DNA = deoxyribonucleic acid
DNA is a polymer, the subunits are nucleotides
...N1-N2-N 3-N4-N5....
A nucleotide is composed of:
PHOSPHATE-SUGAR-BASE
The phosphate and sugar compose the backbone while the base is the variable region.
In DNA sugar = deoxyribose (dR) has 5 carbons
Numbering of atoms in nucleotides:
Sugar: 1', 2', 3', etc.
Base: 1, 2, 3, etc.
Bases:
A Adenine
T Thymine
G Guanine
C Cytosine
The phosphate group has a negative charge which spreads the DNA out into linear formation.
Nucleotides are joined by linking the phosphate on the 5' end of the deoxyribose of the second one to the 3' position of the first.
Linking nucleotides: The 3'-OH of one nucleotide is linked to the 5'-phosphate of the next nucleotide. One way to depict this is by a graphic of staggered linked nucleotides.
DNA base composition
A + G = T + C
A and G are purines which are larger than T and C which are the smaller pyrimidines
Watson and Crick used chemical data and x-ray data to determine the double helix model for DNA. In the double helix G pairs with C and A pairs with T. The two strands of DNA run antiparallel, each base binds with its complementary base on the opposite strand. There is a major groove and a minor groove in the DNA double helix. Regulatory proteins can often fit into the major groove. The repeat unit is 34 Å, 10 base pairs.
Base pairing
A::T has 2 hydrogen bonds
G:::C has 3 hydrogen bonds
Always pairing a purine and a pyrimidine yields a constant width.
DNA conventions
I. DNA is a right-handed helix. If the observer looks down the helix axis (in either direction) each strand turns clockwise as it moves away from the observer.
II. To write a sequence
[P-5']-A-T-C-C-G-[3'-OH]
The 5' end is to the left by convention and above molecule is written: ATCCG
The antisense strand read from the 3' end is: TAGGC
The ultimate secret of heredity:
If you have one strand you can make the other one.
i.e. The strands are complementary due to base pairing.
------------------ sense
------------------ antisense
Structure of pyrimidines (C4N2H 4) and purines (C5N4 H4)
3 possible types of hydrogen bonds
Structure of a short double stranded piece of DNA
Bacterial DNA
1 double stranded circular chromosome, about 3000 genes
Molecular weight = 2.5 x 10*9
Percent of cell material = 3%
E. coli chromosome - Genetic Map
ori = origin
ter = terminus
|
ORGANISM |
GENES |
BASE PAIRS |
| small virus | 3 | 4 x 10*3 |
| large virus | 200 | 3 x 10*5 |
| bacterium | 3000 | 3 x 10*6 |
| yeast | 10,000 | 1 x 10*7 |
| man/fruitfly | 100,000 | 1 x 10*8 |
Average gene of 1000 bp can code for average protein of just over 300 amino acids
In higher organisms up to 95% of DNA is non-coding. Some of it may have originally been viral DNA.
Non-coding DNA of:
bacteria = 10-20%
virus = very little
Chromosome = long DNA double helix
Functional division of bacterial chromosome: intergenic regions, regulatory regions and genes
Intergenic region = in bacteria is 50 base pairs or less
Regulatory region = precedes the gene
Structural gene or coding region = codes for protein
Gene: A region of DNA that is transcribed to produce a piece of RNA
At any particular time a bacteria is only using about 1/3 of its genes
Human chromosomes
23 pairs, the 23rd is the sex chromosome
Standard picture of chromosome is misleading because they are caught in the act of division (3rd picture).
DNA dimensions
1 base pair = 660 daltons
1 helical turn = 10 bp
E. coli (typical bacterium)
has: 3000 genes
3 x 10*6 bp
2 x 10*9 daltons
1 millimeter of DNA
cell = 1/1000 mm long
DNA of prokaryotes is circular = covalently closed circle (CCC)
nicked circle = open circle
covalently closed circles (CCC) are usually supercoiled
a) bring ends of linear double helix together to form circle
b) before joining up, unwind one turn of the double helix, the CCC twists (in a left-handed way) to form a figure-8
c) in typical DNA, one supercoil per 200 bp
Packaging
a) supercoils
b) protein scaffold for 50 giant loops
Reason for supercoiling:
Each unwinding of double helix unpairs 10 base pairs, therefore the helix will rewind to reform the base pairs. This forces a compensatory twist at the higher structural level.
Each unwinding = one left handed supercoil
Each overwinding = one right handed supercoil (not found naturally)
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Last updated: 23-Jan-08 / ksb
