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BIOL 1090
Peter Dawson

Genetics Guy Rohkin Embryos genomes will develop in response to environmental queues in response to their surroundings to assure their survival.  Induced pluripotent cells can be engineered from Adult cells DNA is a polymer  A nucleotide composed of: 3- o A phosphate group PO 4 o 5 Carbon sugar (2-Deoxyribose) o One of four cyclic nitrogenous bases Pyrimidine  Cytosine  Thymine Purine  Adenine  Guanine DNA is double stranded and the strands are antiparallel 3‘ lines with 5‘ and 5‘ lines with 3‘  No covalent bonds  Held together by Hydrogen bonds between bases on opposing strands and hydrophobic interactions between adjacent stacked bases  Adenine base pairs with Thymine (2 Hydrogen Bonds) and Cytosine base pairs with Guanine (3 Hydrogen bonds; more stable)  Higher G-C Content DNA molecule is more stable; harder to break down. DNA strands are POLAR  Each strand has chemical polarity  5‘ end has a free phosphate group  3‘ end has a free hydroxyl group Most common form of DNA is called B-DNA Two grooves of different width: The major groove and minor groove 10 Base pairs per turn Z-DNA left handed helix, 12 base pairs per turn Distance between sugar phosphate backbone varies (Major and minor grooves) The DNA in living cells is supercoiled  Double helix is in turn super coiled  Circular, snip one string and hold one end fixed and rotate the other through 360 degrees in a right or left handed rotation. The circle collapses into a negative or positive supercoil (shaped like an ‗8‘)  Prokaryotic chromosomes are circular pieces of RNA; folded into 40- 50 loops, supercoiled and each loop is independently supercoiled Similar amounts of supercoiling exist in the DNA of bacterial and eukaryotic chromosomes  The DNA found in mitochondria and chloroplasts exists in circular chromosomes that resemble those of prokaryotes  Human Mitochondrial DNA (mtDNA) approximately ~17Kb  Chloroplast genome of the liverwort (121,024 nucleotide pairs) Eukaryotic chromosomes are composed of proteins, DNA & RNA  Most of it is Chromatin o DNA + Nonhistone Proteins + Histones + RNA  Histones are basic proteins (Rich in Arginine and Lysine)  Positively charged o Nonhistone proteins are heterogeneous (mostly acidic)  The rest is Non-Chromatin nuclear constituents The first level of condensation –Packaging DNA as a negative supercoil into nucleosomes (produces an 11 nm fibre) Nucleosome core, 146 nucleotide pairs of DNA wrapped as 1,3/4 turns around an octamer of histones Linker DNA on the other hand varies in length from 8 to 114 nucleotide pairs The linker region is susceptible to digestion by an endonuclease (it is weaker than the nucleosomes) Histone H1 = Linker histone; comes along and connects it  Lots of electrostatic attraction on DNA to attract to a positively charged protein like histone H1  Lots of stability in these nucleosome cores because of attraction The Second level of condensation- an additional folding or supercoiling of the 11nm fibre to produce a 30 nm fibre  This is driven by nucleosomal interactions using histone H1  There is the solenoid and zig-zag models  30 nm fibre is the basic structural unit of the metaphase chromosome (DNA in its most condensed form) [It begins to form the chromosomes; it gets more or less condensed as the chromosome moves through the cell cycle] Starts as DNA, winds around histones which then come together in octamers to then turn into solenoids. Solenoids then bind together to create a polymerized chain and then it gets compacted into a chromosome using a scaffold. How many Free 5‘ ends are present in a chromosome? A) 1 b) 2 answer C) the number varies but between 2 and 10 D) the number varies but between 10 and 20 e) more than 20 Chromosome ends are protected by telomeres.  Resist degradation by DNases (protect the ends from nucleases)  Prevent fusion of chromosomal ends (rather sticky so they tend to attach to other 3‘ and 5‘ ends)  Facilitate replication of the ends of the linear DNA ( Helps the DNA polymerase work closer to the ends)  Every time a cell divides it needs to copy its DNA and the DNA polymerase doesn‘t quite make it to the extreme end of the chromosome  So every time DNA is replicated the telomeres get a little bit shorter at some point the telomere will cross over some threshold such that its no longer functioning as a telomere. Thus it sets a limit on how many times the cell can divide which causes mistakes in mitosis.  Hundreds of thousands of short sequences motifs to make up the ends of chromosomes Telomere loop is coded in these protein complexes  So they bind to only these protein complexes ; This helps to hide these free ends from DNAses and from the ends of other chromosomes  Having long telomeres is one way to protect against this  Making an enzyme called telomerase which will extend telomeres  Early embryonic cells have telomerase but over the course of embryonic and fetal development start to switch off telomerase and now has a finite number of times left to divide before it dies.  Cancer cells have telomerase, they reactivate the telomerase gene and the abhorrently reactivate it and keep their telomeres long. (Blackburn, Greider, Szostak) Centromere -Constricted part of the chromosome (in the center) -Not always in the center, usually displaced towards one or the other end  Centromeres provide the point of attachment of chromosomes to microtubules in the mitotic spindle 3 conserved elements: 2 specific sequences However the central element is very Adenine-Thymine rich (93%) -That is the point where the centromere gets physically attached to the mitotic spindle The centromeres of more complex eukaryotes are more complex and longer.  There is a centromeric sequence that is repeated (5000-15000 of the 171 base pair alpha satellite sequence)  Yeast centromeres are 110-120 BP long  Contains binding proteins involved in spindle attachment, uses CENP protein to bind (related to Histone 3 [H3]) A+T = C+G (False) Ratio A/T = C/G (true) Ratio of A/T or C/G = 1 (true) The Nature of the gene and its expression 2 Versions of Strep. Pneumonia - Infected a mouse with a virulent strain and harvested the living type out of the mouse - Introduced nonvirulent strain with mouse , mouse didn‘t die. - Heat treated virulent cultures into mouse , mouse didn‘t die (bacteria were dead) -put heated virulent cultures and nonheated nonvirulent cells into mouse and mouse died. Nonvirulent bacteria took information from the dead virulent ones. - -Sia and Dawson repeat the experiment except they use culture plates. Take the IIR cells and extract the DNA from heat killed cells and mix it with the live IIR cells + serum to get rid of live cells and plate out cells and the IIIS colonies grow once again. The last experiment they add enzymes that will chew up RNase and Protease to show that the colonies keep growing. However when DNAse is added no colonies are formed because DNA is destroyed. -Attach radioactive phosphate to Viral DNA, put viruses into a blender and smoothie. Radio labeled DNA stays in the bacterial cells in the centrifuge. Reverse can be done aswell by adding sulfur to methionine aminoacid, so when put in a blender and then centrifuge, the bulk of the material is protein. -Some Viruses use RNA as their genetic material and NOT DNA  Tobacco mosaic virus Type B and Type A  Assemble these chimeric virions and use the reconstituted mixed virus on the tobacco leaf  The progeny was made of the RNA conferring the information  Virus progeny ALWAYS matched the RNA, never the protein strain  (Delbruck, Hershey, Luria) Transcription and Translation Genes encode for five known types of RNA  snRNA -> Spliceosomes  rRNA-> Ribosomal Subunits  tRNA  mRNA-> go into Ribosomes for translation  pre-microRNA-> Important molecules in the regulation of gene expression ; How much or perhaps when a gene is expressed and the protein quantities it produces; base pair RNA to RNA; keep gene expression low or off (Not that important, still a recent discovery) Transcription  Synthesis depends on having a DNA template  One strand acts as the template and the other is the nontemplate (doesn‘t participate)  RNA synthesized by RNA Polymerase (5‘ to 3‘)  Occurs from left to right  RNA strand resembles the nontemplate strand (substituting Uracil for Thymine)  Template is the 3‘ to 5‘ end  Double helix Is locally unwound during transcription  Central Dogma  A gene is a segment of DNA the encodes an RNA molecule  What defines where the gene starts and where it ends? What signals the RNA Polymerase to start transcription in a certain location and not just randomly along the chromosome, and similarly whats causing it to start? o o o If you take gene sequences from E. Coli, you learn some sequences that are conserved. The first base on the template strand as +1. At -10 base pairs you will find an AT rich sequence. At -35 sequence there is a more specific sequence that is well conserved; this sequence turns out to be a good binding site for the RNA polymerase. The RNA polymerase which has some affinity for DNA much prefers to bind to the -35 sequence than anywhere else. High affinity binding site from the polymerase and a region at a distance away that is easy to unwind characterizes a bacterial promoter . Termination of Transcription  Rho-independent transcription terminator sequence  Requires a sequence in the DNA  RNA folds back on itself in a hair pin  Consequence of adopting hairpin fold is that the RNA polymerase gets stalled Bacterial genomes  Several protein coating genes are side by side on a single RNA molecule.  Genes that are co-transcribed  Operon is a set of genes, often genes that should be expressed at the same time, so its just efficient to transcribe them all together on one molecule Eukaryote Transcription  Additional complexity  Promoters contain a sequence in the DNA that helps position the polymerase.  A good number of our promoters have something called the TATA Box, an AT rich sequence that acts as a binding site for proteins to help position RNA polymerase.  Other protein binding sequences, GC Box, CAAT box, Octamer box.  Transcription factors : proteins that help regulate the process, not just make it happen, but determine whether it will happen on a particular gene or cell type or time.  Help or prevent the RNA polymerase from getting to a certain part of the gene  Doesn‘t start by the binding of polymerase but TATA-binding protein recognizes the TATA box and brings with it other associated proteins and factors  Complexes of several proteins coming together  RNA polymerase eventually comes in  Polymerase assembles and starts to transcribe  An endonuclease will cut an RNA strand at a defined sequence and it leaves a free 3‘ end  The 3‘ end is recognized by the poly-A polymerase that simply adds A nucleotides . Can consist of a few 100 Adenine nucleotides.  A-Tail gives molecule stability; DNA is very stable.  RNA is unstable; exactly how unstable depends on each specific sequence. Depends on the halflife. Some only persist in the cell for a few minutes, others may hang around for a day or two before they turn over  5‘ is bonded with a cap called 7-methyl-guanosine; helps stabilize and protect the RNA and to protect the 5‘ end from exo-nucleases. (Also used in translation)  Some bits are spliced out (removed). Introns/Exons  Are noncoding sequences located between coding sequences.  Removed from the pre-mRNA and are not present in the mature RNA  Introns are variable in size and may be very large  Everything else that remains are exons  Not all parts of exons are coding (there are some noncoding sequences)  UTR: Untranslated Region; not spliced out but not coding regions (like 3‘ and 5‘ ends) mRNA molecules once processed act as intermediates between the DNA and the cytoplasm.  Used as templates for translation  In prokaryotes an RNA sequence positions the ribosome to begin translation at the beginning of a coding sequence or open reading frame o Shine-Dalgarno sequence (AGGAGG) in the 5‘ end and is complimentary to the 3‘ (UCCUCC). Ecoli ribosome is placed on the 5‘ end , then starts from the first AUG codon a few spots upstream from the Shine-Dalgarno sequence.  In Eukaryotes the RNA sequence around the AUG influences where translation begins and the ribosome sans from the 5‘ end until a suitable start codon is found  GCCRCCAUGG (R= A or G) Changing underlined reduces efficiency 10 fold  Proteins associated with the 5‘ cap that help position the ribosome on the mRNA and it scans along until it finds an acceptable start codon (Not necessarily AUG) Features of the Genetic Code  If AUG is start of translation then there are no spaces between codons and codons do not overlap.  Units of 3 nucleotides in the RNA  Code is degenerate (64 combinations of 3 using 4 difference nucleotides; only 20 aminoacids)  Many codons that code for the same aminoacid  Code is ordered, similar amino acids tend to be coded by similar looking codons  THE GENETIC CODE IS NEARLY UNIVERSAL, Each triplet codon has the same meaning in ALL organisms; except for some mitochondrial RNA  All proteins start with methionine as their first amino acid When Avery Macleod and Mccarty mixed the DNA of a smooth streptococcus with live cells of rough. The progeny was transformed rough to smooth. Which statement best describes the experiment? A) They were using the DNA from smooth Strep B) Strep cells only contain DNA C) They could prevent the transformation with DNase D) Some viruses use RNA as the genetic material Codon (mRNA) is recognized by the anti-codon (tRNA)  It has 3 pockets/binding sites that the tRNA occupies  tRNA uses Inosine  Inosine has relaxed binding specificity  3 Stop codons (UAA, UAG, UGA)  Stop codons are bound by a release factor that binds to the termination codon in the A site of the ribosome and tRNA leaves the E site.  Carboxyl Terminus is placed at on end and Amino-terminus on the other The seq. on the template strand of DNA is 5‘ GGAACCCTGCTTCGGAGTCCCAATTAGCATGA The corresponding RNA molecule begins 5‘-UUGGGAC What is the next nucleotide? (opposite 3‘ – 5‘ ) A) A B) C C) G D) T E) U Mitosis Cellular organelles and cytoplasmic contents are divided more or less equally between daughter cells  ER and Golgi are fragmented and reformed after  Mitochondria and chloroplasts are randomly divided  NUCLEAR CHROMOSOMES MUST BE DUPLICATED EXACTLY AND DISTRIBUTED EQUALLY AND EXACTLY TO DAUGHTER CELLS Phases  G1 (Gap 1) Phase: Growth, cellular metabolism  S (synthesis) Phase – DNA replication (Chromosome duplication  G2 (gap 2) Phase : Preparation for mitosis  M Phase (Mitosis)– Mitosis, chromosomal pas de deux and cytokinesis  Interphase : The time between successive mitoses (G1+S+G2) o Cells that are not actively cycling may exist the cell cycle o From G1 they enter a state called G0; these cells are said to be quiescent( not dividing) Centrosomes  The centrosome cycle, in which centrioles are duplicated progresses along with the cell cycle  During S Phase, the centrioles duplicate  Each pair establishes its own centrosome and organized microtubule spindles  Organize them so that + ends radiate outwards from the centrosomes Chromosomes How many chromosomes do we have in total? 1. 22 2. 23 3. 46 4. 92  Each sister chromatid (2 of which are connected at their centromere) IS a complete chromosome that has been duplicated  Condensin and Cohesin have a role in the formation of mitotic chromosomes.  Condensin keeps it condense  Cohesin loops around to hold it at the centromere Interphase  Duplicate chrosomes called sister chromatids  Joined at the centromere by cohesion  Centrosome is duplicated Prophase  Initiation of spindle formation  Condensation of duplicated chromosomes  Fragmentation of ER and golgi  Nucleolus disappears  Nuclear membrane breaks down  Spindle microtubules invade the nuclear space Prometaphase  Chromosomal microtubules attach to the kinetochores which are on the outer surface of the centromeres.  Bridges are formed between the chromosome and microtubules using corona fibers and motor proteins Metaphase  duplicated chromosomes are aligned midway between the  spindle poles  this equatorial plane is called the metaphase plate  Duplicated chromosomes migrate to the equatorial plane of the cell and the nuclear membrane breaks down (Metaphase) Anaphase  Sister chromatids of each duplicated chromosome move to opposite poles of the cell  Centromeres split and Chromatids separate  Chromosomes move towards opposite spindle poles  Spindle poles move further apart Telophase  Chromosomes cluster at opposite spindle poles  Chromosomes become dispersed and decondense  Nuclear envelope assembles around chromosomes  Golgi and ER reform  Daughter cells form by cytokinesis  Chromosomes decondense and new nuclear membranes form  Cleavage furrow forms Meiosis Karyotype n= number of chromosomes = the haploid state 2n chromosomes = the diploid state n=23  Maternal Chromosome 1 is homologous to Paternal chromosome 1  Heterologous are any different pair or combination. o Chromosome 2 is heterologous to chromosome 3 o Autosomes 22 autosomal pairs (All chromosomes that aren‘t the sex chromosomes) o 1 pair of sex chromosomes Meiosis notes  Bivalents if 2 pair , tetrad is all 4 pairs of chromosomes  Process of pairing up the synapse allows crossing over to occur between homologous sister chromatids that are on the inside of the pairing  Associations between chromatids called chiasmata  Double Stranded DNA on the inside sister chromatid breaks and it breaks in the same place on both sister chromatids and there is a reciprocal exchange of strands.  Two outside chromatids that aren‘t changed, and 2 of the inside are chimeric  The diplotene (in Diplonema; prophase 1) stage may persist for the entire reproductive life of the individual (more than 40 years in female humans)  Bivalents now attached to microtubule spindles and align in the equator  Move to opposite poles and absorbed into nuclear membrane and they split apart  May not even lay down a full nuclear membrane Meiosis II Kinetochore position  Kinetochore position changes between Prophase I and II  Kinetochore must sit on the same surface of sister chromatids so that the sister chromatids separate from the other homologous pair but stay together themselves (in Meta/Anaphase I)  Kinetochores are on opposite phases in Meta/Anaphase II so that the chromatids can split up Spermatogenesis  Spermatogonia undergo mitosis to produce large quantities  One sparmatogonia decides to undergo meiosis and turns into a primary spermatocyte  Primary spermatocyte undergoes meiosis I and produces a secondary spermatocyte and are still joined by a cytoplasmic bridge  Cytoplasmic bridges are attached to spermatids to provide them with cytoplasm  Until the spermatid detaches to produces a sperm cell  Has lots of mitochondria Oogenesis  Oogonium (2n) replicates and turns into primary Oocyte  After meiosis I primary oocyte divides are produces a first polar body with a little bit of cytoplasm and secondary oocyte is left over  Happens in the fetal ovary  Meiosis I begins in the fetal ovary and arrests at prophase I  For ovulation an egg matures to metaphase II and is released into the fallopian tube  First polar body may complete meiosis II or degenerate before it happens. Polar bodies normally degenerate.  The secondary oocyte gets to metaphase 2 and goes through anaphase 2. Does not complete telophase and cytokinesis until it is fertilized.  Egg is produces with an egg coat and cortical coat  2 Haploid genomes to produce a zygote Other Organisms Fungi Plants Genetics During Mitosis, will an allele on one sister chromatid always match the allele on the other sister chromatid? A) Yes or the chromosome wouldn‘t align properly at telophase B)Yes Because sister chromatids are the product of DNA replication C) No because its random D)No only when an individual is homozygous Mendelian Genetics He would take 2 plants that varied in one characteristic and crossed them  I.E: Tall X Dwarf = Tall  Hybrid progeny crossed it with another tall produced a tall and a dwarf thus making a ratio of 3:1  No matter which characteristic he picked, all the ratios were approximately 3:1 All 7 traits behaved similarly  Each trait was controlled by a heritable factor that was either dominant or recessive  Mendel‘s heritable factor = gene  Dominant and recessive forms called alleles  An allele is a difference in DNA sequence (a difference at the locus/position on the strand o Homozygous – both alleles are identical o Heterozygous – the two alleles are different o In a genetic cross, the parents are referred to as the parental (P) generation o Their offspring represent the first filial generation (F1) o Their grand offspring the F2 generation. Principle of Dominance  A dominant allele masks the presence of the recessive allele  Principle of Segregation o Neither allele is typically changed by coexisting with the other in a heterozygote The alleles of different genes assort/segregate independently of each other during meiosis They wouldn‘t assort independently on the same chromosome unless there is a crossover. For 2 genes on the same chromosome if they are a long distance apart, then virtually all the time there will be at least one crossover between the two genes. Therefore at a population level considering thousands of meiosis there will be an equal distribution of the genes. The closer the two genes come to lie on the chromosome the more often there will not be a crossover. Thus they will tend to segregate. The 2 genes and alleles are close enough that they tend to be co-inherited. Chromosome Terminology Punnett Squares For n Different genes assorting independently, There are: n 2x2x2…) n= 2 possible haploid genotypes Therefore for 23 different genes all on a different human chromosome, 2 23 9,399,608 different haploid gamete genotypes.  The closer genes are to one another, the more likely they are to be dependent on each other because they both get crossed over. The dominant alleles for genes encoding height, seed colour, and seed texture in peas are D, G, W respectively. A cross between individuals that are heterozygous at all three loci produces 64 offspring. If the genes assort independently, how many of these would you expect to be tall, green, and wrinkled?  3 tall, green and wrinkled peas  multiply along the forks  use ratios The Rules of probability- the multiplication rule If events A and B are independent, the probability that they occur together is the product of their individual probabilities of their occurrence. The product is the area of their overlap. The situation where both A and B must both occur. P(A) X P(B) The additive rule If the events A and B are independent the probability that at least one of them occurs is the sum of their individual probabilities of occurrence minus the probability of their joint occurrence P(A) + P(B) –[P(A) x P
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