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

Nucleic Acid Structure and Organization Lecture 1 Nucleic Acid Structure and Organization

One Midterm exam worth 35% Final exam: 50% Course Syllabus Text: Molecular Biology Lecture Notes. Hansen B. and Jorde L.B., 2002. Kaplan Inc. (Required) Grades Assignments: 15% One Midterm exam worth 35% Final exam: 50%

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

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

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.

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

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.

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

Comparison of Gene Expression and DNA Replication - Produces all the proteins an organism requires - Transcription of DNA: RNA copy of a small section of a chromosome (average size of human gene 104 - 105 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

The concept of the cell cycle Lecture 2 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.

The Eukaryotic cell cycle

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.

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.

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 !!!

Types of Nucleotides Based on Number of Phosphates Nucleoside = Base + Sugar Nucleotide = Base + Sugar + Phosphate (mono, di, tri) Ribose nucleotides: Adenosine/Guanosine/Cytidine/Uridine What would the names be for ribose nucleotides?

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

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

Nomenclature of the Ribonucleotide Series of Compounds

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

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)

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

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

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'

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.

The B-DNA Double Helix

A-form B-form Z-form

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]

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

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)

Melting and Renaturing DNA

The melting temperature (tm) for A given DNA is when half of the DNA is single-stranded Figure 1-1-10. DNA Melting Point

Tm Curve % G + C Versus Tm

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.

UV absorption of nucleotides

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

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 0.1 0.1 x 50 ug / ml = 5 ug / ml dsDNA

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)