Biochemistry Problem Set

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BIOCHE

IOCHEMISTRY I

MISTRY I

(CHMI 2227 E)

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PROB

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Eric

Eric R.

R. Ga

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hier, P

er, Ph.D.

h.D.

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Department of Che

partment of Chemist

mist ry and Bioch

ry and Bioch emistry

emistry

January 2007

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Note:

This problem set has been prepared for students taking the course Biochemistry I (CHMI This problem set has been prepared for students taking the course Biochemistry I (CHMI 2227E), as offered at Laurentian University. It contains several problems taken from textbooks 2227E), as offered at Laurentian University. It contains several problems taken from textbooks and from the author’s imagination.

and from the author’s imagination.

While the vast majority of the problems found in this book can be relatively easily solved with While the vast majority of the problems found in this book can be relatively easily solved with the help of the class notes, more difficult questions have also been included. Questions marked the help of the class notes, more difficult questions have also been included. Questions marked by a star

by a star (*) will require more work (*) will require more work from the student. As for from the student. As for the questions labeled with the questions labeled with two starstwo stars (**), they constitute a good challenge to any student interested in tackling them.

(**), they constitute a good challenge to any student interested in tackling them.

After the « Problems » section, the complete, detailed solution for every question is found. For After the « Problems » section, the complete, detailed solution for every question is found. For obvious reasons, we strongly encourage students to look at the solutions only as a last resource. obvious reasons, we strongly encourage students to look at the solutions only as a last resource. The list of pKas and pI for the 20 natural amino acids, as well as the table of the genetic code, The list of pKas and pI for the 20 natural amino acids, as well as the table of the genetic code, can be found after the “Problems” section.

can be found after the “Problems” section.

The following texts were consulted while writing this manual: The following texts were consulted while writing this manual: 1) Kuchel, P. W. and Ralston, G. B.

1) Kuchel, P. W. and Ralston, G. B. Biochimistry. Biochimistry.Schaum Series. McGraw-Hill. 1989.Schaum Series. McGraw-Hill. 1989. 2) Lehninger, A. L., Nelson, D. L., Cox, M. M.

2) Lehninger, A. L., Nelson, D. L., Cox, M. M. Principles of BiochemistryPrinciples of Biochemistry. . 22ndnd_{édition. Worth}_{édition. Worth}

Publishers. 1993. Publishers. 1993.

3) Mathews, C. K. et van Holde, K. E.

3) Mathews, C. K. et van Holde, K. E. Biochemistry Biochemistry. 2. 2ndnd_{édition. Benjamin/Cummings}_{édition. Benjamin/Cummings}

Publishing Company, INC. 1996. Publishing Company, INC. 1996. 4) Rawns, J. D.

4) Rawns, J. D. Biochemistry Biochemistry. Editions du renouveau pédagogique. 1990.. Editions du renouveau pédagogique. 1990. 5) Wood, W. B., Wilson, J. H., Benbow, R. M., Hood, L. E.

5) Wood, W. B., Wilson, J. H., Benbow, R. M., Hood, L. E. Biochemistry. Biochemistry. A A ProblemsProblems Approach

Approach. Benjamin/Cummings Publishing Company, INC. 1981.. Benjamin/Cummings Publishing Company, INC. 1981. 6) Zubay, G. L., Parson, W. W., Vance, D. E.

6) Zubay, G. L., Parson, W. W., Vance, D. E. Principles of BiochemistryPrinciples of Biochemistry. Wm. C. Brown. Wm. C. Brown Publishers. 1995.

Publishers. 1995.

More problems and questions can be found in these and other references. More problems and questions can be found in these and other references.

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Problems

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Chapter 1:

Chapter 1: Acid-Base Equi

Acid-Base Equilibrium and

librium and Spectrophotom

Spectrophotometry

etry

1.1

1.1 Acid-Bas Acid-Base Equilibriue Equilibrium :m :

What is the pH of the following solutions? What is the pH of the following solutions?

a)

a) 0.35 M hydrochloric acid0.35 M hydrochloric acid b)

b) 0.35 M acetic acid (pKa = 4.76)0.35 M acetic acid (pKa = 4.76) c)

c) 0.035 M acetic acid.0.035 M acetic acid. 1.2

A weak acid, HA, has a total concentration of 0.20M and is ionized (dissociated) to 2%; A weak acid, HA, has a total concentration of 0.20M and is ionized (dissociated) to 2%;

a)

a) Calculate the Ka for this acid.Calculate the Ka for this acid. b)

b) Calculate the pH for this acidic solution.Calculate the pH for this acidic solution. 1.3

1.3 Acid-Bas Acid-Base Equilibriue Equilibrium :m : Calculate the pH of the

Calculate the pH of the following mixtures:following mixtures: a)

a) 1M acetic acid and 0.5M sodium acetate1M acetic acid and 0.5M sodium acetate b)

b) 0.3M phosphoric acid and 0.8M KH0.3M phosphoric acid and 0.8M KH22POPO44 (pKa=2.14) (pKa=2.14)

1.4

You need to prepare a buffer solution at pH = 7.00 with KH

You need to prepare a buffer solution at pH = 7.00 with KH22POPO44 and Na and Na22HPOHPO44 (pKa=7.21). If (pKa=7.21). If

you use a 0.1M solution of KH

you use a 0.1M solution of KH22POPO44, what would be the concentration of Na, what would be the concentration of Na22HPOHPO44 needed? needed?

1.5

1.5 Acid-Bas Acid-Base Equilibrie Equilibrium :um :

You need to prepare a buffer solution at pH = 7.00 with KH

You need to prepare a buffer solution at pH = 7.00 with KH22POPO44 and Na and Na22HPOHPO44. What would be. What would be

the respective concentration of these substances if yo

the respective concentration of these substances if yo u wished to obtain a final phu wished to obtain a final phosphateosphate concentration ([HPO

concentration ([HPO44-2-2] + [H] + [H22POPO44-1-1]) of 0.3M?]) of 0.3M?

1.6

1.6 Spectrophotometry :Spectrophotometry :

What is the concentration of the amino acid tyrosine (

What is the concentration of the amino acid tyrosine (εε=1 420 L mol=1 420 L mol-1-1 cm cm-1-1) if you obtain an) if you obtain an absorbance of 0.71 with a 1 cm cuvette? With a 0.1 cm cuvette?

absorbance of 0.71 with a 1 cm cuvette? With a 0.1 cm cuvette? 1.7

What would be the absorbance reading of a 37 mM solution of tyrosine? What would be the absorbance reading of a 37 mM solution of tyrosine? 1.8

You wish to determine the concentration of haemoglobin in a blood sample by You wish to determine the concentration of haemoglobin in a blood sample by spectrophotometry. You first create a standard curve of

spectrophotometry. You first create a standard curve of the absorbance at 412 the absorbance at 412 nm of severalnm of several solutions of known haemoglobin concentrations.

solutions of known haemoglobin concentrations. The data for the standard curve iThe data for the standard curve is showns shown below.

below. What is the concentration (in µg/mL) of haemoglobin in your sample if the absorbanceWhat is the concentration (in µg/mL) of haemoglobin in your sample if the absorbance obtained at 412 nm was 0.303?

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Absorbance Absorbance (412nm) (412nm) Concentration of Concentration of standard solution standard solution (µg/ml) (µg/ml) 0.069 1 0.069 1 0.113 2 0.113 2 0.201 4 0.201 4 0.377 8 0.377 8 0.730 16 0.730 16

Chapter

Chapter 2:

2: Amino

Amino acids

acids

* 2.1. Molecular mass of an amino acid. * 2.1. Molecular mass of an amino acid. 1.812 g of a crystallized

1.812 g of a crystallized αα-amino acid (pKa1: 2.4; pKa2; 9.7) has a pH of 10.4 when dissolved in-amino acid (pKa1: 2.4; pKa2; 9.7) has a pH of 10.4 when dissolved in 100 mL of 0.1M NaOH.

100 mL of 0.1M NaOH. Calculate the molecular mass Calculate the molecular mass of this amino acid.of this amino acid. 2.2. Titratio

2.2. Titration curven curve

Calculate the pI of histidine and draw its titr

Calculate the pI of histidine and draw its titration curve. ation curve. Indicate the position of all pKas and theIndicate the position of all pKas and the pI

pI as as well awell as s the the percentages percentages of of each each ionic ionic form aform at t the the start and start and finish finish of of the the titration and titration and at at allall pKas. The

pKas. The list of list of pKas for pKas for all 20 all 20 amino acids amino acids can be can be found at found at the end the end of the of the “Problems” section“Problems” section of this problem set.

of this problem set. 2.3. Net cha

2.3. Net charges of amirges of amino acidsno acids

What is the net charge (+, 0, -) of the amino acids glycine, serine, aspartic acid, glutamine and What is the net charge (+, 0, -) of the amino acids glycine, serine, aspartic acid, glutamine and arginine at:

arginine at: a)

a) pH pH 2.01 2.01 b) b) pH pH 3.96 3.96 c) c) pH pH 5.68 5.68 d) d) pH pH 10.7610.76 2.4. Ionic ex

2.4. Ionic exchange chchange chromatogrromatographyaphy

A mixture of lysine, glycine, alanine, isoleucine and glutamic acid are separated by ionic A mixture of lysine, glycine, alanine, isoleucine and glutamic acid are separated by ionic exchange chromatography.

exchange chromatography. What is the What is the order of elution of order of elution of these amino acids if these amino acids if you use gradientyou use gradient buffer system from pH 10 to pH 2:

buffer system from pH 10 to pH 2: a) with a cation exchange resin? a) with a cation exchange resin? b) with an anion exchange resin? b) with an anion exchange resin?

Which column would give the best separation? Which column would give the best separation? 2.5. Amino a

2.5. Amino acidscids

What amino acids can be converted into another amino acid with gentle hydrolysis, resulting in What amino acids can be converted into another amino acid with gentle hydrolysis, resulting in release of ammonia?

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2.7. Amino a 2.7. Amino acidscids

Phosphoserine is found aft

Phosphoserine is found after enzymatic hydrolysis er enzymatic hydrolysis of casein, a miof casein, a milk protein. lk protein. However, it doesHowever, it does not belong to the 20 amino acids coded during protein synthesis. Give a plausible explanation. not belong to the 20 amino acids coded during protein synthesis. Give a plausible explanation.

*2.8. Ionic exchange chromatography *2.8. Ionic exchange chromatography

Glycine, alanine, valine and leucine can be successfully separated by ionic exchange Glycine, alanine, valine and leucine can be successfully separated by ionic exchange chromatography even though their

chromatography even though their pKas are almost pKas are almost identical. identical. Explain the behaviour Explain the behaviour of theseof these amino acids.

amino acids. 2.9. Peptide 2.9. Peptides.s.

A peptide is hydrolyzed and its

A peptide is hydrolyzed and its amino acid content analyzed. amino acid content analyzed. Hydrolysis destroys the Hydrolysis destroys the amino acidamino acid tryptophan, therefore the content of tryptophan can be estimated with spectrophotometry. tryptophan, therefore the content of tryptophan can be estimated with spectrophotometry. Establish the empirical formula of the peptide with the

Establish the empirical formula of the peptide with the following information.following information.

Amino acids

Amino acids mmol mmol Ala Ala 2.742.74 Glu Glu 1.411.41 Leu Leu 0.690.69 Lys Lys 2.812.81 Arg Arg 0.720.72 Trp Trp 0.650.65 2.10. Peptid 2.10. Peptides.es.

Draw the structure of the following peptide GWYQR. Indicate the ionic form of the peptide at Draw the structure of the following peptide GWYQR. Indicate the ionic form of the peptide at the following pH: the following pH: a) a) pH pH 2.0 2.0 b) pH b) pH 7.0 7.0 c) c) pH pH 10.510.5 CH2-CH2-CH-COOH CH2-CH2-CH-COOH O O PO PO33-2-2 NH NH22

Phosphoserine

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Chapter 3. General properties and purification of proteins

3.1.Protein

3.1.Protein PurificatPurificationion Why do we often

Why do we often use ammonium sulphate precipitation in initial purification steps of proteins?use ammonium sulphate precipitation in initial purification steps of proteins? 3.2. Protein

3.2. Protein PurificatiPurificationon

DEAE cellulose columns are rarely used at pH greater than 8.5. Why? DEAE cellulose columns are rarely used at pH greater than 8.5. Why? 3.3. Protein

3.3. Protein PurificatiPurificationon

6-phosphogluconate dehydrogenase has

6-phosphogluconate dehydrogenase has a pI a pI of 6. of 6. Explain why Explain why the buffer the buffer used for used for aa chromatography on DEAE-cellulose must have a pH greater than 6 but less than 9 in order to chromatography on DEAE-cellulose must have a pH greater than 6 but less than 9 in order to ensure the enzyme is efficiently bound to the column.

ensure the enzyme is efficiently bound to the column. 3.4. Protein

3.4. Protein PurificatiPurification.on.

Would the enzyme, 6-phosphogluconate dehydrogenase bind to a CM-cellulose resin if the same Would the enzyme, 6-phosphogluconate dehydrogenase bind to a CM-cellulose resin if the same conditions as the previous problem were used? Why?

conditions as the previous problem were used? Why? 3.5. Protein

What pH would the buffer need to be in order to permit the dehydrogenase in the previous What pH would the buffer need to be in order to permit the dehydrogenase in the previous problem to bind to the CM-cellulose resin?

problem to bind to the CM-cellulose resin? 3.6. Protein

We load a DEAE-cellulose column adjusted to a pH of 6.5 with the following mixture of We load a DEAE-cellulose column adjusted to a pH of 6.5 with the following mixture of proteins:

proteins: ovalbumin ovalbumin (pI (pI = = 4.6), 4.6), urease urease (pI (pI = = 5.0), 5.0), and and myoglobin myoglobin (pI (pI = = 7.0). 7.0). The The proteins proteins areare eluted first with a buffer of weak ionic strength at a pH of 6.5, and then the same buffer eluted first with a buffer of weak ionic strength at a pH of 6.5, and then the same buffer containing increasing amounts of

containing increasing amounts of sodium chloride is sodium chloride is used to elute the used to elute the proteins. proteins. What order areWhat order are the proteins eluted?

the proteins eluted? 3.7. Protein

An enzyme (MW 24 kDa, pI 5.5) is contaminated with two other proteins, one with a similar An enzyme (MW 24 kDa, pI 5.5) is contaminated with two other proteins, one with a similar molecular mass and a pI of 7.0 while the other has a molecular mass of 100 kDa and a pI of 5.4. molecular mass and a pI of 7.0 while the other has a molecular mass of 100 kDa and a pI of 5.4. Suggest a procedure to purify the contaminated enzyme.

Suggest a procedure to purify the contaminated enzyme. 3.8. Protein

A procedure used to purify 6-gluconate dehydrogenase from E. coli is presented below. A procedure used to purify 6-gluconate dehydrogenase from E. coli is presented below. a) Calculate (1) the specific activity, (2) the percent yield based on the

a) Calculate (1) the specific activity, (2) the percent yield based on the initialinitial quantity of the quantity of the enzyme and (3) the degree of purification for each step (i.e.

enzyme and (3) the degree of purification for each step (i.e. fold fold increase in purification). increase in purification). b) Indicate which step purifies the protein the most.

b) Indicate which step purifies the protein the most.

c) Assuming the protein is pure after gel permeation chromatography (on Bio-Gel A), what c) Assuming the protein is pure after gel permeation chromatography (on Bio-Gel A), what percent of the initial extract contained 6-gluconate dehydrogenase?

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Purificat

Purification stepion step Volume (mL)Volume (mL) Total proteinTotal protein (mg) (mg)

Enzymatic

Enzymatic activityactivity (µg/min) (µg/min) 1-

1- Cellular Cellular extract extract 2 2 800 800 70 70 000 000 2 2 700700 2-

2- Ammonium Ammonium sulfate sulfate 3 3 000 000 25 25 400 400 2 2 300300 3-

3- Heat Heat denaturation denaturation 3 3 000 000 16 500 16 500 1 1 980980 4- DEAE 4- DEAE chromatography chromatography 80.00 80.00 390.00 390.00 1 1 680680 5- CM-cellulose 5- CM-cellulose chromatography chromatography 50.00 50.00 47.00 47.00 1 1 350350 6- Bio-Gel A 6- Bio-Gel A chromatography chromatography 7.00 7.00 35.00 35.00 1 1 120120 3.9 Protein

3.9 Protein PurificatiPurification.on.

Why is SDS omitted when proteins need

Why is SDS omitted when proteins need to undergo isoelectric focusing?to undergo isoelectric focusing? 3.10. Prote

3.10. Protein Purificain Purification.tion.

A series of proteins with known molecular mass and an enzyme of unknown molecular mass are A series of proteins with known molecular mass and an enzyme of unknown molecular mass are separated by chromatography

separated by chromatography on a Sephadex G-200 on a Sephadex G-200 column. column. The elution volume The elution volume (V(Vee) for each) for each

protein is indicated in the table below.

protein is indicated in the table below. Estimate the molecular mass of the unknown protein.Estimate the molecular mass of the unknown protein. Protein

Protein Mr Mr V V _ee (mL) (mL) Blue

Blue dextran dextran 1 1 000 000 kDa kDa 85.0085.00 lysozyme

lysozyme 14 14 kDa kDa 200.00200.00 Chymotrypsinogen

Chymotrypsinogen 25 25 kDa kDa 190.00190.00 ovalbumin

ovalbumin 45 45 kDa kDa 170.00170.00 Serum

Serum albumin albumin 65 65 kDa kDa 150.00150.00 aldolase

aldolase 150 150 kDa kDa 125.00125.00 urease

urease 500 500 kDa kDa 90.0090.00 ferritin

ferritin 700 700 kDa kDa 92.0092.00 ovomucoid

ovomucoid 28 28 kDa kDa 160.00160.00 unknown

unknown ? ? 130.00130.00

*3.11. Protein Purification. *3.11. Protein Purification.

Referring to the previous problem, give a plausible explanation for the bizarre behaviour Referring to the previous problem, give a plausible explanation for the bizarre behaviour ferritin’s elution from the sephadex column.

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3.12. Prote

A student isolates a protein from anaerobic bacteria and analyses the protein by polyacrylamide A student isolates a protein from anaerobic bacteria and analyses the protein by polyacrylamide gel electrophoresis

gel electrophoresis containing SDS (containing SDS (PAGE-SDS). PAGE-SDS). Following protein Following protein staining, a staining, a single bandsingle band appears, which excites the s

appears, which excites the student’s supervisor. tudent’s supervisor. To be certain, the sTo be certain, the supervisor suggests that upervisor suggests that thethe student run a second electrophoresis under

student run a second electrophoresis under nativenative conditions (i.e. non-denaturing, or without conditions (i.e. non-denaturing, or without SDS).

SDS). This gel This gel shows two shows two bands after bands after staining. staining. Assuming Assuming no errors no errors were committed were committed duringduring these experiments, explain the observations.

these experiments, explain the observations. 3.13. Prote

A student from CHMI 2227 analyses bovine serum albumin (BSA) with a polyacrylamide gel A student from CHMI 2227 analyses bovine serum albumin (BSA) with a polyacrylamide gel electrophoresis

electrophoresis (PAGE-SDS). (PAGE-SDS). During During the the experiment, experiment, the the student student forgets forgets to to addadd β β--mercaptoethanol to the

mercaptoethanol to the sample. sample. When comparing his When comparing his sample to sample to those of his those of his classmates heclassmates he realizes that the molecular mass of his BSA sample determined by PAGE-SDS is 57 kDa, while realizes that the molecular mass of his BSA sample determined by PAGE-SDS is 57 kDa, while all the other students (those that added

all the other students (those that added ββ-mercaptoethanol) found a molecular mass of 68 kDa.-mercaptoethanol) found a molecular mass of 68 kDa. Explain this difference.

Explain this difference. 3.14. Polyp

3.14. Polypeptide seqeptide sequencinguencing Consider the following peptide: Consider the following peptide:

A-L-K-M-P-E-Y-I-S-T-D-Q-S-N-W-H-H-R A-L-K-M-P-E-Y-I-S-T-D-Q-S-N-W-H-H-R Indicate the fragments generated after the following digestions :

Indicate the fragments generated after the following digestions : a)

a) trypsin trypsin b) b) pepsin pepsin c) c) protease protease V8 V8 d) d) cyanogen cyanogen bromidebromide

3.15 Polyp

3.15 Polypeptide seqeptide sequencinguencing

Deduce the polypeptide sequence that generated the following results: Deduce the polypeptide sequence that generated the following results: a) acid hydrolysis: (Ala

a) acid hydrolysis: (Ala₂₂, Arg, Arg , Lys, Lys₂₂, Met, Phe, Ser , Met, Phe, Ser ₂₂);); b) Carboxypeptidase A digestion: Ala;

b) Carboxypeptidase A digestion: Ala; c)

c) Trypsin Trypsin digestion: digestion: (Ala, (Ala, Arg)Arg) (Lys, Phe, Ser) (Lys, Phe, Ser) (Lys)

(Lys)

(Ala, Met, Ser) (Ala, Met, Ser)

d) cyanogen bromide treatment: (Ala, Arg, Lys

d) cyanogen bromide treatment: (Ala, Arg, Lys₂₂, Met, Phe, Ser), Met, Phe, Ser) (Ala, Ser)

(Ala, Ser) e)

e) thermolysine thermolysine digestion: digestion: (Ala)(Ala)

(Ala, Arg, Ser) (Ala, Arg, Ser) (Lys

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3.16. Polyp

3.16. Polypeptide seqeptide sequencinguencing A polypeptide is reduced by

A polypeptide is reduced by ββ-mercaptoethanol to yield two peptide fragments with the-mercaptoethanol to yield two peptide fragments with the following sequences :

following sequences :

fragment 1:

fragment 1: A-C-F-P-K-R-W-C-R-R-A-C-F-P-K-R-W-C-R-R-V-CV-C fragment 2: C-Y-C-F-C

fragment 2: C-Y-C-F-C The non-reduced polypeptide is dige

The non-reduced polypeptide is digested with thermolysine and yields the following fragments :sted with thermolysine and yields the following fragments : (A,C,C,V) (A,C,C,V) (R,K,F,P) (R,K,F,P) (R,R,C,C,W,Y) (R,R,C,C,W,Y) (C,C,F) (C,C,F)

Indicate the positions of disulfide bridges in the polypeptide. Indicate the positions of disulfide bridges in the polypeptide.

3.17. Polyp

3.17. Polypeptide seqeptide sequencinguencing

An analysis of the polypeptide Shawi isolated from the bacteria

An analysis of the polypeptide Shawi isolated from the bacteria Chretientus negativiiChretientus negativii, yields the, yields the following results :

following results : a) acid hydrolysis: (Ala

a) acid hydrolysis: (Ala₄₄, Val, Lys, Val, Lys₂₂ , Arg, Gly, Asp, Met, Pro, Trp) , Arg, Gly, Asp, Met, Pro, Trp) b) carboxypeptidase digestion: Lys

b) carboxypeptidase digestion: Lys c) dinitrofluorobenzene treatment: Val c) dinitrofluorobenzene treatment: Val

d) cyanogen bromide treatment: generates two polypeptides: d) cyanogen bromide treatment: generates two polypeptides:

peptide A

peptide A: (Gly, Arg, Trp, Asp, Lys, Ala); Treatment of this peptide with DNBF and: (Gly, Arg, Trp, Asp, Lys, Ala); Treatment of this peptide with DNBF and carboxypeptidase yields :

carboxypeptidase yields : DNFB:

DNFB: Gly Gly Carboxypeptidase: Carboxypeptidase: LysLys

peptide B

peptide B: (Ala: (Ala₃₃ , Lys, Val, Met, Pro); Treatment of this peptide with DNFB and , Lys, Val, Met, Pro); Treatment of this peptide with DNFB and carboxypeptidase yields:

carboxypeptidase yields: DNFB:

DNFB: Val Val Carboxypeptidase: Carboxypeptidase: MetMet e) trypsine digestion: yields three peptides

e) trypsine digestion: yields three peptides

peptide C

peptide C: (Lys, Trp, Ala); Treatment of this peptide with DNFB and carboxypeptidase: (Lys, Trp, Ala); Treatment of this peptide with DNFB and carboxypeptidase yields :

yields :

DNFB: Trp DNFB: Trp

peptide D

peptide D: (Ala: (Ala33 , Val, Lys, Pro) , Val, Lys, Pro)

peptide E:

peptide E: (Met, Asp, Gly, Arg); Treatment of this peptide with DNFB and (Met, Asp, Gly, Arg); Treatment of this peptide with DNFB and carboxypeptidase yields :

carboxypeptidase yields : DNFB: Met DNFB: Met

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Finally, treating peptide D with thermolysine yields the following: Finally, treating peptide D with thermolysine yields the following: Val Val Ala Ala Ala Ala

(Ala, Lys, Pro) (Ala, Lys, Pro)

What is the primary structure of this peptide? What is the primary structure of this peptide?

Chapter 4. Three dimensional structures of proteins

4.1. 3-D Stru

4.1. 3-D Structures of pctures of proteinsroteins

What amino acids among the following would you expect to find a) inside, and b) at the surface What amino acids among the following would you expect to find a) inside, and b) at the surface of a typical globular protein in an aqueous solution of pH 7?

of a typical globular protein in an aqueous solution of pH 7?

Glu Arg Val

Phe Ileu Asn

Lys Ser Thr

4.2. 3-D Stru

According to the structure of urea, deduce how this compound can promote denaturation of According to the structure of urea, deduce how this compound can promote denaturation of proteins.

proteins. 4.3. 3-D Stru

Phenylalanine, a hydrophobic amino acid, is frequently found at the surface of natives and Phenylalanine, a hydrophobic amino acid, is frequently found at the surface of natives and functional proteins.

functional proteins. Give the most probable rGive the most probable role of phenylalanine in this sole of phenylalanine in this situation.ituation. *4.4. 3-D Structures of proteins

*4.4. 3-D Structures of proteins

Aspartic acid, a charged amino acid, is frequently found inside of native and functional proteins. Aspartic acid, a charged amino acid, is frequently found inside of native and functional proteins. Give the most probable role of phenylalanine in this situation.

Give the most probable role of phenylalanine in this situation. 4.5. 3-D Stru

The following table describes the amino acid compositions of three proteins. The following table describes the amino acid compositions of three proteins.

Number of residues per molecul Number of residues per moleculee Amino acids

Amino acids protein 1 protein 1 protein 2 protein 2 protein 3protein 3 Polar residues Polar residues Arg Arg 12.00 12.00 4.00 4.00 7.007.00 Asn Asn 9.00 9.00 6.00 6.00 5.005.00 Asp Asp 14.00 14.00 5.00 5.00 9.009.00

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Number of residues per molecul Number of residues per moleculee Amino acids

Amino acids protein 1 protein 1 protein 2 protein 2 protein 3protein 3 Cys Cys 7.00 7.00 2.00 2.00 6.006.00 Gln Gln 8.00 8.00 7.00 7.00 6.006.00 Glu Glu 11.00 11.00 4.00 4.00 6.006.00 His His 4.00 4.00 2.00 2.00 4.004.00 Lys Lys 22.00 22.00 6.00 6.00 15.0015.00 Ser Ser 20.00 20.00 5.00 5.00 11.0011.00 Thr Thr 15.00 15.00 3.00 3.00 11.0011.00 Trp Trp 2.00 2.00 3.00 3.00 3.003.00 Tyr Tyr 7.00 7.00 7.00 7.00 6.006.00 Non-polar residues Non-polar residues Ala Ala 1144..000 0 2288..0000 2255..0000 Gly Gly 9.00 9.00 9.00 9.00 8.008.00 Ileu Ileu 5.00 5.00 16.00 16.00 9.009.00 Leu Leu 3.00 3.00 19.00 19.00 7.007.00 Met Met 7.00 7.00 11.00 11.00 9.009.00 Phe Phe 9.00 9.00 13.00 13.00 11.0011.00 Pro Pro 8.00 8.00 13.00 13.00 10.0010.00 Val Val 16.00 16.00 29.00 29.00 21.0021.00

Knowing that protein A has a rod-like form, protein B is a monomeric globular protein, and Knowing that protein A has a rod-like form, protein B is a monomeric globular protein, and protein

protein C C is is a a globular globular protein protein with with four four identical identical sub-units, sub-units, deduce deduce the the corresponding corresponding aminoamino acid composition of these proteins.

acid composition of these proteins. 4.6. 3-D Stru

Indicate which secondary structure or structures (

Indicate which secondary structure or structures (αα -helix,-helix, ββ -pleated, random coil) will the -pleated, random coil) will the following peptide adopt in an aqueous solution at pH 7

following peptide adopt in an aqueous solution at pH 7

Ileu-Glu-Asn-Glu-Gln-Asn-Met-Ala-His-Phe-Trp-Tyr Ileu-Glu-Asn-Glu-Gln-Asn-Met-Ala-His-Phe-Trp-Tyr 4.7. 3-D Stru

Gly-Ala-Gly-Ala-Gly-Ser-Gly-Ala-Gly-Ser-Gly-Ala Gly-Ala-Gly-Ala-Gly-Ser-Gly-Ala-Gly-Ser-Gly-Ala 4.8. 3-D Stru

Lys-Gly-Arg-Arg-Lys-Gly-Arg-Gly-Arg-Pro Lys-Gly-Arg-Arg-Lys-Gly-Arg-Gly-Arg-Pro 4.9. 3-D Stru

1 10

Gly-Pro-Glu-Ser-Ala-Tyr-Lys-Thr-Leu-Phe-Asp-Val-Pro-Asp-Asp-Glu-Asp-Gly-Gly Gly-Pro-Glu-Ser-Ala-Tyr-Lys-Thr-Leu-Phe-Asp-Val-Pro-Asp-Asp-Glu-Asp-Gly-Gly

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20

20 2626

Ser-Ala-Gly-Ser-Ser-Gly-Ala Ser-Ala-Gly-Ser-Ser-Gly-Ala

4.10. 3-D St

4.10. 3-D Structures oructures of proteinsf proteins The following t

The following table describes the able describes the amino acid amino acid composition of three composition of three proteins. proteins. Determine whatDetermine what structure these proteins will adopt:

structure these proteins will adopt: αα-helical,-helical, ββ-pleated or a triple helix of collagen.-pleated or a triple helix of collagen.

protein

protein A A B B C C protein protein AA BB C C Ala

Ala 29.40 29.40 5.00 5.00 10.70 10.70 Leu Leu 0.50 0.50 6.90 6.90 2.402.40 Arg

Arg 0.50 0.50 7.20 7.20 5.00 5.00 Lys Lys 0.30 0.30 2.30 2.30 3.403.40 Asp 1.30

Asp 1.30 6.00 6.00 4.50 4.50 Met Met - - 0.50 0.50 0.800.80 Cys

Cys - - 11.20 11.20 - - Phe Phe 0.50 0.50 2.50 2.50 1.201.20 Glu

Glu 1.00 1.00 12.10 12.10 7.10 7.10 Pro Pro 0.30 0.30 7.50 7.50 12.2012.20 Gly

Gly 44.60 44.60 8.10 8.10 33.00 33.00 Ser Ser 12.20 12.20 10.20 10.20 4.304.30 His

His 0.20 0.20 0.70 0.70 0.40 0.40 Trp Trp 0.20 0.20 1.20 1.20 --Hypro -

Hypro - - - 9.40 9.40 Tyr Tyr 5.20 5.20 4.20 4.20 0.400.40 Ileu

Ileu 0.70 0.70 2.80 2.80 0.90 0.90 Val Val 2.20 2.20 5.10 5.10 2.302.30

Chapter 5. Enzymology

5.1. Enzyma

5.1. Enzymatic kinetictic kineticss

With the following enzyme activity results determine: With the following enzyme activity results determine: a) Vmax

a) Vmax

b) why is the velocity v constant at [S] greater than 2 x 10 b) why is the velocity v constant at [S] greater than 2 x 10-3-3_M?_M?

c) what is the free [E] at [S] = 2 x 10 c) what is the free [E] at [S] = 2 x 10-2-2_M?_M?

5.2. Enzyma

The results for enzyme activity analysis can be found below.

The results for enzyme activity analysis can be found below. Without using a graphWithout using a graph,, determine : determine : a) Vmax; a) Vmax; b) Km; b) Km; c) initial

c) initial velocity at velocity at [S] = 1 [S] = 1 x 10x 10-1-1_M;_M;

[S] (mol/L) [S] (mol/L) v (v ( 2 x 10 2 x 10-1-1 60.0060.00 2 x 10 2 x 10-2-2 60.0060.00 2 x 10 2 x 10-3-3 60.0060.00 2 x 10 2 x 10-4-4 48.0048.00 1,5 x 10 1,5 x 10-4-4 45.0045.00 1,3 x 10 1,3 x 10-5-5 12.0012.00 mol/min) mol/min)

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d) the amount of product formed during the first 5 minutes at [S] = 2 x 10

d) the amount of product formed during the first 5 minutes at [S] = 2 x 10-3-3_{M. At a [S] of 2 x 10}_{M. At a [S] of 2 x 10} --6

6_M?_M?

e) what is Km and Vmax if

e) what is Km and Vmax if the free [E] is increased by a factor of the free [E] is increased by a factor of 4?4?

5.3. Enzyma

5.3. Enzymatic kinetictic kineticss The following table descri

The following table describes the results bes the results from an enzymology experimfrom an enzymology experiment. ent. Using a Lineweaver-Using a Lineweaver-Burke plot determine:

Burke plot determine: a) Km;

a) Km; b) Vmax; b) Vmax;

5.4. Enzyma

5.4. Enzymatic kinetictic kineticss We study the effect of

We study the effect of pH on the enzymatic actipH on the enzymatic activity of 6-phosphogluconate dehydrogenase. vity of 6-phosphogluconate dehydrogenase. ThisThis enzyme catalyzes the reaction:

enzyme catalyzes the reaction: 6-phosphogluconate

6-phosphogluconate + + NADP NADP 6- 6- phosphogluconic phosphogluconic acid acid + + NADPHNADPH₂₂ NADPH

NADPH₂₂ absorbs light at 340 nm. The activity of the dehydrogenase is measured absorbs light at 340 nm. The activity of the dehydrogenase is measured

[S] (mol/L) [S] (mol/L) v (v ( 5 x 10 5 x 10--22 00..2255 5 x 10 5 x 10--33 00..2255 5x 10 5x 10--44 00..2255 5x 10 5x 10--55 00..2200 5 x 10 5 x 10--66 00..0077 5 x 10 5 x 10--77 00..0011 [S] (mol/L) [S] (mol/L) v (v ( 1 x 10 1 x 10--33 6655..0000 5 x 10 5 x 10--44 6633..0000 1x 10 1x 10--44 5511..0000 5x 10 5x 10--55 4422..0000 3 x 10 3 x 10--55 3333..0000 2 x 10 2 x 10--55 2277..0000 1 x 10 1 x 10-5-5 17.0017.00 5 x 10 5 x 10-6-6 9.509.50 1 x 10 1 x 10-6-6 2.202.20 5 x 10 5 x 10-7-7 1.101.10 mol/min) mol/min) mol/min) mol/min)

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spectrophotometrically by monitoring the absorbance (A) at

spectrophotometrically by monitoring the absorbance (A) at 340nm, which is proportional to the340nm, which is proportional to the concentration of NADPH concentration of NADPH₂₂.. [S] x 10 [S] x 1044 M M Increase in AIncrease in A at pH 7.6 at pH 7.6 Increase in A at pH Increase in A at pH 9.0 9.0 0.174 0.074 0.034 0.174 0.074 0.034 0.267 0.085 0.047 0.267 0.085 0.047 0.526 0.098 0.075 0.526 0.098 0.075 1.666 0.114 0.128 1.666 0.114 0.128 4.000 - 0.167 4.000 - 0.167

At what pH will the enzyme have more affinity for the substrate? At what pH will the enzyme have more affinity for the substrate? 5.5. Enzyma

The following results describe the effect of an inhibitor on enzyme activity of an enzyme. The following results describe the effect of an inhibitor on enzyme activity of an enzyme. Determine:

Determine:

a) Vmax in the presence and the absence of an inhibitor a) Vmax in the presence and the absence of an inhibitor b) Km in the presence and the absence of an inhibitor b) Km in the presence and the absence of an inhibitor

c) Ki c) Ki

d) type of inhibition d) type of inhibition

[S] (mol/L)

[S] (mol/L) Without inhibitorWithout inhibitor v ( v ( With inhibitor With inhibitor [I] = 2,2 x 10 [I] = 2,2 x 10-4-4_M_M v ( v ( 1 x 10 1 x 10-4-4 28.00 28.00 17.0017.00 1,5 x 10 1,5 x 10-4-4 36.00 36.00 23.0023.00 2x 10 2x 10-4-4 43.00 43.00 29.0029.00 5x 10 5x 10-4-4 65.00 65.00 50.0050.00 7,5 x 10 7,5 x 10-4-4 74.00 74.00 61.0061.00 5.6. Enzyma

5.6. Enzymatic kinetictic kineticss A biochemist studies

A biochemist studies the properties of the properties of a metabolic enzyme she has a metabolic enzyme she has just isolated. just isolated. She obtainsShe obtains kinetic data in the presence and in the absence of

kinetic data in the presence and in the absence of two different inhibitors two different inhibitors (A and B). (A and B). The identityThe identity of the inhibitors is unknown but we know that one of these is an substrate analog while the other of the inhibitors is unknown but we know that one of these is an substrate analog while the other is an alkylating agent. is an alkylating agent. mol/min) mol/min) mol/min) mol/min)

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Determine: Determine:

a) Km and Vmax of the enzyme ; a) Km and Vmax of the enzyme ;

b) which inhibitor is the substrate analog? Which is the alkylating agent? b) which inhibitor is the substrate analog? Which is the alkylating agent?

c) Ki for both inhibitors; c) Ki for both inhibitors;

d) what would be the Vo for this enzymatic reaction at [S] = 3 x 10

d) what would be the Vo for this enzymatic reaction at [S] = 3 x 10-4-4_{M and in the presence of the}_{M and in the presence of the}

inhibitor [

inhibitor [A] = A] = 2 x 2 x 1010-5-5_M?_M?

[S] (mol/L)

[S] (mol/L) Without inhibitorWithout inhibitor v (µmol/min) v (µmol/min) With inhibitor A With inhibitor A [I] = 5 x 10 [I] = 5 x 10-4-4 M M v (µmol/min) v (µmol/min) With inhibitor B With inhibitor B [I] = 3,2 x 10 [I] = 3,2 x 10-6-6 M M v (µmol/min) v (µmol/min) 5 x 10 5 x 10-4-4 1.25 1.25 0.82 0.82 0.480.48 2,5 x 10 2,5 x 10-4-4 0.87 0.87 0.49 0.49 0.330.33 1,7 x 10 1,7 x 10-4-4 0.67 0.67 0.36 0.36 0.250.25 1,2 x 10 1,2 x 10-4-4 0.54 0.54 0.26 0.26 0.200.20 1 x 10 1 x 10-4-4 0.45 0.45 0.23 0.23 0.170.17 5.7. Enzyma

5.7. Enzymatic catalystic catalysisis

The effect of pH on the activity of an enzyme is demonstrated in the following graph : The effect of pH on the activity of an enzyme is demonstrated in the following graph :

How would you explain the effect of pH on enzyme activity? How would you explain the effect of pH on enzyme activity?

5.8. Enzyme

5.8. Enzyme catalysiscatalysis

Several enzymes show a dependance on pH similar to the one shown in the previous problem. Several enzymes show a dependance on pH similar to the one shown in the previous problem. However, the optimal pH

However, the optimal pH varies a great deal varies a great deal from one enzyme to from one enzyme to another. another. What side chainsWhat side chains would you expect to

would you expect to find on active sites of enzymes if the find on active sites of enzymes if the optimal pH is:optimal pH is: a) pH 4 a) pH 4 b) pH 11 b) pH 11 pH pH E E n n z z y y m m e e a a c c t t i i v v i i t t y y ( ( % % ) )

(17)

5.9. Alloste

5.9. Allosteric enzymric enzymeses We study the kinetic

We study the kinetic properties of two properties of two enzymes (A and B). enzymes (A and B). From the resultFrom the results shown below,s shown below, determine if they constitue

determine if they constitue an ordinary enzyme or an allostan ordinary enzyme or an allosteric enzyme. eric enzyme. Explain the shape of theExplain the shape of the curves representing the velocity, v, in relation to the concentration of substrate, [S].

curves representing the velocity, v, in relation to the concentration of substrate, [S].

Chapter 6. Structure and properties of

Chapter 6. Structure and properties of nucleic acids.

nucleic acids.

6.1. Nucleic acid structure.

Consider the following polynucleotide: Consider the following polynucleotide:

AUUACGUGGUGCACUCGGGAACAUCCCGAGUGCACCACGUAAUGGA AUUACGUGGUGCACUCGGGAACAUCCCGAGUGCACCACGUAAUGGA Draw the two most stable

Draw the two most stable intramolecularintramolecular secondary structures this polymer can adopt secondary structures this polymer can adopt *6.2. Nucleic acid structure.

*6.2. Nucleic acid structure.

A solution of double stranded DNA is heated and then cooled to room temperature for two A solution of double stranded DNA is heated and then cooled to room temperature for two minutes.

minutes. Predict, qualitatively, Predict, qualitatively, the variation the variation in absorbance in absorbance at 260 at 260 nm in nm in the followingthe following conditions:

conditions:

a) the solution is heated to a temperature slightly above Tm before being cooled; a) the solution is heated to a temperature slightly above Tm before being cooled; b) the solution is heated to a temperature way above Tm before being cooled; b) the solution is heated to a temperature way above Tm before being cooled;

c) suggest the structure of two polynucleotides (synthetic or natural) which will result in an c) suggest the structure of two polynucleotides (synthetic or natural) which will result in an absorbance profile following a cooling which is the perfect inverse of the pattern absorbance profile following a cooling which is the perfect inverse of the pattern obtained in (b).

obtained in (b). 6.3. Nucleic acid structure. 6.3. Nucleic acid structure.

Explain why, RNA, and not DNA, is hydrolyzed under basic pH conditions. Explain why, RNA, and not DNA, is hydrolyzed under basic pH conditions.

[S] [S] (x 10 (x 1033_M)_M) _v_{v (enzyme}_{(enzyme A)}_A) (( v (enzyme B) v (enzyme B) (( 0.00 0.00 00..0000 00..0000 0.50 0.50 88..8800 00..3300 1.00 1.00 1144..0000 11..0000 2.00 2.00 1199..0000 44..7700 3.00 3.00 2211..5500 1122..4400 4.00 4.00 2222..8800 1199..0000 5.00 5.00 2222..3300 2211..8800 6.00 6.00 2233..5500 2222..8800 8.00 8.00 2233..6600 2233..3300 mol/min)

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6.4. Nucleic acid structure. 6.4. Nucleic acid structure.

The following results were obtained during a denaturation/renaturation experiment of a simple The following results were obtained during a denaturation/renaturation experiment of a simple nucleic acid (polyA :polyU). How would you interpret these results?

nucleic acid (polyA :polyU). How would you interpret these results?

6.5. Nucleic acid structure. 6.5. Nucleic acid structure.

IMP (inosine monophosphate) is present in chez

IMP (inosine monophosphate) is present in chez E. E. colicoli as an intermediate of biosynthesis of as an intermediate of biosynthesis of purines and

purines and it is poit is possible to ssible to incorporate IMP incorporate IMP to DNA to DNA if the if the ITP (inosine ITP (inosine triphsophate) is presenttriphsophate) is present in t

in the reaction he reaction medium. medium. However, in However, in nature, nature, IMP IMP is is never prnever present iesent in DNAn DNA. . Propose Propose anan explanation.

explanation.

6.6. Nucleic acid structure 6.6. Nucleic acid structure What are

What are the products the products of the of the digestion of digestion of the oligoribonucleotide the oligoribonucleotide 5'pACGAUGCUAUC3' 5'pACGAUGCUAUC3' byby each of the following enzymes:

each of the following enzymes: a) pancreatic ribonuclease; a) pancreatic ribonuclease; b) T2 ribonuclease; b) T2 ribonuclease; c) T1 ribonuclease; c) T1 ribonuclease;

6.7. Nucleic acid structure 6.7. Nucleic acid structure Lets proceed to the

Lets proceed to the analysis of an analysis of an RNA molecule. RNA molecule. Its global base Its global base composition is 2A, composition is 2A, 2C, 1U,2C, 1U, 1G.

1G.

Its treatment with the serpent venom phosphodiesterase yields pC. Its treatment with the serpent venom phosphodiesterase yields pC.

Its hydrolysis by pancreatic ribonuclease yields 1C, a dinucleotide containing A and C, and Its hydrolysis by pancreatic ribonuclease yields 1C, a dinucleotide containing A and C, and a trinucleotide containing A, G, and U.

a trinucleotide containing A, G, and U.

Temperature ( Temperature (ooC)C) A A b b s s o o r r b b a a n n c c e e ( ( 2 2 6 6 0 0 n n m m ) )

Solution cooled rapidly Solution cooled rapidly

Solution cooled slowly Solution cooled slowly

Tm Tm

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The action of RNase T2 yields pAp, a dinucleotide containing U and C and a trinucleotide The action of RNase T2 yields pAp, a dinucleotide containing U and C and a trinucleotide containing A, G and C.

containing A, G and C.

What is the primary structure of this RNA? What is the primary structure of this RNA?

6.8. Nucleic acid structure 6.8. Nucleic acid structure

Let’s proceed to the analysis of an RNA molecule whose global base composition is 2A, 4C, 2G, Let’s proceed to the analysis of an RNA molecule whose global base composition is 2A, 4C, 2G, 1U.

1U.

Pancreatic ribonuclease treatment yields 2Cp, two dinucleotides, one containing G and C and the Pancreatic ribonuclease treatment yields 2Cp, two dinucleotides, one containing G and C and the other containing A and U, and a trinucleotide containing A, C and G.

other containing A and U, and a trinucleotide containing A, C and G.

A mixture of RNase T1 and RNase T2 yields C, Ap, pGp and two trinucleotides, one containing A mixture of RNase T1 and RNase T2 yields C, Ap, pGp and two trinucleotides, one containing A and C and the second containing CG and U.

A and C and the second containing CG and U. The serpent venom phosphodiesterase yields pC. The serpent venom phosphodiesterase yields pC. What is the formula of this RNA?

What is the formula of this RNA? 6.9. Nucleic acid structure

6.9. Nucleic acid structure

What is the global charge of the trinucleotide ApGpUpC at neutral pH? What is the global charge of the trinucleotide ApGpUpC at neutral pH? 6.10. Nucleic acid structure

6.10. Nucleic acid structure

Why does a circular double stranded DNA renature more rapidly than a linear double stranded Why does a circular double stranded DNA renature more rapidly than a linear double stranded DNA?

DNA?

6.11. Nucleic acid structure 6.11. Nucleic acid structure Why does DNA denature in p

Why does DNA denature in pure water, that is where the ionic ure water, that is where the ionic strength is close to zero?strength is close to zero? 6.12. Nucleic acid structure

6.12. Nucleic acid structure The size of the E. coli chrom

The size of the E. coli chromosome is 4000 kpb. osome is 4000 kpb. What length of DNA does What length of DNA does it contain?it contain? 6.13. Nucleic acid synthesis.

6.13. Nucleic acid synthesis.

During an experiment similar to that performed by Meselson and Stahl, you grow bacteria for 3 During an experiment similar to that performed by Meselson and Stahl, you grow bacteria for 3 generations (instead of 2 as in the classic experiment) in a mixture containing only generations (instead of 2 as in the classic experiment) in a mixture containing only 1414_N._N.

Following DNA isolation and analysis by analytical centrifugation, what proportion of heavy Following DNA isolation and analysis by analytical centrifugation, what proportion of heavy DNA, hybrid DNA and light DNA will you

DNA, hybrid DNA and light DNA will you obtain?obtain? 6.14. Nucleic acid synthesis.

6.14. Nucleic acid synthesis.

An isolated strand (+) of DNA (base composition: 10% of A, 20% of G, 30% of C and 40% of An isolated strand (+) of DNA (base composition: 10% of A, 20% of G, 30% of C and 40% of T) is r

T) is replicated by E. coli eplicated by E. coli DNA polymerase DNA polymerase into a complimentary into a complimentary starnd (-). starnd (-). The double-The double-stranded DNA is then used as a model for the E. coli RNA polymerase which transcribes the (-) stranded DNA is then used as a model for the E. coli RNA polymerase which transcribes the (-) strand.

strand.

Indicate the base composition of the

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*6.15. Nucleic acid synthesis. *6.15. Nucleic acid synthesis. The time required t

The time required to completely synthesise o completely synthesise the E.coli genome is the E.coli genome is 40 minutes. 40 minutes. However, it takesHowever, it takes only 20 minutes for these

only 20 minutes for these bacteria to produce one generation. bacteria to produce one generation. Can you explain this paradox?Can you explain this paradox?

**6.16. Nucleic acid synthesis. **6.16. Nucleic acid synthesis. You are the first scientis

You are the first scientist to successfully analyze a micro-organism t to successfully analyze a micro-organism found on Mars. found on Mars. Because thisBecause this bacterium

bacterium contains contains double-stranded double-stranded DNA DNA as as genetic genetic material, material, you you decide decide to to analyze analyze usingusing Meselson-Stahl techniques.

Meselson-Stahl techniques. You obtain You obtain the following the following results:results:

a) how would you interpret these results? a) how would you interpret these results? b) in order to better understand this phen

b) in order to better understand this phenomenon, you isolate the components implicated in omenon, you isolate the components implicated in DNADNA replication in

replication in this organismthis organism. . You identify You identify :: -

- a a RNA RNA polymerase polymerase activity;activity; -

- a a DNA DNA polymerase polymerase which which functions functions only only on on single-stranded;single-stranded; -

- a new a new enzyme which enzyme which can generate can generate a product a product sensitive tsensitive to DNAse o DNAse in the in the presence ofpresence of NADH and a product insensitive to DNase and resistant to heat.

NADH and a product insensitive to DNase and resistant to heat.

According to this information, deduce the mechanism by which this micro-organism replicates According to this information, deduce the mechanism by which this micro-organism replicates its DNA.

its DNA.

6.17. mRNA and transcription 6.17. mRNA and transcription

Differently than DNA polymerase, RNA polymerase does not

Differently than DNA polymerase, RNA polymerase does not proofread and edit its products.proofread and edit its products. a) Why does this absence of proofreading/correction in the synthesis of RNA not threaten the a) Why does this absence of proofreading/correction in the synthesis of RNA not threaten the cells’ viability?

cells’ viability? b)

b) How How would would an an enzyme enzyme using using RNA RNA as as a a template template for for DNA DNA synthesis synthesis modify modify the the rate rate ofof mutations for an organism?

mutations for an organism? *6.18. mRNA and transcription *6.18. mRNA and transcription

The great majority of mRNAs have a very short half life – in the order of 3 minutes in bacteria. The great majority of mRNAs have a very short half life – in the order of 3 minutes in bacteria. What caused evolution to form mRNA molecules so unstable?

What caused evolution to form mRNA molecules so unstable?

Generations after Generations after1414NN transfer transfer 0 0 1 1 2 2 L LL L HHLL HHHH

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6.19. mRNA and transcription 6.19. mRNA and transcription

If RNA polymerase lengthens RNA at a speed of 35 to 70 nucleotides per second and if each If RNA polymerase lengthens RNA at a speed of 35 to 70 nucleotides per second and if each molecule of polymerase binds to 70

molecule of polymerase binds to 70 base pairs of DNA :base pairs of DNA :

a) What is the maximum speed of transcription per minute where a gene of 6000 base pairs is a) What is the maximum speed of transcription per minute where a gene of 6000 base pairs is transcribed into RNA molecules?

transcribed into RNA molecules? b) What

b) What is the is the maximum number maximum number of of molecules of molecules of polymerase that polymerase that could could be be found found bound bound to thisto this gene at any given time?

gene at any given time? 6.20. Protein coding 6.20. Protein coding

Consider the following mRNA: Consider the following mRNA:

AGU CUC UGU CUC CAU UUG AAG AAG

AGU CUC UGU CUC CAU UUG AAG AAG GGG AAG GGGGGG AAG GGG a) indicate the amino

a) indicate the amino acid sequence which would be coded acid sequence which would be coded (read from 5’ (read from 5’ to 3’). to 3’). The tableThe table containing the genetic code can be found in the appendix.

containing the genetic code can be found in the appendix. b) you obtain

b) you obtain mutations which consist of addmutations which consist of additions or deletions of one itions or deletions of one nucleotide. nucleotide. If we insert GIf we insert G between the third and

between the third and forth nucleotide, and we eliminate the forth nucleotide, and we eliminate the 1010thth nucleotide from the right (it is a nucleotide from the right (it is a G), what would be the peptide sequence?

G), what would be the peptide sequence? 6.21. Protein coding

6.21. Protein coding..

The amino acid sequence from part of lysozyme isolated from a wild type and a mutant The amino acid sequence from part of lysozyme isolated from a wild type and a mutant bacteriophage T4 is given below:

bacteriophage T4 is given below: wild type:

wild type: -Tyr-Lys-Ser-Pro-Ser-Leu-As-Tyr-Lys-Ser-Pro-Ser-Leu-Asn-Ala-Ala-Lys- n-Ala-Ala-Lys-mutant: mutant:

-Tyr-Lys-Val-His-His-Leu-Met-Ala-Ala-Lys-a) can this mutant be the result of a change in a single base pair in the DNA of phage T4? If not a) can this mutant be the result of a change in a single base pair in the DNA of phage T4? If not how was this mutant produced?

how was this mutant produced? b) what is

b) what is the base sequencthe base sequence of the e of the mRNA which codes mRNA which codes for the five afor the five amino acids in mino acids in the wild typethe wild type which are different than those of the mutant type?

which are different than those of the mutant type? 6.22. Protein coding

6.22. Protein coding..

A strand of DNA has the following sequence A strand of DNA has the following sequence::

5' TCGTTTACGATCCCCATTTCGTACTCGA 3' 5' TCGTTTACGATCCCCATTTCGTACTCGA 3' a) what is the sequence of its complementary strand?

a) what is the sequence of its complementary strand?

b) what is the base sequence of mRNA transcribed from the first st b) what is the base sequence of mRNA transcribed from the first strand?rand?

c) what is the coded amino acid sequence? c) what is the coded amino acid sequence?

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d) what is the coded amino acid sequence if the second T from the 3’ end of the DNA is deleted? d) what is the coded amino acid sequence if the second T from the 3’ end of the DNA is deleted? 6.23. Genetic engineering

6.23. Genetic engineering..

Give the restriction fragments obtained following digestion of the following nucleic acid with the Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme EcoR I:

enzyme EcoR I:

5’ATGCTCGATCGATC

5’ATGCTCGATCGATCGAATTCTATAGCCCGGGAATTCTATAGCCCGGGGCTGGATCCAGGTACGGCTGGATCCAGGTACCAAGTTAAGCTTG3’CAAGTTAAGCTTG3’ 3’TACGAGCTAGCTAG

3’TACGAGCTAGCTAGCTTAAGATATCGGGCCCTTAAGATATCGGGCCCCGACCTAGGTCCATGCCGACCTAGGTCCATGGTTCAATTCGAAC5’GTTCAATTCGAAC5’

6.24. Genetic engineering 6.24. Genetic engineering..

Give the restriction fragments obtained following digestion of the following nucleic acid with the Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme BamHI:

enzyme BamHI:

5’ATGCTCGATCGATC

Give the restriction fragments obtained following digestion of the following nucleic acid with the Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme Sma I:

enzyme Sma I:

5’ATGCTCGATCGATC

Give the restriction fragments obtained following digestion of the following nucleic acid with the Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme KpnI and Hind III:

enzyme KpnI and Hind III:

5’ATGCTCGATCGATC

You want to map the genome of the

You want to map the genome of the λλ bacteriophage bacteriophage (a (a double double stranded stranded linear linear DNA). DNA). ToTo accomplish this, you label the genome of phage

accomplish this, you label the genome of phage λλ (total length of 48 500 bp) at the 5’ end with a(total length of 48 500 bp) at the 5’ end with a radioactive phosphorous (

radioactive phosphorous (3232P). P). You then digest You then digest the marked genome the marked genome with different with different restrictionrestriction enzymes under conditions

enzymes under conditions which will which will permit partipermit partial digestion al digestion of the DNAof the DNA. . You analyze theYou analyze the resulting fragments by agarose electrophoresis and then visualize the bands with resulting fragments by agarose electrophoresis and then visualize the bands with autoradiography. The results are shown in the table below.

autoradiography. The results are shown in the table below. a)

a) Calculate the length of each restriction fragment obtained.Calculate the length of each restriction fragment obtained. b)

b) Create the restriction map of theCreate the restriction map of the λλ phage. phage.

DNA

DNA standard standard Apa Apa I I Pvu Pvu I I BamH BamH II Length

Length (bp) (bp) DistanceDistance migrated (cm) migrated (cm) Distance Distance migrated (cm) migrated (cm) Distance Distance migrated (cm) migrated (cm) Distance Distance migrated (cm) migrated (cm) 23 23 130 130 3.5 3.5 2.76 2.76 2.76 2.76 2.762.76 9 9 416 416 4.1 4.1 4.12 4.12 3.02 3.02 2.892.89

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6 6 557 557 4.5 4.5 3.29 3.29 3.063.06 4 4 361 361 4.9 4.9 3.98 3.98 3.243.24 2 2 320 320 5.25 5.25 3.433.43 2 2 027 027 6.15 6.15 4.654.65 560 6.7 560 6.7

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pK

pKa

as and p

s and pI

I V

Va

alu

lue

es f

s for

or

C

Common Amino

ommon Amino Acids

Acids

pKa1

pKa1 pKa2 pKa2 pKR pKR pIpI

G G 2,34 2,34 9,60 9,60 5,975,97 A A 2,34 2,34 9,69 9,69 6,016,01 V V 2,32 2,32 9,62 9,62 5,975,97 L L 2,36 2,36 9,60 9,60 5,985,98 II 2,36 2,36 9,68 9,68 6,026,02 P P 1,99 1,99 10,6 10,6 6,486,48 F F 1,83 1,83 9,13 9,13 5,485,48 Y Y 2,20 2,20 9,11 9,11 10,07 10,07 5,665,66 W W 2,83 2,83 9,39 9,39 5,895,89 S S 2,21 2,21 9,15 9,15 13,60 13,60 5,685,68 T T 2,63 2,63 10.43 10.43 13,60 13,60 5,875,87 C C 1,71 1,71 10.78 10.78 8.33 8.33 5,075,07 M M 2,28 2,28 9,21 9,21 5,745,74 N N 2,02 2,02 8,80 8,80 5,415,41 Q Q 2,17 2,17 9,13 9,13 5,655,65 D D 2.09 2.09 9,82 9,82 3,86 3,86 2,772,77 E E 2,19 2,19 9,67 9,67 4,25 4,25 3,223,22 K K 2,18 2,18 8,95 8,95 10,79 10,79 9,749,74 R R 2,17 2,17 9,04 9,04 12,48 12,48 10,7610,76 H H 1,82 1,82 9,17 9,17 6,00 6,00 7,597,59

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↓

↓ UU CC AA GG ↓↓

U U

Phe Ser Tyr Cys

Phe Ser Tyr Cys UU

Phe

Phe SSeerr TTyyrr CCyyss CC

Leu

Leu Ser Ser SSttoopp SSttoopp AA

Leu

Leu Ser Ser StopStop TrpTrp GG C

C

Leu

Leu PPrroo HHiiss AArrgg UU

Leu

Leu PPrroo HHiiss AArrgg CC

Leu

Leu PPrroo GGllnn AArrgg AA

Leu

Leu PPrroo GGllnn AArrgg GG A

A

Ile

Ile TThhrr AAssnn SSeerr UU

Ile

Ile TThhrr AAssnn SSeerr CC

Ile

Ile TThhrr LLyyss AArrgg AA

Met

Met TThhrr LLyyss AArrgg GG G

G

Val

Val AAllaa AAsspp GGllyy UU

Val

Val AAllaa AAsspp GGllyy CC

Val

Val AAllaa GGlluu GGllyy AA

Val

(26)

A

(27)

Acid-Base Equilibriu

Acid-Base Equilibrium and Spectroph

m and Spectrophotometry

otometry

1.1

1.1 Acid-base Acid-base equilibriequilibrium :um : a)

a) Since HCl is a strong acid, Since HCl is a strong acid, it will completely dissociate when in solution:it will completely dissociate when in solution: HCl

HCl  H H++ + Cl + Cl

--Stoichiometry tells us that, since the initial HCl concentration is 0.35

Stoichiometry tells us that, since the initial HCl concentration is 0.35 M, the finalM, the final concentration of H

concentration of H++ in the solution will also be 0.35M. This gives us: in the solution will also be 0.35M. This gives us: pH = - log[H

pH = - log[H++] = -log 0.35 = 0.46] = -log 0.35 = 0.46 b)

b) Acetic acid will also dissociate in solution:Acetic acid will also dissociate in solution: CH

CH33-COOH -COOH CH3-COOCH3-COO-- + H+ H++

However, since it is a weak acid, it will not completely dissociate, and we have to take into However, since it is a weak acid, it will not completely dissociate, and we have to take into account the association constant (Ka) in our calculations. This constant is described as account the association constant (Ka) in our calculations. This constant is described as follows: follows: pKa = - log Ka pKa = - log Ka Ka = 1/10 Ka = 1/10 pKa pKa = 1.74 x 10 = 1.74 x 10-5-5MM We can now easily determine the H

We can now easily determine the H++concentration:concentration: Ka = [H Ka = [H++] [CH3COO] [CH3COO--]] [CH [CH33COOH]COOH] 1.74 x 10 1.74 x 10-5-5M = [HM = [H++] [CH3COO] [CH3COO--]] 0.35 M 0.35 M 1.74 x 10 1.74 x 10-5-5M M x 0.35M x 0.35M = [H= [H++] [CH3COO] [CH3COO--] = [H] = [H++]]22 [H [H++] = (6.09 x 10] = (6.09 x 10-6-6 M M22))1/21/2 = 2.47 x 10 = 2.47 x 10-3-3MM Finally: pH = - log [H

Finally: pH = - log [H++] = - log 2,47 x 10] = - log 2,47 x 10-3-3 = 2.61 = 2.61

c)

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1.2

1.2 Acid-base Acid-base equilibriequilibrium :um : a)

a) We have a weak acid. The acid-base equilibrium is:We have a weak acid. The acid-base equilibrium is: HA

HA HH++ + A + A --We can determine Ka as follows:

We can determine Ka as follows:

Ka = [H

Ka = [H++] [A] [A--]] [HA] [HA] The question stipulates that this acid is only 2%

The question stipulates that this acid is only 2% ionised (or 0.02, that’s the same thing).ionised (or 0.02, that’s the same thing). This allows us to obtain the respective concentrations of species HA, H

This allows us to obtain the respective concentrations of species HA, H++ and A and A-- : : [H [H++] = [A] = [A--] = 0.20M x 0.02 = 0.004M] = 0.20M x 0.02 = 0.004M [HA] = 0.2M – [H [HA] = 0.2M – [H++] = 0.196 M] = 0.196 M Therefore : Therefore : Ka = [H Ka = [H++] [A] [A--]] [HA] [HA] Ka = 0.004M x 0.004M Ka = 0.004M x 0.004M 0.196 M 0.196 M Ka = 8.16 x 10 Ka = 8.16 x 10-5-5 M M b)

b) The pH of this solution is: pH = - log [HThe pH of this solution is: pH = - log [H++] = - log 0.004M = 2.39] = - log 0.004M = 2.39

1.3

1.3 Acid-base Acid-base equilibriuequilibrium:m: a)

a) This mixture is a buffer solution made of aceThis mixture is a buffer solution made of acetic acid and its conjugated batic acid and its conjugated base, sodiumse, sodium acetate:

acetate:

CH

CH33COOH COOH HH++ + CH + CH33COOCOO-- Na Na++

This pH of this type of solution can be determined with the Henderson-Hasselbach equation: This pH of this type of solution can be determined with the Henderson-Hasselbach equation:

pH = pKa + log [Conjugated base] pH = pKa + log [Conjugated base]

[Acid] [Acid]

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pH = 4.76 + log 0.5M = pH = 4.76 + log 0.5M = 4.464.46

1 M 1 M

b)

b) We have the following acid-base equilibrium :We have the following acid-base equilibrium : H

H33POPO44 HH++ + H + H22POPO44-- K K ++

Using the same procedure as in (a), we Using the same procedure as in (a), we get:get:

pH = pKa + log [H pH = pKa + log [H22POPO44--]] [H [H33POPO44]] pH = 2.14 + log 0.8 M pH = 2.14 + log 0.8 M 0.3 M 0.3 M pH = 2.57 pH = 2.57 1.4

1.4 Acid-base Acid-base equilibriequilibrium :um :

We have the following equilibrium: We have the following equilibrium:

H

H22POPO44-- HH++ + HPO + HPO44-2-2

And the pKa for this equilibrium is 7.21. And the pKa for this equilibrium is 7.21.

The Henderson-Hasselbach equation gives us: The Henderson-Hasselbach equation gives us:

pH = pKa + log [HPO pH = pKa + log [HPO44-2-2]]

[H [H22POPO44--]] 7.00 7.00 = = 7.21 7.21 + + log log [x][x] 0.1 M 0.1 M -0.21 = log x – log 0.1 M -0.21 = log x – log 0.1 M -0.21 + log 0.1 M = log x = -1.21 -0.21 + log 0.1 M = log x = -1.21 x = 10 x = 10logxlogx= = 0.062 0.062 MM 1.5

1.5 Acid-base Acid-base equilibriuequilibrium:m:

We have the following acid-base e

We have the following acid-base equilibrium:quilibrium: H

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According to the question, we have: [H

According to the question, we have: [H22POPO44--] + [HPO] + [HPO44-2-2] = 0.3M] = 0.3M

Hence: [H

Hence: [H22POPO44--] = 0.3 M - [HPO] = 0.3 M - [HPO44-2-2]]

From the Henderson-Hasselbach equation, we have: From the Henderson-Hasselbach equation, we have:

pH = pKa + log [HPO pH = pKa + log [HPO44-2-2]]

[H [H22POPO44--]] 7,00 = 7.21 + log [HPO 7,00 = 7.21 + log [HPO44-2-2]] [H [H22POPO44--]] 7,00 = 7.21 + log [HPO 7,00 = 7.21 + log [HPO44-2-2]] [H [H22POPO44--]] -0.21 = log [HPO -0.21 = log [HPO44-2-2]] [H [H22POPO44--]] 10 10-0.21-0.21 = [HPO = [HPO44-2-2]] [H [H22POPO44--]] 0.616 = [HPO 0.616 = [HPO44-2-2]] [H [H22POPO44--]] 0.616 x [H 0.616 x [H22POPO44--] = [HPO] = [HPO44-2-2]] Which is identical to : Which is identical to : 0.616 x (0.3M - [HPO

0.616 x (0.3M - [HPO44-2-2]) = [HPO]) = [HPO44-2-2]]

0.185 – 0.616 x [HPO

0.185 – 0.616 x [HPO44-2-2] = [HPO] = [HPO44-2-2]]

0.185 = 1.616 x [HPO 0.185 = 1.616 x [HPO44-2-2]]

0.114 M = [HPO 0.114 M = [HPO44-2-2]]

And the concentration in H

And the concentration in H22POPO44-- will be : will be :

[H

[H22POPO44--] = 0.3M - [HPO] = 0.3M - [HPO44-2-2] = 0.186 M] = 0.186 M

1.6

1.6 SpectrophotometrySpectrophotometry

The relationship between the absorbance and the concentration of a solution is given by the The relationship between the absorbance and the concentration of a solution is given by the Beer-Lambert equation:

Beer-Lambert equation:

A = A = εεclcl

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Where:

Where: A A = = absorbanceabsorbance

εε = Molar extinction coefficient (units: litres x mol = Molar extinction coefficient (units: litres x mol-1-1 x cm x cm-1-1)) c = concentration (units : mol/l =

c = concentration (units : mol/l = M)M) l = light path (thickness of the c

l = light path (thickness of the cuvette; units: cm)uvette; units: cm)

We get the following: We get the following:

0.71 = 1.420 L mol 0.71 = 1.420 L mol-1-1 cm cm-1-1 x c x 1 cm x c x 1 cm c = 0.71 mol cm c = 0.71 mol cm 1420 1420 L L x x 1 cm1 cm c = 5 x 10 c = 5 x 10-4-4 M M If we use a cuvette where c=0.1 cm, we get:

If we use a cuvette where c=0.1 cm, we get: 0.71 = 1.420 L mol 0.71 = 1.420 L mol-1-1 cm cm-1-1 x c x 0.1 cm x c x 0.1 cm c = 0.71 mol cm c = 0.71 mol cm 1420 1420 L L x x 0.1 cm0.1 cm c = 0.005 M c = 0.005 M 1.7 1.7 SpectrophotometrySpectrophotometry

With the Beer-Lambert equation, we have: With the Beer-Lambert equation, we have:

A = A = εεclcl Therefore: Therefore: A = 1420 L mol A = 1420 L mol-1-1 cm cm-1-1 x (37 x 10 x (37 x 10-3-3M) x 1cmM) x 1cm A = 52.54 A = 52.54 1.8 1.8 SpectrophotometrySpectrophotometry

We first have to graph the standard cuve of the absorbance as a function of the concentration of We first have to graph the standard cuve of the absorbance as a function of the concentration of the haemoglobin standards. This graph is shown on the following page.

the haemoglobin standards. This graph is shown on the following page.

Since the unknown has an absorbance of 0.303, we can use the standard curve to determine the Since the unknown has an absorbance of 0.303, we can use the standard curve to determine the corresponding haemoglobin concentration, in this case 6.31µg/mL.

corresponding haemoglobin concentration, in this case 6.31µg/mL. A more accurate value can be obtained by using the familiar equation: A more accurate value can be obtained by using the familiar equation: y = mx + b

y = mx + b where

where y = y = value on value on the y the y axisaxis x = value on the x axis x = value on the x axis m = slope