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Biomedical Engineering
Y U Liu

BME 50A 1 Week 1 • Eukaryotes - animal/plant cell o nuclear envelope o nucleus (holds DNA) o plasma membrane - interacts w/outside world and ECM - (extra cellular matrix)  double lipid bilayer  membrane bound proteins o protien trafficking and modification  endoplasmic reticulum  closer to the nucleus  makes proteins (factory)  golgi apparatus  ships things out (fedex)  vesicles  ribosomes  catalyzes protein synthesis (reads RNA and makes proteins) o cyyoskeleton  microtubules/centrosome - cell division  actin - focal adhesian  intermediate filaments o mitochondria - ATP  double lipid bilayer  phagocytosis  it is believed that eukaryote cells engulfed bacteria. this is supported by mitochondria having:  its own DNA  plasma membrane for energy regeneration o Plant cells only  chloroplast - light ->energy  vacuole - storage of water  cell wall • Prokaryotes - bacteria/archaea o bacteria cell  flagellum  DNA  cell wall  ribosomes  plasma membrane • sickle cell anemia o hemoglobin in red blood cells do not carry oxygen like they are supposed to • function, genes, and protiens o genes ->protiens = molecular biology (1930’s-50’s) o protiens + genes (DNA) ->function o recombiant DNA = messing w/genes (1970s+) Bio chem - chemical bonds • molecules that make up 99% of life BME 50A 2 o hydrogen o carbon o nitrogen o oxygen • zwitterionic - both + and - charge in the same molecule • valence electrons - outermost electrons, where chemical bonds take place • orbitals o S orbital (2 e-) o P orbital (8 e-) • covalent bond o e- sharing between 2 atoms o very strong, 90 kcal/mol to break o polar vs nonpolar covalent bonds  electrons shift toward negative pole  Oxygen atoms have a higher affinity for electrons than hydrogen – they are more electronegative  relatively equal electro-negativity = nonpolar o hydrogen bond  week electrical interaction between polar molecules  1~4 kcal/mol (weak)  raising temp can easily break it  amine groups (nitrogen containing) can attract hydrogen • ionic bonds o transfer of e- between 2 atoms o bond strength is dependent on how far away atoms are  1/r^2  3-80 kcal/mol • van der waal's interaction o weak electrical interactions between nonpolar molecules o very weak (.1 kcal/mol), gets its strength in #'s o transient dipole  usually nonpolar but can have a dipole • hydrophilic - water loving/hydrophobic - water hating o aliphatic (totally water hating, stings of just H and C) gets water caged (lower entropy) o amphipahtic (has a hydrophilic head and hydrophobic tail) groups in a circular pattern  doesnt like water or fat o amphiphilic - water and fat loving Molecules in Cells BME 50A 3 • • lipids o polysacchicides/polymer - repeating units  carbs stack on themselves and form long repeating chains o saturated - no double bonds (better packing: hard margarine) o unsaturated - double bond (loose packing: soft margarine) o Fatty Acid   phosphatidylcholine- 1 satuarted and 1 unsaturated strand of fatty acid BME 50A 4  hydrophobic tail, hydrophilic head o esterfication vs. condensation reaction  o Lipid bilayer   appear to not follow 2nd law of thermodynamics, but they do thanks to water-cages  membrane permeability  ions (na, k, ca) need ion channels and pumps to pass the membrane  water needs aquaporins to pass  proteins, nucleic acids, and sugars require specialized transports  cholesterol fits in between phospholipids and increase packing density and stiffness   florescence recovery after photo-bleaching is a method used to show membrane fluidity • carbohydrates BME 50A 5 o o 5 carbon  pentose  forms ribose o 6 carbon  hexose  forms glucose o disaccharides - condensation reaction  sucrose = glucose + frucose  takes sucrase to break  lactose = glucose + galactose  takes lactase to break o carbs in cells  glycogen stores sugar/energy • proteins • nucleic acids BME 50A 6 Week 2 • proteins o are pretty much amino acids o there are 20 different amino acids in our body o basics makeup of an amino acid   amine group (N-H) is N-terminus when bonded  carboxylic acid (C-O/OH) is C-terminus when bonded o a charged hydrophilic amino acid's charge is pH dependent o interactions  noncovalent interactions  hydrophobic/hydrophilic   ionic BME 50A 7  Van der Waals   covalent interactions  disulfide bonds  o Peptide bond  joins amino acid groups together with a condensation reaction  side chain does not affect bonding  peptide - short chain of amino acids (<10)  polypeptide - long chain of amino acids (>10) o structures of protiens BME 50A 8  primary structure  amino acid sequence (ex. FML)  secondary structure  folded structures w/in a segment of polypeptide (3D shape which side chains determine)  alpha helix (spiral)  beta sheet (wavelike)  arrows point toward C-terminus  if all arrows don't point the same way said to be antiparallel  globular  filamentous  tertiary structure - full 3D conformation of a polypeptide chain (includes 2nd and primary)  quarternary structure - more then 1 peptide  contains 4 similar polypeptide - tetramer  like hemoglobin, which is expressed by red blood cells and carries oxygen o Linus Pauling  father of biochemistry  hydrogen bonding/electro-negativity o transmembrane proteins  o what proteins can do!  enzymes catalyze reactions  lysozyme can cleave a polysaccharide (sugar molecule)  structural or motor proteins allows muscles cells to exert forces  myosin (motor protein) binds to actin  Green Fluorescent Protein (GFP)  first found in jellyfish  structure is a beta barrel w/a chromophore in the middle that makes it glow o in a protein, structure is very important in its function • Nucleic Acids! BME 50A 9 o Fred Griffith (1928)   S strain = smooth bacteria  R strain = rough bacteria  this experiment demonstrated the transforming principle o Avery, Macloed, & McCarty (1944)  BME 50A 10  tested* - combined w/R strain o Watson & Crick (1953)  discovered double helix o Nucleotide   Phosphate + Base + Sugar (nitrogenous)  nucleoside  base + sugar  the nitrogenous bases!!  purine (big guys)  Adenine  Guanine  pyrimidine (small guys)  Cytosine  Thymine (DNA only)  Uracil (RNA only) o DNA structure   Chargaff's rule (1950) A:T G:C BME 50A 11  A:T - 2 hydrogen bonds  C:G - 3 hydrogen bonds  10.5 base paris to make a full revolution  sugar/phosphate backbone is - charged and <3's water  DNA forms w/a condensation reaction  polymerization occurs at the 3rd carbon  there is a major and minor groove  sequence specific binding proteins usually bind to the major groove  C:G rich DNA has higher melting tempt because of more hydrogen bonds  DNA's structure allows it to be copied (replication)  DNA can bond w/RNA to make a complimentary form w/out destroying the original strand (transcription) o difference between DNA and RNA  DNA is more stable  RNA has a ribose where DNA has a deoxyribose  RNA folds and binds to itself (called intramolecular binding) and can be shown by the hairpin model  translation occurs when RNA makes proteins in ribosomes BME 50A 12 Week 3 Energetics and Binding Models • laws of thermodynamics - o first - energy is conserved o second - all systems tend toward entropy o cells increased order on their insides will release heat which in turn will increase disorder on the outside • free energy - energy of a molecule that can be used to do useful work o aka gibbs free energy (G) o measured in kcal/mol o change in energy (delta G) must be smaller then 0 for a reaction to occur o delta G = delta Gnot + (gas constant)(temperature)(ln([x]/[y])) o delta Gnot - standard free energy change (free energy change of a reaction when the concentrations are set to 1M) o free energy depends on concentration • ATP provides energy o straight fructose to glucose will not occur because delta Gnot = + 5.5 kcal/mole o ATP will attach itself to glucose which will make delta Gnot = -1.8 kcal/mole which will occur • At equilibrium, the forward and reverse reaction rates are equal, and the ratio of product to reactant is constant (Keq) o Keq = product/reactant:[C]/[A][B] = kon/koff o dissociation rate: koff[C] o association rate: kon[A][B] o koff[C] - kon[A][B] = 0 at equilibrium o BME 50A 13 o o [ ] means concentration which = mole/L Genetic stuff • Gregor Mendel (mid 1800s) - pea plant • definitions! o gene - factors of inheritance controlling trait o allele - alternative forms of a gene (R, r) o phenotype - appearance (round, wrinkled) o genotype - pair of alleles carried out by an individual (RR, Rr) o homozygous - 2 copies of the same allele (RR, rr) o heterozygous - 2 alternative forms (Rr) o pure breeding RR + rr = round o recessive (r) dominant (R) o test cross - cross w/homozygous recessive to see if one parent was RR or Rr o incomplete dominance - pure breeding strains result in 3rd (blue + red = purple) o co-dominance - pure breeding strains result in co-expression (blue + red = blue/red)  like blood types o punnett square - o u know... • chromosomes (colored things) (we have 23 pairs, so 46 in all) • BME 50A 14 o o o individual traits are inherited individually BME 50A 15 Week 4 • TH Morgan (1910) studied fruitflies (they only have 4 chromosomes) o wild type (+++ normal) vs. black and vestigial (short winged) o showed the difference between having the trait on the same chromosome vs. not on same o o experiment showed they were on the same chromosome, but some nonparental phenotypes showed up • recombination - crossover during meiosis 1 o chiasma - the place where the gene crosses BME 50A 16 o o diploid - recessive genes are masked o haploid - you can see recessive genes o recombination rate - how often the crossover occurs (#of occurances/total number of babies)  bigger number, farther away the genes are, and the more likely it will happen  the # = map units, or how far away 2 genes are o double cross - take the map units of the genes and multiply them to see likelihood % • sex linkage BME 50A 17 o • yeast o prototroph (wildtype) - grows on minimal media, can make its own stuff o auxotroph - mutants that have lost the ability to grow on minimal media  auxotroph library - a nice group of mutants • biochemical pathway o o tests to find pathway  recessitivity  the mutant gene must be recessive for other tests to work  cross mutant w/wild type to see if it lives  complementation  cross mutants to see how many groups/mutant genes there are   epistasis (double mutants)   in the biochemical pathway above, B is epistasis to A  B 'stands on' A, and A is masked by B Genome Project • 3 billion base pairs in human gene • started in 1995 • Venter - shotgun sequencing (faster then old method) BME 50A 18 • drafts o initial - 2001  covered 90%,  250k gaps, many erros o final - 2004  covered 99.7%  300 gaps, not many errors • 250 eukaryotes and 4k bacteria are sequenced • GWAS - genome wide association studies o so many more diseases found o locci - position on the gene • Sangar sequencing - what we use today • having everyone sequenced could be dangerous o work may not hire, lawsuits ect. o Genetic Information Nondiscrimination Act  passed in 08  health insurance and employment can't discriminate • Francis Collins - leader of genome project BME 50A 19 Practice Problems #1 cross 1: tail-less mouse ____Tt________ X normal mouse ______tt______ gives F1: 10 tail-less 9 normal cross 2: tail-less F1 X tail-less F1 gives F2: 10 normal 21 tail-less 9 dead The breeding of two black horses gives the following progeny: Black: 75%, Chestnut: 25%. What are the most likely genotypes for the parents Parent 1: Ee Parent 2: Ee A second gene works along with the Extension gene: the Cream dilution gene (Cr or N) can reduce the intensity of color. A chestnut horse can be diluted to palamino (golden) or cremello (white) by the presence of Cream alleles. Breeding two palamino (golden) horses together gives progeny with the following ratios: Palamino: 50%, Chestnut: 25%, Cremello: 25%. Likewise, a black horse can be diluted to smoky black (not quite black) or smoky cream (off-white) by the presence of Cream alleles. Breeding two smoky black horses together gives the following progeny: Smoky Black: 50%, Black: 25%, Smoky Cream: 25%. Queen Isabella of Spain loved palamino horses. For her birthday, you wish to present her with a newborn palamino foal (baby horse). In order to guarantee the birth of a palamino, which two horses would you breed with each other? Chestnut X Cremello (i.e. NN x CrCr = NCr, or eeNN x eeCrCr = eeNCr) The breeding of two horses produces progeny at the ratios shown below. Fill in the genotype or genotypes for each phenotype, listing alleles for BOTH genes. Phenotype Frequency Genotype(s) Smoky Cream 25% EECrCr or EeCrCr Cremello 25% eeCrCr Smoky Black 25% EENCr or EeNCr Palamino 25% eeNCr min +A +B +C +arg + + + + + m1 - - + - + m2 - - - - + m3 - + + - + m4 - + + + + pathway: m4 C m3 A m1 B m2 ARG Genotype Color Phenotype m1+ m2+ m3+ red (wild-type) m1- m2+ m3+ orange m1+ m2- m3+ yellow BME 50A 20 m1+ m2+ m3- colorless m1- m2- m3+ orange m1+ m2- m3- colorless m1- m2+ m3- colorless pathway: colorless m3 OJ m1 YW m2 RED BME 50A 21 Week 5 Molecular Bio (genes to proteins) • Hershey and Chase (1952) o virus' are made of only protein and DNA o these guys attached radiolabled D to proteins and radiolabled P to DNA o e.coli was mixed with radiolabled virus' and they showed DNA (means virus injects DNA into bacteria) • • • Meselson and Stahl (1954) o bacteria was put into radiolabled 15N and 14N then separated (w/centrifuge) o the 2 different DNA weights combined to middle ground weight (conservation model was discharged) o they used this semi-heavy DNA and broke the Hydrogen bonds and centrifuged them again o half were heavy, half were light (means Semi-conservative Model is correct) • Arthur Kornberg put: o template - DNA strand to be copied o primer - short sequence of DNA complentetary to template that helps process start o polymerase - enzyme to polymerize DNA (goes after primer) o dNTP (monomers) - dATP, dTTP, dCTP, sGTP o into a test tube and showed thats all you need in DNA replication • new strands always grow in 5'-3' direction • helicase - en
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