May 4, 1970 - tablished by the overlaps in sequence provided by unique chymo- tryptic and peptic peptides (Table VIII in Reference 4). The following analysis ...
Vol.
245,
THE JOURNAL OF RIOLO~ICAL CHE>~IISTRY No. 17, Issue of September 10, pp. 45704582, Printed
The
in
1970
U.S.A.
Complete
Amino
Acid
Sequence
of Bovine
a-Lactalbumin* (Received
I.
1
0.76(l) 0.90(l) 0.71(l) 2.80(3) 1.95(2) 0.70(l) 1.00(l) 0.95(l) 0.98(l)
1.400) 3.00(3) I.9401 1.05(l) 0.940) N.D.0(1) 0.85(l) ...................
that
in italics
are t,he molar
ratios
6
1.10 1.13 1.09 2.00 1.66 1.0 0.89 N.D. N.D.
1.07 1.09 1.15 1.97 1.68 1.0 0.10 N.D. N.D.
0.98 1.0 0.83 1.18 1.54 1.0
0.98
0.99 1.0 0.86
0.49 0.90 0.87
0.70 1.1
0.69 1.1
0.89
N.D. N.D.
N.D. N.D.
ND. N.D.
N.D. N.D.
GUY
Tyr
Gly
Val
Ser
Pro
in these
of T5,6
and t,he values
T5.6
Total
no. of residues.
Yield
(7;)
a Includes * Estimated c Contains
L
N.D. ND. GUY
in parentheses
(Glu, Trp, Val, Cys)
peptides
derived from
are the assumed
Acid
number
of thermolysin
T5,6-Th-2
Enzymic
T5,6-Th-6
Enzymic
Acid
TS,bTh-8
Enzymic
3.0(3) 1.1(l) 1.0(l) 2.2(2)
1.1(l) 1.0(l) 3.6(4)” 1.1(l)
1.3(l) 1.0(l) 1.2(l) 1.3(l)
1.1(l) 1.0(l) 0.96(l)
1.1(l)
0.8(l)* 0.1
0.84(l)
0.84(l)
1.0(l) 0.8(l)
1.1(l) 0.9(l)
3.2(3)
3.3(3)
1.1(l) 1.1(l)
0.97(l) 1.0(l)
0.90(l)
0.90(l)
41
11
1.0(l)
G9
T6,6 of residues.
peptides
6.3(G) 5.0(5) S.S(.) 5.1(5) 1.1(l) 5.1(5) 1.0(l) 2.6(3) 1.9(2) 2.3(2) 3.2(3) 2.0(2) +(I) 1.0(l) 1.0(l) 0.8(l)
11
1.0 0.48
Iv
and thermolysin
T
TS,GTh-1 Acid
Aspartic acid .......... Threonine ............. Swine. ................ Glutumic acid. ........ l’roline. ............... Clycine ................ Alanine. ............... Valine. ............... Isoleucinc ............ Leucine ............... Tyrosine .............. Phenylalanine ......... Tryptophanb. .......... J,ysine. ................ Histidine. ............. Carboxymethylcysteine
1.1 0.89 0.67 1.6
analyses.
Composition Amino acid
degradation 5
acid was not determined
acid composition
of residues.
4
TABLE
The values
number
Gly-Tyr-GlJr-Gly-Val-Ser-Leu-Pro
Amino
17
3
..............
the amino
243, Fo.
2
........................................... indicates
are the assumed
Steps of Edman
Enzymic hydrolysate
1.06(l) 1.18(l)
Vol.
of T6
in parentheses
acid composition
Acid hydrolysate
Serine................ Glutamic acid. Proline. Glycine Valine Leucine Tyrosine . Tryptophan . Aminoethylcysteine..
and the values
of a-Lactalbumin
7
7
55
04
Enzymic
:\cid
2.20(2) f .0(l)
1.4(l)
1.0(l)
Acid
TJ,6-Th-9
1 .6(2)1&
I /
1.0(l) 1.0(l)
w
1.0(l)
1.0(l)
0.84(l)
0.90(l)
: /
1.0(l) 1.0(l)
6”
4
4
68
serine, glutamine, and asparagine. qualitatively. two peptides as described in the text.
Xlep 1, Gly, 0.18; quenced by Edman degradation as follows. Leu, 1.04; Phe, 0.98. Step .Z, Leu, 0.38; Phe, 1.0. CiO. Residues S2 to 36: His-Thr-Ser-Gly-Tyr-Four steps of Edman degradation established the sequence of this peptide. Step 1, Thr, 1.0; Ser, 1.1; Gly, 1.0; Tyr, 1.0; His, 0.1. Step 2,
Thr, 0.17; Ser, 0.99; Gly, 1.05; Tyr, 1.0. Step S, Ser, 0.37; Gly, 1.27; Tyr, 0.73. Step 4, Gly, 0.2; Tyr, 0.74. C2. Residues 44 to 50: Asx-Asx(Gln ,Ser , Thr, Asp) Tyr-Two
steps of Edman degradation established 2 residues of this peptide as follows. 0.95; Ser, 0.89; Glu, 1.31; Tyr, 0.84.
0.96; Ser, 0.89; Glu, boxypeptidase
1.22; Tyr, 0.81.
the sequence
of t,he first
Xtep 1, Asp, 2.05; Thr, Step 2, Asp, 1.10; ‘lb, Digestion of C2 with car-
A for 150 min gave tyrosine
ot,her amino acids. The amide distribution estimated by analysis of T5,6-Thl. Cl7. Residues 8 to 60: Lys-Ile-Trp-This
in a 98% yield
and no
in this region peptide
was
was se-
Issue
of September
10, 1970
K. Brew, F. J. Castellino,
T. C. Vanaman,
quenced by Edman degradation as follows. Step 1, Lys, 0.12; Ile, 1.0; Trp, not determined. Step ,%‘, only tryptophan found in unhydrolyzed peptide. Pl. Residues 41 to 49: Ile-Val-Glx (Asx , Asx ,Glx ,Ser , Thr , Asx)-Three steps of Edman degradation gave the following results. Step 1, Asp, 3.11; Thr, 1.01; Ser, 0.96; Glu, 1.98; Val, 0.80; Ile, 0.05. Xtep d, Asp, 3.0; Thr, 0.84; Ser, 0.87; Glu, 2.0; Val, 0.14. Slep S, Asp, 3.0; Thr, 0.84; Ser, 0.87; Glu, 1.54. Analysis of T5,6-Thl gave the amide distribution in this region. P4. Residues 50 to 5.2: Tyr-Gly-Leu-Three steps of Edman degradation established the sequence of this peptide as follows. Step 1, Gly, 1.16; Leu, 1.0; Tyr, 0.02. Step 2, Gly, 0.26; Leu, 1.0. P5. Residues 50 to 65: (Tyr ,Gly ,Leu, Phe)-Although this peptide was not sequenced its composition was identical with P4 except for 1 additional residue of phenylalanine, which appears to
and R. L. Hill
4575
follow leucine at Residue 52 in accord with the sequence of C5 (Table II). Pl la. Residues 26 to 31: (Trp , Val, Cys(Ae) , Thr , Thr , Phe)This peptide was not sequenced but its composition indicated that it could correspond only to the residues shown, in accord with the sequence deduced for this region with other peptides. P6. Residues 32 to 40: His-Thr(Ser ,Gly , Tyr , Asx , Thr)GluAla-This peptide was sequenced partially by Edman degradation as follows. Step 1, Asp, 1.04; Thr, 1.89; Ser, 1.0; Glu, 1.1; Gly, 1.0; Ala, 1.0; Tyr, 1.02; His, 0.10. Step 9, Asp, 1.07; Thr, 1.0; Ser, 0.95; Glu, 1.1; Gly, 1.2; Ala, 1:05; Tyr, 0.09. Digestion of the peptide with carboxypeptidase A for 2 hours gave only alanine and glutamic acid in a yield of 15% and lo’%, respectively. P2Db,S,4. Residues 27 to 31: Val-Cys(Cm)-Thr-Thr-PheThis peptide from peptic digests of cr-lactalbumin was sequenced as described in the following paper (5) and confirmed the alignment of T5 and T6. T5,6. Residues 17 fo 58-This 41-residue peptide was isolated as the insoluble fraction from tryptic hydrolysates of S-carboxymethyl-a-lactalbumin and gave the composition listed in Table IV. It was digested with thermolysin and the resulting digest was fractionated as shown in Fig. 1. Although not all of the thermolysin peptides were sequenced, several proved of value in deducing the amino acid sequence and, in particular, the amide distribution of Residues 17 to 57. Those thermolysin peptides analyzed were as follows. T6,6-Thi. Residues 41 fo 51: Ile-Val-Glu-Arm-Asn-Gln-XerThr(Asp , Tyr ,GZy)-Analysis of acid and enzymic hydrolysates of this peptide (Table IV) showed a total of 3 residues of asparagine plus glutamine and 1 residue each of glutamic acid and aspartic acid per molecule. The peptide was submitted to Edman degradation, as shown in Table V, to give the sequence of the first 7 residues. After five steps of Edman degradation, an aliquot of the degraded peptide was digested with aminopeptidase and the total enzymic hydrolysate gave the following composition: Asp, 1.1; Thr, 1.0; Ser, Gln and Asn, 1.7; Gly, 0.9; Tyr, 1.0. These analyses show that the peptide contains a single residue of glutamine, which must be Residue 46, because only 1 of the 2 residues of glutamic acid in the peptide is amidated (Table IV) and the single amide found in the degraded peptide cannot be asparagine because 1 residue of aspartic acid was also detected
111111 2
l
200
400
600 Volume
600 (ml)
1000
I200
2000
FIG. 1. Chromatographic separation of a thermolysin digest of T5,6. T5,6 (4 pmoles) was digested with thermolysin and the digest was chromatographed on a 0.9- X 15-cm column of Dowex 50-X8 equilibrated at 50” with 0.01 M pyridine-acetate, pH 2.0. The column was developed at a flow rate of 45 ml per hour with two successive gradients. The first gradient was formed with 500 ml each of the equilibration buffer and 0.5 M pyridine-acetate, pH 3.7. The second gradient was formed with 500 ml each of 0.5 M pyridine-acetate, pH 3.7, and 2.0 M pyridine-acetate, pH 5. The column was monitored automatically as described earlier (6).
V
TABLE
acid sequence of T5,6-Thl
iimino
The values in italics
are the molar
ratios
and the values in parentheses
are the assumed
number
of residues.
Steps of Edman degradation Amino acid composition
Amino acid
2
1
Aspartic acid. ............. Threonine .................. Serine ...................... Glutamic acid ............... Glycine, .................... Valine ...................... Isoleucine ................ Tyrosine .................... Residue Sequence.
removed.
..........................
.............................
5.0(3) 1.1(l) 1.0(l) 2.2(2) 0.8(l) 1.1(l) 1.1(l) 1.0(l)
3.0 0.9 0.8 2.1 1.0 1.1
