Chapter 12 - DNA Structure & Replication - Welcome to Citrus · PDF file ·...

Chapter 12 - DNA Structure & ReplicationHistory of DNA

1869 Miescher

Analyzed the contents of the nucleus & discovered a substance (containing nitrogen & phosphorus) which he called nuclein

1909 Garrod

Linked protein deficiencies to unusual phenotypes.

Could proteins be the genetic (hereditary) material?

1929 Griffith

Experimented with two phenotypically different bateria: R (rough coated) which caused no disease & S (smooth coated) which would cause pneumonia

Mice injected with the R bacteria did not dies of pneumonia

Mice injected with the S bacteria died of pneumonia

When S bacteria were heat treated (to kill them), they no longer caused pneumonia

Makes sense, but here’s the kicker!

Chapter 12 - DNA Structure & Replication

Chapter 12 - DNA Structure & Replication1929 Griffith (continued)

When heat treated S bacteria was co-injected with live R bacteria, the mice DIED of pneumonia & LIVE S bacteria was found in the dead mice! What HAPPPENED?

Chapter 12 - DNA Structure & Replication1930s Avery, MacLeod, McCarty

Something in the heat-treated S bacteria transformed the R bacteria into live S bacteria

Mice were injected with a mixture of heat-treated S bacteria, live R bacteria & either:

1. A protease which destroys proteins

2. A DNAse which destroys DNA

Mice given the protease mixture died of pneumonia & live S were recovered

Mice given the DNAse mixture lived

Conclusion: S type DNA is capable of transforming (changing) live R into S type

Confirmation: They isolated DNA from the heat treated S bacteria & co-injected it with live R into mice = mice died & contained living S bacteria

Chapter 12 - DNA Structure & Replication1950 Hershey & Chase

Confirmed that DNA & not protein is the genetic material

Used viruses because:

Virus structure is simple: nucleic acid core surrounded by a protein coat

Once a virus infects a cell it causes the host cell to produce more viruses (this is the phenotypic change that will indicate a transformation of the host cell’s genotype)

What part of the virus gets into the host cell to induce this transformation?

Experiment

You can differentially mark proteins & nucleic acids using radioactive isotopes of sulfur & phosphorus. These isotopes give off radiation which can be detected.

Proteins were labeled with S35 which made them “glow” – Think Homer Simpson

Nucleic acids were labeled with P32 which made them “glow”

Chapter 12 - DNA Structure & Replication1950 Hershey & Chase

They incubated their host bacteria cells with either S35 labeled viruses or P32 labeled viruses & here are the results

Both sets of bacteria cells were transformed into cells that produced more of the viruses

In the S35 labeled experiment the radioactivity was localized outside the bacteria cells. Proteins were not injected into the bacteria cells & the cells were still transformed

In the P32 labeled experiment the radioactivity was localized inside the bacteria cells. Nucleic acids were injected into the bacteria cells & the cells were transformed

Conclusion: Injection of nucleic acids induces transformation

Chapter 12 - DNA Structure & ReplicationStructure of DNA

The basic building block of a DNA molecule is the nucleotide

1. Deoxyribose – 5 carbon sugar – forms part of the backbone structure

2. Nitrogenous base – Information is coded in the sequence of these bases

A. Purines – Double ring structure – Adenine (A) & Guanine (G)

B. Pyrimidines – Single ring structure – Thymine (T) & Cytosine (C)

3. Phosphate group – forms part of the backbone structure


Nucleotides are arranged end-to-end in a very specific orientation. The backbone of the DNA molecule is formed by the phosphate group & the sugar, while the nitrogenous bases extend perpendicular to the backbone

Each DNA molecule is actually made up of two strands of DNA. The two strands associate with each other via their nitrogenous bases. Hydrogen bonds form between them, stabilizing the molecule into its familiar double helix form

Adenine always forms hydrogen bonds with thymine

Guanine always form hydrogen bonds with cytosine

These introduces us to Chargaff’s rule

Regardless of species

1. The number of purines in DNA always equals the number of pyrimidines

2. The amount of adenine always equals the amount of thymine, while the amount of guanine always equals the amount of cytosine


In order for the nitrogenous bases to form these hydrogen bonds, the orientation of the backbones must be specific.

DNA strands must be anti-parallel to each other, meaning that the sugar-phosphate backbones must be arranged in opposite directions to each other

The specific directionality is determined by the position of specific carbons & is referred to as 5’ to 3’

Why is this important?

Important for understanding the intricacies of DNA replication

Why is DNA replication necessary?

When does it occur?

Chapter 12 - DNA Structure & ReplicationOverhead diagram of replication – simple version

Chapter 12 - DNA Structure & ReplicationDNA Replication

Origin of replication is where it all begins

Helicases bind to the origin & begin unraveling the strands

Single-strand binding proteins keep the strands separated

Topoisomerases relieve tension placed on the DNA molecule caused by this unraveling, by creating transient cuts which allow the DNA to untwist a little & then repair the cuts

Once the strands have separated, the strands become templates for replication

Do you remember how the bases are matched up?

New DNA daughter strands cannot be synthesized without some “priming”. As soon as the parental strands have separated, RNA primase adds a short sequence of RNAnucleotides to begin the replication process

Once the RNA primer sequence has been laid down, DNA polymerase will begin adding DNA nucleotides to the new daughter strand of DNA until the strand is completed

Chapter 12 - DNA Structure & ReplicationDNA Replication

Since each of the parental strands is used as a template for replication, the process is referred to as semi-conservative replication

How do we know that this process works in this fashion?

1957 Meselson & Stahl

Chapter 12 - DNA Structure & ReplicationDNA Replication – DETAILS!

DNA replication must occur in a specific direction: 5’ to 3’

Since DNA replication must occur in this direction, a problem arises due to the antiparallelism of the parental strands

DNA replication is continuous on one parental strand & discontinuous on the other

DNA polymerase reads & lays down the new DNA on the daughter strand

DNA ligase is important for joining the discontinuous DNA fragments on the “lagging”daughter strand & well as the entire DNA molecule as you will see

Eukaryotic DNAs have multiple origins of replication which will reduce the time it takes for replication to occur

Chapter 12 - DNA Structure & ReplicationOverhead diagram of replication – leading & lagging strand version

Chapter 12 - DNA Structure & ReplicationPRACTICE QUESTIONS

1. How did the following individuals contribute to our knowledge of the importance of DNA as the hereditary material: Miescher, Griffith, Avery et al, Hershey & Chase?

2. What is Chargaff’s rule?

3. What 3 components make up a DNA nucleotide? RNA nucleotide?

4. What is the difference between DNA & RNA?

5. What enzymes are used during the steps of replication?

Chapter 12 - DNA Structure & Replication - Welcome to Citrus · PDF file ·...

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