1. Define the following terms and
utilize these terms in an appropriate context:
adenine
antiparallel chain
complementarity
cytosine
deoxyribonucleic acid
guanine
nitrogenous base
nucleotide
purine
pyrimidine
replication
ribonucleic acid
thymine
transcription
transformation
transgenic
uracil
semi-conservative
2. Discuss the characteristics that the genetic material must have inorder to be an effective material for the job.
3. List and explain the meanings of the primary functions ascribed to the genetic material: replication, expression, storage and mutation.
4. Discuss the reasons why proteins were generally favored over DNA as the genetic material before 1940.
5. Contrast the various contributions made to an understanding of transformation by Griffith with those of Avery and his coworkers.
6. Explain the transformation of IIR pneumococci into IIIS pneumococci, including the identification of the "transforming substance".
7. Describe the experiment performed by Hershey and Chase. Discuss why 32P and 35S were chosen for use in their experiment. Discuss the rationale and conclusions of this experiment.
8. Discuss the "Central Dogma of Molecular Genetics" making observations which are consistant with DNA serving as the genetic material in living organisms. List any exceptions to this dogma.
9. Draw the chemical structure of the three components of a nucleotide and then link the three together.
10. Show by diagrams how a phosphate group joins the 3' carbon of one deoxyribose to the 5' carbon of the next deoxyribose.
11. Describe the various characteristics of the Watson-Crick double-helix model for DNA.
12. Explain the statement that DNA consists of antiparallel and complementary strands, which read 3'-5' on one strand, but '5-3' on the other.
13. Describe the bonding between the nucleotide bases on one DNA strand and the nucleotide bases of the antiparallel strand.
14. Discuss the contributions of Erwin Chargaff to the development of rules pertaining to DNA.
15. List three main differences between DNA and RNA.
16. Discuss the chemical basis of molecular hybridization.
Resources: Text Chapter 10, Cartoon
Guide pgs. 97-107; pgs. 120-128
Chemistry of the Gene
I. Search for the genetic
material
A. Requirements
1. Must be able to store info used to control development and metabolic
activities of the cell or organism.
2. Must be stable so that it can be replicated with high fidelity during
cell
division and be transmitted from generation to generation
3. Must be able to undergo rare changes called mutations in order to generate
variability that is acted upon during evolution (changes must be
inheritable)
4. Must be chemically diverse
5. Must be capable of being used.
B. Early researchers tabbed protein as genetic material
-Swiss chemist Meischer, removed nucleus from cells and discovered nuclein,
rich in phosphorous but no sulfur (diff. from protein)
-later named nucleic acids, discovered DNA consisted of only four repeating
units,
Therefore, no variability, can't be genetic material
C. 1928 Frederick Griffiths pneumonia bacteria
S strain ---> produces protective coat (virulent)
R strain ---> no protective coat (avirulent)
Transformation in bacteria
mice ---> S strain ---> die
mice ---> R strain ---> live
mice ---> heat killed S strain ---> live
mice ---> mix heat killed S and live R ---> die
Bacteria recovered from dead mice were living S strain. Therefore,
avirulent
transformed to virulent
Therefore, some virulent trait passed from dead S to live R and this trait
continued
to be passed from generation to generation.
-assumed that change in phenotype due to change in genotype
D. Oswald Avery (1944)- worked with R and S strains in vitro -used numerous
chemical techniques to further identify what molecule was transferring
trait
1. DNA caused transformation
2. Enzymes that degrade protein had no effect on transform.
3. Enzyme digestion using DNAase did prevent transformation
4. Molecular weight of transforming substance was very large, containing
1600 nucleotides, enough for some variability
**Despite these findings being very convincing, not accepted
E. Hershey and Chase (1952)
-work with bacteriophage, viruses that attack bacteria
-used phage T2 conposed of DNA and protein coat ~50:50
-reasoned that which ever of these enters bacteria and controls phage reproduction
must be genetic material.
-protein contains sulfur; DNA does not DNA contains phosphorous; protein
does not
1. 2 Experiments
a. radioactively labeled P32 was incorporated into the core DNA. Bacteria
grown
in P32 rich environment
b. radioactively labeled S35 was introduced to protein coat. Second type
grown
in S35 rich medium.
c. phages allowed to infect bacteria
d. after short time phages broken off and seperated from bacteria
e. in one experiment they found that most of the P32 labeled DNA remained
behind with the bacteria.
f. in the other they found most of the S35 remained in phage coat
***indicates DNA passed on and DNA genetic material
Structure of DNA - Watson, Crick, Chargaff, Franklin
Nucleotide basic unit,
3 essential components:
1. a nitrogenous base (A, T, C, G)
2. a pentose sugar (5 carbon, in DNA deoxyribose, in RNA ribose)
3. a phosphate group
A. Nitrogenous bases
- 2 kinds
1. nine membered double ring- purines adenine (A), guanine (G)
2. six membered single ring- pyrimidines cytosine (C), thymine (T), uracil
(U)
**DNA contains base T, RNA contains base U, RNA and DNA both use A, C,
G
B. Pentose Sugars -
2 kinds
1. RNA uses ribose
2. DNA uses deoxyribose (missing one hydroxyl at C2 position)
Nucleoside = nitrogenous base + pentose sugar
C. Phosphate OH O- O = P - OH ---> O = P - O- OH O-
Nucleotide = nitrogenous base + pentose sugar + phosphate
Phosphate attaches to 3 carbon of one base sugar and the 5 carbon of the next
5 ---> 3 assembly
**polynucleotides ---> DNA and RNA
Levene- tetranucleotide
hypothesis repeating units had four bases (nucleotides)
Therefore, each nucleotide
would occur equally to the others
Rather ---> long chains of nucleotides poly w/ 1000 nucleotides has 41000
different arrangements
-Phosphodiester- phosphate binds to the 5 carbon on one and the 3 of its
neighbor,
creates the phosphate/sugar backbone
Watson/Crick- Double Helix
Erwin Chargaff 1949-1953 - chromatographic studies to determine amounts of four bases.
Chargaffs Rules
1. Base composition of DNA differs from species to species
2. In each species, however, the # of adenine (a) equals the # of thymine
(T).
Also # of G = # of C
3. sum of purines (A+G) = sum of pyrimidines (T+C)
4. ratio of A+T:C+G does not necessarily equal one.
**Complementary Base Pairing
-purines = pyrimidines Total purines [A]+[G] = Total pyrimidines [C]+[T]
-concentration of adenine and thymine are equal as are guanine and cytosine
Therefore, A pairs with T and C pairs with G in DNA
-bonds between polynucleotide chains are hydrogen bonds A- T two bonds
C- G three bonds
Double Helix
1. polynucleotide chains coiled around central axis. Coiling is plectonic,
meaning coils can only be seperated by completely unwinding them.
2. two chains antiparallel, theri C5 to C3 orientations run in opposite directions.
3. bases stacked and are 3.4 angstroms apart
4. nitrogenous bases on opposite chains form hydrogen bonds. Only A-T and
C-G pairs allowed
5. each complete turn of the helix = 34 angstroms (10 bases per turn)
6. alternating major grooves (wide) and minor grooves (narrow)
Loose Ends
-if DNA of an organism
is 22% cytosine, what are the %s of the other bases?
-if the left hand strand
is: What is the complimentary strand? Which base will be
complimented
first? (anti-parallel)
DNA Denaturation
(melting)
-temp at which hydrogen bonds between nitrogenous bases break
-organic solvents
- C-G composition effects physical properties such as melting point, such
findings are unique to different species
DNA Renaturation
(annealing)
-hybridization
DNA-RNA (ID genes responsible for specific proteins)
DNA-DNA (determine similarities between species)
DNA Replication
-genetic continuity
between parental and progeny cells to be maintained, DNA
must replicate
w/ few mistakes
-Consider human genome:
(copy present in every cell) has some 3 billion base
pairs,
An error rate of 10-6 (one in one million) would create 3000 errors per
replication.
. . excessive and unacceptable -imagine!
Meselson and Stahl- semiconservative replication; viruses, prokaryotes and eukaryotes
Mode of Replication
-each DNA strand serves as a template
-when completely unwound each strand would have an affinity for its compliment
of
nucleotides
-each new DNA molecule would consist of one "old" strand and one "new"
Proposed modes:
Conservative
Semi-conservative
Dispersive
Meselson and Stahl
-E. coli grown in N15 medium ("heavy")
-after many generations all nitrogen containing molecules contain heavy
isotope
-DNA containing N15 can be distinguished from N14 by use of sedimentation
equilibrium centrifugation
-Dense N15 will settle out closer to bottom of tube than N14 (N14 DNA vs.
N15 DNA)
-uniformly labeled N15 cells transferred to normal N14 growth medium
-E. coli allowed to replicate for several generations w/cells removed at
various intervals
-After one generation isolated DNA was at an intermediate density
Centrifuge After one gener.