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Structural hierarchy in proteins

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Title: Structural hierarchy in proteins


1
Structural hierarchy in proteins
2
Color conventions
3
Protein Geometry
CORN LAW amino acid with L configuration
4
Greek alphabet
5
The Polypeptide Chain
6
Chapter 5 Covalent structures of
proteins Proteins function as 1.
Enzymesbiological catalysts 2. Regulators of
catalysis-hormones 3. Transport and store i.e.
O2, metal ions sugars, lipids, etc. 4.
Contractile assemblies Muscle
fibers Separation of chromosomes etc. 5.
Sensory Rhodopsin nerve proteins
7
6. Cellular defense immuoglobulins Antibodies
Killer T cell Receptors 7. Structural Collagen
Silk, etc. Function is dictated by protein
structure!!
8
There are four levels of protein structure
1. Primary structure 1? Amino acid sequence,
the linear order of AAs. Remember from the
N-terminus to the C-terminus Above all else this
dictates the structure and function of the
protein.
9
There are four levels of protein structure
2. Secondary structure 2? Local spatial
alignment of amino acids without regard to side
chains. Usually repeated structures Examples
a helix, b sheets, random coil, or b turns
10
3. Tertiary Structure 3? the 3 dimensional
structure of an entire peptide. Great in detail
but vague to generalize. Can reveal the detailed
chemical mechanisms of an enzyme.
11
4. Quaternary Structure 4? two or more peptide
chains associated with a protein. Spatial
arrangements of subunits.
Chapter 5.3 is how to determine a proteins
primary structure. Protein Chemistry
12
Example of each level of protein structure
13
Insulin was the first protein to be sequenced
F. Sanger won the Nobel prize for protein
sequencing. It took 10 years, many people, and
it took 100 g of protein! Today it takes one
person several days to sequence the same
insulin. 1021 AA b- glactosidase 1978
14
Steps towards protein sequencing
Above all else, purify it first!! Chapter 5.3
then 5.1 and 5.2 1. Prepare protein for
sequencing a. Determine number of chemically
different polypeptides. b. Cleave the
proteins disulfide bonds. c. Separate and
purify each subunit. d. Determine amino acid
composition for each peptide.
15
Bovine insulin note the intra- and inter- chain
disulfide linkages
16
2. Sequencing the peptide chains a. Fragment
subunits into smaller peptides ? 50 AAs in
length. b. Separate and purify the fragments c.
Determine the sequence of each fragment. d.
Repeat step 2 with different fragmentation
system.
17
3. Organize the completed structure. a. Span
cleavage points between sets of peptides
determined by each peptide sequence. b.
Elucidate disulfide bonds and modified amino
acids. At best, the automated instruments
can sequence about 50 amino acids in one
run! Proteins must be cleaved into smaller pieces
to obtain a complete sequence.
18
End Group Analysis
How many peptides in protein? Bovine insulin
should give 2 N-terminii and 2 C-terminii N-termin
us 1-Dimethylamino - naphthalene-5-sulfonyl
chloride Dansyl chloride Reacts with amines
N-terminus Lys (K) side chains
19
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20
Disadvantage with the Dansyl-chloride method is
that you must use 6M HCl to cleave off the
derivatized amino acid, this also cleaves all
other amide bonds (residues) as well.
Edman degradation with Phenyl isothiocyanate, PITC
21
Edman degradation has been automated as a method
to sequence proteins. The PTH-amino acid is
soluble in solvents that the protein is not. This
fact is used to separate the tagged amino acid
from the remaining protein, allowing the cycle of
labeling, degradation, and separation to
continue. Even with the best chemistry, the
reaction is about 98 efficient. After sufficient
cycles more than one amino acid is identified,
making the sequence determination error-prone at
longer reads.
22
Demonstration of Edman degradation
Use your CD disk- install it and run chapter 5
Edman degradation.
23
Carboxypeptidase cleavage at the C-terminus
Carboxypeptidase A Rn? R, K, P Rn-1? P
If the Tyr-Ser bond is more resistant to cleavage
than the Leu-Tyr, the Ser and the Tyr will
appear simultaneously and the C-terminus would
still be in doubt.
24
Cleavage of disulfide bonds
Permits separation of polypeptide chains Prevents
refolding back to native structure
Performic acid oxidation Changes cystine or
cysteine to Cystic acid Methionine to Methionine
sulfone 2-Mercaptoethanol, dithiothreitol, or
dithioerythritol Keeps the equilibrium towards
the reduced form -S-S-
2SH
25
Amino acid composition
The amino acid composition of a peptide chain is
determined by its complete hydrolysis followed by
the quantitative analysis of the liberated amino
acids.
Acid hydrolysis (6 N HCl) at 120 oC for 10 to 100
h destroys Trp and partially destroys Ser, Thr,
and Tyr. Also Gln and Asn yield Glu and Asp
Base hydrolysis 2 to 4 N NaOH at 100 oC for 4 - 8
h. Is problematic, destroys Cys Ser, Thr, Arg
but does not harm Trp.
26
Amino acid analyzer
In order to quantitate the amino acid residues
after hydrolysis, each must be derivatized at
about 100 efficiency to a compound that is
colored. Pre or post column derivatization can
be done.
These can be separated using HPLC in an automated
setup
27
Amino acid compositions are indicative of protein
structures
Leu, Ala,Gly, Ser, Val, Glu, and Ile are the most
common amino acids His, Met, Cys, and Trp are the
least common. Ratios of polar to non-polar amino
acids are indicative of globular or membrane
proteins. Certain structural proteins are made of
repeating peptide structures i.e. collagen.
28
Long peptides have to be broken to shorter ones
to be sequenced
Endopeptidases cleave proteins at specific sites
within the chain.
Trypsin Rn-1 positively charged residues R, K
Rn ? P Chymotrypsin Rn-1 bulky hydrophobic
residues F, W, T Rn ? P Thermolysin Rn I,
M, F, W, T, V Rn-1 ? P Endopeptidase V8 Rn-1
E
29
Specific chemical cleavage reagents
Cyanogen Bromide Rn-1 M
Cleave the large protein using i.e trypsin,
separate fragments and sequence all of them. (We
do not know the order of the fragments!!) Cleave
with a different reagent i.e. Cyanogen Bromide,
separate the fragments and sequence all of them.
Align the fragments with overlapping sequence to
get the overall sequence.
30
How to assemble a protein sequence
1. Write a blank line for each amino acid in the
sequence starting with the N-terminus. 2. Follow
logically each clue and fill in the blanks. 3.
Identify overlapping fragments and place in
sequence blanks accordingly. 4. Make sure
logically all your amino acids fit into the
logical design of the experiment. 5. Double check
your work.
31
1 2 3 4 5
6 7 8 9 10 11 12
13 14 H3N-_-_-_-_-_-_-_-_-_-_-_-_-_-_-COO

K

F - A - M - K

K - F - A - M
Q - M - K
D - I - K - Q - M
G - M - D - I - K
Y - R - G - M Y - R
Trypsin cleaves after K or R (positively charged
amino acids) Q - M - K G - M - D - I - K F - A -
M - K Y - R
Cyanogen Bromide (CN Br) Cleaves after Met
i.e M - X D - I - K - Q - M K K - F - A -
M Y - R - G - M
32
There are a variety of ways to purify peptides
All are based on the physical or chemical
properties of the protein. Size Charge Solubility
Chemical specificity Hydrophobicity/
Hydrophylicity
Reverse Phase High Pressure Liquid Chromatography
is used to separate peptide fragments.
33
Peptide mapping digest protein with an
appropriate agent, then separate using two
dimensional paper chromatography
Digested Peptide from normal (HbA) and Sickle
cell anemia (Hbs) hemoglobins HbA V - H - L - T
- P - E - E - K HbS V - H - L - T - P - V - E -
K ? 1 2 3 4 5 6 7 8 Beta
chain position 6 contains altered amino acid
34
Red blood cells (a) normal (b) sickle
cell Electrophoretic separation of hemoglobins
35
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36
Deoxyhemoglobin aggregates and deforms cell.
Primary structure changes dictate quaternary
structure. Why did the problem not die
out? Homozygotic Heterzyatic
Homozygotic normal sickle cell
trait sickle cell gets malaria
resistant gets
sickle cell to malaria
dies dies
37
Species variation in homologous proteins
The primary structures of a given protein from
related species closely resemble one another. If
one assumes, according to evolutionary theory,
that related species have evolved from a common
ancestor, it follows that each of their proteins
must have likewise evolved from the corresponding
ancestor.
A protein that is well adapted to its function,
that is, one that is not subject to significant
physiological improvement, nevertheless continues
to evolve.
Neutral drift changes not effecting function
38
Homologous proteins (evolutionarily related
proteins)
Compare protein sequences Conserved residues,
i.e invariant residues reflect chemical
necessities. Conserved substitutions,
substitutions with similar chemical properties
Asp for Glu, Lys for Arg, Ile for Val Variable
regions, no requirement for chemical reactions
etc.
39
Amino acid difference matrix for 26 species of
cytochrome c
Man,chimp 0 Rh. monkey 1 0 Average
differences Horse 12 11 0 Donkey 11 10 1 0
10.0 cow,sheep 10 9 3 2 0 dog 11 10
6 5 3 0 gray whale 10 9 5 4 2 3 0
5.1 rabbit 9 8 6 5 4 5 2 0 kangaroo 10 11
7 8 6 7 6 6 0 Chicken 13 12 11 10 9 10 9
8 12 0 penguin 13 12 12 11 10 10 9 8 10 2 0
9.9 Duck 11 10 10 9 8 8 7 6 10 3 3 0
14.3 Rattlesnake 14 15 22 21 20 21 19 18 21
19 20 17 0 12.6 turtle 15 14 11 10 9 9 8
9 11 8 8 7 22 0 Bullfrog 18 17 14 13 11 12
11 11 13 11 12 11 24 10 0 Tuna fish 21 21 19
18 17 18 17 17 18 17 18 17 26 18 15 0
18.5 worm fly 27 26 22 22 22 21 22 21 24 23 24 22
29 24 22 24 0 silk moth 31 30 29 28 27 25 27 26
28 28 27 27 31 28 29 32 14 0 25.9
Wheat 43 43 46 45 45 44 44 44 47 46 46 46 46 46
48 49 45 45 0 Bread mold 48 47 46 46 46 46 46 46
49 47 48 46 47 49 49 48 41 47 54 0 47.0
Yeast 45 45 46 45 45 45 45 45 46 46 45 46 47 49
47 47 45 47 47 41 0 Candida k. 51 51 51 50 50 49
50 50 51 51 50 51 51 53 51 48 47 47 50 42 27 0
Man,chimp monkey Horse Donkey cow,sheep dog gray
whale rabbit kangaroo Chicken, penguin Duck Rattl
esnake turtle Bullfrog Tuna fish worm
fly silkworm Wheat Bread mold Yeast Candida
40
Phylogenetic tree
Indicates the ancestral relationships among the
organisms that produced the protein. Each
branch point indicates a common
ancestor. Relative evolutionary distances between
neighboring branch points are expressed as the
number of amino acid differences per 100 residues
of the protein. PAM units or Percentage of
Accepted Mutations
41
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42
PAM values differ for different
proteins. Although DNA mutates at an assumed
constant rate. Some proteins cannot accept
mutations because the mutations kill the function
of the protein and thus are not viable.
43
Mutation rates appear constant in time
Although insects have shorter generation times
than mammals and many more rounds of replication,
the number of mutations appear to be independent
of the number of generations but dependent upon
time
Cytochrome c amino acid differences between
mammals, insects and plants note the similar
distances
44
Evolution through gene duplication
  • Many proteins within an organism have sequence
    similarities with other proteins.
  • These are called gene or protein families.
  • The relatedness among members of a family can
    vary greatly.
  • These families arise by gene duplication.
  • Once duplicated, individual genes can mutate into
    separate genes.
  • Duplicated genes may vary in their chemical
    properties due to mutations.
  • These duplicate genes evolve with different
    properties.
  • Example the globin family.

45
  • Hemoglobin
  • is an oxygen transport protein
  • it must bind and release oxygen as the cells
    require oxygen
  • Myoglobin
  • is an oxygen storage protein
  • it binds oxygen tightly and releases it when
    oxygen concentrations are very low

46
The globin family history
1. Primordial globin gene acted as an
Oxygen-storage protein. 2. Duplication occurred
1.1 billion years ago. lower oxygen-binding
affinity, monomeric protein. 3. Developed a
tetrameric structure two a and two b chains
increased oxygen transport capabilities. 4.
Mammals have fetal hemoglobin with a variant b
chain i.e. g (a2g2). 5. Human embryos contain
another hemoglobin ?2e2. 6. Primates also have a
d chain with no known unique function.
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48
Protein Evolution is not organismal evolution
  • Chimpanzee ? human are about 99 the same amino
    acid sequences in proteins!
  • However
  • Rapid divergence with few mutational changes
    suggest altered control of gene expression.
  • Controlling the amount, where, and when a protein
    is made.
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