Molecular Genetics
Unit 3
Unit 3 MOLECULAR GENETICS
INTRODUCTION
Part I DNA, RNA and PROTEIN SYNTHESIS
A] HISTORY OF DNA
Figure 11, pg 215
NOTABLE SCIENTISTS:
____________________________ : Discovery that the Nucleus is the source of inheritance
______________________ and __________________ : Showed that DNA actually controls
heredity
____________________________ : Crystallography Structure of DNA
____________________ and _____________________ : Mathematical modeling of DNA Structure
B] DNA STRUCTURE
1] Parts of a nucleotide
____________________
HOH2C
O
_________________
O
II
HO—P—OH
I
OH
OH
______________________
pyrimadine
___________
___________
purine
OH
O
II
HO—P—OH
I
O
CH2
___________
___________
O
combined to form
______________
OH
1
2] arrangement of nucleotides
a] the __________ bonds to the last _______________ at carbon _____
and so the sugar end of DNA is called the _______ end
b] the ______________ bonds to the next ___________ at carbon ____
and so the phosphate end of DNA is called the _____ end
c] the two strands are arranged as shown below
3. Hydrogen bonding
- adenine forms _____ hydrogen bonds with thymine
- cytosine forms _____ hydrogen bonds with guanine
Since A – T and C – G are the usual bondings,
if the amount of A does not equal T or the amount of
_____ end
l
P
____ end
l
l
sugar—base……
base—sugar
l
l
P
P
l
l
sugar—base……
base—sugar
l
l
P
P
l
l
sugar—base……
base—sugar
l
l
_____ end
P
l
____ end
C does not equal G, then the DNA is not
double stranded
Sections of the DNA with more ________________ bonds, are more stable.
[_____ H-bonds vs. ______ H-bonds]
4. DNA shape – the stands are arranged in a ________________________________
5’
3’
5’
Each Human Cell contains approximately ___________________________ base pairs.
These are arranged into roughly ______________________ different genes.
In addition, up to __________________ does not code for any protein.
In total, DNA from 1 cell could be _________________ long.
___________________ is very stable but ______________ is very easily __________________.
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C. DNA REPLICATION
Introduction
a. Before a cell _________________ it must first copy all its DNA
b. It must do this while keeping the amount of ______________ to a minimum.
c. Also, the DNA will have to be __________________ to allow access to the bases.
[The bold numbers following the bold terms refer to the DNA picture of replication on the next page]
1. Separating the strands
a] DNA helicase
[1]
– breaks___________ between the strands at an “origin of replication”
b] DNA topoisomerase [2]
- releases _____________ by unwinding
c] replication fork [3]
- an area where the two strands are _________________
d] ssB’s
- single-stranded binding proteins keep strands from_______________
[4]
2. Building complementary strands
a] starting
-DNA primase [5] - creates a _____________ strand of RNA
called an RNA-primer[6] at the exposed______end of the replication fork
b] building
-the enzyme DNA polymerase III [7]- adds complimentary bases that are
_________________ moving along the old DNA in the ________ direction
- the energy released by removing ______________ allows the bonding
- the strand that is copied continuously __________ strand is
called the leading strand [8]
c] other strand
DNA polymerase III only moves__________________
- so the lagging strand [9] has to be built_________________________
- these ____________________ pieces are called Okazaki fragments [10]
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d] finishing
DNA polymerase I [11] removes the __________________________
& replaces them with the appropriate DNA bases
DNA ligase [12] attaches __________________ fragments
by bonding sugar and ___________________________
3. Picture – see next page
DNA Replication
3
1
3’
4
5’
7
2
8
6
5
7
11
10
7
5
5’
’
9
33’
12
3’
5’
5’
___ DNA Helicase
___ DNA Topoisomerase
___ DNA Ligase
___ DNA Polymerase I
___ DNA Polymerase III
___ Lagging strand
___ Leading strand
___ Okazaki fragments
___ DNA Primase
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___ Replication Fork
___ RNA primer
___ ssB’s -
D. RNA
5. RNA – ribonucleic acid
RNA Structure
a] _________ sugar + phosphate + nitrogenous base but ____________ replaces Thymine
b] __________ stranded molecule
c] much _______________ than DNA
5’
l
A
P—sugar—P—sugar—P—sugar—P—sugar—P—sugar—P—sugar
l
l
l
l
l
C
G
C
U
G
3’
RNA features
unlike DNA, RNA can _______________ in the cytoplasm if properly prepared
Types of RNA
1. mRNA – messenger RNA
- varies in _______________, but always much _______________ than DNA
- carries the ______________ message for protein building from DNA to the _____________
2. tRNA – transfer RNA
- very short, _________________ bases only
- acts to transport the ________________________ to the mRNA at the ribosomes
3. rRNA – ribosomal RNA
- varies in length, but ______________ than mRNA
- binds the large and small __________________ to form a functional _________________
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E. TRANSCRIPTION – the process of making mRNA
1. Initiation
a. ______________________ binds to a specific region of the DNA
that is _______ rich and so easy to separate
b. the DNA is _________________ and the two strands are _________________
2. Elongation
a. only the ___________________ DNA strand is used � _________________ strand
b. the unused strand [5’ to 3’] is called the_________________ strand
c. free floating nucleotides A, U, C & G are __________________
to their complimentary bases by ______________________
d. RNA polymerase moves along the DNA _________________& _______________
as it goes while the DNA winds back up behind it
e. the ____________ hangs free off the side of the DNA strand
attached only at a few base pairs
GCCTAATCGTCACTGCAA
C
3’—C G C A T T
5’—G C G T A A
G
CGGATTAGCAGTGACGTT
T
G C G T A A C G—5’
C G C A T T G C—3’
A
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3. Termination
a. RNA polymerase encounters a sequence of bases at the end of the gene
called a ___________________________________
b. this results in RNA polymerase and the _______________ being released from the DNA
4. Post-transcriptional modifications
a. a ______________ is added to the mRNA to protect it from cytoplasmic enzymes
b. a ________________ of 200 or so adenines is added by poly-A-polymerase to the 3’ end
c. this results in the _____________________ being ready for release from the nucleus
d. in eukaryotes__________________ remove all the non-coding regions [_____________]
and ____________ the coding regions [_________________]
to form the ____________________________________
INTRON
EXON
INTRON
EXON
INTRON
EXON
INTRON
EXON
INTRON
EXON
INTRON
EXON
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F. PROTEIN SYNTHESIS [or DNA translation]
converting the RNA message into protein
1. The code mRNA
a. the message on mRNA is written in 3-letter ____________ called ______________
b. there are ________ or ____________ possible codons
c. pg 248 shows the codons – you will have to learn the ______________
but NOT memorize this with the exception of the __________ and the ___________ codons
d. notice that UUU & UUC both code for _____ and CUU, CUC, CUA & CUG all code for_____
this indicates that the __________ letter is the least important in the code
e. only two amino acids have a unique code UGG � _______ and AUG � _________
f. all proteins must start with ________ as AUG is the __________code; this may be removed later
g. there is__________ amino acid that is coded by UAA, UAG or UGA
and so the protein breaks here and these are called _______________codes
2. The ribosomes [rRNA]
a. the rRNA holds together the ____________ subunit [60s] and the___________ subunit [40s]
to form the active ribosome [___________]
b. the _________ [CAP region] adheres to the active ribosome
c. the ribosome will move the _________ through it
starting at the _________ end and going toward the _________ end
3. The tRNA
a. each tRNA has a_________________________ structure [pg 251]
b. at the base there is an _______________ which is the compliment of the _____________ on
the mRNA
eg the mRNA has the codon ________ for met � the tRNA—met has the anticodon ______
c. at the 3’ end of the tRNA is the _______________that holds the amino acid
d. it requires 1 ATP to attach each amino acid to the correct tRNA to make an ________________
e. there are ______ possible tRNA for the 20 aa, therefore a cell needs ____ enzymes for this
4. Protein Synthesis
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STEPS IN PROTEIN SYNTHESIS
1. The mRNA, rRNA and tRNA are ____________ from DNA in the ___________. They randomly drift around
in the _______________where the large and small subunits of the _____________ are also drifting.
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
UAC
______ subunit
of ribosome
_______ subunit
of ribosome
2. The mRNA ____________ the nucleus and encounters the _______________ of the ribosome.
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
3. The________________ attaches the large subunit to the small subunit that is attached to the mRNA.
The tRNA’s are ____________ by the attachment of an amino acid, by the use of an _____________.
Gly
Gln
Met
Ile
CCG
UAC
Pro
GUC
Val
UAG
GGU
CAU
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
P-site A-site
4. The tRNA with ____________ attached, randomly enters the _____________ on the ribosome.
Because its ________________ matches the _____________ in the P-site, this tRNA is held in place.
Pro
Gln
Ile
Met
Gly
Val
GGU
GUC
UAC
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
UAG
CCG
CAU
9
Val
Gly
5.
The tRNA for the next
CAU
Met
Ile
Gln
Pro
CCG
amino acid _______________
UAG
GUC
moves into the ______________
UAC GGU
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
E-site P-site
P-siteA-site
A-site
Gln
6. The tRNA that brought Met now
Val
GUC
UAC
Gly
is ____________ after Met is _________ to
Met-Pro
CAU
Ile
the Pro. The ______ is now moved through
CCG
the ribosome until the ___________ with
GGU UAG
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
Met-Pro is in the ________. The tRNA for
P-siteA-site
A-site
E-site P-site
the next code now fits into the _______
UAC
E-site P-site A-site
GGU
CCG
7. The processes of step 5 & step 6
are repeated until one of the ________
codes is in the _____________.
UAG
Met-Pro-Ile-Gly-GlnVal
GUC
CAU
CC GAG CUC AUG CCA AUC GGC CAG GUA UAG GGG ACC
At this point all of the amino acids have been bonded into a________________
E-site P-site A-site
chain conected to the last tRNA in the _______________.
CCA AUC GGC CAG G
8.
UA UAG GGG ACC
Because there was no tRNA to
CC GAG CUC AUG CC
match _________________, the
polypeptide chain is ____________
and the ribosome ______________.
Met-Pro-Ile-Gly-GlnVal
________________
The mRNA will be _____________.
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From the piece of DNA below, make the complimentary strand of DNA, transcribe the appropriate
strand, package it for the cytoplasm and then translate it as described in the steps below.
In row A; make the complimentary DNA strand;
In row C, transcribe the DNA into RNA;
In column 1 and 28, place the appropriate 3’ or 5’; In column 2 & 27, ‘package’ the RNA in row C;
In row D, underline the ‘words’ in the RNA;
1
2
3
4
5
6
7
8
9
1
0
1
1
In row E, translate the RNA into amino acids
using the 3-letter codes for them
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
2
0
2
1
2
2
2
3
2
4
2
5
2
6
2
7
2
8
A
B 3
5
G G T A C G C T A C C G C A A G G A C T C T C G
C
D
E
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G. GENE CONTROL MECHANISMS
1. Genes ON or OFF?
a. each cell in a multicellular organism has the_______________ genetic information
b. yet each cell makes a ____________ set of proteins & these may change over time……HOW?
c. therefore, most genes in most cells are _________________ most of the time
2. Types of control
pg 255 Table 1
3. Transcriptional control parts
a. promoter – a sequence of DNA that is rich in _____&_____ to which RNA polymerase _______
this is located ‘________________’ from the TAC ‘start’ signal on the DNA
b. operator – a sequence of DNA beside the promoter to which a ______________________ binds
if the repressor protein does _____________ to the operator,
the RNA polymerase can______________ to the promoter
and so ________ transcription occurs
c. repressor protein – binds to the ________________ unless it first binds to the ______________
binding to the inducer changes the shape of the _____________
and now it ___________________ binds to the operator
therefore the RNA polymerase can bind to the _________________
and so now ________________________ can occur
d. inducer – a molecule that binds to the repressor and___________________
it from stopping transcription
4. trp operon in bacteria
[see pg 257]
a. trp is____________ in the environment
and cells need it so normally the gene to make trp is ________________
b. if trp is ________________, it binds to the repressor protein
and changes its ________________ so that it does bind to the promoter and blocks
the RNA polymerase from binding which turns the gene ________________
c. because it helps bind, trp is called a ________________________
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5. lac operon in bacteria
a. lactose is a ___________ sugar [food source] in the environment
so the lac genes to breakdown lactose are usually ____________________
b. if lactose is present, it is the _______________ that binds to the repressor protein
which now does NOT bind to the ___________________
c. when _______________ is present, the gene is ON
three -galactosidase genes
PROMOTER
OPERATO
R
lacZ
lacY
lacA
genes are ______________
three -galactosidase genes
PROMOTER
OPERATO
R
lacZ
lacY
lacA
genes are_________
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Part II: Manipulating DNA
A. Isolating a specific DNA Segment
a. This would allow researchers to __________________ DNA samples
b. It may allow researchers to ________________ DNA from one organism to another.
Why would we want to do this?
B. Cutting the DNA Segment: _________________________________
a. Enzymes that cut _______________ at a specific recognition site
b. These recognition sites are ______________________________
c. eg.
GTGCAC
CACGTG
Sal 1
GAATTC
CTTAAG
EcoR1
AAGCTT
TTCGAA
Hind III
d. the restriction endonucleases often cut the DNA unevenly [as shown above]
and leave _______________________ which are short pieces of ssDNA
that will be able to __________________to similar ssDNA pieces
e. generally a piece of DNA will have a __________________ for any one restriction
enzyme every _____________________ bp or so, spaced at random along the DNA
[eg 2000, 9000, 1400, 16000 23000]
f. the restriction enzymes are named after the bacteria that they are extracted from
Factor
EcoR1
Hind III
Genus
species
strain
isolate
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C. Methylases
a. These enzymes add –_____________ groups to the nitrogenase bases
b. A methylated base will __________________ be recognized by a restriction enzyme
& the DNA will NOT be _________________
c. This allows researchers to ______________ sections of DNA before using the restriction
enzymes
DNA Ligase
d. This enzyme attaches ssDNA
_______________________________________
e. After, the ‘sticky ends’ form _______________________________ to hold the bases together
eg.
uncut DNA
unprotected DNA cut with
restriction enzymes
protected DNA [red] cut
with restriction enzymes
CH3
CH3
protected DNA after being
cut and first 2 segments
ligated
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D. Gel Electrophoresis
a. a procedure for _____________ the cut pieces of DNA
to allow researchers to select the one they want
b. a gel is a _____________________ substance made of starch, agarose or polyacrylamide
c. electrophoresis refers to separation based on _________________________
d. ______________ pieces of DNA will move further than large pieces of DNA
– separation by________
e. so that researchers can find the DNA on the gel – ___________________________ is added
this chemical fits into the twists of DNA without altering it and __________________
under UV light
f. DNA pieces can now be cut, separated & isolated from the gel and checked for the
___________ gene
Using DNA fingerprinting
[which does not involve _________________________]
1. Paternity tests
the child’s DNA has to match both __________________ and _________________
Mary
Bob
Larry
Child
2. Criminal investigations
eg. ________________ where the victim and the ___________ are both sampled
eg. ________________ where matching the _____________ is the key
6. Polymerase chain reactions
make notes from web presentation D.
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Increasing the DNA �
Polymerase Chain Reactions [PCR]
[Refer to Youtube -NOT ON POWERPOINT]
Why do researchers do this?
They wish to amplify _______ copies of DNA to ____________ copies.
What section of DNA is amplified?
Only a _______________ section of the DNA is amplified.
It is chosen because it is highly ____________________ and is called the ___________ DNA.
The idea is that if we look at __________________ of these sections, any one person should have a
________________ pattern.
Procedure:
Start with the sample of DNA and add:
1. _____________________: short [20 nucleotide] pieces of DNA of known sequence that
will bond to both the _____________ and the _________________ strands.
2. _____________________: the basic building blocks of DNA
3. _____________________: an enzyme from a thermophilic bacteria that lives in
_______________ and so is adapted to very warm temperatures.
Cycles:
a. heat to _____oC: The heat _____________ the DNA breaking ___________ bonds.
b. cool to _____oC: To allow the primers to _____________ to complimentary sequences
on both of the strands.
c. heat to _______oC: To allows Taq polymerase to _______________ DNA by adding
complimentary bases.
Results:
one cycle � _______ copies
two cycles � _______ copies
three cycles � _______ copies [general formula ________________________]
twenty-five cycles � __________________ copies
The amplified sample is now run on a ____________ to separate the DNA pieces based on size and
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to establish the _____________ identity of the sample.
GENETIC ENGINEERING
A. Overview
1.
_________________ the DNA segment with the desired gene
2.
_________________ the DNA segment from the genome
3.
__________________________ to place the desired gene in a new organism
B. Finding the DNA segment
1. Based on starting with a __________________
a. -make a synthetic ___________________ based on the amino acid sequence of the protein
b -allow a radioactively labeled mRNA to __________________
to heated, single-stranded DNA [ssDNA]
c. – the labeled mRNA will stick to piece of DNA with the_______________ gene
2. Based on starting with _________________
a. –find a cell that is active in synthesis of the protein so that it’s making lots of ____________
b. –isolate the mRNA from the cytoplasm and label it radioactively then allow it
to __________________ to heated ssDNA as above
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Locating a complete gene on a segment of DNA
1
2
3
4
5
6
7
8
9
10
1
cut with ______________ enzymes
4
2
8
add a piece of _______
5
mRNA for the
desired gene
5
the mRNA ________ to
the piece with the
target gene sequence
6
separated after
being run on the
_____________
gel
6
3
9
10
2
1
4
7
3
heat the DNA
to make it
________
+
if the piece of labeled mRNA anneals to piece 6, 7 as well as 8 it may be because of exons
and so a different restriction enzyme would have to be used
segment
segment
6
exon 1
segment
7
exon 2
8
exon 3
STEPS TO SOLVING LOCATION of GENE PROBLEM
STEP 1 – Write the segment numbers of the digested DNA on the samples of the gel grid, based on
relative lengths of the segments.
STEP 2 – Based on step 1, write the segment letters on the digested DNA grid corresponding to the
few samples for which protein synthesis data was obtained.
STEP 3 – Identify the segment of DNA that produces a normal protein. This segment contains the
gene, and likely more DNA on both ends, the next steps will allow you to eliminate
the excess DNA that is not part of the gene.
STEP 4 – Any segment that produces a smaller, not useful, protein is missing the stop codon at the
end of the gene.
STEP 5 – Any segment that produces no protein until a promoter is added, has the start codon, but is
missing the promoter that is found just upstream from the gene.
STEP 6 – The final segment of DNA that is actually needed should include the shortest segment that
contains the stop codon and the start codon plus a tiny upstream segment.
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A] Samples of identical DNA digested by restriction enzymes
1
2
6
3
7
8
4
5
9
EcoR I
Hind III
10
11
12
13
14
15
16
Sol I
B] Gel electrophoresis samples
EcoR I
Hind III
L
A
B
F
G
H
C
I
D
E
Sol I
J
K
M
N
O
P
Band ‘C’ made a normal protein.
●
Band ‘F’ made a smaller protein that is not useful.
●
No Sol I band made a protein, but when a promoter was added,
●
band ‘M’ did make a normal protein.
On the grid at the top, show where the gene is located, showing your “calculations” on the gel grid, the
digested DNA and on a separate page if necessary.
eg.
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D. Vectors – to get the gene into other cells
1. Plasmids
a. these are small circular pieces of _____________ found in bacteria
b. bacteria have restriction enzymes to allow them to cut their own large ______________________
and integrate ___________________ into it
c. most _____________________ resistance genes occur on plasmids
d. there are genetically engineered plasmids that have known restriction enzyme __________________
and known antibiotic ________________________
e. the _______________ [segment of DNA selected from the electrophoresis procedure and annealing
with the mRNA] is now placed into the _____________________ using the restriction enzymes
f. the plasmid is now placed into bacteria and if the bacteria ____________________ the plasmid
1� it will have the _________________________
2 � it will resist the ________________________
g. researchers screen _________________ of bacteria to find the few that have integrated the plasmid
h. however, without the _______________________ region, the target gene will not work in the bacteria
i. these plasmids may also be used to infect _________ and integrate the target gene into the _______
2. Viruses
a. a virus is used as a ___________________ instead of a plasmid
b. the virus is modified to remove all of the ___________________ genes, if possible
c. these must be _______________ viruses that undergo a _____________________ cycle
d. the virus is then used to infect ____________________ cells
and introduce the target gene into the __________________________ cell
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