Download Chapters 13-16, Molecular Genetics

AP Biology – Molecular Genetics
(Chapters 13-17)
Chapter 13: DNA Structure and
Functions
I. Evidence for DNA as the genetic
material
A. Griffith (1928): Bacterial
transformation experiment –
virulent vs non virulent forms
1. non virulent bacteria could be
changed to virulent (virulent
bacteria killed by heat were
injected into mice with
nonvirulent bacteria – some
mice died and contained
virulent forms)
2. What chemical substance was
causing this change?
Something was being passed on
from the dead virulent forms to
the living non virulent forms
changing their genetic
characteristics.
B. Avery (1944)
1. transforming agent was DNA
2. What else could it be?
(proteins)
3. DNA appears to be the genetic
material in prokaryotes, but
what about eukaryotes?
C. Hershey and Chase (1952)
1. used bacteriophages (viruses
consisting of a protein coat and
DNA inside)
2. Bacteriophages attack bacteria,
take over a bacterium’s cell
machinery, and cause it to
produce many more
bacteriophages (viruses)
3. lysogenic vs lytic cyle of viruses
4. distinguished between proteins
and DNA using radioactive
isotopes (S35 – proteins, P32 for
DNA)
5. New bacteriophages contained
P32
6. Would not have worked with
eukaryote cells whose
chromosomes are made of both
protein and nucleic acid.
D. Quantity of DNA in cells
1. sex cells or gametes contain ½
the amount of DNA as body or
somatic cells
2. Amount and types of protein
differ significantly in the type of
body cell and is not necessarily
lower in gametes
3. DNA is more stable chemically
whereas proteins are constantly
being destroyed and rebuilt in
cells
E. Proportions of nitrogen bases in
DNA
1. adenine, guanine, cytosine, and
thymine
2. Chargaff (1940’s) – four bases
occur in different proportions
in different species, however,
adenine = thymine and guanine
= cytosine
II. Structure of DNA (deoxyribonucleic
acid)
A. Basic structure (Watson and
Crick, 1962)
1. repeating units = nucleotides
(nitrogen base, phosphate, and
5 carbon sugar – deoxyribose
sugar)
2. nitrogen bases form the “steps
of the ladder”
a. single ring pyrimidines:
thymine and cytosine
b. double ring purines:
adenine and guanine
c. adenine always bonds with
thymine (2 hydrogen bonds)
and cytosine with guanine (3
hydrogen bonds)
3. sugar – phosphate backbone
forms the “sides of the ladder”
and run antiparallel (53
versus 35)
4. overall shape = double helix
(twisted ladder) was discovered
by Rosalind Franklin (1952) via
x-ray diffraction
a. 2 complementary strands
are twisted around each
other making a complete
turn every 10 nucleotides or
3.3 nm
b. each “step” of the ladder
occupies 0.33nm
c. width of DNA is 2 nm
d. DNA is billions of
nucleotides long
B. DNA organization in cells
1. prokaryotes – single double
helix of DNA folded many
times, occupying a nuclear area
about 1/10 of the cell’s volume
2. eukaryotes – organized into
chromosomes, each
chromosome contains one super
coiled DNA molecule wrapped
around protein clusters called
histones
III. DNA Replication
A. Occurs prior to cell division when
DNA makes a copy of itself
1. during “S” phase of the cell
cycle (mitosis, cytokinesis,
G1, S, G2)
2. semi-conservative: one
strand of original DNA
combines with new
nucleotides to form the
double helix (one new strand,
one old strand)
3. Meselson and Stahl
demonstrated “semiconservative DNA replication
using isotopes of Nitrogen =
N14 and N15)
B. Process of replication
1. involves 20+ enzymes
2. DNA unwinds at specific
points called replication
origins
3. DNA helicase enzymes bind
and move along the double
helix in both directions as the
strand opens at replication
forks, separating the two
strands
4. DNA polymerase adds new
nucleotides to the 3’end of
each new strand (it is
5.
6.
7.
8.
directional: works in the 53’ direction)
new tri-phosphate
nucleotides, present in the
nucleoplasm, bind to
complimentary bases via
hydrogen bonds, attached
phosphate groups provide
binding energy
500-1000 nucleotides are
added per second in
prokaryotes; 50 nucleotides
per second in eukaryotes
Since DNA polymerase is
directional, gaps are created
in the duplication of DNA (it
is not continuous).
New “lagging” strands
starting from separate
replication forks are made in
sections called Okazaki
fragments; whereas new
“leading” strands are made
continuously.
9. DNA fragments are joined to
form a continuous strand by
DNA ligase.
10. Twenty or more “repair”
enzymes check for mistakes
and make repairs if
necessary
IV. DNA Organization and Structure
A. The order of the nitrogen bases in
a DNA molecule determines the
genetic code of life
B. Mistakes in the genetic code can
produce inheritable changes
called mutations
1. somatic vs germ mutations
2. point vs chromosomal
mutations
3. mutations are generally
harmful
C. Eukaryote cells contain linear
chromosomes
1. each chromosome is a single
DNA molecule tightly coiled
(condensed) and folded many
times around protein
(histones) clusters called
nucleosomes
2. chromatin = loose, unraveled
DNA: contains proteins,
DNA, and RNA
D. Prokaryote cells have a single
DNA molecule formed in a “loop”
and attached to the cell
membrane.
1. additional rings of DNA
called plasmids lie outside
the nuclear area
Chapter 14: Protein Synthesis or How
Genes Work
I. Overview
A. DNA contains the genetic code of
life based upon the order of the
nitrogen bases
1. transcription – DNA code is
copied and rewritten in the
form of mRNA (messenger
RNA)
2. mRNA leaves the nucleus and
travels to the ribosome where a
protein is built
3. translation = building of a
protein with the help of rRNA
(ribosomal RNA) and tRNA
(transfer RNA) which carries
amino acids
4. information flow in a cell
DNA  mRNA  protein
5. What is a gene?: a gene is a
piece of information found
within a DNA molecule that
codes for a protein (structural,
functional, or regulatory)
B. RNA (ribonucleic acid)
1. sugar is ribose, not deoxyribose
2. uracil replaces thymine
3. single strand
4. some RNA (tRNA) is folded
over on itself because of base
pairing producing “hairpins”
II. Transcription = copying the code of
DNA into RNA
A. Process
1. RNA polymerase binds to a
special area of DNA called the
promoter, DNA strand opens up
2. one of the DNA strands serves
as the template to build mRNA
3. RNA polymerase works in the
3-5’ direction along the template
strand, make the mRNA from the
5 to 3’ direction
4. As RNA polymerase moves
down the template, it adds RNA
nucleotides
5. when RNA polymerase reaches
a termination sequence, it leaves
and so does the mRNA
6. three different types of RNA
polymerase
7. pre-RNA (precursor RNA or
transcript RNA) is made prior to
mRNA which must be modified
before forming mRNA that is
exported out of the nucleus
8. snRNA (small nuclear RNA)
found in spliceosomes inside the
nucleus helps modify pre-RNA
9. “introns” are removed and
“exons” are spliced together
10. A “cap” (7 methyl guanosine)
and poly AAAA “tail” are added
B. Genetic Code of Life
1. triplet code
a. 20 different amino acids:
4 bases = 4 amino acids, can
not be a one base = one
amino acid code
b. two base code (42) = a two
base code will only produce
16 combinations, not enough
c. three base code (43) = a three
base code will produce 64
possibilities, more than
enough
2. codon = 3 bases found on
mRNA
a. 3 “stop” codons
b. 1 “start” codon
c. third base in the codon is
often less specific than the first two
d. several codons can code for
the same amino acid (degenerate)
3. genetic code of life is universal
III. Translation = building of the protein
A. tRNA (transfer RNA)
1. one type of tRNA for each
different amino acid
2. shape = 3 main hairpins with an
anticodon and amino acid
attachment side
B. Ribosomes
1. contain 4 rRNA molecules of
different sizes and over 70
different proteins
2. subunits: one large, one small
3. ribosomes are produced in the
nucleus (nucleolus)
4. has two attachment sites: P,A
C. Process
1. initiation
a. mRNA binds to small
subunit of ribosome,
anticodon of tRNA carrying
methionine binds with AUG
(start) codon at the P site
b. large ribosomal subunit
attaches
2. elongation
a. binding of the next tRNA
molecules carrying a second
amino acid to the A site
b. amino acid from the P site is
bonded to the amino acid at
the A site
c. mRNA moves from the A
site to the P site using
energy from GTP
(translocation)
d. this continues as the protein
is built by adding amino
acids
3. termination
a. ribosome reaches a “stop”
codon
b. releasing factor (protein)
binds to the stop codon and
pushes the mRNA off the
ribosome
4. as the 5’ end of the mRNA
leaves the ribosome, it may
bond to another ribosome
forming a polysome or
polyribosome enabling multiple
copies of the polypeptide to be
built very quickly
5. polypeptide (protein) is
processed further in the golgi
apparatus
Chapter 15: Genome Organization and
Gene Expression (What is a gene and how
do genes work?)
I. Basic differences between
eukaryotes and prokaryotes
A. Prokaryotes – gene expression
is controlled by proteins encoded by
regulatory genes
B. Eukaryotes – some undergo
embryonic development during
which cells mature and differentiate
from each other
1. Different genes are switched on
and off in different cells
resulting in cell specialization
and division of labor among
cells.
2. Differentiation is a series of
events in which a gene or
groups of genes are switched on
or off followed by others in an
orderly progression of events
3. Hierarchy of “gene control”
exists in complex eukaryotes
a. Homeotic genes
b. Regulatory genes
c. Structural genes
II. Gene expression in prokaryotes
A. Genes code for proteins. The
process of protein synthesis can be
“switched” on and off at various
points in that process.
B. Protein synthesis is usually
controlled by regulating
transcription.
C. Control of transcription in
prokaryotes
1. regulatory genes code for
transcription factors (proteins)
which bind to regulatory sites
(operators) and inhibit or
initiate transcription.
2. Jacob and Monod (1961)
studied the Lac Operon system
in Escherichia coli (bacteria).
When we drink milk, these
bacteria produce enzymes that
break down lactose into glucose
and galactose.
a. bacteria usually contain a
repressor protein that
inhibits transcription
b. a regulatory gene
manufactures this protein
which binds to the
“operator” region.
c. “operator” is “upstream”
from the structural genes
and the “promoter” where
RNA polymerase binds.
d. Repressor protein blocks
the binding of RNA
polymerase – thus the
structural genes are “off”
e. Repressor protein can be
removed by a cofactor,
allolactose (a form of
lactose).
f. Allolactose binds to the
repressor, causing an
allosteric change in its
shape, removing it from the
operator.
g. RNA polymerase is now
free to bind to the
promoter, transcribe the
structural genes, which are
turned “on”
h. When allolactose (lactose)
is no longer present and its
concentration drops, the
repressor binds to the
operator and the genes are
turned “off.”
i. Lactose = inducer;
inducible operon
3. Other regulatory genes may
produce “activator” proteins
which initiate transcription
rather than inhibit it
a. CAP (Catabolic activator
protein) must be present
for bacteria to metabolize
glucose
b. A cofactor (cyclic AMP)
must be present for CAP to
bind and for the genes to be
transcribed (genes are
“on”)
c. When glucose is plentiful,
the concentration of cyclic
AMP decreases, it
dissociates from CAP, and
returns CAP to its inactive
form. CAP leaves the
operator and the genes are
now “off.”
4. Trp Operon – normally exists
in the “on” rather than “off”
condition
a. Regulatory gene codes for a
“repressor” protein –
unable to attach to
operator
b. RNA polymerase binds to
the promoter
c. Structural genes are
expressed
d. However, if tryptophan is
present, it binds to the
repressor, changing its
shape so it allows the
repressor to bind to the
operator
e. Structural genes are then
turned “off”
f. Tryptophan = corepressor
repressible operon
5. antisense RNA – RNA
polymerase may transcribe the
noncoding DNA strand.
a. antisense RNA can bind
with mRNA preventing it
from being translated at
the ribosome.
III. Gene control in multicellular and
complex eukaryotes
A. Homeotic genes  regulatory
genes  structural genes
1. many regulatory genes (five or
more) may act upon one or
more structural genes
2. transcription requires that
RNA polymerase and several
other proteins assemble into an
RNA polymerase complex
bound to the promoter
B. Regulation is possible at four
different points in the protein
synthesis pathway
1. transcriptional control:
organization of chromatin and
use of transcription factors
a. DNA tightly wound around
histones (heterochromatin)
is inactive genetically
(genes are off)
b. DNA “opened up” or
“exposed” is genetically
active (genes are on)
c. Lamp brush chromosomes
and chromosome “puff” –
active DNA
d. Several transcription
factors (regulatory
proteins) as well as
“enhancers” may be
necessary for transcription
to occur
e. Some transcription factors
work in pairs called
“dimers” whose shapes
enable them to bind to
certain portions of the
DNA helix (leucine zippers
and zinc fingers)
2. post transcriptional control:
differential processing of
mRNA via spliceosomes before
export out of the nucleus
3. translational: life expectancy of
mRNA, ability to bind to
ribosomes, RNA modifications
a. Hormones may extend the
“life expectancy” of certain
mRNA molecules – creating
more protein product
4. post translational control:
polypeptide modification,
feedback control and inhibition