains both. Disposable cuvettes are recommended.

Procedure: Reagents 1.

Bradford reagent: Dissolve 100 mg Coomassie Brilliant Blue G-250 in 50 ml 95% ethanol, add 100 ml 85% (w/v) phosphoric acid. Dilute to 1 liter when the dye has completely dissolved, and filter through Whatman #1 paper just before use.

2.

(Optional) 1 M NaOH (to be used if samples are not readily soluble in the color reagent).

The Bradford reagent should be a light brown in color. Filtration may have to be repeated to rid the reagent of blue components. The Bio-Rad concentrate is expensive, but the lots of dye used have apparently been screened for maximum effectiveness. "Homemade" reagent works quite well but is usually not as sensitive as the Bio-Rad product.

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Assay: 1.

Warm up the spectrophotometer before use.

2.

Dilute unknowns if necessary, to obtain between 5 and 100 µg protein in at least one assay tube containing 100 µl sample

3.

If desired, add an equal volume of 1 M NaOH to each sample and vortex (see Comments below). Add NaOH to standards as well if this option is used.

4.

Prepare standards containing a range of 5 to 100 micrograms protein (albumin or gamma globulin are recommended) in 100 µl volume. Add 5 ml dye reagent and incubate 5 min.

5.

Measure the absorbance at 595 nm.

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Review Questions What amino acids does the Bradford Protein Assay primarily measure? The Bradford Protein Assay primarily measures the Basic Amino Acids. For Example: Arginine, Lysine and Histidine. What color change occurs when proteins combine with Coomassie dye under acidic conditions? When proteins combine with Coomassie dye under acidic conditions, the color changes from Red-Brown to Blue. What is the standard curve equation for this BSA standard? We can use any complete protein as standard, but usually bovine serum albumin (BSA) is used as standard because it is inexpensive and easy to get through. Step 1: Prepare at least 5 dilutions of the BSA standard. For Example: BSA dilutions may be 5, 10, 25, 50, 75 and 100 micrograms per milliliter. Step 2: Apply the reagent (which includes the acid and Coomassie dye) to the dilution of the BSA Step 3: Incubate for 5 minutes to 1 hour. Step 4: Calculation of absorbance, with spectrophotometer set at 595 nm. Be sure to allow the instrument to warm up for at least 15 minutes prior to use. Plot absorption in accordance with the theoretical concentration of each BSA standard. The plot is intended to be a linear one. Determine the best fit of the data on a straight line in the form of the equation "y = mx + b" in which y = a...