The composition of chromatin and comparison of genome sizes very broadly for viruses prokaryotes and eukaryotes level discussed in lecture chromatin DNA proteins Viruses Prokaryotes E Coli Human 5 1 kb base pairs 4 000 kb base pairs 2 900 000 kb base pairs 0017 mm long 1 36 mm long 990 mm long cell only 2 10 6 by 8 m nucleus is 10 10 6 m 1 histones 7 nonhistone for 1 g DNA about 1 g histone nearly 1 g nonhistone HU bending of chromosome basic proteins that stabilize structure supercoiling due to 2 topoisomerases simple cut apart strands and unwind then religate forced to twist on itself torsion created The classes of histones and the general traits features of these proteins highly conserved in all animals basic due to amino acid side chains positive at neutral pH attracted to negative DNA lysine and arginine 8 molecules 2 of H2A B 2H3 2H4 make up core H1 is separate and helps tighten binding nucleosome is 110 and wraps 147 bp DNA The folding or levels of condensation of a eukaryotic chromosome starting with DNA and through the architecture of a metaphase chromosome plus the experimental basis behind the information at the level of details discussed in lecture and related reading DNA double helix beads on a string chromatin solenoid or zig zag section of chromosome in extended form condensed section of chromosome entire mitotic chromosome most highly condensed result each DNA is packaged into a mitotic chromosome and condensed 10 000 fold how to examine the nucleosome 1 isolate unfolded chromatin by digestion with nucleases that cut between nucleosomes 2 exposed DNA between nucleosomes linker regions is degraded 3 dissociate DNA from core via high ionic solutions 4 measure length of the DNA associated with that nucleosome 5 run the sections of DNA on a gel DNA from 2 nucleosome will travel 2x as much as DNA from 1 nucleosome etc low ionic concentration beads on a string beads of 100 higher ionic concentration 300 fiber solenoid or zig zag high ionic concentration DNA dissociates completely from histone core looped domains scaffold which is nonhistone protein How would a sample of cells obtained from an individual be examined for potential chromosomal abnormalities using karyotype analysis details of the assays and the reasons behind each of step when cell is about to enter mitosis chromosomes condense spindle fibers enrich for cells in metaphase 1 extract fluid from growing fetus want to look at their chromosomes the cells in mitosis few cells that are proliferating are in mitosis mitosis is only 20 min of 20 24 hour long process 2 increase of cells in mitosis by colchicine 3 fix cells in methanol acetic acid 4 trypsin for 10 15 min 5 stain with giemsa 6 smash cells in mitosis so chromosomes will spread at random by dropping at a specific height 7 take picture and cut them apart by size and banding pattern organize them cholchicine incubate for 20 hr chromosomes are stuck in metaphase binds to monomers of microtubules and keeps them from polymerizing so spindle fibers don t form spectral karyotyping chromosome painting procedure combines elements of above process up until staining chromosome does not use giemsa 46 chromosomes chose one at random sequences on each chromosome is known complementary nucleotide sequences oligonucleotides covalently attach a fluorescent dye that can appear in computer incubate isolated chromosome with mixture of oligos instead of gimesa oligios specific to parts of the chromosome after incubation different fluorescent markers can bind used to identify structural chromosome aberrations in cancer cells and other disease conditions when Giemsa banding or other techniques are not accurate enough Give examples of different forms of facilitated diffusion that allow the movements of molecules through a an aqueous channel water soluble molecules water not ions glycerol urea glycine b via a carrier ligand specific carrier glucose translocator protein glycolysis maintains unidirectional entry of glucose in the cell makes it appear as though there is less glucose in the cell so direction is to move inside unidirectional flow through hexokinases Discuss the selectivity of each type of translocator and contrast the structural differences of each class of translocators 6 alpha helices span the membrane and have 2 dangling loops green alpha helices yellow is core of channel where substances can pass GLUCOSE TRANSPORTERS 1 first level of regulation type of translocator families of genes that code for proteins GLUT GLUT 1 3 GLUT 2 GLUT 4 brain heart smooth muscles liver pancreas striated muscles fatty cells kM 1mM kM 15 20 km kM 5mM blood glucose concentration is 5mM function maximally even in small concentrations high affinity for glucose only when glucose if very high will the carriers be functioning optimally different GLUT ensure there is adequate distribution of glucose to all the specific tissues 2 second level of regulation 12 alpha helices provide wall of a cylinder larger than aquaporin slight amino acid differences affect affinities determine specificity as glucose enters channel H bonds between H and Oxygen on glucose and some of the side chain molecules of amino acids which provide specificity How would you determine whether the transport of the amino acid glycine into red blood cells occurs by simple diffusion or by facilitated diffusion If glycine were to be imported by facilitated diffusion how could you determine whether or not another amino acid such as methionine moves into the cells by the same carrier 1 add to suspension a concentration of amino acid whose diffusion you want to measure do this at different concentrations of amino acids 2 run the experiment at different concentrations 3 measure over time intervals 4 finally pellet the cells and count how many amino acids there are in them label the amino acids with radioactive tags so they can be counted ex amino acid methionine labeled as 35S methionine finally graph the results and the trend will suggest the type of diffusion facilitated diffusion is faster because there is an actual hole for it to go through as opposed to working its way through the lipid bilayer RBC ghosts 1 put RBC into water so they will expand and contents will leak out 2 place them in isotonic solution and resume original shape 3 vacant cell is created you know exactly what it contains no internal contents to interfere with quantitative analysis more accurate telling of Vmax you can alter initial loading of the RBC now 1
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