1. Agenda 2/22 History and Structure of DNA Lecture
DNA structure modeling activity
Homework
1. Chp 16 reading and notes
2. DNA replication video and notes

2. As your warm up, write down at least 8 things you know about DNA

3. The History and Discovery of the Structure and Role of DNA
DNA Part I
The History and Discovery of the Structure and Role of DNA

4. Fred Griffith
Fred Griffith demonstrated that bacteria could be “transformed” from one strain to another
He took ‘genetic factor’ from one virulent strain, put it in a non virulent strain
The non virulent strain became virulent
He did not know the molecule was DNA
Fred Griffith ( ) 1928-Fred Griffith performed an experiment with 2 different strains of Pneumococcus. One was virulent and the other was not. The virulent strain had a smooth polysaccharide capsule which protected from the immune system. This allowed to caused pneumonia in mice and killed them. The other strain did not have the capsule and was "rough.” This Strain could not cause pneumonia in mice. When Griffith injected the rough strain of bacteria in mice they lived, and when the smooth strain of bacteria was injected into the mice they died. He killed some of the smooth bacteria by heating them and then injecting them into the mice. The mice lived. He then took some of the killed smooth bacteria and some of the rough bacteria and mixed them together. This bacteria then had the ability to kill mice. This is because the rough bacteria had been "transformed" by taking up some of the DNA from the smooth bacteria. Griffith did not identify DNA as the molecule that was taken up but that some genetic factor has transformed the bacteria.

5. Griffith’s Experiment
Fred Griffith ( ) 1928-Fred Griffith performed an experiment with 2 different strains of Pneumococcus. One was virulent and the other was not. The virulent strain had a smooth polysaccharide capsule which protected from the immune system. This protection allowed the bacteria to caused pneumonia in mice and kill them. The other strain did not have the capsule and was "rough.” The R strain could not cause pneumonia in mice. When Griffith injected the rough strain of bacteria in mice they lived, and when the smooth strain of bacteria was injected into the mice they died. He killed some of the smooth bacteria by heating them and then injecting them into the mice. The mice lived. He then took some of the killed smooth bacteria and some of the rough bacteria and mixed them together. These bacteria then had the ability to kill mice. This is because the rough bacteria had been "transformed" by taking up some of the DNA from the smooth bacteria. Griffith did not identify DNA as the molecule that was taken up but that some genetic factor has transformed the bacteria.
S strain: virulent
R strain: not virulent
Transformed R strain: virulent

6. Avery, MacLeod, McCarty
Avery, MacLeod and McCarty examined the various molecules found in the S-strain Pneumococcus cells to prove that DNA was responsible for the transformation of the bacterial cells.

7. Avery, MacLeod, McCarty They analyzed the molecules in each strain
DNA was the molecule extracted from the S-strain and ‘transformed’ R strain
1944-Avery, MacLeod and McCarty tried mixing the rough strain with different isolated molecules from the S strain and it was found the DNA extracted from the smooth-strain and transformed the rough strain.
***Students need to know this experiment and how it was conducted and its conclusion. It needs to be pointed out that it was extending the work of Fred Griffith.
Many scientists did not support the conclusion of Avery, MacLeod and McCarty and their work was ignored by many scientists.

8. Experiment of Hershey and Chase
Alfred Hershey and Martha Chase demonstrated the genetic material is DNA by using viruses that infect bacteria (bacteriophage).
Bacteriophage are composed of protein and DNA.
They eliminated protein as the genetic material DNA
*** Students need to be familiar with this experiment.

9. Experiment of Hershey and Chase
Trial 1
Replicate phages with radioactive sulfur (labels nucleotides) which is found in proteins and not DNA. These are now labeled phages for their radioactive proteins.
Mix radioactively phages with bacteria. Allow them to infect cells.
Put mixture in blender to separate infected cells from the viruses on the outside of the bacteria.
Centrifuge the mixture to separate and layer viruses and infected bacteria. Determine if the shell of the virus is radioactive OR if the bacteria are radioactive.
Conclusion
If the bacteria are radioactive, then proteins are the molecules of heredity. The bacteria were not radioactive the empty viral coats were radioactive.
Trial 2
Repeat the experiment above but this time replicate the viruses and with radioactive phosphorus. This labels the DNA and not the proteins. These viruses are now labeled phages for their radioactive DNA.
When the experiment is complete and if the bacteria are radioactive, then DNA are the molecules of heredity. The bacteria were radioactive and the empty viral coats were not radioactive.
Results
In the second trial, the bacteria were radioactive. So it can be concluded that DNA is the molecule of heredity.
This is a great experiment because the experiment took advantage of differences the structure of DNA and proteins.
It demonstrated that DNA is the material that genes are made of and not protein.

10. Chargaff's Rule -> A+G=C+T=50%
Chargaff’s Experiment
Chargaff's Rule -> A+G=C+T=50%
Percentage of Various Nucleotides in Genome
Organisms A T G C
Humans
30.9
29.4
19.9
19.5
Wheat
27.3
27.1
22.7
22.8
Sarcina lutea
13.4
12.4
37.1 T7
26.3 26
23.8
23.9
Erwin Chargaff ( ) was a Austrian biochemist that immigrated to the U.S. during W.W. II. He was a professor at Columbia University. Using paper chromatography and using a ultraviolet spectrophotometer, Chargaff was able to demonstrate that in a given organism the number of adenine nucleotides was approximately the same as the number of thyamine nucleotides and that the number of cytosine nucleotides was approximately the same as the number of guanine nucleotides. This is called Chargaff’s first rule. His second rule was based on the observation that these percentages were unique for various species.
Students are responsible for this information.
Based on the observations above, two rules can be deduced
A+G=C+T=50%.
The percentages of the nucleotides vary for different species

11. Work of Rosalind Franklin
Rosalind Franklin worked in a lab with Maurice Wilkins. She was investigating the structure of DNA using x-ray crystallography. She determined that the phosphate groups were on the outside of the double helix. She did not determine the base of DNA pairing of DNA or their role in the structure of DNA.
Rosalind Franklin used x-ray crystallography to determine that DNA was double stranded, a helix, with phosphates were on the outside
3 distances, 2.0 nm, .34 nm, and 3.4 nm showed up in a pattern over and over again in the diffraction pattern.

12. Work of James Watson and Francis Crick
Based on the rules of Chargaff and the information from the work of Franklin, James Watson and Francis Crick, determined the structure of DNA by making models.
Determined that the sugar and phosphates were on the outside.
2. Determined that the nitrogenous bases were forming the rungs of the ladder.
Watson and Crick determined that 2.0 nm was the distance from one strand to the other nm was the distance from one base pair to another and finally 3.4 nm determined that there were 10 bases to a complete twist in the helix. So with 2.0 nm from one strand to the other, it was determined that the a purine had had to be base paired with a pyrimidine. Next problem to investigate is why did adenine base pair with guanine and why did cytosine base pair with thymine? The answer had to do with the hydrogen bonding, as adenine base paired with thymine because they could form two hydrogen bonds and guanine base paired with cytosine because they could form three hydrogen bonds.
Watson, Crick, and Wilken received the Nobel Prize in Unfortunately, Franklin died of cancer at age 38 in Watson, Crick, and Wilkens received the Nobel Prize in 1962.

13. Determining the Nitrogen Base Pairing
Based on the work Franklin’s x-ray crystallography, Watson and Crick found the bonding;
two purines are too wide and would overlap.
two pyrimidines are too far apart to form the hydrogen bonds.
a purine and a pyrimidine however, are just right!
Also due to the hydrogen bonding, A and T forms two hydrogen bonds and C and G forms three hydrogen bonds. That is the only way they hydrogen bonds will fit.

14. Structure of a Nucleotide
DNA is the longest molecule found in the cell, yet its structure is quite simple. The human cell contains 5-6 feet of DNA in every cell autosomal cell. The basic building blocks of nucleic acids are the nucleotides. There are 3 billion base pairs or 6 billion nucleotides in a human cell.
Have the students compare and contrast the differences between purines and pyrimidines and point out the differences. Purines have two rings whereas pyrimidines only have one ring. Adenine and guanine are purines and cytosine, thymine, and uracil are pyrimidines
Have the students compare and contrast the differences between ribose and deoxyribose. Deoxyribose is missing an atom of oxygen.
Be sure to go over how the carbons are numbered and emphasize that # 3 and # 5 are important and termed 3’ or three prime and 5’ or five prime.
The significance of the various carbons are listed below:
Carbon #1 is where the nitrogenous base is attached.
Carbon #2 is what differentiates between ribose (C5H10O5) which has a hydroxyl group attached (OH) and deoxyribose (C5H10O4) has only hydrogen (H). Hence the name deoxyribose makes sense.
Carbon #3 is where the next nucleotide will attach.
Carbon #5 is the phosphate group is attached.
Compare and contrast purines and pyramidines
Compare and contrast deoxyribose and ribose

15. Carbon #3 is where the next nucleotide will attach. (3’)
Carbon #5 is where the phosphate group will attach (5’)

16. Sides of the Ladder
DNA is a double stranded and analogous to a ladder. The sides of the ladder are composed of alternating sugars (deoxyribose) and phosphate groups that run antiparallel to one another. On the left side (next card) the first carbon found on the strand is #5 and moving on down the last carbon is carbon # 3. This side is said to be 5'-3'. The opposite side is upside down compared to the other side. The right hand side, the first carbon found on the strand is #3 and moving on down the last carbon is carbon # 5. This side is said to be 3'-5'.

17. Hydrogen Bonding and Nitrogenous Bases
The nitrogenous bases form the rungs of the ladder. Thymine will base pair with adenine on the opposite side, which is a pyrimidine base paired with a purine. This will form 2 hydrogen bonds. Hydrogen bonds are weak but millions of them together will keep the two strands together.

18. Hydrogen Bonding and Nitrogenous Bases
Guanine will base pair with cytosine on the opposite side. This is a pyrimidine base paired with a purine. This will form 3 hydrogen bonds instead of 2 hydrogen bonds.

19. Hydrogen Bonding and Nitrogenous Bases
This will continue for billions of base pairings forming a molecule of DNA.

20. Hydrogen Bonding and Nitrogenous Bases
The DNA molecule once formed will make a double helix. There are 10 base pairs in one complete turn of the DNA molecule.

21. Forming the Double Helix
There are 10 base pairs in one turn of the DNA molecule. The length of one complete turn is 3.4 nm. The radius of the spiral is 1 nm. The distance between nitrogenous bases is .34 nm.
The spiral shows a large groove called the major groove and a small groove call the minor groove. Students do not have to know the measurements or the information about the grooves.

22. DNA Forming Chromosomes
Structure in eukaryotes.
the DNA is wrapped around proteins called histones forming nucleosomes.
This forms a fiber known as chromatin.
This forms a coil within a coil.
The final structure is a chromosome which can only be seen during mitosis. During the rest of the cell cycle, the DNA is in chromatin state.

23. DNA Paper Model Questions
Directions: Create your paper model by cutting out the template and gluing the correct base pairs together. Answer these questions on the back of your model
Label: anti-parallel ends, purines, pyrimidines, hydrogen bonds, phosphate, deoxyribose sugar
Reflection Questions:
1. Why does one purine always bind with one pyrimidine?
2. How does your model show the anti-parallel structure?
3. Why are hydrogen bonds significant between the base pairs?
4. Why does the amount of adenine always have to equal the amount of thymine?