• -single cell: algal cell
  • -three regions
    • -cap, stock, foot-nucleus
    • -unique to
• -performed experiment to discover control center
• -Algal cell belongs to genus Actasularia
  • -medieranea (med) A. crenulata (cren)
• -the two are distinguished by their foot (handout)
• -Reciprocal experiment
  • -whatever is done to A is done to B
  • -removed stock and cap and started with a med foot
  • -graphed cren stock and waited for a new cap to grow
  • -two sources of genetic information
  • -second cap was identical to foot
  • -cren foot - second cap cren
  • -nucleus was control center because of deduction
  • -Nuclear center determines the phenotype in eukaryotic cells

• -Canalization - plinko?
  • -development of germ layers/ systems from germ layers
• -progressive loss of DNA during canalization
  • -a single cell could only have specific characteristics
  • -207 different cell types in body
• -Theory lasted 70 years

• tested nuclear differation theory
• took ovary from frog
• irradiated ovary (UV)
• took intestianal cell
• homogonized them and retrieved a nucleus (diploid)
• enjected nucleus into irradiated egg
• approx. 1% resulted in mature frog
• if weissman was correct in his theory the end result should be intestianal cells
  • -only intestianal dna would have been present a whole frog could not have been formed

• -the other line will give rise to somatic cells
• the chromosomes in the germ cell line remain intact
• the chromsomes in the somatic cell line expereince a loss of DNA
  • -the chromatin fragments and is loss
• -remember that this is not inclusive
  • -unusal and does not happen in all organisms

• all diminutized somatic cells give rise to diminutized somatic cells
  • -the diminutized somatic cells are composed of heterochromatin
• second division same thing happens
• at the 32 stage only 2 have a full compliment

• -remove phophate (sugar and base) create a nuceoside
• -bases
  • -purines(AG) double ringed structures)
  • -pyrmidine(TUC) single ringed structures)

• hypo-methylated (less than all cytosines are methylated)

• -fertilized egg develops from imprinted genes
• -body develops
• -when body produces gametes orginial imprints are erased
  • -the bodies own genetic code for de-methylation and methylation take over
• -the body's new imprints are passed own to second generation

• "Parental Imprinting of Genes"
• -nuclear transfer
• -worked with mice
• -working with the fertilized eggs prior to fertilization
  • -micropipet
• -removed male pronucleus/female pronucleus/both
  • -created gynogenotes/androgenotes
    • -neither develop more than a few cells
    • -some do however
  • -if male pronucleus is removed and replaced with female pronucleus from same donor
  • -gynogenotes
  • -the contribution of parent has been imprinted identically
  • -the placenta (yolk sac) developed very poorly and fetus developed normally but does not make it to birth
• -if female pronucleus is removed and replaced with male pronucleus from same donor
  • -androgenotes
  • -the contribution of parent has been imprinted identically
  • -the placenta developed properly but fetus did not
• -these are not equivalent (sex linkage cytoplasmic)

• -human
• -we find that there is stunted growth, servere mental retardation, extreme obesity, adnormally small hands and feet because of the fact that these individuals end up obtaining chromosome 15 from the mother and not the father; both chromosomes come from mother

• -Cell Cycle varies between species and taxonomies
• -as we age the cell cycle changes
• -M: mitosis
  • -involves four stages (prophase, metaphase, anaphase, telophase)
• -Interphase (G1, S, G2)
  • -G1 phase: longest stage in animals
    • -all enzymes nessacary for DNA replication are being prepared
    • -preparation for cytocynesis is being made
  • -S phase: DNA replication
  • -G2 phase: some cells once they develop they do not divide again (G1 phase)
    • -GO phase is this

• -cannot do anything until it binds with cyclin B

• any kinase takes a phosphate away from an ATP molecule

• -if phosphates are clear of the active site it can perform function
• -phosphorylation of nuclear membrane
  • -leads to the breaksdown of the nuclear membrane
• -stimulates spindle formation
• -cyclosome breaks down cyclin B and CDK1 is inactivated
  • -p34 is altered and is phosphoralyated so that it can rebind to cyclin B
• -MPF dephosphorylates CDK1 and mitosis is ready to begin again

• -haploid: a single set
• -diploid: a double set of paired chromosomes
• -homologous chromosomes: paired chromosomes
• -somatic cell line most of the time
• -does exist in germ cells also

• -centrosome divides (centrioles)
  • -sit outside nucleus

• -mitotic apparatus
  • -achromatic figure
  • -centrioles
    • -half spindles (interpolar spindles)
    • -whole spindles
    • -astral rays
  • -chromsomes
    • -as long as they are together
      • -two chromotids
    • -once apart they become single chromosomes
  • -condensation: packing the chromosomes back into compact bodies

• -condensation
• -spindle formation
• -membrane breakdown

• -mitotic apparatus
  • -centrioles and associated microtubules
  • -form aster and/or interpolar spindles
    • -both are polarized microtubules

• -kinesin: go towards the plus end (microtubular motors)
• -dynein: go towards the minus end(microtubular motors)

• -attach to the primary constrictor
• -attaches only during cell division

• -kinetochore disaggrates the tubulin subunits and the chromatids is drawn towards the minus end of the microtubule which is attached to the centriole
• -the attachment between spindle and kinetochore occurs during prophase
• -difference between mitosis and meiosis is that kinetochore attaches to each chromatid in mitosis and does not in meiosis I

• nucloli (the construction site for ribosomes) disappears

• -metaphase plate is the area in the cell where the chromosomes are going to line-up
• -chromosomes are going to line up and the interpolar spindles are going to be constructed towards them
• -centrioles are found on opposite poles of the cell and the interpolar spindles are attached
• -we are ready for the chromosomes to be separated

• -the kinetochore has pulled the chromatids apart and pulled them to opposite poles
• -two sets are still present

• -cytokinesis (everything before is called karyokinesis)
  • -the cytoplasm is separated and a plasma membrane is drawn together by furrowing (balloon twisting) (plant: cell plate formation)
• -nuclear membrane reappears
• -centrosomes move to the outside of the nuclear membrane
• -nuclelous are seen in great numbers
• -both cells are now diploid

• -(reduction/division)
• -Monad: helix (chromatid)
• -Dyad (Valent): two heli (chromosomes)
• -Tetrad (bivalent): four parts
• -Reduction Division (Meiosis I): tetrads are separated into dyads
• -Equational Division (Meiosis II): dyads are separated into monads

• -Leptotene
  • -condensation of chromosomes
  • -telameres attach to the nuclear membrane
• -Zygotene
  • -synapsis (dyads come together to form tetrads)
    • -telameres move around nuclear membrane so that dyads can join
    • -synaptonemal complex (Handout):
      • -lateral element: protein rod
      • -lysine/arginine are present in lateral elements
      • -lateral elements come together when dyads come together 500 angstroms
      • -central element is formed
      • -central region 1000 angstroms
    • -monads are simple threads of serially connected genes
    • -the synaptonemal complex is important because as the tetrad is formed the complex will make sure that every gene is lined up perfectly with it corresponding gene
    • this allows for crossing over to occur in metaphase and for variation to exist
    • monad from one set to a monad of another set
    • 2000 agnstroms apart (uniform distance)
  • refered to as zygotene because synapsis causes and "yolk" to appear
  • fig. 15
• Pachytene
  • -recombination(crossing over)
• Diplotene(Dictyotene)
  • -repulsion begins
    • -the tetrads are beginning to separate back into dyads
    • -slipping
  • -Dictyotene is female only
    • -meiosis ceases until sexual maturity and obulation begins
    • -oogonia (one million present)
    • -only about 400 will be used
    • -atresia
• Diakinesis
  • -Terminalization
    • -the slipping points (chiasmata) move towards the telemars
• the same things that happen in mitosis: prophase happen in meiosis prophase
• Prophase I only one spindle is attached to the kinetochore of the dynad

• -the tetrad, composed of four monads(one helix), or two dyads(one set),
• -spindles are attached to the kinetochore
• -tetrads are moved into the central region of the cell along the equator
• -polarization of sets
  • -independant assortment
    • -the tetrads randomly align so that paternal set may go to the north while the maternal set goes south, but the next tetrad may do the opposite way
    • -allows for a 2^23 different varations

• -the tetrads are separated into dyads

• -nuclear membranes redevelop
• -cytokinesis
  • -furrowing leads to the division of the cytoplasm
• -haploid

• -not much different from mitosis except for haploid
• -Interphase II (Interkinesis)
  • -a brief resting phase
• -nuclear membrane disappears
• -mitotic apparatus develops
• -spindles form and attach to the kinetochore
• -two kinetochore are formed

• -alignment of dyads along equator

• -separate monads to opposite poles

• -two cells
• -nuclear membrane redevelops
• -haploid

• -imprinting
• -independant assortment
• -crossing over

• -there is a 1/128 chance or .5^7

• ½ to the x

• -what happens to the chromosomes in the cell

• -position of centromere
  • -primary constriction (holds kinetochore: spindles)
• -presence of secondary constrictions
• -lengths of arms
• -banding patterns
• -time sequence of replication

• mathematical illistration Fig. 4.1

• -are nucleolar organizers
• -appear as unstainged gaps in chromosomes
• -most often near chromosome end separating a small segment (satellite) from main chromosome body
  • -connecting secondary constriction called satellite stalk
• (nucleolous is location of rRNA manufacturing)
• satellite stalk is the portion of DNA that links the satellite with the arm
• stalk is so small it is considered unstained (20 angstroms)

• -the short arm is designated p (petite)
• -the long arm q (next letter in alphabet)
• -arm ratio (q/p) number will always be larger than one
• -centromere index [(p/p+q)=_ x 100] (a percetn that expresses p as a percent of the total length of the chromosome)
• -length of chromosome + arm ratio and centrosome index somtimes sufficient to identify chromosome

• -'The' contemporary approach to chromosome identification
• -Q (quinicrine), G (giemsa) and R (reverse of G) most commonly used bands (treatments)
• -G/Q bands relatively rich in Adenine-Thymine
  • -when G is light Q is dark
  • -when Q brightly flourescent G is dark
• -R bands rich I Guanine-Cytosine

• -segments that replicate proteins are going to replicate first
• -Individual bands show a constant replication order
  • -G/Q regions replicate late in S period of cell cycle
  • -R regions replicate early in S period
    • -location of most active genes

• -landmarks
  • -telomeres, centromeres, prominent bands
• -region
  • -section of chromosomes between two landmarks
  • -numbered 1, 2, 3, 4 etc, in both directions, starting from centromere
  • -bands within regions likewise numbered
• -ISCN (International System for Human Cytogenetic Nomenclature)
  • -1p21
    • -chromosome 1, p-arm, 2nd region, 1st band
  • -sub-bands
    • -numbered sequentially from centromere
    • -number preceeded by decimal
    • -14q32.3
• -Exra-missing chromosome indicated using + or - before number of affected chromosome
  • -47, XX, +13 (trisomy)
  • -45, XY, -21 ( monosomy)
• -Addition or loss of segments (use - or + after arm designation)
  • -46, XX, 5p-
  • -46, XY, 4q+

• -very simple experiments were done
• -diplococcus pneumonia
• -two strains
  • -one with a polysaccaride capsule (S or smooth)
  • -one that could not produce capsule and were rough (R or rough)
    • -mutants
• -if R strain was injected into another mouse the mouse lived
• -if S strain was injected into another mouse the mouse died
  • -the capsule was the protective layer of the strain
  • -the capsule protected the strain from the body's immune system
• -heat killed S strain type and injected the mouse
  • -mouse came out healthy
• -placed both heat killed S with alive R and the mouse died
  • -S type was recovered
• -theory was that transformation occurred
  • -somehow some substance was carried from one of the cells into the heat killed S
  • -did not know what the substance was
  • -left it at that

• -tried to identify substance
• -Handout 9.2
• -In Vitro experiments
• -started with the R cells
• -fractionated S cells and recovered all types of compounds or bodies
• -did this to find out what was being transported
• -place a single part into a solution of R cells and after a certain amount of time would add an anti-R serum to remove any R cells
• -would shake up the solution and try and see cloudy ness
• -did not see anything until they got to DNA
• -to make sure that their results were corrected added DNA and a enzyme that destroyed DNA for doulbe comformation
• -the purpose was to discover what was causing the transformation that Griffith saw
  • -what is the hereditary source for the nonvirulent R

• -Fig. 7.1
• -another name for a virus is phage
• -viruses have a protein capsule that contains nucleic acid core
  • -RNA and DNA viruses
• -tail fibers are the point of attachment
  • -cells have to have a receptor before the virus can bind
  • -some people do some people don't
• -important for two reasons
  • -provided additional proof that DNA was the hereditary substance
  • -revealed for the first time; the mechanisms for phage infection
• -experiment
  • -p282 (look at side chains) cystine/methyanine (S)
    • -these two amino acids appear in most protiens
    • -important for the folding process of the protiens
  • -knew that P only occurred in nucleic acids and did not contain S
  • -used P32 and S35 (isotopes) as labels
  • -used E. coli
  • -only source of S was the isotope which meant that the bacteria was already labeled
  • -another dish would have P and the bacteria would be labeled
  • -recovered T2 phage and spinkled them on the two sets of petri dishes
  • -holes began to be formed
  • -cold (non-radioactive) when they went in hot (radioactive) when they came out
  • S -with the hot phages new cells are infected
    • -ghost: hollow capsule; after the DNA is injected into cell
  • -placed cells in blender and cut the ghost
  • -the ghost is lighter than the new phages
  • -the ghosts and new phages are seperated through centerfuge
  • -found that most of the radioactive isotopes detected with the ghost and not the new phages
  • P -with the hot phages new cells are infected
    • -ghost are formed again
  • -found the radioactivity in the new phages (they only found hot phages)
  • -confirming step
  • -reason: confirming DNA as genetic material
  • -first experiment with viruses
  • -most viruses are DNA based but some are RNA (sub-group)

• -two purposes
  • -can viruses be degraded and reconstituted
• -found that this could be done
  • -called reconstituted virus (the orginial coat is removed and replaced with new)
• -tobacco virus requires only rubbing to be infected
• -two didfferent strain of viruses
  • -common
  • -Type M (mask)
• -Franco-Conrad performed experiment
• -first experiment using RNA viruses
• -using the reconstituted viruses
• -they would rub on the tobacco leaves and then later collect samples
• -common core and M-protein coat
  • -rubbed on leaf
  • -core was always the same as the core that was started with
• -found that whether DNA or RNA nucleic acids are the inherited substances
  • -prions: mad-cow disease ; cronic wasting disease
  • -all it takes is the aquasistion of protein to acquire disease

• -characteristics of inherited substance
  • -must be able to replicate itself with low mutablity
  • -must be found in the nucleas of eukarytoic cells
    • -must be found in the nucleoid of prokaryotic cells
  • -must direct the control of protein synthesis
    • -must do so with high efficiancy

• deoxribose has only a H on the #2 carbon
• -sugar-base combination is reffered to as nucleoside
• -all nucleosides exist as nucleoside triphosphates
• -double ringed: purines
• -single ringed: pyrimidines

• -Adenine Triphosphate (ATP)
• -nucleosides + phosphates = nucleotides
• -polarity of the sugar determines the anti-parallel nature of the helix
  • -lining up a bunch of 3' sugars with a 5' phosphate group
  • -other side is opposite
• -double comes from polymer of nucleotides or two strands
• -phosphodiaster linkage: 5' phosphate group is bonded to the 3' hydroail group
• -1953: two page article was published
  • -threw the scienctific community into a mess
  • -Crick was an English scientist
  • -Watson was an American
  • -Watson lied to get money to get over there
  • -became friends
• -Roslynn: x-ray diffraction
  • -did not know what it looked like until the photograph was done
• Fig. 913
• -the crystalligraph told them the dimensions of the DNA
• -had to be anti-parallel
• -the radius of DNA is 10Angstroms
• -10 base pairs in one complete turn of the helix
  • -the complete turn was 34Angstroms
• -1 base pair is separated from its neighbor by 3.4Angstroms
• -B-DNA is the typicall DNA (right handed) clockwise
• -there are other forms of DNA
  • -if we increase water percentage to 75% in cell (normal 70%)
    • -DNA is shifted 180 degrees
    • -A-DNA
  • -anti-oxidants (Vitamen C, D, etc.)
    • -protects DNA from oxidants
  • -if salt is added
    • -C-DNA
  • -if the DNA is methylated
    • -imprinted
    • -left handed (counter clockwise)
  • -Z-DNA when the DNA is hypermethylated
    • -some is right and some is left handed

• -B-DNA is the functional form (euchromatic)
• -Z-DNA heterochromatin

• -two groups of bases (purines(double ring)=pyrimidines)
  • -A & G = T & C
  • -A=T: double bond
  • -G=C: triple bond
• -the ratio of AT is not equal to GC
• -Denature: the heating that causes the ladder to becomes unstable
  • -another name is denaturation
  • -is an indicator of the ratio
  • -the higher the temp. the greater the ratio of GC (triple bond harder to break)
  • -mammals have a high denaturation temp.

• -with the information from Chargaff they modeled the structure

• -substance had to be self-replicating and have low mutability
• -high on the list of beliefs about duplication was that it was "semi-conservative"
  • -start with parent strand: the two strands of the parent will then produce to daughter strands so that the daughter strands have new and old
• -second most popular belief was a "conservative"
  • -start with parent strand: the strands separate and two daughter strands are created but then the daughter strands sepearte and the old strands of DNA return together and the new meet up so that the originial parent strand and a new daughter strand are formed
• -the least popular theory was "Dispersive"
  • -start with parent strand: fragmentation of the parent strand when replication begins and we have peices of old and pieces of new mixed together to form two daughter strands

• -started out with
• -ultracentrifugation: turned on ultracentrifuge and left on for hour s
• -centrifugal tubes full of cesium chloride
• -ran the tubes on the centrifuge
  • -two forces: increasing density on the outside
    • -diffusion: movement from high to low concentrations
• -a range of concentrations called concentration bands
• -they would place DNA into the tubes and the DNA would seek out the density that was equal to that of DNA
• -the use of UV light showed where the DNA was because of florescent light given off
• -N14 is normal weight of nitrogen in DNA found from E. Coli grown with N14 containing agar
• -N15 band was determined by growing E. Coli with N15 as their only source of nitrogen
• -band in the middle of the two
• -NO HYBRIDS Conservative model

• -Original DNA was N15 DNA one parent molecule
  • -centrifuged and found the band at N15
• -First Generation in a N14 medium
  • -centrifuged and found the band at the center of N15 and N14
  • -dispersive (multiple bands)
  • -conservative (one N14 and one N15 two bands)
• -Second Generation
  • -centrifuged and a N14 band is found but the N14/N15 band is still present
• -Third Generation
  • -centrifuged and N14 band is getting thicker but the N14/N15 band is still present

• -Fig. 9.22
• -used tritiated thymidine (isotopic form of thymine which incorporates H3)
  • -heavy hydrogen
• -allowed E. Coli to grown on a plate with tritiated thymidine as only source
• -used large slides and were covered with a photographic emulsion that contained silver grains
• -as the tritiated relaesed the beta particles they only traveled one micron so you could find them
• -grew the e. coli and recovered the DNA and let the beta particles release and a picture was made(exposed the silver)
• -autoradiographic
• -since the beta particles only went one mircon they could count the 'dots' and see the progression of DNA replication
• -the chromosome for all prokaryotics is circular
  • -theta structure is an intermediate stage
  • -OriC: Origin Site
    • -as you move in any direction from that you create two directions of replication (replication forks)
  • -used the rules of complintarity

• -polymerase: forms polymers (assembles nucleoside and phosphate group)
• -nuclease: the ability to dissassemble nucleotide
  • -Endo-nuclease: can break-up phosphodiaster linkages in the middle
  • -Exo-nuclease: can break off nucleotides from the ends of the helix

• -don't have to know specific locus but understand that it is a protein
• -fills in short gaps
• -can remove RNA primers

• -can also act as polymerase I
  • -only does so when you have a defective template or parent strand

• -does majority of work
• -has proofreading capabilities
  • -as the polymerase is moving along and mistakes are made the polymerase will back up and replace the mistake with the correct base
• mutator or antimutator
  • -mutator is a polymerase III that has flawed proofreading capabilities
  • -anti-mutator is a polymerase III that is too perfect and instead of skipping every few mutations they correct everyone and the pressure from natural selection begins to select against the anti-mutator

• -creates small little 10-12 nucleotide primer that consist of RNA
• -DNA dependant RNA polymerase ( has to have DNA to create the primer)
• -created on both strands
• -primer will have a 3' OH group and a 5' Phosphate

• -discontinous process and continous process occur at the same time

• 1. Primer Synthesis
• 2. Elongation
• 3. Primer Removal with or/and Gap filling
• 4. Formation of phosphodiaster linkage (Ligation)
  • -DNA Ligase

• -helicase is important because it recruits everything needed to make a replisome
• -unzips the helix

• -makes sure that the DNA polymerase III stays on strand

• -different bases (U)
• -different sugar (Ribose)
• -not imprinted

• if l=50 then there is equal number of turns on each strand
• if l > 50 then there is a positive supercoiled condition
• if there were not enzymes that corrected this replication would stop
• topoisomers:
• if l < 50 then there is a negative supercoiled condition
• Topoisomerase: is responsible for reliving the tention(also separates circles of DNA)
• Topoisomerase I, II, III, IV
• Topoisomerase I (a group of two I and III): clips one of the strands and passes the other through it and then seals the strand back together
• Topoisomerase II(a group of two I and IV): clips both strands and passes the helix through then seals the strands back together (only brought in to fix major problems)
• if l=48 then Topoisomerase I will be called in mostly likely act on the helix twice

• -repeated hundreds and hundreds of times

• not just foreign DNA, nucleases destroy any DNA and that is why telomers are present to protect our chromosomes from being destroyed by the nucleases

• linear chromosomes
• replication takes place with multiple eyes being formed until the end is reached
• complementary base pairs
• replication forks are going in opposite directions
• eventually they will reach another fork
• at some point replication forks will reach the telomere
• at the end there will be a primer that has to be removed on each end
• causes each end to have unpaired bases which could cause them to form a circle
• could cause the combination of two chromosomes
• telomerase is relied upon to prevent this
• ribonucleoprotien (RNA part and protein part)160 base sequence
  • pairs with telomers forms DNA/RNA hybrid
• Fig. 14.27
• templete of RNA is from in ribonucleoprotein which allows for the development of DNA
• on the extra
• telomere is elongated
  • -translocation ( RNA template is shifted down so telomere will be elongated; this can happen multiple times
  • uses telomere end RNA as primer to fill gap
  • Gap filling using DNA polyermase Beta
  • then ligation occurs through ligase
  • TTAGGG

• -composed of DNA and assosiated proteins
• -Histone octamer
  • -four dimers(pairs) of proteins
  • -H2A, H2B, H3, H4 (Lysine and arginine rich) makes these positive so they can bond to the negative
  • -H1 is the ninth protein but does not help form the core
• -two full turns equals 166 base pairs

• -the space between two nucleosomes
• -important promoters

• metaphase chromosomes

• -conserved sequence is a sequence that is found in every promoter region
• -consensus sequence is sequences that differ by a few bases in several different organisms

• -TATAAT

• Prosthetic group (inorganic part)
• -Holoenzyme is complete enzyme

• -sigma factor recongizes promoters
  • -if sigma factor was not present the Core enzyme would attach to any spot on the DNA strand

• -(-10) TATAAT
• -(-35)
• -(-50)

• -must be positioned for the right type of RNA to be coded for

• first base transcribed is reffered to as +1

Fig. 10.9

• there are different sigma's because there are different types of RNA and different proteins being made

• -Alpha components recongize -50
• -beta components recongize

• -if histones where in the center, other protiens would interact with the DNA
  • -this occurs because the other protiens are created to interact with histone protiens

• - prokaryotic are very different from eukcaryotic

• -SOS response

• -alpha control rate
• -beta controls location

• TATAAT: only two bonds compared to GC (three)

• -prokaryotes

• -rho protein

• -called rho-independant termination
  • -does not involve rho
• -called rho-dependant termination
  • -does involve rho

• -is recongized by NusA
• -stem loops
  • -inverted repeats are complentary
  • -A's and T's form loops

• -poly A region sequence exist (multiple A)
  • -easier to destablize
• -poly A sequence is important because the A's bond with U in RNA
• -spontaneous seperation of RNA and the DNA
  • -brownian motion

• -rho binds and moves along the RNA transcript at the same rate of movement along the DNA
• -does not have the poly A region
• -both cases the stem loop configuration is present
• -starts with leader region and keeps pace with polymerasization
• -motors along until it reaches the stem loop
• -rho has helicase properties and unwinds and breaks the strand off
• -the stem loop is also unwound
• -ATP dependant

• this is the message that will be read to form protiens

• -developed a system of isolation of ribosomal components
• -used a sucrose density gradient centrifugation
• -the rate at which an object falls depends on wieght and shape
• -put a hole in the tube
• -place each drop in a new tube
• -the 70s would be found at the top
• -the 20s would be found on the bottom

• -each has a different shape and mass
• P 256

• -120 bases in length to begin with (primary transcript)
• -80 after leaders and trailers are removed(secondary transcript)
• -polarity Alanine is attached to 3' ACC always at the 3' end
• -at the 5' end there is always GGG
• -modification of the bases occurs and the unusual bases are added
  • -the unusual bases are characteriestic of tRNA

• T-loop
  • -t is an unusual base: ribothymidine
  • -the ribosome recongizes this particular loop
  • -the ribosome will hold the tRNA just right so that the next one is lined up correctly

• Anti-codon loop
  • -codons: triplet of bases in an mRNA strand
    • -complementary bases with template DNA strand
  • -anti-codon's relationship with the codon exist here
    • -the stop codons cannot be recongized by a tRNA
  • -5' side uracil always will be present
  • -3' side methylinosine (MI) will always be present

• D-loop
  • -just one of many places on the tRNA molecule by Aminoacyl tRNA Synthetases
  • -Dihydrouridine is what characterizes the D-loop
  • -the D-loop and other places on the tRNA is recongized by Aminoacyl tRNA Synthetases
  • -these Aminoacyl tRNA Synthetases are what load or attach amino acids to the tRNA
  • -an uncharged tRNA is not carrying an amino acids
  • -a charged tRNA is carrying an amino acid
  • -there is a specific AtRNASynthetase for each of the 20 amino acids
  • -AtRNASynthetase will recongize more than one tRNA because there are more tRNA than there are amino acids
• -the purpose of using altered bases such as ribothymidine is to keep the loops open
  • -the altered bases cannot bond with any other bond

• -RNA polymerases I
  • -transcribes nuclelo organizers
  • -satalitte stalks are nucleol organizers
  • -these contain repeated sequences
  • -they code for 5.8s, 18s, 28s
• -RNA polymerase II
  • -does most of the work
  • -most of the genes are read by this polymerase
• -RNA polymerase III
  • -5s RNA comes from a different region of DNA
  • -tRNA
  • -??

• -consist of multiple proteins
• -is a transcription factor TFIID (much like the sigma factor)(just one protein making up the entire polymerase)

• -there is a concensus sequence found here
• -TAT then four A's
• -this box is recongized by TFIID
• -primary promoter

• -if it didn't waste of time and energy

• -enhancer: enhances
• -silencer: silences

• -one pole recongizes DNA
• -other pole recongizes TAF
• -enhancer regions are recongized by activator
• -the DNA is set perfectly so that transcription can occur properly

• -proteins are bound to the silencer
• -stops transcription

• -constituitive operon
  • -on and never shuts down(later test)

• -two functions
• -acts as a kinase
  • -kinases phosphorales
  • -the phosphorylation starts everything
• -acts as a helicase
  • -helicase unwinds DNA

• -consists of RNA polymerase II, TFIID(for a short time), TFIIH

• -some linkers are longer and those are important promoters
• -these hypersensitive sites are important because they lie outside the histone spool
• -the polarity if the histone spool is changed which allows for the loosening of DNA

• -the 5' end of the transcript (any of the three types) we will have a 7-methyl guanosine cap
• -this particular modification occurs most frequently in mRNA
• -this is the end that the ribosome recongizes to bind and begin translating a transcript
• -things happen in the right sequence
• -on the 3' end the polymerase will read beyond(20 bases) the stop codons and then some how a endonuclease is recruited and the strand is clipped

• -once it is clipped a Poly A Tail is added and consists of 20 to 200 adenine nucleotides
• -the Poly A Tail is used as a guide to get the transcript into the cytoplasm
  • -recongized by proteins and is then taken out
• -the Poly A Tail also stabilizes the transcript and protects the transcript from nucleases within the cytoplasm

• -heterogeneous nuclear RNA (hnRNA): composed of functional and disfunctional segments
  • -functional: exons
  • -disfunctional: introns
• -there exist sequences that are transcibed and translated these are exons
• -there exist sequences that are only transcribed and these are introns
• -in order to make a functional gene there must be a modification to the sequence
• -splicing: the removal of the introns and splicing together of the exons

• -Group I: Self splicing: is unique: this is the first time we will refer to RNA as having enzymatic properties, these segments are called ribozymes
  • -requires guanine
  • -fig. 10.30
  • -These introns will always begin with an A and end with a G
  • -the A is displaced from U by the bonding of G to it
  • -Then U displaces G from U and then the two Us splice
• -Group II: Self splicing
  • -The adinine froms three phosphodiaster linkages
    • -one with G, and two with other G
  • -no guanine required
  • -known to occur in mitochondrial forms of transcripts
  • - ATP independent
• -Protein mediated splicing
  • -Fig. 10.35
  • -with a majority of transcripts, most introns end with variant sequences
  • -5' end will be a GU
  • -there will also be a branch point (UACUAACA)
  • -3' end will be a AG
  • -the GU and AG being where they are is referred to as Chambon's Rule
  • -named because it relies upon the spliceosome
    • -which consist of five ribonucleoproteins
      • -these will be composed of uricial rich RNA (100-215 bases)
      • -U1, U2, U4, U5, U6
      • -they are small and referred to as snRNP's
        • -small nuclear ribonucleoproteins
    • -U1: recongizes the 5' end and initiates the splicing
      • -binds to GU
    • -U2: binds to the branch point
    • -while this is going on a complex is being formed
    • -U4: brings U5 and U6 together
    • -the complex of U5 and U6 displaces U1
    • -U1 and U4 leave
    • -U6 is in position of U1 and is bound to GU
    • -these have a rolling effect and brings U2 closer
    • -U5 can then splice the exons
    • -Adinine forms the three phosphodiaster linkages
    • -both self and protein mediated will form larets
    • -Fig. 10.36

• -this is called gene shuffling
• -remove the introns and shuffle the exons to produce multiple proteins from a single strand

• -these are the primers that are needed to match the flanking sequence

• >>Primers bind to DNA strands (30-65C, 30s)
  • >> Polymerase synthesizes new DNA strands (65-75CC, 2-5min)
    • >>Denature to separate DNA strands (94C, 30s)
      • >>back to Primers bind to DNA strands....etc

• -An operator and structural genes are an operon

• -the repressor protein has its own operon located somewhere on the genome
• -the repressors are build using the Central Dogma and manufactured for specific operators
• -the regulator gene has its own operator
  • -produces the repressor
• -Allolactose binds to repressor protein and causes it to fall free of the operator
  • -this is called the allosteric effect
  • -doesn’t necessarily require allolactose

• -negative impact: the transcription wants to but can occur

• -regulator gene will produce a repressor protein
• -the repressor protein will affect the operator
• -there are two things in competition to effect the activity
  • -RNA polymerase: starts the transcription
• -this picture shows that operator is flanked by promoters
• -if not for repressor protein the RNA polymerase could bind to the strand
• -in the absence of lactose this operon is going to be repressed
• -this is a lac operon, if there is no lactose then the proteins needed to use the lactose will not be needed so the operon must be repressed
• -the active repressor protein binds prevents the RNA polymerase from binding and energy is saved
• -this is an repressed operon

• -put the e. coli in a medium containing lactose
• -lactose is a disaccoride and binds to the repressor protein
  • -the allosteric effect causes the repressor protein to change shape
• -the new shape is not able to bind to the operator and therefore RNA polymerase can bind to the strand and transciption can occur
• -the inducer is what changes the shape: the inducer is lactose

• -a transcript

• -Beta glatosidase:
• -Permease: brings glucose right in for use
• -Transacetylase

• -instead of using something to activate the metabolic pathway (previously we used lactose) it is going to use an end product called the corepressor
• -we have an inactive repressor
• -as long as we have no end product the repressor will stay inactive
• -eight esstential amino acids: each is going to have their own metabolic pathway
  • -cannot get from diet must make ourselves
• -an accumulation of amino acids for protein synthesis forms
• -these amino acids then act as corepressors and bind with the inactive protein
• -the allosteric effect occurs and the repressor protein becomes active
• -the repressor protein binds to the operator and stops transcription
• -the correpressor and inactive repressor combine, change shape, and bind to the promoter

• -breaks down ATP and forms Cyclic AMP (adenine monophosphate)

• -not all operons have both controls
• -these are important because glucose is so vitally important

• -the cyclase form cAMP (a second order communicator)

• -activated when there is no glucose
• -bind at a CAP which is further upstream from the RNA polymerase
• -cAMP recongizes and binds to the CAP protein
  • -CAP cannot bind to transcript unless cAMP has bound to CAP
• -cAMP and CAP can bind to the CAP site
• -the CAP bends DNA about 90 degrees and allows sigma to find itself
• -CAP is important because it folds the DNA so that the activators can be bound to the enhancer regions
• -positive operon
• -if the bending does not take place then nothing can happen

• -why waste the energy to form proteins to break galactose
• -adenyl cyclase is not activated
• -sigma and RNA polymerase do not bind and nothing happens

• -activation
• -charging

• -Activation reaction
  • -energizes the amino acid
• -now we can charge the tRNA
  • -aminoacyl tRNA synthetases will charge the tRNA
  • -they recongize the correct matchup
• -the amino acid is bonded to the tRNA and AMP is displaced
• -thus far we have used 1 ATP

• -initiation
• -elongation
• -termination

• -IF3: Initiation Factor 3 (proteins)
• -three steps
• -IF3 binds to the small svedberg unit (30s subunit): prokaryotic systems
  • -if the IF3 doesn’t bind to the 30s subunit it cannot bind to the mRNA
• -Shine-Dalgarno Sequence: 5AGGAGGU3
  • -found just upstream of AUG
  • -important because it is what makes sure translation occurs at the right spot
• -Induced fit takes place: two proteins come together and the charges allow for a change in shape
• -Shine-Dalgarno Seq. is part of the 16s and IF3 binds to it

• -IF2 binds to tRNA (fMet) with the aid of GTP
  • -anytime GTP is involved and hydrolized there is a change in shape
    • -does not provide energy
• -complex is formed

• -Two complexes come together
• -Initiation complex is formed
• -GTP gives up phosphate and changes shape of initiation complex
• -IF3 and IF2 leave and the 70s ribosome is complete

• -the first charged amino acid to enter the ribosome is the first to enter:
  • -n-formylmethionine-tRNA
  • -it is AUG and is the only amino acid that can enter the P site
  • -all other amino acids will enter the A site

• -to do so it must pick up a GTP (hydrolosis)
• -once it is bound to the GTP the EF-Tu(protein) will pick up a charged tRNA
• -moves towards the 70s ribosome
• -the anti-codon on the tRNA should matchup with the mRNA codon
• -if the anti-codon is match then the tRNA is left behind and the EF-Tu moves on
• -a phosphate has been lost and now the EF-Tu is bound with a GDP
• -EF-Ts(protein) displaces the GDP, now EF-Tu is bound to EF-Ts
• -EF-Ts is displaced by GTP and the process starts over again
• -p292-293

• -now there are two amino acids on the A-site tRNA
• -EFG: translocase
  • -it moves the mRNA and the attached tRNA one codon in opposite direction
  • -now the tRNA with two amino acids is in the P site and nothing is in the A site
  • -GTP is involved; changes the shape of the ribosome in order that it may move
  • -Fig. 11.17 p296
  • -water is involved (condensation reaction)
  • -process starts over again with elongation

• -the nonsense codons are not three bases but four
  • -the three codons are usually followed by adenine
  • -20seconds to form prokaryotic protein
  • -Fig. 11.20 P. 298
• -Termination factors or releasing factors (RF)
• -RF1, RF2, RF3, IF3

• -p307-308
• -mRNA strand has polarity of 5’ to 3’
• -tRNA is aligned anti-parallel to the mRNA strand
• -the outside base on the tRNA coming from the 5’ end is stretched to form (three fingers)
• -this allows for mutation to occur more easily
• -fig. 11-31
• -mixed families have a greater chance to mutate because the third base matters
• -the most critical amino acids will have unmixed families

• -does what the shine-dalgarno sequence does in prokaryotes

• -brings in the appropriate machinary

• -once the signal peptide is formed translation slows down
  • -Polysomes: clustered ribosomes
  • -if you have polysomes you don’t go through this process
• -SRP recongizes the signal peptide (signal recongition particle)
• -SRP binds to the signal peptide chain
• -brings the complex to the DP (Docking protein)
• -Translocation Channel is found close to the DP
• -DPs seem not to be specific
• -what determines whether or not the protein belongs is the glycoprotein on the membrane
  • -the glycoprotein is called a Signal Seq. Receptor
  • -if the matchup is good it stays if not it leaves to find another
  • -if the matchup is good then translation continues
• -SP cuts the signal peptide chain free
• -the polypeptide chain is then taken to the interior ER and then vacoules take over the final production of the final product
• -the ribosome is released from the DP and ribosome is reused

• chains
• -alpha bond
• -polypeptide chains are orientated in different directions
  • -an amino and carboxyl end

• -polypeptides

• -alpha helix and beta sheets

• -alpha and beta domain
• -monomeric polypeptide molecule: multiple alpha and beta domain
• -folding takes place on the chaperone protein
• -the chaperone proteins are very important
• -the chaperones have a configeration
• -once the protein is folded into its functional form the chaperone leaves
• -if the chaperone is mutated then the folding may be wrong and protein cannot work

• -dimers, tetramers
• -dimeric protein molecule (two monomeric polypeptides)

• eukaryotes is number of mutations per gamete per generation

• -Intragenic: (within the gene) C(A)TCATCAT; if A is deleted it is a point mutation and sequence now becomes CTCATCAT and causes different amino acids to be coded for; CTCATCA(+)T; if the distanct is short between deletion and addition then the consequences are not as bad because the amount of codons affected is small; if however the distance is great then the consequences will be great because of greater number of codons affected
• -Intergenic: (substitution)mRNA strand >> UAC is present; UAC is aa tyr and polar
  • DNA strand >>> ATG is present
  • if A in mRNA is changed to U to make UUC (aa: phe; nonpolar)
  • -UAC is polar and UUC is not
  • -the precise folding of the protein is critical to the survival of the cell.
  • -by replacing a polar aa with a nonpolar then the folding of the protein will be changed
  • -the tRNA will mutate to compensate for the mutated codon on the mRNA

• -box genes control developmental pathways
• -if a mutation occurs in the box genes a pyramid of genes are going to be altered
• -creeper chickens

• -evolutionary development for functioning at specific temp. ranges
• -SOS response; distress response
• -a whole new set of proteins; heat shock proteins; come into play

• -deletion: removal of base
• -addition: insertion of base
• -substitution: bases are substituted
  • -least harmful unless nonsense turns to sense (read through) or sense turns into nonsense (premature termination)

• -refer to polymerase III (prokaryotes)
• -exonuclease activity will be altered and mistakes are not created
• -mutations are passed down and as long as the organism survives the mutations grow expontially

• -polymerase is too perfect and corrects all mistakes
• -organism will not be able to survive in new environemnt because of lack of change

• -by shifting the codoning strand or template strand ?????????

• -Amino group to an Imino group; Keto group to a Enol group
• -movement of a proton within one of the bases
• -body temp. can cause tautomerization
• -polymerase can proof-read

• -template strand has A bonded with T
• -replication occurs and A becomes bonded with C
• -this is a transisition mismatch
• -template strand is fine, but new nucleotide is mismatched
• -G and C are bonded
• -A bonds to the template C and forms a mismatch

• -A and T are bonded
• -A becomes bonded with A during replication
• -glycosidic bond occurs bewteen base and sugar
• -this bond shifts and allows for the AA to bond

-adenine is converted to hypoxanthine; can bond to cytosine
• -guanine is converted to xanthine; can bond to cytosine

• -Apurinic-Apyrimidinic Acids
• -Removes Adenine from DNA
• -Continued mutagenesis occurs
  • -if removed A is replaced by an A then Restoration occurs
  • -if removed A is replaced by a G and template T is mutated; Transition
  • -if removed A is replaced by a T and template T is mutated; transversion
  • -if removed A is replaced by C and template T is muated; transversion

• -they insert themselves into the host DNA

• -Damage Reversal
  • -present only in E. Coli
  • -damage is located and reversed
  • -UV-induced thymine dimer formation
    • -caffine, nicotine
    • -thymines bond to form dimers
    • -causes the adenines on opposite side to be unbonded
  • -relys upon DNA photolyase
  • -this happens in the dark
  • -photolyase recongizes the thymine dimers
  • -binds to the dimer and waits until light occurs
  • -uses the light energy to release the bonds forming the dimer
  • -if the gene that codes for photolyase is mutated then problems happen
• -Excision Repair
  • -occurs in all organisms
  • -Base Excision Repair
    • -deals with mismatches
    • -if a base does not belong in DNA such as Uracil
    • -there is a glycoslyase for all the bases
    • -the glycoslyase acts on glycosidic bond and allows for the excision of U
    • -Aprymindinic site because there is no pyrmindine present
    • -AP endonuclease breaks one side of the wrong base
    • - is on the 5' side
    • -less expensive to repair a single base as opposed to a sequence
    • -DNA polymermase fills in gap
    • -base is broken off by exonnuclease
    • -ligase comes and finishes bonding the bases
    • -all controled by operons
  • -Nucleotide Excision Repair
    • -deals with prymidine dimers
    • -dimers occur on the sam side
    • -dimers appear
    • -6 protiens involved
    • -UvrA2 (UV damage repair)
    • -UvrA2 is a dimer and combine with UvrB
    • -they go down strand and look for bulges
    • -leaves the UvrB on the 3' side approx. 3-5 bases from the dimer
      • -bends the DNA
      • -initates the binding of UvrC to the strand 8 bases away from the dimer
    • -UvrC and UvrB cut the phosphodiaster bonds
    • -UvrD comes in and unwinds DNA
    • -endonucleases within the cell will cut the base sequence that was cut
    • -DNA polymerase I going in a 5' to a 3' direction will fill the gaps
    • -DNA ligase will follow and reform the phosphodiaster links
    • -ABC Exinuclease (UvrA, UvrB, UvrC)
    • -Fig. 16.29
  • -Mismatch Repair
    • -approaches 100% of the repair mechanisms are of this type
    • -fig. 16.30
    • -template strand: methylated strand (methylase has not done its work yet)
    • -progeny strand: non-methylated strand
    • -A mismatch has occurred and G is bonded to T
      • -polymerase III has been mutated
    • -MutS (Mutator S) discovers the mismatch and recruits MutL
    • -MutL combines with MutS and they recruit MutH
    • -MutH endonuclease nicks DNA at GATC site
      • -GATC can be 100s bases away from mismatch
    • -MutU unwinds nicked progeny strand (helicase properties)
    • -Exonnucleases digest the nicked progeny strand
    • -DNA polymerase III fills the gap and ligase finishes up
      • -DNA polymerase I fills in the smaller gaps
      • -majority of the time the segment in question is long

• -Postreplicative Repair
  • -RecA-Dependant Repair (Fig. 16.35)
  • -polymerase motors along transcribing the DNA until it hits a dimer
  • -it falls off and moves to the next primer where it will start again
  • -now we have unpaired DNA with dimer present
  • -recA is a protein that is always maintained in some quantity (repressable operon)
    • -recA: recombination A
    • -two functions
    • -on it's surface it has an active site that binds the incomplete and complete strands
      • -site I: incomplete strand
      • -site II: complete strand
    • -it moves the incomplete strand along the complete strand until it finds a region of homology and establish complementarity
    • -Endonuclease breaks the now paired incomplete strand and leaves the complete strand incomplete
    • -DNA polymerase I fills the gap on both the now incomplete strand and now complete strand; DNA ligase then creates the phosophodiaster linkages
    • -ABC exinuclease removes dimer
    • -Autocatalysis of LexA occurs and LexA dimers are broken down into monomers
      • -the monomers cannot compete for control of operon
    • -No repression occurs
    • -a box operon controls a host of other operons
    • -recA is not activated, then LexA can bind to the repressor and no transcription occurs because the monomers cannot compete with RNA polymerase
    • -the process will continue until all DNA is fixed
  • -LexA triggers SOS boxes which allows for repair responses
  • -LexA dimer triggers operator for SOS boxes

  • -p53 (us) can be equated to recA
  • -apoptosis: suicide mechanism
  • -p53: 2366 nucleotides
  • -cig. smoke changes G to T and C to A
  • -p53 can gain function, stimulates cell division
  • -40% of cancers are not p53 dependant

• -p53 is involved
• -if p53 cannot fix cell
• -a kinase is activated
• -Apoptitic Protein Activating Factor 1 (Apaf-1)
• -kinase phosopholates the Apaf-1
• -Caspases are proteins that chop up the cells; zymogen: golgi ap
• -the cytochrome c is released from mitochondria
• -this allows Apaf-1 to bind to the Caspases which then activates the destruction of the cell
• -shape of the caspase is altered and the cell is dead

• transcription
• -mRNA (codon: G1)(codon:G2)(codon:G3)(codon:G4)
• translation
• -Enzyme::(enz 1)-------(enz 2)----------(enz 3)---------(enz4)
• -Initial:subA>>>>>subB>>>>>subC>>>>subD>>>END PRODUCT(Phenotype)

• -if substrate A is not converted into substrate B then the block a block is present and the following genes are inconsequenctial
• -If addition somewhere close to the end of the pathway then individual will be marginally fit

• -if phenylalanine(AA) is found in excess mental retardation can occur
• -if found early then can be treated by a diet void of phenylalanine and rich in tyrosine

• -no such thing as post-adaptive

• A4 is very successful under the given environemental conditions
  • -A4 will maximaze the fitness of the individual
  • -does not matter where
  • -"normal"
• This is compared to A1
  • -minimizes fitness
• Heterozygous

• -this constitutes a box
• -the farther up the box a mutation occurs the worse the mutation is going to be

• -establishes a set of ground rules that must be met in order that there be no change in a allelic frequency
• -if a condition is not met then the allelic frequency will change and evolution has occured
• -these conditions must be met:
  • -random mating
    • -Disassortive Mating: demonstrating a preference
    • -Positive Assortive Mating: a preference for traits exist
    • -Negative Assortive Mating: some traits are not prefered to other traits
  • -frequency of alleles in gametes must equal the freq. of alleles in the pop.
    • -to achieve this there must be random mating
  • -offspring within a community of individuals must be equally viable
    • -offspring must have equal fecumdity or ability to reproduce as much as anyone
    • -they share an equal level of fitness
  • -all factors producing gene freq. change are absent
    • -mutations
    • -selection
    • -genetic drift
    • -emigration (leaving)
    • -immigration (entering)
  • -all populations must be infinitely large

• -we have a situation in which two genes are equally viable under certain conditions
• -red, pink, and white oraclocks??
• -brown, red, white cattle color
  • -some cells have white gene while others have both brown and white gene
• -Numbers of Individuals
  • -401 samples collected of phenotype A: Red; Genotype:A1A1
    • -number of A1 alleles: 802
  • -106 samples collected of phenotype B: Pink; Genotype:A1A2
    • -number of A1 alleles: 106
    • -number of A2 alleles: 106
  • -46 samples collected of phenotype C: White; Genotype:A2A2
    • -number of A2 alleles: 92
  • -Total number of samples collected: 553
  • -Total number of alleles: 553(2)=1106
  • -Frequency of Alleles is determined by dividing # of A* alleles/Total # of alleles
    • -Frequency of A1: 908/1106=82%
    • -Frequency of A2: 198/1106=18%
  • -The next generation you come back and different numbers of alleles are found then evolution has occurred because the allelic freq. has changed
    • -If no change is found the all conditions of Hardy-Weinburg Equilibruim have been met

• -two phenotypic classes
• -a black and white mouse are placed on table
  • -we know that the white mouse is homozygous recessive
  • -the black mouse can either be homozygous or heterozygous dominant
• -binominal expansion
  • -assuming that there are only two alleles for any given trait in a pop.
    • -this isn’t true
    • -if three alleles existed trinominal expansion would be used
• -B? (black mouse)dominant
  • -don’t care what ? is because we have a dominant B
  • -could be BB or Bb
• -bb (white mouse)recessive
• -Frequencies
  • -Freq. of B(dominant) = P1
  • -Freq. of b (recessive) = P2
• -To determine freqencies use equation
  • -(P1+P2)^2 =1
• -Step One: Square Root of recessive allele
  • -if 4 out of 100 mice are white then 4/100=.04
  • -Square Root of .04 = freq. of recessive
• -Step Two: Multiply (P1+P2)^2 = 1
  • P1^2 +2P1P2 +P2^2
  • (.8)^2 +2(.8)(.2) + (.2)^2 = 1
  • 64% +32% + 4% = 100%
  • BB Bb bb
• -Step Three: Resample and determine new percentages
  • -if percentages change then evolution is occuring

• -females can be carriers (homozygous or heterozygous)
  • -binomianl expansion
• -males express what ever the X chromosome has
  • -determined by simple counting

• -mutations in genes that may affect an enzyme somewhere other than the active site and may not be harmful enough to decrease performance of the enzyme
• -SubA >> Sub B >> Sub C >> End Product
• -in pop. individuals will have genes that are slight mutations that will not detramentally affect the function of the proteins that are created using the mutated genes

• -selection pressure effects not only one allele but several different ones
• -as selective forces at upon a region, one allele is not selected over all the rest; several alleles are favored
• -Selectionist Theory: there is a genotype that is best and if a mutation occurs there will negative selection pressure exerted upon the mutated gene

• -Biston betularia: peppered form
• -favored because pepper form blended into the background of tree trunks
• -the white form could be seen on tree trunks
• -tons of sut was emitted into the environment and lichens died
• -peppered form died off drastically because they did not blend into the background
• -Carbonaria became the dominant type because they blended into the background
• -the environment is changed and now does not favor one extreme of allelomorphs
• -the normal curve is shifted away from those allelomorphs which are incapable of surviving

• -Sickle Cell Anemia
  • -hemoglobin: 2alpha and 2beta
  • -normal hemoglobin: HbA
    • -Glutamic Acid in Beta Chain (6th position)
      • -Acidic amino acid
  • -sickle cell hemoglobin: HbS
    • -valine is found
      • -non-polar amino acid
  • -Plasmodium falciparum cannot adequately digest HbS
  • -Individuals
    • -HbAHbA: die of malaria
    • -HbSHbS: hemolytic anemia; die
    • -HbSHbA: can survive
• -both extremes are selected against and the normal curve is pushed in to include less allelomorphs

• -Geographic region with X miles
  • -microhabitat systems exist within the geographic regions
    • -microhabitats exist in varying sizes
• -organisms within the habitat are best suited for their habitat
• -one allele will be favored over other neutral alleles
• -the normal curve will be pressed in the center because of selection pressure on certain allelomorphs

• -an expression of abiotic and biotic selection forces
• -the sum of all positive and negative selection pressures
• -if a hypervolume has a greater positive selection pressure then expansion occurs
• -as we expand, the hypervolume moves toward negative selection pressure
• -pre-adaptation allows for individuals to move into new environments
• -comfort zones exist
  • -as we get hot we sweat
  • -as we get cold we shiver
  • -the fewest ATP is used in the comfort zone
  • -the ATP that is saved goes towards fitness
• -”liveable space”

• -involves a geographic area
• -original pop.
• -barrier is formed (river)
• -new groups will have different selection pressures and some allelomorphs will be favored over others
• -the two new species will have individuals that will enter into the area occupied by the other species because of allelomorphs

• -originial pop.
• -a new niche is opened up
• -the niche is isolated
• -new species evolves because of selection pressures
• -the two new species will have individuals that will enter into the area occupied by the other species becuase of allelomorphs

• -originial pop. (fleas)
• -polymorphism occurs
• -within the same pop. new species can develop because of barrier to gene flow

• -storm

• -the different finches have evolved to exploit different niches
• -in the absence of competition experimentation can occur
• -those individuals who survived past their genes on to the next generation and through successive generations post-adaption occurs
• -some finches survive on large seeds
• -these finches do not have to worry about the finches who eat larvae
• -if competition is high; negative selection pressure; fitness decreases

• -will occur less in frequency
• -lies on the edges of normal curve

• -will occur in higher frequency
• -lies in the center of the normal curve
• -is most fit for given environment

• -ratio is .25:.5:.25 = 1:2:1

• -homologue: one member of a pair

• -Dd: d expressed alone is an allele

• -DD or dd
• -homozygote: referring to any member of a pair that is alike??

• -heterozygote (hybrid):

• -used to determine genotype of dominant parent

• -dominance or sex-linkage is determined
• -if the this deals with dominance then the same outcome no matter which parent has the different genotypes (genotypic ratio: 1:1)
• -if sex-linkage then different outcomes will occur when genotypes are switched
  • -genotypic ratio: first cross 1Aa:1a? second cross 1Aa:1R?

Product Rule: using a monohybrid cross

Bb x Bb	B(.5)										b(.5)
B(.5)	BB(.25)										Bb(.25)
b(.5)	bB(.25)										bb(.25)

• -all traits are determined by what he called inherited factors (genes)
• -these genes exist in pairs (except sex-linkage: hemizygous{pseudo-dominance})
• -Law of Segregation:
  • -involves a monohybrid condition
  • -members of a pair will segregate or separate so that each gamete only gets one member of that pair
  • -any violation of this law is reffered to as non-disjunction
    • -dyads fail to separate and one cell ends up with two
  • -meiosis
• -Law of Independent Assortment
  • -involves multiple gene pairs
  • -anytime more than one pairs of genes exist they are going to separate independantly of one another so that each gamete gets one member of each of the pairs

• Y?T? = Yellow and Tall: IIIIIIIII
• Y?tt = Yellow and drawf: III
• yyT? = Green and Tall: III
• yytt = Green and drawf:I

• Y?T? = Yellow and Tall: I
• Y?tt = Yellow and drawf: I
• yyT? = Green and Tall: I
• yytt = Green and drawf: I

• Y?T? = Yellow and Tall: I
• Y?tt = Yellow and drawf:
• yyT? = Green and Tall: I
• yytt = Green and drawf:

• (1/4)^(4 genes) = 1/256

• (3/4)^4 = 81/256

• 2^4

• 2^4

• 3^4

• monohybrid A: ½ aa
• monohybrid B: ½ bb
• A x B = ½ x ½ = ¼

• monohybrid A : ½ aa
• monohybrid B : ½ bb
• monohybrid D : ½ dd
• A x B x D = ½ x ½ x ½ = 1/8

• ½

• ¼ B x ½ C

• ½ A x ¾ B x ½ C = 5/16

• ½ A x ½ B x ½ C = 1/8

• ½ A x ½ B x ½ C = 1/8

• ½ A x ¼ B x ½ C= 3/16

Know material in text pertaining to figures 5.18; 5.19; 5.21; 5.23

• H (.5) H (.5) = .25
• Combinantions of coin toss
  • HH = .25
  • HT = .5
  • TH = .5
  • TT = .25

• -used if a series of terms is wanted
• -"at least" or "no fewer than"

• -in each term of a expanded binomial 'a' decreases by one and 'b' increases by one
• -the sum should always equal one
• (a + b)^n = 1
• a^n
• (a + b)^4
• (a^4) + (4a^3b) + (((4x3)/2)a^2b^2) + (((6x2)/3)ab^3) + b^4
  • the 2 comes from the number of terms

• - AaCc x aaCc

• culry hair: ¼ normal: ½ childs prob. is 1/8

• wavy hair: ½ astigmatic: ½ child's prob. is ¼

• -determines the amount of deviation of observed vs. standard phenotypic ratios
• -the phenotypic ratios come from previous crosses done over and over again
• -if the deviation is within reason (p77)
• -Table 4.4 is the deviation squared
• -degrees of freedom are simply the number of classes minus one
• -we know that as the chi^2 increases the difference between observd and standard is increasing which in turn tells us that the probablity of the observed happening decreases
• -if the chi^2 value exceeds .05 probability then we reject

• 2/3 * 300 = 200
• 1/3 * 300 = 100

• = (210-200)^2/200 + (90-100)^2/100
• = .5 + 1
• = 1.5

	progeny	Ho	chi^2
yellow(dominant) x green seeds	6022: 2001	3:1	.015
green (dominant) x yellow pods	428: 152	3:1	.45
violet (dominant) x white flower	47:40	1:1	.56
round (dominant) x wrinkled pods	31:26:27:26	1:1:1:1	.618

• -one gene pair can preclude the necessity of another gene pair
  • Guinea Pigs
  • C? = Prod of pigment
  • cc = no pigment
  • B? = Deposition of much pig: black
  • bb = deposition of moderate pig : brown
• P1: outcross of CCBB x ccbb
• F1: CcBb (selfed)
• F2: genotypic ratios: 9:3:4
  • phenotypic ratios:

• -RRpp(rose) x rrPP(pea) = RrPp(walnut) x RrPp (walnut) = 9:3:3:1 ratio
• -Fig. 2.21 cross a walnut and single and see what happens
• -if a problem is given that requires us to cross a walnut and a pea
• -single will always be homozygous recessive
• -R?P? = walnut
• -R?pp = rose
• -rrP? = pea
• -rrpp = single

• AABBCCDDEE = 5'7"
• A'A'BBC'DD'EE = 5'11"
• average height = 5'7"

• -if a gene is expressed in every generation then it is assumed to be dominant
• -some ppl who had a gene failed to express it (epistasis)
• -if they do express the gene they have a slightly curved pinky finger and a little webbing
• -most extreme expression was to have three fingers conjoined
• -phenotype expressed in whole range
• -those who failed to express the trait are catorgorized into "zero penetrance"
  • -the percentage of individuals in a pop. which failed to express the gene they posess for that trait
• -define "100% penetrance"
  • -everyone with the gene expressed the trait
• -define "70% penetrance"
  • -70% of individuals with gene expressed the trait

• -normal rabbits raised at 25C
• -white extremities: totally white at 30C
• -flank cooled to below 25C and a black spot develops

• CC = normal
• Cc = creeper
• cc = lethal (not expressed and alters ratio)

• -what portion will be normal ½

• -neutralism: many alleles for same trait and many of them have equal chance of survival
• -co-dominance of alleles
• -blood types come from the gene I which attaches a sugar to a lipid chain on the cell membrane of red blood cell

  -IaIa or Iai = A	: B antibodies
  -IbIb or Ibi = B	: A antibodies
  -IaIb = AB (co-dominance):	(universal recipient)
  -ii = O	: A and B antibodies (universal donor)

• -if asked what genotypes must the parents for the offspring to have all four blood groups in their offspring
  • -parents must be Iai and Ibi

• -including recessive traits

• - X/A ratio

• -this gene is influenced by two different sets of numerator and denominator elements
• -numerator elements
  • -if they influence the sex lethal gene the outcome is female

• -If the denominator elements influence the sex lethal gene the outcome is a male

• -this was thought that this was true for humans
• -there are genes on the X that can cross over they are found in the pseudoautosomal region; synapsis will occur but crossing over may occur or may not occur

• -gene that produces protein that has everything to do with sex determination
• -the SRY protein influences two other operons
  • -p450 aromatase: its protein converts testosterone into estrodial (negative)
    • -if this happens female develops
  • -mulleriam-inhibiting substance
    • -stimulates testes development (positive operon)
    • -it also surpresses female tissue development
  • -p450 will be produced if SRY protein is not present
  • -if SRY is present it acts to repress the p450 operon so that testosterone can accumulate; the stimulation of testes development is also seen
    • mulleriam-inhibiting substance operon is activated

• -if TDF functions then testes develop and male is the outcome
• -if TDF is absent then overian development occurs and female is the outcome

• "Hairy ears" - Indian people thought to be holandric

• -if offspring receive the condensed X from both parents then cells compensate for the doulbe condensation with communication from other cells

• -most of the time secondary and primary sex characteristics are limited to one sex
• -however, if females take steroids they can obtain some attributes as men (muscles)
• -breast development is limited most of the time to one sex, but 5% of American men die of breast cancer

• ZbWRRpp x ZbZbrrpp
  • -what is the genotypic ratio 1:1
  • ½ ZBZbRrpp
  • ½ ZbWRrpp

• M? or M = normal
• m or mm = MD
• the young man who died had MD that was sex-linked recessive: XmY
• the genotype of the parents is male ( XMY) female (XMXm)
• living children: girls XX, XX, XX (1/2 will be normal ½ will be carriers)
  • boys XY (received normal gene)

• -seperate from the X
• -in the heterozygous condition the dominant gene is expressed in males, but repressed in females

• -syntenic: all the genes in this particular case are coupled within the same linkage group
• -all members of a chromosome make-up the same linkage group
• -trance arrangement: recessive genes are in different sets

• -the monads must be from two different sets

• -the first cross over often affects the succesful outcome of the second cross

• +fz+ has more in common with +++
• the gene that is different you always put in the middle
• the gene sequence is: h-fz-eg or eg-fz-h (the outside two does not matter right now)

• -Region I is found between 1st and 2nd genes
• -Region II is found between 2nd and 3rd genes

• +fzeg(58) and h++(80)

• ++eg(28) and hfz+(30)

• cross over in both regions occurs
• +h+ (1) and h+eg (1)

• -the number of doulbe cross over should be equal the product of region I and region II
• C of C = .002/.0084 = .238 or 23.8% of the time actual double cross overs occurred

h

bl

+

hbl+	301
+bl+	19
+bld	6
hbld	172
++d	305
h+d	21
h++	8
+++	160

• Gains or loses of segments of chromosomes can occur
• A rearrangement of genes within a chromosome to establish new gene linkage relationships between genes
• A gain or lose of complete chromosomes

• -a plant has tetraploid condition with 2 dominant and 2 recessive genes
• -an example is seedless fruit; an increased number of ploids can prevent seeds from developing

• -tetraploid: alfalfa, potato, peanut
• -triploid: banana, seedless watermelon
• -these individuals are sexually sterile
  • -failure of normal segregation of chromosomes
• -every now and again a 2 and 2 develop into a tetraploid which can be propogated

• -Horses and Burros produced mules

• AW	>>(colchacine)	AA	WW	(amphidiploid or allotetraploid)
  BX	>>>	BB	XX
  CY	>>	CC	YY
  DZ	>>	DD	ZZ

• -cotton, tobacco, wheat, strawberries, rasberries
• -instant speciation?????

• -Down Syndrome: 3 #21 chromosomes trisomy
  • -flattened nose, protruding tongue, retardation, fingers are shortened, stature shortend
• -Tetrasomy: 4 and so on
• -2n +1 : trisomy
• -Monosomy: 2n -1
  • -Turner syndrome: should be a female (XO)
  • -short stature, sheild shaped chest, tissue from ears to shoulders
  • -reduced longevity, retardation, failure to develop a gonad
  • -reduced breast development
• -Nullisomy: 2n-2 complete loss of a pair
  • -probablitily of survival: 0
  • -loss of two chromosomes from two different pairs

• -problem arises because of females; mieosis is stopped at diplotine stage and doesn't resume until ovulation occurs; these eggs may be dorminant for 50 years and DNA can become excessively brittle which can cause problems
• -tall, thin, female breast-like development, reduced development of the gentitals, retardation, secondary sexual characteristics are blended

• -something that occurs way down the line
• -such as eye color spots or hair spots
• -combinations of ploidy within the same organism; localized usually

• -if you like the monoploid then you can use chochcine or something and create a new organism
• -grasses can survive this because they don't have to reproduce to grow and expand
  • -bermuda grass

• -the target site is melted and broken to create sticky ends; the transposon then inserts itself
  • DNA polymerase I and ligase then complete the insertion by filling the DNA gaps
• -the transposon can remove themselves from the genome and replace the orginial codons, or it may make copies of itself and propogate; if the transposon were to insert itself into a regulator gene that control the active repressor of an operon then an inactive repressor will be created; if the transposon inserts itself in a gene that occurs at the end of the genes that will create a protein then only a minor variation will happen, if the insertion is at the beginning there will be great variation
• -this is referred to as "position effect"

• -semi-sterile state: 50% of division will not be able to reproduce
  • -oligospermia

• -Alternate Segregation: the only combination that gives a normal genetic outcome
  • -1/6 will be normal 1/3 will be viable

• -supression of crossing over occurs with the formation of the inversion
• -in inversion loops beneficial genes are often times protected in inversion loops to the point that they are seldom broken up; those clusters are called super-gene clusters; these clusters regulate major patterns of develop within an organism box genes control the development of limbs and organs; these are subject to other types of variations such as polymerases

• -isochromosomes: arms of equal length; occurs because of misdivision of the centromere on a horizontal plane instead of a vertical plane; the results are two chromosomes with equal arm lengths
• -fundamental number; (NF): counting the number of arms of species that are very similiar but have differnet numbers of chromosomes
• -there are number of location that chromatin can be broke, the locations of the breaks will determine whether or not the chromosome will be viable; if the chromosome is not viable the individual will not live

• -plasmids: independent self-replicating pieces of chromotin about 1/50 the size of a genophore
  • -plasmids are present; they may contain a sex factor (F+) and is said to be a donor
  • -the F+ can exist in a plasmid or an integrated plasmid called a episome
  • -F- cells are recipitants
  • -episomal sex factors are referred to as HFR (High Frequency Recombination)
  • -F+ cell produces Pilus (Pili) that are hollow (tube-like) they grown from the surface of an F+ cell to a F- cell and pull the two together
  • -Conjugal Bridge is formed (Conjugation occurs)
  • -endonuclease nicks one of the strands and creates a 5' end
  • -one of the strands slips through the bridge so that each cell has a template
  • -each cell is now F+ because they have sex factors

• -endonuclease nicks the DNA and creates a polarized structure
• -F portion of the DNA is the last portion to leave
• -the bridge is broken by the size of the HFR and the F portion is lost
• -if there exist complentary segments between the old and the new strands genetic exchange will occur
• -the new strands portion that have complentary with the old will bond to the old and the strand that is being replaced will be biodegraded

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