exam 1 study guide.2

8/12/2019 Exam 1 Study Guide.2

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BIOCHEMISTRY

1. GENETIC MATERIAL

1. TO DESCRIBE THE CHEMICAL STRUCTURE OF DNAAND RNA,INCLUDING THE MAJOR ANDMINOR GROOVES,THE N-GLYCOSIDIC LINKAGE,THE DIFFERENCE BETWEEN RIBOSE ANDDEOXY-RIBOSE,THE PHOSPHODIESTER LINKAGE,THE POLARITY IN THE LINEAR STRUCTURE

OF POLYNUCLEOTIDES,THE ANTIPARALLEL ARRANGEMENT OF THE TWO STRANDS IN THE

DNADOUBLE HELIX AND THE NATURE OF THE INTERACTIONS BETWEEN SPECIFIC BASE

PAIRS IN THE DNA DOUBLE HELIX

a. A polymer of 2 deoxyribonucleotides connected via 3 to5phosphodiester linkages.

i.Nucleotides: nitrogenous bases in DNA; derivatives ofeither:

1. Purines (adenine or guanine)2. Pyrimidines (cytosine or thymine)

ii.Nucleotides have three components1. Bases linked via an N-glycosidic bondto..2. A 5 carbon cyclic sugar

a. Nucleoside= base + sugar3. A phosphate esterifiedto the hydroxyl on

carbon 5 of the sugar.

a. Nucleotide= base + sugar + phosphateiii.The polynucleotide chain has a 5 (phosphate) and a 3

(Hydroxyl) terminus1. p to the left of a base indicates 5 linkage, to the right of a

base indicates 3 linkage2. phosphate makes nucleic highly charged at physiologicalpH

iv.DNA is written from 5 base to 3 base (direction of synthesis; newbases are added to 3 side)

v.Double helix: two polynucleotide chains wound about each other inantiparallel manner

1. Hydrophobic bases oriented inside, sugar and phosphate onoutside

2. Held together by hydrogen bonds between bases of oppositestrands

a. Weak ionic non-covalent interactioni. Oxygen is hydrogen donor; N is acceptor

b. Complementary base pairingi. AT 2 hydrogen bonds

ii. GC 3 hydrogen bonds3. Structure suggests how DNA prevents information loss when damage occurs; the

strand containing the damage can be corrected using the complementary strand

b. The 2 position differentiates DNAfrom RNA1. In RNA there is an OHin place of the H


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2. TO OUTLINE BASIC ASPECTS OF DNATOPOLOGYa. Supercoiled DNA: when DNA becomes twisted around its own axis

i. negative supercoiling: twisted about an axis in direction opposite tointrinsic turns of right-handed double helixii. when the phosphodiester backbone is broken, the ends are free to rotate and the confirmationis changed to relaxed

b. Topoisomerases: enzymes that catalyze changes in DNA supercoiling(relaxes)

i. Catalyze a three step process:1. Cleavage

a. Topo I cleaves just one strand of DNAb. Topo II cleaves both strands

2. Passage of DNA strand(s) through the break3. Resealing of the DNA break

ii. Many antitumor drugs act as specific poisons for DNAtopoisomerases; inhibiting topo inhibits cell growth

1. Partially selectively hitting cancer cells since they are much more active than normalcells

2. Camptothecin targets Topo I3. M-AMSA targets Topo II

3. TO DESCRIBE ASPECTS OF THE ORGANIZATION OF DNAIN EUKARYOTIC GENOMES AND TOBEGIN TO APPRECIATE THE LINK BETWEEN DISEASES AND ALTERATIONS AT THE DNA

LEVEL.

a. Melting temperature of DNAi.

More stable(higher Tmvalues) when rich in GC (due to the 3hydrogen bonds)

ii. When cooled, complementary strands of DNA reanneal (hybridizeinto complementary double strands)

1. Highly specific; takes place only if base sequences arecomplementary

2. Basis for hybridization of nucleic acid probesa. Probes are single stranded segments of DNA (or RNA) that bind to a

nucleic acid of interest via complementary base pairing; usually labeled(radioactive, fluorescent)

b. Organization of Eukaryotic DNAi. Human haploid genome(23 chromosomes, one copy of each autosome and a single

sex chromosome)

contains roughly about 3 x 10

9

base pairs

ii. Functional genes constitute ~1% of human genome1. Are not randomly distributed

a. Genes associated into gene families are thought to have arisen byduplication and variation from an ancestral gene (i.e., -globin genecluster)

2. Genes encoding abundant products are often tandemly repeated and clusteredtogether

iii. Additional genomic elements (JUNK DNA)


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1. Pseudogenes: closely related to functional genes and arose bythe same duplication processes that generate gene families but

they no longer code for normal gene productsa. Processed Pseudogenes: formed when DNA copies

(reverse transcripts) of RNA molecules are inserted

backinto an organisms genomeb. Exons: encode; sequences converted into mRNAc. Copy of gene missing introns means reverse

transcriptase was involved

2. Proviruses: DNA copies of retroviruses inserted intochromosomes

a.

~8% of human genome is derived from proviruses (containing ~1000ancient viruses)

3. Repetitive DNA sequences:repeat sequences can be widelydispersed or tandemly arrayed

a. Transposable elements: sequences capable of insertingcopies of themselves into new genomic locations

i. Elements that transpose through reinsertion ofthe products of reverse transcriptase

1. SINEs: Short, interspersed repeatelements (e.g. Alu repeats)

a. ~10% of human genome2. LINEs: Long

a. Encode reverse transcriptaseb. ~20% of genome

ii. Transposase elements: not active (fossils);transpose directly through DNA copies and

encode their own transposase

1. ~ 3% of genome Elements thatb. Simple Sequence Repeats (SSRs): tandem repeats;

comprise ~3% of human genome

RNAase eats up the RNA strand, leftwith just DNA

DNA polymerase makes double standedDNA

Integrase integrates into the genome; ifits a sex cell than it can be passed to the

next generation


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i. Minisatellites1. Longer repeat units (14-500 bp)

ii. Microsatellites1. Short (2-5 bp) repeats present at

thousands of loci in genome

c.

Satellite sequences: highly repetitive, short DNAsequencesi. Large clusters in non-transcribed regions of chromosomes

(centromeres & the Y chromosome of telomeres)

c. Single Nucleotide Polymorphism (SNPs)i. Serve as Biological markers

d. Telomeresi. A satellite sequence consisting of a hexameric element (TTAGGG)

organized in tandem arrays, found at the ends of every chromosome

ii. Functions in chromosome protection and replicationiii. Mitotic clock: shortening of the length of telomeres with each cell

division

1. Sufficiently short telomere may be signal for senescence in normal cells2. Maintenance of telomere length is required for cells to proliferate indefinitely(Telomerase is active in immortal cells)

iv. Synthesized by enzyme telomerase(a reverse transcriptasetakesRNA template to use RNA to get the sequence of the telomere)

1. Seems to be repressed in somatic tissues but reactivated intumor cells

4. TO DESCRIBE THE STRUCTURE OF THE NUCLEOSOMEa. Multiple levels of DNA packaging

i. DNA + Histones nucleosomes (DNA+protein) chromatin fiberchromatid

2. DNA REPLICATION, REPAIR AND MUTAGENESIS.

1. TO DESCRIBE THE MODE OF ACTION OF THE ENZYMES INVOLVED IN DNAREPLICATION.a. Initiation: at Origin of Replication

i. Helicase:an enzyme that disrupts the hydrogen bonds holdingduplex DNA together using the energy of ATP hydrolysis

1. T-ag is used by SV40 at the replication fork2. Defect Syndromes:

a. Werners: Premature agingi. Associated gene appears to encode a helicase (defect of

helicase)ii. Dwarfism, cataracts, scleroderma like skin changes, alopecia

(baldness)

b. Blooms: increased risk of malignanciesi. Deficient cellular helicase

ii. RPA(Single Strand Binding protein)


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1. Stabilizes single-stranded DNA (once it is separate, it is coded by RPR soit cannot reanneal into the duplex)

iii. Topoisomerase I1. Changes topology of DNA; unwinds supercoiling

2. TO DESCRIBE THE MECHANISM OF DNAREPLICATION,INCLUDING THE ROLE(S)OF ORIGINSOF REPLICATION,THE NEED FOR SHORT RNAPRIMERS DURING DNASYNTHESIS AT THE

GROWING FORK AND THE NATURE OF LEADING AND LAGGING STRANDS

Elongation

a. DNA primase(Pol -primase) synthesizes a primer(~10 nt of RNArequired to start synthesis)

i. A primer is necessary because DNA polymerases cannot initiate DNA synthesesb. DNA polymerases

i. Need the push of an RNA primer to start1. use to be an RNA world, not a DNA world

ii. All polymerases move in 5 to 3 directioniii. Catalyzes 3 OH group nucleophilic attack on next nt5 Phosphate Creates phosphodiester bond and splitting out PPiiv. Types:

1. Pol /a. Synthesizes 5 3 but

with a 3-5

exonuclease forproofreading

b. high fidelitypolymerases

2. Pola. Synthesizes RNA

primersfor pol /3. Pol

a. For mitochondrial DNAsynthesis

4. Pola. For repairmechanisms

5. Lagging Stranda. Okazaki fragments 5-3

b. Primers are removed and filled in by pol /c. Replication fork:loads polymerases onto DNA using RFC (Replication

Factor C)

i. PCNA:clamps the polymerase onto the DNAii. Ligase:fills in the nicks created from filling in gaps from the RNA

primers on the lagging strand

d. Leading strand: formed bypol moving in 5 to 3 directioni. Very few errors are made due to 3 to 5 exonuclease proofreading

activity of Pol e. Lagging strand: second nascent strand


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i. Strand grows from 3 to 5 by polymerasemoving in oppositedirection of growing fork

ii. Synthesized in fragments (using the Okazaki fragments) in the 5 to3 direction

iii. Pol is only polymerase known to be associated with a primase1. Pol -primase complex plays critical role in initiation of every Okazaki fragment;

required to initiate leading strand synthesisiv. After RNA primer is removed via 5 to 3 exonuclease, the gaps

between Okazaki fragments are filled in by Pol

v. The Okazaki fragments are covalently linked together via DNAligase

3. TO EXPLAIN WHY THE 3'-EXONUCLEASE ACTIVITY OF CERTAIN DNAPOLYMERASE(S)ISIMPORTANT FOR THE FIDELITY OF THE REPLICATION PROCESS .

a. Role of the 3'-to-5' exonuclease is to EDIT DNA (remove incorrectlypolymerized nucleotides)

i. Increases the accuracy or fidelity of DNA synthesis by a factor of 10to 1000.b. Errors due to incorporation of the wrong bases by the DNA polymerase are low because it must escape

screening by two systems:

i. The base pairing rules recognized by the 5' to 3' catalytic siteii. As well as the 3' to 5' exonuclease proofreading site

4. TO OUTLINE BASIC CONCEPTS ABOUT THE CELL CYCLE.


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a. Interphasei. G1: Begins immediately after division

ii. S: DNA synthesisiii. G2iv. G0 (quiescent): Not actively cycling (e.g. neurons)v. Driven by cyclins & kinases (Cdk)

b. Mitosisc. Tumor suppressors and apoptosis

i. P53 functions as a tumor suppressor at the G1/S checkpoint1. Helps prevent cells from entering S phase if they contain

damaged DNA

2. Normal p53 stops cell cycle and tries to repair DNA; if itcannot repair it then undergoes apoptosis

3. >50% of people with cancer have a mutation in P53a. Lack functional P53 and G1/S checkpoint may break down

b. DNA that shouldnt be entering replication is not stoppedc. Leads to no cell growth arrest, inefficient DNA repair, genetic instability,

and malignant potential

4. Li-Fraumenti Syndromea. Inherit mutated version of p53 tumor suppressor gene

and yield tumors at multiple sites

5. TO OUTLINE BASIC FEATURES OF THE DNAREPAIR PROCESSES AND TO ILLUSTRATE THERELATIONSHIP BETWEEN DNAREPAIR AND DISEASE.

a. DNA Repairi. Mismatch repair

1. (In E.Coli) parent strand is methylated, so unmethylated strandhas the mismatch (since newly replicated DNA is

unmethylated)2. MutS recognizes GATC site

a. Human homologue: hMSH3. Mut H incises strand


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4. Mut L interfacesa. Human homologue: hMLH

5. Following incision, entire incorrect strand is removed(DNAhelicase & exonuclease) and resynthesized by DNA

polymerase (III) and ligase

6.

Defects can increase risk of hereditary non-polyposis cancer(HNPCC); increased mutation ratea. Defects may be related to expansion of trinucleotide repeats

7. This mechanism stabilizes trinucleotide repeats like CAG inHuntingtons

a. More CAG repeats = earlier onset of Huntingtonsii. Excision repair:

1. Four distinct steps:a. Incision

i. Specific for type of damage presentb. Excisionc. Resynthesis

i. DNA polymerase I (pol I) uses the 3 OHend of the nicked strand as a primerd. Ligation

i. The 3 end of the newly synthesized DNAis joined to the original DNA strand byligase

2. Proteins required for replication and synthesis are alsorequired for excision repair

3. Bulky lesion generated on DNA by sun damagea. Pyrimidine dimers form which do not fit into double

helix and thus, block gene expression

b. Recognized by uvrABC in E. coli4. Xeroderma Pigmentosum

a. Skin cancer correlated with defectiveexcision repair

b. Defect is in the nuclease that cleaves DNAnear pyrimidine dimers

b. Removal of Uracil from DNAi. Cytosine spontaneously deaminates to form uracil

1. Potentially mutagenic since uracil base pairs with adenine and newlyreplicated DNA will contain an AU base pair rather than the originalGC pair

ii. This mutation is prevented by a repair system using theuracil-DNA glycosidaseenzyme to recognize uracil inDNA and creates an AP (apyrimidinic) site where an APendonuclease recognizes the AP site and nicks the

backbone adjacent to the missing base

1. This nick triggers additional repair events similar toexcision repair

c. Trinucleotide repeats and disease


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2. Endonuclease: cleaves at the non-methylated sequence on theinvading DNA

a. Bacterial DNA is rapidly methylated after synthesis, allowing it todiscriminate between self and invading DNA

ii.Restriction endonuclease cleavage sites are palindromic (same reading left to right and right toleft)

b. Gel Electrophoresis: Electrophoretic Analysis of restriction fragmentsi.Cut DNA can be sorted by lengthii.Restriction endonucleases generate a very large number of restriction fragments (can be on the

order of 106fragments with a range of 0.520 x 103bp)1. Average gene is a few kb long (103bp = kb) so many genes will be located on a

single fragment, which can be separated using pulsed-gel electrophoresis

c. Southern blotsi.Used to identify DNA fragment(s)

1. Can identify restriction sites and also deletion & insertion mutations if they alter arestriction site (as in hemoglobin S)

2. Can provide a restriction map showing the distribution of various restriction sites ona gene and the size of the restriction fragmentsii.Run gel, develop with radioactive DNA probe

iii.E.g. fragment of DNA containing four cleavage sites for some restriction endonuclease X1. After restriction with X and gel electrophoresis, the three fragments will be located

at different positions in the gel (based on size)

2. Whole contents of the gel are transferred, by blotting, onto a membrane that has theproperty of binding DNA very tightly

3. During the transfer, DNA is denatured (via sodium hydroxide) so the single strandsare bound to the membrane

4. Fragments bearing sequence of interest are recognized by hybridization with a 32P-labeled single stranded DNA or RNA probe

a. Reaction involves complementary base pair formation between thedenatured restriction fragments on the membrane and the radioactive DNA

of the probeb. Therefore, only the fragment bearing the specific DNA sequences under

study will become labeledc. (Probes are frequently derived from cloned DNA)

iv.Point mutations1. Can be detected on Southern blots if the base substitution or

deletion is in a sequence that is recognized by a restriction

enzyme2. E.g. Sickle Cell Anemia

a. Mutation leading to hemoglobin S involves a change in codon # 6 of the globin gene from GAG to GTG (A T), resulting in loss of a restriction

site.

b. Since the mutation affects only one of the alleles (one of severalalternative forms of a gene at a given locus) in a heterozygote (an

individual with different alleles) the DNA will show the normal fragment,

as well as the mutant fragment.

3. This is a method of di rect detection of a poin t mutation thatgives rise to a disease (versus RFLPs indi rectly)

d. Restriction fragment length polymorphisms (RFLPs)i.Detects restriction site polymorphisms

ii.This technique requires no knowledge of the location or nature of thedisease-producing mutation


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e. Allele-specific oligonucleotide probes(ASO)i.Can identify mutations not interfering with restriction sites

f. Fluorescence in situ hybridization (FISH)i.Used to check for the presence, absence, or copy of a chromosomeii.Similar to Southern blot except not running gel

iii.Probe hybridizes to certain chromosomes based on complementary basepairing

g. Northern blotsi.Used to detect RNA molecules (usually the size and amount of specific

mRNA [carry genetic message from chromosomes to ribosomes])ii.RNA is separated by size via electrophoresis through a denaturing gel

1. Denaturing gel keeps RNA single stranded due to the disruption of hydrogen bondsand hydrophobic interactions

iii.Then transferred to a solid support (nitrocellulose membrane)iv.RNA of interest is localized by hybridization with radiolabeled single stranded DNA or RNA,

followed by autoradiography

h. Western blotsi.Detects proteins (size & amount)

1. This technique is to proteins what Southern blotting is to DNA andNorthern blotting is to RNA

ii.Proteins are separated on SDS-polyacrylamide gel1. SDS is an anionic detergent that denatures proteins and gives them all an uniform

negative charge can then separate based on size (small proteins move rapidly,

larger ones remain in upper sections)

iii.Gel-fractioned proteins then transferred to a membrane (polymer sheet) and probedi. Reliance:

i.Western blots rely on protein-protein interaction as probesii.Southern and Northern blots rely on nucleic acid- nucleic acid interaction

j. Studying Protein-DNA Interactionsi.DNA Band Shifts

1. Used to determine if a protein interacts with a particular DNAfragment

2. Run DNA and DNA+Protein on gel, if bound then there will bea band shift due to increased molecular weight changing the

mobility

ii.DNase I Footprinting Assay1. Used to establish where a given protein binds to DNA2. DNA asymmetrically labeled (at one end) and DNAse I is

added (which digests DNA)3. Regions of DNA covered by a bound protein will be protected

from digestion; endonuclease will cut before and after the

proteins location, resulting in reveal of where proteinis bound


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4. Band corresponding to cleavage at these points will be absent2. TO DESCRIBE THE POLYMERASE CHAIN REACTION .

a. Used to geometrically amplify up and get a lot of DNA of interest; canamplify just the region of interest

b. Repetitive cycling of three reactions:i. Denaturation (heat)

1. The resulting single stranded DNA fragments can potentially reanneal to any DNAmolecule that contains complementary nucleotides

ii. Annealing1. Upon cooling, two oligonucleotides (a.k.a. primers) are annealed to their

complementary sequences on opposite strands of the denature SS DNA2. The pair of oligonucleotides defines the limit of the region of DNA undergoing

amplification

a. Need to know the sequence of the primers that flank the segment ofDNA you want to amplify

iii. Extension1. Synthesis of new strands of DNA complementary to the parental strands2. Catalyzed by a heat stable DNA polymerase3. Newly synthesized DNA strands contain primers at their 5 ends, followed by a run

of nucleotides that are complementary to parental templateiv. The whole process is repeated, and all the previously synthesized products act as templates for

new primer-extension reactions in subsequent cyclesv. In general there is a very low probability of primers finding a second region of homology

vi. Fast and convenient method for characterizing small DNA samples1. Technical limitation because PCR can only amplify up to

5000 base pairs between primers, so larger mutation will not

show

vii. RT-PCR: modification


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1. Amplifies RNA sample by using a RTase (reversetranscriptase) to make cDNA, than PCR as usual

2. E.g. BCR-Abl fusion proteinis expressed as mRNA in leukemia cellsa. Can search for mRNA to see if any leukemia cells are present and

expressing the BCR-Abl cancerous protein in body

c. Single Strand Conformation Polymorphism(SSCP)i. Mutations cause different conformations in ssDNAii. PCR products are chemically separated into single strands using formamide and run on a

native polyacrylamide gel

iii. ssDNA migrates in a gel according to its conformation1. If a mutation is present, the conformation is altered and the

band pattern will differ from the normal PCRproduct

3. TO DESCRIBE DNASEQUENCING.Sanger Sequencing

a. Dideoxy analog is a chain terminator (no 2 and 3 hydroxyl)i. The 3 hydroxyl group is required to catalyze a nucleophilic attack during phosphodiester

bond formationb. By adding labeled ddNTPs at low concentrations to dNTPs and DNA

polymerase I, can sequence segment of DNA of interest on gelc. E.g. End up with a population of molecules where the strands were stopped opposite a G

i. First position past primer had to be a Gii. The next position has to be another G

iii. Helps figure out the sequence when you repeat it with different base pairs4. TO OUTLINE A BASIC UNDERSTANDING OF GENE CLONING

a. Insertion of restriction fragments into self-replicating vectorsi. Vector and insert have compatible sticky ends (ends that can base pair

owing to complementary sequences)ii. Ligate insert into recombinant plasmid vector via restriction sites +

ligase

iii. Grow bacteria with new plasmidb. Replication of recombinant plasmidsin bacteria

i. Only cells infected with recombinant plasmids (that confer drugresistance and therefore a growth advantage) will be able to survive ina medium containing the antibiotic being selected against

1. Surviving cells generate colonies derived from single cells(clones)

c. DNA librariesi. Large population of bacteria containing plasmids with DNA of an

organismii. Genomic library: constructed directly from genomic DNA and used to

isolate specific fragments of genomic DNA


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iii. cDNA library: contains only cDNA derived from mRNA (the mRNA isconverted to DNA by reverse transcriptase and then cloned)

4. RNA SYNTHESIS

Background information: RNA differs from DNA in two respects:

The pentose sugar in RNA is ribose The thymine (T) (5-methyluracil) is replaced by uracil (U)in RNA

RNA cannot form a double helix of the B-DNA type because of steric interference by the 2 hydroxylgroups of its ribose units

RNA can have the same sequence as a DNA coding strand RNA and DNA chains will form hybrid double strands if their sequences are complementary. Adenine forms a base pair with uracil just as well as with thymine

1. TO DEFINE WHAT IS MEANT BY A TRANSCRIPTION UNIT AND BY A PRIMARY TRANSCRIPT.a. Transcription unit

i.A DNA coding (+) sequence is read by RNA polymerase, whichproduces a complementary, antiparallel RNA strand

ii.Mechanism: ribonucleotides derived from nucleoside triphosphates are added sequentially to the3 end of a growing chain

iii.RNA polymerase catalyzes process of RNA detaching from template strand1. Unlike DNA, RNA polymerases are capable of initiating new chains

iv.The rRNA transcription unit is transcribed as a single RNA chainb. Primary transcript

i.An RNA molecule that has not undergone any modification aftersynthesis (e.g. pre-mRNA); not functional, needs to be processed (e.g.via methylation or addition to the termini)


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and S factor activated & promote binding of pol I to the regulatory

region

b. Pol IIi.Located in nucleus

ii.Carries out transcription of the genes responsible for various cellularproteins

iii.Makes pre-mRNA(heterogeneous nuclear RNA) by promotors andenhancers

c. Pol IIIi.Located in nucleus

ii.Makes small RNA including 5S rRNA (in splicosome), tRNA, snRNPsiii. Activated by TFIIIA, TFIIIB, & TFIIIC

1. B is recognized by RNA Pol III for transcriptiond.

The different Eukaryotic RNA polymerases can be distinguished by theirsensitivities to -amanitin

i.Pol I = resistantii.Pol II = sensitive

1. E.g. eating a mushroom where pol II is particularly abundant will shut down all themRNAs that are responsible for making proteins

iii.Pol III = intermediate4. TO DESCRIBE THE SEQUENCES UPSTREAM OF THE INITIATION SITE THAT ARE USUALLY

REQUIRED FOR EFFICIENT TRANSCRIPTION

a. Activation region for rRNA lies within the geneb. Assembly factors bind to this region and enable the binding of transcription

factors

c. Pol III recognizes these transcription factors and transcription begins5. TO DESCRIBE HOW THE PRIMARY TRANSCRIPT OF THE RIBOSOMAL GENE IS CONVERTED

INTO THE MATURE RIBOSOMAL RNASPECIES.

a. RNA polymerase III catalyzes formation of 5S rRNAb. The regulatory unit of this smaller RNA is within the body of the gene (versus

having the promoter upstream)

i.The promoters often lie within the genesc. Binding of TFIIIA and TFIIIC to the promoter permits recruitment of TFIIIB

and finally pol III

d. Transcription beginse. The primary transcript is processed at the 3 end by removal of several bases

from the 3 end

6. TO DESCRIBE THE SEQUENCE OF EVENTS IN THE FORMATION OF FUNCTIONAL MRNAFROMTHE PRIMARY TRANSCRIPT


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a. RNA polymerase II catalyzed messenger RNA synthesisb. Processing

i.Capping1. m7G cap added 5 5 to 5 end. This blocks degradation

while transcriptiontranslation take placeii.Polyadenylation1. Poly-A tailAt termination site, poly-A polymerase adds ~250

A residues2. Poly-A site is AAUAAAand recognized by CPDF3. The poly A addition process in vivo is highly selective; only RNA chains destined to

become mRNA receive a poly A tail

iii.Splicingcleave out introns1. SpliceosomesnRNPs and pre-mRNA

a. GU-----------A-----AGIntron structureforms lariatvia spliceosome and is cleaved out

b. Catalyzes exon 3 nucleophilic attack on next exons 5end

2. Alternative Splicingmultiple exons can be included/cleavedbased on final protein function; generates mRNAs that aretemplates for different forms of a protein

a. E.g. whether an antibody is membrane bound or not is afunction of alternative splicing

iv.Correct splicing of pre-mRNA is medically relevant1. E.g. The thalassemiasare hereditary abnormalities of

hemoglobin production in which the primary difficulty is a

quantitative deficiency of either -globin (-thalassemia) or -

globin (-thalassemia)

a. Approximately 25% of the mutations in the human globingenes underl ying the thalassemias are in sequences

requir ed for correct spli cing

7. TO DEFINE THE TERMS INTRONS AND EXONS.


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a. Introns: part of primary transcripti.Non-coding segments that interrupt the amino acid coding region

ii.Not found in mature mRNAb. Exons: part of eukaryotic gene coding for RNA which will ultimately be

spliced into functional mRNAi.Are thought to contain individual functional domains Inhibitors of transcription

o Actinomycin D Inhibitor of eukaryotic DNA-dependent RNA synthesis Intercalates itself into double-helical DNA between successive G-C bp and distorts DNA Prevents it from being used as a template for RNA synthesis

o Rifampicin Blocks initiation of bacterial RNA synthesis Potent antibacterial agent

cDNA cloningo Used to clone individual mRNA molecules; it is important to start with cells

that express the gene of interest

o Isolate mRNA by using poly-T column (poly-A tail will bind leaving onlymRNA)

o Reverse Transcriptase creates cDNA from bound mRNA templateo Remaining mRNA can be degraded w/ alkali leaving ss cDNAThis is

quickly turned into ds cDNA

o Hairpin loop can be cut by S1 nucleaseo Resulting cDNA can be cloned into bacteria via recombinant plasmid

Eg: clone insulin by getting rid of the introns Get a cDNA of the pancreas Intron information does not end up in the proteins anyway

5. CONTROL OF GENE EXPRESSION1. TO DESCRIBE THE RELATIONSHIP BETWEEN CHROMATIN STRUCTURE AND GENE

EXPRESSION.

a. Configuration of genes in chromatin:i. Heterochromatin: transcriptionally inactive and not accessible to

limiting amount of DNase I1. The inactive state of a gene is determined by its association with this condensed

heterochromatin

ii. Euchromatin: transcriptionally active and accessible to limitingamounts of DNase I

1. Genes active in transcription are associated with this less condensed chromatiniii. Hypersensitive sites: sections of chromatin (usually in control

regions) that are very sensitive to DNase I1. Very accessible sites on genes in their upstream control regions


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a. Wikipedia:

i. Enhancer function underlies regulatory processes by which cells establish patterns of geneexpression.

ii. Recent results suggest that many enhancers are specified by particular chromatin marks inpluripotent cells, which may be modified later in development to alter patterns of geneexpression and cell differentiation choices.

iii. Enhancers work as cis-regulatory elements to mediate both spatial and temporal control ofdevelopment by turning on transcription in specific cells and/or repressing it in other cells.Thus, the particular combination of transcription factors and other DNA-binding proteins in adeveloping tissue controls which genes will be expressed in that tissue.

5. TO DESCRIBE THE ANATOMY OF THE TRANSCRIPTION APPARATUSa. Basal transcription factors

i. Essential for initiation of transcription in all genesii. Assembly depends on presence of activators

b. Activatorsi.

Bind enhancers to increase transcriptionc. Coactivatorsi. Are linked in a tight complex to the TATA binding protein (TBP)

ii. Activator/ coactivator communication enables the basal factors toposition RNA polymerase II at the start of the protein-coding regions,

and to set the polymerase in motion

d. Repressorsi. Bind silencer regions of DNA to repress transcription


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ii. Interfere with the functioning of activators and block initiation events6. TO OUTLINE THE CONCEPT OF INDUCIBLE GENE EXPRESSION .

a. Transcription is affected by protein-protein interactions and covalentmodification (phosphorylation, etc.)b. Induction of gene transcription by various agents:

i. Heat shock1. Heat-shock factor (HSF) protein is present in cells and changed

to an active form that binds to a specific DNA site up-stream of

certain genes

2. Binding (and subsequent phosphorylation of the bound factor)results in increased transcription

ii. Steroid hormones1. Affects transcription by binding to specific intracellular steroid

receptors that are site-specific DNA-binding molecules2. The binding sites on DNA for the hormone-steroid receptor complex are

termed response elements

a. Binding is specific, so different hormonal signals will affect onlythose genes that contain the appropriate response element

7. TO DESCRIBE A RELATIONSHIP BETWEEN ABERRANT LEVELS OF TRANSCRIPTION ANDCERTAIN CANCERS.

a. DNA microarraysi. Powerful technique used to determine difference in mRNA

population (& gene expression)between two cell types (normal

versus malignant)

ii.

Primers for many different genes attached to glass slidemeasures mRNA expressioniii. If a particular mRNA is present at abnormal levels in one cell type, it

is easily detected (the spot will be one of the primary colors, rather

than a blend of the two colors used for each mRNA)

1. Normal and abnormal samples fluorescently labeled green or red if under/overexpressed, dot wont be yellow

b. Clinical example:i. Microarray used to detect differences in gene expression in normal

versus breast epithelial cells

ii. Oncogene ErbB2(HER-2/neu) is expressed at abnormally high levelsin malignant breast tumor cells

1. ErbB2 is a growth factor receptor in the epidermal growth factor receptor familyc. Antisense therapy

i. Oligonucleotide complimentary to mRNA binds and doesnt allowtranslation

1. Eliminate a given mRNA by introducing nucleic acid complementary to it(introduce a complementary strand of ssmRNA to the ssmRNA already present;

binding results in dsmRNA that blocks ribosomes from translating it)ii. In order to inhibit the degradation of the antisense nucleic acid, sulfur atoms are substituted for

oxygen on the phosphate links between nucleotides


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d. RNAi (RNA interface) via siRNA (small-interfacing RNA)i. Want to stop translation of a specific gene so the protein it encodes

never gets made using dsRNAii. Mechanism of sequence-specific, post-trascriptional gene slicing initiated by double stranded

RNAs (dsRNA) homologous (complementarity) to the gene being suppressed1. Make a dsRNA that is complementary2.

Protein RISC comes and recognizes it; only binds to dsRNA, opening it up,hybridizing to a sequence, and destroying whatever piece of RNA it complements

with

iii. Being used to treat autosomal dominant diseases

6. PROTEIN SYNTHESIS

Essential Components for Translationa. Ribosomes (different for Eukaryotes and bacteria)b. mRNA and the genetic code

i. Codons encode for amino acids61 for aa, 3 for stopc. tRNA complexed with appropriate amino acidi. Many different kinds, each aa will bind one or multiple tRNAs to 3 CCA sequence

Translation Processa. Polypeptide chain initiationb. Polypeptide chain elongationc. Termination of protein synthesis

1. TO DESCRIBE THE STRUCTURAL ORGANIZATION OF RIBOSOMES AND TO UNDERSTAND THATTHE RRNADOES NOT HAVE ANY CODING INFORMATION.

a. Ribosomes: ribo-nucleoprotein particles associated with RNA that assemblepolypeptide chains in the cytoplasmb. Consist of two subunits of unequal size (60S and 40S)

i. Large and small cycle between being together and being separate as different subunitsc. Are found either free in the cytoplasm or associated with the endoplasmic

reticulumi. ER associated ribosomes synthesize proteins to be exported from the cell and also to become

part of cellular membrane

d. mRNA carries the information for the amino acid sequence of proteinsi. Not called mRNA until after RNA splicing

2. TO REVIEW THE ADAPTER FUNCTION OF TRANSFER RNAAND THE CRUCIAL ROLE OF THEAMINOACYL -TRNA SYNTHETASES IN THE ACCURACY OF GENE EXPRESSION.

a. tRNA molecules can interact specifically with both an amino acid and a codonof mRNA

i. It acts as an adaptor, bringing specific aminoacids into position where they can beassembled into a polypeptide chain ofdefined sequence

ii. Translation = process by which apolypeptide chain is formed using an mRNAtemplate


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b. tRNAs are single stranded but about half of the residues are base paired in any molecule, resulting infolding

c. The molecule always has a CCA at the 3 terminus; the final Ais the site of attachment of the aminoacid

d. Before an amino acid is incorporated into a protein chain, it is attachedby its carboxyl end to the 3 of an appropriate tRNA molecule. This:

i.

Covalently links the amino acid to a tRNA containing the correct anti-codonii. Attachment activates the amino acid (generates a high energy linkage at its carboxyl end)

e. Aminoacyl-tRNA synthetasei. Catalyzes the activation and transfer steps for an amino acid

1. There is a different Synthetase for every Amino acid (20)ii. The synthetase transfers the amino acid to the 3 hydroxyl group of the terminal nucleotide of

a specific tRNA, to give an aminoacyl-tRNA

iii. The synthetase must recognize both the specific amino acid and the correct tRNA1. Each enzyme must have two specific binding sites, one for the amino acid and one

for the tRNAiv. Codon/ anticodon interaction utilizes complementary base pairing

3. TO REVIEW THE MEANING OF CODONS IN MRNA,ANTICODONS IN TRNAAND HOW THETWO INTERACT DURING TRANSLATION.

a. Codon: a trinucleotide sequence that determines an amino acidi. 61 codons are used to specify 20 amino acids; the three remaining codons function as chain

termination signals

ii. Most amino acids are specified by more than one codon1. This redundancy makes the genetic code degenerate

b. Anticodon: a trinucleotide sequence that binds to a codon in mRNA bycomplementary base pairing

i. A given tRNA contains an anticodon complementary to a codon for its amino acidii. Wobble hypothesis: The 3 base of the codon, and the 5 base of the anti-codon are

spatially oriented such that the more relaxed bonding takes place at this position

1. This allows a single tRNA to recognize several codons(thisis why there are fewer tRNAs (anticodons) than codons)

2. E.g. leucine can be: CUU; CUC; CUA; CUG; UUG; UUA; GACa. Need three sets of tRNAs

b. Only need one tRNA synthetase per amino acidi. The tRNA synthetase for leucine would be able to make all three

tRNAs


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4. TO REVIEW THAT MRNAHAS 3'AND 5'NON-CODING REGIONS AND TO APPRECIATE THEFACT THAT THESE REGIONS ARE CONCERNED WITH THE CONTROL OF M RNAFUNCTIONS,

SUCH AS METABOLIC DECAY AND TRANSLATION.

a.

Coding region makes the proteini. Contains start site for protein synthesis and stop siteii. Between 5cap and start codon is a UTR

iii. UTRs control the stability of the mRNA; they are not turned intoprotein

5. TO REVIEW THAT TRANSLATION BEGINS A T ANAUG CODON.TO UNDERSTAND THAT THEAUGIS SELECTED AS THE FIRST CODON BECAUSE THE MET-TRNAFOR INITIATION IS THE

ONLY TRNATHAT CAN BIND TO THE 40S RIBOSOMAL SUBUNIT DURING THE INITIATION

PROCESS.

6. TO REVIEW THE CRUCIAL ROLE OF THE INITIATION FACTOR EIF-2IN THE SELECTION OFTHE AUGCODON.

7. TO REVIEW THE FUNCTIONS OF THE TWO DIFFERENT TRNABINDING SITES AND OF THEPEPTIDYL-TRANSFERASE ON THE RIBOSOME.

8. TO REVIEW THE SEQUENCE OF EVENTS IN THE STEPWISE ADDITION OF AMINO ACIDS TO THEGROWING POLYPEPTIDE CHAIN,AND THE ROLE OF THE TWO ELONGATION FACTORS (EF-1

AND EF-2)IN THIS PROCESS.

9. TO REVIEW THE PROCESS OF POLYPEPTIDE CHAIN TERMINATION ,AND WHY THE CODONSUGA,UAAAND UAGFUNCTION AS TERMINATION SIGNALS.

Ribosomal subunits, not ribosomes, are the starting point for translation

Translation:a. Initiator Met-tRNAbinds to soluble protein eukaryotic initiation factor 2

(eIF-2) and GTP

b. This ternary complex binds to the small 40S ribosomal subunit (occurs in theabsence of mRNA)

c. eIF4Frecognizes and associated with the m7G cap of mRNAd. The 40S subunit migrates in a 5 to 3 direction, searching for the AUG

initiation codon


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c. Proteins fold to maximize weak, non-covalent forces (e.g. electrostatics, H-bonds, van der Waals)

i.A partially-folded protein (often with a mutation) behaves similar to a molten globule and isprone to aggregation (disease)

1. Two hydrophobic patches will come together to stabilize each otherii.For a water-soluble protein, the folded protein form with the most hydrophobic groups buried is

the state least prone to aggregation

d. Larger proteins frequently need chaperone proteins (influence the pathway but not thefinal structure)

i.Helps proteins avoid pitfalls, such as molten globule statee. helixhas a very specific geometry to it

i.Hydrogen bonds not from the R group can cause helical shape; occursin membrane spanning sections

ii.If two A.A. residues adjacent in sequence have ionized side chains of the same charge, the helix is not as energetically favorable as the sheet would be

iii.Proline: 5 membered ring that cannot adopt the conformation needed tofit into the middle of a helix is a helix breaker

iv.The helix of p53fits into the major grove of DNA and exhibitshydrogen binding

f. sheeti.Hydrogen bonding not from the R groups can cause a flat sheet

1. The h-bonding is between stretches of A.A. residues (rather than within a continuousstretch)

ii.Polypeptides are much more extended2. TO DESCRIBE THE COMPONENTS OF A PROTEINS STRUCTURE

a. The A.A. sequence (1ostructure) is determined by mRNA & specific for eachprotein

i.Proteins with similar sequences have similar functions1. Thep73protein has a sequence that is similar to that ofp53. It appears to have a

similar function-it acts as a tumor suppressor

b. Specific activity of a protein is the amount of activity per amount of proteini.To measure the amount of purified protein: Use UV light at 280 nm

ii.Tyrosineand tryptophanabsorb the strongestc. Each domain of a protein behaves much like a separate protein (with a

separate biochemical function)

3. TO RECOGNIZE THE SPECIAL PROPERTIES OF GLYCINE,PROLINE,TYROSINE,ANDTRYPTOPHAN.

a. Glycinei.Glide has the smallest/lightest R group (a hydrogen)

ii.Allows proteins to form sharp bends & for chains to come in closecontact to one another

iii.e.g. hairpin bends of p53b. Proline

i.o only amino acid that forms a ring on its side chain involving thebackbone nitrogen

ii.Limits its flexibility & makes it incompatible with helix formation


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c. Tyrosine & Tryptophani.Absorb UV light at 280 nm

ii.Makes proteins easy to detect in the labd. Cysteine

i.Can form disulfide bonds (covalent bonds within or between proteins)in the oxidizing environment outside the cell

4. TO DESCRIBE THE EFFECT OF MUTATIONS ON PROTEIN STRUCTURE AND FUNCTIONa. The function & specificity of a protein are dependent on the amino acid

sequence and the three-dimensional structure

i.A hydrophobic for hydrophilic AA charge change or catalytic sitechange can be dramatic

b. Mutations effect:i.Three classes

1. A: no effect2. B: affect only function

a. Failure to bind DNA by these mutants can be attributable to a loss of acritical DNA interaction

3. C: affect structure/ stability and functiona. Loss of DNA binding by a different class of mutants can be attributed to

defects in structure & stability

c. Recognition of DNA by p53i.A part of the protein, an helix fits neatly into major groove of DNAii.Each domain recognizes the sequence

1. uuuCATGyyy2. yyyGTACuuu

iii.The placement of atoms is preciseiv.Tumor suppressor protein p53plays a critical role in protecting cells against cancer

1. Loss or malfunction of p53 leads to development of about half of all cancersa. In breast cancer, the p53 mutation is associated with more aggressive

disease and worse overall survivalb. >60% of lung cancers have mutated p532. Frequent missense mutations cluster in the DNA-binding domain

a. E.g. The hairpin bend of p53 is disrupted by a glycine (residue 245) toserine mutation resulting in an unstable protein

5. TO DESCRIBE THE BIOCHEMICAL BASIS OF PRION-RELATED DISEASESa. Prion protein

i.Can exist in either a mainly helical benign, soluble form, or a sheetaggregated form

ii.Consist only of proteinand do not contain RNA or DNAb. Prion Disease: Transmissible spongiform encephalopathies caused by

misfolded prion protein

i.Creutzfeldt-Jakob disease1. Inherited form2. Single A.A. mutations switch protein structure from helical to

aggregated form, associated with neurological disease

3. The presence of a small amount of sheet form can act as apathogen and trigger the helical form to convert

a. Exponential increase of infectious form


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4. New-variant CJD results from coming into contact with the sheet form whichconverts the patients wild type

ii.Mad Cow1. Thought to originate from feed supplements containing sheep parts

iii.Scrapie (sheep)iv.Kuru

6. TO STATE THE CHANGE IN CHARGE OF A MOLECULE WITH PHAND HOW THE MOLECULEWILL BEHAVE IN ELECTROPHORESIS

a. Some side groups are acidic or basic; altering pH can cause proteins to notfunction properly

i.The charge on several A.A. residues change in response to pH1. Many biological processes produce acid (e.g. muscle exertion)

b. Buffers prevent changes in pHi.Proteins are effective buffers because they contain amino acid residues

with different pKa valuesii.pKa= where 50% is in acidic/basic form; multiple pKas per protein for multiple acid/base side

groupsiii.When the pH is close to the pKa, the pH is most resistant to change

1. Large additions of acid or base produce only small changes in pH (maximalbuffering capacity) (flat region on graph)

iv.When pH=pKa, half of the molecules are in the ionized/acidic (A-)form and half are in the non-ionized/basic (HA) form

v.Major buffer inside cells is phosphate; pKa ~7vi.Major buffering system of the blood is bicarbonate

1. pKa of carbonic acid/bicarbonate is 6.1. Although far from physiological normal, theability to regulate the components of the buffering system makes it an optimal

system

2. Hyperventilatinga. Increases O2b. Removes CO2from the lungsc. Lowers amount of carbonic acid (H2CO3) in blood

i. Carbonic acid is in equilibrium with bicarbonate(HCO3

-)

d. Increases pH

c. Proteins work as buffers


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i.Wont function properly if not at correct pH because protein will have wrong charges on someof its side-chains

ii.E.g. chloride channels activity drops off at pH values below 7; is the channel disrupted in cysticfibrosis

1. Metabolic acidosis causes problems in protein function

d. Charges change on the protein side chain by:i.+1if pH is lowered

1. pH < pKaacid form predominatesii.-1if pH is raised

1. pH > pKabase form predominates

e. Isoelectric point: pH where protein has zero charge (electrically neutral) andwill not move in an electric field

i.Found by averaging the two pKs surrounding the isoelectric species


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7. TO STATE THE INFLUENCE OF CHARGE ON A MOLECULE S ABILITY TO CROSS A MEMBRANEa. Only non-ionized molecules cross cell membrane passively

b. Polarity of a molecule is dependent on ionizationc. E.g. aspirin

i. Weak acid at low pH with a pKa of 3.51. Low pKa = acidic; more likely to donate its proton/ positive

charge

ii. When pH = pKa, then half of the aspirin are negatively charged andhalf are neutral

iii. When pH < pKa then there are more protons in solution(the lowerthe pH the more protons)

1. Will have more neutral (protonated) aspirin molecules2. Stomach is acidic with pH of 1.5 (pHstomach< pKaaspirin),

whereas intestine has pH of 6.5

a. Aspirin will be largely non-ionized in the stomach andnegatively charged in the intestine

iv. It is important that aspir in is neutr al (base form) at stomach pHbecause it is easier f or uncharged molecules to go through the cell

membrane1. More aspirin will be absorbed into the blood through cells

lining the stomach than the small intestine

8. OXYGEN TRANSPORTERS

1. TO DESCRIBE THE STRUCTURE OF MYOGLOBIN AND HEMOGLOBIN IN TERMS OF OVERALLSHAPE

a. Myoglobini.Oxygen transporter in muscle

ii.Single chain subunitiii.Heme is located in a deep, non-polar crevice

and is surrounded by nonpolar A.A. residues1. Keeps Fe in the +2 required for O2transport2. Easily oxidized (looses electrons) to +3 when in

polar environment such as water & cannottransportmetmyoglobin

iv.FE is stabilized by 2 histidines and 4 N, E7 hispreferentially binds O2over CO

v.Undergoes a conformation change on binding O21. Heme flattens out; iron comes into plane (due to change in its size after binding); F

helix gets pulled down2. Since myoglobin is a monomer, this change in structure has little consequence

vi.Oxygenation follows a simple hyperbolic curve representing anequilibrium between protein molecule and oxygen molecules

1. Functions well in muscle tissue where the partial pressure of oxygen is low (from 5up to 30 mmHg)


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2. Would not be an effective oxygen transporter between lungsand tissues since little oxygen would be released

b. Hemoglobini.A globular dimer (tetramer)

1.

Four subunits w/ four heme groups 2, 22. Structure of myoglobin is nearly identical to the structure of the individual subunitsof hemoglobin

ii.An allosteric (other site) proteiniii.Shape of oxygen dissociation curve is sigmodial

2. TO DESCRIBE THE ROLES OF THE HISTIDINE RESIDUES SURROUNDING THE HEME GROUPWITH RESPECT TO OXYGEN BINDING AND INTERACTIONS WITH IRON

a. Two histidines in heme pocketi.One is involved in coordinatingthe heme

ii.Residue in the active site decreases the occupancy of heme by CObyforcing it to bind in the same bent geometry as O2(120

o

) versusperpendicularly; this allows continued oxygen binding

3. TO BRIEFLY DESCRIBE THE MECHANISM OF COOPERATIVITY OF HEMOGLOBIN IN BINDINGOXYGEN.

a. Cooperative: binding of O2to one binding site affects the O2affinity of othersites

b. Mechanism of cooperativityi. Initial trigger is movement of the Fe2+ion leading to a conformational

change in one of the subunits

ii. The subunit contacts the subunitiii. Movement of one subunit causes the other subunits to take on theoxygen-binding conformation

1. Taut statea. Poor affinity for oxygenb. More stable than R form in the absence of oxygen

i. More electrostatic & hydrogen bondsc. Likely to be present in peripheral tissues

2. Relaxed statea. Oxygen bound

3. R & T states differ in arrangement of the subunits, although the subunits themselvesbarely change structure

4. Cooperativity is mediated by changes in quaternary structure (structure formed bythe interaction of separate polypeptides)

5. Separated and subunits do not exhibit cooperativity; each functions likemyoglobin

4. TO DEFINE THE BOHR EFFECT,INCLUDING THE EFFECT OF CO2AND PHON OXYGENAFFINITY.


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ii. More oxygen can be transported per cycle (although hemoglobin is notfully oxygenated in the lungs)

6. TO DESCRIBE OXYGEN BINDING PROPERTIES OF FETAL HEMOGLOBIN.a.

Fetal hemoglobin has 2 subunits and 2 subunits (22)b. Has higher affinity for oxygen

i. Binds BPG more weakly1. subunits have fewer positively charged amino acid residues in BPG binding site

and bind negatively charged BPG less well

ii. Oxygen flows from maternal oxyhemoglobin to fetal deoxyhemoglobin7. TO INTERPRET THE HILL COEFFICIENT OF A HILL PLOT.

a. Hill plot: determines extent of cooperativityi. Myoglobin slope = 1: nothing fancy going on, just binding & releasing; no sigmoidal character

and no cooperativity

ii. Hemoglobin slope >1: positive cooperativity1. The heme groups communicate through conformational changes transmitted across

subunit interfacesiii. Slope


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e. Treatmenti. Antibiotic therapy (to prevent secondary infections)

ii. Hydroxyurea1. Stimulates production of HbF(chain formation)2. Replacing chains with chains alleviates issue of polymer

formationiii. Bone marrow transplantation1. Stimulates RBCs that will form normal hemoglobin2. Replaces HbS with HbA

iv. Gene therapy9. TO DEFINE METMYOGLOBIN AND METHEMOGLOBIN AND TO DESCRIBE THE BIOCHEMICAL

DEFECT.

a. Metmyoglobini. Myoglobin oxidized (looses electrons) to +3 when in polar

environment such as water & cannot transport O2b. Methemoglobinemiai. Hemoglobin carrying +3 (ferric) iron; cannot carry oxygen

ii. Results from a single substitution of a tyrosine for a histidine F8,which is in one of the coordination position of iron

1. Causes a structural change leading to subsequent oxidation of iron to +3iii. Ingestion of nitrates and nitrites (oxidizing agents) can promote HbM

1. Defects can also arise from the lack of maintenance of reducing agent in RBCsiv. There are no homozygous individuals with both defective genes

1. That would mean there is no functional hemoglobinv. Heterozygotes have reduced oxygen-binding capability

vi. Reducing agents (Vit C & methylene Blue) can help treat the disease Thalassemiasubunit imbalance aka not enough or is produced or is

defective

9. PROTEIN PROCESSING

1. TO DESCRIBE THE PRODUCTION OF INSULIN AND THE SPECIAL PROPERTIES OF CYSTEINEInsulin:bioactive peptide by proteolytic processing

a. Preproinsulin: contains N-terminal signal sequence and c-peptideb. Proinsulin: signal sequence cleaved off, c-peptide still therec. Insulin: c-peptide cleaved off,

i.The two polypeptide chains of mature insulin are held together bydisulfide bonds formed from two sulfur atoms of two cysteine residues

ii.Outside the cell, including plasma and ER lumen, (oxidizing conditions)residues form covalent bonds with the sulfur of another cysteine

1. Disulfide bonds impart high stability to proteins


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2. TO DESCRIBE THE SYNTHESIS,FOLDING,AND ASSEMBLY OF COLLAGEN INCLUDING ALL THESTEPS IN THEIR PROPER ORDER;TO KNOW THE ENZYMES AND COFACTORS REQUIRED FOR

EACH STEP.

a. Collagen:fibrous protein made by the fibroblast that provides structuralintegrity to tissues & organs; does not dissolve in wateri.Not only provides structural integrity, but, in other tissues provides the

matrix for calcium phosphate deposition to make bones (type I)

ii.All types share the collagen triple helix1. made of three polypeptide chains (all called chains) and

mostly hydroxyproline, proline, glycine in addition to ~100modified amino acids (hydroxyproline and hydroxylysine);

iii.chainsusually gly-pro-Hypro repeated, 1 and 2 are very similar but different genes(allows for structural diversity in different tissues)

b. Types:i.Collagen type IBone, Skin, Tendon

1. type ONE is prevalent in bONEii.Collagen type III Reticulin (supporting mesh in soft tissues)

c. Biosynthesis Stepsi.(Pre) Pro (polypeptide) chain synthesis; soluble, does not contain

hydroxylated AAs

1. Signal is cleaved when delivered to lumen of ERii.

Hydroxylation1. Several prolines (and some lysines) are hydroxylated in a post-

translational modification2. Requires prolyl hydroxylase(a.k.a. procollagen-proline

dioxygenase) and lysyl hydroxylaseand the cofactor ascorbic

acid (Vitamin C)

iii.Glycosylation1. Hydroxylysines are attachment sites for sugar molecules

iv.Disulfidebond formation at C-terminus (in the oxidizing ER lumen)


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b. Ehlers-Danlos Syndromei.Hyperelastic joints & skin flexibility

ii.Occurs from structural defects in collageniii.Many types; usually autosomal recessiveiv.Two kinds of a wide variety of molecular defects

1. Improper processing of collagen caused by lack of a processing enzymea. Heterozygosity is probably OK (sufficient collagen processing enzyme

[50%] produced)i. The abnormal allele is recessive relative to the normal allele

b. Most defects occur with homozygous recessive mutations2. Gene defects in the collagen genes (as described for OI)

a. Heterozygosity is not okayi. Even though they synthesize equal amounts of normal and

abnormal collagen chains, the structurally defective chain isdominant

c. Osteogenesis Imperfectai.Results from mutations in Type I collagen

1. Point mutation in collagen gene, resulting insubstitution of acysteine for glycine

2. Glycine is critical for triple helix development; mutation leadsto defective assemblyand subsequent degradation of collagen

3. Lethal mutationa. Only heterozygotes live but are severely affected because proteins with

multiple subunits have dominant mutations

b. If either alpha 1 or alpha 2 has defect in Type I collagen, patient will havedisease

ii.Symptoms: bone deformities leading to brittle bones & teeth; fragile blood vessels

Other fibrous proteins:

Elastin is less ordered than collagen, little hydroxylation, four lysines are linkedforming a desmosine; Rich in gly, ala, val

Keratintougher fiber, mostly helices


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10. ENZYMES AS CATALYSTS

1. TO LIST THE GENERAL CHARACTERISTICS OF ENZYMES IN TERMS OF THEIR ACTIVE SITESAND ABILITIES TO CATALYZE A REACTION.

a. An enzyme binds substrate at its active site, and catalyzes conversion ofsubstrate to product

b. Active sitei.Active site comprises A.A. residues that:

1. Bind (recognize) the substrate2. Catalyze the reaction; carry out hydrolysis

a. E.g. serineii.The catalytic residues are not the same as substrate-binding residues (think of Professors farm-

chicken-catching example)

iii.A mutation in a residue may affect catalysis but not affect the strength of bindingiv.Substrate binds by variety of weak interactionsv.Specificity pocket recognizes specific side-chains of a suitable peptide

and positions the substrate in the correct orientation for catalysis

vi.Enzyme is held in place by hydrogen bonds between the peptidebackbone and the protease backbone

2. TO EXPLAIN ENZYME CATALYSIS IN THERMODYNAMIC TERMS .a. Energeticsreaction occurs spontaneously with -G, if G=0 than rxn is at

equilibrium (death)

G = G + RT ln Q where G is the standard free energy

When G = 0, Q is called Keqor equilibrium constantb. If G is very negative the reaction is essentially irreversiblec. Steady state: Not equilibrium, but substrates and products remain at a steady

concentration

i.Driven by the production/reduction of both substrate and productd. G can not predict reaction rate, energy barrier (of activation) Gcan be

lessened by catylystsi.G cannot be calculated from Go(or Keq) unless the reactant and substrate concentrations are

known

e. Catalysts do not change Gof the substrate or products,i.they lower Gor activation energyby transition site stabilizationii.Stabilization of the transition state by a catalyst (enzyme) increases the concentration of the

reactive intermediate that can result in product formationiii.Changes the rate but not the equilibrium constant

f. Differences between chemical and biochemical reactionsi.Reactions in body are not at equilibrium

ii.Reactions can be driven by reducing product concentration by removing it in pipeline effect(product formed from one reaction can form the substrate for another reaction)

iii.Reactions proceed under very non-standard conditions; hinder the accurate prediction of Gfrom Go

iv.Some processes do not take place in solution, but are solid phase reactions


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3. TO DESCRIBE AND USE COUPLED REACTIONS.a. ATP mediates energy flow from catabolism of fuel to work, in coupled

reactions and phosphate transfers using a common intermediate i.Hydrolysis of one high energy bond G ~= -7 kcal/mol (large amt of free energy)

ii.Large release of free energy can be used to assemble a different high energy bond (peptide,phosphate add)

1. Enough free energy to drive reactions that otherwise would be thermodynamicallyunfavorable (e.g. phosphorylation)

a. An enzyme can have more than one substrateb. E.g. glutamine synthase binds both ATP and glutamate

i. The intermediate is activated by phosphorylationiii.Coupled reactionone reaction w/ +G paired with one w/ more-G so that both can proceed

b. Catabolism of food yield ATP; energy from ATP is used to power the bodys anabolic reactionsc. Other compounds, such as cleaving a peptide bond, releases a small amount of energy

i.Alternatively, it takes a small input of energy to reverse the process (form a peptide bond)d. Many events can trigger huge requirements for increased energy (food intake)

i.AIDS (and other infections); the repair associated with surgery; competitive swimming4. TO DEFINE THE PROPERTIES OF AN ENZYME AND THEIR MEASUREMENT ,SUCH AS KMAND

VMAX

5. TO USE MICHAELIS-MENTEN AND LINEWEAVER-BURK PLOTS TO DETERMINE ENZYMEPROPERTIES.

Michaelis-Menten plots, V vs [S]

a. Vmaxoccurs when an enzyme is saturated w/ substratei.Speed of a reactions depends on substrate concentrations

1. When [S] = 0 then v = 02. When [S] is infinite then V = Vmax3. When [S] = Km then V = Vmax

ii.As enzymatic reactions proceed, the rate of product formationslows for two reasons:

1. Substrate is being used and less is available to theenzyme

2. Product inhibitioniii.Velocity is proportional to amount of enzyme present (more enzyme

high Vmax)

1. E.g. double enzyme concentration results in:a. Vmax increase (Adding more enzyme means

more spots are available for substrate to bind)b. Km remains unchanged (Km is characteristic of

an enzyme; will not change regardless of amount

of enzyme present)2. If you just change the [S] then nothing will change

b. Low [S] V ~ [S]i.At low substrate concentrations:

1. [S]


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c. High [S] V ~ [E]i.At high substrate concentrations: all enzyme molecules are saturated with substrate

1. [S]>>Kmand velocity of reaction is constant and independent of substrateconcentration

a. Rate cannot be increased by adding more substrate2. High, saturating substrate levels are suitable for assays of enzyme concentrations

since velocity is directly proportional to the amount of enzyme

d. Kmcan describe enzyme affinity to its substratei.Is a property of the binding site; can be changed with indirect inhibitors or mutations

1. Most enzymes have the right Km according to what they need to do; allows theenzymes activity to be controlled by amount of substrate present

ii.Km= [S] at Vmax (small Km= high affinity)1. E.g. chymotrypsinhas a low Km for a peptide such as Gly-Gly-Trp-Gly,

and a high Km for a peptide such as Gly-Gly-Lys-Gly

Lineweaver-Burk: Michaelis-Menton plot, 1/V vs 1/[S]a. Used to determine Vmax/Kmb. Y-intercept is 1/Vmaxc. X-intercept is -1/Km (the negative of a negative results in a logically positive value for Km; Km

can never be negative- can never have a negative substrate value)d. slope = Km/Vmaxe. Vmaxis: Intercept at zero of 1/V and 1/[S]

a. Effective therapy from analysis of enzymatic activityiii. Genetic diseases are often biochemically heterogeneousiv. Methylmalonic acidemia

1. Isomerase reaction converts MethylmalonylCoA to succinylCoA, whichenters TCA cycle2. If conversion cannot take place, excess methylmalonylCoA is

hydrolyzed to methylmalonic acid3. Excess methylmalonic acid in blood causes methylmalonic

academia: poor feeding, progressive lethargy, muscular

hypotoniafailure to thrive & developmental delay


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4. The appropriate treatment depends on knowing the biochemicaldefect

a. Patient may be lacking the conversion enzyme(methylmalonyl-CoA mutase)

i. Assayed by Western blotb. May have enzyme defective for binding a substrate

i. Requires Vitamin B12 coenzymeii. Initially, treat by increasing Vit B12, if no response,

problem is with methylmalonyl mutasec. Lack of substrate

6. TO OUTLINE THE MECHANISM OF A SERINE PROTEASE INCLUDING STABILIZATION OF A

TRANSITION STATE.

Serine proteases: proteolytic enzymes; take a protein (polypeptide) and break it

into smaller bits;i.Enzymes contain an active site, which is usually a crevice on the enzyme surface

ii.Two specificities:1. For binding a peptide (i.e., not a lipid, not a carbohydrate)2. For having a specificity pocket for preferential binding to

particular amino acid sequencesiii.E.g. digestive enzymes, blood coagulation enzymes (thrombin), immune response, cell

differentiation

a. Enzymes:iv.Trypsin

1. Negatively charged group (asparate) in pocket2. Binds to peptides that are positively charged(arginine,

lysine)

v.Elastase1. Smaller, narrow, hydrophobic pocket2. Binds elastin (lung) and smaller A.A.s(glycine, alanine,

valine)vi.Chymotrypsin: endopeptidase (cuts in middle of protein);

1. Hydrolase: catalyzes peptide bond hydrolysis on C-terminalside of bulky hydrophobic A.A.s

2. Specificity pocket is large & hydrophobica. Like binds like

Enzyme Group Reaction

Oxidoreductases Oxidation-reduction reaction

Transferases Transfer of a chemical group

Hydrolases Lysis by water at neutral pH; can show acid orbase catalysis

Lyases A cleavage reaction not using water

Isomerases Change of molecular configuration

Ligases Joining of two compounds


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3. Mechanism for action of chymotrypsin:a. Substrate is bound in correct orientationb. Attack by a serine (catalytic machinery)

i. Serine 195 (a base) is set up in a catalytic triad(charge-relay)

1.Aspartic acid 102Histidine 57Serine 1952.Located at enzymes active site in a particular

configuration3.Oxygen behaves like an alkoxide ion[CH2O- instead of

CH2OH]a. Extremely electron rich; makes it a good

nucleophile that wants to attach carbonylpeptide bond

c. Transition state stabilizationi. During catalysis it is stabilized by the enzyme

ii. E.g. the oxyanion formed in the transition state is stabilized bychymotrypsin by hydrogen bonds through A.A.s located inoxyanion hole

d. Product release (hydrolysis)i. Half of the peptide (the C-terminal half with newly generated

NH2) is free to goii. Attack by a water molecule results in final release of all products

& regeneration of enzyme

11. ENZYME REGULATION

1. TO LIST THE WAYS THAT ENZYME ACTIVITY IS REGULATED BY CELLULAR LOCATION .a. Enzymes are expressed in only certain tissues, not others,

i.Increased blood conc. of intracellular enzymes ~ tissue/organ damage(liver - ALT, heart - creatine kinase)

1. ALTs appearance is indicative of viral hepatitis or liver damage; Another indicatorof live damage is a low level of alpha-1 antitrypsin (an enzyme inhibitor)

b. Enzyme zymogensi.Proteolytic cleavage at a specific activation site on the chain activates a

zymogen (inactive protein) by aligning the active site residuesc. Phosphorylation or dephosphorylation modifies the charge of an amino acid

residuei.Protein kinases phosphorylate; protein phosphatases dephosphorylate

ii.If serine, threonine, or tyrosineis in or near the enzymes active site,the enzyme activity may be changed

iii.E.g. phosphorylation in the activation loop of a MAP kinase leads to a thousand-fold increase incatalytic activity (increases substrate binding which occurs between the domains that undergo

a conformational change)d. Regulation by substrate levels

i.Changes in the substrate concentration change enzyme activity2. TO DESCRIBE THE SYNTHESIS OF THROMBIN FROM PROTHROMBIN AND THE BIOCHEMICAL

BASIS FOR ATTENUATING BLOOD COAGULATION

a. The blood clotting cascadeis a zymogen activation of serine proteasesi.Injury triggers activation of several serine proteases


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ii.These proteases activate a series of other proteins to become proteasesin a cascadeof ever more abundant proteins

1. Pro-thrombin is cleaved on a membrane surface by anotherenzyme to make pre-thrombin,

2. Pre-thrombina. The left half has no enzymatic activity; is important for sitting

prothrombin down on membrane surfaceb. Right half has potential for serine protease activity

3. Pre-thrombin is converted to thrombin by cleavage betweenresidues 15 & 16

a. In thrombin, residue 16 reaches over the enzyme and forms a stabilizinginteraction with residue 194, whose neighboring residue serine 195 is theactive nucleophile in a serine protease

i. This allows the proper formation of the catalytic triad, correctactive site, and a now active enzyme

4. Thrombin cleaves the N- and C-termini of fibrinogen to makeit into fibrin; stick ends self assemble; crosslinking occurs to

provide stability

a.

Similar to collagen synthesisiii.End result is conversion of abundant fibrinogen to fibrin, leading toblood clot formation

b. Activity of thrombin is controlled by locationi.Thrombin is water soluble but N terminal is -carboxylated causing

Ca2+

binding and anchoring to membrane1. -carboxylationis a post-translation modification that does not directly regulate

enzyme activity, but is required for formation of active enzyme; requires Vitamin

K cofactor; Vitamin K promotes the binding of prothrombin to membranes

a. Warfarin(Vitamin K analogue) is used in treatment ofthrombosis (blood coagulates too readily); it

inactivates the -carboxylation reaction, reducing the

location of prothrombin on the membrane, thus reducing the amount ofactive thrombin

c. Inactivation of thrombin occurs through the high affinity binding ofantithrombinwhich binds tightly to the active site

i.Binds so tightly that the enzyme-inhibitor complex is degraded as a whole and the inhibition isirreversible

ii.Heparinis an anticoagulant that promotes the binding of antithrombinto thrombin, hencereducing blood clotting

3. TO DESCRIBE THE BIOCHEMICAL BASIS OF EMPHYSEMA.Emphysema

a. Inhibition of elastasein lung tissueb. Neutrophils (WBC) contain elastase (a digestive serine protease) and release it into surrounding lungtissue

c. Unregulated elastase activity would result in degradation of the elastin fibers of the lungs, andeventually emphysema, but an elastase inhibitor called 1-antitrypsinprotects the tissues

i.Mutations in this elastase inhibitor leads to unregulated elastaseactivity(also trypsin and collagenase) and cleaved (destroyed) elastin

ii.Most evident in homozygous individuals (two defective genes) and


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heavy smokers (which oxidizes 1-antitrypsin)iii.Treatment is through intravenous administration replacement therapy of 1-antitrypsin

(which diffuses from blood into lung)

4. TO DESCRIBE COMPETITIVE AND NON-COMPETITIVE INHIBITORSa. Competitive

i.Occupy binding siteii.Reversible; effects end when no longer administered

iii.Compete with the substrate for the active site1. The more inhibitor you use, the more competition there is with

the substrate

2. Have a larger Kmthan the substrateiv.Competitive inhibitors do not alter Vmax

v.E.g ethanol competes with ethylene glycolfor its metabolismvi.E.g. chemotherapeutic agents

1. Selectively block essential pathways in bacteria or rapidly growing tumor cellsa. Sometimes the infection or tumor can respond by raising the substrate

level, rendering the competitive inhibitor uselessb. Infinite amounts of substrate out-competes the inhibitor

vii.Can also see product-inhibition

b. Non-competitivei.Catalytic machinery only

1. Affect the amino acid residues of the enzyme involved in catalysis of the reactionii.A type of allosteric (other site) inhibitor

iii.Do not affect Km; does not interfere with substrate bindingiv.Reduce Vmax(it cannot be reached no matter how much substrate is added)

1. Ef fectively reduce the amount of enzyme presentv.Can be reversible or irreversible

1. E.g. irreversible inhibitors such as 1-antitrypsin andantithrombin

2. E.g. DIEP(a.k.a. nerve gas) is a potent irreversible inhibitor ofserine proteases; only one serine out of many serines are modified;results in improper neurotransmitter function; functionally reduces the

emzyme velocity present (Vmax)


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5. TO DESCRIBE ALLOSTERIC ENZYMESa. Allosteric effectors: an effector modifies the V vs [S] curve such that [S]

binds at different affinitiesi.Inhibitors or activators

1. Activators drive enzyme to relaxed form; inhibitors to constrainedii.Can easily be modified by [S]

1. Causes a change in reaction velocityiii.Protein alternates between two conformations, relaxed and constrained

1. Substrate binds with high affinity to constrained form2. The binding of substrate increases the likelihood that the

allosteric protein is in the relaxed form (increases the

affinity to bind more substrate)iv.Usually contains multiple subunits (e.g. hemoglobin)v.

Does not usually follow Michaelis-Menten kinetics;can define an apparent

Km which is the substrate concentration at which you get Vmax1. A steeper dependence activity then M-M on the substrate concentration around the

Km

vi.Usually has a regulatory site for binding of effector moleculesvii.Often regulates a reaction pathwayby catalyzing the first step of a

dedicated, committed reaction sequence


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1. One of the effector molecules is the end product of thesequence, resulting in feedback inhibition(different from

product inhibition)

viii.E.g. ATCase1. Catalyzes formation of carbomyl-L-aspartate from carbamoylphosphate and L-

asparatea. carbamoylaspartate is a pre-cursor unique to making pyrimidines

(dedicated)

b. Is the first committed steptopyramidine formation

2. CTP (end product) inhibits this stepa. Acts as a negative allosteric

effector if pyrimidine synthesis isnot needed in cell

b. ATP acts as a positive allostericeffector if DNA replication andrapid production of pyrimidine

nucleotides is required

12. THE BIOCHEMISTRY OF CANCER


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1. TO DESCRIBE THE RELATIONSHIP BETWEEN TUMOR SUPPRESSOR GENES AND CANCER .a. Retinoblastomais caused by mutations that inactivate the RB genelocated

on chromosome 13i.Different types of mutations (large deletions, point mutations, mutations leading to

splicing defects, and promoter mutations) can result in loss of RB function

ii.The product of RB gene is a tumor suppressor proteinin the nucleusiii.Mechanism of action of pRB protein

1.

When pRB lacks phosphateit can actively block cell cyclingdue to its affinity for transcription factors like E2F (makes E2F inactiveinnormal cells)

2. As levels of cyclins riseduring the cell cycle, they interact and activatecyclin dependent kinases (Cdk) whichtransfer a phosphatefromATP to pRB, this releasing molecules such as E2Fso it isavailable to stimulate the production of proteins required for S phase

3. In the absence of pRB, regulation of E2F is lost and cellgrowth is abnormal, resulting in tumor growth

iv.Two forms of the disease, hereditary and sporadic, but need two hits; An unusually smallnumber of mutations are required to form retinoblastomas

1. Mutation to both alleles in one cell leads to uncontrolled growthv.The RB gene is frequently missing in several other types of cancers

b. p53 is the substrate for many stress-related phosphorylation events (a process thathelps stabilize the protein)

i.Many proteins monitor what is going on in the cell; if they recognize a problem theyphosphorylate p53 which causes it to go to work

1. Genes regulated by p53 have regulatory regions termed response elements2. Once p53 is phosphorylated, it turns on different genesto

deal with the problems encountered by the sensors

ii.p53 has two responses to DNA damage:1. G1/S arrest


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a. P53 transcription factor can turn on p21 gene whichinhibits cyclin/Cdk complex

i. In absence of p53, p21 is not induced, the cellcycle is not halted, and cells will replicated

damaged DNA

2.

Apoptosisa. Cannot happen without p53b. E.g. CML: WBC count will be through the roof; have

not entered blast stage; no p53, and no apoptosis

c. Additional tumor suppressor genesi.NF1neuroblastoma tumor

ii.APC colon & stomach canersiii.BRAC1breast canceriv.Increased methylation of DNA may be associated with the silencing

of particular tumor suppressor genes

2. TO DESCRIBE PROTO-ONCOGENES AND ONCOGENES.a. Proto-oncogenes

i.Genes that normally encode growth promoting proteins (e.g. growth factors and their receptorsinvolved in signal transduction and nuclear transcription factors)

b. Oncogenesi.Mutated proto-oncogenes that normally encode a growth promoting factor

ii.Only one of the cells two gene copies needs to be altered (a domin anteffect); unlike many tumor suppressors

iii.Tumor cells generate many of their own growth signals3. TO DESCRIBE BASIC FEATURES OF SIGNAL TRANSDUCTION PATHWAYS.


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iv.Altered regulation of transcription factors1. Fosand Junare transcription factors that form a heterodimer

(AP1 complex) prior to binding to DNA

a. AP1binds promoterand enhancer elements containing the sequence5 TGAGTCA 3(TATA box)

b. This interaction plays a critical role in normal cellsc. Elevated levels of fos and jun results in increased

and continuous expression and cell growth

2. Burkitts Lymphomaa. MYC gene (chromosome 8) is translocated to an Ig gene

(chromosome 14),i. C-myc gene is overexpressed incells that

produce IgH chains (B lymphocytes) causing

these cells to become cancerousb. The Myc gene encodes a transcription factor that regulates expression of

~15% of all genesc. It binds to enhancer sequences (E-boxes) and recruits histone

acetyltransferases (HATs) which remove positive charges fromnucleosomes, making negatively charged DNA more accessible

d. Mutated forms of Myc (where it is constitutively expressed) lead to theup-regulation of many genes, some which are involved in cell proliferation

v.Cyclin dysregulation1. Excessive or inappropriate expression of various cyclins is

related to diverse malignancies

4. TO DESCRIBE HOW DNATUMOR VIRUSES TRANSFORM CELLS.a. Tumor cells can arise via infection with a tumor virusb. Two distinct types of tumor viruses

i.Those with DNA genomes1. Simian virus 40, papilloma, and adenoviruses2. Protein-protein interactions play a key role in transformation

by DNA tumor viruses3. In cells transformed by SV40, the viral protein T-antigen is

complexed to p53 and also pRB, making them both unable

to do their joba. By binding RB it releases E2F and the cell enters S phase so the virus can

replicate

4. In HPV,two viral genes (E6 and E7) are actively expressed upon insertion ofHPV DNA into hosts genomea. E7 binds pRB, releasing E2F

b. E6 complexes with p53 and is able to induceproteolysis of p53

ii.Those with RNA genomes1. HIVand human T-cell leukemia viruses (HTLVs)


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exam 1 study guide.2

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