Genetics test 1 sudying new PDF

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Genetics test 1 studying: CHAPTER 10 – DNA structure and analysis Properties of genetic material  Replication- the processes of mitosis and meiosis. Complementary base pairing allows replication of DNA and RNA.  Storage of information- storing complete DNA component vs. gene expression. Polynucleotide chains explain the ability of DNA to store large amounts of information. A polynucleotide of X nucleotides can be arranged in 4 x ways.  Expression of stored information- the central dogma of molecular genetics  Variation through mutations. Levene’s tretranucleotide hypothesis.

Griffiths transformation of bacteria:  Griffeith tested Streptococcus pneumoniae. He inserted living virulent (encapsulated- smooth colonies) strains a mouse and it died but when he put living avirulent (noncapsulated- rough colonies) strains into a mouse the mouse surived. When he inserted heat killed virulent strains into the mouse it survived but when he inserted the avirulent strain and heat-killed virulent strains into the mouse it died. This is due to transformation.

Avery et al. proved DNA transformed. Used IIIS (Virulent) and IIR (Avirulent) cells.

Hershey-Chase Experiment:  Hershey-Chase provided more support for the hypothesis that DNA is the genetic carrier – phage T2 experiments. Used Bateria phage T2 and E.coli. Labelled the Phosphorus ( ) in DNA in one experiment and labelled the Sulfur ( ) in protein in another experiment. Knew: -DNA contains phosphorus and protein contains sulfur; 50% DNA and 50% protein; Phage injects genetic material into host to replicate.

The process of infection by only the viral nucleic acid = Transfection. This proves that DNA alone contains all necessary information for production of mature virus. Life cycle of T-even phage = phage attaches to bacteria wall, injects genetic material into bacterium, phage reproduction in bacterium happens and bacteria lysis occurs and new phages are released. Eukaryotes— Initial indirect indications = DNA distribution, amount and levels of ploidy, UV mutagenesis (action vs absorption spectra). Later direct proof = recombinant DNA technology- expression of human DNA fragments (genes) in bacteria, e.g. expression of human insulin gene in E.coli.

RNA as genetic material=> (proof found in viral studies) Tobacco mosaic virus (TMV) – viral RNA spread on tobacco leaves cause symptoms characteristic of disease. FraenkelConrat’s TMV hybrid viruses – hybrid viruses constructed by using the protein coat of one TMV strain to cover the RNA of another strain. Symptoms on leaves always corresponded to the strain that contributed the RNA component of the hybrid. Phage QB- in vitro replication with RNA replicase. Retroviruses: RNA serves as a template for RNA-dependant DNA polymerase. Central dogma.

The chemical composition of nucleic acids: Nucleotides are building blocks of nucleic acids and consists of 3 components: nitrogen base, pentose sugar and a phosphate group. There are 2 types of nitrogen bases- Purines (double ringed) or Pyrimidines (single ringed). Purines are Adenine and Guanine. Pyrimidines are Cytosine, Thymine (DNA) and Uracil (RNA). The pentose sugar for DNA is deoxyribose and for RNA is ribose. Nucleoside = nitrogen base + pentose sugar Depending on the amount of phosphate group that are attached to nucleoside, the molecule is called a nucleoside- monophosphate (NMP), diphosphate (NDP) or triphosphate (NTP). Polynucleotide= 2 nucleotides are joined to each other by a phosphate group linking the two sugars (phosphodiester bond). The joining of 2 nucleotides forms a dinucleotide, when 3 are joined it’s a trinucleotide, less than 20 is a oligonucleotide and more than 20 is a polynucleotide. DNA structure (Chargaff et al.) Base compositions studies (the arrangement of the bases & sugar helps with stability) showed: A=T and G=C; (A+G) = (C+T) i.e. purines = pyrimidines. Watson and Crick model: (Rosalind Franklin)  DNA consists of 2 right handed helical polynucleotide chains, spiralling around a central axis.  The 2 chains are antiparallel (C-5’ to C-3’ run in opposite directions).  Nitrogen bases are paired by hydrogen bonds (A+T and C+G)  Nitrogen bases are flat structures, stacked perpendicularly to the axis and separated from each other by a distance of 3.4Å (1nm = 10Å).  Each complete turn of the helix is 34Å long and contains 10 bases.  Double helix has a diameter of 20Å.  Larger “major grooves” alternate with smaller “minor grooves” throughout the total length of the DNA molecule. There are different types of DNA (from A-DNA to E-DNA as well as Z-DNA) B-DNA is the most important. RNA structure Ribose instead of deoxyribose. Uracil instead of Thymine. It is a single stranded, linear (but not actually as the bases are hydrophobic and so it fold in on its self) molecule. There are 3 types of RNA: ribosomal, messenger and transfer. Analysis of nucleic acid:  Absorption of UV light – nucleic acids absorb UV light within a range of 254-260nm. This can be used to calculate the approximate concentration of nucleic acids in a soln.  Sedimentation behaviour – use density gradient centrifugation to separate nucleic acids of different densities in gradient (e.g CsCl). Sedimentation equilibrium centrifugation and sedimentation velocity centrifugation .  De- and renaturation of nucleic acids – denaturion induced by heat or chemical treatment. Hyperchromic shift increase in UV absorption and buoyancy of DNA melted by heating. Used to determine Tm.Melting profiles are GC dependent.  Molecular hybridisation – based on renaturation (reassociation) of complementary single strands of nucleic acids. Most powerful analytic techniques in molecular biology. Used for Fish, PCR, microarrays, northern blotting and southern blotting.  Gel electrophoresis – separation of different-sized fragments of DNA in a molecular matrix under influence of an electric field. DNA, RNA and proteins are negatively charged and would move towards the anode. Matrices made of agarose of polyacrylamide that separates bases. It is very specific.

CHAPTER 11- DNA replication and recombination

DNA replication is an essential function of genetic material. It has an error rate of

10−6 .

Watson & Crick model predicts that genetic continuity between parental and progeny cells is maintained by semiconservative replication. Mode of DNA replication— each DNA strand is a potential template. Acenylic acids attract thymidylic acids, cytidylic acids attract guanidylic acids. Each replicated DNA molecule consists of one old and one new string = Semiconservative replication. 2 other possible mechanisms are conservative and dispersive replications. Conservative DNA helix unwinds and rewinds conserving the original helix and forming a new helix with 2 new strands binding. Dispersive each strand consist of both old and new DNA. Involves cleavage of parental strands during replication.

Meselson-Stahl experiment: Proved semiconservative replication is used by bacterial cells to produce new DNA molecules. (lighter isotope of Nitrogen) cells were In the experiment- labelled 15 ❑N transferred to a medium containing only 14❑N H 4 Cl . All new synthesis of DNA had 14❑N . It was allowed to replicate over several generations with a sample removed after each replication. DNA was isolated by sedimentation equilibrium centrifugation. After the 1st generation semiconservative band and replication occurred. The 2nd generation showed one pure 14 ❑N one intermediate. After the 3rd generation there was an increase in the 14 band. They then denatured and separated the DNA sample strands ❑N by heating them. Once tested they found that each strand was either 14❑N or and dispersive replication.

15 ❑

N . This ruled out conservative

Taylor et al. showed that semiconservative replication occurred in eukaryotes when using Broad bean Vivia faba. The sister chromatids were labelled with ❑3 H -thymidine. They used autoradiography to monitor replication. (Autoradiography is a common technique that pinpoints the location of a radioisotope in a cell). An unlabelled chromosome proceeds through the cell cycle in the presence of ❑3 H -thymidine. As it enters mitosis, both sister chromatids of the chromosome are labelled by autoradiography. After a second round of replication in the absence of ❑3 H -thymidine. Only one chromatid of each chromosome is expected to be surrounded by grains. Except where a reciprocal exchange occurred between sister chromatids.

Origin of replication: Unidirection replication (single direction away from origin) occurs when DNA strands unwind to form a replication fork (@any point where replication is occurring the strands of the helix are unwound). The fork then moves along the DNA strand as replication proceeds. Bidirectional replication occurs when 2 replication forks are present migrating in opposite directions away from the origin of replication. The length of DNA that is replicated following one initiation event at a single origin is called a replicon.  For most prokaryotes (where only a single circular chromosome is present) there is only one origin of replication, therefore only 1 replicon. In E.coli the point of origin is called oriC. Replication starts at oriC and proceeds bidirectionally with 2 replication forks. It terminates at a termination region (ter in E.coli).  In Eukaryotes replication also proceeds bidirectionally, but from multiple points of origin (therefore having multiple replicons) with various replication forks migrating along the chromosome. Eventually all forks merge.

DNA synthesis in microorganisms  Kornberg et al. Isolated an DNA polymerase I from E.coli that had the ability to synthesize DNA in a cell-free (in vitro) system.  3 major requirements for invitro DNA synthesis to be successful: all 4 deoxyribonucleoside triphosphates (dATP, dGTP, dCTP, dTTP = dNTP) a primer and template DNA need to be there.  During chain elongation (5’ to 3’ direction) 2 terminal phosphates from 5’-C of each dNTP are removed by DNA polymerase I. The remaining phosphate then binds to the 3’-OH group of the last dNMP in the chain.  Kornberg et al. proved that DNA polymerase 1 is capable of synthesizing the complete DNA of a small phage in vitro, and that this DNA could be used to successfully transfect E.coli protoplasts. There was a mutant strain without DNA Polymerase I and it was still able to replicate but was sensitive to DNA damage by UV light. Therefore DNA polymerase I is not needed for replication but stops damage to DNA. -> At least 1 more enzymes that is capable of DNA synthesis in vivo must be present in E.coli. DNA polymerase I may serve only a secondary function in DNA synthesis. Roles of the 5 polymerases in vivo:  DNA polymerase III – responsible for DNA synthesis during replication. 3’-5’ exonuclease activity (backwards) serves as a proofreading mechanism. Auto-correct function that removes last nucleotides.  DNA polymerase I – removes RNA primers and fills the gaps left after their removal. Has 5’-3’ (can remove nucleotides in front of it) exonuclease activity serves as a proofreading mechanism here as well.  DNA polymerase II, IV and V – probably involved in repair of DNA that has been damaged by external forces

 The DNA Pol III holoenzyme – DNA Pol III is a very large and complex enzyme; the active form (holoenzyme) is a dimer consisting of 10 polypeptide subunits, each with specific function during DNA synthesis; comprise of core enzyme (2X), sliding clamp loader and sliding clamp; the holoenzyme + other proteins at replication fork known as replisome (analogous with ribosome).

Critical steps for the successful replication of DNA 1. A mechanism must exist by which the helix can be unwound locally to form a replication bubble. It must stay open for replication to occur on both strands. 2. As the helix unwinds, tension develops further down the strand. This tension must be reduced. 3. A primer must be synthesized for DNA polymerase III to be able to commence with the polymerisation process. Polymerase can’t initiate de novo synthesis. 4. DNA replication is always performed in 5’->3’ direction and the strands of the parental molecule are orientated antiparallel to each other, continuous synthesis of complementary DNA strands by DNA polymerase III is possible on one strand (leading) while the synthesis is discontinuous on the other strand (lagging). 5. Primers must be removed, and gaps left filled up before replication is over. 6. The DNA synthesized to fill the gaps caused by primer removal, must be ligated to (join) adjacent DNA strands. 7. Even though DNA polymerases are very accurate, occasional errors are made- a proofreading mechanism must exist to correct errors.

Unwinding the helix:  oriC has 245 base pairs. It’s characterised by repetitive sequences of 9 (X5) and 13 (X3) bases and are AT-rich.  A protein DnaA (gene- dnaA), initiates unwinding of the helix by first binding 9mer and then 13mer regions. DnaB and DnaC then bind helix as well to continue unwinding and destabilizing it.  The proteins that require energy by hydrolysis of ATP in order to break hydrogen bonds and denature double bonds are called helicases.  Once the helix has been been unwound, single-stranded binding proteins (SSBPs) bind to the complex and stabilize it.  As unwinding proceeds, tension is created ahead of the replication fork- the result is supercoiling of the helix.  Tension is relieved by the action of gyrase which belongs to the family of proteins called topoisomerase. Initiation of DNA synthesis  Initiation occurs as soon as a small portion of the helix is unwound.  DNA polymerase III requires the free 3’ end of a primer (3’OH) to start elongation of a polynucleotide chain.  RNA primers (5-15 bases) are synthesized on the DNA template (parental strand) by an RNA polymerase called primase (de novo; template dependant).  Elongation is performed by DNA polymerase III.  Synthesis continues until another RNA primer is encountered. DNA polymerase I (5’-3’ exonuclease activity) removes RNA primers and fills the gap with dNTP’s complementary to the parental strand. Continuous and discontinuous synthesis:  DNA strands are unwound and the replication fork progresses down the helix. DNA strands run antiparallel. One strand runs 5’-3’ and the other runs 3’-5’. DNA polymerase III synthesises DNA only in 5’3’ direction. The DNA synthesis on replication fork occurs in different directions on 2 strands.  Only one strand (leading strand) can serve as a template for continuous strand synthesis.  Various initiation points are needed on the other strand (lagging strand) –DNA synthesis is therefore discontinuous on this strand.  Discontinuous DNA fragments (1000-2000 nucleotides) are called Okazaki fragments.  Okazaki fragments are joined together later by an enzyme – ligase. Concurrent synthesis on the leading and lagging strand  DNA polymerase III holoenzyme is a dimer. Concurrent synthesis at the replication fork is achieved by forming a loop in the lagging strand. The orientation of the lagging strand is thus inverted – the biochemical direction is unaffected however. Each parental strand can now be replicated by one of the monomers of the DNA polymerase III holoenzyme. Β-subunit clamp keeps the core enzyme in position. Proofreading sometimes non-complementary nucleotides are incorporated. Compensation is made for inaccurate insertions by providing both DNA polymerase I and III with 3’-5’ exonuclease activity. This provides them with the ability to detect and correct mismatched nucleotides. Proof reading is done by epsilon (ε) subunit. DNA synthesis in Eukaryotes: Similar to synthesis in prokaryotes— requires all 4 dNTPs, template and primer. Replication forks move bidirectionally. There is semi-conservative replication. Controlled by DNA Polymerases. Functions of helicases and same sequence base pairs are the same. RNA primers are also present and Okazaki fragments too.

BUT different: due to larger size (more DNA) and complexity (with nucleosomes- histones) andlinear chromosomes of the eukaryotic genome, the process is much more complex and slower (rep speed went from 100 to 5kb/nmin).  Multiple origins of replication and 14 different polymerases are found in eukaryotes: α- responsible for initiation of DNA synthesis during rep & RNA/DNA primer synthesis β- responsible for base-excision DNA repair. γ- DNA synthesis and repair during replication in mitochondria (has 3’-5’ exonuclease activity) δ- lagging strand synthesis, DNA repair, proofreading (high processivity; 3’-5’ exonuclease activity) ε- leading strand synthesis, proofreading, 3’-5’ exonuclease ζ- translesion DNA synthesis (of damaged regions) Translesion synthesis (TLS) is a DNA damage tolerance process that allows the DNA replication machinery to replicate past DNA lesions such as thymine dimers or AP sites. It involves switching out regular DNA polymerases for specialized translesion polymerases

DNA synthesis at the ends of linear chromosomes:

 Eukaryote chromosomes are linear. A problem arises when the last RNA primer of the lagging strand is removed. There is now no free 3’-OH group for the polymerase to attach a new nucleotide to and this gap can’t be filled. It is solved by telomers as they preserve structure and integrity of chromosome ends.  The 3’ end of a eukaryote leading chromosomes strand contain short tandem repeats that are G and T rich (pic). They are highly conserved and the G-rich sequences are capable of bonding to each other forming G-quartets and T-loops. This closes off the ends of the chromosomes  Telomerase has dual functions. It stabilises chromosome ends by forming a hairpinloop of G-G bonds.  Ribonucleoprotein facilitate replication at telomeres.  (a & b) Contains segments (TERC) of RNA complementary to TTGGGG which binds 3’ G-rich tail (TTGGGG) and segments(TERT) that synthesizes more TTGGGG repeats by reverse transcription.  (c) Telomerase translocated and repeats the process extending the chromosome end.  (d) Telomerase dissociates. Primase and DNA polymerase fill the gaps.  (e) The primer is removed and DNA ligase repairs the phosphodiester bonds.

Cell recombination recombination occurs between DNA strands with high sequence homology in these steps: (reliant on DNA complementarity and enzymatic processes) (a-b) Single-stranded nick made in 2 identical positions of 2 homologous DNA duplexes by endonuclease (b-c) The loose ends of strands created in this manner are displaced and subsequently pair to their complementary sequences on the opposite duplex. Strand displacement results in a cross-bridged configuration. (c-d) A ligase then joins the loose ends to their neighbouring nucleotide polymers, thus creating hybrid heteroduplex DNA molecules. (d-e-f) These cross-bridges move down the chromosome by branch migration resulting in an increased length of heteroduplex DNA on both homologs.

(f-g) Separation of duplexes is initiated by a 180° rotation of one half of the duplex, causing an intermediate planar structure called a die chi-form/ holliday structure.

(g-h) The 2 homologous strands that were previously uninvolved are nicked by endonuclease.

(h-i) Ligation completes process of recombinant duplex formation.

Rec A protein—promotes exchange of reciprocal single stranded DNA molecules in E.coli. enhances H-bonding during strand displacement. Supports cell recombination Enzymes that cleave strand, help unwind helices and ligand strands = RecB, RecC, RecD.

Gene conversion—non-reciprocal exchange

CHAPTER 13 – The genetic code and transcription DNA have a direct interaction with ribosomes. DNA in nucleus- ribosomes in cytoplasm. mRNA discovered by Jacob and monod. Sydney Brenner said that 4 letters specify 20 words. Frame shift mutation is the first proof for triplet nature of code. (Crick et al.- Ecoli K12, Phage T4 and proflavin) Deciphering the genetic code:  The genetic code is linear, using ‘letters’ as the ribonucleotide bases that compose mRNA molecules. The ribonucleotide sequence is derive...