Download Friedrich Miescher (1844-1895) was a Swiss chemist

Slide 1
DNA Part I
The History and
Discovery of the
Structure and Role of
DNA
Slide 2
DNA –How its structure was discovered
2
Slide 3
Identifying DNA as a unique molecule.
1869- Friedrich Miescher was
a Swiss chemist and was the
first to identify DNA as a
unique molecule.
3
Friedrich Miescher (1844-1895) was a
Swiss chemist. He studied tissue
chemistry under Felix Hoppe-Seyler at
the University of Tubingen, Germany.
His field of study included the
molecules found in blood cells. He
showed that when pepsin (an enzyme
that digested) proteins was used on
the nucleus of cells a strange
phosphorous-containing material
remained. He also found that there
was general ratio of phosphorous to
nitrogen in this molecule. He called
this molecule nuclein. Mieischer did
not know that DNA or nuclein (later
called nucleic acid and finally
deoxyribonucleid acid) was the
molecule of heredity. He still thought
that proteins were the molecules of
heredity.
Students are not responsible for
information about Miescher
Slide 4
Staining of DNA reveals somatic cells have the same
amount of DNA and half as much as gametes.
1914-Robert Feulgen, a German
chemist, found a staining
technique that stains more or less
strongly based in the amount of
DNA present (called Feulgen
stain). He found that all cells in
an organism had the same
amount of DNA except gametes,
which had half the normal
amount.
4
Robert Feulgen (1884-1955) was a
German chemist.
It is possible to use an instrument
known as a microspectrophometer to
actually measure the intensity of the
pink stain for the nucleus. Using this
procedure, it was easily determined
that interphase cells were composed
of two populations, those with haploid
DNA in gametes and those with
diploid DNA in the somatic cells. The
nuclei looked identical, but the
somatic cells contained twice as much
DNA as the gametes.
Later on it was found that cells that
just completed mitosis had half the
amount of DNA as those cells that
were just about to enter mitosis. This
supports the cell cycle and the DNA is
replicated in interphase (S) and not
during mitosis itself.
Students are not tested on this
historical information.
Slide 5
Staining of DNA reveals somatic cells have the same
amount of DNA and half as much as gametes.
Cells stained
with Feulgen
stain. It is the
DNA and not
the proteins that
are visible
under the
microscope.
5
Slide 6
History of DNA
Fred Griffith demonstrated that
bacteria could be “transformed”
from one strain to another by
transferring genetic factor from
one organism to another. He used
two different strains of the same
bacteria. One could cause
pneumonia and the other could
not.
6
Fred Griffith (1879-1941) 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.
Slide 7
Griffith’s Experiment
The conclusion was that the bacteria had incorporated heredity
factor from a source and in doing so expressed a new smooth trait.
7
Fred Griffith (1879-1941) 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.
Slide 8
Extending Griffith’s Experiment and Identifying
DNA as the Transforming Factor
http://www.accessexcellence.org/RC/
VL/GG/ecb/DNA_genetic_material.ph
p
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.
8
Slide 9
DNA is the Molecule of Heredity.
When various isolated
chemical components of
the S-strain
Pneumococcus cells
was mixed the R-strain
Pneumococcus cells, it
was shown that the
DNA from the S-strain
cells, that caused transformation.
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.
Slide 10
*** Students need to be familiar with
this experiment.
Experiment of Hershey and Chase
Alfred Hershey and Martha Chase demonstrated the genetic
material is DNA by using viruses that infect bacteria. These
viruses only stay on the outside of the cell when infecting the
cells. Also viruses are composed of protein and DNA. It is
known that the virus injects its genetic material into the
bacterium which had to DNA or proteins.
10
Slide 11
Experiment of Hershey and Chase
It demonstrated that DNA is the material that genes
are made of and not protein.
11
Experiment
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.
Conclusion
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.
Slide 12
Experiment Chargaff
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
37.1
T7
26.3
26
23.8
23.9
Based on the observations above, two rules can be deduced
1. A+G=C+T=50%.
2. The percentages of the nucleotide vary for different species
12
Erwin Chargaff (1911-2005) 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.
Slide 13
Work of Rosalind Franklin
Rosalind Franklin used x-ray crystallography to determine
that DNA was double stranded, a helix, phosphates were on
the outside and three distances, 2.0 nm, .34 nm, and 3.4 nm
showed up in a pattern over and over again in the diffraction
13
pattern.
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.
Slide 14
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.
1. Determined that the sugar and phosphates
were on the outside.
2. Determined that the nitrogenous bases were
forming the rungs of the ladder.
14
Watson and Crick determined that 2.0
nm was the distance from one strand
to the other. .34 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 1962.
Unfortunately, Franklin died of cancer
at age 38 in 1958. Watson, Crick, and
Wilkens received the Nobel Prize in
1962.
Slide 15
Determining the Nitrogen Base Pairing
Based on the work Franklin’s xray 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!
15
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.
Slide 16
Chargaff’s Snub
Chargaff felt there had been an
injustice done when he did not receive
the Nobel Prize in 1962 along with
Watson, Crick and Wilkins. Wilkins’
contribution to the structure of DNA
was to show James Watson the work of
Rosalind Franklin without her
permission. Franklin did not share the
Nobel Prize as she passed away from
ovarian cancer in 1958 and
posthumous nominations are
forbidden.
16
Slide 17
Structure of a Nucleotide
17
Chargaff is quoted as saying, “I told
them all I knew. If they had heard
before about the pairing rules, they
concealed it. But as they did not seem
to know much about anything, I was
not unduly surprised. I mentioned our
early attempts to explain the
complementarity relationships by the
assumption that, in the nucleic acid
chain, adenylic was always next to
thymidylic acid and cytidylic next to
guanylic acid...I believe that the
double-stranded model of DNA came
about as a consequence of our
conversation.*”
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.
Slide 18
Sides of the Ladder
18
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'.
Slide 19
Hydrogen Bonding and Nitrogenous Bases
19
Slide 20
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.
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.
20
Slide 21
Hydrogen Bonding and Nitrogenous Bases
21
This will continue for billions of base
pairings forming a molecule of DNA.
Slide 22
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.
22
Slide 23
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.
Forming the Double Helix
23
Slide 24
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.
24
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.
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.