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Final

BIOB11H3 Final: BIOB11 Final


Department
Biological Sciences
Course Code
BIOB11H3
Professor
Dan Riggs
Study Guide
Final

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BIOB11: MOLECULAR CELLULAR AND GENETIC PRACTICAL WINTER 2018
BIOB11
Lecture 1
THE SEARCH FOR THE HEREDITARY MOLECULE
1900-1040
Biochemists Characterize cellular macromolecules
DNA is a polymer that has four nitrogen bases
Proteins are polymers made of 20 amino acids
o Since proteins were more complex, they thought that was the genetic material
They were both abundant in the cell
Evidence that DNA is the Genetic Material
Avery, McCarthy and McLeod (1940)
o Studies of virulent strains of bacteria showed that a transforming principle could be passed from one bacteria to another and
that DNA may be the transforming principle that carries the information
o They said the transforming principle was DNA based on:
Chemical properties of the transforming principle were consistent with DNA
No other material could be detected in the preparation
o Experiment using a Direct Test
1. Encapsulated strain
2. Lyse cells, filter to obtain cell free extract (had all the insides)
3. Use enzymes that degrade a specific macromolecule (DNAse, proteinase etc.) into the extract
4. Inject mice with the altered extract to determine if it retained the transforming property
o The result was that only the DNAse enzymes could inactivate the transforming principle, so DNA is therefore the genetic
material
o If you used the protease, the mouse still got the infection, therefore the transforming principle was still present so it couldn’t
have been proteins
Hershey and Chase (1950)
o Studied viruses that infect bacteria and proved that phage DNA was responsible for the viral infection
o Bacteriophage virus that infects bacteria
Contains a protein coat that encases the DNA
It is released from a host cell and finds another new host cell that it
can inject the DNA into and it uses that cell to reproduce until it
bursts and releases more phage
o Experiment
The bacteriophage displays heritable properties
They wanted to know if the protein or DNA directed the replication
of the new bacteriophage
They radiolabelled the protein or DNA of a bacteriophage and
infected the bacteria cells
When the DNA was labelled, there was radiolabel found in the
bacteria/ new phage DNA
When the protein was labeled, the radiolabel was found in empty phage particles
o They too concluded that DNA was the genetic material because that is what caused the infection
Griffiths
o Came before the Avery et al experiment
o Showed that heat killed lysed streptococcus pneumonia bacteria contained a transforming principle
o The encapsulated(s) strains were virulent, the capsuleless (r) strains are not virulent
Inject live s cells mouse dies from infection
Inject live r cell mouse lives
Inject heat killed s cells mouse lives
This shows that heat can kill the contagious aspect
Heat killed s cells and live r cells mouse dies from infection
When you combined the two non contagious trials, the mouse still died…. Why
There must be a soluble factor that is released from the s cells to the r cells to transform them into a
virulent cells
o The transforming principle transformed harmless cells into virulent cells
o Therefore there must be some factor that can be taken up by the live r cells to transform them into the virulent s cells
transforming principle

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BIOB11: MOLECULAR CELLULAR AND GENETIC PRACTICAL WINTER 2018
DNA Structure
Nucleotide phosphate group, sugar, and nitrogenous base
Nucleoside the structure without the phosphate group
The phosphate and sugar are connected by phosphodiester bonds
Nitrogenous Bases
o Purines arginine and guanine
o Pyrimidines cytosine, thymine and uracil
Early Work on DNA: Physiochemical Approaches
Base pairing rules established by analyzing base compositions
DNA absorbs light in ultraviolet range
Renaturation experiments defined complexity of genome
Chargaff (1950s)
Base compositions differ between organisms
He notices that there were equal amounts of A to T and for C to G, there for they must be paired together, but there were not equal
amounts of AT to GC
AT bond with 2 hydrogen bonds
CG bond with 3 hydrogen bonds
Absorbance Measured by Spectrophotometer
Ring structure of DNA absorbs light in the ultraviolet range
DNA absorbs maximum 260 nm
Absorbance can be used to determine DNA concentration
The phototube helps determine how much light was absorbed vs how much passed through
The Hyperchromic Effect get fig 10.16
o You take a solution of DNA and put it in a water bath and turn up the heat in the water bath and take samples out and look at
them through the mass spectrometer
o Absorbance increases about 40% if DNA is denatured
At the temperature where the heat over comes the attractive forces of the hydrogen bonds holding the helix together
and you get single stranded DNA
o Tm is the temperature at which DNA is denatured, when half of the bonds have been broken
o Thus changes in absorbance reveal % double stranded vs. % single stranded. Why…
As temp goes up single stranded DNA goes up and double goes down, when the temp goes down vise versa
Nitrogenous bases are aromatic rings that exhibit resonance
Single stranded DNA is flexible
o There is no stacking of bases
o No hydrogen bonding to limit the resonance of rings
o There is more room for the electrons to accept the energy since they are delocalized
Double stranded DNA is more rigid due to base stacking and hydrogen bonding with complementary strand
o Therefore the resonance is limited so absorption is limited
Therefore absorption of light is more limited in DS DNA
Complexity
How many unique sequences are there in the genome
o The more unique sequences, the more complex
o Unique reanneals slower than repetitive
A measure of the number of unique (vs repetitive) sequences that exist in a genome
DNA reannealing (and later use of hybridization techniques) is useful for determining aspects of complexity
Renaturation (reannealing) of denatured DNA depends on many factors, like ionic strength, length of fragments, temperature and time
o You boil them so they denature, then you lower the temp and they reanneal
o Factors that affect reannealing
The new temperature matters, has to be lower than the temperature that denatures hydrogen bonding
The ions influence
Long fragments can get tangled
If there is a limited time, not much will happen
Procedure for reannealing of denatured DNA
1. Purify DNA, shear to average size of 1000-2000bp
2. Denature (heat), then allow to reanneal at lower temperature
3. Measure % reassociated (by absorption) over time, by using a mass spectrometer over a varying amount of time
4. The length of time required for half of the molecules to reanneal is referred to as the Cot ½ (concentration, time) and is a
measure of complexity of the DNA sample

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BIOB11: MOLECULAR CELLULAR AND GENETIC PRACTICAL WINTER 2018
a. If it is high it takes a long time for the items to come together
Kinetics of Renaturation of Bacterial and Viral DNAs
Remember there are many copies of a genome
o 1µg of DNA= 2.1 x 10^8 E. coli genomes, 1.8 x 10^11 MS-2, T4 genomes
This shows that there are many copies of the molecules in that
1microgram of DNA
o The phage genomes are smaller than the e coli genomes so there are more
copies with in the same measurement of the DNA types
Complete reassociation over two logarithms of COT indicative of unique (non-
repetitive) sequences, so these genomes consist of unique sequences (none or very
few copies/genomes
o The curves on the graph are characteristic of genomes that are unique and don’t have many repetitive sequences
o The MS-2 and T4 phages have lower COT ½ values that E.coli because they
were smaller gemones, so in the same measured amount of DNA taken,
there were more copies of the genome, so it was easier for them to find a
copy
Repetitive has a lover COT value than complex
Faster reannealing by MS-2 and T4 vs E.coli due to ease of finding a pairing partner
because there are more copies of the genome
The phages have a lower number because they have more copies, so it is easier for
them to reanneal than it is for e coli
o They can find complementary patterns faster
o These two phage genomes were more simple
More complex the longer it takes to find a match
Complex eukaryotic genomes have three discrete fractions
o Part anneals really quickly because there are some repeated sequences, and there is some that take a long time to reanneal
because it is unique
o The repetitive sequence families make up half of our genome
o The nonrepeated sequences are the genes that are the functional part of the genome
Genome Complexity of Model Organisms
Genomes and organism complexity are not strictly related to genome size
The number of protein coding genes is generally reflective of organismal complexity
Plants have a smaller genome, but much more genes relative to that ratio of other organisms
o The mustard plant is ridiculous
o Plants have had to evolve systems to deal with environmental stresses
Lecture 2: Genome Evolution and Polymorphisms
EVOLUTION OF GENOMES
Unequal Crossing Over
When the genes do not separate equally
Divergence of function driven by mutations/selections of the duplicated
genes
Generally occurs during meiosis (formation of gametes in mammals
wherein sperm and egg are produced
Leads to one chromosome having a duplicated segment and the other
missing a segment, this is caused when the chromosomes are slightly
miss aligned
Normal alignment of homologous chromosomes (one pair from father
and mother) occurs precisely so that reciprocal genetic exchange
generates complete sets of potentially recombinant chromosomes
Implications
o Increase gene number (duplication: create gene family)
There is multiple copies of the same gene that undergo mutations that can alter their function
o Also may result in loss of genes (deletion: often lethal)
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