AP Biology
Molecular Genetics Unit
Chapters 16 & 17
Chapter 16 Objectives
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Explain how Griffith and Avery’s experiment with pneumococcus bacteria
contributed to the understanding of molecular genetics.
Define transformation. Use Griffith & Avery's results to explain how it occurs.
Using Hershey & Chase's procedure, use given hypotheses to predict results.
Evaluate each hypothesis based upon the actual results of the experiments.
Identify the component parts of nucleic acids. Describe their organization into
DNA and RNA.
Explain how Chargaff's rule supports the base pairing rule.
Explain how the two strands of a DNA molecule are complementary.
Explain how DNA replication is semiconservative.
Outline the process of DNA replication. Include the names and roles of any
enzymes involved in the process.
Explain why one strand of DNA is replicated continuously, while the other
strand is discontinuous.
Define Telomere. Explain why the mechanism of replication results in the
shortening of telomeres.
Explain how DNA condenses into chromosomes in eukaryotic cells.
What part of the cell controls heredity?
• Actual traits are not inherited. Traits are developed.
Inheritance must come from something that is in the gametes
and in the resulting zygote
• Since the gametes and zygote are cells, the control center of
heredity must have some cellular origin
• Cells have 3 fundamental regions
• Cell membrane
• Cytoplasm
• Nucleus
• One of these 3 regions must control heredity & determine
how traits will develop in the organism
What part of the cell controls heredity?
• It follows logically that hereditary control is centered either in
the membrane, the cytoplasm, or the nucleus.
• We will assume two alternative hypotheses:
1.
2.
The nucleus controls heredity
The cytoplasm controls heredity
• An experiment to test these must allow for simple
manipulation of cell parts and simple observation of
developing traits
• Manipulating cell parts would be simplified by using a single
celled organism, but it must be large enough to work with and
observe easily, and must have some distinct inherited trait.
1943 - Hammerling’s Experiment
• The giant
unicellular algae of
the genus
Acetabularia are an
ideal experimental
subject.
• They are large,
easy to
manipulate, and
easy to observe
Hammerling’s Experiment
• All Acetabularia
species have a base
(which contains the
nucleus), a stalk,
and a cap.
• The base and stalk
are similar
between species,
but the cap varies
greatly
Hammerling’s Experiment
• A. mediterranea has
an umbrella shaped
cap.
• A. crenulata has a
crenulated (fringed)
cap.
• If you cut off the
cap, it will
regenerate. The
regeneration is
under hereditary
control
Experimental Design:
• Cut off the cap of each alga
• Remove a section of the stalk (which contains cytoplasm,
but not the nucleus)
• Place a stalk from A. crenulata on a base (containing a
nucleus) of A. mediterannea (Experiment 1)
• Place a stalk from A. mediterannea on a base of A.
crenulata (Experiment 2)
• Observe the regeneration of the cap on each “hybrid”
alga
Exp. #1 – Stalk C on Base M
• If the cytoplasm
controls heredity . . .
• The cap that
regenerates will
follow instructions
from the cytoplasm
in the stalk.
• The cap should
resemble type C
• If the nucleus
controls heredity . . .
• The cap that
regenerates will
follow instructions
from the nucleus in
the base
• The cap should
resemble type M
Exp. #1 – Stalk C on Base M
• The algae
regenerated
type M caps,
suggesting
the nucleus
in the base of
the cell
controlled
the
development
of the cap
Exp. #2 – Stalk M on Base C
• Similarly,
algae with a
nucleus from
type C
regenerated
a crenulated
cap.
• In both
experiments,
the nucleus
controlled
regeneration
The Nucleus as Control Center
• If the nucleus is the
center of hereditary
control, there must
be something inside
of it that actually
directs the
development of
traits.
• The nucleus consists
almost entirely of:
• Proteins
• Nucleic Acids
• DNA
• RNA
What material in the nucleus
controls heredity?
• Nucleic Acids?
• DNA and RNA
• Complex polymer
made of nucleotides
• Nitrogenous bases are
variable
• 4 types (AGCT)
• Protein?
• Complex polymer
made of amino acids
• Amino acid side chains
are variable
• 20 different types of
essential amino acids
commonly found in
protein
• With more variability,
protein was the
favored hypothesis
What material in the nucleus
controls heredity?
• To test these hypotheses, we need to separate the proteins
from the nucleic acids, treat a test subject with each, and
observe the results.
• Viruses are made entirely out of nucleic acids and proteins.
• Viruses inject their genetic material into a host cell, take over
the hereditary machinery of the host, and use it to reproduce
itself.
• We can use viruses as a test subject if we can find a way to
easily identify or manipulate the proteins and the nucleic
acids.
Tobacco Mosaic Virus &
Holmes Ribgrass Virus
• Tobacco Mosaic Virus (TMV) was the first virus discovered and
isolated.
• TMV infects plants, and forms characteristic lesions that are
identifiable as tobacco mosaic disease. Holmes ribgrass virus
(HRV) also infects plants. The lesions formed in Holmes
Ribgrass disease are distinctly different from TMV lesions
• Viruses consist of a protein coat surrounding a core of nucleic
acids (RNA in the case of these 2 viruses)
• Viruses inject their host with their genetic material, and take
over the host for the purpose of replication
Heinz Fraenkel-Conrat 1955
• In the first of a series of experiments, Fraenkel-Conrat
enzymatically digested the protein coat from TMV viruses and
isolated the RNA core (and vice versa)
• He then infected tobacco plants with only the protein he
derived from the virus. The plants did not develop tobacco
mosaic disease
• He also infected tobacco plants with only the RNA from his
viruses. These plants developed tobacco mosaic lesions
• These results suggest that nucleic acids control the heredity of
the virus and the characteristics associated with the viral
disease
• Remember, protein was the favored hypothesis for the
hereditary material (due to its greater structural variability)!
Fraenkel-Conrat; Hybrid Viruses
• In this experiment,
he created hybrid
viruses consisting of
the protein coat from
HRV and an RNA core
from TMV
• He infected tobacco
plants with the
hybrid viruses
Fraenkel-Conrat; Hybrid Viruses
• If protein was the
hereditary material,
the plants should
exhibit the symptoms
of HRV
• If the RNA is the
heredity material,
the plants should
form lesions
characteristic of TMV
Results
• Not only did the plants show
disease symptoms
characteristic of tobacco
mosaic disease, but viruses
collected from the infected
plants were fully formed
tobacco mosaic viruses.
• The hereditary material from
the TMV core caused the
disease symptoms, was
replicated in the host cells,
and directed the formation
of TMV protein coats in the
offspring viruses.
• That core material was RNA,
not protein as hypothesized
Hershey and Chase 1952
• In another classic experiment to identify the hereditary
material, Alfred Hershey and Martha Chase used radioactively
tagged viruses (bacteriophages – viruses that infect bacteria)
• Both protein and DNA contain carbon, hydrogen, oxygen and
nitrogen, but each contains one element that the other does
not.
• Protein contains sulfur, while DNA contains phosphorus.
• Viruses grown in a culture containing radioactive sulfur will tag
the proteins in a way that can be identified and tracked in the
lab.
• If viruses are grown in a culture containing radioactive
phosphorus, their nucleic acids can be tracked
Hershey/Chase Experiment
• Produce
radioactively
tagged viruses
• Allow them to
infect bacteria
• Agitate, wash, and
centrifuge the
cultures
• Test the wash
solution and the
bacterial cells for
radioactive residue
Predictions:
• If protein is the
hereditary material,
then the wash
solution should
contain radioactive
phosphorus residue
and radioactive sulfur
should be detectable
in the bacterial
cultures
• If DNA is the
hereditary material,
the wash solution
should contain sulfur,
while the bacterial
culture should
contain the tagged
phosphorus
Results and conclusions:
• The cells that
were infected
contained
residues of
radioactive
tagged
phosphorus.
• DNA was injected
into the host cells
• DNA is the
hereditary
material
Frederic Griffith 1927
• Pneumonia is a deadly disease. There are various causes,
including both bacterial and viral types
• The causes and treatment of bacterial pneumonia has been of
enormous importance to medical science for generations
• To determine the pathogen responsible for infectious disease,
a researcher will adhere to a series of logical concepts called
Koch’s Postulates:
• The bacteria must be present in every case of the disease.
• The bacteria must be isolated from the host with the disease and
grown in pure culture.
• The specific disease must be reproduced when a pure culture of
the bacteria is inoculated into a healthy susceptible host.
• The bacteria must be recoverable from the experimentally
infected host.
Smooth vs. Rough Strains
Streptococcus pneumoniae exists in two distinct
strains. One which forms rough textured pinpoint
colonies and another which forms smooth,
spreading colonies
Smooth = Capsulated
We now know
that the
smooth strains
are smooth
due to a slime
layer, or
“capsule”
which
surrounds the
outside of the
cell wall
Experiment #1
• Inject rough type
pneumococci into
healthy mice
• Result:
• Mice remain healthy
• No pneumonia
symptoms are
observed
• No bacteria are
recovered from blood
• Inject smooth type
pneumococci into
healthy mice
• Result:
• Mice exhibit
pneumonia symptoms
• Smooth type bacteria
are recovered from
the blood of the
infected mice
Follow-Up; Experiment #2
• Only the smooth
type cause the
disease, but is it a
result of the action of
the cells themselves
or is it a result of
some poison in the
capsule?
• Mice were injected
with heat killed
smooth type bacteria
• Results:
• Mice remain healthy
• No pneumonia
symptoms are
observed
• No bacteria are
recovered from blood
OK, so what now?
• Clearly it is the cells
themselves, and not
the capsule that
cause pneumonia,
but the capsule must
have some
significance
• The third experiment
in the series involved
taking living rough
type cultures and
growing them in a
medium containing
the remains of heat
killed smooth type
bacteria
Experiment #3 - Predictions
• Live rough type
hadn’t caused
pneumonia in the 1st
experiment
• Heat killed smooth
didn’t cause
pneumonia in the 2nd
Experiment
• The only logical
assumption would be
that since neither
caused pneumonia
by itself, there would
be no reason to
expect them to cause
pneumonia together
Experiment #3 - Results
• Combined live
rough/dead smooth
cultures are injected
into healthy mice
• Mice develop
pneumonia
symptoms
• Smooth type bacteria
are recovered from
the blood of the
infected animals
How do we account for that?
• So not only did we
get pneumonia from
2 things that were
proven not to cause
pneumonia, but it
sure looks like the
smooth type cells
came back from the
dead like slime
covered microscopic
pneumonia zombies
Not Likely
• OK, so scrap the zombie hypothesis
• What appears to have happened is that hereditary material
from the heat killed smooth cells was absorbed by the living
rough cells
• Even though the cells were dead, the hereditary material was
still operational, and could cause the rough cells to become
“transformed”
• The transformed cells expressed the “smooth” genes they
picked up from the dead cells. They gained the ability to make
a capsule, and because of that they were able to cause
pneumonia in the mice
• It turns out that the capsule doesn’t harm the mice, but it
does protect the bacteria from the mouse’s immune system,
allowing them to stay alive, multiply, and cause the disease
Avery 1943
https://www.youtube.com/wat
ch?v=RWFc8Iqz4Jg
Frederic Griffith and Oswald Avery
DNA Structure
• DNA is a polymer
consisting of
nucleotide subunits
• Nucleotides have 3
parts:
• Phosphoric Acid
(phosphate)
• Pentose sugar
(deoxyribose)
• Nitrogenous Base
• In DNA nucleotides,
the phosphate and
the deoxyribose are
constants
• There are 2
categories of
nitrogenous base:
• Purines
• Adenine and Guanine
• Pyrimidines
• Cytosine and Thymine
Pentoses, Purines, and
Pyrimidines
DNA Polymer
• There are 2
fundamental
structural
components of a
DNA strand, the
Sugar/Phosphate
backbone and the
Nitrogen Base sidechains
Some Vocabulary Clarification
•
•
•
•
•
The deoxyribose is a 5 carbon sugar (pentose)
Structural positions are designated by numbering the carbons
The nitrogen base is at position 1 (1’, read as “1 prime”)
The phosphate is at the 5’ position
The bottom corner (3’) will become the point of attachment
of the next nucleotide in the polymer
Rosalind Franklin - 1951
• Franklin’s x-ray crystallography photographs of
DNA demonstrated that DNA consisted of 2
strands twisted in a “double helix”
Erwin Chargaff 1950
Chargaff enzymatically digested DNA from a
variety of organisms and determined the
relative proportion of each base. With great
regularity, the proportions of A and T are
equal to each other, as are G and C
Base Pairing
• Chargaff’s rule (#A = #T; #G = #C) is explained if the 2
DNA strands align with bases paired across between
the strands, A to T and G to C
• Note the formation of hydrogen bonds between the
base pairs
The Double Helix Illustrated
• The polymer is formed
by the sugarphosphate backbone
• The “phosphodiester”
links are covalent
• The 2 strands are
“antiparallel”
• The strands are joined
by base pairing
• The hydrogen bonds are
weak
• They can separate and
rejoin easily
Watson and Crick - 1953
• James Watson and
Francis Crick won the
Nobel Prize for
publishing the
structure of DNA
• Neither of them is
particularly good
looking
Bozeman Videos
• DNA and RNA part 1
• http://www.youtube.com/watch?v=qoERVSWKmGk&list=PLFCE4
D99C4124A27A&index=34
• DNA and RNA part 2
• http://www.youtube.com/watch?v=W4mYwsr9gGE&list=PLFCE4D
99C4124A27A
• Gene Regulation
• http://www.youtube.com/watch?v=3S3ZOmleAj0&list=PLFCE4D9
9C4124A27A
• Signal Transmission and Gene Expression
• http://www.youtube.com/watch?v=D-usAds_lU&list=PLFCE4D99C4124A27A
MIT OpenCourseWare Videos
• DNA Structure and Classic Experiments, excerpt 1
• https://www.youtube.com/watch?v=P-Ry4rRdDbk
• DNA Structure and Classic Experiments, excerpt 2
• https://www.youtube.com/watch?v=YCeKtM6Hnmc
• DNA Replication: Fundamentals of Biology
• https://www.youtube.com/watch?v=DRBREvFL19g
• Transcription and Translation, excerpt 1
• https://www.youtube.com/watch?v=tMr9XH64rtM
• Transcription and Translation, excerpt 2
• https://www.youtube.com/watch?v=uBRdfsz_YB4
Replication
• “Replica” = exact
copy. Replication is
copying of the DNA
• DNA is copied in
preparation for cell
division
• (mitosis/meiosis/binary fission)
• Replication occurs
in the S (synthesis)
phase of Interphase
• In replication, the
entire genome is
copied
Models for Replication
• It makes sense that a
chromosome must
be copied before
mitosis so that each
“daughter cell” will
receive a complete
copy of each
chromosome . . .
• But what actually
goes to each
daughter cell?
• Does one daughter
cell get the original
while the other gets
the copy?
• Do both daughter cells
get a mix of original
and copy DNA?
• A random mix or a true
split mix?
First, some vocabulary
• To “conserve”
something is to allow
it to remain intact or
unchanged
• Remember, the DNA
consists of 2 strands
• After replication we’ll
have 2 sets of DNA,
each with 2 strands
• If the resulting of
replication is 2
original strands
staying together,
replication is
“conservative”
• If 1 strand is original
and the other strand
is new, replication is
“semiconservative”
Replication Models Compared
• The diagram shows
the difference
between the
different models of
Replication
• Semiconservative
• 1 original, 1 new
• Conservative
• Originals stay together
• Dispersive
• Random mix
Meselson and Stahl – late 50’s
Meselson and
Stahl did
experiments
to evaluate
the different
models of
replication by
measuring
the results of
replication
using heavy
isotopes of
nitrogen
• read
pp.311-312
Meselson – Stahl Experiment
• Bacteria were grown in
cultures containing
heavy isotopes of
nitrogen (15N)
• These bacteria were
transferred to cultures
containing normal,
light nitrogen (14N)
• After each replication,
samples were collected
and centrifuged to
determine their
relative density
• The density of the
resulting DNA varied
depending upon the
proportion of DNA
containing the original
heavy nitrogen-15
Predicted Results - Molecular
Predicting Results - Observed
• The “parent” DNA had
2 strands containing
15N, so would be
consistently very dense
• Semiconservative
hybrid DNA would
have 1 heavy and 1
light strand, so would
be consistently of
intermediate density
Predicting Results - Observed
• Any subsequent
replications occurring
in a 14N environment
would produce a
small amount of
hybrid DNA (from the
2 original conserved
strands) and
progressively larger
amounts of low
density DNA
Results and Conclusions
The results of the experiment are precisely what we
would predict for Semiconservative replication
https://www.youtube.com/watch?v=JcUQ_TZCG0w
Semiconservative Replication
• The 2 original DNA
strands are
conserved but
separated
• Each original strand
serves as a template
for producing the
replica strand
• The process reflects the
structure of the DNA
molecule. Note:
• The sugar phosphate
backbone
• Base pairing
• “Antiparallel” strand
orientation
• https://www.youtube.com/watch?v=zdD
kiRw1PdU&feature=related
• https://www.youtube.com/watch?v=yqE
SR7E4b_8&feature=related
Replication Process Sequence
STEPS
Unzip
form RNA primer
ENZYMES
Helix destabilizing protein, DNA
helicase, single stranded binding
proteins, topoisomerase
base pairing
RNA primase
no enzyme required - base pairs align
by shape and polarity
attach individual DNA nucleotides to
the primer
DNA polymerase III
attach individual DNA nucleotides to
the growing chain
replace the primer with DNA
nucleotides
attach okazaki fragments
DNA polymerase III
DNA polymerase I
DNA ligase
Unzipping the strands
• The area that unzips forms a “replication fork”
• Unzipping is a more complex process than it might seem,
largely because of the helical nature of DNA
• The two major players in unzipping are the Helix Destabilizing
Protein and the enzyme DNA Helicase
• The Helix Destabilizing Protein does exactly what its name
suggests. It interferes with the twisted shape of the double helix,
creating some physical stress between the base pairs
• This creates space for the binding of the DNA Helicase, which will
act like a wedge to separate the base pairs from each other
• Remember, the sugar-phosphate backbone is permanently linked
together with covalent phosphodiester bonds, but the base pairs are
only attracted to each other by hydrogen bonding. Hydrogen
bonding is strong as polar attractions go, but weak in comparison to
covalent bonds.
Resolving complications
• Once unzipping begins, the action of the single strand binding
proteins and topoisomerase become necessary
• Single strand binding proteins hold the strands apart so that they
don’t immediately rejoin. Remember, the base pairs are held
together by hydrogen bonding. If they come back into contact they
will stick back together
• Remember that DNA is a double helix. Disrupting the helix at the
replication fork creates stress further down on the molecule. Picture
uncoiling an extension cord. Every coil you unwrap compresses a coil
you already unwrapped, and pretty soon you have a tangled mess.
Add to that the fact that you are turning one double helix into 2
double helices. The enzyme topoisomerase resolves these stresses
by cutting the DNA every once in a while to relieve the stress and
then bonding it back together afterwards
• Don’t try that with your extension cords. They don’t go back together.
Replication
Leading and Lagging Strand
• Each of the steps of replication is directed by the action of specific
enzymes
• Enzymes and substrates must fit together according to shape and
polarity (“induced fit”)
• Nucleotides are not symmetrical, the central deoxyribose is a
pentose. The phosphorylated end is designated as 5’, the flat end
(3’) will be the point of attachment of the next nucleotide in the
polymer.
• The primary enzyme of replication, DNA polymerase, can only attach
single DNA nucleotides to the 3’ end of the chain.
• The strands of DNA are antiparallel, so replication occurs in opposite
directions on the two strands
• One strand (the leading strand) will replicate in the same direction
as unzipping. The lagging strand replicates opposite the direction of
unzipping
Replicating the Leading Strand
• The process of replication begins with unzipping. This separates the
base pairs, exposing the genetic code.
• Unzipping requires a series of proteins and enzymes
• Once the strands are separated, free triphosphorylated nucleotides
will align according to the base pairing rule.
• The first nucleotides to align will be RNA nucleotides, forming an
RNA primer. The growing DNA replica will be built off of the 3’ end
of the primer
• An enzyme called primase controls the formation of the primer
• Base pairing continues with triphosphorylated DNA nucleotides.
• Each nucleotide is attached to the 3’ end of the chain by DNA
polymerase III
• On the leading strand replication is continuous, the polymerase
follows the helicase without interruption through to completion
Replicating the Lagging Strand
• Replication of the lagging strand is largely the same as the
leading strand, but opposite the direction of unzipping.
• The lagging strand replicates in shorter segments (okazaki
fragments). Each fragment has an RNA primer.
• Each RNA primer must be degraded and replaced with DNA
nucleotides. This is accomplished with a DNA polymerase
enzyme (DNA polymerase I)
• Polymerase is not capable of attaching the okazaki fragments
together. This is done by an enzyme called DNA ligase
Replication
• DNA Replication video
• http://www.youtube.com/watch?v=AGUuX4PGlCc&feature=re
lated
• Crash Course: DNA Structure and Replication
• http://www.youtube.com/watch?v=8kK2zwjRV0M
• Bozeman: DNA Replication
• http://www.youtube.com/watch?v=FBmO_rmXxIw
Biocoach Tutorial Links
• DNA Structure and Replication
• http://www.phschool.com/science/biology_place/biocoach/dnarep/intro.html
• Transcription
• http://www.phschool.com/science/biology_place/biocoach/transcription/intro.html
• Translation
• http://www.phschool.com/science/biology_place/biocoach/translation/intro.html
• Lac operon
• http://www.phschool.com/science/biology_place/biocoach/lacoperon/intro.html
• Restriction enzyme digest
• http://www.phschool.com/science/biology_place/biocoach/red/intro.html
Transcription and Translation
• Animation
•
http://www.youtube.com/watch?v=41_Ne5mS2ls&feature=related
• Bozeman
•
http://www.youtube.com/watch?v=h3b9ArupXZg
• MIT
•
http://www.youtube.com/watch?v=uBRdfsz_YB4
Chapter 17 Objectives
1.
2.
3.
4.
5.
6.
7.
Describe Beadle and Tatum’s experiment and explain how it led to
the “one gene – one enzyme” hypothesis.
Extend the “one gene – one enzyme hypothesis into its more
modern version, “one gene – one polypeptide”. Relate this to the
“central dogma of molecular genetics”.
Explain the process of transcription. Note all molecules needed for
transcription as well as the molecules produced.
Describe the role of eukaryotic promoter sequences in the
initiation of transcription.
Define intron and exon. Explain the processing that occurs to
convert pre-mRNA into mature mRNA.
Explain the process of translation. Discuss the role of mRNA,
tRNA, and ribosomes in translation.
Define mutation. Discuss several types of mutations and evaluate
their effects on the polypeptides produced.
What is a gene?
• When we introduce the study of heredity, the emphasis is on
the inheritance of traits. Early experiments and observations,
beginning largely with Frederick Griffiths discovery of a
“transforming principle”, recognized that heredity has a more
fundamental molecular and cellular basis. Traits develop as
the result of structural and metabolic changes within cells and
tissues.
• We now know that genes code for the production of proteins,
and that those proteins are what more directly influence the
formation of observable traits.
• Most of those proteins are enzymes which will control steps in
metabolic pathways.
Beadle and Tatum 1958
One gene – One enzyme
• George Beadle and Edward Tatum performed a series of
experiments with the bread mold neurospora, for which they
received the Nobel Prize in 1958. They used X-rays to mutate
the mold, and grew pure cultures of the mutated molds. They
attempted to grow sample cultures on “minimal media” which
did not contain a full nutrient complement (all 20 essential
amino acids)
• When they found a strain that could not grow on minimal
media, they tested it in media that was supplemented by
single amino acids to determine what they needed to survive
and grow.
• The pattern of results they observed suggested that the molds
had the ability to convert one amino acid into another, and
that the mutations disrupted that ability
Lame but easy to visualize:
• So you plan an evening in Chicago with friends from different
towns in Northwest Indiana. One from Munster, one from
Griffith, one from Merrillville, one from Valpo and one from
Michigan City.
• You hear on the radio about an overturned truck that shuts
down traffic on Interstate 94, but you miss the part where
they say where it happened.
• Only the friends from Munster and Griffith show up
• Where was the highway blocked?
Now for real:
• Imagine a metabolic pathway, each step controlled by a different
enzyme, which converts an organic precursor molecule into one
amino acid, then to another, and so on down the line.
• Suppose that the pathway involved the amino acids citrulline,
arginine, and ornithine.
• 3 mutant strains of Neurospora exist which cannot grow on minimal
media. All 3 mutant strains can grow on media supplemented with
arginine
• Strain 1 can grow on media supplemented with any of the three
amino acids, but not on the precursor alone
• Strain 2 can grow on media supplemented with arginine or with
citrulline, but not ornithine
• Strain 3 can grow with arginine supplements, but not with any of the
others
• Hypothesize a sequence of events for the metabolic pathway and
pinpoint the enzyme mutated in each strain
What can we surmise?
• The precursor must be the first
step in the pathway because
strain 1 can’t grow with it alone
but can grow with any of the
other 3 supplements
• Arginine must be the final
product in the pathway
because all three strains could
grow on media supplemented
with arginine (if they were fed
arginine, they didn’t need to
make arginine), plus strain 3
could only grow with the
arginine supplement
How about the middle steps?
• Strain 2 can grow on an
arginine supplement or with a
citrulline supplement but not
the others
• Strain 2 must have the
necessary enzymes to
produce arginine from
citrulline, but not from
ornithine
• Citrulline must be after the
roadblock, and ornithine
before
What about the mutations?
Strain 1 can grow on media
supplemented with any of the
three amino acids, but not on
the precursor alone
The roadblock must be
between the precursor and all
of the other enzymes (before
ornithine) .
• The strain 1 mutation
disrupted the production of
enzyme A
What about the mutations?
Strain 2 can grow on media
supplemented with arginine or
with citrulline, but not ornithine
The roadblock must be after
ornithine but before citrulline.
• The strain 2 mutation
disrupted the production of
enzyme B
What about the mutations?
Strain 3 can grow with arginine
supplements, but not with any
of the others
The roadblock must be after
ornithine and citrulline but
before arginine.
• The strain 3 mutation
disrupted the production of
enzyme C
Conclusion:
• The genes that were mutated by the X-Rays disrupted the
ability of the cells to produce particular enzymes
• The “one gene – one enzyme” hypothesis is derived from
these experiments.
• After Beadle and Tatum we now understand that genes aren’t
so much about traits as they are about enzymes that control
metabolism and development.
• Traits result from the actions of these enzymes because the
enzymes control the materials that the cell produces
• Modern genetics takes this a step further, recognizing that
enzymes are made of protein. So technically, one gene is the
instructions for producing one polypeptide
Oh, yeah. One more thing
• Genes are made of DNA
• DNA is in the nucleus
• The machinery used to actually synthesize polypeptides are in
the cytoplasm
• The nucleus and cytoplasm are separated by a nuclear
envelope
• The nuclear envelope has pores, but those pores are too small
to allow DNA to pass between the nucleus and the cytoplasm
• So there must be another molecule, small enough to exit the
nucleus but able to mimic DNA’s genetic code, that carries the
genetic information from the nucleus out to the cytoplasm
• Go ahead, say it, you know what it is . . .
The Central Dogma of
Molecular Genetics
The Central Dogma of
Molecular Genetics
• The copying of a DNA gene into mRNA is called Transcription
• The term transcription is appropriate because a transcript is a
copy of the necessary information for a desired task
• The process of decoding that gene to form a polypeptide is
called Translation
• “Translation” because it is converting from the language of DNA
and RNA (nucleotide sequences) into the language of protein
(amino acid sequences)
Transcription
Fundamentally transcription is much like replication, but with several
significant differences.
• First the DNA strands must unzip
• But they must unzip in the proper place to transcribe a particular
gene. Replication copied the entire genome. Transcription copies
the instructions for a single protein
• Then RNA nucleotides must align according to the base pairing rule
• Remember, RNA has U instead of T, so the base pairing rule is slightly
different
• Since the messenger will be RNA, there is no need for an RNA primer
• Then those nucleotides must be joined together to form a polymer
• These are RNA nucleotides, so the enzyme that binds them will be
RNA polymerase, not DNA polymerase
Transcription
• We can think of transcription as having 3 phases: Initiation,
Elongation, and Termination. Basically it Starts, Happens, and
Stops.
• Many of the enzymes used in Replication are not a part of
transcription
• DNA helicase is unnecessary because the RNA polymerase serves
a dual function. It both unzips the strands and forms the polymer
• No RNA primer, so no primase
• No topoisomerase because the area unzipping is so small that
there is no stress in the helix to release
• Since only one strand is transcribed, there are no Okazaki
fragments and no need for DNA ligase
Initiation
• Since transcription only copies a single gene the process must
be initiated at a very specific spot on the chromosome.
• Transcribing too much before the actual coding sequence
would be enormously wasteful. Remember, each nucleotide is
triphosphorylated. Every nucleotide used but not needed is a
waste of energy
• Transcription begins at a particular region “upstream” from
the coding sequence called the Promoter region
• The Promoter region will facilitate the binding of the RNA
polymerase, either directly or indirectly by binding
“transcription factors” which then bind the polymerase
• In eukaryotes, the promoter region generally begins with a
series of alternating T and A nucleotides called the TATA box
that binds transcription factors
Elongation and Terminaton
• Elongation means to add length. RNA polymerase adds RNA
nucleotides to the 3’ end of the mRNA transcript, making the
chain grow longer.
• Termination means stop or end. After the coding sequence
transcription must end. Every unnecessary nucleotide added
is a waste of energy, so a specific stopping point is of great
adaptive value
• In eukaryotes, the coding sequence is followed by a
“polyadenylation signal” (many A’s) that serve as a stopping
point
The Transcription Unit
The Transcription Unit
•
•
•
•
•
Where is the actual gene?
Where does the RNA polymerase bind?
Where do the transcription factors bind?
Where does transcription begin? End?
Which strand gets copied?
Prokaryotes vs. Eukaryotes
• Prokaryotes (bacteria) do not have a nuclear envelope, so there is
not barrier between the processes of transcription and translation.
• In bacteria, ribosomes will bind to the mRNA even before
transcription is completed
• In eukaryotes, however, a series of modifications will take place
before the RNA leaves the nucleus:
• Addition of a 5’ cap
• The cap consists of a modified G nucleotide.
• Prevents degradation and facilitates binding of the ribosome
• Addition of a 3’ poly A tail
• 50-250 Adenine nucleotides added to the 3’ end
• Prevents degradation
• Splicing
• Removal of Intervening Sequences (“introns”)
Why Degradation Matters:
• Once the mRNA leaves the nucleus, it almost immediately
starts to fall apart (degrade). Degradation begins at the ends
• If the coding sequence began with the first nucleotide,
degradation of even one nucleotide would prevent the protein
from being produced correctly. A longer UTR (untranslated
region) means that the mRNA will continue to function for a
longer period of time before the code begins to be degraded
• What that means is more protein produced from a single
transcript, and remember that transcription is very energy
expensive
• On the flip side, too much protein from one transcript is
wasteful too, not of energy but of amino acids – if you devote
amino acids to making protein you don’t need, they aren’t
available for building more useful structures or enzymes
Eukaryotic RNA splicing
• In eukaryotes, only part of what I represented as the coding
region will actually be translated into an amino acid sequence
• The parts of the sequence that will be expressed in the protein
are called Exons. Intervening sequences that will be spliced
out and will not be expressed are called Introns.
• Whoever came up with those names should be hung up by
their thumbs because they sound like the opposite of what
they actually mean, and scientists shouldn’t do stupid stuff
like that.
• The Ex in Exon doesn’t mean “out” like it usually would. It
means Expressed, so it stays In.
• The In in Intron doesn’t mean it stays in, it means intervening,
so it gets cut out.
Spliceosomes
• A spliceosome is
a particle of RNA
and protein that
will bind to the
pre-mRNA and
cut out the
introns
Spliceosomes, more detailed
Alternative Splicing
• The significance of eukaryotic gene splicing is enormous
• A single gene can result in the production of a wide variety of
proteins by transcribing the same gene and processing it
differently
• A cell can treat a particular part of the RNA as an Intron and
cut it out, or treat it as an Exon, leave it in and express it
• A human cell has about 20,000 genes, but may produce as
many as 100,000 different proteins.
• A great deal of genetic variability can be attributed to
differential RNA splicing.
• Producing the same enzyme with variations in the allosteric
parts of the protein, for example, would serve the same
metabolic function but have a great effect on the optimal
conditions of the enzymes activity
Translation