Genetics & Molecular Biology K322 Exam 2

Q9: A novel organism has a genome complexed with three histone-like proteins. Nuclease digestion give bands of 700 bp, 1400 bp and 2100 bp. What is the likely explanation?

700 bp of DNA wrap around each histone **Could be a and b possibly but a is more likely b) The histones are different sizes and bind different lengths of dna

Plasmids sometimes integrate into the genome

9.12 The F factor is integrated into the bacterial chromosome in an Hfr cell. F+ cell - bacterial chromosome - f factor --> crossing over takes place between F factor and chromosome --> Hfr cell The F factor is integrated into the chromosome **Crossing over = recombination **Analogous **Once integrates, becomes HFR cell

Integrated plasmids sometimes pop back out and can take genomic DNA with them

9.14 An Hfr cell may be converted into an F' cell when the F factor excises from the bacterial chromosome and carries bacterial genes with it. Hfr cell - lac - bacterial chromosome with integrated F factor --> crossing over takes place within the Hfr chromosome --> F' cell -lac - bacterial chromosome when the F factor excises from the bacterial chromosome, it may carry some bacterial genes (in this case, lac) with it --> F' cell and F- cell --> during conjugation, the F factor with the lac gene is transferred to the F- cell --> producing a partial diploid with two copies of the lac gene **Conjugation produces a partial diploid **Plasmid removes itself from host genome **Red is integrative plasmid **Undergoes reverse recombination- f prime + cell and goes recombination with another cell **Recipient cells receives lack gene **You can have a cell that has a gene and shuffles over to another

5.) Describe lytic and lysogenic bacteriophage lifecycles

9.20 Bacteriophages have two alternative life cycles: lytic and lysogenic. 1. Host DNA - attached phage with phage DNA LYSOGENIC CYCLE: 2. Phage DNA enters the host cell 3. The phage DNA integrates into the bacterial chromosome and becomes PROPHAGE 4. The prophage is replicated as part of the bacterial chromosome. This replication can continue through many cell divisions 5. The prophage may separate from the chromosome and the cell will enter the Lytic cycle. LYTIC CYCLE: 3. The host DNA is digested 4. phage DNA replicates 5. The host cell transcribes and translates the phage DNA, producing phage proteins 6. Assembly of new phages is complete. A phage-encoded enzyme causes the cell to lyse 7. New phages are released to start the cycle again **lysogenic **Enters host cell and integrates bacterial genome - prophage **Good way to replicate a genome

Experimental Results

Avery, MacLeod, and McCarty used heat-killed S bacteria and live R bacteria and infected mice The extract of heat-killed S bacteria was divided into aliquots and treated to destroy either DNA, RNA, proteins, or lipids and polysaccharides All aliquots killed the mice except the one with the DNA destroyed - DNA is inheritable CONCLUSION: transformation is not disrupted by the removal of lipids, polysaccarides, proteins, or RNA; therefore, none of these is the transformation factor. CONCLUSION: DNA is the hereditary molecule required for transformation

Why duplications and deletions affect phenotype

DUPLICATIONS increase gene dosage - Why increased dosage is a problem is not always well understood Effects of DELETIONS - Imbalances in gene product due to haploinsufficiency - Expression of a normally recessive gene (pseudodominance) **Being haploid isn't enough to make the cell behave normally **Duplicating or deleting changes protein complexes which affects how the organism looks - small head, hand abnormalities, intellectual disability

DNA Repair Mechanisms

Damaging agent: UV-light DNA lesion: 6-4 photoproduct Repair pathway: nucleotide excision repair (NER) Damaging agent: X-rays DNA lesion: Abasic site (SSB) Repair pathway: base excision repair (BER) Damaging agent: ionizing raditation DNA lesion: DNA double-strand break (DSB) Repair pathway: double-strand break repair (DSBR) Damaging agent: replication and recombination errors DNA lesion: base mismatch Repair pathway: mismatch repair (MMR) **For each kind there is a cellular pathway --> they all fix one of the two strands of a double helix **Excision repair!!!!

2.) How to assemble a strand of DNA from dNTPs --> Base pairing to add the second strand --> Antiparallel orientation of the strands

Deoxynucleotide triphosphates (dNTPs) are assembled into DNA - nucleoside triphosphate (NTP) - attaches with adenosine triphosphate (ATP) --> polymerize with other NTPs

Trinucleotide Repeat (TNR) Expansion

Disorder: Huntington's disease TNR sequence: CAG Region of repeat: Coding of Huntingtin gene Normal repeat length: 6-34 Disease repeat length: 36-121 Symptoms: abnormalities in movement, cognitive discipline, psychiatric features **Replication slippage events **Genes cause diseases where expansions of repeats **normally repeat of 6-34 repeats.... **Then start to see phenotype

Inversions in meiosis

Individuals homozygous: no problems arise during meiosis Individuals heterozygous: - Homologous sequences align only if the two chromosomes form an inversion loop - Abnormal gametes form if recombination occurs in the inverted regions

Base Excision Repair (BER)

Primarily responsible for removing small, non-helix-distorting base lesions from the genome. Oxidized bases: - 8-oxoG, 2,6-diamino-4-hydroxy-5 fomamidopyrimidine (FapyG, FapyA) Alkylated bases: - 3-methyladenine, 7-methyguanine Deaminated bases: - hypoxanthine formed from deamination of adenine. - Xanthine formed from deamination of guanine. **Uracil inappropriately incorporated in DNA or formed by deamination of cytosine

8.20 Primary Down syndrome is caused by the presence of three copies of chromosome 21

Primary Down Syndrome (Trisomy 21) Extra chromosome 21 changes phenotype in individuals - craniofacial abnormalities - Caused by nondisjunction

Q1: If Avery, Macleod and McCarthy had found that samples of heat-killed bacteria treated with Rnase and Dnase transformed bacteria, but samples treated with protease did not, what conclusion would they have made?

Protein is the genetic material

Q5: Mismatch repair requires the ability to distinguish between template and newly synthesized DNA strands. How can E coli. Distinguish between these two strands?

Template DNA is methylated

3.) Define terms related to bacterial transformation

Transformation: - A bacterium takes up DNA from the medium. 1. Competent cells: cells that take up DNA 2. Transformants: cells that receive genetic material 3. Cotransformed: cells that are transformed by two or more genes

Q4: Which is NOT an affect of chromosomal translocations?

Translocations can cause regions of the chromosome to be duplicated

3.) Describe causes and effects of anueploidy

Variations in copy number: aneuploidy and polyploidy Causes of aneuploidy: - Deletion of centromere during mitosis and meiosis - Robertsonian translocation - Nondisjunction during meiosis and mitosis **Aneuploidy Is an Increase or Decrease in the Number of Individual Chromosomes Autosomal aneuploids: (Trisomy 21: Down syndrome) - gain of chromosome 21 - Primary Down syndrome, 75% random nondisjunction in egg formation - Familial Down syndrome, Robertsonian translocation between chromosomes 14 and 21 Sex chromosomes - XXY, XXX, XO

Organisms Use DNA Repair Systems to Counteract Mutations

Want all forms of dna damage to be resolved

The DNA Double Helix

Watson and Crick 1953 Double helix with sugar phosphate backbones on the outsides and nucleotide bases arrayed in complementary pairs toward the center

Q2: All DNA polymerases require a primer with a 3' OH group to begin DNA synthesis. The primer is _____.

a short stretch of RNA nucleotides **Need to start with 3' hydroxyl that already there **Made with rna primer... **Primer is small chunk **Enzyme adds to template strand

Supercoiling

a) - replication fork - template strand - daughter strands - template strand - replication fork - replication bubble - supercoiled DNA b) - replication fork - DNA supercoil 1. topoisomerase cuts DNA strand 2. DNA strand rotates to remove the coils 3. topoisomerase rejoins DNA strand **As pull the two strands apart... **Topo binds then cuts it... straighten it back up **Happens in linear chromosomes as well

18.36 Many incorrectly inserted nucleotides that escape proofreading are corrected by mismatch repair (Ecoli specific method)

a) New DNA (GATC)/Old template DNA (CTAG) 1. In DNA replication, a mismatched base was added to the new strand 2. Methylation at GATC sequences allows old and newly synthesized nucleotide strands to be differentiated; a lag in methylation means that, immediately after replication, the old strand will be methylated but the new strand will not - CTAG --> methyl group --> mismatch repair complex b) - nick, GATC - CTAG, methyl group - mismatch repair complex 3. the mismatch repair complex brings the mismatched bases close to the methylated GATC sequence, and the new strand is identified c) - GATC - CTAG, methyl group 4. exonucleases remove nucleotides on the new strand between the GATC sequence and the mismatch --> DNA bases d) - GATC - CTAG, methyl group 5. DNA polymerase then replaces the nucleotides, correcting the mismatch, and DNA ligase seals the nick in the sugar-phosphate backbone **Wants to remove wrong sequence and put in the right sequence **Mechanism in ecoli uses methylation **Brings the error that it recognizes around the complex to determine which area will be removed

The answer to telomere creation - Telomerase

a) elongation b) translocation **telomerase- protein but within is RNA hybridized within the sequence

Q5: Replication of eukaryotic chromosomes presents several challenges that are not found in prokaryotic cells. These challenges include all of the following EXCEPT

eukaryotic DNA polymerases are, in general, much more error-prone

Q4: DNA replication in eukaryotes differs from replication in prokaryotes in that _____.

eukaryotic chromosomes have many separate origins of replication, whereas prokaryotic chromosomes have a single origin of replication

Q2: The 3'-5' exonuclease activity of DNA polymerase allows the correct nucleotide to be added to the 5' end of the DNA molecule.

false **Always adding 5'-3' **Exonuclease goes in reverse- still add the next nucleotide to 3' end

Q1: Autosomal mutations are heritable

true **somatic are body cells **but all have chromosomes

Barr Bodies are

heterochromatin

Q3: A DNA molecule 300 bp log has 20 complete rotations. This DNA molecule is

negatively supercoiled 300/10 = 30 20, should be 30.. so neg. supercoil

Q4: A DNA molecule 500 bp long has 50 complete rotations. This DNA molecule is ____.

relaxed 500/10 = 50

Q2: Suppose that Hershey and Chase found that phage ghosts contained 32P label but the label was absent from infected E. coli. Furthermore, suppose they found 35S lacking in the ghosts and present in the infected E. coli. They would have concluded _____.

that protein was the genetic material in phage

18.21 Pyrimidine dimers result from ultraviolet light.

(a) Formation of thymine dimer - thymine bases - 3' - phosphate - 5' (sugar phosphate backbone) (covalent bonds)--> UV light T=T (b) Distorted DNA (double helix)

9.26 A retrovirus uses reverse transcription to incorporate its RNA into the host DNA.

(a) Structure of a typical retrovirus. Two copies of the single-stranded RNA genome and the reverse transcriptase enzyme are shown enclosed within a protein capsid. The capsid is surrounded by a viral envelope that is studded with viral glycoproteins. (b) The retrovirus life cycle. 1. Virus attaches to host cell at receptors in the membrane 2. The viral core enters the host cell 3. Viral RNA uses reverse transcriptase to make complementary DNA, and viral RNA degrades 4. reverse transcriptase synthesizes the second DNA strand 5. The viral DNA enters the nucleus and is integrated into the host chromosome, forming a PROVIRUS 6. On activation, proviral DNA is transcribed into viral RNA, which is exported to the cytoplasm 7. in the cytoplasm, the viral RNA is translated 8. viral RNA, proteins, new capsids, and envelopes are assembled 9. an assembled virus buds from the cell membrane **Reverase transcriptase enzyme --> double stranded dna integrated into human genome (hunk of viral dna) **Where integrates now a mutation site **Genome can sit and become activated into viral rna...all proteins needed to make new HIV virus

Eukaryote Replication Origins

- 1. Saccharomyces cerevisiae (yeast) has the most fully characterized origin-of-replication sequences - The multiple origins of replication are called autonomously replicating sequences (ARS) - ARS organization and sequence is similar throughout the yeast genome - Replication origins of other eukaryotes are less well characterized **After every 300 Kbp of DNA, to ensure that DNA replication occurs continuously throughout the chromosome. **In mammalian chromosomes no specific sequences of origin. The zones where initiation of replication occurs can span between 500-50,000 bp. **Driven not by sequence but another process

Meselson-Stahl Experiment

- A centrifuge tube is filled with a heavy salt solution and DNA fragments - It is then spun in a centrifuge at high speeds for several days - A density gradient develops within the tube. Heavy DNA (with 15N) will move toward the bottom; light DNA (with 14N) will remain closer to the top ** Masses will separate out **Light DNA floats higher

Chargaff's rules

- A is stoichiometric with T - A=T - G=C - If you have % of C.. you can find % of G

After the replication fork is past, there are additional error-seeking processes

- A mismatch in base-pairing will make a distortion in the DNA double helix - Such distortions are recognized, though the mechanisms are still poorly understood - They identify a "Mismatch", which induces mismatch repair, or MMR. - This adds another 10-2 to the error probability in DNA replication - The frequency with which mistakes get through all this is ~10-9, so the genome is usually fine (1 mistake per billion base pairs)

You have a lot of DNA

- About 3 billion base pairs split between 23 pairs of chromosomes - Humans have about 20,000 genes that together account for ~2% of the base pairs - How about the rest?

Multiple DNA Polymerases in Eukaryotic Replication

- DNA Pol a used to make primers for Okazaki fragments a) Has primase but no 3'5'-proofreading - DNA Pol e used in leading strand synthesis - DNA Pol d used in lagging strand synthesis a) Comparable to bacterial DNA Pol III b) Has 3' --> 5'- proofreading **Alpha, epsilon, delta **E- leading strand making dna in 5' to 3' but proofreading can reverse it: slows down but increase fidelity? **A- 2 parts: 1st is primase: embedded in dna alpha produces dna primer; 2nd part: extend primer with dna; makes mistakes; leaves chuck that is e or d

Phage DNA, Not Protein, Is Responsible for Infecting the Bacteria

- After infection, in both experiments, agitation by a blender separated the empty phage particles from the infected bacteria - In the protein labeling experiment, the radioactivity was detected in the empty phage particles (ghosts) - In the DNA labeling experiment, the radioactivity was detected inside the infected bacteria STEPS: 1. Label phage DNA by growing phage in 32P containing medium 2. infect new unlabeled bacteria with 32P labeled phage 3. After infection, agitation in a blender separates the empty (ghost) phage particles from bacteria 4. Centrifuge blended mixture of bacteria and ghosts. Bacteria from pallet at bottom; ghosts remain suspended in liquid. CONCLUSION: 1. Almost all the 32P label is in the pallet and is contained in infected bacteria 2. DNA is the hereditary molecule passed by the infecting phage into the host cell and inherited by the progeny phage STEPS: 1. Label phage protein by growing phage in 35S containing medium 2. Infect new unlabeled bacteria with 35S labeled phage 3. After infection, agitation in a blender separates the empty (ghost) phage particles from bacteria 4. Centrifuge blended mixture of bacteria and ghosts. Bacteria from pallet at bottom; ghosts remain suspended in liquid CONCLUSION: 1. Almost all 35S label is in the supernate and remains with the ghost particles. **Made phage that had radioactive DNA **Proteins had sulfur **Phage turns into ghost phase **Centrifuge bacteria to bottom of tube and phage will float in the palate **What is injected in the bacteria? Cell pellets of bacteria P32- DNA was injected and protein S32 never injected

1.) Describe the types and features of repetitive DNA

- Are repeated many times within eukaryotic chromosomes - Several categories of repetitive DNA CHART: --> Repetitive DNA 1. Highly repetitive a. satellite DNA 2. Middle repetitive a. tandem repeats - multiple copy genes; rRNA genes - mini satellites; VNTRs - micro-satellites; STRs b. interspersed retrotransposons - SINEs; Alu - LINEs; L1 - It is important to remember that mammalian genomes have a lot of repetitive DNA and the general types

DNA strands serve as template

- Arrangement and nature of nitrogenous bases allow DNA strands to serve as templates - Complementarity of DNA strands allows each strand to serve as template for synthesis of the other **Double stranded dna denature **Base pairings makes duplicate copy **Add to 3' always

2.) Describe supercoiling, its effect on DNA and whether a molecule will be positively or negatively supercoiled

- B form DNA has 10 bp per turn when relaxed - If over or under twisted, it will supercoil if the ends can't rotate - Topoisomerase enzyme can sever DNA and rotate ends before rejoining a) Linear DNA b) Relaxed circular DNA: L = (+20) c) Underwound circ. DNA: L = (+18) d) Supercoiled DNA (2 supercoils form) **B form-most commonly exists in cellular environment **A- 20 terms (10bp per term) **"telephone cord"

B- form Major and Minor Grooves

- Base-pair stacking creates gaps between the sugar phosphate backbones that partially expose the nucleotides - The major groove, approximately 12Å wide, alternates with the minor groove, approximately 6Å wide - These grooves are regions where DNA binding proteins can make direct contact with nucleotides - The diameter of the molecule is 20Å, where 1Å is 10^-10 m - Each 360 degree turn takes 10 base pairs, or 34Å, which equals 3.4 nanometers **Sugarphosphate backbone is white and pink **Bases are in the middle **Takes 10 bases for one twist **34A per 10 bases **Molecule is thin --> dna area is 20A **B form produces major and minor groove

Double Strand Break (DSB) Repair

- Caused by ionizing radiations, topoisomerase inhibitors, chemical agents. - Both strands of the double helix are severed. - Can lead to genome rearrangements. **Piece of chromosome stuck to another- so need breaks

Karyotyping

- Chromosomes prepared from actively dividing cells - Halted in metaphase, one way is to disrupt the spindle - Chromosomes are spread out on a slide - Chromosomes stained, photographed and then digitally arranged according to size - human karyotype consists of 46 chromosomes

7.) Define CpG islands, DNA methylation and describe how it affects gene expression

- CpG is the DNA sequence 5'-CG-3' and not a GC base pair a.) Lots of these together is called a "CpG ISLAND" - Methylation of cytosine by DNA methyltransferase usually represses gene activity **CpG- p: phosphodiester bond between C and G **Mythelated dna is inactive **Is dna methylation lead to gene activation or repression? Repression because silencing dna

RNA Primers Are Needed for DNA Replication

- DNA polymerase elongates DNA strands by adding nucleotides to the 3' end of a pre-existing strand - They cannot initiate DNA strand synthesis on their own - RNA primers are needed; these are synthesized by a specialized RNA polymerase called primase - Primase and some additional proteins join DnaA at oriC to form the primosome **RNA primase synthesized ontop of dna primase **Make initial segment our of rna **So 3' hydroxyl is the same so dna on top of rna **Primase makes chunck out of rna

DNA Replication Terminology

- DNA replication begins at the ORI (origin of replication) - At site of replication, helix is unwound, creating replication fork - Replication is bidirectional; therefore, there are two replication forks - Replicon: Length of DNA replicated (stuff being synthesized)

Origin of Replication in Bacterial DNA

- DNA replication is most often bidirectional, proceeding in both directions from a single origin of replication in bacterial chromosomes - Eukaryotic chromosomes have multiple origins of replication - John Cairns reported the first evidence of bacterial origins of replication in 1963

Review of Replication

- DNA replication is semiconservative a) Each strand of template DNA is being copied. - DNA replication is bidirectional a) Bidirectional replication involves two replication forks, which move in opposite directions** - DNA replication is semidiscontinuous a) The leading strand copies continuously b) The lagging strand copies in segments (Okazaki fragments) which must be joined

11.5 The nucleosome is the fundamental repeating unit of chromatin.

- DNA wrapped around histones is protected from nuclease enzyme - Linker regions between nucleosomes are easily cleaved - Tail region associated with H1 is moderately susceptible to cleavage a.) core histones of nucleus + linker DNA "beads on a string" view of chromatin 1. "Nuclease": A small amount of nuclease cleaves the "string" between the beads b.) 2. releasing individual beads attached to about 200 bp of DNA 3. "Nuclease": More nuclease destroys all of the unprotected DNA between the beads 4. leaving a core of proteins attached to 145-147 bp of DNA **Nucleosomes tells you what the core particle is **Crystal structure of DNA associated with nucleosome- wraps around histone twice **The colorful pieces are histone tails-change how dna associates with core particle

Types of DNA Damage

- Deamination (C → U or A → Hypoxanthine) - Depurination (A or G lost) - Alkylation: an alkyl group gets added to bases; chemical induced, some harmless, some cause mutations by mispairing during replication or cause polymerase stalling- bases are chemically modified by multiple reagents **bases or strands damaged

Conjugation

- Direct transfer via connection tube is one-way traffic from donor cells to recipient cells. - It is not a reciprocal exchange of genetic information. - F+ cells: donor cells containing F factor - F- cells: recipient cells lacking F factor - Sex pilus: connection tube 9.10 A sex pilus connects F+ and F− cells during bacterial conjugation.

Initiation of Replication

- DnaA first binds the 9-mer sequences, bends the DNA, and breaks hydrogen bonds in the A-T rich sequences of the 13-mer region - DnaB is a helicase that uses ATP energy to break hydrogen bonds of complementary bases to separate the strands and unwind the helix - DnaB is carried to the DNA helix by DnaC - The unwound DNA strands are kept from reannealing by single-stranded binding protein (SSB) - Unwinding of circular chromosomes will create torsional stress, potentially leading to supercoiled DNA - Enzymes called topoisomerases catalyze controlled cleavage and rejoining of DNA that prevents overwinding **oriC- bound by dnaA **Protein recognizes then associates it **dnaB is carried to dna helix by dnaC **Once dna is denatured, its bound by SSB to fix

Herpes virus has latent and lytic phases

- Double stranded DNA enveloped virus - Infects epithelial cells, then associated sensory neurons - Viral latent phase is in neurons where viral is episomal - Variety of factors reinitiate lytic phase **Lipid membrane that surrounds it (envelope) **Good at infecting epithelial cells **Infects the cells first but then gets into neurons that touch the cells **Viruses get into neurons and travel up to body **Latent- episomal (f factor plasmid) **Lytic phase- reinfect epithelial cells

Simultaneous Synthesis of Leading and Lagging Strands

- Each replisome complex carries out replication of the leading and lagging strand simultaneously - The DNA pol III holoenzyme contains 11 protein subunits, with the two pol III core polymerases each tethered to a different copy of the tau (T) protein - The tau proteins are joined to a protein complex called the clamp loader; two additional proteins form the sliding clamp **Synthesizing leading and lagging strand **Structural organization **Dna pol3 molecules attached by proteins that circles dna template held on by sliding clamp **Leading strand its okay because makes dna **Discontinuous strand, dna in same direction but loop.... - 2 green kidney bean (pol III core polymerase) - tan antler looking (T protein) - big yellow circle (clamp loader) - small purple ring (sliding clamp)

Aglets of our genome - TELOMERES

- Ends of DNA contain a tandem repetitive sequence called "telomeres" - Human sequence 5'-TTAGGG-3'(~ 2000 repeats) - Repetitive sequence ensure that incomplete chromosome replication does not affect vital genes

Histone Acetylation

- Enzyme histone acetyltransferase (HAT) - Addition of acetyl group to positively charged amino group on side chain (lysine) changes net charge of protein by neutralizing positive charge - Usually promotes gene activity **Transfer of acetyl group attached onto lysin

Silenced X is passed to daughter cells during mitosis

- Epigenetic changes are heritable a.) All descendant cells (mitotic) have same X-inactivation b.) Example: Calico cats and fur color/patterns **Each fur color is a region derived from an initial founder cell that subsequently divided to form region of the skin **Other tissues are the same, for example female carriers for color blindness have retinas with patches of color sensitive and insensitive receptors **Changes dna expressed but not sequence of dna itself

11.6 Chromosomal puffs are regions of relaxed chromatin where active transcription is taking place.

- Expanded regions are where there is active transcription - Chromatin need to be in a less compacted form to allow for access **Compact and puffy regions- transcription happens

X inactivation is a good example of chromosomal regulation

- Females are XX while males are XY, so gene dosage compensation is required - Inactive X chromosome, highly condensed (Barr body) ** stained because inactivation

Eukaryotic Chromatin

- Filamentous complex of DNA, histones and other proteins - At interphase, eukaryotic chromosomes uncoil and decondense into a form called chromatin dispersed throughout the nucleus - During cell division, chromatin coils and condenses back into visible chromosomes - Chromatin appear as beads on a string ** Chromatin are a mix of DNA and proteins

The Transformation Factor

- Frederick Griffith identified two strains of Pneumococcus: S, which caused fatal pneumonia in mice, and R, which did not - A single nucleotide change can convert the R (rough) strain into the S (smooth) strain - These strains occur in four antigenic types (I, II, III, and IV) that cannot be altered by mutation alone

Banding

- G bands: Giemsa stain (white background; appears as regular karyotype) - Q bands: quinacrine mustard (black background; spread out) - C bands: centromeric heterochromatin (white background; spread; distinct centromeres) - R bands: rich in cytosine-guanine base pairs (black background; colorfull chromosomes)

Nondisjunction

- Gives rise to variation in chromosomal number - Paired homologs fail to disjoin during segregation - Nondisjunction during meiosis I or II - Can also happen in mitosis if sister chromatids fail to separate ** error of DNA between cells

Nucleosomes organize into a 30 nm filament

- H1 is the 5th histone - Core 4- 8 nucleosomes in each region

Simplified overview of homologous recombination (HR) and non-homologous end-joining (NHEJ)

- HR is an extremely accurate mechanism that relies on homology with a sister chromatid to direct DNA synthesis based repair - NHEJ is an error-prone mechanism in which ends are processed to make them compatible and then ligated together (A) HR utilizes a homologous stretch on a sister chromatid to accurately repair the DSB. DNA ends are first processed in order to create single-strand overhangs, a process that is likely mediated by the MRN (Mre11/Rad50/Nbs1) complex. Rad51, Rad52, and RPA associate with these overhangs, followed by the formation of a joint molecule by the damaged and undamaged strands. Template guided DNA synthesis and resolution of the two strands then complete repair of the DSB. (B) NHEJ brings the ends of the broken DNA molecule together by the formation of a synaptic complex, consisting of two DNA ends, two Ku70/80 and two DNA-PKCS molecules. Noncompatible DNA ends are processed to form ligatable termini, followed by repair of the break by the ligase IV/XRCC4 complex. **Don't ask questions about individual enzymes **HR- seals sequence = perfect repair **NHEJ- two broken ends and mashes them together... many insertions and deletions (error-prone)

Satellite DNA

- Highly repetitive, hundreds of thousands to millions of copies - consists of short repeated sequences, often less than 10 base pair sequences (bp) - Rarely transcribed into RNA **The name satellite DNA comes from an early experiment - will not test over this ** Unknown effect- regulatory

Chromatin is not static

- Histone code - Idea that modifications of histones control availability of DNA for transcription - Modifications can be inherited --> Epigenetics **Acetylation **Methylation **Phosphorylation **Amino acids which are modified **Modifications are part of epigenetic: change gene expression (phenotype)

Electron micrograph of circular plasmid DNA

- Increasing amount of supercoiling - Rleaxed = floppy circle - Supercoiled = compact, twisted circle --> line means extremely supercoiled

Assembly of Polynucleotide Chains

- Individual nucleotides are assembled into chains by the enzyme DNA polymerase - It catalyzes the formation of a phosphodiester bond between the 3' hydroxyl group of one nucleotide and the 5' phosphate of an adjacent one - Each polynucleotide chain has a sugar-phosphate backbone, consisting of alternating sugar and phosphate groups **DNA is a polymer Polmer= additional monomers **Start triphosphate and go to single phosphate chain

Features of Hereditary Material

- Localized to the nucleus, component of chromosomes - Present in stable form in cells - Sufficiently complex to contain information needed for structure, function, development, and reproduction of an organism - Able to accurately replicate itself so that daughter cells contain the same information as parent cells - Mutable, undergoing a low rate of mutations that introduces genetic variation and serves as a foundation for evolutionary change

Mutations classified by PHENOTYPE

- Loss-of-function mutations: protein that was there doesnt do its job anymore - Gain-of-function mutations: causes new feature - Visible (morphological) mutations - Nutritional (biochemical) mutations - Behavioral mutations - Regulatory mutations - Lethal mutations - Conditional /temperature-sensitive mutations: black bunny ears and nose

Griffith's Experimental Results

- Mice infected with strain S developed pneumonia and died - Mice infected with strain R or with heat-killed strain S survived - Mice infected with heat-killed strain S and live strain R developed pneumonia and died - live-type S bacteria were recovered from the mice CONCLUSION: Hereditary molecule transformed RII bacteria into SIII bacteria

Skeletal muscle cells have multiple nuclei

- Myoblasts fuse as muscle cells form - Individual cells have many nuclei, so they have several copies of the genome - Individual nuclei supply transcripts to a zone of the cell **Muscle fiber= muscle cell **Fusion of myoblast into one muscle cell **Muscle cell needs to make a lot of protein- RNA- more copies of gene **Huge cells

8.11 Chromosome 4 differs in humans and chimpanzees by a pericentric inversion

- No genetic information gained or lost but there are often phenotypic effects - Genes might be broken in half or fused with another gene - Disrupt position dependent gene regulation **Where they break is one thing **Changes expression level of adjacent genes

Nucleotide Base Stacking

- Nucleotide base pairs are spaced along the DNA duplex at intervals of (3.4Å) - This tight packing is due to base stacking, the offsetting of adjacent base pairs so that their planes are parallel 1. Buries hydrophobic residues in the core of the molecule 2. Permits base pairing (hydrogen bonding) 3. This leads to a twist in the DNA backbone and a helical structure **3' hydroxyl is where nucleotide goes **This is the 3d structure **Sugarribose phosphate is linear part **Bases attach

Types of DNA Damage

- Oxidative damage (guanine oxidizes to 8-oxo-guanine), also cause single and double strand breaks. - T-T and T-C dimers: bases become cross-linked, T-T are more prominent, caused by UV light {[UV-C <280nm] and UV-B (280-320nm) - Double strand breaks: induced by ionizing radiation, transposons, topoisomerases, mechanical stress on chromosomes, single strand nick in a single strand region (e.g. replication or transcription) **Situation where double strand is fractured into two separate pieces

18.19 Oxidative radicals convert guanine into 8-oxy-7, 8-dihydrodeoxyguanine.

- Oxidative damage (guanine oxidizes to 8-oxo-guanine), also cause single and double strand breaks. Guanine (oxidative radicals) --> 8-Oxy-7,8-dihydrodeoxyguanine ** may mispair with adenine

Phage Infection of Bacteria

- Phages must infect bacterial hosts to reproduce - Infection begins when the phage injects DNA into the bacterial cell and leaves its protein shell on the surface - The phage DNA replicates in the bacterium and produces proteins that are assembled into progeny phage - these are released by lysis of the host cell STEPS: 1. T2 phages infect bacteria by injecting DNA through hallow tails 2. Phage DNA circularizes, and replication of phage chromosome occurs 3. Under the direction of phage genes, transcription and translation produce new phage protiens 4. DNA and proteins are assembled into progeny phages 5. Progeny phages are released by lysis of host bacteria

Chromosome Structure Changes

- Polytene chromosome: created by repeated rounds of DNA replication with no cell division - Chromosomal puffs are regions where the chromosome is made less compact a.) These regions are more sensitive to DNase I cleavage b.) Correlates with gene activity **salivary glands... cells get stacked upon eachother

Hershey and Chase Experiments

- Proteins contain large amounts of sulfur but almost no phosphorus - DNA contains large amounts of phosphorus but no sulfur - Hershey and Chase separately labeled either phage proteins (with 35S) or DNA (with 32P) and then traced each radioactive label in the course of infection proteins = 35S = radioactive sulfur DNA = 32P = radioactive phosphorus

Tautomers

- Purines and pyrimidines exist in tautomeric forms—alternate chemical forms - Increase chance of mispairing during DNA replication a) standard base pairing arrangements - thymine(keto)=adenine(amino) - cytosine(amino)=guanine(keto) b)anomaious base pairing arrangements - thymine(enol)=guanine(keto) - cytosine(imino)=adenine(amino)

Inversion (3)

- Rearrangement of linear gene sequence - No loss of genetic information - Segment of chromosome turned 180 within chromosome - Requires two breaks in chromosome, and reinsertion inverted segment - May arise from chromosomal looping - Paracentric DO NOT include the centromere while pericentric inversions DO include the centromere -->A(EDCB)F **180 flip in dna *Pericentric do include centomere *Paracentric dont include centromere

9.6 The F factor, a circular episome of E. coli, contains a number of genes that regulate transfer into a bacterial cell, replication, and insertion into the bacterial chromosome.

- Replication is initiated at ori. Insertion sequences (see Chapter 18) IS3 and IS2 control insertion into the bacterial chromosome and excision from it. - left side 1. These genes regulate plasmid transfer to other cells (OJALEKBCUNFHGSDI) - right side 2. These sequences regulate insertion into the bacterial chromsome - IS3 - IS2 - bottom 3. These genes control plasmid replication - ori (origin of replication) - inc - rep - oriT (origin of transfer)

Bacterial Replication Origins

- Replication origins of bacterial species have similar (conserved) but not identical sequences - Comparison between species leads to identification of consensus sequences, the nucleotides found most often at each position of DNA in the conserved region - The 13-mer and 9-mer sequences of oriC are conserved—they play an essential role in replication **oriC- Mostly As and Ts... because G-C have higher bonds... A-T are easier to denature

Nucleotide Excision Repair (NER)

- Responsible for bulky, helix-distorting lesions - Thymine dimers and 6,4 photoproducts - In-born genetic mutations in NER proteins cause severe diseases - Xeroderma Pigmentosa (50% have skin cancer by age 10) ex) Cockayne's syndrome **Children of the Night **1:250,000 Japanese have a high rate. XP **Cockayne ERCC6 **Set of enzymes that chop out or repair

Frameshift Mutations

- Result from insertions or deletions of base pair Loss or addition of nucleotide causes shift in reading frame - Frame of triplet reading during translation is altered 1. Point mutation (substitution): change of one letter - ORIGIONAL- The cat saw the dog - the bat saw the dog - the cat saw the hog - the cat sat the dog 2. Frameshift mutation (deletion): loss of one letter - ORIGIONAL- The cat saw the dog - the ( )ats awt hed og: loss of c 3. Frameshift mutation (insertion): gain of one letter - ORIGIONAL- The cat saw the dog - the c(m)a tsa wth edo g: insertion of m **Adding or removing a base or two or more **Frameshift if not a group of 3 **If you lose a letter, sentence turns to gibberish

8.17 Fragile sites are chromosomal regions susceptible to breakage under certain conditions

- Shown here is a fragile site on the human X chromosome - Once broken gene can be lost from genome

Influenza

- Single stranded negative sense RNA genome - Most cases influenza A: divided into subtypes based upon expression of hemagglutinin (HA) and neuraminidase (NA). - Replication is error prone - Mutations result in ANTIGENIC DRIFT **Influenza a- HA and NA (H1N1) **Change is antigenic drift- arises from mutations during replication

Mutations classified by LOCATION

- Somatic mutations occur in any cell except germ cells; are not heritable - Germ-line mutations occur in gametes; are inherited - Autosomal mutations occur within genes located on autosomes (any chromosome except for your sex chromosome) - X-linked and Y-linked mutations occur within genes located on X and Y chromosome, respectively

Role Of Telomeres In Aging: The Hayflick Limit

- Telomeres shorten from 10-15 kb (germ line) to 3-5 kb after 50-60 doublings (average lengths of TRFs) - Cellular senescence is triggered when cells acquire one or a few critically short telomeres. **Normal cells only have certain amount of dividing = hayflick limit; once gone, cell cant divide anymore **Why is tomomerase shut off in some cells? Could that be the fountain of youth if turned back on **Might turn into tumor cell if not controlled....

Eukaryotic DNA Replication Is Similar to Bacterial Replication but Differs in Several Aspects

- The location of DNA replication within the nucleus: DNA polymerase is fixed in location and template RNA is threaded through it - Replication at the ends of chromosomes: a) Telomeres and telomerase a) circular DNA - replication around the circle provides a 3'-OH group in front of the primer; nucleotides can be added to the 3'-OH group when the primer is replaced. b) linear DNA 1. in linear DNA with multiple origins of replication, elongation of DNA in adjacent replicons provides a 3'-OH group for replacement of each primer 2. the terminal primer is positioned 70-100 nucleotides from the end of the chromosome 3. leaving a gap that is not replicated CONCLUSION: In the absence of special mechanisms, DNA replication would leave gaps at the end of the chromosome, and the chromosome would shorten each time the cell divides **Bacterial chromosome is circular and linear in eukaryotes **Differ: issue at end of chromosome... gap that's not replaced

Telomerase Function

- The template RNA of telomerase allows new DNA replication, to lengthen the telomere sequences - Once telomeres are sufficiently elongated, the alpha polymerase synthesizes additional RNA primers - New DNA replication then fills out the chromosome ends - Telomere sequences in most organisms are quite similar a) the telomerase has protruding end with a G-rich repeated sequence b) the RNA part of telomerase is complementary to the G-rich strand and pairs with it, providing a template for the synthesis of copies of the repeats c) Nucleotides are added to the 3' end of the G-rich strand d) after several nucleotides have been added, the RNA template moves along the DNA e) more nucleotides are added - the telomerase is removed f) synthesis takes place on the complementary strand, filling in the gap due to the removal of the RNA primer at the end CONCLUSION: Telomerase extends the DNA, filling in the gap due to the removal of the RNA primer **Can grow the end of the chromosome **Apoly can make new primer then feed back into original strand **Eventually chromosome gets shorter and shorter

Bidirectional DNA Replication

- The theta structures Cairns observed are consistent with both unidirectional and bidirectional replication - In bidirectional DNA replication, new DNA is synthesized in both directions from the single origin, creating an expanding replication bubble - At each end of the replication bubble is a replication fork; replication is complete when the replication forks meet 1. double stranded DNA unwinds at the replication origin 2. producing single stranded templates for the synthesis of new DNA. A replication bubble forms, usually with a replication fork at each end. 3. The fork proceed around the circle 4. Eventually two circular DNA molecules are produced CONCLUSION: The products of theta replication are two circular DNA molecules **Double strand to single strand and then have replication (theta structure) --> replication fork

6.) Describe how chromatin is remodeled

- To accommodate DNA-protein interactions, chromatin structure must change - To allow replication and gene expression, chromatin must a.) relax compact structure b.) expose regions of DNA to regulatory proteins c.) have a reversal mechanism for inactivity ** need to go back between relaxed and condensed structures

Non- Homologous End Joining

- Utilizes short homologous DNA sequence called "microhomologies" to guide repair. - microhomologies are often present in single-stranded overhangs on the ends of double-strand breaks. - resection inhibits NHEJ and commits the break to be repaired by recombination. - NHEJ is active throughout the cell cycle, but is most important during G1 when no homologous template (no sister chromatids) for recombination is available. **most important during G1

Duplications (1)

- repeated segment of chromosome - single locus is present more than once in genome - can produce compensation loop - arise from UNEQUAL CROSSING OVER between synapsed chromosomes during meiosis Meiotic products segregated to gametes: --> ABDEF --> ABCCDEF EX. of unequal crossing over: chromosomes do not align properly, resulting in unequal crossing over (results in DUPLICATION & DELETION) --> red green / red green --> red green green (DUPLICATION) --> red (DELETION) --> one resulting chromosome has two green opsin genes (a duplication) and a red while the other only has a red opsin gene

Deletion (2)

- segment of chromosome missing Deletions can also arise independently: 1. origin of terminal deletion --> one gene lost --> BCDEF --> A (lost) 2. origin of intercalary deletion --> two genes (loop) lost --> ABEF - deleted chromosome --> CD (lost)

Nucelosomes

- the beads on a string - nucleosomes "bead" (8 histone molecules + 146 base pairs of DNA) - protiens: H2A, H2B, H3, H4 (four different types of proteins - nucleosomes are made of histone proteins (4 histones x 2 = 8 total proteins)

4.) Define virus and understand that they can be based on single or double stranded DNA or RNA genomes

- viruses are simple replicating systems -->Virus - Small infectious agent that only replicates inside the living cell of another organism - Genome can be DNA or RNA; single or double stranded - Sometimes enveloped, which means they are wrapped in a small piece of the host cell's membrane

1.) Describe why replication is semiconservative, bidrectional and discontinuous.

--> 3 modes of DNA replication 1. Conservative: two newly synthesized stands come together-- origional helix is conserved 2. Dispersive: parental strands are dispersed into two new double helices 3. Semiconservative: each replicated DNA molecule consists of one "old" and one new strand **Conservative- original molecule is reserved but produces new molecule **Dispersive model the parental strands are randomly dispersed **Disperse- pieces/chunks of one disperse and end in the new one **Semiconservative- Two full strands in the original separate and then form together to make the new one

3' to 5' exonuclease activity (KNOW THIS)

--> 5' primer strand --> 3' template strand --> 3 -OH 1. rare tautomeric form of C (C*) happens to base peir with A and is thereby incorporated by DNA polymerase into the primer strand 2. rapid tautomeric shift of C* to normal cytosine (C) destroys its base pairing with A 3. unpaired 3'-OH end of primer blocks further elongation of primer strand by DNA polymerase 4. 3' to 5' exonuclease activity attached to DNA polymerase chews back to create a base-paired 3'-OH end on the primer strand 5. DNA polymerase continues the process of adding nucleotides to the base paired 3'-OH end of the primer strand **inserts correct one and goes about making more DNA

6.) Distinguish antigenic drift from antigenic shift

--> Antigenic shift can result in big changes - More than one type of influenza virus infecting a single cell can produce a hybrid virus a) Birds harbor the most viruses b)Swine can be infected by both bird and human strains **Shift- when you take different type of influenza and mix together (rapid change of combined viruses) --> Change is antigenic drift- arises from mutations during replication

Breast Cancer Susceptibility Genes BRCA1 and BRCA2

--> BRCA1 and BRCA2 in Homologous Recombination - BRCA1 activates checkpoint - BRCA2 stimulates recombination

Bacterial Replication Requires a Large Number of Enzymes and Proteins

--> Bacterial DNA replication - Initiation: a) 245 bp in the oriC (single origin replicon) b) an initiator protein (DnaA in E.coli) - Unwinding: a) Initiator protein b) DNA helicase c) Single-strand-binding proteins (SSBs) d) DNA gyrase (topoisomerase) **Initiation and bacteria rely on nucleus chromosome **dnaA binds

18.14 Depurination (the loss of a purine base from a nucleotide) produces an apurinic site.

--> Base Changes- depurination a) - DNA sugar-phosphate backbone - 5' - bases: T pyrimidine - G purine - G - OH -3' b) DNA: - TGGC - ACCG Depurination (loses a G) - T(apurinic site)GC - ACCG Strand Separation - ACCG --> Replication - ACCG -TGGC: normal DNA molecule (NO MUTATION) - T( )GC 1. in replication, the apurinic site cannot provide a template for a complementary base on the newly synthesized strand --> Replication - A(A)CG 2. a nucleotide with the incorrect base (most often A) is incorporated into the newly synthesized strand -T( )CG Strand Separation - A(A)CG: Template strands 3. At the next round of replication, this incorrectly incorporated base will be used as a template --> Replication - MUTANT!!!! - T(T)GC - A(A)CG 4. leading to a permanent mutation -T( )GC --> Replication - T( )GC - A(A)CG 5. A nucleotide is incorporated into the newly synthesized strand opposite the apurinic site **Double stranding with backbone, just missing the base **Strand separate and comes to blank spot... usually sticks in an ADENINE and the adenine will persist

Transitions and Transversions

--> Base substitutions - Transitions: Pyrimidine replaces pyrimidine, or purine replaces purine - Transversions: Purine and pyrimidine are interchanged

3.) Base stacking and properties of the double helix

--> Complementary DNA Nucleotide Pairing - The two polynucleotide chains of a double helix form a stable structure that follows two rules: 1. The bases of one strand are complementary to the bases in the corresponding strand (A pairs with T and G pairs with C) 2. The two strands are antiparallel, with respect to their 5' and 3' ends **A = T **G = C Base pairs antiparallel

1.) Describe bacterial plasmids and how they can be transferred by conjugation

--> Conjugation - Transfer of the F Factor plasmid **9.11 The F factor is transferred during conjugation between F+ and F− cells. a) -->F+ cell (donor bacterium) - large green top circle: bacterial chromosome - small red bottom circle: f factor -->F- cell (recipient bacterium) - large green top circle: bacterial chromosome b) 1. during cinjugation, a cytoplasmic connection forms c) 2. one DNA strand of the f factor in nicked at an origin of transfer and separates 3. replication takes place on the f factor, replacing the nicked strand 4. The 5' end of the nicked DNA passes into the recipient cell d) 5. where the single strand is replicated 6. producing a circular, double stranded copy of the f plasmid e) the F- cell now becomes F+ **Positive cell that has both circular chromosome and f factor plasmid **Genes stimulate gene of sex pilus **Nick, single strand separates and injects into another **2 f+ cells: resulting factor

Repair of mistakes in DNA polymerase action is rarely necessary

--> DNA polymerase is accurate in its own right. Mistake frequency is ~10-5 - So 1 error per 100,000 bases --> Processive DNA polymerases include a proof-reading function that makes them MORE accurate still: mistake frequency on the second go is ~10-2 --> This implies a net error rate of 10-7 - 1 error per 10,000,000 bases **catalytic activity

Interspersed repeats

--> Dispersed throughout genome --> Constitute 1/3 of human genome a.) Short interspersed elements (SINES) - Alu, 300bp sequence present more than a million times (13% of the genome!) b.) Long interspersed elements (LINES) - Line1, several thousand base pairs - Remnant of transposable elements, which are mobile sequences that can hop around within genome **Scattered throughout genome **13% of DNA **Transposable elements- mobile elements that can pop out and into the genome

5.) Define euchromatin and heterochromatin

--> Euchromatin - Uncoiled and transcriptionally ACTIVE - Appears UNSTAINED during interphase **good; available to do work (active) **so dispersed that dye cant fix to it --> Heterochromatin - Condensed areas are mostly INACTIVE - Appears STAINED during interphase

4.) Distinguish between prokaryotic and eukaryotic polymerases.

--> Genome size and rate of synthesis - Eukaryotes work SLOWER with polymerases; LARGER in size (Bbp) --> E coli - size: 4.6 Mbp - speed: 30Kb/min - S Phase: 40 mins - Origins: 1 --> Humans - size: 3.2 Bbp - speed: 3Kb/min - S Phase: 7 hours - Origins: >10,000 ?

Example of nuclease digest giving a ladder of bands

--> Increasing digestion - with no enzyme, theres a blob - linker region gets chewed away once reaches singles (tells you DNA is associated with each core particle) - triples ( above 500) -doubles (below 500) - singles (below 200); most digested

Base Excision Repair (BER)

--> Key enzymes in BER - DNA glycosylase a) Recognizes and removes damaged base b) Leaves an apyrimidinic or apurinic site - AP endonuclease a) Removes apyrimidinic or apurinic nucleotide b) Leave a gap in the sugar-phosphate backbone - DNA polymerase a) Adds new nucleotide - Ligase a) Seals the sugar-phosphate backbone a) DNA; damaged base --> DNA glycosylase 1. each DNA glycosylase recognizes and removes a specific type of damaged base, producing an apurinic or an apyrimidinic site (AP site) b) AP site --> AP endonuclease c) 2. AP endonuclease cleaves the phosphodiester bond on the 5' side of the AP site 3. and removes the deoxyribose sugar --> DNA polymerase --> NTPs and Deoxyribose phosphate + dNMPs d) 4. DNA polymerase adds new nucleotides to the exposed 3'-OH group --> DNA ligase e) new DNA 5. The nick in the sugar-phosphate backbone is sealed by DNA ligase, restoring the original sequence

Mechanism of NER (know process)

--> Key enzymes in NER - Large complex a)Recognizes lesion and and removes damaged bases b)Surrounding region is melted into single stranded DNA c)Stretch of ssDNA is removed around the lesion to leave a big gap - DNA polymerase a)Adds new nucleotides - Ligase a) Seals the sugar-phosphate backbone a) damaged DNA 1. damage to the DNA, distorts the configuration of the molecule b) 2. an enzyme complex recognizes the distortion resulting from damage c) 3. The DNA is separated, and single-strand binding proteins stabilize the single strands d) 4. An enzyme cleaves the strand on both sides of the damage e) 5. Part of the damaged strand is removed f) DNA polymerase, DNA ligase, and new DNA 6. and the gap is filled in by DNA polymerase and sealed by DNA ligase

All DNA Replication Takes Place in a Semiconservative Manner

--> Linear eukaryotic replication: - Direction of replication a) DNA polymerase adds nucleotides only to the 3' end of growing strand b) Replication can only go from 5' --> 3' c) Continuous and discontinuous replication 1. because two template strands are antiparallel 2. and DNA synthesis is always 5' --> 3' 3. DNA synthesis proceeds from right to left on one strand 4. and from left to right on the other strand **Bottom leading strand: 3' **3': Top lagging strand (has to be in discontinuous nature

All DNA Replication Takes Place in a Semiconservative Manner

--> Linear eukaryotic replication: - Requirements of replication a) A template strand b) Raw material: nucleotides c) Enzymes and other proteins 1. New DNA is synthesized from deoxyribonuclease triphosphate (dNTPs) 2. In replication, the 3'-OH group of the last nucleotide on the strand attacks the 5'-phosphate group of the incoming dNTP. 3. Two phosphates are cleaved off 4. a phosphodiester bond forms between the two nucleotides **5' to 3' direction **5 is bottom and 3 is top where added Nu **New strand is flipped

Gene Mutations

--> Mutation - An alteration in DNA sequence a) Any base-pair change in sequence b) Single base pair substitution c) Deletion or insertion of base pairs d) Major alteration in chromosomal structure --> May occur in somatic or germ cells --> May occur in coding or noncoding regions

Mutations Occur Spontaneously and Randomly: Spontaneous or Induced

--> Mutations are either spontaneous or induced - Spontaneous mutation: Changes in nucleotide sequence that occur naturally - Arise from normal biological or chemical processes that alter nitrogenous bases --> Spontaneous mutation rates vary, but are exceedingly low for all organisms **Consequence of unchangeable chemistry --> Induced mutations - Result from influence of outside factors, either natural or artificial - Radiation - UV light - Natural and synthetic chemicals

Classification of Mutations

--> Mutations: Classified by molecular change - Point mutation or base substitution: Change from one base pair to another - Missense mutation: Results in new triplet code (codon) for different amino acid - Nonsense mutation: Results in triplet code (codon) for stop codon (translation terminated prematurely) - Silent mutation: New triplet code still codes for same amino acid

2.) Understand that antibiotic resistance genes can be transferred by conjugation

--> Natural gene transfer and antibiotic resistance - Antibiotic resistance comes from the actions of genes located on R plasmids that can be transferred naturally. - R plasmids have evolved in the past 60 years since the beginning of widespread use of antibiotics. - The transfer of R plasmids is not restricted to bacteria of the same or even related species.

4.) Describe analysis and structure of a histone

--> Nucleosomes - Core (4) Histones are H2A, H2B, H3 and H4 a.) Basic proteins (rich in arginine and lysine, which carry a (+) charge) so they bind well to negatively charged DNA b.) Highly conserved between species 1.) H4 is 2% different between peas and cows - There is a 5th histone present at 1 per octomer - Micrococcal nuclease digest of chromatin releases 200bp fragments, with further digest yielding fragments of 147 bp **For every bead, you have a histone **Nuclease cuts DNA and RNA

Primary infection vs Recurrance

--> PRIMARY INFECTION - Herpes simplex 1 virus a) mild pharyngitis or stomatitis 1. small spots on face - Varicella virus a) chickenpox **Viral transit up peripheral nerve and to Latent virus: neuron in dorsal root ganglion then to spinal cord -->RECURRANCE - Fever, sunlight to face, menstruation a) cold sore - Age X-irradiation (act via depressed CMI) a) Zoster (shingles) **activation of virus in neuron then viral transit down peripheral nerve --- ** Latent virus is small spots ** Once latent infection there is no way to kill it/get rid of it ** Best way to avoid reoccurance... vaccine

Mismatch repair - Errors During DNA Replication

--> Polymerase errors - Incorporation of mismatches - DNA polymerase α (pol α) has a high mismatch incorporation rate --> Microsatellite instability - GC rich trinucleotide repeats (TNR) - Tend to expand at substantial rates and lengths - Expansions are thought to occur due to errors during DNA replication - Even though many models have been presented suggesting how expansions occur, all agree that they are a result of looped DNA intermediates, which can form during DNA replication **Address errors in dna replication **Microsatellites- good for fingerprinting people- tend to undergo insertion and deletion

Replication Origins

--> Prokaryotes - Parental duplex - replication forks - daughter duplexes (2) --> Eukaryotes 1. each chromosome contains numerous origins (3) 2. at each origin, the DNA unwinds. producing a replication bubble 3. DNA synthesis takes place on both strands at each end of the bubbles as the replication forks proceed outward. 4. eventually, the forks of adjacent bubbles run into eachother and the segments of DNA fuse 5. producing two identical linear DNA molecules (newly synthesized DNA) CONCLUSION: The products of eukaryotic DNA replication are two linear DNA molecules **DNA is Circular

Replication slippage

--> Replication Slippage - If loop occurs in template strand during replication, DNA polymerase misses looped out nucleotides, and small insertions and deletions occur - Replication slippage is more common in repeat sequences (hot spots) a) Hot spots for DNA mutation b) Contribute to hereditary diseases 1. Fragile-X, Huntington disease **Some spots in genome where spot slippage happens (hot spot or deletions?) **Hereditary disease where this occurs and causes a phenotype

HIV retrovirus has a dormant phase

--> Retrovirus - RNA genome - Reverse transcriptase synthesizes DNA from RNA template. - HIV is an example

10.16 DNA can assume several different secondary structures.

--> Specifics of the DNA helix can vary - Salt - Hydration - Sequence (long stretches of a particular nucleotide) --> B form is the most common in living cells **Z form twists in opposite direction = dehydrated form **We are focusing on b form = found in cells **A form: condensed, counterclockwise **B form: spread out, counterclockwise **Z form: spread out, clockwise

Moderately Repetitive DNA

--> Tandem repeats - appear one after the other in clusters at particular regions on the chromosome a.) Multicopy genes - one example beta-globin genes b.)Variable number tandem repeats (VNTRs) - Repeating DNA sequences 15 to 100 bp long - Found within and between genes c.)Microsatellites (short tandem repeats, STRs) - Dispersed throughout genome - Consist of di-, tri-, tetra-, and pentanucleotides **Vary between individuals so they can be used as markers - forensics applications, linkage mapping **Tandem means one after the other

6.) Understand why telomeres act as an age determinant.

--> Telomeres, Aging, and Cancer - Telomere length is important for chromosome stability, cell longevity, and reproductive success - Telomerase is only active in germ-line cells and some stem cells in eukaryotes - Differentiated somatic cells and cells in culture have virtually no telomerase activity; such cells have limited life spans (30 to 50 cell divisions) **Our cells cannot divide infinitely

Bacteria - Genome and Plasmids

--> The bacterial genome: - Mostly single, circular DNA molecule/chromosome - Plasmids: a) Extra chromosome, small circular DNA b) Episomes—freely replicating plasmids 1. Example: "F (fertility) factor" c) Sometime integrate into the genome (reversible) **Bacteria houses plasmids **Doubles stranded DNA **Helps plasmids shuffle through dna

4.) Most important is to understand how the chemical structure of DNA relates to function

-->5' and 3' ends have different chemistry - 3' end has an OH that can react with phosphates on NTPs to form a phosphodiester bond - This means that new NTPs are added to the 3' end (this will be a recurring theme) -->Complementary base pairs joined by hydrogen bonds - Strands can denature and anneal - Single stranded DNA can serve as a template = replication and transcription **3' - hydroxyl group **Add to 3' end; not 5'

Bacteriophage have a lifestyle choice

-->Bacteriophage: bacterial infection virus - Virulent phages reproduce through the lytic cycle and always kill the host cells. - Lysogenic phages: phage DNA integrates into bacterial chromosome where it remains as inactive prophage. Sometimes called temperate phages **Lyse the cell and spill out more offspring **Or intergrate the cell and live there

There are at least five DNA polymerases in E. coli

-->DNA Polymerase I is abundant but is not ideal for replication - Rate (600 nucleotides/min) is slower than observed for replication fork movement - Has low processivity - Its primary function is in clean-up --> DNA Polymerase III is the principal replication polymerase --> DNA Polymerases II, IV, and V are involved in DNA repair **Klenow fragment commonly used **Rna primer and extend it to make dna.... **Dna polymerase3 discovered 3rd but does most of replication and is fast **Dna polymerase1 is slow and doesn't do majority of the work bc replication forks work faster and has low processivity **Poly3 has high processing... cleans up dna

Must fit a lot of DNA in a little package

-->Human cell has about 1 meter of DNA - (3 billion base (haploid) pairs x 0.34 nanometers) - Nucleus is about 5 micrometers, so a lot of folding is required --> Human intestine is ~22 feet long --> Google says there are 40,000,000,000,000 cells in a person (40 trillion) --> 400,000,000 meters to the moon --> 2,735,884,800,000 (2.7 trillion) meters to Uranus **Every 10 base pair = 1 nm **Every 1 = 0.34 nm

Histone Methylation and Phosphorylation

-->Methylation - Enzyme is called a methyltransferase - Adds methyl groups to arginine and lysine residues in histones - Can repress or activate --> Phosphorylation - Enzyme is called a kinase - Adds phosphates groups to hydroxyl groups of amino acids serine and histidine **Each histone has lysine resude that can be monomethylated

Direct Evidence

-->Recombinant DNA technology - Segments of DNA corresponding to specific genes isolated and then manipulated 1. Mutated to change the protein 2. Expressed in a different host 3. Tagged with a fluorescent marker -->Trace a mutant phenotype to a change in the DNA sequence Green florescent protein-can isolate the gene that makes that protein and inject it in mice Florecent light injected in DNA

Continuous and Discontinuous Strand Replication

-->Two strands of double helix are antiparallel: 5'-3' and 3'-5' - DNA Pol III ONLY synthesizes 5'-3' --> Continuous DNA synthesis - Leading strand --> Discontinuous DNA synthesis - Lagging strand **Poly3 and 1 **Replication fork... rna primer and have extended chunk (polym3 extending ends)

Q7: One species has 32 chromosomes and another species has 24 chromosomes. In an allotriploid of these two species, how many chromosomes would there be?

-32 --> 16 = 1n 32 = 2n -24 --> 12 = 1n 24 = 2n 16 + 26 = 40 (3n) 32 + 12 = 44 (3n) -->44 or 40 **Allotrip is from 2 species

Q6: Heterochromatin is characterized by all of the following EXCEPT that it ____.

-contains genes that are transcribed at high levels -remains highly condensed throughout the cell cycle **A is true- inactive barr bodies (is present all over that inactive X chromosomes in female mammals) B- is not true- low levels (contains genes that are transcribed at high levels) C- true: (is present on most of the y chromosomes of male mammals) D- false: self condensed (remains highly condensed throughout the cell cycle) E True: (is present at centromeres and telomeres

Mismatch Recognition

1. - small insertion/deletion - small base: base mispair --> induced bending and strand-separation -Msh2 Domain I - Msh3 MBD (bent and strand-separated target) MIDDLE -Msh3 MBD - Steric "wedge" - stabilization of bent DNA 2. - large insertion/deletion --> recognition of intrinsically bent and strand-separated DNA - bent, strand-separated, and nucleotide-flipped target

8.25 Autopolyploidy can arise through nondisjunction in mitosis or meiosis.

1. Autoploidy through Mitosis - diploid (2n) early embryonic cell - replication - separation of sister chromatids - no cell division - autotetraploid (4n) cell 2. Autoploidy through Meiosis - diploid (2n) - replication - non-disjunction GAMETES: MEIOSIS II - Non-disjunction in meiosis I produces a 2n gamete - 2 (2n) FERTILIZATION (1n) ZYGOTES - that then fuses with a 1n gamete to produce an autotriploid - triploid (3n) **During mitosis- duplication of genetic info - 2n to 4n (all chromosomes duplicate and so you have 4 of each) **Meiosis 2, and chromosomes are diploid **Polypoid (whole genomes)

Homology Directed Repair or Non-homologous end Joining? Which to choose?

1. Cell cycle phase: Homologous recombination requires sister chromatids (limited to S and G2). - Cell Cycle Dependent homology driven repair - Difficult to perform in bulk chromatin in interphase cells 2. If Homologous Repair is NOT suppressed outside of S-G2.... Mutations will be more frequent as weak homology may be selected. - As diploids: can recover sequence off an allele but if heterozygous, the parental allele may differ. 3. Simple, DS breaks with flush ends are rapidly re-ligated since the NHEJ pathway is rapid and recruited quickly to needed sites. (homology directed repair is big and complex = slow) - "Difficult" breaks may be harder to fix and slower to re-ligate (?)

Remove primer added by primase

1. DNA nucleotides have been added to the primer by DNA polymerase III 2. DNA polymerase I replaces the RNA nucleotides of the primer with DNA nucleotides 3. After the last nucleotide of the RNA primer has been replaced, a nick remains in the sugar-phosphate backbone of the strand 4. DNA ligase seals this nick with a phosphodiester bond between the 5'-P group of the initial nucleotide added by DNA polymerase III and the 3'-OH gorup of the final nucleotide added by the DNA polymerase I **Remove primer added by primase **dna polymerase has excess dna activity 5' to 3' (opposite direction) same direction on strand but adding dna bases on 3' strand end

RNA Primer Removal and Okazaki Fragment Ligation

1. DNA pol I binds to a single-stranded gap between DNA and an RNA primer 2. Pol I removes an RNA primer nucleotide 3. and fills the gap with a DNA nucleotide 4. Pol I removes each RNA primer nucleotide ** Poly1 removing primers and putting dna in its place ** Primase put rna primer and polymerase replaces it 5. and replaces it with a DNA nucleotide 6. when primer removal is complete, DNA ligase replaces pol I at DNA-DNA single stranded gaps and 7. catalyzes formation of a phosphodiester bond to join Okazaki fragments ** Falls off and replased by dna ligase: phosphodiester bond between oz fragments

Homologous Recombination

1. DSB double-stranded ends are changed to single-stranded ends. The 3'-ending strands pairs invades the homologous chromosme to form a pair of Holliday junctions. 2. DNA synthesis and ligation to fill gaps and seal 3. Both Holliday junctions are resolved. ---- 1. Two double stranded DNA molecules from homologous chromosomes align 2. A double strand break occurs in one of the molecules 3. nucleotides are enzymatically removed, producing some single stranded DNA on each side 4. A free 3' end invades and displaces a strand of the unbroken DNA molecule 5. the 3' end then elongates, further displacing the original strand 6. the displaced strand forms a loop that base pairs with the broken DNA molecule 7. DNA synthesis is indicated at the 3' end of the bottom strand, the displaced loop being used as a template 8. Strand attachment produces two "HOLLIDAY JUNCTIONS," each of which can be separated by cleavage and reunion **Holliday junctions **Template donor- red **Ends are exposed and all have same sequence **Holliday junctions- cleave so two sepester strands of dna **No test questions over holliday junctions

1.) Define terms related to chromosomal rearrangement and copy number

1. Duplication: In a chromosome duplication, a segment/chunk of the chromosome is duplicated/doubled (creates a longer chromosome) 2. Deletion: In a chromosome deletion, a segment of the chromosome is deleted (creates a shorter chromosome) 3. Inversion: In a chromosome inversion, a segment/section of the chromosome is turned 180 degrees (flips order on chromosome) 4. Translocation: In a translocation, a segment of a chromosome moves from one chromosome to a nonhomologous chromosome, or to another place on the same chromosome (taking a piece on one chromosome and switching it out with a piece on a different chromosome; or placing that cut out section on the same chromosome but in a different location)

Eukaryotic Replication Fork

1. Leading strand -PCNA -RFC 2. DNA polymerase e 3. DNA helicase 4. DNA polymerase a -RPA 5. Lagging strand 6. PCNA -RFC - DNA polymerase d **Dna helicase to unwind **Replication fork after unwind dna a on leading strand **Dna polyA .... You then find delta and epsilon **A lot on lagging strand

8.18 Aneuploids can be produced through nondisjunction in meiosis I, meiosis II, and mitosis.

1. Nondisjunction in meiosis I --> ZYGOTES - 2 trisomic (2n +1) - 2 monosomic (2n -1) 2. Nondisjunction in meiosis II --> ZYGOTES - 1 trisomic (2n + 1) and 1 monosomic (2n - 1) - 2 normal diploids (2n) 3. Nondisjuction in mitosis --> SOMATIC clone cells --> daughter cell is monosomic (2n - 1) = 4 --> will all rest are trisomic (2n + 1) = 4 The gametes that result from meiosis with nondisjunction combine with a gamete (with blue chromosome) that results from normal meiosis to produce the zygotes. Monosomic usually results in death... lethal

Translocations (4)

1. Nonreciprocal translocation of (AB) --> nonhomologous chromosome: ABCDEFG + HIJKLM --> nonreciprocal translocation --> CDEFG + (AB)HIJKLM **piece of one chromsomes goes to the other and gets NOTHING IN EXCHANGE 2. Reciprocal translocation of (AB) and (HIJ) --> nonhomologous chromosome: ABCDEFG + HIJKLM --> reciprocal translocation --> (HIJ)CDEFG + (AB)KLM **move a piece from one to the other and vice versa 3. Robertsonian translocation --> the short arm of one acrocentric chromosome is exchanged with the long arm of another

5.) Describe how lagging strands are processed to form a fully functional strand.

1. On the lower template strand, DNA synthesis proceeds continuously in the 5' --> 3' direction, the same as that of unwinding 2. on the upper template strand, DNA synthesis begins at the fork and proceeds in the direction opposite that of unwinding; so it soon runs out of template 3. DNA synthesis starts again on the upper strand, at the fork, each time proceeding away from the fork 4. DNA synthesis on this strand is discontinuous; short fragments of DNA produced by discontinous synthesis are called "OKAZAKI FRAGMENTS" -Lagging strand: discontinuous DNA synthesis - Leading strand: continuous DNA synthesis **replication fork moves forward as DNA denatures

The End Replication Problem

1. Parental duplex <-- to centromere --> to telomere 2. Parental strand - Lagging strand (3' to 5') a) single stranded overhang left by RNA primer removal at telomere - Leading strand (5' to 3') a) RNA primer removed and replaced by DNA **Linear chromosome = leading strand **Lagging you have RNA primer and extended which makes it become shorter and shorter

8.3 Chromosome Mutations Include: Rearrangements, Aneuploids, and Polyploids

1. Rearrangements: (duplication) 2n = 6 2. Aneuploids: (trisomy) 2n + 1 = 7 Aneuploidy- Gain or loss of any individual chromosomes 3. Polyploids: (autotriploid) 3n = 9 Polyploidy- Gain or loss of one chromosomes

Fork Diagram: Enzymes (Show picture and match enzyme on fork or define enzyme *** for test)

1. Supercoiled DNA is relaxed by DNA topoisomerase (blue loop) 2. Helicase (DnaB protein) breaks hydrogen bonds to unwind DNA strands (purple circular ring) 3. SSB coats single-stranded DNA to prevent reannealing (pink beads) 4. Primase synthesizes RNA primers on leading and lagging strands (red grape) 5. DNA polymerase III synthesizes DNA 5' --> 3' continuously along leading strand and discontinuously along lagging strand (green kidney bean) 6. DNA polymerase I removes nucleotides of RNA primers, replacing them with DNA nucleotides (yellow square) 7. DNA ligase catalyzes phosphodiester bonds to join DNA segments (blue circle) (LEFT)- leading strand 3' --> 5' (RIGHT)- lagging strand 3' --> 5' Okazaki fragment- 5' --> 3' PROTEIN/ROLE - DNA topoisomerase: relaxes supercoiling - Helicase (DnaB): unwinds the double helix - SSB: prevents reannealing of separated strands - Primase: synthesizes RNA primers - DNA pol III: synthesizes DNA - DNA pol I: removes and replaces RNA primer with DNA - DNA ligase: joins DNA segments

Trombone Model

1. The lagging strand loops around so that 5' --> 3' synthesis can take place on both antiparallel strands 2. As the lagging strand unit of DNA polymerase III comes up against the end of the previously synthesized Okazaki fragment with the first primer 3. the polymerase must release the template and shift to a new position farther along the template (at the third primer) to resume synthesis CONCLUSION: DNA must form a loop so that both strands can replicate simultaneously **Leading strand **Lagging strand making dna in same direction but need to flip to feed into polymerase? **Poly extending primer making loop get fed through **End result is two dna molecules???

Base Changes - deamination

1. cytosine loses amine (NH2) - C=O connecting uracil and adenine 2. adenine loses amine and leads to wrong base pairing reaction - C=O connecting hypoxanthine and cytosine

Agarose gel electrophoresis separates DNA based on size, charge and shape

1. each sample is loaded into a well in the gel 2. fragments of DNA move TOWARD the POSITIVE electrode 3. SMALLER fragments move FASTER (and therefore FARTHER) than larger fragments --> SUPERCOILED DNA runs faster on a gel then circular or linear DNA -supercoiled runs quickest through the gel and appears thicker and brighter - load only full-length plasmid DNA

8.9 In an individual heterozygous for a deletion, the normal chromosome loops out during chromosome pairing in prophase I.

1. the heterozygote has one normal chromosome --> ABCDEFG 2. and one chromosome with a deletion --> ABCDG --> formation of DELETION LOOP during pairing of HOMOLOGUES in PROPHASE I 3. in prophase I, the normal chromosome must loop out in order for the homologous sequences of the chromosomes to align (loop appears as big butt)

8.12 In an individual heterozygous for a PARAcentric inversion, the chromosomes form an inversion loop during pairing in prophase I.

1. the heterozygous has one normal chromosome --> ABCDEFG 2. and one chromosome with an inverted segment --> AB(EDC)FG 3. in prophase I of meiosis, the chromosomes form an inversion loop, which allows the homologues sequences to align --> formation of inversion loop

8.15 In a Robertsonian translocation, the short arm of one acrocentric chromosome is exchanged with the long arm of another. Acrocentric- where centromere is near one creating a short arm Bigger metacentric chromosomes and a short fragment chromosome lost when the cell divides

1. the short arm of one acrocentric chromosome.. 2. is exchanged with the long arm of another --> Robertsonian translocation 3. creating a large metacentric chromosome 4. and a small fragment that often fails to segregate and is lost **Acrocentric- where centromere is near one creating a short arm **Bigger metacentric chromosomes and a short fragment chromosome lost when the cell divides

Q8: If a piece of chromatin contained 200 copies of the histone H4, then how many nucleosomes would be present?

100 **For every 2 copies- 1 nucleosome = 100

Strains of influenza virus responsible for major flu pandemics

1918- Spanish flu: H1NI 1957- Asian flu: H2N2 1968- Hong Kong flu: H3N2 2009- Swine flu: H1N1

Q5: A diploid organism has 2n = 36 chromosomes. How many chromosomes will be found in a trisomic member of this species?

2n+1 = 37 **Trisomic- one more (down syndrome) so, 37

8.13 In a heterozygous individual, a single crossover within a PARAcentric inversion leads to abnormal gametes.

3. in prophase 1, an inversion loop forms 4. a single crossover within the inverted region --> crossing over within inversion 5. results in an unusual structure 6. one of the four chromatids now has two centromeres 7. and one lacks a centromere --> Anaphase 1 8. in anaphase I, the centromeres separate, stretching the dicentric chromatid, which breaks. The chromosome lacking a centromere is lost. **Diacentric bridge - not well for anaphase when they try to pull apart --> Anaphase II 9. Two gametes contain NONRECOMBINANT chromosomes: one wild type (normal) and one with inversions 10. The other two contain RECOMBINANT chromosomes that are missing some genes; these gametes will not produce viable offspring CONCLUSION: The resulting recombinant gametes are nonviable because they are missing some genes.

Q3: In humans, 20% of bases in DNA are cytosine (C). What percentage of the bases are expected to be thymine (T)?

30% if 20% are Cytocine then 20% are Guanine =40% and 30% are thymine with 30% being adenine =60%

Q4: How many nanometers long is a 1000 base pair strand of DNA?

340 10 base pairs = 3.4 nanometers 1000 b.p. = 340 nanometers

DNA Nucleotides

A DNA nucleotide is composed of a sugar, one of four nitrogenous bases, and up to three phosphate groups Deoxyribose is the sugar of DNA nucleotides; it has five carbons, identified as 1', 2', 3', 4' and 5' A nucleotide base is attached to the 1' carbon, an OH (hydroxyl) group is attached to the 3' carbon, and a phosphate is attached to the 5' carbon **Deoxyribose sugar attached to 1' carbon **5' add a phosphate

The Avy allele and the spectrum of Avy phenotypes

Avy from transposon? - brought promotor with which turns mouse (yellow); - if promotor off- you get full aguti mouse (brown) Midgestation and germ-line effects of methyl donor supplementation. (A) F1 phenotypes. Shown is the effect of methyl donor supplementation on mid-gestation supplemented offspring. Dams were supplemented with methyl donors from E8.5 to E15.5, and offspring phenotypes were scored as in Fig. 1B. Results are expressed as the percentage of supplemented offspring with the same phenotype as unsupplemented controls, and the percentage of offspring of each coat color is shown beneath each graph. (B) F2 phenotypes. Shown is the heritable/grandparental effect of methyl donor supplementation. Phenotypes of F2 offspring from pseudoagouti female mice shown in A are expressed as a percentage of control offspring (offspring from pseudoagouti dams that had never been supplemented) with the same phenotype. (C) A schematic diagram illustrating the effect of methyl donor supplementation in the germ line. Epigenetic changes to Avy in primordial germ cells exposed to methyl donors during differentiation (Left) are maintained throughout gametogenesis and embryogenesis. Thus, pseudoagouti F1 mice that are genetically and phenotypically identical but were subject to different diets in utero (Left vs. Right), can produce phenotypically different F2 offspring.

Q4: Which of the following types of DNA repair would involve the enzyme AP enonuclease?

Base-excision repair

Bacterial Chromosome organization

Bacterial chromosomes - Circular, double-stranded DNA compacted into nucleoid **Ours are linear **Bacteria don't have nucleus- they have NUCLEOID - Their DNA is associated with HU and H-NS DNA-binding proteins (Histone-like Nucleoid Structuring Protein) **Structure depends on HU and HNS **Histone-like means like histone but DO NOT HAVE HISTONES

Q8: Briefly explain why sex-chromosome aneuploids are more common than autosomal aneuploids in humans and mammals.

Barr bodies

Table 12.5 DNA polymerases in eukaryotic cells

CELL FUNCTION a (alpha): inititiation of nuclear DNA synthesis and DNA repair; has primase activity d (delta): lagging strand synthesis of nuclear DNA, DNA repair, and translesion DNA synthesis e (epsilon): leading strand synthesis **All 5' to 3' polymerase activity ** a (alpha) has no 3' to 5' exonuclease activity

11.8 Variation in DNA methylation at the agouti locus produces different coat colors in mice.

Coat color in mice - Auguti: turns protein on and off... 5th mouse (dark brown) - 1st= yellow allele Avy - 3 middle intermediate- (speckeled) genetically, auguti can be same as yellow mouse AND female auguti will look more likely here and yellow would be the same

Basis of Complementary Pairing

Complementary base pairing combines one purine with one pyrimidine The chemical basis of the pairing is the formation of stable hydrogen (H) bonds between the bases on the antiparallel strands Two H bonds form between A and T; three H bonds form between G and C **Complementary pairs between bonds G --> C

Q1: After the first round of replication, Meselson and Stahl saw only one DNA band of density intermediate to DNA containing only 15N or 14N. After this observation, which hypothesis for DNA replication could be eliminated?

Conservative **Eliminates conservative because that would end in two bands.... One old and one new

Overview of the Repair Process

DNA gets damaged by chance or an external agent 1) Damage is identified 2) A repair device is mobilized: recruit repair device (enzymes) so it can be fixed.. cut out bit thats broken and place correct info 3) The damaged strand is cut out and the undamaged strand is used to provide information that will allow the right thing to be put in its place 4) Everything is sealed up and damage is repaired

DNA Damage Response

DNA repair** **Errors in replication- spontaneous **Viral infection, radiation, chemical exposure, metabolism **Cell needs to resolve in some way **Apoptosis- fakes its own death Why do we care? - genetic disorders - cancer 1. DNA damage - DNA repair - genetic stability 2. DNA damage - defective/incomplete DNA repair - genetic instability a) cancer, hereditary disease b) genetic divergence **we care because ability to repair DNA keeps us alive

23.12 A reciprocal translocation between chromosomes 8 and 14 causes Burkitt lymphoma.

Example of disease-causing reciprocal translocation Burkitt lymphoma - Cancer of B cells, which are the lymphocytes that make antibodies - c-MYC stimulates cell division - Lymphocytes highly express immunoglobin genes (because that's what antibodies are made of) - Translocation puts c-MYC behind the immunoglobin promotor - High expression of c-MYC drives out of control cell division **Lymphoma- cancer of b cells **Reciprical translocation **8- c-MYC gene (important for cell division) **14- immunoglobin gene (antibodies- produces on this gene)

8.21 The translocation of chromosome 21 onto another chromosome results in familial Down syndrome.

Familial Down syndrome - Robertson translocation causes trisomy of 21 even though the individual has only 45 chromosomes Long arm of chromosome 21 attached to chromosome 14 Remaining short arm was lost This individual is a carrier Here, the long arm of chromosome 21 is attached to chromosome 15. This karyotype is from a translocation carrier, who is phenotypically normal but is at increased risk for producing children with Down syndrome. Can be inherited- 21 is acrocentric.... Long arm attaches to chromosome 14..... Karyotype, instead of 46 you have 45... 21 attaches to 14 **NORMAL phenotype but carrier of downsyndrome

Several ways to transfer DNA between bacteria

Gene Transfer in Bacteria - Conjugation: direct transfer of DNA from one bacterium to another - Transformation: bacterium takes up free DNA from environment - Transduction: bacterial viruses take DNA from one bacterium cell to another

Gene Duplication

Gene duplication may play a role in evolution - Gene duplication hypothesized to be major source of NEW GENES - Hypothesis supported by discovery of genes with substantial amount of DNA sequence in common, but distinct gene products - Examples: Genes encoding digestive enzymes trypsin and chymotrypsin Family of genes encoding actin **Actin is a skeletal protein Duplication = opportunity for evolution

Q2: Which is NOT a way that deletions can cause abnormal phenotypes?

Haploinsufficient genes can result because too much gene product is produced

DSB Repair promotes gene editing

Homology directed repair CRSPR-Cas9 Molecule cuts dna and harnessed to express Unique- complexes with sgRNA Use to break a gene of choosing and if repaired- you get deletions of Indel (frameshift) - DSB/HDR- precise gene editing? - Cas9/NHEJ- Indel mutation and premature stop codon

8.22 Translocation carriers are at increased risk for producing children with Down syndrome.

Inheritance pattern of familial Down syndrome 1. A normal parent who is a carrier for a 14-22 translocation has a normal phenotype 2. gametogenesis produces gametes in these possible chromosome combinations. --> GAMETES - 14-21 and 21, 14 - 14-21, 21 and 14 - 14-21, 14 and 21 3. If a normal person mates with a translocation carrier... 4. 2/3 of their offspring will have a normal phenotype-- even the translocation carriers 5. but 1/3 will have Down Syndrome 6. other chromosomal combinations result in spontaneously aborted embryos **Nonviable fetus- cant just have one chromosome of 21 **2/3 of live births normal **Inheritance pattern of normal downsyndrome

Q3: The expression of some genes is dependent upon their chromosomal position. Which kind of chromosome rearrangement would most likely cause a position-effect phenotype?

Inversion

11. 7 DNase I sensitivity is correlated with the transcription of globin genes in chick embryos.

METHOD: - DNA sensitivity to DNase I was tested on different tissues and at different times in development - The U gene encodes embryonic hemoglobin; the αD and αa genes encode adult hemoglobin. The red arrows indicate sites of DNase I digestion. 1.) Chicken DNA RESULTS: 2.) Erythroblasts (first 24 hrs) - before hemoglobin synthesis, none of the globin genes are sesnitive to DNase I digestion 3.) Erythroblasts (5 days) - after globin synthesis has begun, all genes are sensitive DNase, but the EMBRYONIC GLOBIN GENE (U) is the MOST sensitive 4. Erythroblasts (14 days) - in the 14-day-old embry, when only adult hemoglobin is expressed, ADULT GENES are MOST sensitive and the embryonic gene is insensitive. 5. Brain cells throughout development - globin genes in the brain-- which does not produce globin-- remain insensitive throughout development CONCLUSION: Sensitivity of DNA to digestion by DNase I is coorelated with gene expression, suggesting that chromatin structure changes in the course of transcription **Physical state of chromosome changes- how easily can it be cut **Chicken dna- globin gene, U is embryotic and then adult regions later on in genes switch **3- cleave embryotic **4- change confirmation in adult 5- cant make either

Q6: Trisomy and monosomy can be a product of nondisjunction in ____.

Meiosis I, Meiosis II, Mitosis

Q2: Which statement is true regarding negative supercoiled DNA?

Negative supercoiled DNA is under-rotated and allows for easier strand separation during replication and transcription

18.12 Insertions and deletions may result from strand slippage.

Newly synthesized strand: - 5' TACGGACTGAAAA 3' Template strand: - 3' ATGCCTGACTTTTTGCGAAG 5' 1. newly synthesized strand loops out - 5' TACGGACTGAA(A)A 3' - 3' TGCCTGACTTTTTGCGAA 5' 2. resulting in the ADDITION of one nucleotide on the new strand - 5' ACGGACTGAA(A)AAACGCTT 3' - 3' TGCCTGACTTTTTGCGAA 5' (or) 3. template strand loops out - 5' TACGGACTGAAAA 3' - 3' TGCCTGACTT(T)TTGCGAA 5' 4. resulting in the OMMISSION of one nucleotide on the new strand - 5' ACGGACTGAAAACGCTT 3' - 3' TGCCTGACTT(T)TTGCGAA 5' **addition or omission slippage

How polyploidy can arise in animal cells

Normal physiology - cell fusion: 2 normal cells and join together in same cell membrane 4n (2n + 2n) fusion DNA Duplication: - Changes in dna during mitosis- endoreplication (4n) - Errors during mitosis where intentional - single nucleus - mitotic slippage (4n) - Failure in cytokinesis (2n + 2n)

Types of Anuploidy

Nullisomy: loss of both members of a homologous pair of chromosomes; 2n − 2 - no copies Monosomy: loss of a single chromosome; 2n − 1 - 1 copy Trisomy: gain of a single chromosome; 2n + 1 - 3 copies Tetrasomy: gain of two homologous chromosomes; 2n + 2 - 4 copies

8.23 The incidence of primary Down syndrome and other aneuploids increases with maternal age.

Older mothers are more likely to give birth to a child with down syndrome (1 in 12: age 50) than are younger mothers (1 in 2000: age 20) **Eggs are formed and stages in meiosis **As females age, eggs age

Q1: What is the outcome of a Robertsonian translocation?

One metacentric chromosome and one chromosome with two very short arms

5.) Describe an example of polyploidy seen in healthy tissue

Polyploid cells in mammals --> physiological conditions - bone marrow - liver - heart - placenta --> pathological condtions - colon cancer - esophageal cancer - human papiloma virus (hpv) - vascular smooth muscle cell **Any given cell will have more than 2n DNA

Polyploidy

Polyploidy Is the Presence of More Than Two Sets of Chromosomes Autopolyploidy: - From single species Allopolyploidy - From two species The significance of polyploidy: - Increase in cell size - Larger plant attributes - Evolution: may give rise to new species - Many crop plants are polyploid

Meselson-Stahl Experiment

QUESTION: Which model of DNA replication-- conservative, dispersive, or semiconservative-- applies to E. coli? METHOD: - (a) 15N medium (spin) - (b) transferred to 14N; one round of replication (spin) - (c) second round of replication in 14N medium (spin) - (d) additional rounds of replication in 14N medium (spin) RESULTS: 1. DNA from bacteria that had been grown on medium containing 15N appeared as a single band (heavy 15N- bottom) 2. after one round of replication, the DNA appeared as a single band at intermediate weight (middle) 3. after second round of replication, DNA appeared as two bands, one light and the other intermediate in weight (middle and light 14N- top) 4. samples taken after additional rounds of replication appeared as two bands, as in part c (middle and light 14N- top) CONCLUSION: DNA replication in E. coli is SEMICONSERVATIVE (most will be in light band)

2.) Diagram chromosomal rearrangement and explain potential impacts on phenotype

The Bar phenotype in Drosophila melanogaster results from an X-linked duplication. (a) Wild-type fruit flies have normal-size eyes B+B+ --> 1 barr region/ 1 (b) Flies heterozygous bar B+B --> 1 barr region/ 2 (c) homozygous for the bar mutation have smaller, bar-shaped eyes BB --> 2 barr region/ 2 (d) Flies with double Bar have three copies of the duplication and much smaller bar-shaped eyes B+BD --> 1 barr region/ 3 **As you duplicate the region you change the phenotype

Q1: You are conducting an experiment where you must denature a strand of DNA that is very G-C rich. How would the melting temperature ™ of your DNA sample differ from that of a sample that is very A-T rich?

The Tm of your sample would be greater than that of a sample that is very A-T rich

Q3: Which phrase describes the first step in the lytic cycle of bacteriophage?

The phage attaches to the host cell and prepares to insert DNA **Phage has to attach to the cell **Then dna has to go inside the cell

Q1: Which statement is true of plasmids?

They replicate independently of the bacterial chromosome

Q4: HIV is a retrovirus. How are retroviruses unique from other types of viruses?

They use reverse transcription to make DNA from RNA **most rapid changes by shift

Chromosomal mutations are common in cancer cells

Why can tumor cells live with extra chromosomes? - Huge variety of rearrangement - Differences in number... extra which is aneuploidy - Pieces can be deleted or added

Q3: When depurination of a nucleotide occurs, an apurinic site is created. A nucleotide with a/an _____ base is typically inserted across from the apurinic site during DNA replication

adenine

Q5: Influenza antigenicity can change rapidly by

antigenic shift

Q7: Genes that are transcriptionally active will ___.

be located in regions of relaxed chromatin **More sensitive.. B is true False- less active so inactivated- less methylated DNA

Q6: A eukaryotic cell containing defective DNA polymerase (a) would have difficulty performing which of the following?

laying down an RNA primer to start DNA replication

Q6: All of the following DNA-repair mechanisms use an undamaged complementary DNA strand as a template for replacing the excised nucleotides except for ______.

nonhomologous end joining

Deoxyribonucleotides

ribose sugar + base - ribose: 2 - OH groups - 2-deoxyribose: OH and H group Pyrimidine ring (1 structure) - cytosine (NH2 at top) - uracil (double O; H left/right) - thymine (double O; H3C left) Purine ring (2 structures) - guanine (double O; NH2 bottom R) - adenine (NH2 at top) **Deoxy is has no OH at the 2' **Chargoff found that A and T were present in same amounts in DNA, as was G and C. **Pyrimidines pair with purines.