CLASS: 10:00 – 11:00
Scribe: Rachel Tucker
DATE: September 9, 2010
Proof: Karin Tran
PROFESSOR: Chesnokov
DNA STRUCTURE
Page 1 of 5
I. DNA STRUCTURE [S1]
a. Subject of the lecture is the structure of DNA
i. nucleotide building blocks
ii. primary, secondary, tertiary structure
iii. how DNA is assembled into chromosomes
II. GENOME SIZES COMPARED [S2]
a. Sizes of genomes in different organisms
b. Large variation: from 106 nucleotide pairs per haploid genome for some bacteria to 1012 for others
c. Generally, more complex organisms have more DNA, but there are some exceptions
i. Some protists and plants have significantly larger genomes than some mammals
ii. The amount of DNA content does not necessarily estimate the complexity of the genome.
d. READ SLIDE
III. CENTRAL DOGMA OF MOLECULAR BIOLOGY [S3]
a. READ SLIDE
b. This constitutes the central dogma of molecular biology, which describes information transfer in cells.
IV. FIGURE: CENTRAL DOGMA [S4]
a. Illustration of the central dogma
b. Replication of DNA generates two identical DNA molecules that are identical to the parent molecule. This
process ensures the transfer of genetic information from parent to daughter cells.
c. Information encoded in DNA can be translated in the form of messenger RNA, which is subsequently read by
ribosomes during the process of protein synthesis.
V. ESSENTIAL QUESTIONS [S5]
a. Essential questions to ask during the lecture
b. What are the structures of the nucleotides, the building blocks for DNA?
c. How are nucleotides joined together to form nucleic acids?
d. What is the higher-order structure of DNA? This includes the secondary and tertiary structural information of
eukaryotic chromosomes.
e. What are the biological functions of nucleotides and nucleic acids in cells?
VI. OUTLINE (1/2) [S6]
a. Outline of the lecture
b. READ SLIDE
VII. OUTLINE (2/2) [S7]
a. READ SLIDE
VIII. STRUCTURE AND CHEMISTRY OF NITROGENOUS BASES [S8]
a. The building blocks of nucleic acids are nucleotides, but the bases of nucleotides and nucleic acids are
derivatives of either purines and pyrimidines.
b. This is a summary of purines and pyrimidines found in nucleic acids.
c. Pyrimidines: some are found both in DNA and RNA; some are specific only for DNA or RNA
d. Purines: both can be found in DNA and RNA
IX. FIGURE: GENERAL STRUCTURE OF PYRIMIDINE AND PURINE RINGS [S9]
a. READ SLIDE
b. Pyrimidine ring has 2 N atoms at positions 1 and 3
c. Purine ring: pyrimidine ring is on the left; imidazole ring is on the right
X. FIGURE: PYRIMIDINE BASES [S10]
a. Important to know the names of these structures
i. cytosine can also be called 2-oxy-4-amino pyrimidine.
ii. Uracil can be called 2-oxy-4-oxy pyrimidine.
iii. Thymine can be called 2-oxy-4-oxy-5-methyl pyrimidine.
XI. FIGURE: PURINE BASES [S11]
a. Listed are short names and specific chemical names.
XII. WHAT ARE NUCLEOSIDES? [S12]
a. When a base is linked to a sugar, it forms a nucleoside
b. This is a summary of the facts you need to know about nucleosides.
c. The sugars present in nucleosides are always pentoses.
d. The difference between the RNA and DNA pentose is the presence of a –OH at position 2 in RNA, and a –H
at position 2 in DNA
i. The presence of the 2’-OH dramatically affects secondary structure and stability of nucleic acids
e. READ SLIDE
CLASS: 10:00 – 11:00
Scribe: Rachel Tucker
DATE: September 9, 2010
Proof: Karin Tran
PROFESSOR: Chesnokov
DNA STRUCTURE
Page 2 of 5
XIII. FIGURE: TYPICAL SUGARS FOUND IN NUCLEOSIDES [S13]
a. Shown are D-Ribose and D-2-Deoxyribose (found in DNA)
b. These sugars in ring form are called furanoses.
c. The presence of the additional –OH in RNA makes RNA significantly more susceptible to hydrolysis
i. This makes sense because the lifetime of RNA is significantly lower than that of DNA
ii. DNA is significantly more stable because it serves as a repository of genetic information.
iii. RNA is made to perform certain functions and then is subsequently destroyed.
XIV.
FIGURE: -GLYCOSIDIC BONDS LINK NITROGENOUS BASES AND SUGARS… [S14]
a. Sugars and bases are connected through glycosidic bonds
b. The C at position 1 is linked to the N at position 9 of the purine base (bottom illustration), or the N at position
1 of the pyrimidine base (top illustration)
XV. FIGURE: COMMON RIBONUCLEOSIDES [S15]
a. Very important to know these names: cytidine, uridine, adenosine, guanosine
i. These four are most commonly found
b. Inosine
i. an uncommon nucleoside which is somewhat homologous to the base guanosine
ii. found in the nucleic acids of some plants
XVI.
FIGURE: ROTATION AROUND THE -GLYCOSIDIC BOND IS POSSIBLE [S16]
a. With rotation around this bond, syn and anti conformations can be formed
b. Pyrimidine bases (pink structure) adopt the anti conformation because the O of the pyrimidine ring sterically
hinders the syn conformation.
c. Both syn and anti conformations are possible for the purine bases.
i. This affects secondary structure of DNA
ii. syn conformation is found in the Z structure of DNA
XVII. STRUCTURE AND CHEMISTRY OF NUCLEOTIDES [S17]
a. READ SLIDE
b. Most nucleotides in the cell are ribonucleotides because there is significantly more RNA in the cell.
c. All nucleotides have acidic properties due to the addition of the phosphoric group; thus, they are called
nucleic acids
XVIII. STRUCTURES OF FOUR COMMON RIBONUCLEOTIDES [S18]
a. AMP, GMP, CMP, UMP
i. In the case of AMP, this stands for adenosine 5’-monophosphate; can also be called adenylic acid
ii. This is the same for the other four nucleotides
b. The orange structure on the bottom right is the uncommon 3’-AMP nucleotide. This is a product of hydrolysis
of nucleic acids which also can be found in the cells.
c. Deoxyribonucleotides have the same color, with the exception of the –OH group at position 2 (the right –OH
at the bottom of the furanose ring).
XIX.
FUNCTIONS OF NUCLEOTIDES [S19]
a. These are facts to remember
b. Nucleotides (NTPs) and deoxynucleotides (dNTPs) are substrates for RNA and DNA, respectively.
c. The most common functions of the nucleotides ATP, GTP, CTP, and UTP are listed
d. The most important thing to remember is that NTPs are carriers of energy, which is stored in phosphoric
bonds.
e. READ SLIDE
XX. FIGURE: PHOSPHORYL AND PYROPHOSPHORYL GROUP TRANSFER REACTIONS [S20]
a. This type of group transfer results in the release of energy which is utilized in subsequent biological reactions.
b. Many proteins involved in DNA metabolism contain sites for binding and hydrolyzing ATP, which is the prime
source of energy for biological work.
XXI.
FIGURE: CYCLIC NUCLEOTIDES [S21]
a. READ SLIDE
b. In the case of cyclic nucleotides, phosphoric acid, which is normally bound only at position 5, is esterified to
the second available –OH group as well, at position 3 of the furanose ring.
c. Cyclic nucleotides are regulators of many metabolic processes found in the cell are found in pretty much in all
eukaryotic and prokaryotic cells.
XXII. WHAT ARE NUCLEIC ACIDS? [S22]
a. Nucleic acids are essentially polar nucleotides.
b. READ SLIDE
c. It’s very important to know shorthand notations
CLASS: 10:00 – 11:00
Scribe: Rachel Tucker
DATE: September 9, 2010
Proof: Karin Tran
PROFESSOR: Chesnokov
DNA STRUCTURE
Page 3 of 5
XXIII. FIGURE: RNA AND DNA [S23]
a. RNA is on the left, DNA is on the right
b. The difference between the two is the presence of the second –OH on position 2 of the furanose.
c. READ SLIDE
XXIV. FIGURE: SHORTHAND NOTATIONS FOR POLYNUCLEOTIDE STRUCTURES [S24]
a. READ SLIDE
b. Furanoses or pentoses are represented by vertical lines.
c. Sometimes, in describing a sequence, only the base sequence is shown without the other structures
represented.
XXV. WHAT ARE THE DIFFERENT CLASSES OF NUCLEIC ACIDS? [S25]
a. There is just one type of DNA, with one purpose: the storage and propagation of genetic information.
b. Several different types of RNA are found in cells. These will probably be discussed in other lectures.
i. Ribosomal RNA: the basis of structure and function of ribosomes
ii. Messenger RNA: carries genetic information for translation into proteins
iii. Transfer RNA (tRNA): carries amino acids during the synthesis of proteins
iv. Small nuclear RNA and small non-coding RNA perform various cellular functions.
XXVI. THE DNA DOUBLE HELIX [S26]
a. Now, concerning the secondary structure of DNA.
b. Two major discoveries led to the formulation of the DNA double helix model by Watson and Crick.
i. First: Erwin Chargaff found that the number of purine residues is equal to the number of pyrimidine
residues in all organisms.
ii. Second: Rosalind Franklin’s x-ray fiber diffraction data which shows that DNA has a helical structure.
XXVII. TABLE 10.3 [S27]
a. Molar ratios that led Chargaff to formulate his famous Chargaff rules.
b. The ratio of adenine to guanine, for example, can be very different
c. However, the ratios of adenine to thymine and guanine to cytosine are always close to 1. These are ratios of
purines to pyrimidines.
d. The amounts of purines and pyrimidines in the cell are equal.
XXVIII. FIGURE: WATSON AND CRICK BASE PAIRS: ADENINE AND THYMINE [S28]
a. Chargaff’s rules and Rosalind Franklin’s x-ray diffraction data led Watson and Crick to formulate their model.
b. DNA bases interact via hydrogen bonds
i. Two H bond form between adenine and thymine
ii. Three H bonds form between cytosine and guanine.
XXIX. FIGURE: WATSON AND CRICK BASE PAIRS: CYTOSINE AND GUANINE [S29]
XXX. A MODEL OF THE DNA DOUBLE HELIX [S30]
a. Interaction of bases results in the formation of the DNA double helix.
b. Nucleotides of each strand are linked covalently by phosphodiester bonds.
c. Two DNA strands are held together by H bonds (shown in the blown up box to the right)
d. Bases can pair only if two polar nucleotide chains are arranged antiparallel to one another.
e. Each strand contains a sequence exactly complementary to that of its partner strand.
XXXI. STRUCTURE OF DNA SUMMARY [S31]
a. READ SLIDE
XXXII. COMPARISION OF A, B, Z DNA [S32]
a. There are several secondary structures that a DNA helix can adopt.
b. B structure is the most common and is found in all cells
c. A structure can be created artificially from dehydrated DNA; most likely doesn’t exist in vivo.
d. Z structure is found in GC rich regions
XXXIII. FIGURE: COMPARISION OF A, B, Z FORMS [S33]
a. Two strands wrap around each other to form major and minor grooves.
b. In dehydrated DNA, A structure is wider and the major and minor grooves are essentially close in size to each
other.
c. Z structure forms a left-handed helix. This is because bases in G-C rich regions are often flipped to syn
conformation. This makes the structure thin, left-handed, and rather long.
XXXIV. HOW DO SCIENTISTS DETERMINE THE PRIMARY STRUCTURE OF NUCLEIC ACIDS? [S34]
a. Sequencing: the process of determining the primary structure of nucleic acids
b. Sanger’s chain termination method is the most commonly used.
c. READ SLIDE
CLASS: 10:00 – 11:00
Scribe: Rachel Tucker
DATE: September 9, 2010
Proof: Karin Tran
PROFESSOR: Chesnokov
DNA STRUCTURE
Page 4 of 5
XXXV. FIGURE: CHAIN TERMINATION METHOD [S35]
a. DNA polymerase, in the presence of the four nucleotides, can extend radioactively labeled primers to
synthesize a complementary strand of DNA. This is done in a 5’ to 3’ direction.
b. When you incorporate chemically modified nucleotides that cannot be extended, DNA polymerase will extend
the complementary strand until it uses one of these chemically modified nucleotides, at which time it will stop
DNA synthesis.
XXXVI. CHAIN TERMINATION METHOD [S36]
XXXVII. FIGURE: CHAIN TERMINATION METHOD OF DNA SEQUENCING [S37]
a. The chain termination method is show in more detail here.
b. Anneal single stranded DNA to a specific, radioactively labeled primer.
c. Run four reactions. Each reaction contains the four deoxy nucleotides and a small amount of one of the four
nucleotides that is chemically modified to a dedeoxynucleotide.
d. Dedeoxynucleotide doesn’t have a hydroxyl group at position 3. Therefore, once it’s incorporated into the
growing chain, nothing can be added further and the sequence is terminated.
e. A small amount of dedeoxynucleotide in the mix results in a number of terminated strands which correspond
to the parent strand.
f. These terminated strands can be separated in a polyacrylamide gel, and due to the presence of the
radioactively labeled primer, they may be visualized by autoradiography and their sizes determined.
g. Because four reactions are run, each with only one of the four dedeoxynucleotides, the position of each
specific base may be determined.
XXXVIII.
CHEMICAL CLEAVAGE METHOD [S38]
a. This method is a little bit more complicated, but has certain specific uses that the Sanger method cannot
provide.
b. Start with single stranded DNA that is radioactively labeled on one end. The strand is then cleaved by a
chemical reagent
c. Chemical reagents are selected that can modify a specific base and then cut at that position.
d. This results in many fragments which can be electrophoresed and read.
XXXIX. FIGURE: PHOTORADIOGRAPH OF A SEQUENCE [S39]
a. Here, the reaction is run four different times with four reagents that each cut at one of the four bases.
b. This technique may be used to determine sites of protein binding to DNA.
i. Bind the protein to the DNA then subject the DNA to chemical modifications
ii. Bound protein will protect its binding site from being cut.
iii. In a sequencing reaction, the site where the protein is bound will appear as a protected area on the
sequence.
iv. This how the binding sites of many transcription factors were found. This method is still often used in the
lab.
XL. CAN THE SECONDARY STRUCTURE OF DNA BE DENATURED AND RENATURED? [S40]
a. Denaturation and annealing of DNA is the most common method to study the secondary structure of DNA.
b. This is useful to the study of genome complexity.
c. When DNA is heated to 80+ degrees C, its UV absorbance increases 30-40%. This is because with the rise in
temperature, nitrogenous bases are released, no longer stacked together, and can thus absorb more light.
d. READ SLIDE
XLI.
FIGURE: STEPS IN THE THERMAL DENATURATION AND RENATURATION OF DNA [S41]
a. When the DNA is heated, the parent strands separate. They reanneal when the temperature is lowered.
b. Annealing occurs in two steps
i. Nucleation: the bases find their counterparts on the complementary strand
ii. Zippering: when DNA is essentially renatured.
XLII.
FIGURE: RATES OF REASSOCIATION OF DENATURED DNA FROM VARIOUS SOURCES [S42]
a. Denaturation and annealing is important to study DNA complexity.
b. It’s very easy to anneal DNA that is very simple. A poly A strand will anneal to a poly U strand very quickly
(black line).
c. In the case of complex DNA, as in that isolated from a nonrepetitive fraction in calf cells (purple line), it takes
an extremely long time for the DNA to reanneal.
d. By doing these experiments, it’s easy to find repetitive sequences in DNA. For example, mouse satellite DNA
(pink line) anneals very quickly because it contains a large amount of AT rich repetitive sequences.
e. Sequences which are enriched with unique genes, like E. coli (orange line), or some sequences derived from
eukaryotic DNA, will take significantly longer for renaturation and annealing.
CLASS: 10:00 – 11:00
Scribe: Rachel Tucker
DATE: September 9, 2010
Proof: Karin Tran
PROFESSOR: Chesnokov
DNA STRUCTURE
Page 5 of 5
XLIII. TERTIARY STRUCTURE OF DNA [S43]
a. Duplex DNA can form a tertiary structure.
b. An example of this is sometimes found in circular DNA
i. In it’s secondary (B) structure, it sometimes contains more or less than 10 bp per turn.
ii. This is called a supercoiled state, which can be underwound (fewer than 10bp per turn) or over wound
(more than 10 bp per turn).
XLIV. FIGURE: VARIOUS TERTIARY STRUCTURES OF DNA [S44]
a. (a) toroidal structure
b. (b) interwound DNA supercoil
c. (c) both of these structures are found in eukaryotic chromosomes
i. DNA is attached to certain sites in the nuclear matrix
ii. Both interwound and toroidal varieties of DNA structure are found there.
XLV. FIGURE: SIMPLE MODEL FOR THE ACTION OF BACTERIAL DNA GYRASE (TOPOISOMERASE II) [S45]
a. Gyrase can introduce supercoils to circular DNA.
b. READ SLIDE
XLVI. FIGURE: SUPERCOILED DNA IN TOROIDAL FORM [S46]
a. READ SLIDE
b. This structure is found in the formation of nucleosomes.
XLVII. FIGURE: DIAGRAM OF THE HISTONE OCTAMER [S48]
a. DNA wraps around a histone octamer to form the toroidal tertiary DNA structure
XLVIII. EM OF NUCEOSOMES [S49]
a. Top: partially denatured eukaryotic chromosome
b. Bottom: nucleosomes are quite visible as “beads on a string”, which is partially unfolded chromatin
XLIX. FIGURE: STEPS OF EUKARYOTIC CHROMOSOME FORMATION [S50]
a. READ SLIDE
b. Pay attention to the number of base pairs within each structure; he mentioned these.
L. STRUCTURE OF EUKARYOTIC CHROMOSOMES [S47]
a. READ SLIDE
LI. FIGURE: EACH DNA MOLECULE THAT FORMS A LINEAR CHROMOSOME MUST CONTAIN… [S51]
a. He just talked about what will be discussed in his other two lectures.
[End 40:54 mins]