1940's-> Lab that was working with auxotrophic bacteria.

                  -> 2 strains of bacteria that were auxotrophic for different things

                 ->Plate them separately on minimal media

->Both fail to grow

->Mix strands together and allow time for growth

->Plate on minimal media

->See prototrophs line every 10^7 cells

-> This suggests that the bacteria have the ability to excchange information.

F+ cell becomes F-

AND

F-  cell becomes F+

-Conjugation requires a sex pilus

-Tube from the F+ cell that contacts the F- cell

-DNA travels through the tube

It remains F+

This is valuable for attempting to map traits.

These contain 2 regions:

1.) r determinant: antibiotic resistance

2.) Resistance transfer factor (allows resistance to be transferred)

1. Transforming DNA lands on the receptor site on the component bacteria.

2. The DNA passes through the membrane and the entering strands of the DNA separate.

3. One of the strands binds to the homologous regions on the host chromosome (2nd strand is degraded)

4. The transforming strand replaces the homologous region on one strand of the host chromosome.

5. The cell is duplicated (after duplication, you have 1 transformed cell with new dna and 1 parental cell).

IN THIS PROCESS WE CAN PICK UP 10,000 to 20,000 BASE PAIRS OF DNA

An E.coli phage is composed of protein and DNA (the DNA is packed in the phage head).

The protein has an icosohedral structure (polyhedron with 20 faces).

The phage has tail fibers which allow it to specifically dock on the host cell.

It also has a contractile sheath which functions to poke holes in the host cell membrane.

1. Dock on the host cell and inject DNA.

2. Degrade the host chromosome and hijack the DNA's replication, transcription, and translation.

3. Start to produce new phage parts.

4. Start to mature phage.

5. When the critical mass of about 200 phages has been reached, the host cell is lysed and the process is repeated.

Define "lysogenic stage".

The experimenters were working with strains of salmonella auxotrophs.

Using the U-tube, they put LA2 on one side and LA22 on the other side.

The tube has a glass filter at the apex which allows for media to pass but not cells. After incubation, they noticed that the LA2 side had no growth while the LA22 side grew prototrophs.

In 1927, Frederick Griffith was working with diplococcus pnemoniae.

There are two strains: 1) III's (Smooth cells ; virulent) 2) II R cells (Rough cells, avirulent)

He injected IIIs cells into the mouse and found that it died.

He injected IIR cells into mouse and found that it lives. He injects IIIs cells that were heat killed and injects the mouse and the mouse lives. Finally, he injected IIR cells and heat killed IIIs cells and found the mouse died. The results of the final test showed IIIs cells in the dead skin tissue of the mouse. HE MADE A TRANSFORMING PRINCIPLE FROM THIS!!

-It is an extension of Griffith's mouse work.

-Start with IIIs cells and then heat kill, homogenize, and extract the lipids and carbohydrates. Then, you mix them with the IIR cells. This is where transformation occurs.

Next, you take the IIIs filtrate and treat it with protease. This functions to degrade the protein mix with IIR cells. Transformation occurs.

Next, you take the IIIs filtrate and treat it with ribonuclease to degrade the RNA. When you mix this with the IIR cells, transformation occurs.

Finally, you treat IIIs cells with DNAse to degrade the DNA and mix that with IIR cells to get the % transformation.

Take a bacteriophage (virus)

and a T2 phage-> E.Coli

-50% Protein

-50% DNA

You take two cultures of E.Coli

-One of them is labeled S35 (Protein)

-The other one is P32 (DNA)

What you want to do is to expose the phage to each of these cultures.

Next, you expose the labeled phages to E.Coli.

The theory behind this is that the virus will inject its genetic material into the cell during infection.

RESULTS:

-From the P32 labeled culture, recover the phage with the label. (This indicates that the labeled phage had entered the cell)

From the S35 culture, he was unable to recover phages.

CONCLUSION:

DNA is the genetic material

RNA uses the sugar

-in DNA, Deoxyribose uses the sugar

1.)5 carbon sugar

2.) Phosphate group

3.) Nitrogenous Bases

Single 6 membered ring

(Cytosine, Uracil, Thymine)

9 member in two rings

(Adenine, Guanine)

RNA= Adenine, Guanine, Lytosine, Uracil

DNA= Adenine, Guanine, Cytosine, Thymine

They published their structure of DNA. They did that thanks to 2 pieces of data.

1.) Chargaff's Base Composition Studies

-Hyrdrolyze DNA by analyzing the % of each base in the sample.

-The % of A was = or close to %T

-The % of G + C were also similar

2.) Maurice Wilkins + Rosiland Franklin's X Ray Diffraction Data

-This stated that DNA has a 3.4Angstrom width (like coins stacked atop one another)

Watson and Crick proposed their model of the double helix. The outside of the helix contains the sugar phosphate backbone (consists of sugars and phosphates bound together by covalent bonds).

Inside the DNA, there are nitrogenous bases held together by hydrogen bonds.

In addition, adenine pairs with thymine to form 2 hydrogen bonds.

Guanine pairs with cytosine to form 3 hydrogen bonds.

Purines always pair with pyrimidines.

-This explains Chargaff's data because the two strands of DNA are complimentary and they double helix has a diameter of 20Angstrom.

-Equal to the number of turns for circular DNA.

-To calculate this, take the total number of base pairs and divide that by 10.4 (this is your theoretical # of turns)

1. In interphase, DNA exists as chromatin (loosely packed relaxed DNA)

2. In mitosis, DnA exists as mitotic chromosomes.

E.Coli= 1200 um long

Human= 19,000 to 73,000 um long

1.) Histones (plus charge)

2.) Nonhistone proteins to assist the histones.

1.) Treat packed DNA with endonucleases

2.) Microscopal analysis

3.) Histone proteins have a tendency to self associate

4.) Further endonuclease digestion

1.) DNA forms a double helix with a width of 2um and starts to pack the DNA

2.) 11um fiber. it has a width of 11um of DNA. It is referred to as the beads on a string model of packing where the beads are the proteins and the string is DNA. The basis of this structure is the nucleosome core particle that has 146 base pairs of DNA wrapped around histone proteins.

3.) 30um fiber. It has a 30 um width with a number of nucleosome core particles coming together. They are brought together by H1 histones that pull the nucleosomes into a structure called the solenoid structure.

1.) It is genetically inactive and can not be transcribed.

2.) These regions are replicated later in the S phase of the cell cycle.

The site where the spindle fibers attach during mitosis. It is responsible for holding chromosome pairs together during mitosis. Also, it has a repeated region called the "CEN" region (Has 3 particular sequences).

1&3 regions are similar in all chromosomes.(third region is the most important to function).

2 regions are the chromosome specific regions.

GGGATT

This is repeated a number of times. Telomere lengths vary in individuals.