NAME
1. - - - - I 48
4. - - - - I 8
2. - - - - I 9
5. - - - - I 5
Scoring:
3. - - - - I 30
Total Score: - - - - I 100
1. (48 pts)
a. L-Phosphoserine is a modified amino acid that is
generated in proteins by phosphorylation of serine
residues. The amino acid side chain has two acidic
protons, which exhibit different pKa values, as
shown at right. What would be the structure and
charge state of phosphoserine under extremely
acidic conditions, at pH = 1? Then, how would the
structure change with increasing pH?
,
~troJP ~
r/> pO;rJ'I_l
1
F 10-rAL
0
=-P- Ol-\
'o
o c~fr)
OH
L -phosphoserine
-
~o
--..........0
I
I
0 =P
I
pKa=2.1
pKa =7.2
e
-o
'
O l~
Ot~
~£
l for
pl= B
b. What would you predict for the
isoelectric point (pl) of phosphoserine?
©
~j
2 . l -2 Jl 3)
c. If you analyzed a mixture of phosphoserine and serine by ion exchange chromatography,
using a solvent gradient of increasing pH, would you expect phosphoserine to elute
~©
later
relative to serine? (Circle one answer.)
2
or
at the
same time
'9-'\J ....>er
NAME
1. - - - - I 48
4. - - - - I 8
2. - - - - I 9
5. - - - - I 5
Scoring:
3. - - - - I 30
Total Score: - - - - I 100
1. ( 48 pts)
a. L-Phosphoserine is a modified amino acid that is
generated in proteins by phosphorylation of serine
residues. The amino acid side chain has two acidic
protons, which exhibit different pKa values, as
shown at right. What would be the structure and
charge state of phosphoserine under extremely
acidic conditions, at pH = 1? Then, how would the
structure change with increasing pH?
1f
10-rA L
O
(~~r)
OH
pKa
pKa = 2.1
=7.2
L-phosphoserine
,
~rv-oJP ~
rp pottJ'l.l
'o
0
==..P -
Ot-\
I
01.+
~t..
~©
b. What would you predict for the
isoelectric point (pl) of phosphoserine?
pl= ~
(for ~j ~ "\J.....>e.("
2 . l -2 ~ 3)
c. If you analyzed a mixture of phosphoserine and serine by ion exchange chromatography,
using a solvent gradient of increasing pH, would you expect phosphoserine to elute
~©
later
relative to serine? (Circle one answer.)
2
or
at the
same time
d.
become more
positively
charged?
or
retain the same
charge state?
(Circle
one.)
e. Ingested,
phosphoserinedon't
containing
proteins
survive our stomachs. under
HCI
acidic conditions, phospho)
serine residues on the surfaces
(pH= 2)
of proteins undergo acidI
O=P-OH
catalyzed,
nucleophilic
OH
I
I
phosphoacyl substitution, and
OH
O=P-OH
are hydrolyzed back to serines.
I
Draw a mechanism (using
OH
"electron pushing") for this
process. Draw each molecule
and mechanistic step explicitly; don't cheat by combining multiple processes in a single
step.
'o
Mechanism:
0
1-i
l\
(
t-~~c-"'
u
0
r-~~
.
8
~L
+
c.-)
-~o•+
(?;. ~µ
~ ~ o=- \:>- 01-t
J-1~
&\
I
Of.j,
f
O'°' ~
0: p-014
'
I
01-J
3
8otJ D
f.::. O
fOIL
P.., G.-- - PT A1L- ' ~C..
G. ft-0 J f'
f :;; O
g a ,J 0
t_,,f./+ J t N 6. ,
{ c,4-.,J
g
'-"f-A-4 r tJ~
B E-
2-
.J
otJ £.... )
®
fo<2-
P~ro,.JA-7'1A.Jtc..
~ <.) G-~f i~ ( ~
~
G
'FO t..
A-.J 0
( £) ( iJ t;,
p=o
&<J,.JD
6&f-a-LC.
~nt-e..t<:.S "
,..s f--C. A-r Ii' v 6
1t~k/Jl.1J,I\ ,~.i;u~
~.J.. 1
rUJ J.
ofl-
f.
Phosphopeptides-peptides incorporating a
H
.
phosphoserine residue-can be synthesized
via solid-phase peptide synthesis using the
Fmoc-protected phosphoserine ("Fmoc-PS")
o
'o
reagent shown at right. In the box below,
I
propose a multi step synthesis of the dipeptide
O=P-OH
I
phosphoserinyIglycine (H2N-PS-Gly-COO H),
OH
starting from Wang resin (drawn in the box .
below). You do not need to draw any
Fmoc-PS
chemical structures to answer this problem;
you can refer to molecules by name or
chemical abbreviation. In addition to the starting materials I've drawn, you can use any
reagents and reactions we've learned about in class.
OYNYCOOH
@
Wang resin
-----'l
®""""" t'A 1j -
I . ~D CJLz._
f
2. '
F
,,w\.D (., -
le.~
FW\O(,
00
l~
btje,_.
(f
•fV~J.'At-)
o/
0.-" ~
"""e-1 e... /A.
~Afk~r
orJ.e,r .
[
, • P: r u ~
J C:"' e...
Cb)
(Je..rn;,hc:.t-c fMoc)
lf
&.-------------~-------------~-------~-~
2, a.~'~ (r-tf or 'lfA)
CI)
(du.."~
fr.vN't
2. (9 pts) Inosine is a non-natural nucleoside
that is formed when a hypoxanthine base is
attached at the 1' position of a ribose sugar.
Inosine can be incorporated into DNA and
RNA by both biological and synthetic
methods.
SJf
for+)
0
~N0NH
)l_.J
H
N
hypoxanthine
4
a. DNA
polymerase
can
incorporate 2' -deoxyinosine
into DNA when it accepts 2'deoxyinosine triphosphate as
a
substrate.
Draw
the
structure of 2' -deoxyinosine
triphosphate.
2'-deoxyinosine triphosphate
H
o-e - D
'oe')
I
o=P - o"e
~
o
I
0
b. Inosine can base-pair with
l
any of the four typical DNA
O.=P - Oe
bases, though the DNA
\
"'"ck""'-IA..{
double helix accommodates a
pyrimidine across from
'--~~~~~~~---t-~~+-..........~-"-'-~~ cp~+-e,
mosme better than a
hypoxanthine-pyrimidine base pair
c..tf
purine. Draw a base pair
1
between the hypoxanthine
base of inosine (already
drawn in the box for you)
N
o -··
I
and either of the two
pyrimidine
bases
typically found in DNA.
;N-H
""'
Illustrate
hydrogen
bonding in your drawing.
You do not need to draw
the sugar or phosphate
parts of DNA; just draw
the base.
oe
o
(')r-{
N~
Nd
...,_N
I
)~,
'-N
O~
3. (30 pts) Lactose is a disaccharide formed from glucose and galactose; it is a glucosyl
glycoside of galactose.
- - - - - --
a. o-Galactose is the C-4 epimer of
o-glucose, shown at right. Draw
a Fischer projection of Dgalactose in the box provided.
b. Acyclic D-galactose equilibrates
with a cyclic, a-galactopyranose
form. Draw both a Haworth
projection and the most stable
chair conformer of that cyclic
form in the boxes on the next
page.
H
~ OH
-
-
·
-
-
-
H~O
l
l..\
t+ D
OH
CH20H
o-glucose
5
-
0
'
acyclic form of o-ga/actose
(Fischer projection)
I-\
D 1-l1-4-
D
t-\
l-+
or.\-
©
1,.J ~CA-\
~o')l,
cyclic a-o-ga/actopyranose
(Haworth projection)
ED
cyclic a-o-galactopyranose
(chair conformer)
fJ(l.... e4-W~
1rJ (.A)~"I
C-~O,J
~-r~u+Pv1 ~
01~
ett ""h'·" ' 01.+
() ')C ,'4 \
c. Although the a-anomer of galactose that you drew is less stable than the P-anomer, the aanomer is still stabilized by the "anomeric effect", a specific molecular orbital interaction.
Illustrate that molecular orbital interaction on your chair drawing above, drawing lobes for
orbitals.
d. Cyclic a-galactose (a hemiacetal) spontaneously forms glycosides (acetals) with other
alcohols, like glucose, in the presence of an acid catalyst. Using my cartoon illustrations
below, draw a mechanism on the next page for the formation of glycoside from galactose.
R-OH
0
)
H
OR
OH
Mechanism:
d~ @Gf14.
(±)
Of)
'@
~
/0"1.+
H@
rl.../
c;,
@
.ft, r
u..c.h.. c,- f
..>J.
11.; ""j
J. +ef 6
~r e,. cJ\. ~ k"""'-e-clcAA-e. r
(?)
'
o, 14
(~
Mlp)
e. What is the timescale of the reaction on the previous page? Does it take place
spontaneously over
©
minutes?
weeks?
~
or
..
(Circle one.)
4. (8 pts) Each of the reactions below is drawn with two possible products. Circle the preferred
product. If the two products are produced equally, circle "BOTH". If neither product would
result from the reaction, circle "NEITHER". Circle one answer only.
2~
hv
)
Remember: This number
means, "two of this molecule".
BOTH
NEITHER
(equally)
Ph
Ph~
)
BOTH
NEITHER
(equally)
5. (5 pts) For the reaction below, fill in the empty box corresponding to product. Give only one
answer. If you expect the reaction to yield multiple products, draw one major product. If the
reaction yields multiple enantiomers, draw only one enantiomer in the box, and include the
note "+ enantiomer".
0
1. KH,
~
)
~l ,
- 2..
~r +r? " c'~ l
e.,rro r .
7
©