BIOL 450: Modern Genetics (Module 1 Lecture 1 & 2- Chap. 7: DNA Structure & Replication)
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Nature of Biological Information
- - Organisms develop from a single cell that contains information specifying steps of development and biological properties of that organism
- - Organism made of trillions of cells -> each cell has a nucleus with 2 copies of chromosomes -> chromosome pair -> each chromosome is a long DNA strand with functional regions called genes -> DNA is a double helix
- - 2 pairs of chromosomes in each nucleus
- - each copy is the genome
How did we figure out that DNA is genetic material?
- - Needed to determine what are the cellular macromolecules (carbs, lipids, proteins, nucleic acids: DNA & RNA)
- - Whicch is the hereditary material?
- - Determined by Avery's experiment with Diplococcus pneumoniae (1944)
What was known in the early 1940s?
- - Genes are associated with specific traits (mutations alter gene function, but how?)
- - Genes are protein structures
- - Genes are on chromosomes
- - Chromosomes consist of DNA and protein
What was Griffith's experiment in 1928?
- - Used Diplococcus pneumoniae and mice
- - Used 2 strains:
- - Smooth colonies which are "hidden" from immune system and are virulent (mouse dies)
- - Rough colonies which are not "hidden" so are avirulent (mouse lives)
- - Observation:
- - If S-strain is heat-killed then injected, mouse lives
- - If S-strain is heat-killed but contains transforming factor that can convert non-virulent (R-strain) to lethal form and then both are injected, mouse dies
- - Hypothesis: what chemical component is the transformming factor
How would you test Griffith's hypothesis?
What was Avery's experiment (1944)?
- - Using Diplococcus pneumoniae and mice, he tested Griffith's hypothesis: If he destroys the transforming factor in the heat-killed virulent cells, then it will not transform the live avirulent cells
- - Method:
- - Treat extract of heat-killed virulent (S) cells
- - Mix treated extract with avirulent (R) cells
- - Inject into mice
- - 6 treatments: control than either polysaccharides, lipids, RNA, proteins or DNA destroyed
What was the results of Avery's experiment (1944)?
- - The transforming factor is destroyed when the S-strain extract is treated with DNAse
- - Conclusion: the transforming factor is made of DNA
What was the Hershey-Chase experiment?
- - Studies viruses that infect bacteria
- - Hypothesis: if they label viral DNA and proteins with different radioisotopes, the genetic material will carry only that label into the bacterial cells (DNA contains phosphorus, the proteins contain sulfur)
- - Experiment outline
- - Grow 1 batch of virus in radioactive S-35 (label proteins)
- - Grow 1 batch of virus in radioactive P-32 (label DNA)
What was the results of the Hershey-Chase experiment?
- - S-35 labeled viral proteins were found outside the bacterial cells
- - P-32 labeled viral DNA was found insidde the bacterial cellls
- - Conclusion: DNA is genetic material in virus
3 key properties of DNA
- - Faithful repliication
- - Informational content (encodes proteins)
- - Mostly stable, but mutable
What 2 empirical rules did Chargaff establish?
- - Amount of A = amount of T and amount of G = amount of C
- - Amount of (A plus G) = amount of (C plus T)
What critical experimental result did Rosalind Franklin find?
- - She was first to photograph DNA in its natural state
- - Important insights:
- - DNA is helix
- - Backbonne is on outside
- - Diameter 2nm
- - Repeat every 0.34 nm and 34nm
Who put together the first model of DNA?
- - Watson and Crick
- - They were supposed to be studying proteins
- - They put Chargoff's numbers and Franklins structural data together
Primary structure of DNA
- - Nucleotides are the building blocks (monomers) of the DNA polymer
- - Each nucleotide is composed of 3 chemical groups:
- - 5 carbon sugar (deoxyribose)
- - Phosphte group(s) (1-3 linked to each other)
- - 1 nitrogen base (4 possible)
Structure of the ribose sugar
- - 1'C: attachment site for nitrogen base
- - 2'C: OH attachment- sugar is a ribose (RNA); H attached- sugar is deoxyribose (DNA)
- - 5'C: O is attachment site for phosphate group
- - 3'C: O is attachment site for phosphate group from next nucleotide when polymerizin DNA
A DNA nucleotide contains 1 of 4 possible nitrogen bases
- - Adenine, Guanine, Thymine, Cytosine
- - Adenine and Guanine are Purines (double ring)
- - Thymine and Cytosine are Pyrimidine (single ring)
- - Each nucleotide contains a unique nitrogen base
- - Nitrogen bases face inward: lie flat like stair steps
What is dNTP?
- -The triphosphate form of each nucleotide, used in the synthesis of a DNA polymer
- - Any of the 4 nucleotides: dATP, dTTP, dCTP, dGTP
Primary structure of DNA: monomers to polymer
- - Dehydration synthesis, removes water
- -Removes -OH from 3'C of one nucleotide and -OH from phosphate group of other nucleotide
- - Phosphodiester bond forms between 3'C and phosphate group
The sugar/phosphate backbone is linear
- 1.) Nucleotides are joined by covalent phosphodiester bonds. Covalent bonds are not easily broken
- 2.) Polynucleotide chain is polar
- - at 5' end, free phosphate groups
- - at 3' end, free hydroxyl group
- 3.) DNA chain can be tens of millions nucleotides long
- 4.) Sugar/phosphate bonds form linear, unbranched, 'backbone' of DNA chain
Double helix is stabilized by hydrogen bonds
- - Hydrogen bonds form between slightly electronegative atoms (=O and =N in DNA) and slightly electropositive H's (-OH, -NH and -NH2)
- - Hydrogen bonds are weak (3% strength of covalent bonds, easily broken by heat)
- - 3 H-bonds between C and G
- - 2 H-bonds between A and T
Dimensions of a double-stranded DNA helix
- - Always consistent, always 2.0nm
- - Distanes between base pairs is 0.34nm
- - 10 bases per turn of helix
- - Each turn of helix is 3.4nm
- - Helix is right-handed
Hydrogen bondds in DNA
- - Long double-stranded DNA helix is stabilized by an enormous number of H-bonds
- - DNA is not permamently destroyed when H-bonds are broken
- - DNA is destroyed when covalent bonds in sugar phosphate backbone is broken
Implications of DNA secondary structure
- - Complementarity: if know the sequence and polarity of one strand in the double helix, can predict the sequence and polarity of the complementarity strand
- - If know the amount of 1 nitrogen base in the double helix, can predict the amount of the other 3
- - % purines in 1 strand = % pyrimidines in other
- - % purines in both = % pyrimidines in both of other
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