*frequency of recombination events = 2x recombination fraction-1map unit = 1 recombinant chromatidChapter 1: Genetic Approach to Biology

1.2- Molecular basis of genetic informationTranscription: process of converting DNA into mRNA. The difference between RNA and DNA is that the sugar in RNA is ribose not Deoxyribose. Also Thymine is not present in RNA, rather Uracil is. Transcription takes place in the cell nucleus

- the transcript is not always a functional mRNA strand because many of the nucleotides do not code for amino acids.

- Transcripts serve three main purposes1) increases # of copies of genetic info available to cell2) relieves traffic congestion as it leaves nucleus and enters cytoplasm.3) Act as a control on how much of a protein is produced

Translation: production of amino acids based on a sequence of nucleotides in mRNA.- every three amino acids (codon) codes for a specific amino acid.- There are 64 different codon combinations however there are only 20 amino acids- UAG= stop codon- tRNA actually are the correspondence btw codons and amino acid formation.

tRNA have anticodons that compliment codons. They also have the amino acid on there tail so when they meet up with appropriate codon the ribosomes act to add the amino acid to the growing polypeptide.

Gene Regulation: occurs both in the attachment of the RNA polymerase to the beginning of the DNA sequence that is to be read and in the initiation of its movement along the DNA sequence.

1.3 Program of Genetic InvestigationForward Genetics1) genetic investigation that begins with observation of a variant phenotype

a)one such type of forward genetics involves looking for normal natural variation within the phenotype (ex. Varying degrees of insecticide resistance in insects)

b) another form involves looking an abnormal variant in character for which all individuals have a normal form. -normal= wildtype variant=mutant type2) make crosses between strains3) note offspring ratios4) identify genes5)identify molecular and developmental differences between individuals of diff genotypes6) the last stage in forward genetics is to characterize the DNA of different variant alleles

Reverse GeneticsSTARTS FROM AN OBSERVED NORMAL DNA SEQUENCE OF UNKNOWN FUNCTION MANIPULATES THE GENE TO CREATE GENETICALLYMODIFIED ORGANISMS, SUCH AS A KNOCKOUT, FACILITATING STUDY OF WHAT THE GENE DOES AND HOW IT PARTICIPATES IN FORMATION OF A

PHENOTYPE-ex) if don’t want to induce mutations can study very similar species

1.4 Methodologies Used in Genetics1) isolation of mutations affecting biological process under study2) analysis of progeny of controlled matings3) genetic analysis of cells biochemical processes4) microscopic analysis 5)direct analysis of DNA

-genomics: study of structure, function, evolution of whole genomes-bioinformatics: computational analysis of the content of genomes

Probing: A) Southern Blotting (DNA)1)piece of DNA is cut using restriction enzymes: produces many fragments2) fragments are separated by size using gel electrophoresis3)gel soaked in alkali to denature the fragments4)fragments transferred to positively charged membrane that mirrors distance they traveled during electrophoresis5)Random DNA is added to adhere to all fragments but the one of interest6)then treated with probe which is in excess so it will bind to a complimentary strand

B) Northern Blotting (RNA) similar process to southern blot

C) Western Blotting (protein)1) run protein thru SDS polyacrylamidegel using electrophoresis2) Run gel perpendicular to run protein out into a membrane3) Block for nonspecific antibody sites4) incubate with primary antibody for protein of interest5) Use secondary antibody that recognizes tail of first one. With artificially attached enzyme we can view if protein is present because it gives off chemiluminescent light

1.6) Genes, the environment and the Organism

Model 1: Genetic Determination-virtually all differences between species are determined by the differences in their genomes (ie. A lion will never give birth to a lamb)

Model 2: Environmental Determination-in this model the genes impinge on the system giving general signals for development, but the environment determines the actual course of development

Model 3: Genotype-Environment Interaction

-what an organism will become critically depends on its present state and on the environment it encounters during that moment

Developmental Noise-genotypes are fixed throughout life while phenotypes are determined by the interaction of a phenotype and a specific environment-random events in development that lead to variation in phenotype

Chapter 2: Single Gene Inheritance2.2 Mendel’s Law of Equal Segregation-studied seven traits in Garden Peas: Pea color, pea shape, pod color, pod shape, flower color, plant height and position of flowering shoot

-for each character he studied two contrasting phenotypes

MITOTIC CELL CYCLE:1)Interphase: composed of G1, S and G2 phases. Cohesin is introduced during this phase and it is a complex that includes 2 SMC proteins that are linked to 2 non SMC proteins2)Prophase: cohesion still holds sister chromatids together and nuclear envelope breaks down3)Metaphase: condensin is introduced which further compacts the DNA together. Mitotic spindle forms and bivalent attachment occurs at centromeres. Sister chromatids attach to opposite kinetochore microtubules4)Anaphase: cohesion is cleaved but condensin is still in contact and it holds DNA compacted as segregation starts to occur 5) nuclear envelope reforms as telophase occurs and splits into two cellsMEIOSIS CELL CYLCE- gametes formed from meiocytesMeiosis 1: homologous chromosomes separateMeiosis 2: sister chromatids separate

Null alleles: the proteins encoded by them completely lack the function of the non-mutated proteinLeaky alleles: mutant alleles that have a reduced level of the protein function

Bakers Yeast Mating:1) MATa and MATalpha cross2) Become a diploid3) Chromosomes undergo replication to become a meiocyte (tetrad)4) Undergo meiosis to produce 4 haploid cells in the ascus

Haplosufficient: a heterozygote with one copy of mutant allele and one wild-type. The wild-type still provides enough protein product for normal functionHaploinsufficient: When the wild-type doesn’t provide enough protein for normal functionTest Cross: cross of a heterozygous parent with a homozygous recessive parent (tester)

2.5 Sex-linked single gene inheritance patterns

-female is the homogametic sex (XX)-male is the heterogametic sex (XY)

Sex-linkage: genes in differential regionsa) X-linkage: mutant alleles in the differential region of the X chromosome b) Y-linkage: mutant alleles in the differential region of the Y chromosome

Pseudoautosomal Regions: The X and Y chromosomes have two short homologous regions at each end

-sex-linked inheritance occurs at the differentiated regions

*in chickens and moths the female is the heterogametic sex(ZW) and the male, homogametic (ZZ)

X-Linked Recessive Disorders:1) Many more males than females show the phenotype because the female offspring

can only exhibit the phenotype if she inherits a copy from both mom and dad2) None of the offspring of an affected male show the phenotype3) None of the sons of an affected father show the phenotype

X-Linked Dominant Disorders:1) Affected males pass on trait to all daughters but no sons2) Affected heterozygous mothers pass trait to half of sons and half of daughters

Y-Linked Inheritance:-only occurs in sons

Chapter 3: Independent Assortment of Genes3.1- Mendel’s Law of Independent Assortment-For two genes on different chromosomes: A/a;B/b-For two genes on same chromosome: AB/ab-If location is unknown: A/a*B/b

Mendel’s First Law: law of segregation. Each gamete only receives one copy of the geneMendel’s Second Law: Different Gene pairs assort independently in gamete formation.-Holds true for genes on different chromosomes

1-((1-(odds of desired genotype))^n)=% you are sure

Chi-Squared-df=Number of variable-1-if P<.05 then the results are not compatible with the hypothesis

Assumptions:1)segregation laws in effect2)assume all genotypes have same efficiency of fertilization

3)all fertilizations give rise to peas with equal probablility

Synthesizing Pure Lines:Start with heterozygous gene pair (A/a). After selfing the progeny should be ¼ AA, ¼ aa, and ½ Aa. After selfing these f1 generations the total probability of heterozygosity should once again be cut in half since the homozygous individuals from the F1 generation will only produce Homozygous progeny

3.3-Chromosomal Basis of Independent Assortment-carothers observed, through the experiments with grasshoppers, that chromosomes segregate independently of eachother. She found this because there was a heteromorphic pair of chromosomes and also a chromosome that wasn’t paired at all. She observed that the unpaired chromosome segregated with the other chromosomes with equal frequency.

Neurospora (Bread Mold):1) MATa and MATA cross2) Become a diploid3) Chromosomes undergo replication to become a meiocyte (tetrad)4) Undergo meiosis to produce 4 haploid cells in the ascus5) Undergo extra mitosis to produce 8 ascospores

Recombination:-a recombinant is defined as any meiotic product that has a new combination of alleles in reference to the haploid genotypes that formed the meiocyte

3.4 Polygenic Inheritance-polygenes are the interacting genes underlying hereditary continuous variation such as skin color or height

3.5 –Organelle Genes: Inheritance independent of the Nucleus-Not all genes are found in the Nucleus.

-Some are found in the mitochondria, and in the chloroplasts of plants.-mitochondria and chloroplasts are located in the cytoplasm and contain a circular chromosome

-mtDNA(oxidative phosphorylation) and cpDNA(photosynthesis)-Organelle Genes are inherited almost solely by the mother since the egg contributes a lot of cytoplasm while the sperm contributes essentially none (Maternal Inheritance)

Cytohets: Cells containing mixtures of Mutant and Normal Organelle chromosomes-by chance after the cells divide, distinct organelle types (mutant or Normal) will segregate together into progeny cells causing completely mutant or normal progeny cells.

BOVERIStudies with Sea Urchin’s:

Took a look at sea urchin eggs that are fertilized by 2 sperm. The centrosome that is responsible for formation of mitotic spindles after fertilization are inherited only from the father. So for 2 sperms there are 4 spindles formed. After the 54 chromosomes duplicate to become 108 (18 from egg, 36 from two sperms) 27 each attach to 1 spindle. This led to abnormal development. Boveri then did an experiment and fertilized the egg with two sperm again. This time at fertilization he shook the culture which resulted in only 3 spindles formed. This would theoretically allow 36 chromosomes to attach to each spindle pole. He found that most zygotes still developed abnormally but many more developed normally than did those with 4 spindle poles.

Devised an experiment to test the probability that normal development would occur in 3 spindle case as compared to 4 spindle case. Made boards with 4 compartments and 3 compartments respectively. Got 54 balls and 3 balls had 1 specific #. Found that 11% of the time with the 3 spindle situation each compartment had 1 of each type of ball. Never got a successful trial with a 4 spindle situation. This helped him conclude that only a full set of chromosome types could sustain normal development, and that each chromosome had individual characteristics.

Chapter 4- Mapping Eukaryote Chromosomes By RecombinationWhy Genetic Maps are important:1) gene position is crucial for building complex genotypes2) allows you to pinpoint a genes structure and function3) useful in interpreting mechanisms of evolution

4.1- Diagnostics of Linkage-found that some Genes are linked because when performing a self cross of a dihybrid, they did not see a 9:3:3:1 phenotypic ratio that mendel’s second law suggests.-also when he performed a cross of a dihybrid with a homo recessive tester he did not observe a 1:1:1:1 ratio that independent segregation predicts

How crossover produces Recombinant:-during meiosis when two dyads form a bivalent a chiasma occurs between two nonsister chromatids.

Evidence that crossing over is a breakage and rejoining process:-Plant was a dihybrid in the cis formation. One of the chromosomes was longer and had a nob like structure while the other was shorter and normal. However they found that after recombination the long chromosome no longer had the nob on it because it had crossed over.-must occur at a 4 chromatid stage because if it occurred when the chromosomes had been unduplicated there could only ever be 2 genotypes when you isolate the tetrad (in fungi). (4 is what is commonly seen)-crossover can occur between 3 or even 4 chromatids

4.2- Mapping by Recombinant Frequency-recombinant frequencies greater than 50% are never observed

-as recombination frequencies approach 50% it is difficult to distinguish whether they are located far apart on the same chromosome or are on different chromosomes

Three-point Testcross-cross a trihybrid with a triply recessive tester*always use the sum of the two smallest distances and not the longest distance alone. Because it doesn’t take into account the double crossover

Without recombination frequencies we can see use inspection to see if three genes are linked by the following observations:

A) Two genotypes at high frequencyB) Two at an intermediate frequencyC) Two at another intermediate frequencyD) Two rare frequency

Interference-if double crossovers are independent then the frequency of double crossover would be the product of the two single crossovers in the adjacent regions around the middle locus.

I=1-c.o.c=1-(observed # of double recomb)/(expected number of double recomb)-expected number of double recomb is equal to the product of the two single crossovers times the total # of progeny.

Ratios as diagnostics:Monohybrid testcross: 1:1Monohybrid selfed: 3:1Dihybrid Testcrossed(independent): 1:1:1:1Dihybrid selfed (independent) 9:3:3:1Dihybrid testcrossed (linked) P:R:R:PTrihybrid testcrossed (independent) 1:1:1:1:1:1:1:1

4.3 Mapping with Molecular Markers-Loci of molecular heterozygosity are called molecular markers

-useful in detecting genes of interest because they act as milestones

1) Test cross a dihybrid (1 is a gene that is hybrid and the other is a molecular marker)

2) Extract DNA and sequence it3) and if you find a P:R:R:P ratio you can determine the distance of the gene alleles

from the molecular markers

SNPS-Single Nucleotide Polymorphisms-there are theoretically 4 different alleles possible, however there are only ever 2 found in a population.

1) Silent within genes: have no apparent effect on phenotype

2) SNP that causes a mutant phenotype showing single gene inheritance3) SNP in polygenes that help to discover polygenes that contribute to a phenotype

showing continuous variation.4) Intergenic SNPs: have no phenotypic effect but are useful as milestones5) RFLPs: SNP’s located at Restriction enzyme target sites

-Useful because don’t require sequencing to determine locationA) select a restriction site with two morphs of a hybrid organism and then

southern blot it to examine differencesB) perform a test cross, extract DNA from progeny and add Restricition

enzymesC) southern blot the DNA and look for recombinant frequency to use for

mapping

Mapping by SNP Haplotypes:Haplotype: chromosomal segment defined by a series of SNP’s generally between 10 and 50kb’s-This works by the concept of Linkage disequilibrium which is when cross-over does not occur. You can determine where a gene is by recognizing if it is flanked by the same SNP’s over and over again and if it keeps showing the mutated gene. You can identify where the gene is because SNP’s are generally 1kb apart.

Simple Sequence Length Polymorphism (SLLP’s)-repeated short sections of DNA-are of importance because the number of these repeats varies between individuals of a population because of slippage events

1) Minisatellite markers- generally a repeating unit (15-100 bp long)-the whole repetition lasts about 1-5 kb pairs in humans-to recognize the difference you cut with restriction enzymes the flanking regions

of the minisatellite and use gel electrophoresis with a probe.2) Microsatellite- repeating unit (2bp long)

-detect the size differences by using pcr to amplify the target SLLP and plating it using gel electrophoresis using a probe

Making Human Genetic Map1) need Markers that span the genome of unaffected individuals (about 1 marker for

every 3000 centimorgans) 1cm=1Mbp2) location of these markers with respect to eachother must be mapped from normal

individuals3) polymorphisms that span the genome must be chosen (pedigrees from affected

families needed)4) Can pinpoint location of disease genes from markers by using pedigrees and

recombination analysis

4.5- Chi-square test to confirm Linkage-in this case we actually test the hypothesis of no linkage. And if our value rejects the hypothesis then we can infer linkage (null hypothesis)

-look at book to see how to find expected values-df is calculated by taking the amount of variables in a row subtracting by 1 and then multiplying it by the amount of variables in the column minus 1.

4.6-Lod ScoresUsed to estimate distance of a certain gene to a marker when the sample size is very small. Must assume the family is a dihybrid testcross where one hybrid is of the gene of interest and the other is of a molecular heterozygosity. -first find the probabilities of each gamete genotype if independent assortment is assumed. Then you find probabilities of each genotype for certain RF values. -you have to pick out which kids are Recomb and which are non recomb. Then you multiply each probability of each kid by eachother and use the ratio of each specific degree of linkage to the assumption of independent assortment. Whichever lod value is the highest is most likely the accurate answer.

Mapping Function1-(RFx2)=e^-m mx50=map unitsRelationship of Physical and Recombinant MapsThe physical map show’s a genes possible action at the cellular level, whereas the recombination map contains info related to the effect of the gene at the phenotypic level. From pedigree analysis we know of the area where the gene is because our markers flank it, but not exact location. We know some gene functions on the physical map by work with other organisms.1) FIND MARKERS BY RANDOM CLONING OF SMALL DNA PIECES AND SEQUENCING2) SELECT MICROSATELLITES3) FIND THOSE THAT SHOW VARIATION4) MAKE A GENETIC MAP OF VARIABLE MARKERS5) CHOOSE A MARKER SET TO COVER GENOME6) DO PEDIGREE STUDIES ON AFFECTED FAMILIES7) FIND MARKERS THAT FLANK THE DISEASE GENE8) FIND MARKER SEQUENCES IN GENOME9) DISEASE GENE MUST BE IN BETWEEN

Chapter 6-Gene Interaction-Interacting Genes are members of the same pathway or a connected pathway.

-biosynthetic-signal transduction-developmental

6.1-Interactions between alleles of a single Gene: Variations of dominanceComplete dominance-an allele is dominant when only one copy is present and it is expressedDominant mutants:

A) Null mutation: when the null mutation causes the body to not produce enough of a given enzyme to allow the body to function

B) Dominant negative: aka as spoiler, happens when a mutant enzyme inhibits the ability of the wild type enzyme

Incomplete Dominance:When a heterozygote expresses an intermediate phenotype to the homozygous dominant and homozygous recessive genotypes

Codominance:when two separate alleles produce their own unique effect ex) blood types, there are three types of alleles

sub-lethality:-when lethality is expressed only in some but not all homozygous individuals.

6.2- Interaction of Genes in PathwaysArchibald Garrod- Discoverer of inborn error of metabolism-Studied children with black feces and urine.Found that they turned black because of oxidation of homogentisic acid.-induced that tyrosine was involved in the formation of homogentisic acid in patients with alkaptonuria.

-in 1902 he found 9 families with a total of 48 children that had 9 affected children.-40% affection rate

-Also in 60% of the families parents were first cousins-in England only 3% of all marriages were between first cousins

-He postulated that alkaptonuria was an inborn error of metabolism that would convert tyrosine to homogentisic acid through fermentations.

-Like mendel no one gave any thought to Garrod’s studies for 39 years

Beadle and Tatum’s 1941 Paper experiment -rediscovery of Garrod’s studies-*asexual spores produce hyphae

1) mutagenized asexual spores of neurospora and crossed with wild type of opposite mating type

2) grew haploid ascospores on complete media(glucose,malt, yeast extract, inorganic salts) so they will grow no matter what

3) Took some of the ascospores and put them on minimal media4) If it didn’t grow on minimal media take the original ascospores and plate to

the following minimal mediasA) Minimal controlB) Minimal + amino acids

C) Minimal+ vitaminsD) Complete media

5) If it grew on amino acid media then that meant there is a mutation that doesn’t allow the fungi to produce amino acids

6) So they then plate the original ascospores into tubes that each have a different single amino acid

Were able to conclude that biochemical reactions promoted by enzymes in metabolism could be determined by a single gene

15 Mutants were obtained that could not grow without argininePossibilities:

1) All mutants represented the same DNA mutation2) Each mutation occurred in a different gene

Used complementation test to determine what it wasAssume two different strains are each homozygous for a recessive trait that doesn’t grow arginingCross the two strains together and see what happens

-if you cross them and the progeny still shows the same trait then there is no complementation and the mutations must have occurred within the same gene

-if you cross them and the progeny shows wild type then it shows complementation which means the mutations are on different genes

Method for Interpreting Results of Complimentation Test-Put all the genes of interest around a circle and connect the genes that show no complementation (this means that they are in the same complementation group)

-In the beadle and tatum group they found that there were 7 groups that behaved as a single mendelian gene.-members of three groups were analyzed -each group had ornithine, citrrulline, and arginine separately to see if they would grow.

Make these inferences by making dihybrid self cross 9:3:3:1 = no gene interaction9:7= genes in the same pathway

-no matter which wild-type is absent the same pathway fails9:3:4= recessive epistasis (when a double mutant shows phenotype of one mutant but not the other) results from genes being in the same pathway12:3:1=Dominant epistasis9:3:3= synthetic lethal because the double mutant is not viable and dies offSuppressor: a mutant allele of a gene that reverses the effect of a mutation of another gene, resulting in wild type or near wild type phenotype. These organisms that express this mode of suppression are called psuedorevertants

Penetrance: the proportion of individuals with a given allele who exhibit the corresponding phenotype. Has 3 Factors

1) Influence of environment2) Influence of interacting genes3) Subtlety of mutant phenotype

Expressivity: The degree to which a given allele is expressed at the phenotypic level. Measures intensity

Chapter 5- Genetics of Bacteria and Their Viruses-phages= nonliving because they have to parasitize bacteria and use their cell machinery to propogate

5.1- Working with MicroorganismsPrototrophic- wild type bacteria, grow on minimal mediaAuxotrophic-cells that need some sort of nutrient to grow

5.2- Bacterial ConjugationDiscovery: had two different ecoli strains that were both auxotrophs for different nutrients. When plated on minimal media separately they couldn’t grow. When they were mixed however they regained the ability to grow. Some believed that some of the cells would leak out substances that the other cells would absorb. This was ruled out by the U shaped experiment which proved the cells needed to come into contact with eachother.

Discovery of Fertility factor (F+)One cell is always a donor (F+) and donates part of its genome to the recipient(F-)-F+ is a plasmid which replicates in cytoplasm by rolling circle replication which reels out a single stranded copy-the replicated version then moves thru pore into recipient cell.

Hfr strains-created by integration of plasmid F+ into chromosome-when crossed with F- strains none of the F- strains gained the F+ factor, instead the Chromosome of the Hfr stain inserts completely (RCR) into the donor cell and then engages in recombinationLinear Transmission of Hfr-used interrupted mating (put in blender after specific times) and found that certain genes were always found to be transferred into F- strain first (exconjugants). This suggested that the F plasmid always integrated into a specific point on the chromosome (O) so the origin was always transferred into donor cell first and F last.*there were different integration sites however and different strains because during interrupted mating experiments sometimes different genes were transferred at different times

*Hfr single strands however after they have been copied and sent to a donor can only recombine with donor genome by double stranded crossover.

Bacterial RecombinationDoesn’t necessarily take place between two full chromosome sets. One, the F-, is full and called the endogenote, while the other is derived from the Hfr and called the exogenote. The cell is called a partial diploid or merozygote. Since the exogenote usually dies after subsequent cell division only one recombinant product survives.

Chromosome Mapping using Recombination-gotta have shown thru interrupted mating experiments that three different alleles were incorporated into the F- bacteria that had recessive alleles in its own chromosome.-then you have to select for stable exconjugants that picked up the last allele into its chromosome so you can test it by recombination to the other donor alleles.

F’Plasmid:Sometimes the F plasmid can liberate itself from Hfr strain and sometimes it carries with it some of the bacterial genome. Hence F’ plasmid

R plasmidsPlasmids of shigella genus that transfer drug resistant genes to similar bacterium species-sometimes theses resistant genes are contained within transposons of the plasmid, hence these can then be spread and transferred into other cells thru conjugation

5.3- Bacterial Transformation-when bacteria adds fragments of DNA from an external medium. Not a recombination event because the bacteria doesn’t exchange DNA

-Frequency of Double transformants is equal to the product of the single transformants for genes that are far apart. If genes are close the proportion of double will be greater than that of the product of the singles

5.4- Bacteriophage Genetics-contain chromosome covered by protein. Injects its DNA into the bacterial cell which commandeers the cell machinery to produce more phage DNA and Phage protein. Then when that is done the new phages break out of the Bacteria in a process called lysis (virulent phages)-clear plaques are formed when lysis occursMapping phage chromosomesh-r+ x h+r- RF=((h+r+) + (h-2r-rapid))/total plaques

5.5 Transduction-when a phage carries genes from one bacterial cell to another-discovered this because when they used two different auxotroph strains in a youtube experiment they found that something was carrying the correct genes through the filter

and recombining to make a prototroph. These are temperant phages because they don’t kill the host

A) can either integrate into genome of bacteria (Prophage)B) or just be harbored outside the nucleus (lysogenic)

Generalized Transduction-when phage can carry any part of bacterial genome-happens when a phage causes lysis of cell, sometimes it can incorporate some of the broken up pieces of the bacterial cell genome in the phage head.cotransductants – when multiple genes are picked up into phage head and both incorporated into a new bacterial cell genome. Greater percentage of cotransduction means closer genetic markers

Chapter 7- DNA Structure and Replication

7.1- The Genetic Material Discovery of TransformationGRIFFITH: used cap S gene to prove it. S (capsule) strain kills mouse. R (nocapsule) doesn’t kill mouse. Heat killed S strain doesn’t kill mouse. Heat killed S plus R kills mouse AVERY: postulated that DNA was genetic material. Deoxyribonuclease destroyed transforming activity but ribonuclease and protease did notHERSHEY and CHASE: labeled lytic bacteriophage (T2) protein with 35S and DNA with 32P. found that as the phage mixes with host cells only the dna enters the cell. Found that this was true because after the viral chromosome duplicates and produces new phages that exit the cell the new phages only had the 32P still. No 35S

7.2- Structure of DNA1) must allow faithful replication2) must have informational content3) must be able to change

-DNA is made up of linked nucleotides which are made of Phosphate, Deoxyribose and Nitrogenous bases.

Wilkins/Franklin: photographed DNA- nucleotides linked together by 3’,5’ phosphodiester linkages Watson/Crick: found double helix shape held together by hydrogen bonds on opposite bases that run in anti-parallel fashion. Chargaff’s rule: A=T, G=C, A+G (purine)=C+T(pyrimidine)

G-C pairs have 3 hydrogen bonds and A-T have 2.- DNA is not symmetrical in all places there is a major and minor grooves which allow different proteins and enzymes to have different accesses to the DNA

7.3- Semiconservative ReplicationMESELSON: Strand Separation semi-conservative: tagged DNA with 15N. after one generation DNA was HL after 2 generations some was LL and other was HL. Separated by centrifuge found that there was no HH DNA(no conservative). If it was distributive then all generations would be mixed HL

Replication Fork-postulated that during replication there would be replication zippers. Cairns incorporated tritiated thymine into the bacterial cells replication. He then lysed the cell and placed it on photographic emulsion because tritium decays and emits a beta particle that can be detected.

DNA Polymerases-Polymerases added nucleotides to the 3’ end of the growing chain. Synthesis is 5’-3’.-the Substrates for DNA Polymerases are the triphosphate forms of the deoxyribonucleotides.

7.4- Overview of DNA Replication-DNA Pol I was shown not to be main polymerase in replication activity because it moved to slow. DNA Pol III was shown to be the one that was present at most replication forks.-There is a leading and lagging strand. On the lagging strand DNA Pol forms short segments (Okazaki)-DNA Pol’s can extend a chain but cannot start one.Solution: Primosome made up of RNA Polymerase Primase starts the replication then DNA Pol III extends the chain.Completion of DNA synthesis:

1) RNA primer must be removed to make the molecule completely DNA2) The last RNA base is removed by a 5’-3’ exonuclease of DNA Pol 13) The gap is basically a primer template junction so DNA Pol I comes in and closes

the gap but it doesn’t connect the 3’OH to the 5’ phosphate Resulting in Nick 4) so DNA ligase comes in and uses energy of ATP to form phosphodiester bond

7.5- The Replisome: A remarkable Replication MachineReplisome coordinates activities at replication fork.Pol III Holoenzyme- consists of 2 DNA Pol III and 2 B clamps that keep polymerase attached to leading and lagging strands

Unwinding the Double HelixHelicases- unwind the double strandedTopoisomerases- (Gyrase) prevents overwinding of DNA.Single Strand Binding Proteins- stabilize single stranded DNA

Replication Initiation (Prokaryotic)1) DnaA proteins bind to AT rich sequence in oriC and unwind the double strands2) Two Helicases (Dna B) unwind further in 5’-3’ direction

3) DNA Pol III Holoenzyme and primase is recruited to start replication

7.6 Replication in Eukaryotic OrganismsEukaryotic Replisomes are much more complex than Prokaryotic because Eukaryotic chromosomes consist in the cell as Chromatin and the basic unit of chromatin is the Nucleosome (DNA wrapped around histones)*thus the replisome not only has to Replicate the DNA but also disassemble the nucleosome - CAF-1 is a histone chaperone. During replication PCNA is bound to a polymerase to facilitate synthesis of DNA. After Synthesis PCNA unbinds to polymerase but still surrounds DNA. This signals CAF-1 to attach H3/H4 tetramer to start assembling nucleosome onto new DNA strand

Eukaryotic Origins of ReplicationUnlike prokaryotic chromosomes each eukaryotic chromosomes have multiple origins of replication. These origins are also AT rich sequences similar to the oriC of prokaryotic cells

DNA Replication and Yeast Cell cycle1) ORC binds replicator 2) ORC recruits Cdc6 and Cdt1 which are helicase loaders3) Cdc6 and Cdt1 recruit Mcm helicase which pumps DNA thru in leftward manner

to unwind DNA by using energy in ATP bonds*Assembly and activation of Pre-RC depends on cdk’s and ddk’s

-these levels are low in G1 phase but high in S phase-high concentration of kinases activate existing Pre-RC’s to begin replication but

prevent new assembly because they phosphorylate Cdc6 and Cdt1 which triggers proteolysis of these proteins. Activates existing Pre-RC by phosphorylating Mcm 2-7.

-low concentration of kinases activates formation of new Pre-RC’s but inhibits activation of existing complexes

7.7- Telomeres and Telomerase: Replication Termination-On the lagging strand there is an inherent problem with replication at chromosome ends. Once the primer is degraded there is nowhere for the DNA Pol to bind and thus after-Humans have telomere sequence of TTAGGG and the 3’ end of the one strand far outstretches the 5’ end of the other strandTelomerase is a ribonucleoprotein that uses a small RNA molecule as a template for reverse transcription. It extends the 3’OH of the DNA end by reverse transcription. Telomerase translocates to the new end and repeats the process.-now this 3’ end is a center for Okazaki fragment processing. But there are telomere binding proteins so there is always a 3’ overhang

Telomere, cancer and aging-somatic cells have no telomerase activity but germ cells do.

Chapter 8- Transcription and Processing*-In prokaryotes Translation occurs while transcription is still happening. In Eukaryotes Transcription happens in the nucleus but translation happens in the cytoplasm

8.1- RNAPulse-Chase Experiment:Volkin and Astrachan showed that RNA is intermediate

1) Introduced radioactive isotope of uracil (red) so any RNA synthesized in the cell would show up.

2) Then chased out the radioactive uracil with normal uracil and the radioactive uracil is found in the cytoplasm

Properties of RNA1) RNA is single stranded and more flexible so it can form 3-D shapes2) RNA has ribose sugar that has hyrdroxyl group on 2’ carbon as well3) Uracil is in place of Thymine. It can also bind to Guanine4) Some RNA’s are ribozymes which mean they can catalyze biological reactions

Classes of RNA1) mRNA- encodes information that makes polypeptide chains. Are the

intermediary between DNA and protein2) tRNA- molecules that are responsible for bringing correct amino acid to mRNA3) rRNA- major components of ribosomes, which guide assembly of amino acid

chain by mRNA and tRNA. (Most abundant in the cell)4) snRNA- are small parts of ribonucleoprotein spliceosome that removes introns5) miRNA- regulate the amount of protein produced by eukaryotic genes6) siRNA- protect integrity of animal and plant genome. Inhibit production of

viruses and prevent spread of transposable elements

8.2 Transcription -Transcription or adding of ribonucleotides always happens in 5’-3’ manner, so the template strand is always read in 3’-5’.-Only one strand is ever used as the template strand in any given Gene-The nucleotide chain is elongated by RNA pol which also unwinds the DNA and rewinds it after those nucleotides have been transcribed

ProkaryotesInitiation*we refer to genes in the 5’-3’ direction (nontemplate strand)1) RNA pol binds to promoter sequence

-RNA pol Holoenzyme scans for promoter sequence. -Two alpha subunits help assemble the holoenzyme-B subunit is active in catalysis-B’ subunit binds DNA-w(omega) has roles in assembly and gene regulation

-Sigma 70 binds to the -35(TTGACAT) and -10 (TATAAT) sequence and unwinds the DNA2) after core enzyme begins RNA synthesis the sigma 70 unbinds

Elongation-adds to 3’ end of transcript

TerminationTranscription continues beyond protein coding region resulting in 3’ UTRRho-Independent- 3’UTR is GC rich (40 bases) followed by at least 6 U’s.This results in a loop formation and RNA Pol stops processing as it backtracks to stabilize the RNA-DNA hybrid.Rho-Dependent- Termination signal is mainly C rich with less G’s and very little U’s. Upstream from this are is a rut site which binds Rho protein. This protein uses energy from ATP hydrolysis to help the RNA Pol disassociate from the RNA

8.3 Eukaryotic Transcription-More complicated because there are more eukaryotic genes. There is more non-coding regions in eukaryotic genomes. The presence of the Nucleus means transcription and translation are spatially separated.-RNA Pol I used for rRNA -RNA Pol III used to transcribe tRNA-RNA Pol II used for regular genes. First makes pre-mRNA which is spliced to regular mRNA

Initiation:-RNA Pol II and GTF’s (TFIIA,B,D etc) form pre-initiation complex1) TBP which is part of TFIID binds TATA box (-30)2) this recruits the other GTF’s and the RNA Pol II core enzyme3) Carboxyl Terminal Domain of B subunit of RNA Pol is phosphorylated by GTF’s and ends the initiation phase 4) the GTF’s remain at promoter so new RNA Pol II can come and simultaneously transcribe the same gene

Elongation-large subunit of RNA Pol II has sequence of 7 amino acids repeated 52 times

-serines are located at position 2, 5, and 7 of sequence1) phosphorlation of serines unloads GTF’s and loads TFIIS and hsPT5 (xscritpion elongation factors)2) serines are dephosphorylated3) serine 2 is phosphorylated which loads splicing factors4) serine 5 is phosphorylated which loads capping factors

Co transcriptional Processing1) When nascent RNA strand emerges from RNA Pol a cap (7-methylguanosine) is

added to 5’ end to protect from degredation by interaction with CTD

2) When RNA Pol hits termination sequence AAUAAA at 3’ end it disaccociates and adds 150-200 Adenine nucleotides

8.4- Functional RNA’sSmall Nuclear RNA’s (snRNA’s)Spliceosome Splicing (Pre-mRNA mRNA)-happens while RNA is still being transcribed and after the cap has been added.1) -U1 binds to 5’ splice site (requires SR Protein)2) U2 binds branch point (A)3) U4, U5, U6 join in as a complex4) U1, U4, U6 fall off5)U5 brings 5’ splice site into contact with internal A and U26) 3’ splice site is cleaved and exons join together

Self-Splicing-this suggests that RNA is kind of an enzyme since it can catalyze the removal of its own intron

Small interfering RNA’s (siRNA)Injected dsRNA into elegans which was complimentary to a neurological gene.-the ds RNA somehow silenced the gene. One strand was sense RNA (coding) the other was complimentary or antisense1)dicer binds to dsRNA and chops it up into siRNA (short interfering RNA)2) siRNA get taken up by RNA induced silencing complex (RISC) and get broken into single strands3)the RISC and singlestranded RNA find the mRNA of the gene and break down the mRNA causing it to not be able to translate into the required protein.

-Conserved in both plants and animals

CHAPTER 9: Proteins and their SynthesistRNA- Translate the codon of the mRNA into corresponding amino acid which is brought by the tRNA to the ribosomerRNA- major component of ribsome that assemble amino acids to form proteins

-these functional RNA’s are much higher in concentration in the cell than there mRNA counterparts because they are much more stable.

9.1- Protein Structure-amino acid have N-terminus and C-terminus which combine to form peptide bonds.Primary structure: Linear sequence of amino acidsSecondary Structure: Local region of polypeptide chain (alpha helix, B-pleated sheat)Tertiary Structure: Folding of secondary structureQuaternary Structure: combination of 2 or more folded tertiary structuresGlobular Proteins=compactFibrous Proteins= More linear

9.2- Colinearity of Gene and Protein-correspondence between the linear sequence of the gene and that of the polypeptide

9.3- The Genetic CodeProof that codons are three letters longSuppressor Mutations:

1)intragenic: where a mutation at a different point on the same gene can reverse the first mutation(2 missense mutations can restore the structure and function sometimes)

2)intergenic: a mutation on a different gene can reverse the first mutation(change the way the mRNA template is read) changes a specific tRNA-amino acid anticodon to a stop anticodon that can read the mutated stop codon on mRNA.

1) Silence Mutation: where one base is switched but leads to no change in protein sequence. UUU -> UUA because both code for same AA

2) Missense Mutation: causes change in one specific amino acid in sequence3) Nonsense: forms a premature stop codon which produces an incomplete

polypeptide chain4) Frameshift: insertion of a base pair that alters the reading frame5) Deletion: one base pair is deleted which causes shift in reading frame

Cracking the Code-use Polynucleotide Phosphorylase (RNA Synthesizing enzyme) and nucleotides to make synthetic RNA strands. By using specific Nucleotides you can make any sequence you want and observe which type of protein subsequently forms. (in vitro in ecoli)

Stop CodonUAG (amber), UAA(ochre), UGA(opal) are all stop codons. Found UAG stop codon by analyzing mutants of T4 phage that had shorter head proteins. All that was needed to form these mutants was a single mutation that would form a premature UAG codon

9.4-tRNA the adaptorRibosomes

1) a charged tRNA (aminoacyl, amino acid attached to cca-3’) comes into contact with traveling ribosome at A- site

2) the amino acid with tRNA still attached is connected to growing peptide held at P-site

3) the tRNA now without the amino acid is sent to the E-siteAminoacyl Transferases: attach the amino acids to the 2’ OH(class1) or the 3’OH(class2) of the ribose of the tRNA. There are 20 different kinds, 1 for each amino acid

Wobble1)certain charged tRNA’s can bring their specific amino acids to one of several codons because of a loose kind of base pairing at 3’ end of codon and the 5’ end of anticodonIsoaccepting tRNA: has same anticodon as other tRNA’s but differ in nucleotides elsewhere in the molecule. They pair with the same codons as the normal tRNA’s

9.5- Ribosomes-protein synthesis takes place when mRNA and tRNA molecules associate with ribosomes-consists of 2 subunits which each are made up of protein and rRNAProkaryotes: 30s (16sRNA, 21 proteins), 50s(23sRNA,5sRNA, 31 proteins) = 70sEukaryote: 40s(18sRNA, 33proteins), 60s(28sRNA, 5.8sRNA, 5sRNA, 49 proteins)=80sDecoding center-small subunit which makes sure the anticodon and codon matchPeptidyl-transferase center- in the large subunit. where the peptide from the A site gets added to the growing polypeptide in the P site-Both act as ribozymes because they are made up of entirely RNA and not protein

TranslationProkaryotic Initiation

1) 3’ end of 16s rRNA of 30s subunit binds with shine-Delgarno sequence (AGGAGGU) of mRNA to position initiator codon in P site

2) IF3 binds to 30s subunit to keep 50s from associating 3) IF1 and IF2 recognize eachother and make sure only N-formylmethionine

initiator tRNA enters P-site (mRNA, tRNA and 30 s form initiation complex)4) 50s subunit associates with the Initiation complex and releases the initiation

factors

Eukaryotic Initiation1) mRNA is transported to the cytoplasm with many proteins coating it2) eIF4 A,B,G remove the proteins3) eIF4 A,B,G associate with the 5’Cap and with the 40s subunit and initiator tRNA

to form initiation complex4) because eukaryotic mRNA is sometimes base paired with itself the complex

unwinds the mRNA5) AUG codon is found and aligned with tRNA6) 60s subunit attaches7) Initation factors disassociate

Elongation1) EFTu + tRNA-amino acid (ternary complex) enters the A site on the traveling

ribosome. (In eukaryotes, eEF1 is analogous to EFTu.)2) Peptidyl transferase (part of 23S ribosomal RNA) attaches the incoming amino

acid to the polypeptide.EF-G + GTP allows translocation of the incoming charged tRNA from the ribosomal “A” site to the “P” site and the ribosome to the next codon

Termination1) RF1 (class1) recognizes UAA and UAG 2) RF2 (class 1) recognizes UGA and UAA 3) RF3 assists RF1 and RF2

-tripeptides in the RF’s recognize the stop codons and place a water molecule in the peptidyl transferase center which causes disassociation of the polypeptide chain from the tRNA in the P site

9.6-The Proteome-complete set of proteins that can be expressed by genetic material of organism25000 genes but over 100000 proteins in body

Alternative SplicingAllows creation of multiple types of proteins by the same gene

Post-Translational EventsProtein folding- happens after the protein is released from ribosome and it is called the native conformationChaperones (GroE) are a class of proteins that assist in the folding of other proteins

Modification of Protein side chainsKinases-attach phosphate groups to hydroxyl groups of serine, threonine, and tyrosinePhosphates are negatively charge and thus cause a change in protein conformationPhosphatase- removes phosphates

Ubiquitin- adding ubiquitin to amine of lysine residues targets protein for degradation by a protease known as 26S proteasome

Protein TargetingMost proteins have a series of 15 amino acids at the N terminal of their chain that targets them for where they belong in the cell

Chapter 10-Gene Regulation in Bacteria and their Viruses

10.1 Gene Regulation-Regulation in Bacteria is very much the same as it is in eukaryotes-bacterial cells have evolved to shut down transcription of certain enzymes that are not needed at a given time so they don’t expend unnecessary energy.

Genetic Switches in General1)RNA Pol binds to promoter sequencePositive Regulation-If activator binds activator site which is just downstream of promoter sequence then transcription occurs-If activator doesn’t bind then transcription does not occurNegative Regulation-If repressor binds operator(just upstream of promoter) then transcription does not occur-If repressor doesn’t bind then transcription does occur

-these regulatory proteins (repressor, activator) each have two sites: DNA-binding domain, and an allosteric site-Allosteric effector’s bind to allosteric site and turn the regulatory proteins either on or off. In activator’s binding of effector allows activator to bind to DNA. In repressor’s the binding of the allosteric site inhibits repressor from binding to DNA. In both cases it allows for transcription

Lactose Operon: i-CAP-P-O+1-ZYAOperon- DNA that has multiple genes coded into same mRNA molecule and that includes regulatory sites. *I gene is not considered part of operonLac Regulatory Circuit-Z codes for B-galactosidase which breaks down lactose into glucose and galactose-Y codes for permease which transports lactose into the cell-A codes for transacetylase which has nothing to do with metabolism of lactose

Regulatory Components1) I codes for lac repressor protein which blocks this transcription2) Lac promoter site coded by P 3) Lac operator site which is bound by lac repressor protein

*Inducer=effector Lactose and its analogs can bind to lac repressor protein and inhibit it from binding to DNA causing transcription of operon genes

10.2 Discovery of Lac system: Negative Control-used radioactively labeled amino acids to show that the addition of an inducer would provide translation of three different enzymes (ZYA)Evidence for operator and Repressor-used IPTG which acts like lactose but cannot be broken down by B-galactosidase-Found that several different mutations that altered expression of structural genes and wanted to see if they were dominant or recessive. Since bacteria are haploid they needed to insert an F’ plasmid into the cell so it would be diploid for the structural genes.

-Found that the Z+ Y+ wild types were dominantConstitutive Mutations also found two regulatory mutations that caused structural genes to be expressed regardless of whether there was inducer present Oc (cis acting) and I-

I+ is dominant over I- but Is is dominant over I+

-Used DNase to show that the operator sequence was a specific sequence because it would cleave everything not bound to the repressor and the repressor protein covered up the operator sequence protecting it from degredation

Polar Mutations- mutations in one operon gene that show that other genes are not transcribed. Proves that the genes are all part of the same transcript

10.3- Catabolite Repression of the lac Operon: Positive ControlCatabolite= breakdown product-Lactose is really broken down because it makes glucose which the cell uses for energy, so when glucose and lactose are present the cell will not go through energy expensive process of breaking down lactose

1) CAP is activator protein2) cAMP is effector protein3) High glucose doesn’t allow ATP to form into cAMP so it doesn’t bind to CAP

which then cant bind to Activation site and thus transcription does not occur

*CAP binding site and Lac operator are both two fold in symmetry because it allows binding of the regulatory proteins, which are normally composed of multiple identical subunits, to have more sites to interact with the DNA

10.4- Dual Positive and Negative Control: The Arabinose Operon(C-0-Initiatior or activator site-P-BAD)-BAD are functional genes that break down arabinose-C codes for activator protein-arabinose is inducer or effector

-In order for transcription to occur both arabinose-activator protein complex and CAP-cAMP complex must bind to initiator region.-If arabinose is absent then Activator protein (C) binds to both I site and O site and forms a loop inhibiting transcription.

10.5- Metabolic Pathways and Additional Levels of Regulation: Attenuation(O-attenuator region-EDCBA) Repressor gene is elsewhere in the genome-the functional genes are in order of which they act in the metabolic pathway of synthesizing tryptophanattenuation- when regulation acts after transcription initiation has already occurredTryptophan Operon: R-P-O-1/2/3/4-E/D/C/B/A-transcription occurs when tryptophan levels are low1) trp repressor protein binds trp then binds operator sequence which turns off transcription 70 fold2) attenuation 8-10 foldlow levels: ribosomes stall at attenuation region 1 at UGG (trp codons) sites so region 3 snaps back on region 2(palindromic). 4 left free and transcription continueshigh levels: ribosomes don’t stall at region 1, but continues behind RNA Pol. After 3 transcribed it cant snap back on 2 cause of ribosome so 4 snaps back on 3 and causes rho independent termination.

10.6- Lamba Phage: Lytic or LysogenicInt-cIII-N-PL-CI-PRM-OR3-OR2-OR1-PR-Cro-PRE-CII

1) Host is infected by phage. Transcription occurs at PL and PR promoters.

-PL allows transcription of N gene which keeps RNA pol transcribing to the left (CIII) and even allows transcription of genes to the right (CII)-CII encodes activator protein which binds to another leftward promoter (PRE) which allows transcription of CI lamba repressor protein which prevents lytic growth by binding to OR1 and OR2. It also allows transcription of Integrase gene

2) If resources are abundant then CRII is degraded and Cro, which is transcribed from PR, binds to OR3 which blocks transcription of Prm which helps transcribe CI lamba repressor. This causes lytic cycle to propogate

Lysogenycell damage creates DNA s.s. gapsactivation of RecA proteinauto-cleavage of CI proteinlysis

Binding of these regulator proteins bind throught helix turn helix motifs.Their specific affinities to DNA sites has to do with the specific amino acids that are in these recognition helix’s.

10.7-Alternative Sigma Factors Regulate Large Sets of Genomes-Under stress bacterium form spores that are heat and dessication resistant-The bacterium divides asymmetrically with the small forespore being nurtured by the rest of the cell (mother cell) which ends up dying at the end of the process-in normal bacterial cells sigma A and H are active-During sporulation sigma F is activated in the forespore which activates group of 40 genes one of which activtes Pro-sigmaE which is sigma factor in the mother. It also activates Sigmas-K and G.*the activation of these sigmas allows coordinated transcription of different sets of genes, regulons

Chapter 11- Regulation of Gene Expression in Eukaryotes

11.1- Overview of Transcriptional Regulation in Eukaryotes-basic mechanism at work involves molecular signals inside or outside the cell lead to binding of regulatory proteins to DNA-Eukaryotic DNA is packaged into nucleosomes (DNA wrapped in Histones)*ground state of prokaryotes is on, in contrast to Eukaryotic where transcription can only occur if regulatory proteins are present

11.2 Lessons from Yeast: The Gal System7-10-1-2- in between each set of genes is UAS (upstream Activator Sequence)7,10 are transcribed to the left, while 1 and 2 are to the right-GAL1,2,7,10 genes encode enzymes that help breakdown galactose into glucose-GAL4 is the key regulator protein

Gal4 Protein has separable DNA binding domain and Activator Domain-Found this through experiment in which they were able to bind the binding domain to the Gal4 site without the activator domain and transcription was turned off. They made this site upstream from LacZ gene which if transcribed would bind to a certain substrate and produce a fluorescent color. Also showed that the Gal4 activator domain could bind to LexA site and have the same effect.*shows that activator domain helps recruit transcription machinery to the promoter

GAL4 is physiologically regulated-When galactose is low in the cell Gal80 protein binds to activator domain of Gal4 and inhibits transcription-When galactose is high in the cell Gal3 occupies Gal80 and keeps it from binding to Gal4 thus allowing transcription to occur.

-Gal4 activator domain also can directly interact with TBP which is bound to TFIID. This brings RNA Pol closer and promotes transcription. Gal4 also interacts with Mediator complex which in turn interacts with RNA Pol II, thus further enhancing likelihood of transcription.-Since Mediator Complex neither binds to protein nor is part of transcription machinery it is a Coactivator

11.3 Dynamic Chromatin and Eukaryotic Gene RegulationEach histone contains 150bp of DNA wrapped twice around it

Chromatin Remodeling and Gene Activation-Structure of Nucleosome: Octamer of two of each: 2A,2B,3,4. Each of the pieces of the histone has its N-terminal end sticking out from the nucleosome. Histone Tails-Lysines at specific positions on the tails can be molecularly modified to signal for remodeling-Acetylation of lysines signals the histones for remodeling and gene expression-Histone Actetyltransferases are the enzyme that are responsible for this

-ex) GCN5- binds to regulatory sequences on DNA and acetylates histone tails-Histone Deacetylases remove acetyl groups

-Tup1- deacteylase that is recruited by repressor Mig1 so it is a corepressor

11.4- Enhancer Action-Transcriptional levels are finely adjustable due to the clustering of binding sites into enhancers-the binding of many regulatory proteins to many binding sites in enhancer leads to formation of enhanceosome, which is a large protein complex that acts synergistically(greater than additive) to activate transcription.

B-Interferon Enhanceosome-antiviral gene1) transcription factors bind to same face of DNA 100bp upstream from TATA box to form enhanceosome

2) They Recruit GCNG which acetylates Nucleosomes, and then leaves3) This recruits CBP, a coactivator, which also recruits transcriptional machinery. CBP adds more acetyltransferase activity4) RNA Pol II holoenzyme is recruited, and so is SWI-SNF chromatin remodeling complex which moves aside nucleosome to expose TATA box5) TBP binds and allows start of transcription

Control of Yeast Mating type-yeast is haploid and there are two types a and alpha. They mate with eachother and each cell type is determined by the MAT locus

MATa locus-encodes for a1 regulatory protein which only has effect in diploid cells. In haploid cells MCM1 regulates expression of structural genesMATalpha locus-encodes for alpha1 acts in accordance with MCM1 to express alpha specific genes.-alpha2 represses transcription of a-specific genesDiploid Yeast-a1 protein finally has part to play. It binds with alpha2 and binds upstream of haploid specific genes to regulate them.

Enhancer Blocking insulators -Because enhancers are sometimes thousands of bp’s away from the genes which they regulate it could cause the transcription of more nearby genes. Thus insulators have evolved to stop such promiscuous activity. When positioned between an enhancer and promoter it blocks transcription

11.5- Genomic Imprinting-When only one copy of the gene is expressed even when there are two alleles present, thus creating monoallelic inheritance-Paternal imprinting: when only the allele from the mother is expressed (H19)-Maternal Imprinting: When only the allele from the father is expressed (igf2)

The other gene is turned off by methylation of that site-Because the actually DNA sequence is not changed this is an example of epigenetic inheritance

-most imprinting genes are located in clusters as is the H19 and the igf2. Between these two genes is the Imprinting control Region (ICR)

-in the paternal allele these site is methylated and doesn’t allow binding of regulatory protein CTCF. This means the igf2 gene is transcribed and the H19 is not

-in the maternal allele ICR is not methylated and CTCF can bind and transcribe H19 but not igf2

11.6- Chromatin Domains and Their InheritanceYEAST: switching mating types

-There are 3 locations on yeast chromosome with mating type genes-MAT: can be either a or alpha-HML is always alpha-HMR is always a

1) HO endonuclease recognizes sequence adjacent to MAT locus and creates double strand break2) 5’-3’ resection occurs deleting the MAT gene on one of the strands3) if a type was original at MAT then invasion will occur with HML. If alpha was original at MAT then invasion will occur at HMR

PEV-Position effect Variegation- when certain genes are rearranged to become closer to the heterochromatic region and are thus silenced is some cells and consequently in the products of their cell division. This for example led to patches of red and white on fly eyes

Genes necessary for heterochromatin formationSu(Var) genes: when mutated these genes reduce the spread of heterochromatin

-ex) Heterochromatin 1 gene (HP-1) Ex) Histone Methyl Transferase (HMTase)

Barrier Insulators: stop spreading of heterochromatin by binding Histone Acetyltransferases

Dosage Compensation1) X-inactivation: When one of the female sex X chromosome’s has its histones

methylated and also its DNA2) Hypertranscription: happens in drosophilia when MSL complex binds along

entire length of males X chromosome and it has acetyltransferase activity3) Hypotranscription: When both X chromosomes on the female are semi methylated

Chapter 12: Genetic Control of Development

12.1-The Genetic Approach to DevelopmentDrosophila- very fast life cycle, fast development (7 days), and cytogenetics (study of function of cell) and its previous genetic analysis (morgan)-abnormality in body plan of a mutant is identifiable even in larvae stage

12.2- Genetic Toolkit for Drosophila DevelopmentHousekeeping genes: Genes that encode proteins that function in essential processes in all cells of the bodyToolkit Genes: Genes that are concerned with building and development of organs and tissues

Homeotic Genes and Segmental IdentityHomeotic Genes: When one body part is replaced by another normal body part

-Can cause loss of homeotic gene function where it normally acts-can also cause homeotic gene function where it normally doesn’t act

Hox genes: genes that affect identity of segments and their appropriate appendages -Located in two Gene Complexes

1) Bithorax complex- contains 3 genes2) Antennapedia complex- contains 5 genes

-Able to study hox genes and their functions by being able to make cDNA clones of the mRNA transcripts (in situ hybridization) and the protein they encode and using visual technology to monitor them during embryo development after injecting them into fixed embryos

*In absence of Hox genes, segments will form but will all have the same identity

Homeobox-the cluster of hox genes into complexes suggests that they are due to tandem duplication of an ancestral gene. Proven by fact that they all have similar sequences and hybridize to eachother. This region of similarity is dubbed Homeobox. This box encodes a protein domain of 60 Amino acids and is known as the Homeodomain*-Because the Hox proteins have a Helix-turn Helix motif similar to lac repressor and Cro proteins it was found that they were DNA binding proteins-Thus Hox genes bind to regulatory regions of other genes to activate or repress their expression

*Homeodomains are astoundingly similar between species suggesting that they play a fundamental role in development of most animals

12.3- Defining the Entire Toolkit-other toolkit genes are important for the correct development of larvae and embryos-they are also genes that code for transcription factors and thus have effect on succeeding set of genes. Can also affect gene regulation indirectly as signaling pathway proteins.

-maternal required genes-genes with products provided by the female to the egg. Only needs one mutant type from the female to show expression of phenotype in offspring. *-mutant offspring arise form a homozygous mutant mother

Embryonically active1) Maternal Required Genes- Bicoid: effect embryo development

Zygotically acive2) Gap genes: affect formation of continuous blocks of segments3) Pair-Rule genes: act as double segment periodicity. Mutants miss part of each

pair4) Segment Polarity genes: Affect patterning within each segment4b) Hox genes: do not affect number but appearance 5) Genes for segment specific features

12.4-Spatial Regulation of Gene Expression and Development-requires much of the same transcription machinery as physiological control but just more of them.Example)*Combinations of maternal-effect and gap proteins control individual pair-rule stripe formation. Thus the eve stripe 2 element contains multiple sites for maternal bicoid protein and the hunchback, giant, and kruppel gap proteins.

-bicoid and hunchback activate expression over wide range, while kruppel and giant repress expression to just a few cells wide.

Making Segments DifferentHox proteins from Ubx gene and abdominal-a geneUbx is expressed in abdominal segments 1-7 and abdominal-a in 2-7-these proteins bind to distal-less gene to repress expression in the abdomen and promote expression in the thorax. They do this in collaboration with 2 segment-polarity genes known as Slp and En, and two other genes, extradenticle and homothorax

12.5-Post Transcriptional Regulation of Gene Expression in Development-Transcriptional regulation is not only means of restricting expression of gene products. Alternative Splicing is another mechanism-Regulatory sequences in RNA are recognized by splicing factors, mRNA binding proteins, and miRNA’s.

RNA splicing and Sex determinism in DrosophilaDsx (double sex gene) when mutated causes males and females to develop ambiguouslyDsx^m contains a c terminus of 150 amino acids not found in Dsx^f. the Male version represses certain genes in males that the female version activates.-These male and female versions are examples of alternative splicing of the same original transcript by tra gene alternative splicing factor

Regulation of mRNA translation and cell lineages in C. Elegans-simple construction, rapid life cycle and transparency has made it a valuable model.-In blastomere stage mRNA of toolkit genes is present in all cells, but translation of these mRNA’s differ in each cell.glp-1 mRNA is only translated in ABa and ABp anterior cells-the posterior cells have GLD-1 protein which binds to the SCR (spatially controlled region) of the glp-1 mRNA to repress expression of it

miRNA control of developmental timing in C. elegans and other species-Development is temporally ordered as well as spatiallyMutations of heterochronic genes alter the timing of events-the regulatory molecules that arise from these genes (lin-4, let-7) are actually not proteins but RNA products. These RNA products are complimentary to 3’ untranslated regions of developmentally regulated genes such as lin-41. Mutations in this gene cause precocious development from larvae into adult C. elegans

12.6- Many Roles of Individual Toolkit GenesShh gene is expressed in ZPA (zone of polarizing activity) region. Mutations in this gene caused some mice to grow up cyclopic.

12.7- Development and DiseasePolydactyly- development of extra digits on hand or feet due to mutation in the Shh gene.These mutations are not in coding region rather in the regulatory sequences upstream from the coding region.

1) Because they are mutations in cis-acting regulatory elements the phenotypes are dominant

2) Also other gene functions are likely to be completely normal

Holoprosencephaly- mutations in coding region of Shh gene. Since Shh is a ligand, this disease is not exclusively caused by mutation to Shh gene.

Chapter 13- Genome and Genomics

13.1 The Genomics RevolutionThe construction of genomes of species can help aid in identifying disease causing genes-need PCR and bacterial plasmids to clone the genes of interest

The Shotgun Method:Logic is sequence first, map later

1) Cut genome into random fragments2) Make library of cloned fragments3) sequence reads of clones by using primers4) map the sequence reads by finding homologous overlapping segments from

different clones to form contigs5) Use paired end reads. If one read is from a certain contig and the other is from a

second contig then this insert spans the gap and forms a scaffold

Ordered Clone MethodLogic is map first, sequence later

1) Use restriction enzymes and order large insert clones by overlap fingerprints to create a physical map

2) Select clones with minimal overlap3) Divide into subclones by further treating with restriction endonucleases4) Sequence subclones5) Assemble subclones to create genome sequence

13.3 Bioinformatics: Meaning from Genome Sequencing-study of information hidden in the genomeInformation includes:

1) Sequences that encode proteins2) Sequences that encode RNA3) Sequences that are binding sites that govern time and place of their actions

How to deduce Proteome1) ORF’s- there are 3 possible reading frames on each strand so there are 6 possible

reading frames all together. Computer scans area where DNA would produce a mRNA where there is 5’ start codon and 3’ stop codon.

2) cDNA- create cDNA clones that are complimentary to mRNA. These cDNA will only be made of exons and they can anneal to genomic DNA and will show only where exons are and from this you can deduce where the introns are

3) EST’s: Expressed Sequence Tags are cDNA’s that are only made of the 5’ and 3’ ends of transcripts and they can thus be used to find where the boundaries of transcripts are

4) BLAST: BLASTn are commonly known sequences that code for certain proteins and computers use this knowledge to look for areas in genomic DNA where there is similarity to previously known sets of Data. BLASTp is the same thing but is amino acid sequence instead.

5) Codon Bias

13.4-Human Structure of the Genome-Transposable elements and repetitive sequences make up 45% of genomic DNA-only 3% of DNA are exons and only half of those encode for polypeptides because introns are much larger than exons-3 splice variants per gene on average thus there are 3 fold greater proteins than genesPseudogenes: genes that have acquired mutations and activity that make them inactive.

13.5 Comparative Genomics-Genes are conserved over long periods of time because mutations to genes that decrease fitness are selected outHomologs- closely related genes

-orthologs- homologs at same genetic locus inherited from common ancestor-paralogs- homologs that are at different sites due to gene duplication

Synteny- where the order of genes within variously sized blocks is the same between two species

Comparative genomics with Chimpanzee-35 million single nucleotide differences (1.06%)-5 million insertions or deletions from 1b-15kb (3% difference) lie outside coding region-29% of orthologous proteins are completely identical-80 genes present in ancestor have lost function in humans-170 duplicated genes in humans and 90 in chimps

13.6 Functional Genomics and Reverse GeneticsTranscriptome- sequence and expression pattern of all transcripts

Proteome- sequence and expression pattern of all proteinsInteractome- interactions between proteins and RNA, proteins and DNA, and between proteins

Microarrays to study Transcriptome-microarrays are chips of several DNA sequences. It is exposed to a sample of labeled RNA from the cell. RNA complimentary to certain DNA sequences will hybridize and indicate which genes are actively transcribed under given conditions

Two-Hybrid test to study interactome1) Place Gal-4 binding domain gene next to DNA of protein of interest into one

yeast vector (bait)2) Place Gal-4 activation domain next to another DNA of protein of interest into

another yeast vector (target)3) Finally look for activation of transcription of Gal-4 regulated reporter gene (lacZ)

Chromatin Immunoprecipitation Assay to study interactome1) Cross link proteins to DNA2) Break chromatin into small pieces3) Add antibody to target protein and purify4) Reverse cross links to separate protein and DNA

Reverse Genetics-start with observed normal DNA, mRNA, or protein and induce changes into it and see effects that it has

Phenocopying1) RNA interference: naturally protects cells from foreign DNA. dsRNA is made

(introduce reverse repeat into genome) with sequences homologous to gene of interest. The RISC then degrades any mRNA that is complimentary to the dsRNA

2) Chemical Genetics:A) Using Forward Genetics: add one compound per well to yeast colonies

-find compound that produces phenotype of interest-Identify protein target of compound

B) Using Reverse Genetics: Find Protein of interst - screen for compounds that bind to protein - treat cells with molecule that binds to protein

- assay for phenotype

Chapter 14- Transposable Elements-can insert into genes and inactivate them-account for 50% of our genomic material

14.1 Discovery of Transposable Elements-Barbara McClintock discovered that in Maize chromosome 9 would very frequently break at a specific locus. Required 2 Factors:

1) Ds for disassociation was located at the site of the break2) Ac for activation was needed to activate the break but was located at a different

site. Also activates the transposition of Ds.-postulated that they were mobile elements when she found that Ac of different plants were located to different locationsUnstable phenotype- resulted from movement of Ds away from the C gene

Autonomous Elements- elements like Ac that do not require other elements for their mobilityNonautonomous elements- elements like Ds that do require other elements for their mobility

14.2 Transposable Elements in ProkaryotesIS elements (insertion sequence elements)

- All IS elements encode protein transposase- All begin and end with short inverted repeats

Transposons1) Composite: contain variety of genes that reside between two identical IS elements

that are oriented in opposite directions. The IS elements cannot transpose on their own without the genes inbetween because of a mutation in their inverted repeats

2) Simple: flanked by IR sequences that do not code for transposase. Instead transposase gene is included in the group of genes that are flanked by the IR.

Mechanism of Transposition:1) Replicative: when a new copy of the transposable element is generated by

transposition event. This events produces a cointegrate which is the fusion together of 2 plasmids before they get separated. This allows the replication of the Transposable element

2) Conservative: Cut and paste method, where the transposable element is removed from one place in the genome and placed in a new position.

14.3 Transposable Elements in EukaryotesClass 1: Retrotransposons-Studying HIS4 mutants in yeast. Found that 2 of the mutants of HIS4 were unstable phenotypes that were more than 1000x likely to revert back to wild-type than were the other mutants. Found that the cause was a large insertion into the HIS4 gene. It did not look like IS elements however instead looked like animal retrovirus.Provirus: double stranded DNA copy of retroviral genomeContains:

1) Gag gene: involved in maturation of viral genome2) Pol gene: encodes the reverse transcriptase gene

3) Env gene: encodes structural protein that surrounds viral Genome4) Flanked by LTR’s or long terminal repeats

Similar to Ty elements but Ty elements don’t have Env gene so they cannot leave the cell, rather they just insert into different parts of the genome within the same cell.

Ty elements transpose through RNA intermediate:Placed galactose inducible promoter near Ty element and also inserted an intron into its coding region. Found that the new site where the Ty element was inserted lacked the intron meaning that it had been spliced from the pre-mRNA transcript of the original Ty element.

Ex of class I element: Copia like elements of Drosophila

Class 2: DNA transposons-similar to mechanisms of prokaryotic transposition

P elements in DrosophilaContains transposase gene with 4 exons and 3 introns flanked by short IR’s.

Female M x Male P causes hybrid dysgenesisReciprocal cross produces viable offspring-this happens because the female cytoplasm (mitochondria) contains a repressor that represses transcription of the transposase of the p element.

Using P Elements to insert GenesInsert a gene into P element that has had transposase gene deleted but still has the IR sequences that bind transposase. Add this plasmid along with a normal P element plasmid into multinucleated embryo. The developing fly will not show the phenotype you are looking for but its offspring will because the P-elements mobilize only in germ cells.

14.4 The Dynamic Genome: More Transposable Elements than imaginedC-Value Paradox- when the size of the genome does not correspond to the complexity of the organism

Transposable Elements in the Human Genome1) LINEs (autonomous)- Long interspersed Elements: similar to retrotransposons

because they encode reverse transcriptase but do not include flanking LTR’s2) SINEs (nonautonomous)- short interspersed elements that do not include gene for

reverse transcriptase. Instead are mobile because of reverse transcriptase of LINEs elsewhere in the genome. Ex) alu

3) DNA transposons

Insert into Introns and Exons-those inserted into introns have no negative effect because they are spliced out of the mRNA product

-those inserted into exons are subject to negative selection

Chapter 15: Mutation Repair and RecombinationMutation and Recombination are two major processes that are responsible for genetic variation.

15.1- Phenotypic Consequences of DNA MutationsPoint Mutations: mutation to a single base pair or small number of adjacent base pairs

1) Base SubstitutionsA) Transition: replacement of base by another base of the same

categoryB) Transversion: replacement of base by another base of a

different category2) Insertion/ deletion

Synonymous: mutation changes one codon for an amino acid into another codon that codes for the same amino acid (silent)Nonsynonymous: condon for one amino acid is changed into codon for another amino acid (missense)Nonsense: condon for an amino acid is changed into a stop codon

Conservative: mutation that encodes for a different amino acid of similar chemical structureNonconservative: mutation that encodes for a different amino acid not of a similar chemical structure

15.2 Molecular Basis of Spontaneous MutationsFluctuation TestUsed to see if mutations to make bacteria resistant to phage attack were induced by the phage or were spontaneous-if spontaneous than each culture should show various number of colonies that are resistantUsed the idea of replica-plating

Mechanism of Spontaneous Mutation1) Errors in DNA Replication

Transitions: can occur because tautomers of nitrogenous bases can occur. Normally they are present in the cell as ketonesCytosine and Adenine can tautomerize into imino’s (2 H bonds)Guanine and thymine can tautomerize into enol’s (3 H Bonds)Transversions: occur when purine-purine pairs form or pyrimidine-pyrimidine pairs formFramshifts-caused by slip mispairing during DNA synthesis

2) Spontaneous Lesions

Depurination- the loss of a purine base due to the interruption of the glycosidic linkage between the base and the deoxyriboseDeamination causes cytosine to become uracil and thus a transition to occur-can also cause 5-methylcytosine to become thymine which is hard to correct

Tri Nucleotide Repeat DiseasesEx) Fragile X syndrome: normal CGG repeats in the FMR-1 gene contain 6-54 repeats but diseased contain hundreds of repeat caused by slipped mispairing

15.3 The Molecular Basis of Induced Mutations1) incorporation of base analogs-5-Bromo Uracil, which is analog of thymine, when it changes to ionized form pairs with guanine-2-amino purine, which is analog of adenine still pairs with thymine, but when it is protonated it will pair with cystine

2)Specific Mispairing- altering a base so that it forms a specific mispair-EMS and NG add alkyl groups to many positions on all 4 bases-most commonly it adds ethyl to Guanine forming, O-6-Ethylguanine which pairs with thymine instead of cytosine-also adds ethyl to thymine forming O-4-ethylthymine which pairs with guanine

3)intercalating agents-Proflavin, Acridine Orange, ICR compounds, are planar molecules that can slip between stacked nitrogen bases causing insertion or deletion of a single nucleotide pair.

4)Base DamageA) UV light causes adjacent pyrimidines to form a cyclobutyl ring, making specific base pairing impossible and thus blocks replication at this pointB)Ionizing radiation can cause water molecules to form radicals, or hydroxide ions, or hydrogen peroxide, which can lead to degredation products of bases causing mutations-can also directly cause breakage of N-glycosidic linkage sites and cause apurinic or apyrimidinic sites

5)Bulky addition products-such as aflatoxin, bind covalently to DNA leads to breakage of bond between sugar and base

AMES Test-Ames found that not all carcinogens directly cause mutations, rather that their metabolite products directly caused mutations.1) broke down rat liver to extract enzymes. 2) added them to S. typhimurium strains that needed histidine to grow. Plated them on plates lacking histidine.3) added enzymes and potential carcinogens to the plate.

-if the bacterium would form colonies knew that mutations were formed causing revertants and thus because mutagenicity and carcinogenicity are correlated could postulate that certain compounds were likely carcinogens

15.4 Biological Repair Mechanism Error Free Repair:Direct Reversal of Damaged DNA:-CPD Photolyase, with the help of UV light, can split a cyclobutyl photodimer back into its original bases-alkyltransferases can remove certain alkyl groups that have been added to bases by DMS and NG. it does this by transferring the methyl group to the cysteine residue in its active site.

Base Excision Repair: targets non-bulky damage to basesMethylation, deamination, oxidation, loss of DNA base

1) Enzyme Uracil Glycosylase cleaves at glycosidic bond btw base and sugar2) AP Endonuclease comes and cleaves abasic site on both ends of backbone3) Deoxyribophosphodiesterase then removes a part of the backbone of the DNA4) DNA Pol acts on 3’OH side and fills in the gap and ligase seals the nick

Nucleotide Excision Repair- corrects bulky adducts that distort the DNA helixA) Global Genomic Repair- activated by stalled replication forks

1) Recognition complex recognizes photodimer2) Attracts TFIIH3) XPD, XPB, subunits of TFIIH are helicases that unwind DNA4) RPA, single stranded binding protein, stabilizes the structure5) The DNA is excised 15-24 bases upstream and 2 bases downstream6) Bypass polymerase fills in the gap with help from PCNA, ligase seals the nick

B) TCNER- activated by stalled transcription1) CSA, CSB recognize stalled polymerase

-all other steps are the same TFIIH comes in, CSA, CSB and RNA pol leave

Mismatch Repair- Post Replication Repair-error rate in replication is 10^-5. Proofreading reduces this to 10^-7, and mismatch repair further reduces it to 10^-9A) MutS scans hemimethylated DNA and detects a bulge in structure B)MutL is recruited which in turn activates MutH endonuclease which cuts a nick in the strand that contains an adenine methylation siteC) UvrD binds at the nick and unwinds using helicase activityD) single strand binding proteins bind to stabilizeE) An exonuclease then comes in and removes the strand containing the mismatchF) DNA Pol III fills in the gap and ligase seals it

Error Prone RepairDNA Translesion Synthesis

1) DNA Pol is synthesizing new strand then comes across a thymine dimer on template

2) It cant synthesize past this so a Rad6 adds Ubiquitin to PCNA causing its conformation to change

3) Bypass DNA Pol comes in and clamps to PCNA. through evolution has learned to just put down A’s

4) After putting on just two bases it leaves and DNA Pol comes back to finish because Rad 18 removes ubiquitin

5) The thymine dimer that was on the template strand is then fixed by nucleotide excision repair

Repair of Double Strand BreaksNon Homologous End Joining: NHEJ-predominant form of repair of dsDNA breaks in multicellular organisms-occurs if replication has not occurred. It can happen because there are many repetitive non expressive sequences in a multicellular organisms genome1) Ku70/80 recognizes ds breaks and binds the broken ends2) Proteins are recruited that trim the strands back3) DNA ligase IV, XRCC4 comes in and connects the two strands

Homologous Recombination- Synthesis-dependent strand annealing (SDSA)Similar to homologous recombination during meiosis. However this repair mechanism occurs between sister chromatids and is error free. Dmc1 is not involved in this process-replicated by conservative mechanism creating single strands 1) Enzymes trim back 5’ ends 2) resulting 3’ overhangs are coated with Rad 513) Rad 51 DNA filament then begins search for sister chromatids to fill in the gaps

15.5 Mechanism of Homologous Recombination-recombination has to occur at meiosis I or else the homologs will not align properly leading to progeny cells getting either too many or too few chromosomes. Don’t develop properly. Called NONDISJUNCTION1) Spo11 contains a tyrosine group which attacks the DNA backbone and covalently bonds to the 5’ Phosphate. Unlike topoisomerase it doesn’t RELIGATE2) MRX complex (analogous to RecBCD) processes the break and forms 3’ overhangs3) Rad51 and Dmc1 (homologs to RecA) coat 3’ overhangs and promote strand exchange btw sister chromatids. Rad51 involved in meiosis and mitosis while Dmc1 is only involved in meiosis

2 important factors1) There are more fewer chiasmata formed than there are breaks by spo112) Meiotic crossover has several different genetic outcomes

15.6- Cancer: an important Phenotypic Consequence of Mutation

Mutations in Cancer Cells

1) Mutations in onco genes: gain of function dominant mutations. Mutation only needs to be present in one allele. Normal gene is a proto-onco gene

2) Mutations in tumor suppressor genes: loss of function recessive mutations. Causes encoded gene to lose its function

Ras Onco Protein: Single base pair substitution converts glycine into valine at AA number 12. Normal Ras protein cycles between active GTP bound state and inactive GDP bound state. The missense mutation produces an oncoprotein that always binds GTP, that constantly promotes cell proliferation

P53 tumor suppressor gene: Undamaged gene prevents the cell cycle until the DNA is repaired, and it also induces apoptosis.

Chapter 16- Large Scale Chromosomal Changes1) Changes in chromosome number2) Changes in chromosome structure

16.1 Changes in Chromosome NumberAbberant Euploidy- the changes in whole chromosome setsEuploidy- is multiples of the basic set of chromosomes

Monoploid- when a normally diploid organism only has one set of chromosomes (different from haploid)

-includes male bees,waspsPolyploids-Autopolyploids: multiple chromosome sets originating from same species

-autotetraploids can be induced by treating diploids with colchicine-allopolyploids: sets from different species. Homeologous because they are partly homologous

-can be induced by breeding 2 different diploid plant species to produce an amphidiploid. The sterile F1 is then selfed to produce the amphidiploids

Agricultural Applications 1) Monoploids can be derived from anthers of plants. Useful if a desired trait is

normally a recessive allele in a diploid. Process involves cold treatment of a cell destined to become a pollen and it grows into an embryoid

2) Autotetraploids are made because they have increased size

Aneuploidy-changes in parts of chromosome setsTrisomic- 2n+1 Monosomic-2n-1Nullisomic-2n-2Disomic-n+1 (for haploids)-cause of aneuploidy is nondisjunction, which is abnormal segregation-normal disjunction is thought to be related to the presence of crossovers (chiasma)

Turner Syndrome: Monosomic for the X chromosome, represented as XO-sterile femalesKlinefelter Syndrome: XXY trisomic males who have low IQ and are sterileDown Syndrome: Trisomy at chromosome 21-females are more likely to be fertile than males. Incidence is related to maternal age-Pautau syndrome (13) and Edwards Syndrome (18) also occurXXX: trisomy which results in phenotypically normal, fertile females

Gene Balance-Abberant Euploidy leads to increase in size of organism but proportions and shape remains the same-Abberant Aneuploidy tends to lead to malformation in size and shape-normal physiology depends on the proper ratio of gene products and aneuploidy in humans can result in haplo-abnormal or triplo-abnormal*the reason monosomy is more severe than trisomy is because any deleterious recessive alleles present on a monosomic autosome are automatically expressed

16.2 Changes in Chromosome StructureUnbalanced:1) lost chromosome segment2) duplicated chromosome segmentBalanced:3) inversion of segment4) translocation of segment

-can either occur by:A) chromosome breakageB) crossover between repetitive DNA regions (Non-allelic homologous Recombination)

-Chromosome rearrangements have to result in chromosomes with a centromere and 2 telomers-chromosomes without centromeres (acentric) will not be dragged to either pole at anaphase-chromosomes lacking a telomere will not replicate properly

Deletions:Intragenic deletions: small in size and are equivalent to null mutations. Never revert back to wild-typeMultigenic Deletions also occur and are more severePseudodominance – when recessive alleles show dominance because their counterpart allele on the other homolog has been deleted-ex) Williams syndrome, Cri du Chat syndrome-animal sperm are functional with deletions, while pollen normally are not

Duplications-tandem duplication- when the region is inserted next to itself

-insertional duplication- when region is inserted elsewhere in the genome

InversionsParacentric: centromere is outside the inversion-results in formation of dicentric bridge and produces an acentric fragmentPericentric: centromere is within the inversion-does not result in creation of a bridge but results in same products

Reciprocal TranslocationsWhen two segments trade trade acentric fragments-create two types of segregations1) Adjacent 1: (T1+N2) and (N1+T2)2) Alternate: (N1+N2) and (T1+T2) both complete and viable-these two are equal in number and thus half the gametes are viable and half are not (Semi-Sterility)Robertsonian Translocation-when the translocation produces an almost complete extra copy of the chromosome(Down Syndrome)

Chapter 17- Population Genetics

17.1 Variation and its ModulationAllele Frequencyp+q= fA/A+ fa/a +fA/a=1 q=1-p

Protein PolymorphismsImmunological-there are n(n-1)/2 different ways of combining n different things two at a timeAmino Acid Sequence-use gel electrophoresis to plate differences in protein structure

DNA Structure and Sequence PolymorphismChromosomal-extra chromosomes, inversions, reciprocal translocations are observed in many populationsRestriction site Variation-causes Restriction Fragment Length Polymorphisms (RFLP’s)Tandem Repeats-A person or population could be polymorphic for Variable Number Tandem Repeats (VNTR’s)Complete Sequence Variation-variation in single nucleotide (single nucleotide Polymorphisms)1) -sequence of coding regions can be translated to reveal exact amino acid sequence differences

-this is superior to electrophoresis studies of proteins because electrophoresis studies can only show how many or which AA’s are different but not what the difference is.2) sequence variation is studied in sequences that don’t change protein sequence, such as regulatory sequences, introns etc. this is important because it is believed that most evolutionary variation is due to variations in these sequences-variation in intron sequence (silent mutations) are much more common because variations in amino acid sequence are commonly selected out

17.2- Effect of Sexual Reproduction on VariationHardy-Weinberg Equilibrium:p^2+2pq+q^2=1

Heterozygosity-can be used as measure of genetic variationHaplotype-combination of alleles of different genes on the same chromosomal homolog

Inbreeding and Assortative MatingInbreeding- when mating btw relatives is more common than by chanceEnforced outbreeding- when mating between relatives is less common than would occur by chancePositive assortative mating- mating of like with likeNegative Assortative mating- mating with unlike partners

17.3 Sources of VariationThree sources of Variation

1) mutation2) Recombination3) Immigration of genes4) Genetic Drift

-Increase in mutant frequency= nonmutant frequency x mutation rate

Variation from migrationM=deltap/P-po P=allele frequency of donor po=original freq of recipients

17.4 SelectionDarwinian Fitness: Probability of survival and rate of reproduction of a phenotype or genotype

Two types of selectionFrequency independent: the fitness of a genotype does not depend on the how rare or frequent it is in the populationFrequency dependent: fitness of a type changes as it becomes more or less frequent

Viability: probability of survival

17.5-Balanced Polymorphismoverdominance: when a heterozygote is more fit than either homozygote

-selection favors an allele when it is rareunderdominance: when a heterozygote is less fit than either homozygote

-selection favors an allele when it is common, not rare

Genetic Drift: Random change in allele frequenciesFounder effect: form of genetic drift when a single generation of sampling of a small number of colonizers from the original large population

Chapter 19-Evolutionary Genetics-mechanism of evolution starts with the variation that exists between organisms of a specific specie. Thus there is a different rate of survival between individuals

19.1- Darwinian Revolution1) principle of variation- variation in morphology, physiology, behavior2) principle of heredity- offspring resemble parents more than other individuals3) Principle of selection- some variants are more successful at surviving than others

Explains two things:Phyletic evolution- successive change of form and function of single continuous lineDiversification- there have existed many different contemporaneous species having different forms and living in different ways

-these are the consequences of heritable evolution

19.2- A Synthesis of Forces: Variation and Divergence of PopulationsExample of the finches on the Galapagos islands is explained by diversity and adaptation

Variation within populations: caused by mutation, migration, balancingVariation between populations: caused by inbreeding, genetic drift, incompatibility, directional selection (basically things that cause homozygosity)

Population will retain variation in alleles and not differentiate from local inbreeding if:m>1/N or u>1/N

19.3-Multiple Adaptive Peaks-arise if intermediate genotypes have higher fitness (Ab/Ab, aB/aB, AB/ab)-not every difference is an adaptive difference (African Rhino’s have 2 horns, Indian have 1)

Exploration of Adaptive PeaksRandom drifts cause a population under selection to ascend an adaptive peak erratically.

-random drift can enhance selection and also counteract it.-there are always unfavorable alleles at certain loci that are fixed because selection intensity is insufficient to overcome random drift to fixation.

-when there are 5 different changes and they can happen in any order there are: 5x4x3x2x1= 120 possible orders.Sign epistasis: when fitness advantage/disadvantage depends on the mutations previously fixed

19.4-Genetic VariationHeritability of Variation-in general behavioral traits have lower heritability’s than morphological traitsCanalized Characters: when there is substantial genetic variation but very little morphological variation

-genetic differences are revealed through phenotypes when you place the variants in stressful environments

Variation within and between populations-in general different human populations show similar allele frequencies for polymorphic genes

19.5 Mutation and Molecular Evolution1/3 of all protein encoding loci are polymorphic-this DNA variation has three effects:1) they may be deleterious, reducing probability of survival 2) they may increase fitness by increasing efficieny3) they may have no effect on fitness (neutral mutations)

-include mutations that have no effect, and those that have a fitness selection that is lower than the reciprocal of population size

Purifying Selection on DNA-if you plot number of nucleotide differences btw 2 species that diverged from common ancestor then against time since divergence then their slopes should be mu or mutation rate.-we see that synonymous substitutions have much higher mutation rate than do non-synonymous substitutions

-is expected because non-synonymous have deleterious effect and are selected out(purifying selection)

19.6- Relating Genetic to Functional Change: Protein EvolutionSignature of Positive Selection on DNA sequences-comparing synonomous and non-syn mutations within a species against synonomous and non-syn mutations between species can detect adaptive evolution of proteins

-if species differences exceeds the number of polymorphisms than there has been adaptive selection

Morphological EvolutionMelanism is an example-eumelanin forms black and brown pigments phaeomelanin forms yellow or red -amounts of these two types of pigments are controlled by two genes MC1R and agouti protein

-Agouti is an antagonist and binds to MC1R to inhibit production of eumelanin-dark mice have mutations in MC1R region that makes it constitutively active bypassing regulation by agouti protein

Gene InactivationExample: different cave populations of fish separately evolved to lose pigmentation and eyes in underground caves.-found to have mutations in the Oca2 gene.

What might account for repeated inactivation of Oca2 gene?1) Mutations appear to cause no serious defects compared to other pigmentation

genes2) Oca2 region is really large and is more susceptible to mutation

19.7- Regulatory EvolutionThere are many animals that have special coloring patterns. This means that pigmentation-gene regulation must evolve by some mechanism that does not disrupt pigmentation-protein function-Mutations in regulatory sequences provide a mechanism for changing one aspect of gene expression while preserving the role of pleiotropic proteins in other developmental processes.

19.8-Origin of New Genes-Old DNA from old genes must be preserved while new genes with new functions are evolving. Where do new genes come from?

1) Polyploidy- frequent polyploidy2) Duplications: has three consequences

a. Polypeptide may simply increase in productionb. General function of sequence is maintainedc. New segment may diverge dramatically and take on new function

3) Imported DNA-when genes from totally unrelated species are incorporated into host genome

a. Cellular organelles have been obtained in this way (mitochondria, chloroplasts. Evidence for their extracellular origin is based on experiments that show that the DNA-RNA code is different in mitochondria than it is for nuclear genes.

b. Horizontal Transfer- Within a genome DNA can be transferred by transposable elements. Can be transferred between genes by retroviruses

19.9- Genetic evidence of common Ancestry in Evolution-we can observe common evolutionary origin through:a) anatomical differencesb) genome and protein differences-If there was no selection divergence within genes would result in loss of similarity and inability to decipher if the two organisms are related. However mutation rates are generally not high enough to cause loss of similarity and most new mutations cause deleterious effects and are selected against.

19.10 Process of SpeciationSpecies-when a group of one species cannot exchange genes with another organismMost new species form as a result of geographical isolation.

-a single migrant between populations is sufficient to prevent populatins from fixing at alternative alleles by genetic drift alone

-selection toward different adaptive peaks will not succeed in causing complete divergence unless it is extremely strong

In order to for populations to diverge enough there has to be a strong enough barrier to stop migration from happening. Such spatially separated populations are allopatric-enough separation will lead to two different enough species that even if they were to come into contact and mate it would cause severe physiological, developmental and behavioral problems. They become biologically isolated.

Two biologically isolating mechanisms:1) Prezygotic Isolation-occurs when there is failure to form zygotes2) Postzygotic isolation- results from failure of fertilized zygotes to contribute

gametes to future generations (common in animals)