Answers to Chapter 5 - Garland · PDF fileIntroduction to Bioorganic Chemistry and Chemical...

Introduction to Bioorganic Chemistry and Chemical Biology 1

Answers to Chapter 5(in-text & asterisked problems)

Answer 5.1

Answer 5.2Introduction to Bioorganic Chemistry and Chemical Biology | A5110Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz

PDB 3ZNF

H27

H21

C8

C5

Introduction to Bioorganic Chemistry and Chemical Biology | A5111Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz

NH

HN

NH

HN

NH

HN

O

O

O

O

O

O

NH

NH2+H2N

O -

NH

NH2+H2N

O

NH

HN

NH

HN

NH

HN

NH

NH2+H2N

O -

NH

NH2+H2N

O

O

O

O

O

O

O

NH

HN

NH

HN

NH

HN

O

O

O

O

O

O

NH

NH2+H2N

O -

NH

NH2+H2N

O

IRGERA

AREGRI

ent-IRGERA

2 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to ChApter 5

Answer 5.3

A For nKDVLrrMKK:

Residue Charge

Arg (r) +1 × 2

Lys (K) +1 × 3

+h3n terminus +1 × 1

Glu (e) –1 × 0

Asp (D) –1 × 1

–o2C– terminus –1 × 1

net charge +4

similar calculations for rFrGDYFAK and KGnIDKFteK yield net charges of +2 and +1, respectively.

B the peptide is expected to bind better to a negatively charged surface through ionic interactions, because opposite charges attract.

Answer 5.4

residues in the peptide are labeled in red; residues in MDM2 are labeled in blue.

Answer 5.5

h383, h387, and e411.


NH

HN

NH

HN

NH

HN

NH

NH2+H2N

O -

NH

NH2+H2N

O

O

O

O

O

O

O

ent-AREGRI

this “retro-inverso“ peptide is the best match of the natural sequence IRGERA


L26

W23 F19L54

I99

L57

V99

F91

I61

M62


H383

H387

E411

Introduction to Bioorganic Chemistry and Chemical Biology: Answers to ChApter 5 3

Answer 5.6

Answer 5.7

Answer 5.8

Answer 5.9

A A1,3 strain is most important for ϕ angles; A1,2 strain is most important for ψ angles.

B

C

Answer 5.10


O

OH A

O

OH+

O

OH

+

H+-A

HO HO HO: :


NH

NO

O

OH

OO

N

HN

O

O

Fmoc-PNA-T-OH


O

N

NHNH

O

O

HN

HN

O O

N

NHHN

O

O

NH

NH

O

O

OEt

Et

HO

OH

Pro and Gly prefer turnsVal, Ile, and Tyr prefer sheets


NH

O

CONHNH

O

H

= 0°

CONH

= 160°

H

side chain

side chain


NH

O

CONHNH

O

H

= 0°

CONH

= 160°

H

side chain

side chain


- OS

OHS+

HS R..

HA

++ A-OHSS

H R

: :

+-A

:

:

OHSS

R

:

:

+:

OH2

SS

R

:

H A

+

SS

R

:+

SHR..

SRS

R

H

+

-A:

SRS

R


Answer 5.11

Answer 5.12

there are seven wD domains in each half of the dimer, leading to a total of 14 wD domains.


PR3:SSCys

CysP

RCys-S

RR

+OH2: P

RCys-S

RRO

H

PR

OR

R

Cys-S:-

H+

PR

Cys-S

RRO

H B:

B:

PR

Cys-S

RRO -..

H B Cys-SH


NHR

HA +H2NR

NH2+

R

-A H

NH2R

enamine

..:

.. NH2+

R

H2NR

H

..HA

-A:

NH2R

H2NR

aldol cross-linkNH

R

R

H+ A-:NH

R

R

HNR

..H A

+H2NR

NH2R

..

H2N RN

RH

H+

H2N RNH

R

.. H A

A-:

+H3N RN

RH

:R

NR H+ A-:

RN

R

imine

NH2R

+H3NR

..

A

B


NH

HN

NH

HN

NH

NNH

+H3NO

O

O

O

O

O

OHN

OO -

O

S

NH2O

NH2

O

S

OHHN O

OO

N

NH

O

O

HN

HN

HO

NH

O

N

O

NH

H

H2+

NNH

HN

NH

HN

NH

O

O

O

O

O

NH

O

HN

NH3+

O -

O

NH3+

NH3+

H2N NH2+

NH3+

+H3NNH

HN

NH

NNH

HN

O -

O

O

O

O

O

O

O

HO

OHNH

HN

OO O

N

HNNH

O

HN

O

SS

H

spinorphin: LVVYPWT

SV40 NLS sequence: PKKKRKV

cyclo-[FGPTLWP]

oxytocin: CYIQNCPLG malformin A: cyclo-[DC DCV DLI]

phakellistatin 13:

Answer 5.13

*Answer 5.14


*Answer 5.15

*Answer 5.17

A

Residue Charge

Arg (r) +1 × 0

Lys (K) +1 × 0


Glu (e) –1 × 0

Asp (D) –1 × 5

–o2C terminus –1 × 1

net charge –5

B

Residue Charge

Arg (r) +1 × 6

Lys (K) +1 × 2


Glu (e) –1 × 0

Asp (D) –1 × 0

–o2C terminus –1 × 0

net charge +9


cyclo-[RRWWRF]

RPKPQQFFGLM

CWLDVC

or cyclo-[RWWRFR]or cyclo-[WWRFRR]or cyclo-[WRFRRW] or cyclo-[RFRRWW] or cyclo-[FRRWWR] or cyclo-[RRWWRF]

A substance P

B antimicrobial peptide

C synthetic integrin antagonist


*Answer 5.19

*Answer 5.23

*Answer 5.24


NH

OO

S

NH

OS -..

NH

- O

H B

S

NH2

+

- O S

H B+

..

:

..

HN

O

HN

O

HN

O

HNH2N

O

HN

SO..

OO

H2NO

HN

- S

O

HB ..

H2NO

HN

HS


H A

O

OH+

O

OHR

+

O+

OH

CF3

A-

OHO

CF3

O

OAsp

:

:

:

Asp

O O

CF3 O

OAsp

:

HA

NO


O

OAsp H A

O

OH+

Asp

O

OHAsp

+ NH

NH

H +-A

NH

:

:

Trp Trp Trp


*Answer 5.26


Ph Ph

R NH

O

Ph H A..

PhPh

R NH

OH+

Ph

OH

NHR+ +CPh3

OH

NHR

O

NH2+R

H A-:

R NH2

O

Ph

Ph

PhHCF3COO

Sii-Pr i-Pr

i-Pr

Ph Ph

PhH

Sii-Pri-Pri-Pr

CF3COO

HSi

i-Pr i-Pri-Pr

O -O

CF3

+

-..

H A..

H APh

R N

NPh

Ph

H A..

Ph

R NH+

NPh

Ph

R NH

N..

Ph Ph

PhH

Sii-Pri-Pri-Pr

CF3COO

+

-i-Pr3SiO2CCF3 + Ph3C-H

R NH

HN+

R O..

H A+

+R OH

H

Sii-Pri-Pri-Pr

CF3COO

+


R OH

+

NH

O

OH+ OH

ONH

+

H A..

RNH

O

O

R R

H

Sii-Pri-Pri-Pr

CF3COO

+-

i-Pr3SiO2CCF3 + Me3C-H

OH

ONH

R ..HA O

ONH2

+

R

H A-: - O

ONH2

+

R

:

NH2R + CO2..

H A

a concerted, one-step E2 mechanism is also plausible

+

..

Ph Ph

Cys S Ph

H A

Ph Ph

Cys S PhH

++ +CPh3

Cys SH

Ph Ph

Ph

H

Sii-Pri-Pri-Pr

CF3COO

+-

i-Pr3SiO2CCF3 + Ph3C-H

Ka' < -8).

A Initial protonation occurs on the amide carbonyl, not the amide nitrogen.

C

D

E

B Protonation of N would disrupt the aromaticity of the imidazole ring. Therefore initial protonation occurs on the N not N .

Note that protonation of thioethers is extremely unfavorable (p


Ph Ph

R NH

O

Ph H A..

PhPh

R NH

OH+

Ph

OH

NHR+ +CPh3

OH

NHR

O

NH2+R

H A-:

R NH2

O

Ph

Ph

PhHCF3COO

Sii-Pr i-Pr

i-Pr

Ph Ph

PhH

Sii-Pri-Pri-Pr

CF3COO

HSi

i-Pr i-Pri-Pr

O -O

CF3

+

-..

H A..

H APh

R N

NPh

Ph

H A..

Ph

R NH+

NPh

Ph

R NH

N..

Ph Ph

PhH

Sii-Pri-Pri-Pr

CF3COO

+


R NH

HN+

R O..

H A+

+R OH

H

Sii-Pri-Pri-Pr

CF3COO

+


R OH

+

NH

O

OH+ OH

ONH

+

H A..

RNH

O

O

R R

H

Sii-Pri-Pri-Pr

CF3COO

+-

i-Pr3SiO2CCF3 + Me3C-H

OH

ONH

R ..HA O

ONH2

+

R

H A-: - O

ONH2

+

R

:

NH2R + CO2..

H A

a concerted, one-step E2 mechanism is also plausible

+

..

Ph Ph

Cys S Ph

H A

Ph Ph

Cys S PhH

++ +CPh3

Cys SH

Ph Ph

Ph

H

Sii-Pri-Pri-Pr

CF3COO

+-

i-Pr3SiO2CCF3 + Ph3C-H

Ka' < -8).

A Initial protonation occurs on the amide carbonyl, not the amide nitrogen.

C

D

E

B Protonation of N would disrupt the aromaticity of the imidazole ring. Therefore initial protonation occurs on the N not N .

Note that protonation of thioethers is extremely unfavorable (p


*Answer 5.28

A KVKVKVKVKVKVK, because β-branched amino acids have high β-sheet propensities and low α-helix and turn propensities.

B GppGppGppGppGppGppGpp, because pro has low α-helix and β-sheet propensities.

C nLeDKAeeLLsKnYhLeneVArL, because it has a leucine zipper motif with Leu at every seven residues.

D GppGppGppGppGppGppGpp, because (Gpp)n forms a collagen triple helix.

E GppGppGppGppGppGppGpp, because pro has low α-helix and β-sheet propensities.

*Answer 5.31

A the calcium-binding loops are underlined: MADQLTEEQIAEFKEAFSLFDKDG-DGTITTKELGTVMRSLGQNPTEAELQDMINEVDADGNGTIDFPEFLTMMARKMK-DTDSEEEIREAFRVFDKDGNGYISAAELRHVMTNLGEKLTDEEVDEMIREADIDG-DGQVNYEEFVQMMTAK

B the first calcium-binding loop is positions 21–32, -DKDGDGtIttKe-. the first and last positions of the calcium-binding loop are the only positions conserved as ani-onic amino acids. therefore Asp21 and Glu32 are most likely to bind as anionic car-boxylates. In the crystal structure for pDB 1CFC, both of the side-chain carboxylate oxygens of Glu32 bind to the calcium ion. Because asparagines are also tolerated at the second and third ligand positions, Asp23 and Asp25 probably coordinate as neutral ligands through the side-chain carbonyls. Because phenylalanine is also tol-erated at the fourth ligand position, it is likely that the backbone carbonyl, and not the side chain, is coordinating to calcium. Because serine is a common ligand for the fifth position, thr29 probably binds through the side-chain hydroxyl. As a result of the high pKa of alcohols compared with carboxylic acids, thr29 probably binds as a neutral oh ligand.

C Glycine is highly flexible and can adopt tight turns. Isoleucine and valine are β-branched hydrophobic amino acids that limit the flexibility of the side chain and loop.

*Answer 5.32

A two types of dimers are known. the dimer based on swapping the s-peptide (pDB 1Bsr) has the s-peptide from the first molecule bound in the cleft of the second molecule, and vice versa.


Ca

OO

O

O

OH

OO

OH

OH

NH

CH3

O

HNOH

HN

HN

HN

O

O

O

NH

O

NH

O

Asp21

Glu32

Asp23

Asp25

Thr27

Thr29


domain-swapped dimer


proteinengineering

(truncate loop)

wild type deletionmutant


B there may be many solutions to this problem. Any mutation to the loop that prevents the helix from binding to the cleft in residues 1–113 will favor the formation of the domain-swapped dimer. Investigators have induced dimerization by generating a deletion variant Δ(114:119) that lacks the flexible loop.

*Answer 5.36

*Answer 5.38

A Cys28 makes contacts with DnA (rendered as a brown surface).

B the nearby cysteine residue is Cys120.

C Cysteine thiols are efficient at conjugate addition reactions.




proteinengineering

(truncate loop)

wild type deletionmutant


O

NH2

OOH

OH

HO

OO

OH

R

R'R"

H

R

R"H

R'

bonds with limited rotation due to A1,3 strain A1,3 strain


O

O

O

H

H

H

HO

S

S

Cys

Cys


O

O

O

H

H

H

HO

S

S

Cys

Cys


*Answer 5.41


X• OH X H

•O

+

R

O

R

•

O

R

•

O

R

•

O

OR

R OH

HOR

RH

H

R

etc

tautomerization

resonance-stabilized

keto tautomers enol tautomers (aromatic)

Answers to Chapter 5 - Garland · PDF fileIntroduction to Bioorganic Chemistry and Chemical...

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