1. Instructor: Prof. Dr. Fadel A. Sharif
Molecular Biology
Instructor: Prof. Dr. Fadel A. Sharif

2. Course Syllabus Quizzes: 10% Midterm exam: 30% Final exam: 60%
Molecular Biology Lecture Notes. Hansen B. and Jorde L.B., Kaplan Inc.
Cell and Molecular Biology. Chander and Viselli Lippicott’s illustrated reviews.
Journals and Internet Resources
Grades:
Quizzes: 10%
Midterm exam: 30%
Final exam: 60%

3. Topics Nucleic Acid Structure and Organization
DNA Replication and Repair
Transcription and RNA Processing
The Genetic Code, Mutations, and Translation
Genetic Regulation
Recombinant DNA
Genetic Testing
Cell Signaling
Abnormal Cell Growth (Cancer)
Apoptosis

4. Nucleic Acid Structure and Organization
Lecture 1
Nucleic Acid Structure and Organization

5. What is Molecular Biology?
"Study of the synthesis, structure, and function of macromolecules (DNA, RNA, and protein) and their roles in cells and organisms"

6. OVERVIEW: THE CENTRAL DOGMA OF MOLECULAR BIOLOGY
An organism must be able to store and preserve its genetic information (stored in the base sequence of DNA molecules)
pass that information along to future generations, and
express that information as it carries out all the processes of life.
Classically, a gene is a unit of the DNA that encodes a particular protein or RNA molecule.

7. The central dogma of Molecular Biology
What genes do? Genes replicate/ Genes direct RNA  protein synthesis/ Genes accumulate mutations
The central dogma of Molecular Biology

8. Gene Expression and DNA Replication
Transcription, the first stage in gene expression, involves transfer of information found in a double-stranded DNA molecule to the base sequence of a single-stranded RNA molecule. If the RNA molecule is a messenger RNA, then the process known as translation converts the information in the RNA base sequence to the amino acid sequence of a protein.
When cells divide, each daughter cell must receive an accurate copy of the genetic information. DNA replication is the process in which each chromosome is duplicated before cell division.

9. Gene Expression

10. Gene Expression
What genes do?Genes replicate/genes direct RNA  protein synthesis/ genes accumulate mutations

11. Comparison of Gene Expression and DNA Replication
- Produces all the proteins and RNAs an organism requires
- Transcription of DNA: RNA copy
of a small section of a chromosome (average size of human gene nucleotide pairs)*
Translation of RNA: protein synthesis
Occurs throughout interphase
- Transcription in nucleus
- Translation in cytoplasm
- Duplicates the chromosomes before cell division
- DNA copy of entire chromosome (average size of human chromosome, 108 nucleotide pairs)
- Occurs during S phase
- Replication in nucleus
*Recently discovered SMORFs

12. Chromosome
Total size (Mb) 1
249.3 2
243.2 3
198.0 4
191.2 5
180.9 6
171.1 7
159.1 8
146.4 9
141.2 10
135.5 11
135.0 12
133.9 13
115.2 14
107.3 15
102.5 16
90.4 17
81.2 18
78.1 19
59.1 20
63.0 21
48.1 22
51.3 X
155.3 Y
59.4
Total
3095.7

13. The concept of the cell cycle
Can be used to describe the timing of some events in a eukaryotic cell.
The M phase (mitosis) is the time in which the cell divides to form two daughter cells.
Interphase is the time between two cell divisions or mitoses. Gene expression occurs throughout all stages of interphase.

14. Interphase is subdivided as follows:
G1 phase is a period of cellular growth preceding DNA synthesis. Cells that have stopped cycling, such as muscle and nerve cells, are said to be in a special state called Go.
S phase is the period of time during which DNA replication occurs. At the end of S phase, each chromosome has doubled its DNA content and is composed of two identical sister chromatids linked at the centromere.
G2 phase is a period of cellular growth after DNA synthesis but preceding mitosis. Replicated DNA is checked for any errors before cell division.

15. The Cell Cycle

16. Cyclins and CDKs
Cell-cycle-stage-dependent accumulation and proteolytic degradation of different cyclin subunits regulates their association with CDKs to control different stages of cell division.
CDKs promote cell cycle progression by phosphorylating critical downstream substrates to alter their activity.

17. Humans have 4 CDKs and 4 cyclins
Humans have 4 CDKs and 4 cyclins. Cyclins: G1 cyclin/G1-S cyclin/S cyclin/M cyclin.
G1/S-Cdk complexes commit the cell to a new cell cycle
S-Cdk complexes promote S phase (and block it in G2!)
M-Cdk complexes trigger entry into mitosis
M-Cdk complexes are inactivated before anaphase

18. Cyclins
Four classes
Defined by phase of the cell cycle in which they bind their cdk
G1/S phase cyclins-bind cdks at the end of G1, commit cell to DNA replication (cyclin E)
S phase cyclins- bind cdks during S phase, required to initiate replication (cyclin A)
M phase cyclins- bind cdks immediately before M phase, initiate early mitotic (or meiotic) events (cyclin B)
G1 cyclins- involved in progression through the checkpoint in late G1 (cyclin D)

19. The cell cycle Checkpoints
Cell cycle checkpoints are control mechanisms that regulate cell division and prevent cells from continuing through the cycle if the events of the preceding phase have not been completed. The main control points in mammals are:
      At the end of G1 phase (restriction point) the cell checks the quality of the DNA, the presence of specific growth factors (p.e. fibroblast growth factor) and cell size. Arrest at that checkpoint allows repair of damaged DNA before the cell enters the replication phase. If DNA can’t be repair the cell induces apoptosis. The phase of arrest in which the cell doesn’t continue its cycle is G0. This checkpoint is controlled by Cyclin D-CDK 4/6 and Cyclin E-CDK 2;
         At the S phase, continual control of the integrity of DNA allows that mutated bases are not replicated and the repair of possible errors that occur during replication. The proteins involved in this damage detection are DNA pol ε, PCNA and RFC. This checkpoint is controlled by Cyclin A-CDK 2;
         At the end of G2 phase , cells check if there are conditions for mitosis (cell size, quality of DNA and existence of nutrients necessary for the mitotic phase) and prevent the initiation of mitosis before DNA replication is completed and until there are conditions for cell division (cell size and existence of necessary nutrients). If there are no conditions, cells enter in a quiescent state where DNA repair or apoptosis can occur. This checkpoint is controlled by Cyclin A-CDK1;
    At the beginning of anaphase, spindle assembly checkpoint stops mitosis if the chromosomes are not well aligned and its centromeres are not properly attached to the microtubules and thus, not prepared for equal distribution. This checkpoint is controlled by the anaphase-promoting complex/cyclosome (APC/C).
The cell cycle Checkpoints

20. Reverse transcription
Produces DNA copies of an RNA, is more commonly associated with life cycles of retroviruses, which replicate and express their genome through a DNA intermediate (an integrated provirus).
Also occurs to a limited extent in human cells, where it plays a role in amplifying certain highly repetitive sequences in the DNA.
Telomerase has reverse transcriptase activity.

21. Basic Structure of Nucleic Acids
Repeating nucleotides linked by phosphodiester bonds
DNA “backbone” = sugar (deoxyribose) + Phosphate
RNA “backbone” = sugar (ribose) + Phosphate
Pentose Sugar
RNA
DNA
Negative (-) charge on Phosphate Units Give DNA/RNA a Uniformly (-) negative charge !!!

22. Types of Nucleotides Based on Number of Phosphates
Nucleoside = Base + Sugar
Nucleotide = Base + Sugar + Phosphate (mono, di, tri)
Ribose nucleotides: Adenosine/Guanosine/Cytidine/Uridine

23. Nitrogenous Bases Provide “Genetic Information”
Order of bases in DNA is the “SEQUENCE”
Two general types of nitrogenous bases
Purines (two rings) Pyrimidines (one ring):
Adenine (A) Cytosine (C)
Guanine (G) Thymine (T)
Uracil (U) – only RNA

24. - Other purine metabolites, not usually found in nucleic acids, include xanthine, hypoxanthine, and uric acid.

25. Linkages to different carbon atoms in sugar:
1`–5` numbering is based on organic nomenclature
This defines orientation of nucleic acids, 5` and 3` ends
Two nucleotides are linked by a 5`, 3`-phosphodiester bond
5` Carbon linked to “Upstream” Phosphate
3` Carbon linked to “downstream” Phosphate

26. Nucleotides Base Pair By Hydrogen bonds
Hydrogen bonds (H-bonds) form between purine and pyrimidine bases in DNA and RNA
Nitrogenous bases pair with complementary bases:
A pairs with T (A-T) = 2 H-bonds
A pairs with U (A-U) = 2 H-bonds (in RNA)
G pairs with C (G-C) = 3 H-bonds (stronger pairing)
H-bonds:
- H atom is shared between two atoms
Typically between oxygen (O) and nitrogen (N) atoms
- Bond is strongest when three atoms are in a line (O-H-N)
- Strength ranges from ~ 2–6 kcal/mol (energy unit/bp)

27. Pairing Between Complementary Bases Promotes Formation of Double-Stranded DNA
“Chargaff Rule” for Base Pairing

28. Using Chargaff's Rules
In dsDNA (or dsRNA)
% A = % T (% U)
%G =%C
% purines = % pyrimidines
A sample of dsDNA has 10% G; What is the %T?

29. Nucleic Acids Nucleotide is linked by 3',5' phosphodiester bonds
Have distinct 3' and 5' ends, thus polarity
Sequence is always specified as 5'3'

31. There are about 10 base pairs per complete turn of the helix.
Most DNA occurs in nature as a right-handed double-helical molecule known as Watson-Crick DNA or B-DNA.
The hydrophilic sugar-phosphate backbone of each strand is on the outside of the double helix. The hydrogen-bonded base pairs are stacked in the center of the molecule.
There are about 10 base pairs per complete turn of the helix.
A rare left-handed double-helical form of DNA that occurs in G-C-rich sequences is known as Z-DNA. The biologic function of Z-DNA is unknown, but may be related to gene regulation.
While no definitive biological significance of Z-DNA has been found, it is commonly believed to provide torsional strain relief (supercoiling) while DNA transcription occurs.[5][12] The potential to form a Z-DNA structure also correlates with regions of active transcription. A comparison of regions with a high sequence-dependent, predicted propensity to form Z-DNA in human chromosome 22 with a selected set of known gene transcription sites suggests there is a correlation

32. Grooves are binding sites for regulatory proteins

34. B-form: sodium salt of DNA at very high relative humidity (92%)
A-form: sodium salt of DNA in reduced humidity (75%). E.g.,
DNA/RNA hybrid
dsRNA
Both A- & B-forms are right-handed
Z-DNA: left-handed, assumed by dsDNA containing strands of alternating purines & pyrimidines e.g., poly[dG-dC].[dG-dC]

35. Different ways to represent DNA sequence
5`-pApCpGpT-3`
5`-ACGT-3`
3`-TGCA-5`
5`-ACGT-3`

36. dsDNA Can be Denatured and Renatured
Denaturing = “melting” = breaking H-bonds
Renaturing = “annealing” = reforming H-bonds
Ways to Denature:
High heat: ~ 95°C will “melt” most DNA
High pH: Concentrated OH-will break H-bonds
Chemicals: Formamide, Urea, DMSO & Formaldehyde
Lowering salt conc. of DNA solution aids denaturation
Renature:
Cool (room temperature) and given time (min-hr)

37. Melting and Renaturation of DNA

38. The melting temperature (tm) for A given DNA is when half of the DNA is single-stranded
DNA Melting Curve

39. DNA and RNA Absorb Ultraviolet (UV) light:
Peak absorbance is at 260 nm wavelength
Damaging UV light (breaks DNA)
DNA & RNA are quantified using this property
Hyperchromic effect: when two strands separate the absorbance rises 30-40%.
Hypochromicity: caused by the fixing of the bases in a hydrophobic environment by stacking, which makes these bases less accessible to UV absorption.

40. UV absorption of nucleotides

41. DNA & RNA have constant UV Absorbance:
Peak absorbance is at 260 nm wavelength
Absorbance at 260 nm (A260) is constant:
Double-stranded DNA (dsDNA):
A260 of 1.0 = 50 ug / ml
Single-stranded DNA (ssDNA):
A260 of 1.0 ~ 30 ug / ml
Single-stranded RNA (ssRNA):
A260 of 1.0 = 40 ug / ml
ssDNA 33 ug/ml. dsRNA ~ 40
1 OD260 Unit = 50ug/ml for double stranded DNA •1 OD260 Unit = 40ug/ml for single stranded RNA •1 OD260 Unit = 33ug/ml for single stranded DNA (ssDNA) •1 OD260 Unit = 20ug/ml for single stranded oligo (ssOligo)

42. Determine dsDNA concentration with A260:
For DNA: 1) Determine A260 with spectrophotometer 2) Use A260 to calculate concentration: Constant: A260 of 1.0 = 50 ug / ml dsDNA
For Example: A260 was determined to be x 50 ug / ml = 5 ug / ml dsDNA

43. Reuniting the Separated DNA Strands
Renaturation: when 2 separated strands, under proper conditions, come back together again.
Annealing: base paring of short regions of complementarity within or between DNA strands. (example: annealing step in PCR reaction)
Hybridization: renaturation of complementary sequences between different nucleic acid molecules. (examples: Northern or Southern hybridization)