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					Jorde Book notes
Saturday, May 24, 2008
10:33 AM

   5/24/2008, 10:34 AM
   Pg 40-52
   DNA Technology
   Detection and measurement of genetic variation
   Blood groups                        • Using antigens on surface of erythrocytes
                                       • 2 bood group systems ABO and Rh
                                       • ABO
                                             ○ 2 antigens possible to be present on erythrocytes (A & B)
                                             ○ Can have A, B, AB, or O (neither)
                                             ○ Have antibodies towards the antigenes we don't have
                                             ○ Encoded by single gene on ch 9
                                       • Rh system
                                             ○ Defined on basis of antigens present on erythrocyte surfaces
   Protein electrophoresis             • Charge differences between proteins with a single AA mutation can be determined
                                       • Good for Sickle cell disease mutation detection
                                            ○ Test for normal (HbA) or sickle cell (HbS) hemoglobin
                                       • Can not detect silent mutations (genetic code still codes for same AA) or AA substitutions which do not alter the
                                         charge
   Restriction fragment length         • Uses bacterial enzymes (restriction endonucleases or restriction enzymes)
   polymorphisms (RFLP)                • search for restriction site polymorphisms (RSPs)
                                       • Cut at specific restriction sites (usually palindromes which will cause "sticky ends")
                                             ○ Also used for genetic engineering human DNA into bacterial plasmids
                                             ○ Vectors which can be used as cloning vehicles include cosmids and YACs (yeast artifical chromosome)
                                                which are able to insert larger and larger strands
                                       • Process:
                                            1. DNA exposed to restriction enzyme for digest
                                            2. Fragments subjected to gel electrophorisis
                                            3. DNA denaturation (exposure to alkaline chemical solutions)
                                            4. Transfer to solid membrane for "southern blot" analysis
                                                     i. Probes are used to determine where specific sequences are located on an array
                                                    ii. Exposure to x-ray film produces an autoradiogram
                                       • Example for sickle cell disease:
                                             ○ Normal individuals have glutamic acid at position 6 of the B-globin polypeptide (DNA should be GAG)
                                             ○ Sickle cell mutation causes GTG which places valine instead of glutamic acid
                                             ○ Restriction enzyme MstII can be used to cut at this sequence in normal code, but not sickle cell
                                             ○ Those with the disease will have a larger, single DNA fragment vs those without having 2 smaller
                                                fragments
                                       • Polymorphisms in disease genes have been found for CF, Huntington disease, and type 1 neurofibromatosis
   Tadem repeat Polymorphisms          • Minisatellites = regions where the same DNA sequence is repeated over and over again, in tandem
                                       • Number of repeats in a given region varies substantially from individual to individual (looking at veriable number
                                         of tandem repeats = VNTRs)
                                       • Process:
                                             1. DNA digested
                                             2. Electrophoresed
                                             3. Denature
                                             4. Transfer to solid medium
                                                    i. Difference is that probes look for only given minisatellite regioins
                                       • Used mostly for mapping genes by linkae analysis
                                       • Microsatellite repeats are smaller di-, tri-, or tetra-nucleotide repeats (2,3,or 4 bp)
                                       • Short tandem repeat polymorphisms (STRPs) are isolated by PCR
                                       • Polymorphism of choice for most gene mapping studies
                                       • Both VNTRs and STRPs are useful for forensic applications
   PCR (polymerase chain reaction)     • Artifical means of replicating a short, specific DNA sequence very quickly so that millions of copies are made
                                       • Requires:
                                              ○ 2 primers corresponding to the DNA sequences immediately adjacent to the sequence of interest
                                              ○ DNA polymerase which is themally stable
                                              ○ Large number of free DNA nucleotides
                                              ○ Genomic DNA which will be copied
                                       • Process
                                             1. DNA denaturing at high temperature
                                             2. Primer hybridization at low temperature
                                             3. Primer extension at intermediate temperature
                                       • Pros to PCR
                                              ○ Able to use extremily small amounts
                                              ○ Much faster then cloning (sickle cell determination takes only 1 day vs 1 week with cloning)
                                              ○ Can use safer, nonradioactive means to detect mutations (since such a high amount of pure DNA is
                                                produced)
                                       • Cons
                                              ○ Primer synthesis requires knowledge of DNA sequence flanking DNA of interest
                                              ○ Extreme sensitivity makes it susceptable to contamination
                                              ○ Must be used for on relativly small sequences

   DNA sequencing
   Dideoxy method • Dideoxynucleotides are are used to terminate chain growth
                  • They are missing the OH group which would normall participate in the phosphodiester bond when a new nucleotide is added
                  • A larger amount of regular nucleotides are used for PCR but the small amount of one of the Dideoxynucleotides will cause chain
                    termination ant various points
                  • This process is performed 4 times, once for each base and the fragments are run sid by side on a gell, allowing us to piece
                    together the sequence



                                                         Biochem Page 1
                      •
                        together the sequence
                      • New automated ways of doing the same thing use fluorescent, chemiluminescent or colorimetric detection systems. These have
                        the advantage of doing just one PCR reaction with specific fluorochromes on each set of ddNucleotides to allow for each to emit a
                        distinct spectrum of light
                      • Capillary sequencing allows for thing tubes to be used instead of electrophoresis
                             ○ This is much faster since it produces no heat
                      • Allows for sequencing of a particular gene to find mutations which may be too smalll to detect in Southern blot or that do no occur
                                                                                                                                                     t
                        at known restriction sites

    Genetic variation is caused by several different effects, most notably:
      ○ Natural selection
              Increases certain favorable mutations based on vitality of subject with mutation
              Example is sickle cell: heterozygous form actually increases survival for those in locations where certain malaria is found ( west africa)
      ○ Genetic drift
              Fluctuations occur at a higher frequency in a smaller population then a larger one
              Also seen as the founder effect
              Example is Ellis-van Creveld syndrome found in the amish community in U.S. and Tay -sachs disease in Jewish populations
      ○ Gene flow
              When migrants from different populations mate
              Lead to more genetically similar populations
              Example is African-americans in US have lower incidence of sickle cell then those in Africa



5/24/2008, 12:53 PM
Pg 160-192
Gene mapping and cloning
Linkage analysis
Linked       • If 2 loci are near each other on the same chromosome they will be transmitted to the same gamete
Loci         • If they are on different chromosomes, there is a 50% chance they will end up in the same gamete, this is known as the princip le of
               indipendent assortments
             • Recombination allows for new allele combinations (haplotypes) by crossover during meiosis
             • centiMorgans measure distance between two loci by means of recombination frequency (how frequently recombinations occur)
                     ○ 50 cM is considered to be unlinked as this would equate to indipendent assortment

Polymorphisms can be used as markers to follow a disease gene throuh a family
Several factors affect crossover rates such as
  ○ 1.5 x more common in oogenesis then spermatogenesis
  ○ More common near telomeres then centromeres
  ○ Recombination hot spots occur which are areas on a chromosome which exhibit substantially elevated crossover rates

LOD score: Logarithm of odds of likelihood of observing a pedigree vs not seeing it
     Use pedigree data to determine likelihood of linkage at a given recombination frequency divided by the likelihood of no linkage
     Scores of 3.0 or higher are taken as evidence of linkage where as scores lower than -2 are seen that there is no linkage

To be useful in gene mapping, linked marker should be codminant, numerous, and highly polmorphic (increases the probaility th at matings will be
informative

Linkage equilibrium occurs when a preferential association between a disease gene and a specific allele is not seen in a larg e collection of families but it
is seen within one family. This is a result of recombination through many generations. Can be used to determine order of ge nes on a chromosome.

Physical mapping and cloning:
Heteromorphisms      Variation in the appearance of a chromosome
Dosage mapping       Looking for gene deletions and duplications by analyzing the change in levels of proteins
Positional cloning   If you know of a linked marker but do not know the gene you can canvas the region around the marker to locate the disease gen e
Functional cloning   Starting with the AA sequence to deduce the DNA secquence and then search for the gene
                            Sequence taged sites (STSs) can aid with this as their location in the genome is known
CG islands           Unmethylated CG dinucleotides in the 5` region most likely indicates a coding gene




                                                              Biochem Page 2
Lec-1
Wednesday, May 28, 2008
3:51 PM


Covalent bonds are considered irreversible bonds because of the change in energy which takes place
  ○ They allow for single, double, and triple bonds as well as resonance structures
  ○ Chemical reactions require making and breaking covalent bonds

Electronegativies: O>N>S≥C≥H
  ○ General rule is the higher and more to the right on the periodic table you are, the more electronegative you will be
  ○ More electronegative atoms will draw electrons to them
  ○ Polar substances tend to have higher boiling points then nonpolar substances

Non-covalent interactions (reversible bonds):
      Hydrogen bonds               H and partial (-) atom, polar groups "share" the H (between N, F, or O only)                           5 kcal/mol*
                                         Length of H bonds can very based on which molecule they are bonded to
                                         There is however a very strong angle dependence with linear bonds being the strongest
      Salt bonds                   Electrostatic interactions between permanently charged residues                                        1 kcal/mol*
      Van der Waals                Transiently partial (-) and partial (+) atoms, induced asymmetric distribution of electrons            ~1 kcal/mol*
      Hydrophobic interactions     Nonpolar domains/groups shielded from water, (the organization of water is energetically unfavorable
      * covalent   bonds are about 71 kcal/mol on average

      These reversible bonds are the basis for macromolecular 3D structure


If water is a product (condensation) the reaction is endergonic (energy requiring)
If the reaction needs water (hydolysis) it is exergonic (energy is released)

Breaking bonds releases energy, building them requires it




                                                            Biochem Page 3
Lec-2
Wednesday, May 28, 2008
3:43 PM


Differences between the following isomers (different compounds with same molecular formula):
Positional (constitutional)   • Different position of atoms of functional groups within molecule
                              • Different physiochemical properties
Geometric (stereoisomers)     • Differ in geometric position of different parts of the molecule (cis/trans)
                              • Different physiochemical properties
Optical                       • Asymmetric carbon has four different substituents
                              • If only 1 asymmetric carbon = enantiomers (mirror image)
                                     ○ Enantiomers have identical physiochemical properties, but they will react with other molecules differently
                                     ○ Different enantiomers can lead to very different biological responses as noted with various drugs and even sweeteners
                                             Because of the possible unwanted effects, all new drugs must be pure (non-racemic) to pass FDA approval

Acid base theroy:
Acid is proton donor (adds H+ to solvent)
Base is proton acceptor (removes H+ from solvent)
       Strong acids and bases will dissociate completily, weak acids and bases will not
pKa       The -log of K a (Ka is the acid dissociation constant)                            Clinic significance
          The larger the pKa, the weaker the acid
          If pH equals pKa the group is half-protonated
          If pH > pKa it is mostly deprotonated
          If pH < pKa it is mostly protonated
pH        Inverse relationship with H concentration                                  Normal blood pH is about 7.4, dropping below 7.35 is acidosis
          Derived from equation where [H] and [OH] concentrations multiplied by each Low pH (high H) leads to:
          other would equal 10-14                                                        ○ cell/tissue damage
                                                                                         ○ Protein alteration
                                                                                         ○ Interference with normal physiological function
                                                                                     High pH (low H) could also cause problems but is rare
buffers   • Aqueous systems which resist changes in pH when small amounts of                Body has several buffers in various locations to maintain that pH does
            strong acids or bases are added                                                 not become out of whack
          • Consists of a weak acid and its conjugate base
          • Best buffering occurs around pKa, (region of maximum slope on a titration
            curve)
                ○ Most effective at pHs +/- 1 pH unit of pKa




                                                                   Biochem Page 4
Lec-3
Wednesday, May 28, 2008
3:45 PM

Define:
Aldose              Monosaccharide with aldehyde group (C bound to -R, -H, and =O)
ketose              Monosaccharide with ketone group (C bound to -R, -R, and =O)
triose              3 carbon monosaccharide
pentose             5 carbon monosaccharide
hexose              6 carbon monosaccharide
hemiacetal          C bound to -R, -OH, -H, and -OR`
hemiketal           C bound to -R, -OH, -R`, and -OR``
Glycosidic bond     bond between anomeric carbon of carbohydrate and some other group or molecule
(glycosidic link)      ○ Example is C-O in disaccharides and C-N in nucleosides

General structure of monosaccharide (open chain and cyclic form)




What is a reducing sugar?
 ○ A reducing sugar is one which has a hydroxyl group on the anomeric carbon
 ○ Carbohydrates with only acetal groups (all glycosidic linkages) are non-reducing sugars
 ○ Reducing sugars will test positive in Tollen's or Benedicts reagents
          All aldoses will test positive
          Ketoses test positive but it is actually the enediol intermediate which is being oxidized

What is mutarotation?
 ○ The change in optical rotation as an equilibruim mixture of anomers forms
 ○ Example is alpha and beta forms of glucose which in aqueous solution will slowly move towards equilibrium

What are structural differences between amylose, amylopectin, glycogen, and cellulose?
 ○ Starch is the term for the glucose based storage molecules in plants (amylose and amylopectin)
 ○ Amylose (alpha 1-4 ) is in a "straight chain" which forms a compact helical arragement.
 ○ Amylopectin (alpha 1-4) has branching points (alpha 1-6) every 20-25 glucose units
 ○ Glycogen (alpha 1-4, branch alpha 1-6) major carbohydrate storage molecule in animals
         Similar to Amylopectin except far more branching (every 16-18)
 ○ Cellulose (beta 1-4) has relatively straight chains (the beta linkage is linear compared to alpha)
         Chains can H-bond (again better because of linear line up with beta link) with each other to give insoluble fibers found in plant cells
         Humans can not cleave the beta linkages so we can not use for energy




                                                            Biochem Page 5
Lec-5
Wednesday, May 28, 2008
3:46 PM


What is the relationship of DNA, genes, chormosomes, and genomes?
      Chromosomes are made up of thousands of DNA which are deoxyribonucleic acids. A gene is a DNA sequence which encodes for prot eins and
      functional RNA (something which is of value to the host). All the genetic material for an organism is known as its genome. In humans, genes make up
      less than 3% of their total genome.

Number of genes in our body: ~30,000 # proteins: many more
How is this possible?
      Each human gene can make many different proteins, by process called alternative splicing. 1 gene could code for up to 1000 proteins.
             Alternative splicing is a process by which different proteins are produced by splicing different sets of exons together from same gene (RNA
             transcript).

What are 3 important findings that the human genome project has provided to us?
 1. There are a relatively small number of human genes (<30,000).
 2. Proteome is complex, 1 gene can code for up to 1000 proteins. There is a variation in gene regulation as well as post transcriptional modification.
 3. Hundreds of genes appear to have come from bacteria.

What are the differences and similarities between genetic healthcare and genomic health care?
     Genetic health care was persued before the HGP and was based on understanding the impact of single genes on diseases. Genomic health care is
     based on understanding the impact that the entire genome- as well as environment al factors- have on disease as well as health. It is built on the
     foundation of the Human genome project. They both try to find the underlying cause of disorders but genetic health care is a much more focused look,
     perhaps missing some of the broader connections. Genomic Health care will one day provide the ability to define disorders by their biological
     causation, rather than by the symptoms. It will also allow for personalized medicine, from pre-screening for certain diseases to personalized
     medications.

What is GWAS and how is it expanding our understanding of the human condition?
      GWAS stands for Genome-wide association studies. They are studies of genetic variation across the human genome, often times SNP (single
      nucleotide polymorphisms). If people with a particular disease have a particular genetic difference when compared with those without the disease, it
      can be assumed that the disease is associated with that genetic difference. GWAS allows us to hone in on the small variations in our genome
      responsible for the many different diseases which have a genetic cause/susceptibility

How is DNA packaged into a metaphase chromosome?
      DNA in the standard double helix is approxamitly 2 nm wide. This is normally wound 2 times around proteins called histones, giving a 1st order unit of
      packing known as a nucleosome. DNA is drawn to Histones because they have several of the positive side chain groups (lys, arn) which attract the
      negatively charged DNA. The nucleosomes are about 10nm wide and contain about 200 bp each (core contains only ~160 bp). Wit h the help of H1, 6
      nucleosomes are wound around themselves giving a 2nd order compaction unit called a solenoid, about 30 nm wide. Solenoids form looped domains
      giving a chromatin fiber which has loops on average 300nm away from the scaffold proteins. This is usually the highest level of packing in an
      interphase cell, but in a metaphase cell it will coil and fold into a traditional chromosome (each chromatid ~700nm wide).

What is the structure and function of a nucleosome?
      Nucleosomes are DNA wrapped 2x around the octamer of core histones (H2a,H2b, H3, H4). The nucleosome is about 11nm wide and contains about
      200 bp after the 2x wrap (Core contains only 160 bp). This initial packing allows for organized condensing of even higher orders, while keeping the
      DNA stable. The nucleosomes keep the DNA negatively supercoiled which aids later in strand seperation.

What are centromers? What are telomeres?
 ○ Centromeres are the largest constriction of the chromosome, they are the site of the kinetichore which is the attachment for spindle fibers which will pull
     the 2 chromatids away from each other during mitosis. They are made up of thousands of 171 pair repeats (called alpha satellite sequences). These
     areas have not been well solved in regards to the genome project because of all the repeats.
 ○ Telomeres are the tips of chromosomes and they usually have repeats of short nucleotide sequences. Every replication shortens telomere which
     destabilizes the chromosome.

What is a karyotype and when should a doctor order one?
      A karyotype is a chart of all the chromosomes in a particular person. The Chromosomes are seen here in their relative sizes, they were named based
      on their size and centromere position. Dark spots on the Karyotype are heterochromatin which are close to nuclear membrane during interphase. They
      are able to tell about specific genetic abnormalities fairly quickly as particular pieces will be either missing or duplicated.




                                                             Biochem Page 6
DNA structure and Replication
Wednesday, May 28, 2008
3:47 PM




What are the components that make up the structure of DNA?
 ○ DNA in the strictest definition would be made up of only a sugar (deoxyribose), a base (A,T,C, or G), and a phosphate group
 ○ A nucleoside consists of a base linked by an N-glycosidic bond to the 1`-carbon of 2-deoxyribose. Nucleotides also have a phosphate group attached to
     the 5`-carbon of the 2-deoxyribose.
 ○ DNA is itself an unbranched polymer of 2-deoxyribonucleoside monophosphates which are linked by phosphodiester bonds between the 3` end of one
     deoxyribose and the 5` end of the next deoxyribose. The principle higher order structure of DNA is a right-handed double-helix.
 ○ There are major and minor grooves in DNA which allow for the base sequence to be recognized by DNA binding proteins even in the helix.

How is DNA replicated?
  ○ DNA is replicated in what is called semi-conservative replication. This means that each newly replicated DNA will have one strand which is new and one
      of the parent strands.
  ○ The basic mechanism for DNA replication is: unwinding of the helix (replication fork) gives 2 strands, synthesizing of new DNA using each old strand as a
      template
  ○ Euchromatin tend to be replicated sooner then heterochromatin
  ○ In S-phase there is recognition of certain origin sequences which are spread across the chromosome by binding proteins which start melting the DNA
      around that area. DNA helicase (a homo-heximer) moves along the lagging strand, unwinding the DNA into 2 separate single strands. SSB (RPA in
      eukaryotes) binds and stabilizes ssDNA. An RNA pol (Pol α in eukaryotes) makes a primer for each strand allowing DNA pol (Pol δ in Eukaryotes) to
      come in an start making DNA, one of the strands will be able to progress continuously, while the other will be a lagging strand, made up of Okazaki
      fragments. These will be pieced together later first by the removal of the RNA primer by an RNAse H, completion of the DNA by DNA pol, then finally the
      phosphate bond connection by Ligase.
  ○ Opening the DNA heliix instills positive supercoils to the DNA

What are the proteins invovled in replications process?
      Prokaryotic                             Eukaryotic                                          Function
     • SSB                                    • RPA (Replication protein A)                       •   Binds to sugar phosphate backbone, leaving bases exposed
                                                                                                  •   prevents DNA from annealing
                                                                                                  •   Required for genomic stability
                                                                                                  •   regulator of DNA damge-induced cell cycle arrest
                                                                                                  •   RPA deficiency will cause spontaneous DNA damage,
                                                                                                      apaoptosis and induction of an ATM dependent G2/M
                                                                                                      checkpoint
     • Primase                                • Pol α                                             • synthesizes RNA primer
     • Pol III polymerase                     • Pol δ                                             • Elongates/copies DNA
                                                                                                  • Not very processive (falls off DNA easily)
     • β2 sliding clamp subunit of Pol III    • PCNA (proliferating cell nuclear antigen)         • clamp which provides stabilization to the polimerase
           ○ Dimer                                 ○ Homotrimer, able to increase DNA             • Clamp holds a lot of H2O and DNA is attracted to H2O
                                                     syntehsis 1,000-fold
     • 3` exonuclease of Pol 1                • RNaseH + FEN1 (flap endonuclease)                 • removes RNA primer on okazaki fragment
     • γ subunit of Pol III (sliding clamp    • RFC                                               • displaces the primase/Pol α and recruits stabilizatoin clamp
       loader)

Compare prokaryotic and eukaryotic DNA replication.
      Prokaryotic                                           Eukaryotic
     • 1000 nt/sec                                          • 50 nt/sec
     • 1 replication origin                                 • Many replication origins (possible 7,000 for 1
     • Circular DNA                                           chromosome)
                                                            • Associated with histones
                                                            • Specialized DNA polymerases
                                                            • Linear chromosome
                                                            • In S-phase there are 4 copies of each gene

      Both organisms replicate polynucleotide chains in the 5`-3` direction and need an RNA primer.

What drugs target DNA replication and how do they interfere with the process?
      Camptothecin                             Topo I inhibitor                 Chemotherapeutic agent
      Doxorubicin                              Topo II inhibitor                Chemotherapeutic agent
      Ciprofloxan                              Topoisomerase DNA gyrase         Antibiotic
                                               (only found in bacteria)
      Telomerase RNA as antisense target       telomerase                       Prevent extension of telomeres, leading to cell death in cancer
      G-tetraplexes as drug target             G-tetraplexes                    Stabilize the G-tetraplex structure inhibiting telomerase activity/DNA replicatoin
          ○ NMM
          ○ TMPyP4
          ○ Telomestatin

What diseases result from defects in DNA replicatoin mechanisms and explain which enzymes are defective
      Disease                     pathology                                      symptoms
      Xeroderma pigmentosum       Nucleotide excision repair gene mutation       2000 fold increase in sunlight-induced skin cancer
      Ataxia telangiectasia       ATM mutation, (DNA damage detection)           Increased risk of X-ray, breast cancer
      Fanconi anemia              Ubiquitin ligase defect                        Increased risk of X-ray, sensitivity to sunlight
      Bloom syndrome              DNA helicase gene mutation                     Increased risk of X-ray, sensitivity to sunlight
      Cockayne syndrome           Defect in transcription-linked DNA repair      Sensitivity to sunlight
      Werner's syndrome           DNA helicase gene mutation                     Premature aging

      Proof reading allows for high fidelity of replication, in order to be able to proof-read the polymerases must always add bases to the 3` end and



                                                                   Biochem Page 7
Proof reading allows for high fidelity of replication, in order to be able to proof-read the polymerases must always add bases to the 3` end and
exonucleases must therefore always cut off this end too, this is the only way a new tri-phosphate base can come in to bind.
    RNA viruses do not have proof-reading since they copy RNA to RNA, leading to a high rate of mutation.
    Proofreading activity by 3`-5` exonuclease increases accuracy of replication by 102-103 fold.
    Methylation machinery is able to improve DNA accuracy after DNA replication
    Longer living organisms have better repair machinery

Topoisomerases: relieve supercoiling by process where covelant intermediate is formed with DNA
   Type 1 cuts only 1 strand and wraps it around the other
   Type 2 makes nicks in both DNA strands allowing untangling of DNA helices as well as relieving supercoiling
   Energy to reseal the chain is kept in the covelant bond with the Topo

Telomers
    Unable to replicate ends in the same fashion due to RNA primer necessity
    Telomerase, enzyme with integrated RNA template
    Binds to repeating sequence (TTAGGG) and allows for extension of chain
    2 possible structures are possible at the ends of the chromosomes: T-loops and G-tetrads
    In Werner's Syndrome (advanced aging) telomers are shorter than normal

Histones split during replication and each new ds gets 1/2, chromatin assembly factors add new ones after replication




                                                        Biochem Page 8
Chromosomes and Chromosome Diseases
Wednesday, May 28, 2008
3:49 PM




Telomere of lagging strand will not be "finalized" because the RNA primer which is required for Pol to work will be degraded by Rnase H, leaving an open single
strand of DNA, this will fold back upon itself to create the T-Loop which is a fold back triple strand structure. This structure attracts certain capping proteins,
preventing the DNA from being mistaken for a virus and degraded.
Telomere contains 3-20,000 bp of TTAGGG repeats. There are other repeats which are more internal which allow for capping of the strand
Telomerase is not normallly active in most somatic cells but in cancer it is
Telomerase is the only enzyme with its template (RNA) carried within it

Chromosomes in nucleus always seem to be in the same respective places

Centromere is the attachment site for mitotic spindle (kinetocore)

To obtain Karyotype:
  1. Obtain cells (blood, bone marrow, embryonic…)
  2. Add phytohemagglutinin to force cell divison (mitosis causes chromosomes to condense.. If obtained in pro-metaphase you can increase the # of bands
      seen)
  3. Culture for 3 days
  4. Add colchicine to interfer with the spindle apparatus so that chromosomes do not get pulled apart
  5. Hypotonic saline will cause cells to flood, making them engourge
  6. Add fixative to stop enzyme activity
  7. Digest w/ trypsin after placing on slide (bursting cell membrane), allows for proteins to be removed and banding to occur
  8. Stain with giemsa to observe bands

If using bone marrow you don't need to culture but it will hurt the patient more. Prenatal tests differ in their risk to thefetus and the time frame in which they can
be obtained (amniotic cells take 2-3 weeks to culture but low risk to preg, also have to weight until around 16th week of preg. Chorionic villus sample does no     t
need a culture and can be taken about the 10th week of preg but at high risk of losing the fetus)

Satellites on certain chromosomes (acrocentric ones, 3 in group D (13-15), 2 in group G (21-22 and Y)) are areas where DNA for rRNA is located. These can be
lost with minimal effect on the phenotype of the patient as there are "backup" copies.

The dark G-bands on the karyotype are representative of heterochromatin which get replicated after the euchromatin. They probably have shorter loop structures
                                                                                                                            e
and are more condensed then euchromatin but they still have active genes (this is why extra barr bodies will still give alter d phenotype). Heterochromatin is
higher in A-T basepairs and found in a condensed state even in interphase. R-banding gives reverse banding pattern.

                                                                                                                             r
If there is a numerical error in the cell, there is no possible way to be developmentally normal, either the fetus will die o developmental problems will persist

                                                                                                                             ost
-ploidy refers to all genes and polyploidy would be multiples of all genes, this is not compatable with life and one of the m common reasons for early preg
loss.

                                                                                                                                x
-somy refers to numbers of one chromosome. Trisomy is 3 copies of one chromosome and is only viable for 13, 18, 21, or the se chromosomes. The only
monosomy compatable with life is Monosomy X. Trisomy (and Monosomy X) is usually an issue with non-disjunction of the maternal gamette during Meiosis I.
                                                                                                                            m
Trisomy 21 and monosomy X are the most viable genetic defects and they still have extremily poor chance of surviving till ter (20% and 1%)

Karyotype abnormalities
(note that heart and brain are the greatest effected by genetic disorders, due to the large amount of proteins which each expresses)
Isochromosome formation: • One chromosome is pulled apart correctly and another ends up with both p arms in one cell and both q arms in the other.
                         • Denoted on karyotype nomenclature as (iso)
Ring chromosome               • If the telomeres at both ends are lost, the chromosome will form a ring to save itself from degradation
                              • Denoted as (r)
Reciprocal translocation      •   Exchange of 2 fragments of chromosome, denoted as (t) in karyotype
                              •   Can be stable (balanced) or not stable ( 2 centromeres on one chromosome one chromosome with none)
                              •   Robertsonian translocation is a form where both q arms of acrocentric chromosomes are combined, the satellite portions are lost
                              •   Translocations occur from unsuccessful repair of chromosome breaks
                              •   If all genetic material is conserved, then phenotype will be normal (a balanced carrier), although offspring have a 50% chanc e of
                                  having partial trisomy and monosomy for the chromosomes invovled (remember only chromosome 13, 18, 21, and the sex
                                  chromosomes can survive trisomy, only X can survive monosomy), and a 25% of being a balanced carrier themselves, 25%
                                  chance of being normal

Chromosome Syndromes
Down              Diag:                                                  Causes
Syndrome              ○ mental retardation                                   ○ Nondisjunction in maternal meiosis I
                      ○ 40% have heart malformations                         ○ 3-4% have translocation (usually robertsonian
                      ○ Hypothyroidism (if not treated will lead to even     ○ 1% mosaicism (caused by non-disjunction during developmental mitosis- not
                        more advanced complications)                           maternal meiosis)
                      ○ Karyotype shows trisomy 21                       Incidence
                  Ageing effects                                             ○ 1/700 live births
                      ○ Early senility (may be due to poor nutrition         ○ More than 60% are spontaneously aborted
                        also)                                                ○ 20% are stillborn
                      ○ APP gene is attributed to Alzheimer's and is on
                        ch 21, extra chromosome may lead to
                        overexpression
Edwards           Diag:                                                     Cause:
syndrome              ○   Prominent occiput                                     ○ Trisomy 18
                      ○   Small chin                                        Incidence
                      ○   Rocker-bottom feet                                    ○ 1:8000, 80% female
                      ○   Heart defects                                         ○ Most die w/in 1 year (most don't survive till term)
                      ○   Failure to trhive
                      ○   Profound mental retardation
                      ○   Hypertonicity
                      ○   Clenched fingers
Patau             Diag                                                      Cause:
Syndrome              ○ Sloping forehead                                       ○ Trisomy 13
                        Forebrain defects                                        Some translocations/mosaics



                                                                  Biochem Page 9
                      ○   Forebrain defects                                   ○ Some translocations/mosaics
                      ○   Cleft lip/palate                                Incidence
                      ○   Polydactyly                                         ○ 1:20000 (affects both sexes equally)
                      ○   Heart defects                                       ○ Most die w/in first year (again most do not make it to term
                      ○   Profound mental retardation
Turner          Diag                                                      Cause:
Syndrome            ○ Short stature                                           ○ 45 X (Monosomy X)
                    ○ Undeveloped/degenerated ovaries                     Complications:
                    ○ Missing development of secondary sexual                 ○ Heart and Renal defects in 30%
                      characteristics                                         ○ 99% do not make it to term, fairly mild post birth though
                    ○ Broad, widely spaced nipples                        Treatments:
                    ○ Normal Intelligence                                     ○ Estrogen given at ~12yo to induce secondary sexual characteristics
                    ○ Webbed skin fold of neck at birth                       ○ Growth hormone at 8yo to increase height
                    ○ Pinky and thumb are sometimes same length               ○ If virilization appears there may be "hidden" Y-chromosomal material and gonands
                    ○ Pre-birth ultrasound will show nuchal                     will need to be removed due to risk of malignancies
                      translucency                                        Incidence
                                                                              ○ 1/2500 to 1/5000
                                                                              ○ Mosaic in 30-40% of patients (could be 45X/46XX or 45X/46XY)
                                                                              ○ Maternal age effect is not seen with Turner syndrome
Klinefelter     Diag:                                                     Causes
Syndrome            ○ Men seeking help with infertility                      ○ 56% extra maternal X
                    ○ Boys with developmental delay or social                ○ 44% extra paternal X
                      maladjustment                                          ○ 15% are mosaics
                    ○ More than average lentgh, long limbs, small         Complications
                      testis                                                 ○ Normal to mental retardation
                    ○ Produce reduced levels of testosterone                 ○ Obesity, diabetes, pulmonary disease
                    ○ Reduced sexual function                                ○ Thyroid function problems
                    ○ 55% experience gynecomastia (increases risk            ○ High pituitary gonadotrophins due to lack of feedback inhibition with non-
                      of breast cancer                                         producing testis (treat with androgens)
Trisomy X       Diag:                                                     Cause:
                    ○ No abnormality to slight mental deficiency              ○ Maternal age effect (non-disjunction maternal MI)
                    ○ There is more advanced mental retardation           Incidence
                      with each additional X chromosome                       ○ 1:1000 females
                    ○ Most are taller then average
XYY             • Increased height
                • Social trouble
Cri-du-Chat     Symptoms                                                  Cause:
                   ○ Distinctive cry in babies                                ○ Deletion from p arm in Ch 5
                   ○ Low birth weigth                                                 Most are spontaneous but ~10% come from parents who are balanced
                   ○ Most are severly retarded                                         reciprocal translocation carriers
                   ○ Psychomotor retardation                                  ○ Contain several genes (contiguous gene syndrome)
                   ○ microcephaly                                             ○ Larger the # genes affected, thre greater the retardation
                                                                                      The cry is itself a gene so some cases (very few) have just the cry
                                                                          Incidence
                                                                              ○ 1/20,000 to 1/50,000 (however it is seen in about 1/100 of severely retarded
                                                                                patients
DiGeorge        Aka catch 22                                              Cause:
Syndrome        Symptoms                                                     ○ variable deletion of ch 22q11 region (>34 genes deleted in most cases)
                   ○ Defects in aortic area of the heart                     ○ 90% show missing areas when undergoing FISH analysis
                   ○ Hypoparathyroid leading to hypocalcemia
                   ○ Reduced immune function
                   ○ Increased risk of psychiatric diseases

                                                                                                                             nt
Mosaics start out as one genotype and there is a problem somewhere during mitosis, chimerism on the other had are two differe genotypes which fuse during
development

                                                                                                                             w
Comparative Genome hybridization (CGH) is a great analysis tool if we want to detect a duplication/deletion but we do not kno the sequence/area in the
genome it is. Great for seeing subtelometric delettions which are too small to be seen on normal karyotypes

Chronic Myelogenous Leukemia (CML):
  ○ Leukemia is over/mal-production of white blood cells
         Can appear on examination as enlarged spleen and sternum tenderness
  ○ This diagnosis can only be confirmed 100% if the philidelphia chromosome (t (9;22)(q34;q11.2)) BCR-ABL fusion gene is present
         This translocation will cause a "fusion" of genes, causing activation of a protein under non-normal conditions
  ○ Patients c/o fatigue, night sweats, and low grade fever, normally presenting in middle age
  ○ 3 phases, initial lasting up to years, then accelerated and last with bone marrow failure complications
  ○ Bone marrow transplantation can cure, otherwise interferon treatments slow disease progression
  ○ New drug (2001) STI571 can control disease in early stages
         Inhibits BCR-ABL activity, could prove to be more realistic treatment then bone marrow transplant

Sex Determination
  ○ SRY gene on Y chromosome will lead to production of male gonad (default path is female)
  ○ Translocation of this gene can show up as 46,XX males, however the other genes on the Y chromosome are necessary for secondary sexual
     characteristics
  ○ Mutations of the SOX9 or AR genes will lead to 46,XY female

      True                  • Has both testicular and ovarian tissue
      hermaphrodite         • Most are 46XX with SRY, rest are mosaics or chimeras
      Mixed gonadal         • Has testis and streak ovary (similler to turner syndrome)
      Dysgenesis            • Often get virilization at puberty
                            • Must be mosaic 45X/46XY in gonad system, other wise it is considered just turner syndrome
      Female pseudo-        • Only has ovaries by shows virilized phenotype
      hermaphrodite         • 2 causes:
                                  ○ Prenatal exposure to progesterone or androgen
                                  ○ Adrenogenital syndrome: Inherited deficiency of 11 or 21 hydroxylase in adrenal cortex (this would prevent feedback inhibitio n
                                    of progestins (corticosteroids are not able to be formed which would normally prevent secretion of ACTH from pituitary) so




                                                                Biochem Page 10
                       there would be a buildup of androgens)
                     ○ Treatment with cortisol allows for feedback inhibition to ACTH
Androgen        • 46, XY
Insensitivity     Type A
syndrome              ○ Aka testicular feminization
                      ○ No Wolffian ducts
                      ○ Androgen receptor defect (X-linked recessive) so no sensitivity to the male inducing factors
                      ○ Sometimes there is non-descended testis which would have to be removed due to cancer risk
                      ○ 3rd most common cause of amenorrhea
                  Type B
                      ○ Wolffian structures present
                      ○ Virilization at puberty
                      ○ Caused by 5α reductase deficiency (autosomal recessive)
                               Does not allow conversion of testosterone to the more active dihydrotestosterone




                                                   Biochem Page 11
Transcription
Wednesday, May 28, 2008
3:50 PM




Transcription is synthesis of a single-stranded RNA molecule using DNA as a template
  ○ 2 high energy bonds are lost for each nucleotide addition during DNA or RNA chain elongation
  ○ The nucleotide being added to the chain contains the energy to form the bond (the phosphates on
     it's 5` end)
  ○ Polymerization occurs 5` to 3` with template read 3' to 5'
  ○ Differences from DNA synthesis
          NTPs instead of dNTPs
          No primer
          No proofreading
          U instead of T
          RNA pol
  ○ 3 steps of Transcription
        1. Initiation (E. Coli)
               □ Promoter region of gene attracts RNA polymerase
                      -35 bp TTGACA region and -10 bp TATAAT region (pribnow box)
                      These will not be included in message, but terminator will
                      The closer the gene matches these sequences, the higher the affinity for sigma
                          factor
               □ RNA polymerase combines with sigma factor to create RNA polymerase holoenzyme
                      RNA pol holoenzyme recognizes promotor, initiates transcription
                      Sigma factor allows for efficient binding and transcription, each sigma factor will
                          recognize a different promoter sequence
                      Binds loosely to the -35 promoter, where the DNA is kept in ds, tightly to the -10
                          promoter untwisting it
                      E. Coli RNA polymerase is multisubunit (core + sigma = holoenzyme)
                               Alpha x2     Interact with other factors that bind upstream of the -35 box
                               Beta x1      Basically the polymerase (forms the phosphodiester bonds)
                               Beta prme    Binds to DNA template
                               x1
                               Sigma x1     • Recognizes the promoter and facilitates initiation
                                            • Many different sigmas for different sets of genes
                                            • Decreases affinity of core for random DNA but improves it
                                              for specific promoters (Ka = 1013-15)
                                            • Falls off once polymerization starts since you then will want
                                              to get away from the promoter region
        2.   Elongation
               □ Once initiation takes place (after 8-9 bp) sigma factor dissociates from core
                   polymerase, allowing core complex to polymerize chain
               □ Nus A comes in to take the place of the sigma factor, aiding in transcription termination
               □ RNA polymerase completes transcription and 30-50 bp/second
        3.   Termination
               □ Prokaryotes have 2 possible termination sequences
                     1. Type I (rho-independent )
                          ◊ Inverse repeats of G-C rich area near the end of the chain cause a hairpin
                              to form.
                          ◊ The geometrical torsion likely rips the poly U region which follows, off of
                              their poly A template of DNA (A-U is the weakest of the base pairs)
                     2. Type II (rho-dependent)
                          ◊ Rho factor proteins breaks the H-bonds between template DNA and RNA.
                          ◊ Rho-dependent terminators lack the poly U stretch after the hairpin


4 different types of RNA
mRNA     • Messenger RNA
         • Encodes AA sequence of a polypeptide
         • In prokaryotes has Shine-Dalgarno sequcence which is poly Purine before AUG. This
           provides for correct placement of the AUG start codon in the ribosome
tRNA     • Transfer RNA
         • Transports AA to ribosomes during translation
rRNA     •   Ribosomal RNA
         •   Forms ribosomal complexes with proteins
         •   Structure on which mRNA is translated
         •   Regions of DNA which code for rRNA are rDNA (ribosomal DNA)
snRNA    • Small nuclear RNA
         • Forms complexes with proteins used in eukaryotic RNA processing/post processing

Rifampicin:
   ○ inhibits RNA polymerase in bacterial cells by binding to beta-subunit of RNA polymerase (the
     essential polymerase enzyme)
   ○ Used to treat TB, Leprosy, MRSA, and bacterial meningitis due to lipophic nature and uniqueness
   ○ Adverse effects include hepatotoxicity
   ○ Potent inducer of CYP-450 (hydroxylates compounds, making them more water soluble, increasing
     their excretion rate)
          Drugs are then filtered out of the body faster, making their potency appear lower
          Grapefruit blocks CYP-450 activity

What are the differences between prokaryotic and eukaryotic transcription?

Prokaryotic mRNA                                                             Eukaryotic mRNA



                                                             Biochem Page 12
Prokaryotic mRNA                                                                 Eukaryotic mRNA
• mRNA transcript is mature and used directly for translation w/o mod            • mRNA transcript off of DNA is considered pre-mRNA, it is not mature and must
• mRNA is translated on ribosomes before transcription is complete (no             be processed
  nucleus to separate translation machinery)                                     • Transcription and translation are seperated, transcription is compartmentalized
• mRNA's are poly cistronic for prokaryotes (contain AA info for more              by the nucleus where as translation takes place in the cytoplasm
  then one gene)                                                                 • Eukaryote mRNAs are monocistronic (contain AA sequences for just one gene)
Prokaryote polymerases                                                           Eukaryote Polymerases
• Have only one RNA polymerase to transcribe mRNAs, tRNAs, and                   • 3 different RNA polymerases
  rRNAs                                                                                1. RNA pol I: transcribes 3 major rRNAs (28S, 18S, 5.8S)
• This polymerase has only 4 subunits + alpha to make holoenzyme                                1 Promoter is upstream of start site (-170 to -110)
                                                                                                Core promoter occurs at -40 to +20
                                                                                       2. RNA pol II: transcribes mRNAs and some snRNAs
                                                                                                Affected by mushroom toxins (alpha amanin)
                                                                                                Made up of over 10 subunits
                                                                                                Has basal element promoters and proximal element promoters
                                                                                       3. RNA pol III: transcribes tRNAs, 5S rRNA, and snRNAs (U6)
                                                                                                3 types, 5S type has promoters at +50 to +90, tRNA type has
                                                                                                 promoter at +8 to +62, U6 type promoter is upchain
Prokaryotic Initiation                                                           Eukaryotic Initiation (RNA Pol II, only for mRNA)
• Promoters                                                               • Promoters
       -35 bp TTGACA region and -10 bp TATAAT region (pribnow box)             ○ Basal elements (near transcription start -25 bp)
• RNA polymerase combines with sigma factor to create RNA                               TATA box = TATAAAA
  polymerase holoenzyme                                                                 AT-rich is easier to denature then GC-rich
       RNA pol holoenzyme recognizes promotor, initiates transcription         ○ Proximal elements (upstream -50 to -200 bp)
       Sigma factor allows for efficient binding and transcription, each               CAT box = CAAT
        sigma factor will recognize a different promoter sequence                       GC box = GGGCGG
       Binds loosely to the -35 promoter, where the DNA is kept in ds,   • Transcription factors are required (similar to sigma factor for prokaryotes)
        tightly to the -10 promoter untwisting it                               ○ Assemble on basal promoter elements
       E. Coli RNA polymerase is multisubunit (core + sigma =                  ○ Numbered to which polymerase they work with (ex TFIID works with RNA
        holoenzyme)                                                               pol II)
                                                                                ○ RNA polymerase + TFs is called a pre-initiation complex (PIC)
                                                                          • Enhancers are anything which makes the polymerase work better
                                                                                ○ Areas of chain which loop around and stabilize the PIC
                                                                                ○ Can be several Kb from the gene
                                                                                ○ Silencer/repressor factors also exist

Ribosomes of Prokaryotes                                                         Ribosomes of Eukaryotes
• 70 S model                                                                     • rRNA genes are tandemly repeated many times on acrocentric chromosomes
      ○ 50S subunit                                                                (ch 13, 14, 15, 21, 22)
             23S                                                                      ○ Contain internal, external, and nontranscribed spacers (ITS, ETS, NTS)
             5S                                                                 • 80S model
             34 proteins                                                              ○ 60S subunit
      ○ 30S subunit                                                                            28S
             16S                                                                              5.8S
             20 proteins                                                                      5S
                                                                                               50 proteins
                                                                                       ○ 40S subunit
                                                                                               18S
                                                                                               35 proteins
                                                                                 • Transcription of rRNA is done by RNA polymerase I in the nucleus
                                                                                 • Synthesis of rRNA requires its own transcription factors (TFs-TFIs)

What are the 3 major RNA processing events which occur in generating mRNA in eukaryotes?
mRNA processing
 ○ 5` UTR and 3` UTR include exons which will be spliced together with the translated region exons to
     form a mature mRNA ready to be transcribed
 ○ C terminal domain (CTD) is an area on the RNA polymerase where proteins involved in processing
     are able to attach
         These include
             capping factors                                  splicing factors                                    polyadenylation factors
             • 1st processing event                           • 2nd Processing event                              • Final processing event
             • Adds 5` to 5` cap after 20-30 nts have been    • Introns always have 5` GU and 3` AG               • AAUAAA consensus site is found in virtually
               synthesized                                      sequence as well as an A at a particular            every protein known
             • Protects RNA from degradation by                 branch point in the middle                              ○ Found about 10-30 nts from the 3` end
               exonucleases, as well as essential for         • Cleavage occurs at 5` end which is then                    (cleavage site)
               binding to ribosome                              attached at the A mid-chain (forms lariat)        • Poly-A Polymerase adds poly A tail (~200 A's)
             • Methylation of 2`OH of nt 1&2                  • Splicisomes (snRNAs and proteins) mediate         • Poly A associates with PBP (poly a binding
             • Finally CBPs (cap binding proteins) bind the     the splicing                                        protein)
               cap to stabilize the mRNA                             ○ U1/U6 bind 5` end, U2 sets midchain        • PBP interacts with Cap binding protein
                                                                       site, U5 holds complex together            • Stabilization is main goal
                                                                     ○ These are associated with the
                                                                       phosphorylated CTD and loaded as
                                                                       mRNA is transcribed
                                                              • Introns get degraded by the cell
                                                              • Spliceosomes assemble during synthesis of
                                                                pre-mRNA (before transcription is
                                                                completed)


Using examples, how can DNA mutations affect the transcription process and cause disease
Hemophilia Factor IX deficiency:
  ○ There are 2 promoter binding sites for the factor IX gene, each one utilized at a different point in life
         AR (-36:-22) = androgen receptor and is used when androgen levels increase at puberty
         HNF4 (-27:-15) = hepatocyte nuclear factor-4 is used early in development
  ○ Mutations at these different binding sites could allow for expression of the gene at different stages




                                                              Biochem Page 13
  ○
      of life
           Mutation at -20 results in Hemophilia B Leyden which would improve at puberty when
              levels of androgen increase (would only effect binding of HNF4, not AR)
           Mutation at -26 results in Hemophilia B Brandenburg which does not improve throughout
              life (both binding sites are covered in the -26 NA)

Diseases involved with splicing mechanism
Retinitis Pigmentosa     ○ Hereditary retinal degenerations caused byt loss of rods in retina
(RP):                           More specifically, there are mutations in the pre-mRNA splicing factor genes encoding for proteins of the U4/U6-U5 tri-
                                 snRNP
                         ○ Night blindness and loss of peripheral visual field are clinical characteristics
Spinal Muscular          • Affects U5, U2, and U1
Atrophy
Myotonic dystrophy       • Trinucleotide repeat affects regulatory complex and interferes with splicing

Most common result of a mutation is skipping of entire exon
Diseases caused from splice site mutation:
Familial Isolated Growth Hormone Deficiency     Fraser Syndrome (WT-1 gene)            Frontotemporal Dementia and Parkinsonism    Atypical cystic fibrosis
Type II (mutations in GH-1 gene)                                                       linked with chromosome 17 (FTDP-17)         (CFTR gene)
• Mutations cause increased alternative         • Mutations inactivate the 5`          • Mutations alter regulatory elements       • Polymorphisms in
  splicing of exon 3                              splice site, resulting in shift of   • Cause diseas by increasing inclusion of     intron 8 affect splicing
• Causes short stature                            isoforms                               exon 10 of MAPT gene                        of exon 9
                                                • Underdevelopment of eyes
                                                  and genitals

2nd most common result is activation of a nearby cryptic site
β-thalassemia            • 2 types, β0 which has no β chain synthesis on one chromosome and β+ which has some synthesis on one chromosome
                         • βo thalassemia occurs when there is a mutation in the acceptor site (3` AG->GG) of Intron 2, causing a cryptic splice site to
                           be used. There is no β-chain synthesis in this form because a sop codon is introduced with this cryptic splice.
                         • β+ thalassemia occurs in 2 cases
                               a. A mutation in Intron 1 causes a new possible acceptor site. Both sites are used with the mutant being used 90% of the
                                  time and the normal site being used 10% of the time. The mutant translation still gives a pre-mature stop codon, but the
                                  10 % normal transcript will be translated providing those with this form some β-chain synthesis.
                               b. A mutation in Exon 1 causes a new Donor splice site. Both sites are used with the mutant being used 40% of the time.
                                  The incorrect splicing causes a frameshift which activates a premature stop codon. Because this mutation occurs in the
                                  exon region even the in the 60% which splice correctly have a mutation where lysine is substituted in for a glutamate,
                                  although they still show some β-chain synthesis.


Alternate splicing allows for multiple different proteins to be made from one gene sequence, based on the
way exons are spliced together

All mitochondria come from mother, a tRNA mutation disease would be MIDD (maternally inherited
diabetes and deafness). A mutation in tRNA which matchs to leu residues. This mutation reduces the
amount of complex I available to the mitochondria for Ox-Phos pathway, this in turn leads to lower ATP
production




                                                              Biochem Page 14
                    Questions to Guide Your Learning of
                      Molecular Biology Techniques
1. What is a restriction enzyme?
        An enzyme which will cut DNA at known locations (usually pallandromes)
2. To what does the “restriction” part of the name refer?
        The prevention of foreign DNA infiltrating the cell, (class def: something which restricts a virus’
    movement from one bacteria to another OR restricting which strain of bacteria a virus could actually
    infect)
3. How does a bacterium protect itself from its own restriction enzymes?
        Its DNA is methylated at the recognition sites
4. What kind of sequence does a restriction enzyme recognize?
        Palindromes
5. How do you identify the size of a restriction fragment?
        Gel electrophoresis
6. In a electrophoretic gel, what electrical pole do DNA fragments migrate?
        The Positive pole
7. Given a set of markers and a picture of a electrophoretic separation of DNA fragments, be able to
    determine what size of DNA fragments are present on a gel.
        Larger fragments will be closer to negative pole (usually displayed on the top)
8. How is a Southern blot performed?
        After electrophoresis, gel is placed in basic solution (NaOH) to denature (melt) the DNA
    fragments. DNA is transferred to Nylon paper using a “paper wick” procedure. DNA is fixed to nylon,
    possibly by UV radiation Radioactively labeled single-stranded DNA probe is washed over the sample
    binding it to complimentary single strands
9. What is a radioactive DNA probe?
        Short single stranded DNA segment of a known sequence which can be used to “show” what
    DNA is contained within the lines of the electrophoresis gel
10. What is a “sticky end” when it comes to a restriction fragment digestion?
        When a restriction endonuclease cuts a dsDNA it will many times make a jagged cut, cleaving
    phosphodiester bonds of each backbone at non-synchronis locations. This causes an overhang of each
    strand where there are unpaired bases, enabling other DNA which was cut with the same endonuclease
    to potentially attach.
11. What is a vector?
        A complete package (usually a plasmid) for everything necessary to transfect a desired gene into a
    host cell. It contains an origin of replication, recognition sequences for restriction enzymes, and
    reporter genes so that successful transfection can be determined.
12. What are the feature that make a vector so useful for cloning?
        They make sure that the newly introduced DNA will be part of a replication unit, ensuring that the
    DNA is propagated to the host cell.
13. What kinds of vectors exist and how much DNA can each vector contain?
        Plasmids for bacteria and yeast are commonly used, Yeast artificial chromosome (YAC) can hold
    a substantial amount of genetic material and be successfully uptaken into the cell.
14. How does one clone a piece of DNA?
        Restriction endonucleases clip out a piece of DNA which is then introduced to a plasmid/vector
    which has been exposed to the same endonuclease. Combined with a ligase, the DNA will be inserted
    into the plasmid which is then injected into a bacterial cell. Part of the vector would contain a reporter
    gene which would allow for selection of only the bacteria which have taken up the vector.
15. How does one “select” out a clone from the many “clones” that are made during a ligation reaction?
        Knowing where the endonuclease will cut will tell you which resistances your vector should have
    when it has the desired DNA added as well as if an “vacant” plasmid gets taken up. Bacteria which
    show resistance that should have been lost by the insertion of new DNA have the vacant plasmid. You
    would select for Bacteria which only have the resistance you expect.
16. How do you make a genomic library?
        Cut up desired genome by endonucleases and combine with plasmids which have been cut with
    the same endonuclease. Grow on medium which restricts for only the recombinante forms and
    separate out/maintain individual colonies which will contain part of the overall library
17. How do you make a cDNA library? What does a cDNA library really represent?
        cDNA (complimentary DNA) library contains DNA strands which are basically only exons. You
    would create this by combining an mRNA template with a transcriptase along with a oligo dT primer
    (to bind to the poly A tail). cDNA will be synthesized by the transcriptase giving a DNA/RNA
    hybrid. The mRNA is removed and DNA polymerase is allowed to synthesize a new complementary
    strand, and thus cDNA.
18. What is a knockout mouse?
        A mouse which has been genetically modified during embryonic development.
19. How is a knockout mouse created?
          A plasmid is inserted into the embryonic stem cell which causes recombination at targeted gene,
    replacing the natural gene with the newly introduced one
20. For what are microarrays and DNA chips used?
        Screening of thousands of sequences at the same time.
21. How does antisense interfere with gene expression?
        Antisense is a process where hybridization of complimentary oligostrands of RNA to mRNA
    prevents translation machinery from acting on the mRNA.
22. How does siRNA interfere with gene expression?
        siRNA is a dsRNA which will melt and bind to mRNA blocking the ability for ribosomes to
translate it and eventually lead to the fragementation of the mRNA.
23. Why would one use a two hybrid system? How does the two hybrid system work?
        A 2 hybrid system allows scientists to determine which proteins interact in cells. It works by
introducing 2 plasmids to a cell, one in which the protein target is known (the bait) and one in which the
target protein is unknown (the prey). There is also Transcriptional activation domains linked to the prey
protein so that if successful binding occurs, a gene will be activated in the cell to tell that the process
worked.
24. What are the advantages of DNA biotechnology when it comes to making new products that are useful
    to mankind.
        Large scale production of eukaryotic gene products are possible with DNA biotechnology. It
allows for the synthesis of proteins which would have been to difficult to obtain in other ways.
“Pharming” allows for animals to produce useful products in their milk or to make transgenic crops which
are already adapted to a particular environment.
25. What are 5 new “drugs” that have been created by DNA biotechnology methods?
        Insulin, Growth hormone, platelet-derived growth factor, Factor VIII, Erythropoitin, Colony-
    stimulating factor, TPA,
26. How can DNA techniques be used in forensic science?
        With PCR a DNA fingerprint can be obtained. Because of polymorphisms in people can add or
    eliminate different recognition sites to endonucleases each persons pattern of “cuts” will be unique.
27. Be able to read a radioactive DNA sequencing gel.
        Larger fragments will not move as far and therefore stay near the top or (-) side of the gel. Small
    fragments travel fast and will be found farther from the starting point (usually at the bottom).
    Remember that if there is a cut you will get 2 strips farther towards the bottom and the loss of a
    “heavy” or large strip near the top.
28. What components must be in a reaction to do “Sanger” sequencing or “dideoxy” sequencing?
        The template DNA, an oligonucleotide primer, a DNA polymerase able to withstand high
temperatures, dNTPs, a relatively small amount of ddNTPs each labeled with a different florescent dye.
29. How do dideoxynucleotides stop DNA synthesis?
        They lack the 3` hydroxyl group which would normally be a place for phosphorylation
30. What are the steps to dideoxy sequencing?
        Strand separation, primer annealing, extension, and termination (by incorporation of one of the
ddNTPs).
31. What is the principle that makes capillary electrophoresis a valuable tool for DNA sequencing?
        It can be completely automated. Separation of the DNA is precise enough to get incremental
strand lengths which are read by computer.
32. What is the difference between hierarchical sequencing and shot-gun sequencing?
        Hierachical sequencing makes use of known STSs to piece the retrieved fragments back together,
where as shotgun sequencing makes use of computer power to analyze the sequences and put the strand
backtogether.
33. What is a STS?
        Sequence tagged site which act as chromosomal landmarks.
34. What does PCR stand for?
        Polymerase Chain Reaction
35. What are the steps in PCR?
        Melting (~95 C), annealing of the primer (~60 C), extension (~72 C), then repeat.
36. What is Taq polymerase? From what organism was it derived?
        Taq polymerase is a DNA polymerase found in hot springs bacteria which can take the heat.
37. How many DNA molecules exist after 1, 2, 3, 4, 5, 6, and 25 cycles of a PCR reaction?
        21, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64, 225 =33,554,432 ,
38. What is a restriction fragment length polymorphism?
        A polymorphism which will change the ability for a restriction endonuclease to cut at a particular
position. This will give different lengths of fragments and thus a different fragment “fingerprint”
39. How frequently would one expect to find a 4 bp, 6 bp and a 8 bp restriction enzyme site in DNA?
        (1/4)4 = 0.3% or 3 out of 1,000, (1/4)6 = 0.02% or 2 out of 10,000, (1/4)8 = 0.0015% or 15 out of
every 1,000,000.
40. Be able to determine if a person has a disease by looking at a RFLP gel result.
        If they have a line which matchs with normal, they are normal, if the line matchs with abnormal, if
they have lines which match with both, they are heterozygous.
       Questions to guide your learning of Transcription
1. List at least 4 kinds of RNA produced by a cell and describe their function.
       mRNA-encodes the AA sequence of a polypeptide
       tRNA- transports AA to ribosomes for translation
       rRNA- forms ribosomal complex with proteins
       snRNA- forms complexes w/ proteins for processing in eukaryotes
2. In what direction does the RNA polymerase read the template? 3`-5`
3. In what direction does the RNA polymerase polymerize the new RNA strand? 5`-3`
4. How does the polymerization of deoxynucleotides compare to that of ribonucleotides?
       Ribonucleotide polymerization uses NTPs instead of dNTPs, there is no primer, no
       proofreading, U takes the place of T and there is a RNA polymerase
5. What three steps occur during transcription?
       Initiation, Elongation, and termination
6. Describe the elements of a typical E. coli promoter.
       -35 bp TTGACA region and -10 bp TATAAT region (Pribnow box)
7. What is the composition and purpose of the Shine-Dalgarno sequence?
       The Shine-Dalgarno sequence is a poly purine group which appears upstream of the AUG
       codon on prokaryotic mRNA. Its binding to complementary pyrimidine areas on ribosomes
       allow for correct placement of the start codon in the ribosome.
8. What recognizes the Shine-Dalgarno sequence during protein translation?
       The poly pyrimidine area of the 16S ribosomal RNA
9. What is the difference between RNA polymerase enzyme and RNA polymerase holoenzyme?
       RNA polymerase holoenzyme is the combination of the RNA polymerase with the α subunit
10. Explain the physical location of the following with respect to their location on a stretch of DNA:
            a. +1 initiation nucleotide
                       The first nucleotide to be transcribed from the 3` end of the DNA sequence, at
                       the 5` end of the promoter region, before the Shine-Dalgarno sequence in
                       prokaryotes. Usually a T or C, (since RNA synthesis is initiated with ATP or
                       GTP as the first nucleotide)
            b. +1 initiation codon
                       First set of three nucleotide to be transcribed from 3`end of DNA.
            c. Pribnow box
                       The TATAAT promoter located at -10 bp for prokaryotes
            d. -35 region
                       Promoter located -35 bp upstream of the +1 initiation nucleotide
            e. Promoter
                       Located upstream of the transcribed region. In eukaryotes, basal (0 to -25 bp)
                       and proximal (-50 to -200 bp) elements vs -35 bp and -10 bp in prokaryotes.
            f. Shine-Dalgarno sequence
                       Poly purine region of mRNA located upstream of the start codon, so would be
                       poly pyrimidine sequence of DNA located 5` (downstream) of promoter region
            g. Coding region
                       The actual translated region, directly downstream of promoter till the
                       termination region (on DNA this would be a 3`-5` strip between promoter and
                       terminator)
            h. Stop codon
                       5` end of DNA strand being transcribed before the terminator
            i. Termination of transcription signal
                       5` end of DNA, final piece of transcription
11. Compare and contrast function of the components of the E. coli RNA polymerase.
       α x2. Interacts with other factors which bind upstream of the -35 box
       β x1. Functional part of polymerase (forms phosphodiester bonds)
       β` x1. Binds to the DNA template
       if holoenzyme with also have σ x1.
12. What is the function of the sigma factor?
       Recognizes the promoter and facilitates initiation
13. What is the function of NusA?
       Takes the place of sigma factor, aiding in transcription termination
14. What is the function of the Rho protein with regard to translation?
       Breaks H-bonds between DNA and RNA for Type II (Rho-dependent) termination in
       prokaryotes, allowing for RNA strand separation from DNA.
15. What structure is involved with transcription termination in E. coli?
       Hairpin
16. What is rifampicin and how is it used medically?
       Lipophilic molecule which inhibits RNA polymerase in bacterial cells by binding to the beta
       subunit of RNA polymerase. This prevents phosphodiester bonds from forming. Used to treat
       TB, Leprosy, MRSA, and bacterial meningitis.
17. What is the potential drug interaction that has to be considered with using rifampicin?
       It will cause an apparent decrease in the available drug concentration of almost all drugs due to
       its upregulation of the CYT-P450 enzyme in the liver.
18. Compare and contrast the similarities and differences between prokaryotic and eukaryotic mRNA.
       Prokaryotic: transcript is mature and used directly for translation w/o modification, usually
       even during transcription. They are also polycistronic
       Eukaryotic: transcript of DNA is considered pre-mRNA since it must be processed (5` cap, ,
       splicing, poly A tail). Eukaryotic mRNA are monocistronic and have AAUAAA sequence 10-
       30 nt upstream of cleavage site.
       Both have their own promoters and termination sequences
19. Compare and contrast the differences between prokaryotic and eukaryotic gene expression.
       Prokayote Genes have promoters, Shaine-Dalgarno sequences, coding sequences, and
       termination sequences and there is no post processing.
       Eukaryotic genes have Enhancers, proximal control elements, promoters, exons, introns, poly-
       A signal sequence and terminator regions. There is post processing.
20. What are the functions of the three eukaryotic RNA polymerases?
       RNA pol I: transcribes 3 major rRNAs (28S, 18S, and 5.8S)
       RNA pol II: transcribes mRNAs and some snRNAs
       RNA pol III: transcribes tRNAs, the 5S rRNA, and snRNAs
21. What three RNA processing events occur in generating a eukaryotic mRNA?
    1. 5` cap placement to protect RNA from degradation by exonucleases and allow binding to
       ribosome
    2. Splicing introns out of sequence
    3. Addition of the polyadenylation tail
22. What elements “boxes” are often found within a eukaryotic promoter?
       Basal elements have TATA box near 0 to -25 bp and Proximal elements have CAT box and GC
       box further upstream, -50 to -200bp.
23. What is the function of the transcription factors?
       They are similar to sigma factor for prokaryotes. They come together with the specific RNA
       polymerase to form the pre-initiation complex (PIC)
24. How are transcription factors named?
       TFs are named according to the RNA polymerase they work with. Ex: TFIID works with RNA
       pol II
25. Explain the physical location of the following with respect to their location on a stretch of DNA:
           a. +1 initiation nucleotide
                      The first nucleotide to be transcribed from the 3` end of the DNA sequence, at
                      the 5` end of the promoter region
           b. +1 initiation codon
                      First set of three nucleotide to be transcribed from 3`end of DNA
           c. Enhancer
                      Pieces of the DNA chain which can be several Kb from the coding region,
                      usually upstream. They loop around to stabilize the PIC
           d. Proximal control elements
                      Located -50 to -200 bp upstream, include the CAT box and GC box
           e. Promoter
                      Located upstream of the transcribed region. In eukaryotes, basal (0 to -25 bp)
                      and proximal (-50 to -200 bp) elements.
           f. Cap
                      If this is referring to the 5` cap there is no DNA equivalent location since this is
                      added on post translationally. It would added to the location where the promoter
                      orignially.
           g. Coding region
                      3` to 5` strip of DNA after promoter and before the termination sequence. Will
                      begin at the start codon and end at the stop codon.
           h. Exons
                      Located within the coding region between the promoter and termination
                      sequence
           i. Introns
                      Located within the coding region between the promoter and termination
                      sequence
           j. Stop codon
                      Located at the 5` end of the template strand before the UTR, Poly A signal
                      sequence, and termination region. It will be the last piece of the coding strand
           k. Poly A tail
                      This is again a sequence which is added posttranslationally to the mRNA. It
                      would match up to the same place as the final intron before the termination
                      region of the 5` end of the DNA.
           l. 5’ UTR
                      Located at the 3` end of the DNA template sequence, just after the promoter and
                      before the start codon.
           m. 3’ UTR
                      Located at the 5` end of the DNA template strand, upstream and including the
                      poly a signal sequence
           n. Poly A signal sequence
                      5` end of DNA sequence, part of the final exon
           o. Termination of transcription signal
                      5` end of DNA, final piece of transcription
26. Which of the above are retained in a mRNA?
      The 5` cap is added, 5` UTR, start codon, coding segment (exons), stop codon, 3`UTR and a
      Poly A tail is added.
27. What is the order of assembly of TFs at a Pol II promoter?
      IID, IIB, RNA pol II, IIF, IIE, IIH
28. What two “promoter” factors are involved in Hemophilia B Leyden and Brandenburg?
      The factor IX gene has 2 promoter sites which are utilized at different points of life. They are
      AR (androgen receptor) located at -36 to -22 and HNF4 (hepatocte nuclear factor-4) located at -
      27 to -15.
29. What would the position where these factors bind to DNA be considered?
      Promoter regions or recognition sites for TFs
30. Which Hemophilia B improves during puberty?
      Hemophilia B Leyden
31. Why? Explain with respect to the utilization of the Factor VIII promoter.
      Hemophilia B Leyden occurs when there is a mutation at the -20 nt which only effects the
      promoter region for HNF4. During puberty the AR receptor becomes more preferential with the
      increase in androgen levels and thus the gene is able to be expressed.
32. How many proteins (approximately) make up RNA polymerase II?
      ~12. The largest subunit (~200 kD) is similar to the β` subunit in prokaryotes, binding to DNA.
      The second largest (~100 kD) is related to the β subunit in prokaryotes, being the functional
      part actually forming the phosphodiester bonds.
33. How does RNA polymerase and DNA interact?
      DNA is fed through a channel in the polymerase, like a factory
34. Which is really moving, the DNA or the RNA polymerase?
      Everything is relative but DNA is said to be moving through the factory
35. Compare and contrast enhancers, activators, silencers and repressors.
           a. Enhancers are regions of DNA which makes the polymerase work better. They can be
               located several kb from the gene and are usually upstream but can be anywhere. They
               loop around and stabilize the PIC through their relationship with activators
           b. Activators allow for high-level transcription by stabilizing the PIC while binding to
               enhancer regions on the DNA
           c. Silencers are regions of DNA which allow for the binding of repressors
           d. Repressors bind to silencer regions and interfere with the functioning of activators,
               slowing transcription

36. What is a UTR?
       UTR stands for untranslated region. They are the extensions of exons at the 5` and 3` ends of
       the coding sequence which will be formed in the final mature mRNA but are outside the start-
       stop codon region and so will not be translated
37. Are 5’ UTRs and 3’ UTRs considered exons?
       Yes, since they will be a part of the final mature mRNA
38. What is the function of the C-terminal domain of RNA polymerase with regard to the RNA
    processing events that are needed to make a mature mRNA?
       The CTD is an area on the RNA polymerase where proteins involved in post-processing are
       able to attach. When this area is phosphorylated and the proteins recognize the RNA sequence
       which dictates the particular post-processing envent they will hop off the CTD to perform their
       job.
39. What the structure and function of the 5’ cap that is put on mRNA?
       There is a 5` to 5` cap applied to the first nt in the mRNA to stabilize the message and prevent
       it from being degraded. The added nucleotide is a guanine and it is attached by a triphosphate
       linkage in a 5` to 5` fashion. Usually methyl groups get added to the 1st and 2nd nts at the 2` OH
       to further protect the mRNA
40. What is the consensus polyadenylation signal?
       AAUAAA, located usually 10-30 nts upstream from the 3` end cleavage site of the primary
       transcript
41. How is the 3’ end of an eukaryotic mRNA generated?
       Protein on CTD recognizes the AAUAAA sequence and clips the mRNA ~20 bases to the 3`
       end of it (downstream).
42. What is the function of a poly A tail?
       It is associated with a poly A binding protein (PBP) which interacts with the Cap binding
       protein (forming a loop) to stabilize the mRNA. PBP stabilizes CBP which prevents message
       from getting degraded.
43. What proteins bind to the poly A tail?
       PBP or polyA binding proteins
44. To what other proteins do the PBPs bind?
       The CBPs or Cap binding proteins
45. Compare and contrast introns and exons.
       Intron is a non-coding DNA sequence between exons in a gene
       Exons are the expressed DNA sequences in a gene which code for AAs, including the 5`UTR
       and 3` UTR
46. What 5 nucleotides are always present in an intron?
       GT(U) at the 5` end an A in a specific branch point in the middle and AG at the 3` end
47. What is a splicosome composed of and what is its function?
       snRNAs and proteins, they cleave the intron at the 3` end and join the exons
48. What is a lariat?
       The structure which is formed when the 5` end of the intron, after cleavage, comes back around
       and binds to the A at the branch point in the middle of the intron by a 2` to 5` phosphodiester
       bond.
49. How is it formed and what is its structure?
       See above.
50. When does splicing occur with respect to when transcription occurs?
       Splicing occurs during the transcription, as the mRNA is being fed out of the RNA pol II
51. How do the snRNAs assist in splicing?
       They assist in cleaving the intron at the 3` end and joining the exons
52. What two diseases are caused by mutations in the pre-mRNA splicing factor genes?
       Retinitis Pigmentosa (RP) and Spinal Muscular Atrophy
53. What disease is caused by an expansion of a trinucleotide repeat within an intron that interferes
    with splicing?
       Myotonic dystrophy
54. Where are the splicing recognition proteins located with respect to the RNA polymerase?
       CTD
55. What is the most common result from a mutation in the conserved nucleotides of an intron?
       Loss of an entire Exon
56. What are four diseases that are known to be caused by a splice site mutation?
       Familial Isolated Growth Hormone Deficiency Type II (GH-1 gene)
       Fraser Syndrome (WT-1 gene)
       Frontotemporal Dementia and Parkinsonism on Chr. 17 (FTDP-17)
       Atypical Cystic Fibrosis (CFTR gene)
57. What is a cryptic splice site?
       A Cryptic splice site is a part of an intron which is similar to the actual 3` cleavage site. When
       splicing occurs at this site there is a small part of the intron which makes its way along with the
       exons as part of the coding sequence.
58. How is a cryptic splice site activated?
       Mutation of the ture splice site on the 3` end of the intron causes the splicisome to look for
       another site which happens to be upstream, in the intron.
59. Why are some splice site mutations cause βo thalassemia while other cause β+ thalassemia?
    • βo thalassemia occurs when there is a mutation in the acceptor site (3` AG->GG) of Intron 2,
       causing a cryptic splice site to be used. There is no β-chain synthesis in this form because a sop
       codon is introduced with this cryptic splice.
    • β+ thalassemia occurs in 2 cases
               a) A mutation in Intron 1 causes a new possible acceptor site. Both sites are used with
                   the mutant being used 90% of the time and the normal site being used 10% of the
                   time. The mutant translation still gives a pre-mature stop codon, but the 10 %
                   normal transcript will be translated providing those with this form some β-chain
                   synthesis.
               b) A mutation in Exon 1 causes a new Donor splice site. Both sites are used with the
                   mutant being used 40% of the time. The incorrect splicing causes a frameshift which
                   activates a premature stop codon. Because this mutation occurs in the exon region
                   even the in the 60% which splice correctly have a mutation where lysine is
                   substituted in for a glutamate, although they still show some β-chain synthesis.
60. What is alternative splicing?
       Combinations of different exons from a particular gene to give different proteins.
61. How does alternative splicing affect the number of proteins produced in a cell?
       One gene can code for many different proteins, although each mRNA is monocistronic in a
       eukaryote
62. Compare and contrast the function of rRNA, tRNA and snRNA.
    tRNA- transports AA to ribosomes for translation
    rRNA- forms ribosomal complex with proteins to hold mRNA/translate protein
    snRNA- combines with proteins to form splicisomes which aid in splicing out introns
63. Compare and contrast the prokaryotic and eukaryotic ribosome.
Ribosomes of Prokaryotes                    Ribosomes of Eukaryotes
70 S model                                  rRNA genes are tandemly repeated many times
   • 50S subunit (23S, 5S, 34                           o Contain internal, external, and
      proteins)                                            nontranscribed spacers (ITS, ETS, NTS)
   • 30S subunit (16S, 20 proteins)         80S model
                                               • 60S subunit (28S, 5.8S, 5S, 50 proteins)
                                               • 40S subunit (18S, 35 proteins)
                                            Transcribed by RNA polymerase I. Synthesis of rRNA
                                            requires its own transcription factors (TFs-TFIs)

64. Compare and contrast the α-amanitin sensitivity of the three eukaryotic RNA polymerases.
      II > III > I
65. Where are ribosomes assembled?
      In the
66. Where is rRNA trasnscribed?
      In the nucleus
67. What is the organization of ribosomal genes?
      They are tandemly repeated many times
68. Where are the rRNA genes located in the human genome?
      rDNA are coding regions in DNA which code for rRNA. These genes are located at the satellite
      sequences of the acrocentric chromosomes (13-15, 21, 22)
69. What are the two sequence components of a Pol I promoter?
      The upstream promoter element which occurs -170 to -110 and the core promter which is -40 to
      +20.
70. What is the sequence of events that allows a RNA pol I to start transcribing rRNA genes?
      Binding by UBF, binding of the core binding factors (TIF-1B, SL1, Rib1), RNA polymerase
      binds to startpoint
71. Why is one polymerase dedicated to transcribing rRNA only?
      rRNA is transcribed only in the nucleolus and is the only thing transcribed in the nucleolus.
      RNA pol I is active in the nucleolus and only transcribes rRNA there.
72. What kinds of RNA processing events occur in the generation of a tRNA?
      The 3` end gets a tinucleatide CCA added to it. This will be the binding site for all AAs.
      tRNAs also get modified bases such as ribothymidine, pseudouridine, dihydrouridine,
      dimethylguanine, and inosine.
73. What part of the tRNA binds to the amino acid and what part binds to the mRNA?
      Loop 2 which contains the anti-codon recognizes mRNA codons during translation
74. What polymerase(s) transcribe snRNAs?
      RNA pol II or RNA pol III
75. Compare and contrast the three types of RNA polymerase III promoters.
      Type 1: for 5S rRNA has A box (+50 to +60), Intermediate element (+67 to +72), and C box
      (+80 to +90)
      Type 2: for tRNA has A box (+8 to +19) and B box (+52 to +62)
      Type 3: for U6 (snRNA) has a distal sequence element (-215 to -240), proximal sequence
      element (-65 to -48), and a TATA box (-32 to -25)
76. What kinds of diseases are caused by tRNA mutations?
      Mitochondrial diseases. Many times symptoms include diabetes, hearing loss, blindness,
      cardiomyopathy
77. What is the underlying reason that the tRNA mutations cause these diseases?
      The mutation reduces the amount of complex I available to the mitochondria for Ox-phos
      pathway which in turn leads to lower ATP production. The reduced ATP prevents insulin
      secretion leading to diabetic state.
        Questions to guide your learning of Translation
1. What is the central dogma of molecular biology?
      DNA => RNA => proteins
2. How often does an aminoacyl-tRNA synthetase incorporate an incorrect amino acid into tRNA?
      1 for every 10,000-100,000
3. What is the start codon?
      AUG (Also codes for Met)
4. What are the stop codons?
      UAA, UAG, UGA
5. Which amino acid is usually “degenerate”?
      Leu (6), Ser (6), Arg (6). Arg has the most even spread
6. What usually determines the codon usage by an organism?
      The relative amount of tRNAs that are expressed
7. What three nucleotides are always added to the 3’ end of all tRNAs?
      CCA
8. Be able to match the modified nucleotide structures with their correct name.
         m2G                 UH2             4-thiouridine         inosine       pseudouridine
   2 methyl groups 2 hydrogens             Sulfur replaces O Deamination of     Base ring is
   added to G          added to U          in U               G (NH2 replaced “rotated”
                                                              with H

9. Which is the most important step in protein synthesis?
        Aminoacyl-tRNA synthetase charging the tRNA with the right AA
10. What is an activated amino acid?
        An AA with AMP attached on the Aminoacyl-tRNA synthetase (the AMP has the energy in
        it to transfer the AA to the tRNA)
11. What is a charged tRNA?
        tRNA with an AA bound to it
12. How many ATP equivalents are used up in charging a tRNA?
        2, there is a PPi lost from an ATP to charge each tRNA
13. How does the aminoacyl-tRNA synthetase know which tRNA to charge?
        It recognizes key shapes and bases on the particular tRNA it is built to bind to
14. What happens to a mischarged tRNA?
        There is an editing site on the tRNA synthetase which will hydrolyse off AAs which are too
        small. AAs which are too big are not given the opportunity to bind b/c they will be rejected
        by the acylation site. If the tRNA makes it off of the synthetase with the incorrect AA, it
        will end up in a protein.
15. Compare and contrast Class I and Class II aminoacyl-tRNA synthetases.
        They have different 3 dimensional structures, they recognized different sides of tRNA, they
        bind ATP differently, and they work on different AA’s
                                  Class I                                  Class II
        - Monomeric                                      - Dimeric
        - Acylates the 2` OH on terminal ribose          - Acylates the 3` OH on terminal ribose
16. What are the two levels of control to ensure that the proper amino acid is incorporated into a
    protein?
        1. Charging of right AA with proper tRNA by tRNA synthetase
         2. Matching cognate tRNA to the mRNA (base pairing)
17. Where does protein synthesis take place? (structure and cellular compartment)
         On ribosomes either free in they cytosol or bound to RER (still technically in cytosol)
18. Compare and contrast the prokaryotic and eukaryotic ribosome?
                       Prokaryote Ribosome                               Eukaryote Ribosome
              - 70S w/ 2 sub units (50S and 30S)               - 80S w/ 2 sub units (60S and 40S)
                                                               - has 5.8S “extra subunit)
19. Which ribosome is the mitochondrial ribosome more like?
         Prokaryote
20. What is the basic organizational scaffold for the ribosome?
         The rRNA, proteins are secondary structural
21. Where are ribosomes assembled?
         In the nucleolus
22. In what part of the nucleolus is the ribosome generated?
         Fibrillar centers
23. Describe the synthesis of the rRNA that is generated in the nucleolus.
       i. RNA Pol I transcribes rDNA repeats (large array at the end of each acrocentric
              chromosome) to produce a 45S RNA precursor.
                  a. Note: 5S subunit is not part of this precursor, it comes in later (uses Pol III)
      ii. 45S precursor is processed/cleaved to mature rRNAs which bind to proteins to generate
              large and small ribosomal subunits
24. What is really the most important part of the ribosome, the RNA or the protein?
         The RNA, since it is the catalytic part. The proteins just support the structure
25. Explain how the prokaryotic ribosome finds the start codon of a mRNA.
         The Shine-Dalgarno sequence of the mRNA is matched up with the Py rich part of a
         ribosome to place the AUG at the correct start site.
26. What is the function of the three initiation factors involved with starting translation?
         IF1 and IF3 keep the small and large subunits apart, helping to guide the “tertiary complex”
         into the P site. IF2 directs the first tRNA (fMet since it is UAC) to the P site.
27. What high energy molecule is involve with translation initiation?
         GTP is hydrolyzed by IF2 to cause the release of the IFs and binding of the large and small
         subunits
28. How many high energy molecules are used up to get the initiator tRNA set up in the P site of the
    ribosome?
         1
29. What are the two parts of the complete ribosome called?
         Large and small subunits
30. Compare and contrast the eukaryotic and prokaryotic ribosomes with respect to size.
         Eukaryote has 80S ribosome which has a molecular weight of about 4.2 million
         Prokaryotes have a 70S ribosome with a molecular weight of about 2.5 million
31. What three tRNA binding sites are there and what is their function?
         A site is the aminoacyl tRNA binding site (all tRNAs will come here 1st except the start one)
         P site is the peptidyl-tRNA binding site (this is where the start codon is bound)
         E site which is the exit site
32. How does the newly synthesized protein exit the ribosome?
         There is a tunnel in the large subunit which allows the polypeptide chain to pass through as
         it grows
33. Where is the P site which contains the growing peptide located on the ribosome?
        The growing sidechain is catalyzed/fed through the large subunit
34. What catalyzes the formation of the peptide bond?
        The large ribosomal subunit (rRNA). More specifically a 28S adenine by acid-base catalysis.
35. What does wobble mean when it comes to translation?
        The 3rd position of the codon (mRNA) or 1st position of the anticodon (tRNA) has a “slop”
        ability in choosing which nt it will bind to. For instance in eukaryotes a U in this position
        on the mRNA could bind to a G or an I on the tRNA
36. Which “unusual” nucleotide is often involved with the wobble base pairing?
        Inosine
37. What elongation factor guides the acylated tRNA into the A-site?
        In prokaryotes EF-Tu guides the new amino-acyl tRNA into the empty A site, in eukaryotes
        it is the eEF-1
38. In order for this elongation factor to leave the tRNA in the A-site, what high energy molecule
    must be hydrolyzed?
        GTP
39. What elongation factor is required to “translocate” the mRNA so that the P and the A sites are
    positioned correctly so that the A site can accept another charged-tRNA?
        In prokaryotes EF-G, in eukaryote eEF-2
40. What high energy molecule is hydrolyzed during the transloction step?
        GTP
41. How many high energy molecules are necessary to incorporate 1 amino acid into a peptide
    chain? (Include charging the tRNA, entry into the A site and the translocation step.)
        2 to charge the tRNA, 1 for entry into the A site, and 1 for translocation = 4
42. What is a polysome?
        Many things doing one thing
43. What interactions usually occur between the 3’ and 5’ end of a message?
        The PBP interact with the CBP. eIF-4s are located here and they direct the small ribosomal
        subunit to the start
44. What does is meant by “transcription and translation” are coupled?
        In prokaryotes ribosomes and the RNA polymerase are linked so that translation starts while
        the mRNA is being transcribed. This is important for them because if there is a pause in the
        ribosome activity and too much of the mRNA is exposed a termination protein will halt
        translation
45. What would happen to the expression of a gene if a mutation occurred where an frequently used
    codon was replaced by an infrequently used codon?
        It would not be expressed at the normal rate since tRNAs would not be readily available, in
        prokaryotes this could cause a stall in the ribosome which would expose the mRNA as the
        RNA polymerase continues, allowing termination proteins to bind.
46. What triggers termination of translation?
        Stop codons which cause release factors to bind. These in turn cause hydrolysis of the
        peptidyl-tRNA bond
47. What high energy molecule needs to be hydrolyzed so that the various components of the
    translation machinery can dissociate? GTP
48. How does a release factor recognize the stop codon?
        It mimics tRNA
49. Does the release of the protein from the ribosome require the hydrolysis of GTP?
        No
50. How many high energy (ATP equivalents) are needed to generate a protein 100 amino acids
    long?
        2 for each charged tRNA, 1 for each placement, 1 for each chain movement, 1 for
        termination, -1 since first one does not need chain movement = 400
51. What is usually necessary for a ribosome to recognize the 5’ end of mRNA?
        A 7-methylguanosine cap (5` to 5` cap)
52. What is the complex of proteins called that binds to the cap structure?
        eIF4 CBPs
53. With what complex does the small ribosomal subunit interact before it binds to the cap binding
    proteins?
        The small ribosomal subunit binds the eIF-2/tRNA/Met/GTP complex (this is just the first
        tRNA with an initiation factor and GTP) and is then attracted to the CBPs
54. Compare and contrast the translational control of ferritin and transferrin.
    - Ferritin’s mRNA has a protein bound (cytosolic aconitase) on it’s 5` end near the start codon
        when there is no iron. This prevents binding of the ribosome and thus no ferritin is
        produced. If iron is present the protein dissociates and translation can occur.
    - Transferrin’s mRNA has the same protein bound to its 3` end (again when no iron is
        present) which stabilizes the mRNA transcript and still allows ribosomes to bind to the 5`
        end allowing synthesis of the transferring receptor. If iron becomes present the protein
        dissociates and the mRNA is at higher risk of degradation
55. What TCA cycle intermediate is the protein that binds the mRNA secondary structures to
    regulate ferritin and transferrin translation?
        Cytosolic aconitase
56. What is the general effect of having a low concentration of amino acid in the serum?
        Downregulation of translation
57. How is the feast/famine phenomena exert its control over translation?
        If there are a lot of nutrients available, many growth factors will be released, causing certain
        cascade of phosphorylations (usually eIF-4)
58. How does phosphorylation of eIF-2 stop translation?
        If eIF-2 has GTP bound to it, it is active and can guide the first AA into the P site of the
        ribosome to start translation. This process uses the GTP and GDP is then attached to the eIF-
        2. This is alright under normal circumstances because the guanine nucleotide exchange
        factor (eIF-2b) will switch out a GTP for the GDP and the eIF-2 can proceed. If however
        eIF-2 gets phosphorylated while it has GDP bound to it (by protein kinase) when eIF-2B
        tries to switch out the GDP it will be trapped and no active eIF-2 will be available. This
        happens during periods of amino acid depirvation
59. How is translation controlled by eIF4E?
        In this situation the eIF-4E is needed as a CBP to initiate translation. There is a binding
        protein, 4E-BP, which will bind to eIF-4E, making it unavailable as a CBP. When nutrient
        levels are high however, growth factors will phosphorylate the 4E-BP which prevents it
        from binding to eIF-4E, allowing the eIF-4E to acts once more as a CBP and help initiate
        translation.
60. What two viruses modify the translational machinery as part of their life cycle?
        Polio virus and Encephalomyocarditis virus
61. How does poliovirus interfere with cellular protein translation?
        Poliovirus as a viral protease 2A which is translated once the virus is brought into the cell.
        This cleaves one of the CBPs (eIF-4G) preventing it from functioning as a bridge between
        the methyl cap binding subunit and the 40S subunit. (eIF-4G binds directly to the 40S)
62. Why are the polio virus messages translated when cellular translation is inhibited?
      The viral RNA has an internal ribosomal entry site (IRES) which can act as a Cap-indepdent
      initiation site (similar to the SD sequence), allowing for initiation at alternate internal start
      codons
63. What is an internal ribosome entry site?
      It is a structure in the mRNA which can bind to the rest of the eIF-4G acting as kind of a
      promoter region. The Poliovirus uses it in addition to other picornaviruses such as rhinovirus
      and enterovirus types.
64. When does the cell tend to use IRES elements?
      During G2/M phase of cell cycle because many 4E-BPs are activated. They take out the eIF-
      4E which is usually a CBP, directing the mRNA through the ribosome to the right start
      location.
65. What is ribosomal frameshifting?
      If the mRNA get misaligned in the ribosome by +/- 1 or 2 nts the reading frame will be
      shifted and completely different AAs will be placed then those planned.
66. What kind of virus will induce ribosomal frameshift as part of its life cycle?
      Retroviruses induce ribosomal frame-shifting (eg. HIV)
67. How do the following antibiotics inhibit translation:
           a. Streptomycin
                       Inhibits initiation and causes misreading of mRNA in prokaryotes
           b. Tetracycline
                       Binds to the 30S subunit inhibiting binding of aminoacyl-tRNAs
           c. Chloramphenicol
                       Inhibits the peptidyl transferase activity of the 50S ribosomal subunit
           d. Cycloheximide
                       Inhibs the peptidyl transferase activity of the 60S ribosomal subunit
           e. Erythromycin
                       Binds to the 50S subunit and inhibits translocation
           f. Puromycin
                       Causes premature chain termination by acting as an analog of aminoacyl-
                       tRNA. Basically sits in the A site and prevents additional translation. Can
                       affect both prokaryotic and eukaryotic translation.
Receptors
Wednesday, June 04, 2008
6:56 PM

In Biochemistry receptors are analyzed on the molecular level, not just at the cellular level.
A receptor is usually a protein which when bound to a particular ligand will induce biological effects
   - Binding is steroselective, saturable, selectively inhibitible and usually not covalent (it is reversible)
   - Enzymes, membrane channels/transporters, transcription and gene expresion are all possible targets of signaling cascades

Binding of receptors is usually reversible
   - can be expressed mathmatically
   - Kd = concentration of ligand at which [RL] = [R] or


Types of receptors:
Steroid: • Steroids are able to pass membrane so binds to receptor in cytoplasm or nuclear space
         • Hormone receptor complex binds to response elements in promoters or enhancers of genes to stimulate or inhibit transcription
         • If cell does not have receptor, it will not show effect to hormone
                ○ Similarly, only genes which contain the specific response element will be affected
                ○ allows for 2 levels of specificity => only desired cells, and only desired genes

Water soluble molecules can not readily pass the membrane so they will act on integral membrane proteins
  - Can act directly on channels, or create secondary messengers

Ligand Gated Ion Channel
   - Many neurotransmitters use these b/c they are fast
   - Ex: Nicotinic acetylcholine receptor
         Located in muscle fibers at neuromuscular junctions
         Ion channel (allows Na to enter cell upon activation, depolarizing cell)
         5 subunits
   - Binding of ligand produces fast effect (there is no cascade for response)
   - Channels are generally closed in resting state, opened with ligand binding
   - Na channels will depolarize the membrane, Cl and K will hyperpolarize it (remember gradients)
         GABA-A receptor in the brain is an example of a Cl channel which has an inhibitory effect
              □ Target for benzodiazepines and barbituates

G-protein Receptors
   - Have 7 TMDs
   - Receptor associates with hetero-trimeric G protein
          α subunit has GDP bound in resting state; β and γ subunits are always together and when bound to α prevent it from activating (binding GTP)
          Binding of ligand on receptor allows α subunit to looose GDP and gain GTP. This causes dissociation from other subunits.
          Activated alpha will dissociate along the membrane and act on an effector. Its GTP is then hydrolyzed to GDP, inactivating th e alpha subunit.
               □ The effector could be many different proteins; the example in class used an α s (Gs ) to activate adenylate cyclase and conversly showed α i
                   (Gi) inhibiting it. There is also a Gq which acts on PI-specific phospholipase C and G12 which acts on ion channels.
   - G proteins stay near the membrane so they must use second messangers which was cAMP in class examples
          Adenylate cyclase converts ATP to cAMP by cleaving off a pyrophosphate (which is quickly hydrolysed, making this irreversible ) and linking the
            final phosphate group to the 3` C in the ribose of AMP
          The most important target of cAMP is protein kinase A (PKA)
               □ PKA under standard circumstances is bound to regulatory subunits which prevent its activity
               □ cAMP binds to the regulatory subunits, which causes the release of the active subunits (takes 4 cAMP to release 2 PKAs)
               □ PKA when active will phosphorylate certain serine and theronine hydroxy groups on specific proteins (very selective process)
                      These proteins could be cytosolic proteins, membrane proteins, or even TFs (such as CREB, the cAMP response element binder)
                      Thus certin genes will be "turned on" in the presence of cAMP
          The degradation of cAMP to AMP is carried out by phosphodiesterases
               □ Phosphodiesterases are inhibited by methylxanthines which include caffiene, theophylline, and aminophylline. Ingestion of the se materials
                   causes cAMP to remain active.
   - Cascade pathway allows for amplification of signal at each step (1 hormone starts many g-proteins, which acivate many adenylate cyclases…)

Different Epinephrine receptors
β-receptors    • Binds to Gs proteins (increases cAMP) in:
                     ○ adipose tissue (fat breakdown)
                     ○ Bronchial smooth muscle (relaxation)
                     ○ Vascular smooth muscle (relaxation)
α2-receptors   • Binds to Gi proteins (limits cAMP) in:
                     ○ Brian (hypotension)
α1-receptors   • IP3/Ca modulator
                     ○ Vascular smooth muscle (contraction)

G-protein diseases:
      Pseudohypoparathyroidism      • PTH (parathyroid hormone) stabilizes blood Ca levels by G s pathway
                                    • PTH deficiency casues:
                                          ○ Hypocalcemia, tetany
                                    • With PHPT patients will present with above symptoms but PTH blood levels are normal or even elevated
                                    • Inhibitory mutations in PTH or Gs protein must then be the cause, leading to inability to activate the adenylate cyclase
                                    • This is an inherited defect and found throughout the body
      Toxic Thyroid nodule          • TSH (Thyroid stimulating hormone) stimulates hormone production and cell proliferation in thyroid by G s pathway
                                      (raising cAMP)
                                    • Benign tumors in thyroid gland overproduce the thyroid hormones which would normally only be produced by
                                      activation w/ TSH
                                    • Excitatory mutations in TSH receptor or Gs protein must be the cause, leading to constant activation and production
                                      of cAMP
                                           ○ Gs mutation is most likely the prevention of the alpha GTPase activity
      Cholera toxin                 • Increases in cAMP in intestinal mucosal cells causes secretion of water and electrolytes
                                    • Infection by vibrio cholerae which secretes a protein toxin which enters cells and catalyze the covalent modification
                                      of the α-subunit of the Gs protein, removing GTPase activity which make the subunit permanently active.




                                                              Biochem Page 15
      Pertussis Toxin             • Bordetella pertussis lives on respiratory epithelium and releases virulence protein which enters the cell
                                       ○ The toxin modifies the α-subunit of the Gi protein, inactivating the protein. Leads to build up of cAMP.

Inositol pathway
   - The inositol pathway is the 2nd most important secondary messaging system in cells after cAMP.
           Will usually act to release Ca to the cytoplasm
   - Phosphatidylinositol (PI) is a combination of di-acylglycerol (DAG) with an inositol esterified to the 3rd C.
           Inositol is a cyclohexane derivative (NOT a sugar, notice there is no O in the ring)
           PI will be bound to the membrane, the inositol is phosphorylated on the 4th and 5th C by PI kinase and PIP kinase, each step requiring ATP
   - Binding of signal to Gq receptor activates an α-subunit which will in turn activate phospholipase C-β
   - PLC-β is responsible for cleaving inositol from DAG giving the secondary messanger IP 3 (inositol 1,4,5-triphosphate)
   - IP3 can act on Ca channels on the ER, releasing Ca to the cytosol
   - Ca can then work with DAG at the membrane to activate PKC, or on Calmodulin or Troponin C in muscles to lead to contraction

Muscle Contraction
  - Ca is vital component for muscle contraction
  - Striated muscle use torponin acting on tropomyosin
  - Smooth muscle use calmodulin working with MLC kinase to phosphorylate the MLC, leading to contraction

Ca regulates basically 2 things: muscle contraction and secretion of H20 soluble molecules (fusion of storage vesicle with pl asma membrane)
Ca is removed from the cell by a Ca ATPase and/or a sodium antiport, it is stored in the ER by a Ca ATPase.
Ca levels can be elevated in the cell by:
   - Hormones acting on the IP 3 path (release from ER)
   - NMDA receptor in the brain (glutamate opens extracellular Ca channel)
   - Voltage gated Ion channles (allow extracellular Ca in)
Cell growth/mitotic division is usualy stimulated by IP3/Ca path

cGMP pathways leading to vascular smooth muscle relaxation
Atrial Natriuretic Factor (ANF)                                             Nitic Oxide
     - Formed in the atrium of the heart to regulate BP by the excretion of      - Increased Ca concentrations in endothelial cells (by muscarinic ACH,
       Na in the kidneys and the relaxation of vascular smooth muscle              Bradykinin, or Hist.) will activate NO synthase
     - Can be released when BP or blood Na is too high                           - NO synthase cleaves an N group from Arg, producing NO which diffuses
     - ANF receptor is the guanylate cyclase                                       to the vascular smooth musscle directly below
     - Binding of ANF to receptor causes cGMP to be created from GTP             - NO activates guanylate cyclase which in turn creates cGMP which
     - cGMP activates Protein kinase G which antagonizes the effect of Ca          causes relaxation (PKG activation)
       (causes relaxation of smooth muscle)

Vascular Smooth Muscle:
  - Ca causes contraction by hormones vasopressin, angiotensin, or epinephrine by means of α 1-receptors (remember this is IP 3 path)
  - cAMP causes relaxation by epinephrine binding to β-receptors
  - cGMP causes relaxation by ANF or NO
         Nitrovasodilator drugs cause relaxation of blood vessels since they are precursors of NO and will be metabolized to it. Important for heart
           troupble, (nitroglycerin..)
         PDE5 inhibitors inhibit cGMP phosphodiesterase in corpora cavernosa (in penis). Inhibition prevents cGMP breakdown, keeping i ntercellular
           levels of the second messenger abnormally high. These would be viagra, cialis, ….

Tyrosine Kinase Receptors
Growth Factor Receptors                                                                         Insulin
    - Mediate mitogenic and growth regulatory events                                                 - Similar to growth factor receptors except inactive
    - Inactive state is monomeric                                                                      state is a tetramer held together by disulfide bonds
    - Upon binding of dimer signal molecule the receptor dimerizes and becomes active,                      Upon activation of receptor, IRS-1 or IRS-2
      autophosphorylating itself on certain tyrosine residues and leading to "true"                           binds to phosphorylated groups and gets
      activated form                                                                                          phosphorylated itself
    - Signalling proteins containing a specific binding domain (SH2 domain) bind to the                     This they attracts signalling proteins which
      phosphorylated receptors becoming allosterically activated upon binding or the                          contain the SH2 domain
      phosphorylation by the receptor

Cascades from Growth factor receptors:
  - One of the activated proteins is PLC-γ which (just like the β subunit activated in G-proteins), generates IP 3 from PIP2
  - Another activated protein is the lipid kinase Phosphoinositide-3-Kinase (PI3K) which leaves the DAG group attached giving PIP3.
        PIP3 is an activator of protein kinase B (PKB)
        This pathway is significant in cancerous cells
  - The Ras protein is a G-protein which is also activated by autophosphorylated receptors.
        Ras activates protein kinases which in turn phosphorylate and activate MAP kinases (mitogen-activated protein kinases).
        MAP-kinases fortify mitosis stimulating TFs constantly inducing mitosis.
        Mutations to the GTPase of Ras is common in cancer cells, causing the inability to deactivate this cascade and constant induc tion of mitosis

Desensitization
  - Some receptors can be regulated in 2 stages, quick temporary inactivation or perminant breakdown by endocytosis to lysosomes
  - Temporary inactivation can occur by phosphorylation of the receptor (in the case of β-adrenergic receptor) which can be accomplished by specific or
     general kinases activated by the secondary signals.
          Phosphorylation of these receptors generally blocks the attachment of their complimentary G proteins, therefore limiting acti vation.
  - In the case of β-adrenergic receptor, if the receptor stays phosphorylated for very long arrestin will bind and cause other phosphorylated receptors to
     cluster together. This clustering will eventually be endocytosed and broken down by lysosomes.




                                                           Biochem Page 16
Viruses
Saturday, June 07, 2008
7:17 PM

A virus:
   ○ an obligatory intracellular parasite which will always cause some damage to a host. There is no such thing as "normal viral lfora"
   ○ Contains only 1 kind of nucleic acid: dsDNA, dsRNA, ssDNA, ssRNA, and only 3 to 250 genes
   ○ Does not have a cellular structure, ribosomes, or ATP synthesis
   ○ Is normally about the size of a ribosome (diameter of ~20-50 nm)
   ○ Can only enter/find new cells by a specific membrane receptor; the need of a receptor limits a viruses range of infection

Virion or virus particle is the extracellular form of the virus.
   ○ Includes the nucleic acid, capsid, and if applicable the envelope (membrane stolen from previous host cell)

T-Bacteriophages
  ○ dsDNA viruses which replicate through the lytic pathway
  ○ There is timed expression of the viral genes
          First viral proteins made is DNAse which cleaves the host DNA but not the hydroxymethylcytosine modified DNA of the virus
          Last genes to be translated are 2 enzymes which will cause lysis of the host cell membrane
  ○ Proteins of the virus never make it to the host cell, just to docking on the membrane/protein, only the DNA makes it and is a that is required to
                                                                                                                                ll
      replicate the virus

Lambda Phage
  ○ Has 2 options after entering the cell and becoming circular (by aid of the bacterial DNA ligase)
        Lytic infection causing destruction of the host cell
        Lysogenic infection where the phage DNA integrates itself into the host cell chromosome
            □ Maintained by the lambda repressor
            □ Always has the potential to go through lytic phase
  ○ Integrase enzyme is encoded by viral gene to integrate the viral DNA into host chromosome (at this point called a prophage)
  ○ Lamda repressor is synthesized at the same time as the integrase
  ○ Lamda repressor is destroyed by the bacteria when the cell is damaged, activating the viral genome causeing destruction of ho cell
                                                                                                                                st

Bacteria can become resistant to bacteriophages relatively easily and defective phages can carry DNA from bacteria to bacteri a (through transduction). This
is why they are not used as antibiotics very often.

Animal viruses:
  ○ Generally do not kill the host
  ○ Are usually enveloped
          Acquired either from plasma or nuclear membrane of host cell
          Contains viral spike proteins
          Since these envelopes are lipid bi-layers, these viruses are very sensitive to their external environment
  ○ Must fight an immune system
          Host will many times coat the extracellular spike proteins (which it uses to dock with new target cells) with antibodies. Th is serves 2 functions,
            one being the virus is more readily attracted by phagocytic cells, and the other that the binding of the virus is now restric ted.
          Cellular machinery is also present to destroy viral nucleic acids and prevent synthesis of viral proteins
          Apoptosis of cells often occurs once they have been found to carry a virus.

Viruses have 3 ways to enter their host cell:
   ○ Uncoating at the plasma membrane
          Viral envelope fuses with the plasma membrane
   ○ Uncoating within an endosome
          Virus envelope stays intact during endocytosis and fuses with the endosome membrane upon entering cell
   ○ Uncoating at the nuclear membrane
          Upon endocytosis endosome membrane is ruptured, leaving viral envelope intact w/in cell. Docking to nuclear membrane causes u ncoating and
            virus to be set free in nucleus

RNA viruses
  ○ Must have RNA-directed RNA polymerase (RNA replicase) encoded in their genome to be able to replicate their genome and make their mRNA
         This replicase does not have proofreading abilities which lead to the increased number of mutations
  ○ (+) RNA viruses are single stranded which serves as mRNA
  ○ (-) RNA viruses are single stranded but complementary to mRNA; they must replicate their strand before viral proteins can be ma   de
         They must carry the RNA replicase within the viral particle
  ○ dsRNA viruses must transcribe mRNA from one of the strands before viral proteins can be made.
         These must also carry the RNA replicase withing the viral particle
  ○ Retroviruses
         Carry 2 copies of (+) ssRNA genome and must go through an integrated DNA intermediate to replicate
         Contain 3 major genes
              □ gag for capsid proteins
              □ pol for reverse transcriptase and integrase
              □ env for spike proteins
         Genes are flanked by LTRs (long terminal repeats) which helps with insertion to the host genome as well as a promoter for gen e transcription
         Uncoating occurs at the plasma membrane. Reverse transcriptase turns the genomic RNA into a ds -cDNA which is then inserted into the host
            DNA by viral integrase. Only then is the viral RNA transcribed from integrated cDNA and translated into proteins. Spike pro teins appear at cell
            membrane while new viral particles are being formed, which will eventually bud off from cell.

Polio virus        • (+) RNA virus
                   • Icosahedral capsid
                   • All 4 subunits which make up cap are traslated as 1 large protein which is then cleaved later
Adenovirus         • More complex coat structure then polio, has 2 different proteins on coat with penton base holding fibers to extend out
                   • dsDNA genome
Influenzaa Virus   • Genome consists of 8 RNA "Chromosomes"
                   • Has envelope
Poxvirus           • Most complex virus known (and largest)
                   • Has 2 envelopes
                   • DNA genome
HIV                • Retrovirus (2 (+) RNA genome, reverse transcriptase, and integrase present in viron)
                   • Viral proteins are synthesized only after integration with host genome



                                                             Biochem Page 17
• Viral proteins are synthesized only after integration with host genome
• Only cells with CD4 and a particular chemokine receptor on their surface will be infected by HIV
      ○ Mutation of Ccr5 (β-chemokine) give partial resistance to HIV




                                         Biochem Page 18
Genetic Mutations
Monday, June 09, 2008
4:09 PM

Types of Mutations
Single-base substitution                      Insertion/deletion                Translocation               Inversion           Duplicaiton
• Aka point mutations                     • Could be a single base pair or • Problem more for            • Problem for         • Occurs due to misaligned
• 2 types                                   many kBPs                        offspring cells if there is   promoter              chromosomes during
      ○ Transitions: Pur->Pur or Py-> Py  • End up as frame shift            unequal distribution of       associatoin           metaphase
      ○ Transversions: Py->Pur or Pur->Py   mutation if not multiple of 3    genetic information
• Lead to
      ○ Missense: different AA coded
      ○ Non-sense: Stop codon
      ○ Silent mutations: same AA due to
        genetic code is degenerate

Somatic mutations are those which arise in cells throughout the body over time due to aging, cancer, or exposure to mutagens
  ○ All mutagens are possible carcinogens
Germline mutations are those which are present at the point of conception and so will affect all cells in the child
  ○ Paternal age increases the probability of germline mutations
  ○ Each child is born with at least 100 new mutations, most of which are harmless.
         Mutational load refers to the accumulation of slightly unfavorable mutations which contribute to polygenic diseases
         Single gene disorders occur rarely due to single mutations at the germline level
         Children are considered "sicker" then their parents due to their inheritence of the parents' mutations as well as their >100 extra mutations

Dominant diseases mutations are phenotypically expressed when heterozygous genes are present
  ○ These are evolutionarily selected out faster then recessive gene diseases due to their mortality and general incombatibility with procreation.
Recessive diseases mutations are phenotypically expressed only when the patient has the gene in its homozygous form

Consequences of mutations will vary based on the location in the DNA where the mutation occurs.
  ○ Junk DNA (which makes up ~90-95% of DNA) mutations will have no consequence
  ○ Promoter or enhancer region mutations will lead to normal proteins being translated, but the rate of expression will be altered (or even stoped)
  ○ Coding sequence mutations can take many forms:
         Silent                Missense               Nonsense                Read-through                   Frameshift                       Splice-site
• Same AA coded for        • Different AA coded • AA changed to a   • A stop codon is           • Entire sequence of AAs is • Generally leads to loss or
  due to degeneracy of       for                  stop codon          converted to another AA     changed due to addition or     "skipping" of exon
  genetic code             • Depending on site • leads to premature • This causes an              loss of 1-2 nts.             • Can also lead to addition
• May affect rate of         could lead to loss   termination of      extension of protein      • All normal function is lost,   of intron sequence which
  translation but will not   of activity          protein and most    length which can have       most likely will hit random    can cause frame-shift or
  affect activity                                 likely loss of      varying degress of effect   stop codon w/in 20 AAs         nonsense characteristics
                                                  function

Basal mutation rate is the rate at which replication errors are incorporated into DNA, absent of any external mutagenic factors. It is usually due to
spontaneous tautomeric shifts and mistakes made by replicaiton/proof-reading machinery.

Tautomeric Shifts:
  ○ Thymine can shift between keto (normal) form which binds with A or the less stable enol form which will bind with G
  ○ Adenine can shift between its amino (normal) form which binds with T or its imino form which again has different binding properties
  ○ Problem arises only if in the abnormal form during DNA replication

Induced DNA damage
   ○ Ionizing radiation (x-rays, radioactivity)
         Energy rich radiation which can penetrate the whole body and thus also cause germline mutations
         Most likely to cause double strand breaks
   ○ UV radiation (sunlight)
         Not energy rich enough to penetrate the whole body or ionize electrons
         Will cause Py dimers on same strand but not germline mutations
   ○ Free Radicals
         Most common mutagen in the body and the reason why mitochondrial mutations are so common
   ○ Base analogs
         A methyl group on 5th C of thymine distinguishes it from Uricil but if there is a bromo group there it will be treated just l ike Thymine. Problems
            arise because the 5-Bromouracil analog has relativily stable enol form which will lead to inappropriate base pairing.
   ○ Deaminating agents
         Deamination will remove the amino group on a base and replace it with a keto group (=O). This causes a mismatch which will h ave to proceed
            through base excision repair to be fixed.
         If the deamination occurs on cytosine it converts it to Uracil which is easily recognized as wrong.
         Deamination of methyl-cytosine (used as gene silencer) will give Thymine which is not recognized as wrong, thus methyl -cytosine locations are a
            mutational hotspot showing a 10 fold mutation rate.
   ○ Alkylating agents
         Covelantly bond alkyl groups to amines of bases
         Methyl bromide or epoxides (ie. ethylene oxide)
               □ Aflatoxin is a mutagen which is found on molds growing on peanuts/bread in hot/humid locations. It is converted from a non-chemically
                   active form into a reactive epoxide in the liver. Once in the epoxide form it can covalently bind to DNA. It is the suspected reason for the
                   noticible increase in Liver cancer/diseases in tropical areas.
   ○ Intercalating agents
         Planar structures which insert themselves b/w base pairs causing changes in strand shape and thus small insertions/deletions.
         Proflavin and ethidium bromide (used as a DNA dye) are 2 examples

Repair mechanisms
  ○ Replication errors are delt with by a 3` exonuclease activity on the DNA polymerase or post-replication mismatch repair systems
         These are important only in dividing cells and only work next to a replication fork
             □ The newly synthesized strand will have nicks in it, (both the okazaki fragments as well as the leading strand) allowing for determination of
                 which strand is new and which is old
                    actually, exonuclease on mismatch repair system can only start "chewing away" at the new strand since it has these cuts
             □ If the mismatch repair machinery finds a problem, it will start at the nearest "nick" and degrade the problematic strand. DNA polymerase
                 and Ligase come in to lay down a new strand.
  ○ Apurinic sites occur spontaneously as the covalent bond between the ribose and purine is somewhat "weak"
         AP endonuclease finds these apurinic sites and cleaves the phosphate bonds around it.



                                                              Biochem Page 19
         AP endonuclease finds these apurinic sites and cleaves the phosphate bonds around it.
         Polymerase Beta will come in and replace the missing nt, with DNA ligase linking the remaining phosphodiester bond
  ○   Base excision repair occurs when bases are determined to be abnormal, (ie alkylation, deamination, ...)
         A DNA glycosylase first creates an apurinic site at the abnormal base
              □ There is a specific DNA glycosylase unique for each type of abnormal base
         After the apurinic site is formed repair proceeds through the same route as other apurinic sites
  ○   Nucleotide excision repair
         Repair system containing 7 proteins which will scan the double helix searching for damage (pyrimidine dimers).
         Once found the single strand is cleaved at 2 positions spanning about 20 -30 nts, including the damaged bases, by a nuclease.
         A helicase separates the strands and DNA polymerase rebuilds the strand with ligase sealing the end.
  ○   Double-strand Breaks
         Once an accidental break occurs, there will be loss of nucleotides due to degradation from the now open ends
         There are 2 possiblities when trying to repair double-strand breaks
              □ Homologous end-joining: If a homologous chromosome is available a copying process can rebuild the lost information and reseal the
                  strand w/out mutation
              □ Nonhomologous end-joining: If there is not a homologous chromosome available then a "quick-fix" will occur where the ends are "stitched"
                  together with some loss of information

DNA Repair Diseases
Xeroderma Pigmentosum                   Cockayne Syndrome                              Hereditary non-polyposis colon     Ataxia-telangiectasia
                                                                                       cancer
• Defect to genome-wide nucleotide      • Defect to transcription-coupled nucleotide   • Defect of post-replication       • Mutations in ATM protein
  excision repair                         excision repair                                mismatch repair                    kinase (required for repair of
      ○ 7 possible types due to 7             ○ Important in tissues which are               ○ Affects rapidly dividing     DNA double-strand breaks)
        different proteins in machinery         terminally differentiated (where               cells most                 • Immunodeficiency,
• Increased risk of skin cancer                 whole genome is not important but      • 80% risk of colon cancer           chromosome breakage,
• Autosomal recessive                           genes being transcribed are)           • Autosomal dominant inheritance     cerebellar ataxia, extreme
                                        • Poor growth and neurological problems                                             sensitivity to ionizing radiation
                                        • Autosomal recessive                                                             • Avg lifespan = 20 years,
                                        • 2 possible types                                                                • Autosomal recessive




                                                             Biochem Page 20
Neurotransmitter Biosynthesis
Thursday, June 12, 2008
1:20 PM

Neurotransmission occurs between a pre-synaptic cell (always a neuron) and a post-synaptic cell which could be any number of cell types
  ○ Nerve cells have only 1 axon but it is possible that the axon is branched
  ○ In order for there to be "transmission" the postsynaptic cell must have a receptor for the particular neruotransmitter (Nt)
          There are 2 types of receptors which mediate the effects of Nts which are either excitatory (make inner cell more pos) or inh ibitory (make inner
           cell more neg). They are:
              □ Ligand-gated ion channels (LGIC)
                      Act fast, usually found in nervous system
              □ G-protein linked receptors
                      Slower acting (more complex), found with many hormone receptors
  ○ Neurotransmitters are synthesized in the presynaptic cell and stored in vesicles near the synapse. Their release is always Ca dependent and usually
     occurs b/c of local depolarization (voltage gated Ca channels).
          All neurotransmitters are small, water soluble molecules

AcetylCholine (ACH)
• Classified as an amine because it is basic                                              • Breakdown occurs in synapse by AcetylCholineEstera se
• Has 2 types of receptors                                                                     ○ Similar to serine proteases, has serine in active site
      ○ Nicotinic (LGIC-Na)                                                                    ○ Mechanism shows ping-pong mechanics with acetyl group
              Found in neuromuscular junction and autonomic ganglia/brain                       transferred to the enzyme, then water allowing for
              Several different isoforms which will have varying sensitivities as well          regeneration of the enzyme/release of acetyl group
               as rates of excition                                                            ○ Inhibited by organophosphates (irreversible binding to
              Inhibited by Curare which leasds to paralysis                                     active serine), which have been used in chemical warfare
      ○ Muscarinic (Gprot-Gi or Gq)                                                              (both against bugs and humans)
              Found in parasympathetically innervated tissues as well as sweat                        Ex: diisopropylfluorophosphate
               glands and in the brain                                                         ○ Breakdown products are reuptaken and used to synthesize
              Atropine and scopolamin inhibit                                                   more ACH
• Synthesized form Acetyl-CoA and Choline by Choline-acetyl transferase
      ○ Choline is found in 2 of the membrane phospholipids (phosphotidylcholine
        and sphingomylin, both on extracellular side)

Biogenic Amines
  ○ Synthesized from aromatic amino acids by decarboxylation
  ○ Amphetamines release dopamine, 5-HT, and N-Epi vessicles at nerve terminals
  ○ Carry a (+) charge at ph 7 (they are amines)
  ○ Can act as Nts as well as paracine messangers and hormones
  ○ 3 main types:
Catecholamines (Dopamine, N-Epi, and Epi)                                                 Indoleamines (5-HT/Serotonin, Melatonin)
    Precursor is Tyrosine (Para OH is required for catechole formation)                      ○ Precursor is Tryptophan
          1st synthesis step (also rate limiting step) forms L-DOPA (dioxyphenylalanine)     ○ 1st step Tryptophan hydroxylase (requires O2) places
           by tyrosine hydroxylase action (places OH in meta, requires O2 to work)              OH group on ring making 5-hydroxytryptophan
          2nd step forms Dopamine by DOPA decarboxylase action which requires                ○ 2nd step decarboxylase (Vit B6 dependent) removes
           pyridoxyl-phosphate (Vit B6)                                                         CO2 giving 5-HT aka serotonin
          Dopamine can be converted to Norepinephrine (addition of OH group to β-C) by       ○ In pineal gland Melatonin can be created from 5-HT by
           dopamine β-hydroxylase which is another O2 requiring enzyme (uses Cu)                acetylation and methylation
          N-Epi can be converted into Epi in the adrenal medula by phenylethanolamine                Only active compound with methoxy group on ring
           methyltransferase with SAM (places Methyl group on amine). Both N-Epi and                  Excreted in urine and can be broken down by liver
           Epi will act on the same receptors but with different affinities (also get                  enzymes (C-P450)
           transported on same receptors and broken down with same receptors).            Histamine
    Note that if you have a cell producing Epi or N-Epi it will have some levels of          ○ Precursor is Histadine
     Dopamine in it as this is a precursor

  ○   Inactivation by MAO or COMT occurs inside the pre-synaptic cell, after reuptake by a Na cotransporter
         Some drugs block re-uptake leading to longer Nt activation/effect
                □ Cocaine and methylphenidate inhibit uptake of dopamine, N-Epi, and 5-HT
                □ Tricyclic antidepressents block re-uptake of N-epi and 5-HT
                □ SSRI's (Prozac) block re-uptake of only 5-HT
         MAO (monoamine oxidase) causes oxidative deamination with the help of FAD (H2O2 is side product of regenerating FAD and must be broken
             down by peroxidases or catalases)
                ◊ Located on the outside of the mitochondria
                ◊ 2 isoforms: A (5-HT) and B (dopamine)
                     ► Histamine is degraded by neither
                     ► N-Epi is degraded by both
                ◊ Degradation products many times will have carboxy groups on them b/c aldehydes will spontaneously convert in the cell
                ◊ MAO inhibitors allow for extended Nt activation and are prescribed as antidipressents
                     ► Major side effect is inactivation of MAO in liver. This causes the patient to have an increased sensitivity to certain compounds such
                         as tyramine (found in red wine and cheese). Without the MAO in the liver tyramine will cause a hypertensive crisis.
         COMT (catechol-O-methyltransferase) causes methylation to the meta -OH group with SAM providing the methyl group
         Each enzyme can work on the products of the other (they each work on different sides of the molecule)

Parkinson's Disease
  ○ Age-related neurodegenerative disease affecting nigro-striatal dopamine neurons
  ○ Signs are rigor, tremor, and akinesia
  ○ Treatment currently uses L-DOPA along with carbidopa
         L-DOPA is given because its formaiton is the rate-limiting step and with it the body can make its own dopamine. Treatment with dopamine would
            not work b/c it would be broken down fast by MAO and it could not pass the BBB.
         Carbidopa is not able to pass the BBB and will saturate receptors outside the CNS which would tie up the L -DOPA. This allows the L-DOPA to
            have a greater affect in the brain, where it is needed.
  ○ Other treatments include deprenyl which is an MAO-B inhibitor which allow for longer presence of dopamine and have the added benefiet of limiting
      H2O2 formation in case this is what is causing the disease
  ○ Anticholinergics may also be given as cholinergics would have an antagonistic effect related to dopamine.

Peptide Nts
  ○ Common in short interneruons, released in conjunction with other Nts
  ○ Synthesized from precursor proteins in perikaryon and transferred in vesicles where actual synthesis occurs by proteases
  ○ Endorphins are an example of peptide Nts which act on opiod receptors




                                                               Biochem Page 21
Amino Acid Nts
  ○ Most important in the CNS (most brain neurons use AA as their major Nt)
  ○ Except for GABA, they do not need a biosynthetic/inactivating enzymes since they are already being made/present
        They do however need a special means of transportation which ends up being packaging into vessicles and high -affinity uptake from synaptic
           cleft
        They are limited to the CNS because they are not able to cross the BBB
  ○ Glutamate is the most important excitatory Nt in the brain, aspartate is another excitatory Nt
  ○ Glycine is an inhibitory Nt
  ○ GABA is most important inhibitory Nt in the brain
        Synthesized from glutamate by glutamate decarboxylase (Vit B6 required), removes the alpha carboxy group but there is still an amino acid due
           to other carboxy group (still a zwitter ion too)
        Degradation of GABA requires GABA transaminase as well as Vit B6, producing succinic semialdehyde
  ○ Enhancement of GABA leads to sedative effects (sleep, anti-epileptic, anxiolytic…)
        2 examples (both work on GABA-A receptor)
              ◊ Barbiturates: high ceiling
              ◊ Benzodiazepines (valium): Low ceiling




                                                         Biochem Page 22

				
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