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Lecture 2

BIOB11H3 Lecture Notes - Lecture 2: Centromere, Chromosome, Thymidine Triphosphate


Department
Biological Sciences
Course Code
BIOB11H3
Professor
Dan Riggs
Lecture
2

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BIOB11H3S
Lecture 2 Genome Organization and Molecular Evolution
- DNA: double-stranded, antiparallel helix
- Complementary base pairing by H-bonding: A-T and C-G
- Early History on DNA PHYSIOCHEMICAL
1. DNA is polynucleotide chain
2. Estimates made of genome size
3. Values were ‘tiny yet enormous’
Humans = 3.5 picograms of DNA/haploid
Genome = 3.2 billion base pairs
4. Base pairing rules established by analyzing base composition
5. DNA absorbs light in ultraviolet range
6. Renaturation experiments defined complexity
- Chargaff (1950s) base composition differs b/w
organisms
A-T = 2 H-bonds and C-G = 3 H-bonds
- Absorbance measured by spectrophotometer
Ring structures of DNA absorb in ultraviolet range
DNA reabsorption MAX = 260 NM
Absorbance can be used to determine DNA concentration
Absorbance increases about 1.5 X if DNA IS DENATURED
Satellite Dish Analogy
ssDNA: Flexible, nitrogenous bases can freely rotate about glycosidic
bonds, PLANAR, MANY ARE IN ODD ANGLES, SIGNLE STRANDED
DNA CAN ABSORB MORE LIGHT THAN DOUBLE STRANDED DUE TO
ORIENTATION
dsDNA: More rigid, nitrogenous bases limited in # of position they can occupy
Complexity: Measure Of # of unique (vs. repetitive) sequences that exist in
genome
- DNA reannealing (& later use of hybridization techniques) is useful for determining
aspects of complexity
- Renaturation/renealling of denatured DNA DEPENDS ON MANY FACTORS (i.e. ionic
strength, length of gragments, temp, time)
- Procedure:
1. Purify DNA, shear to 1000-2000 bp
2. Denature/heat, then allow reanneal at
lower temp
3. Measure % of reassociated by absorption
of time
4. Length of time required for ½ of
molecules to reannel = Cot1/2, measure of
cmplexity
- Similar patterns indicate nique/non-repetitive
sequences
- Cot values refeleccts complexity (genome size, # of each ‘type’ of
sequence)
- Cot cruves and Matching socks
You have 1 black sock, need to fin other one.
1. If drawer has 500 socks, 400 are balck, no problem finding other quickly
2. If drawer has 500 socks, 100 are black, takes longer
3. If drawer has 500 socks, 2 are black, takes much longer
# of sock to # DNA fragments, some match, some don’t
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