Biology 2005N Paper 3 Option H

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Prepared by Lawrence KokOption B, UV/Vis spectrophotometer, protein analysis, and buffer calculation

Absorption spectrum- Colour Formation

Colour seen BLUE Blue reflect/transmit to eyes - Red/orange absorb (complementary colour) Complementary colour

BluetransmittedWavelength - absorbed

Visible light

absorbed

Absorption spectrum- Colour Formation Violet = 410nmE = hf = 6.626 x 10-34 x 7.31 x 1014 = 4.84 x 10-19 JRed = 700nmE = hf = 6.626 x 10-34 x 4.28 x 1014 = 2.83 x 10-19 J

High EnergyLow Energy

Absorption spectrum- Colour Formation

Colour seen GREEN GREEN reflect/transmit to eyes - Red/Blue absorb (complementary colour) Complementary colour

GreentransmittedWavelength - absorbed

Visible light

absorbed

Absorption spectrum- Colour Formation Violet = 410nmE = hf = 6.626 x 10-34 x 7.31 x 1014 = 4.84 x 10-19 JRed = 700nmE = hf = 6.626 x 10-34 x 4.28 x 1014 = 2.83 x 10-19 J

High EnergyLow Energy

Beer Lambert Law Linear relationship bet Abs and conc at fix wavelength

Using UV/vis spectrophotometerProtein Assay Amt /Conc protein solutionColour formation Bind protein with complex dye (Coomassie blue/BSA/Biuret)Blue complex sol formedFind wavelength complex absorb MaxMax 595nm Max absorption at 595nm

Protein sol colourless Complex protein with BSA

Add complex dye BSA

Colour seen BLUE Blue reflect/transmit to eyes - Red/orange absorb (complementary colour)

Wavelength Max 595nm

Bluetransmitted

absorbedUsing UV/vis spectrophotometerMonochromator Cuvette DetectorIoIIo = Intensity incident lightI = Intensity transmitted = molar absorption coefficient (constant for absorbing subI = path length (constant, 1cm)

and I constant

Beer Lambert Law Linear relationship bet Abs and Conc at fix wavelength

Using UV/vis spectrophotometerMonochromator Cuvette DetectorIoI

and I constant

Absorbance- Unknown protein conc determine by interpolation- Abs unknown is 1.11, by interpolation, conc is 0.75mg/l- Must be diluted , at high conc , deviation from LawinterpolationTubeVol,ProteinBSAVolH2OVolBufferDil ProteinconcAbs/540nm10.002.002.000.000.0020.101.902.000.1250.3430.301.702.000.3750.6740.501.502.000.6251.0250.701.302.000.8751.3561.001.002.001.251.65

Tube 1-6 are diluted protein with complex dye BSA, prepared using initial protein conc of 5.00mg/l

Add BSA Add water/buffer Protein colourless Complex protein/ BSA Diluted protein/BSA (5.00mg/l)0.125 0.375 0.625 0.875 1.25 0.340.671.021.351.65Std calibration plot, known protein conc vs Abs

Unknown Absorbance = 1.11 Conc = 0.75mg/l

interpolationAbs = 1.11 Conc = 0.75 mg/lprotein conc

Beer Lambert Law Linear relationship bet Abs and Conc at fix wavelength

Using UV/vis spectrophotometerMonochromator Cuvette DetectorIoI

and I constant

protein concAbsorbanceinterpolation0.125 0.375 0.625 0.875 1.25 0.340.671.021.351.65Std calibration plot, known protein conc vs AbsBeers Lambert Law Apply for diluted solution Absorbance Conc Absorbance,A = log10 (Io/I) = lc

Abs = lcIf and l = constantAmt light absorb depend on = Molar extinction of compound c = Conc l = path length Molar extinction of compound, : Measure strength of absorption of sub Higher = Higher Absorbance Sub with high = effective at absorbing light even when low conc is used.Path Length, l: Longer path length Higher AbsConcentration, c: Higher conc of analyte Higher AbsAbs Conc, c

Molar absorptivity, = A/bc = 0.3554 /(2.10 x 7.25 x 10-5) = 2.33 x 103 L mol-1cm-1 Determine conc of unknown using Beer-Lambert Law7.25 x 10-5M X has absorbance of 0.355 when measured in 2.10 cm cell at wavelength 525nm. Cal Molar absorptivity, 100dm3 contaminated water was reduced by boiling to 7.50dm3. Reduced vol was tested, its absorbance is 2.00. Cal conc Pb2+ (mgdm3) in original sample.

V = 100 dm3M = ?V = 7.50 dm3M = 1.15mg/dm3Amt bef heating = Amt aft heatingMoles bef = Moles aftM x V = M x VM x 100 = 1.15 x 7.50M = (1.15 x 7.50)/100M = 0.0863 mgdm-3

Std calibration plot, Abs vs Conc PbAbsorbance 0.3 0.5 0.7 1.1 1.2 Pb conc

interpolation Conc Pb2+ 1.15 mgdm-3

Std calibration curve, Abs vs Conc

Conc unknown (Pb2+) by interpolation

Abs = 0.340Conc = 0.310123Determine unknown conc of Pb2+ using std calibration plot. If unknown sample has Abs 0.34, find conc PbAbsConc

Determine conc of unknown using Beer-Lambert Law

V = 5 cm3M = ?Amt bef dilution = Amt aft dilutionMoles bef = Moles aftM x V = M x VM x 5 = 0.38 x 100M = (0.38 x 100)/5M = 7.6 mgcm-3

Std calibration plot, Abs vs Conc proteinAbsorbanceo.1 0.2 0.3 0.4 protein conc

interpolation Conc protein 0.38 mgcm-345 cm3 sample protein diluted with buffer to vol of 100 cm3 and analysed with UV. Abs was 1.85Using std calibration plot, determine conc protein in original sample.V = 100 cm3M = 0.38

2 cm3 protein was diluted with buffer to vol of 25 cm3 and analysed with UV. Abs was 0.209Using std calibration plot, determine conc protein in original sample.5 Conc protein 0.310 mmoldm-3V = 2 cm3M = ?

V = 25 cm3M = 0.310Amt bef dilution = Amt aft dilutionMoles bef = Moles aftM x V = M x VM x 2 = 0.310 x 25M = (0.310 x 25)/2M = 3.87 mmoldm-3

interpolation

Chromatography Techniques Separation technique of mix into their pure components Identify sample - mix or pure both quantitatively and qualitatively Interaction of sub bet 2 phase - Stationary phase and Mobile phase Separation based on Partition or AdsorptionChromatography TechniquesSeparation analysisPaper ChromatographyThin Layer ChromatographyAdsorption ChromatographyChromatographyPartition ChromatographyColumn ChromatographyPartition Chromatography Component distribute bet TWO immisible liquid phase Depend on relative solubility bet TWO phase Solutes bond to stationary phase or mobile phaseAdsorption Chromatography Component adsorb on solid stationary phase Depend on polarity of stationary, mobile phase and solutes Stationary phase is polar polar solutes adsorb strongly Stationary phase is non polar non polar solutes adsorb strongly Mobile phase is polar polar solutes stay in mobile phase Mobile phase is non polar non polar solutes stay in mobile phase

Application Detection amino acids in mix Diff dyes in food colouring Separation plant pigmentsY adsorb stronglyApplication Preparative/collection of sample of pigmentsX adsorb stronglyApplication Detection amino acids in mix Diff dyes in food colouring Separation plant pigmentsX adsorb weakly

Chromatography Partition Chromatography components distribute bet 2 immisible liquid phase relative solubility in 2 phase bond strongly to mobile phase move fasterAdsorption Chromatography components adsorp on solid stationary phase

Stationary phase has layer of liq Mobile liq phase containing X and YXXXYYY Stationary Liquid phase

Partition distribution solute X and Y bet 2 liq phase X more soluble in mobile phase (move with mobile liq phase) Y less soluble in mobile phase (stay on stationary liq phase) Stationary phase has layer of liq

YYYYXXXXXX Mobile Iiq phase containing X

Stationary phase solid AI2O3 SiO2O-O-O-O-O-Adsorption solute X and Y adsorb temporary on solid Y adsorb strongly on solid phase, eluted slower X in liq mobile phase, eluted fasterYYYY Mobile liq phase containing X and YXXX

O-O-O-YYYYXXXX Mobile liq phase containing XY adsorb strongly

XXXXYYYYSeparation of X and Y

XXX

Y Y YChromatography Techniques

Paper ChromatographyPartition chromatographyDistribution solute bet both liquid phase Depend on relative solubilityAqueous liq phase on surface of stationary phase (paper)Mobile liq phase - solventSolvent move by capillary action

Stationary phase - Cellulose paper absorb water on its surface Mobile Liquid phase with solute X and YYYYXXChromatography TechniquesComponents separated identified using Rf value Rf = Retention factor for given eluent. Measured distance from original spot to centre of particular component to solvent front

Rf green spot= (3/12) = 0.25Rf blue spot= 6/12 = 0.5

Separation using TLC

TLC techniques step by step

Column Chromatography

Column separation

Electrophoresis

Amino acidZwitterion (Electrically neutral)Amino acid with isoelectric pointIsoelectric pointpH when amino acid is electrically neutralIsoelectric point alanine = 6.02 (Ave of pKa)Separation amino acid based on charges using electric fieldAt pH 6.02 Alanine is electrically neutral (Zwitterion)

pH = isoelectricAlanine ( Neutral)

pH < isoelectricAlanine (+ve)

pH > isoelectricAlanine (- ve)

12

Electrophoresis

Isoelectric pointpH when amino acid is electrically neutralIsoelectric point alanine = 6.02 (Ave of pKa)Separation amino acid based on charges using electric fieldAt pH 6.02 Alanine is electrically neutral (Zwitterion)

pH = isoelectricAlanine (Neutral)

pH < isoelectricAlanine (+ve)

pH > isoelectricAlanine (-ve)

Mix of amino acid (ala, arg, isoleu, asp acid)Spot at center on gel (polyacrylamide)Buffer, pH 6 added.Electric field appliedAmino acid separate based on charges.Ninhydrin applied to identified spots.

Separation based on chargespH = iso point = No movement (zwitterion)/ electrically neutralpH < iso point = + ve charge (cation) = move to ve (cathode)pH > iso point = - ve charge (anion) = move to +ve (anode)

pH = 6

+ ve- ve

13

Electrophoresis

Isoelectric pointpH when amino acid is electrically neutralIsoelectric point alanine = 6.02 (Ave of pKa)Separation amino acid based on charges using electric field

pH = isoelectricAlanine (Neutral)

pH < isoelectricAlanine (+ve)

pH > isoelectricAlanine (-ve)

+ H+

Acidic (pH < pI)Cation/zwitterion- H++ H+- H+

Cation (conjugate acid) Zwitterion (conjugate base)+ H++ H+ Zwitterion (conjugate acid) Anion (conjugate base)Alkaline(pH > pI)Zwitterion/anion

14

Cation (+) Zwitterion Anion (-)

+ H+

Acidic (pH < pI)Cation/zwitterion- H++ H+- H+

Cation (conjugate acid) Zwitterion (conjugate base)+ H++ H+ Zwitterion (conjugate acid) Anion (conjugate base)Alkaline(pH > pI)Zwitterion/anion Acidic Base Buffer Calculation

Find pH 0.8M zwitterion and 0.2M anionic form serine

+ H++ H+- H+- H+Amino acidpK1pK2IsoelectricSerine2.29.15.7

Zwitterion (conjugate acid)Anion (conjugate base)

15

At pH 7, alanine contain zwitterion and anionic formi. Deduce structural formulaii. State eqn of buffer when small amt acid and base added

Zwitterion (conjugate acid) Anion (conjugate base)Alkaline (pH > pI)Zwitterion/anion Zwitterion (conjugate acid) Anion (conjugate base)

+ OH-

+ H+

Acid addedBase added21

Zwitterion (conjugate base) Cation (conjugate acid) Acidic (pH < pI)Cation/zwitterion

Amino acidpK1pK2IsoelectricGlycine2.39.66.0

Amino acid prepared by mixing 0.6 dm3, 0.2M HCI and 0.4dm3, 0.5M glycineFind pH original buffern (HCI) = 0.6 x 0.2 = 0.12 moln (glycine) = 0.4 x 0.5 = 0.20 molTotal vol = 1 dm3Conc (HCI) = Mol/Vol = 0.12/1 = 0.12MConc (glycine) = Mol/Vol = 0.20M

2

Zwitterion (conjugate base) Cation (conjugate acid) Acidic (pH < pI)Cation/zwitterion

Amino acidpK1pK2IsoelectricGlycine2.39.66.0

Amino acid prepared by mixing 0.4 mol zwitterion and 0.16mol cationic glycine in 1 dm3Find pH of bufferTotal vol = 1 dm3Conc (cation) = Mol/Vol = 0.16/1 = 0.16MConc (zwitterion) = Mol/Vol = 0.4/1 = 0.4M

Find pH buffer when - 0.025M of NaOH added to buffer340.160.0250.4- 0.025- 0.025+ 0.0250.1350.425OH ---COOH +Conc (zwitterion) = Mol/Vol = 0.4/1 = 0.4MConc (cation) = Mol/Vol = 0.16/1 = 0.16M

NH3 NH4+ Buffer Solution

Acid partNeutralize each otherSalt partBase part- NH3(weak base) + NH4CI (salt)- NH3+ H2O NH4+ + OH NH3 neutralise addedH+- NH4CI NH4+ + CI NH4+ neutraliseadded OH- Effective buffer equal amt weak base NH3 and conjugate acidNH4+Acidic BufferBasic BufferResist a change in pH when small amt acid/base added.CH3COOH + H2O CH3COO- + H3O+ Acidic Buffer - weak acid and its salt/conjugate baseCH3COOH CH3COO- Conjugate acid base pair

CH3COOHCH3COO- Weak Acid Conjugate BaseBUFFERDissociate fully

Dissociate partially

- CH3COOH(weak acid) + CH3COONa(salt)- CH3COOH CH3COO- + H+ CH3COOH neutraliseadded OH- CH3COONa CH3COO- + Na+ CH3COO- neutraliseadded H+- Effective buffer equal amt weak acid CH3COOH and baseCH3COO-

BUFFER

Add acid H+ Add alkaline OH- Neutralize each otherBasic buffer - weak base and its salt/conjugate acid

NH3 + H2O NH4+ + OH-NH3Weak BaseNH4+ Conjugate acid

BUFFERConjugate acid base pair

Add acid H+

Add alkaline OH- Neutralize each otherNeutralize each otherDissociate partially

Base partSalt partAcid partDissociate fullyBUFFER

How to prepare acidic/ basic bufferAcid Dissociation constantCH3COOH + H2O CH3COO- + H3O+Ka = (CH3COO-) (H3O+) (CH3COOH)-lgKa = -lgH+ -lg (CH3COO-) (CH3COOH)-lgH+ = -lg Ka + lg (CH3COO-) (CH3COOH)pH = pKa + lg (CH3COO-) (CH3COOH)Acidic Buffer Formula Mix Weak acid + Salt/Conjugate base CH3COOH CH3COO- + H+ (dissociate partially) CH3COONa CH3COO- + Na+ (dissociate fully)Basic Buffer Formula Mix Weak base + Salt/Conjugate acid NH3 + H2O NH4+ + OH_ (dissociate partially) NH4CI NH4+ + CI_ (dissociate fully)

pH = pKa - lg (acid) (salt)pH = pKa + lg (salt) (acid)Base Dissociation constantNH3 + H2O NH4+ + OH-Kb = (NH4+) (OH-) (NH3)-lgKb = -lgOH- -lg (NH4+) (NH3)-lgOH- = -lgKb + lg (NH4+) (NH3) pOH = pKb + lg (NH4+) (NH3)pOH = pKb + lg (salt) (base)pOH = pKb - lg (base) (salt)

Basic Buffer Acidic Buffer saltsaltacidbaseHenderson Hasselbalch Eqnmultiply -lg both sides

Henderson Hasselbalch Eqn

Acidic Buffer CalculationHow much 0.10M butanoic acid and solid potassium butanoate needed to make 1.0 dm3, pH 5.00 buffer solution. State assumption used. pKa acid = 4.83Need 0.15 mol in 1 dm3Mass salt = mol x RMM = 0.15 x 126.12 = 19 g salt

Click here video Khan AcademyFind pH buffer made with 0.20 mol CH3COONa(salt) in 500cm3 of 0.10M CH3COOH(acid) Ka = 1.8 x 10-5Find pH buffer - adding 25 ml, 0.10M CH3COOH(acid) 25ml, 0.10M CH3COONa(salt) Ka = 1.8 x 10-51st method (formula)1Convert Ka to pKa2nd method (Ka)21st method (formula)31st method (formula)Molar mass salt = 126.12 gmol-12nd method (Ka)Click here explanation from chem guide

Conc salt

Equal vol and concRatio acid/salt = 1

Assumption used[butanoic acid]eq = [salt] used [acid]eq = [acid] usedNo vol change during mixing

Acidic Buffer CalculationFind pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid) Ka = 1.8 x 10-5Conc CH3COO- = Moles/vol = 0.20/0.5 = 0.40M

Click here video0 Khan AcademyFind conc CH3COONa(salt) added to 1.0dm3 of 1.0M CH3COOH(acid) Ka = 1.8 x 10-5, pKa = 4.74 , pH 4.5Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt)Ka = 1.8 x 10-51st method (formula)4Convert Ka to pKa2nd method (Ka)51st method (formula)Convert Ka to pKa2nd method (Ka)61st method (formula)Conc salt2nd method (Ka)Click here explanation from chem guide

Conc [H+]

Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt)Kb = 1.8 x 10-5Basic Buffer CalculationFind pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base) Kb = 1.8 x 10-5 Mole NH4CI = mass/RMM = 4.28 / 53.5 = 0.08 mol

Conc NH4CI = moles/vol = 0.08/0.25 = 0.32M71st method (formula)2nd method (Kb)1st method (formula)82nd method (Kb)Conc saltFind mass CH3COONa added to 500ml, 0.10M CH3COOH(acid) pH = 4.5, Ka = 1.8 x 10-5, pKa = 4.74Conc CH3COO- = 0.0578M x RMM (82) 4.74g in 1000ml 2.37g in 500ml92nd method (Ka)1st method (formula)

Click here addition base to buffer

Click here addition acid to buffer

Conc [H+]

Given 100ml of 0.05M HCOOH , what vol 0.05M NaOH needed to make pH buffer 4.23. pKa = 3.75Buffer Calculation100ml buffer contain 0.10M butanoic acid and 0.20 M sodium butanoate. What pH change when 2.0cm3, 0.10M HCI added. pKa = 4.821011After adding acid

Click here addition base to buffer

Click here addition acid to buffer

NaOH + HCOOH HCOONa + H2OAll NaOH react to form saltMol NaOH react = Mol salt formMol acid remain = Mol initial mol react

How would adding 100ml DI water affect pH. Before adding acid

After adding acid

Buffer do not change pH on dilution

Acknowledgements

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Prepared by Lawrence Kok

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