GEN 4200 DNA Structure, Replication, and Recombination Learning Objectives

Explain the four characteristics required of genetic material

1. Replication: Accurate copies must be made to ensure stability of cell populations and to pass to next generation

2. Storage of Information: Information is stored in the form of a macromolecule. Information is contained within every cell

3. Encoding of phenotype and a method of transmitting of this information to produce the phenotype

4. Show variation through the process of mutation

Review key experimental results that supported DNA, rather than proteins as the genetic material (Griffith, Avery, MacLeod, McCarty, Hershey and Chase)

• Frederick Griffith (1928): Principle of transformation. Virulent bacteria that are heat-treated can pass virulence to harmless bacteria

• Avery, Macleod, and McCarty (1944): DNA is the agent of transformation. Selected elimination of macromolecules results in transformation only when DNA is present

• Hershey and Chase (1952): DNA is transferred to cells. Proteins and DNA of bacteriophages were labeled with radioactivity. DNA contains instructions for viral replication

Describe the overall structure of the DNA molecule (Double helix, antiparallel, complementary base pairing, nucleotides, etc)

• Nucleotides are the building blocks (subunits) of DNA:

1. Deoxyribose sugar, phosphate, and a nitrogenous base

2. Types of nitrogenous bases: purines (feature a double-ring structure consisting of a six-membered ring fused to a five-membered ring. Adenine (A) and Guanine (G) in both DNA and RNA) and pyrimidines (Feature a single, smaller six-membered ring structure. Cytosine (C) in both DNA and RNA but Thymine (T) in DNA and Uracil (U) in RNA)

3. Nucleotides are linked through the phosphodiester bond

• Because of the linkage and the linear nature of DNA, polarity is automatically established

1. One end will have a free phosphate on the 5’ carbon, and the other a free OH on the 3’ carbon meaning the free phosphate group on the 5’ end isn’t binding to anything

• Two DNA strands (polymers) are joined together to create the functional molecule of heredity

1. Double helix with strands arranged in an antiparallel manner

• Complementary base pairing is key to holding two strands together

1. 3 hydrogen bonds between cytosine and guanine (pyrimidine with purine)

2. 2 hydrogen bonds between thymine and adenine (pyrimidine with purine)

• The DNA double helix model:

1. Two strands are held together through the hydrogen bonds between complementary base pairs

2. The two strands are in an antiparallel orientation

3. The sugar-phosphate backbone is on the outside with base pairs in the middle

4. Two grooves are present: the major groove and the minor groove (these grooves are important for gene expression)

5. Nitrogenous bases are inside and hydrophobic

Describe Semiconservative Replication and its requirements (not part of the learning objectives)

• Because of complementary base pairing, one strand of DNA automatically has the information to build a new strand

1. Original (parent) strand is called the template strand (already contains information)

2. Used to create new daughter strands

• Semiconservative-template strand becomes part of new DNA molecule

1. A single double helix becomes two new daughter helices

2. Mechanism established via Meselson and Stahl experiment (1958)

• Successful replication requires the following:

1. A DNA template strand (must be single strand for replication)

2. Pool of free nucleotide triphosphates (must be accessible and balanced ex: dATP, dGTP, dCTP, dTTP)

3. Enzymes and other proteins (DNA polymerase, primase, single-strand binding proteins, DNA helicase, DNA topoisomerases, ori proteins)

4. A primer with a free 3’ OH

• DNA replication occurs during the S phase of the eukaryotic cell cycle

1. Multiple signaling inputs and regulatory steps dictate the progression of the cell through the stages of the cell cycle

2. Replication must be complete and accurate before the division phase of the cycle

Explain DNA polymerase and how it works (not part of the learning objectives)

• DNA Polymerase catalyzes the formation of new phosphodiester bonds

1. Joins two nucleotides together in the same strand

2. Uses energy from high energy phosphate bonds contained in the free nucleotide triphosphates

• DNA polymerase can only make a new phosphodiester bond when the free 3’ OH is available

1. DNA strands can ONLY be created in the 5’-3’ direction

Describe the key steps in the semiconservative replication of DNA (Review results of Meselson-Stahl experiment)

• Initiation - DNA helix is opened to create two template strands

1. Designated sequences within the DNA serve as initiation site

2. Called origins of replication, eukaryotes have multiple origins per chromosome

• Creates a replication bubble in the DNA

1. DNA helicase unwinds helix

2. Initiator proteins recognize origin sequence

3. Primase synthesizes RNA primer

• Elongation - Synthesis of a new DNA strand

1. DNA polymerase reads the nitrogenous bases on the template strand and adds complementary nucleotides

• New DNA strands are build in the 5’-3’ direction ONLY

1. The parent double helix has two strands oriented in opposite directions

Explain the occurrence of the leading and lagging strands in DNA replication

• The direction of replication is bidirectional along both template strands

• Because template strands are antiparallel, new DNA cannot be synthesized continuously in both directions

1. New daughter strands will be antiparallel to template

• Leading strand: New DNA molecule is synthesized continuously

• Lagging strand: New DNA molecule is synthesized discontinuously

1. Each fragment is called an Okazaki fragment

2. Okazaki fragments are joined together via the enzyme DNA ligase

Define homologous recombination and explain the mechanism (Explain why crossing over may or may not occur)

• DNA replication during meiosis provides the opportunity to create new combinations of alleles

1. Alleles-different versions of the genes

• New allele combinations are created through independent assortment or recombination (homologous recombination, HR)

1. HR is how new allele combinations are made for genes located on the same chromosome

• During meiosis, homologous chromosomes have been replicated

1. Each chromosome has two sister chromatids

• The pairs of chromosomes are aligned during prophase, allowing non-sister chromatids to exchange genetic information

1. Crossover chromosomes are also called recombinant chromosomes or recombinants

• HR requires a physical break in the DNA strands

1. Double strand break is made by specific proteins

2. Exonuclease will degrade 5’ ends of the break to leave 3’ ss trails

• To repair the break, HR relies on the availability complementary sequence from the opposite homologous chromosome

• HR proteins will force the double helix of the opposite homolog to open at regions of complementarity

1. Heteroduplex forms when DNA from one homolog is paired to the other

2. D loop forms due to strand displacement

• The single strand of the D loop pairs with opposite free 3’ tail

• DNA synthesis replaces the DNA that was degraded to create the 3’ tails

1. Backbones sealed by DNA ligase

• Cleaving of the heteroduplexes will result in either a crossover or non-crossover chromosome

1. Cleavage occurs at structures called the Holliday junctions

• Only cells undergoing meiosis exhibit homologous recombination

1. DS breaks initiated by the Spo11 protein

• Mitotic cells do have replicated DNA, so does recombination occur?

1. Yes, called mitotic recombination

2. DS breaks are typically initiated by environmental agents that damage DNA (X-rays, chemicals)

3. Can be useful in laboratory settings to create new mutations

Define site-specific recombination and explain the possible genetic outcomes of this process

• Site-specific recombination is the basis of site-specific gene editing

• The breakage and rejoining of DNA molecules occurs only at particular DNA sequences

1. Target sites are usually less than 200 base pairs long

2. Typically requires only a recombinase protein rather than a suite of many proteins

• Limited to certain organisms, which includes bacteriophages

1. Allows insertion of viral genome into the host genome

2. Other outcomes are possible