Note: This is a brief outline of the topics we have covered that could be tested on the exam. This does NOT mean they WILL be tested. Also, little detail is given here. Consult your notes, the book, or the articles we have used for that. Study for understanding, not memorization. Therefore, do not solely rely on this review sheet.
I. Bonds Important for Molecular Biology and the
Influence of Water on These Bonds
A. covalent, strong, equal
sharing. Same in water or not.
B. ionic, strong unless in the
presence of water, then weak. Forms positive and negative ions.
C. hydrogen, H bonded to an
electronegative
atom (O, N). Induces a dipole with a slight positive H and a
slight negative O (or N as the case may be). Water participates in
H-bonds with other molecules that are polar.
D. hydrophobic interactions,
nonpolar
regions of molecules interact in order to reduce unfavorable entropy
changes
in the presence of water.
E. van der Waall's forces, caused
by electrons moving in their orbitals. Causes transient attractions and
repulsions
that are very weak but collectively contribute to the stability of
macromolecular
structures.
F. We also discussed the importance of spatial
orientation in determining molecular function also, using
polysaccharides (cellulose, from beta glucose monomers
vs. starch/glycogen from
alpha glucose monomers) as on
example and phospholipids
(bylayers in membranes) as another. Reviewed condensation reactions and
the formation of glycosidic bonds and triglyceride
formation also..
II. Amino Acid and Protein Structure
A. Amino
Acid Structure
1. alpha-carbon bonded to H, NH2, COOH and a side chain, R.
NH2 behaves like a base and becomes
positively charged at physiological pH; COOH behaves like an acid and
becomes
negatively charged.
Called a zwitterion.
2. R determines the classification of amino acids,
based
on chemical characteristics
a. hydrophobic
b. polar, but not charged
c. basic, + charged
d. acidic, - charged
e. special: glycine, proline (imino acid), cysteine (can
form S-S). Our books lists these as nonpolar.
3. exists in 2 possible stereoisomeric forms, D and L
a. only L used in cells.
B. Primary
Structure of Polypeptides
1. The peptide bond
a. resonance and planarity; omega angle, psi angle,
phi angle.
b. formed by condensation reactions between the COOH
of one amino acid
and the NH2 of the next
c. Free NH2 end is called the amino terminus; free
COOH
end is the carboxy terminus
d. Sequence of amino acids is called the primary
structure.
C.
Secondary
Structure
1. alpha-helix
a. stabilized by H-bonds between one amino acid
peptide
linkages four peptide linkages apart.
b. R chains project to the outside of the helix.
2. beta-pleated sheet
a. linear arrays of amino acids (beta strands).
Stabilized by
hydrogen
bonding between peptide bonds in different strands of the beta sheet.
Pleated due to maintaining the planarity of the peptide
bonds.
b. R chains project above or below the pleated sheet
c. can run in parallel or in antiparallel arrays.
antiparallel H-bonds linear and a little stronger. Mixed sheets also
exist.
D.
Tertiary
Structure
1. considers all possible interactions of the
side-chains
of the amino acids
2. stabilized by hydrogen bonds, ionic interactions,
hydrophobic interaction, and
van der Waal's forces.
3. disulfide linkages between cysteines can also
influence
the tertiary structure
4. creates structural domains.
5. can also characterize functional domains that
participate
in particular protein functions.
E.
Quarternary
Structure
1. interacting subunits needed for a protein to
function.
(same or different)
. Example: hemoglobin. 2 alpha and 2 beta subunits.
F.
Prosthetic
groups (e.g. heme in hemoglobin) often are part of active sites of
enzymes
or functional parts of
proteins.
They help provide the chemical environment for appropriate function.
III. Protein Folding and Function
A. Chaperones and
Chaperonins
1.
Chaperone proteins bind to exposed hydrophobic side chains on proteins
to prevent
their aggregation until the protein is fully translated.
a. They then use the energy of ATP to release from
the
protein which then can
fold correctly.
b. They also function to help denatured proteins
refold
correctly.
c. Multi step mechanism. Binds hydrophobic side-chain
when ATP bound to the chaperone; second chaperone causes hydrolysis of
ATP
to ADP and the first chaperone changes shape to sequester
the polypeptide and allow it to fold; third protein forces out ADP from
first chaperone and ATP replaces it. Refolded protein is
released.
2.
Chaperonins
are needed to help some proteins fold correctly.
a. They sequester the partially folded protein inside
a barrel-like structure. Then a cap structure covers the barrel
allowing the protein time to fold.
b. They then release it, using the energy of ATP.
B. Protein degredation
1. Some
short-lived proteins or misfolded proteins are marked for degredation
by sequences that are
recognized
by ubiquitinylating enzymes E1, E2, and E3.
a. These enzymes attach a polymer of repeating
ubiquitins
to the amino group attached
tothe side chain of lysine residues in the protein.
First E1 activates
an ubiquitin
by attaching to it through an E1 cysteine side chain. Then it transfers
the ubiquitin to the E2E3 complex (ubiquitin ligase) which then
attaches the ubiquitin to the lysine
described above. This process repeats again and again forming the
polyubiquitin signal.
b. Other factors then bind these ubiquitins and bring
the protein to a proteosome.
c. Within the core of the proteosome, the protein is
degraded by proteases.
C. Importance of 3-D structure to
protein
function
1. The
3D structure of proteins allows the proper chemistry at the bindins
site
or active
site of proteins.
2. One example was the
serine protease family.. Enzymes lower activation energy of reactions.
Do
not effect the delta G.
a. binding pocket for polypeptide determines the
specificity of a particular serine protease.
b. active site has asp, his, and ser. Charge relay
pulls a proton from his to asp. Proton then is pulled from ser to his.
c. activated ser then forms first transition state
with substrate.
3. Lysozyme
example of lowering activation energy.
a. Binding of oligosaccharide chain to lysozyme binding
site induces strain and distortion in the sugar D.
b. Glutamic acid attacks the glycosidic bond while aspartic
acid attackes the C 1 of sugar D. This breaks the glycosidic bond and
covalently attaches
the asp to the C1.
c. Glu causes water to give its H to Glu's COOH group. The
remaining OH from the water then attaches itself to the C1 of sugar D,
causing the
release of asp. Reaction completed. Enzyme back to its
original form.
D. Allosteric regulation
1.
Allosteric
enzymes have both regulatory and catalytic subunits.They exist in
either
active or inactive configurations that are in equilibrium unless bound
by regulators.
2. When
regulators bind to the regulatory subunits of one form, that form is
stabilized,
shifting the equilibrium..
3. Usually
used to regulate metabolic pathways.
4. These
enzymes have quaternary structure.
5.
Single-subunit enzymes can also work similarly, where a regulatory
molecule can favor binding to the active form of an enzyme and
stabilize the shape of
the enzyme that favors substrate
binding (positive regulation) or can stabilize the inactive form and
thus make it less likely for the substrate to bind the
active form
(negative regulation.)
6.
These differ from competitive inhibition which has a similarly
structured competitior molecule that can bind the active site in place
of the substrate if the
relative concentrations of competitor and
substrate are altered.
IV. DNA Structure
A. The nucleotide
1.
pyrimidines
and purines in DNA and RNA (know which)
2. Ribose
in RNA; deoxyribose in DNA
3.
nucleoside
monophosphates, diphosphates, triphosphates.
B. The DNA polymer
1. The
phosphodiester bond
2. 5'end
defined
3. 3'end
defined
C. B-DNA
1.
double-stranded
2. two
antiparallel alpha helices
3.
sugar-phosphates
to the outside
4. bases
to the inside
5. bases
hydrogen bond: A and T; C and G
6. bases
of a strand also interact by hydrophobic interactions
7. about
10 base pairs per turn of the helix
8. major
and minor grooves
D. Evidence for B-DNA structure
1. X-ray
crystallography patterns.
2.
Chargaff's
experiments: conc. of A equals T; conc. of C equals G in all DNAs
3. Linus
Pauling's description of protein alpha-helix stabilized by hydrogen
bonds
4. DNA
loses viscosity when heated (hydrogen bonds involved); density
suggested
two strands.
E. Other structures exist in cells
also.
F. Chromatin structure
1.
Nucleosomes
a. Histone octamer (2 each of H2A, H2B, H3, H4) with
146 base pairs of DNA wound around twice
2. Linkers
of DNA connecting nucleosomes, average of 200 base pairs, variable.
3. HI
bound to linker regions
a. helps fold the chromatin into a 30nm fiber or
solenoid
V. DNA Replication
A. Overall
1.
Semi-conservative
(one parental template strand and one new daughter strand become
double-stranded
DNA molecule.
2. DNA
polymerase is the major enzyme involved.
B. Properties of DNA polymerases
1. Require
a pre-existing 3'OH to which to attach the incoming dNTP.
2. Need
a template strand
a. base-pairs the incoming nucleotide to the template
strand to select the correct one.
3. Can
have three activities
a domain where the 5'-3' polymerizing reaction is
catalyzed
b. domain where the 3'-5' exonuclease activity is
- involved in the proofreading mechanism that removes incorrectly added
nucleotides
c. domain where the 5'-3' exonuclease activity is
- involved in removing the RNA primers and in DNA repair mechanisms. In
eukaryotes, this activity
may be provided by a separate molecule.
C. Types of DNA polymerases
1.
Eukaryotic
a. DNA polymerase alpha: begins the synthesis of the
Okazaki fragments and first leading strands.
b. DNA polymerase delta: synthesizes the leading
strand
and most of the lagging strand.
c. DNA polymerase beta: works during DNA repair
D. Replication Fork
1. New
DNA polymers grow 5'-3'
2. Leading
strand grows continuously
3. Lagging
strand grows discontinuously using Okazaki fragments
4. DNA
helicase melts the hydrogen bonds between the bases to open the fork
5. Primase
makes a short RNA primer to begin each new strand.
6. DNA
ligase connects the Okazaki fragments by making the last, untemplated,
phosphodiester bond.
7. Ligase
reaction is two-step reaction.
a. Pyrophosphate removed from ATP and AMP attached to
the 5'P..
b. AMP removed and phosphodiester bond made between
the
5'P and the 3'OH of the other nucleotide.
8.
Topoisomerase
works ahead of the replication fork, relieving torsional strain
a. Covalently links to the phosphate of one of the
phosphodiester
bonds, thus breaking that DNA polymer
temporarily
b. Helix unwinds through the nick, relieving the
strain.
c. Topoisomerase then reforms the original
phosphodiester
bond.
9.
Single-stranded
DNA binding proteins (RPA) attach to the lagging strand as DNA helicase
melts the hydrogen
bonds, keeping the region single stranded until the replication
apparatus
moves through.
10. PCNA
and RPC work with DNA polymerase delta on the leading strand to
increase
its processivity, allowing
the synthesis to be continuous.
11. It
is believed that DNA polymerase delta actually proofreads both strands
during DNA replication.
12. In
three dimensions, the lagging strand winds around the DNA polymerase to
facilitate the priming of new
Okazaki fragments.
13. We
believe that the primase function associates with DNA polymerase alpha.
E. Initiation of DNA replication
1. occurs
at origins of replication
a. palindromic sequences that interact with initiator
protein.
2.
initiator
protein melts the double helix at the origin of replication allowing
the
helicase- primase-DNA
polymerase alpha complex to load on
3. A
replication
fork begins, moving in one direction from the origin
4.
Shortly,
a second replication fork begins in the opposite direction.
5. DNA
polymerase delta quickly replaces alpha on the leading strands of both
replication forks once the
priming and initial synthesis is done.
6.
This is called bidirectional DNA synthesis and creates replication
bubbles
7.
Replication
continues until two oppositely moving bubbles meet each other.
8. The
ends of eukaryotic DNA chromosomes must be copied by an unusual
reaction
involving DNA telomerase (a type of reverse transcriptase).
a. telomerase contains an RNA with base sequence
complementary
to the telomeres.
b. telomerase adds DNA nucleotides using its RNA as a
template, up to position 35, then displaces the
upstream section of the copy and reanneals to the RNA template.
c. after displacement of the newly made section, the
process repeats sequentially, extending the telomere.
IV. DNA repair
1.
nucleotide
excision repair
in mammals:
a. used to remove bulky lesions such as thymine
dimers(e.g.covalently linked thymines on one strand).
b. proteins recognize and bind the
lesion,
initially distorting the double-helix
c. TFIIH and associated helicases binds and uses its
helicase activity to open
up the helix surrounding the lesion.
d. RPA and XP-G bind, further stabilizing the opened
region.
e. XP-G and XP-F endonucleases cut on either side of
the lesion, releasing a fragment 24-32 bp.
f. DNA polymerase and ligase fill in the gap and
complete
the last bond.
2. base excision repair
a. used when spontaneous deaminations occur.
b. Recognized by a DNA glycosylase that breaks the
N-glycosyl
bond between the base and the C1 of the
ribose.
c. Then APE1 endonuclease cuts the
phosphodiester
bond to the 5' side of the lesion.
d. Then the Deoxyribose phosphodiesterase activity
removes remaining deoxyribose.
e. Then DNA polymeras and ligase fill in the
gap
and complete the last bond.
3.
Mismatch repair
a. Used to repair mismatched base pairs shortly
after DNA replication.
b. In bacteria, MutS and MutL mark the mismatch on
the parental strand by recognizing its methylation. In eukaryotes it
may be recognizing the strand
that contains many nicks due to Okazaki fragments
that have not yet been joined.
Then MutH cleaves the newly
replicated strand and MutS, Mut L, helicase and an exocnuclease chew
away the strand, including the mismatched
region.
c.
Gap filled in by DNA polymerase and ligase.
4. Repair of double-stranded DNA breaks.
a.
Non-homologous recombination method that removes any overhanging
nucleotides, producing blunt ends that are ligated together. Deletions
result.
b. Homologous
recombination method that uses the information on a sister chromatid to
replace the damaged region accurately. Can only occur during cell
cycle stages that have a sister chromatid present.