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Molecular Biology
Transcription and Translation
(Chapter 17)
Genotype and Phenotype
• Genotype - caused by changes in DNA (ss)
• Phenotype - change in protein activity
(Sickle Cell Anemia)
– To observe the phenotype we need to first
express the genes carried on the DNA into
protein.
Central Dogma
DNA (deoxyribonucleic acid)
Transcription
RNA (ribonucleic acid)
Translation
Protein (amino acid)
Figure 17.2 Overview: the roles of transcription and translation in the flow of genetic
information
Central Dogma
DNA (deoxyribonucleic acid)
Transcription
RNA (ribonucleic acid)
Translation
Protein (amino acid)
2 things required for RNA synthesis
(transcription)
1. Single stranded DNA
2. Enzyme and nucleotides
• Use U instead of T
• ribose instead of deoxyribose
Enzyme is RNA polymerase
• Binds to specific DNA sequence called
promoter.
• Only transcribes DNA into RNA in one
direction on gene.
Figure 17.6 The stages of transcription
Figure 17.7 The initiation of transcription at a eukaryotic promoter
Transcription begins at
specific sites called
promoters.
RNA polymerase binds,
unwinds the DNA and
begins to synthesize
RNA.
Unlike DNA replication,
transcription only goes in
one direction.
Figure 17.8 RNA processing: addition of the 5´ cap and poly(A) tail.
In eukaryotes, RNA is processed before leaving the nucleus.
A cap is added to the 5’ end.
A poly(A) tail is added to the 3’ end.
Introns are removed.
Figure 17.9 RNA processing: RNA splicing
Exons are the portions of DNA that will encode protein.
Introns are spacer DNA that need to be removed before translation,
they do not encode the correct protein.
Removal of introns is called splicing
Movie 17-06
Given the b-hemoglobin gene sequence,
draw the mRNA that would be synthesized
during transcription.
Coding Strand
5’ CACCATGGTGCACCTGACTCCTGAGGAGAAG 3’
3’ GTGGTACCACGTGGACTGAGGACTCCTCTTC 5’
Non-Coding Strand
5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’
9. In transcription _____ is used
as a template to form ____.
a. DNA, RNA
b.
c.
d.
e.
DNA, protein
RNA, DNA
RNA, protein
Protein, RNA
10. Transcription uses which
enzyme?
a. DNA polymerase
b. RNA polymerase
c. Ribosome
d. Ligase
e. It doesn't need an enzyme
11. If you added an inhibitor of
transcription to a cell, the formation of
_____ would be blocked immediately.
a. RNA
b.DNA
c. protein
d.a and b
e. a and c
12. RNA polymerase binds to...
a. RNA
b.introns
c. exons
d.a promoter
e. a ribosome
13. In splicing _____ are removed
from RNA.
a. RNA
b.introns
c. exons
d.a promoter
e. a ribosome
14. The portion of a gene that
encodes protein is found on ____.
a. RNA
b.introns
c. exons
d.a promoter
e. a ribosome
Central Dogma
DNA (nucleic acid)
Transcription
RNA (nucleic acid)
Translation
Protein (amino acid)
Translation: RNA  protein
• Translate from nucleic acid language to
amino acid language.
• Uses an enzyme called a ribosome, made up
of ribosomal RNA (rRNA) and protein.
• Occurs in cytoplasm or on surface of
endoplasmic reticulum.
Figure 17.3 The triplet code
Messenger RNA
Figure 17.4 The dictionary of the genetic code
Each amino acid is encoded
by a three letter
combination of nucleotides
called codons.
Figure 17.4 The dictionary of the genetic code
Which protein would be
made with the following
mRNA?
AUG CCU AAU GAU UAA
Met Pro
Asn Asp Stop
Figure 17.2 Overview: the roles of transcription and translation in the flow of genetic
information
Figure 17.11 Translation: the basic concept
Translation occurs in the
ribosome.
A ribosome contains
ribosomal RNA (rRNA)
and protein.
By reading the order of
codons the ribosome
knows which amino
acids to insert into the
growing protein.
Transfer RNA (tRNA)
• The amino acids are transferred to the
ribosome by transfer RNA (tRNA)
• These tRNA molecules can bind to the
mRNA at one end and hold onto an amino
acid at the other end.
Figure 17.12 The structure of transfer RNA
Figure 17.14 The anatomy of a ribosome
3D Structure of a Ribosome
(spaghetti and meatballs)
http://www.bio.cmu.edu/Courses/BiochemMols/ribosome/70S.htm
http://www.umass.edu/molvis/pipe/ribosome/tour/index.htm
Figure 17.15 Initiation of translation
Translation begins at an
ATG codon.
ATG = Methionine
Figure 17.16 The elongation cycle of translation
Figure 17.17 Termination of translation
There are three stop codons that terminate translation.
TGA, TAA, TAG
Figure 17.18 Polyribosomes
Multiple ribosomes can translate a mRNA simultaneously
Figure 17.23 A summary of transcription and translation in a eukaryotic cell
Movie 17-10
15. In translation _____ is used
as a template to form ____.
a. DNA, RNA
b.DNA, protein
c. RNA, DNA
d.RNA, protein
e. Protein, RNA
16. Translation uses which enzyme?
a. DNA polymerase
b.RNA polymerase
c. Ribosome
d.Ligase
e. It doesn't need an enzyme
17. Which are found in a ribosome?
a. DNA
b.RNA
c. Protein
d.A and C
e. B and C
18. A codon contains how many bases?
a. 1
b.2
c. 3
d.4
e. 5
Molecular Biology
Understanding Genetic Diseases
(Chapter 17)
Protein Translation: Reading Frames
I O P T - Nucleotides
I I T I P T O P P O T P O P T O T P I T
I I T I P T O P P O T P O P T O T P I T
I I T I P T O P P O T P O P T O T P I T
I I T I P T O P P O T P O P T O T P I T
If we look at a section of DNA, we don’t
know which strand will be transcribed
into RNA.
5'
3'
ACATTTGCTTCTGACACAAC
tgtaaacgaagactgtgttg
3'
5'
?
5'
3’
ACAUUUGCUUCUGACACAAC
DNA
?
3'
5'
uguaaacgaagacuguguug
RNA
Each strand of RNA could be translated
in three different reading frames. Thus
there are 6 possible reading frames.
5'
3'
ACATTTGCTTCTGACACAAC
tgtaaacgaagactgtgttg
3'
5'
DNA
5'
3’
ACAUUUGCUUCUGACACAAC
3'
5'
uguaaacgaagacuguguug
1
2
3
4
5
6
ThrPheAlaSerAspThr
HisLeuLeuLeuThrGln
IleCysPheStpHisAsn
AsnAlaGluSerValVal
MetGlnLysGlnCysLeu
CysLysSerArgValCys
RNA
Protein
The first three reading frames are
on the upper strand of the RNA.
Each reading frame starts in one
base further than the one before it.
1
ACA UUU GCU UCU GAC ACA AC
Thr Phe Ala Ser Asp Thr
2
A CAU UUG CUU CUG ACA CAA C
His Leu Leu Leu Thr Gln
3
AC AUU UGC UUC UGA CAC AAC
Ile Cys Phe Stp His Asn
The other three reading frames are on the
lower strand of the DNA. Again, each reading
frame starts in one base further than the one
before it. The bases are always read from 5'
to 3', so the first codon in reading frame 4
would be read gtt.
4
ug uaa acg aag acu gug uug
Asn Ala Glu Ser Val Val
5
u gua aac gaa gac ugu guu g
Met Gln Lys Gln Cys Leu
6
ugu aaa cga aga cug ugu ug
Cys Lys Ser Arg Val Cys
Given the b-hemoglobin mRNA
sequence, translate it into amino acids.
5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’
N-Met-Val-HisLeu-Thr-ProGlu-Glu-Lys-C
Mutations
Substitution mutations:
Replace one base with another.
I I T I P T O P P O T P O P T O T P I T
I I T I P T O P P I T P O P T O T P I T
Mutations
Insertions: Gain of one or more bases.
Deletions: Loss of one or more bases.
Frame shifts: Addition or gain of bases can
lead to a shift in reading frame.
T I P T O P I P O T P O P T O T P I T
Insertion
T I P T O P P O T P O P T O T P I T
Deletion
T I P T O P O T P O P T O T P I T
Figure 17.22 Categories and consequences of point mutations
19. How many potential reading
frames are present on a double
stranded DNA?
a. 1
b.
c.
d.
e.
2
3
6
9
20. How many nucleotides in a
codon?
a. 1
b.
c.
d.
e.
2
3
6
9
21. Addition of a single nucleotide to
a DNA sequence can result in?
a. A substitution
b. A deletion
c. A translocation
d. Non-disjunction
e. A frameshift
Figure 17.4 The dictionary of the genetic code
We saw before that the RNA
AUG CCU AAU GAU UAA
was translated to the amino
acids.
Met Pro
Asn Gly Stop
What type of mutation would
be represented in this RNA?
AUG CCU AAC UGA UUA
Met Pro
Asn Stop
FRAMESHIFT
Figure 17.4 The dictionary of the genetic code
We saw before that the RNA
AUG CCU AAU GAU UAA
was translated to the amino
acids.
Met Pro
Asn Gly Stop
What type of mutation would
be represented in this RNA?
AUG CCU AAU CAU UAA
Met Pro
Asn His Stop
SUBSTITUTION
Translate the normal and sickle cell b-hemoglobin genes
Normal
5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’
5’ CACCAUGGUGCACCUGACUCCUGTGGAGAAG 3’
Sickle Cell
N-Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-C
N-Met-Val-His-Leu-Thr-Pro-Val-Glu-Lys-C
Figure 17.21 The molecular basis of sickle-cell disease
22. Sickle cell anemia is caused by?
a. A substitution
b. A deletion
c. A translocation
d. Non-disjunction
e. A frameshift
We have been working with a very short
segment of the b-hemoglobin gene.
How did researchers find the mutation in
DNA that causes Sickle Cell Anemia?
• Sequence the hemoglobin gene
• Translate the DNA into amino acids
• Compare normal and disease causing genes
Figure 20.x3 DNA sequencers
Hemoglobin sequences
>normal B-hemoglobin 626 base pairs
ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAGAAGTCTGC
GGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGG
ACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCA
AGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAGCTGCA
CTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGC
AAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACT
AAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGA
AGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGC
>sickle-cell B-hemoglobin 626 base pairs
ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGTGGAGAAGTCTGC
GGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGG
ACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCA
AGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAGCTGCA
CTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGC
AAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACT
AAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGA
AGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGC
Study by hand?
No Way, Use Computers - Bioinformatics
Biology Workbench
• Free bioinformatics program
• Anyone can generate an account
• http://workbench.sdsc.edu/
Hemoglobin DNA was 626 base
pairs long.
• How large a protein could this DNA
encode?
626/3 = 208 amino acids
Open Reading Frames (ORF)
Have a start codon AUG
and a stop codon UAA, UGA, UAG
Translation Programs
look for Open Reading Frames (ORF)
DNA sequence is entered, and the program translates it into
amino acids.
In this example Frame 3 was the longest ORF.
Why pick the longest ORF?
• By chance alone, how often would you
expect to find a stop codon?
3/64 about every 20 amino acids
• The longest ORF was 147 amino acids
Not likely to occur by chance
• We predicted the gene could encode a
protein of 208 amino acids. Why do we see
a difference?
Normal b-hemoglobin gene sequence
showing start and stop codons
ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGA
CACCATGGTGCACCTGACTCCTGAGGAGAAGTCTGCGGTTACTGCC
CTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGG
GCAGGCTGCTGGTGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTC
CTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAG
GTGAAGGCTCATGGCAAGAAAGTGCTCGGTGCCTTTAGTGATGGCC
TGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGA
GCTGCACTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTC
CTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGCAAAG
AATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGG
TGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGCTTTCTT
GCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACT
ACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTG
CCTAATAAAAAACATTTATTTTCATTGC
23. An open reading frame has
which of the following?
a. A start codon
b. A stop codon
c. An even number of bases
d. A and B
e. A and C
24. If the distance between the start
and stop codons are 300 nucleotides,
how many amino acids will be in a
protein?
a. 300
b.
c.
d.
e.
200
150
100
50
Translated Hemoglobin Sequences
Where is the mutation?
normal B-hemoglobin
MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFE
SFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFAT
LSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQK
VVAGVANALAHKYH
sickle-cell B-hemoglobin
MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFE
SFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFAT
LSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQK
VVAGVANALAHKYH
Study by Hand?
Alignment Program
Program aligns amino acid
sequences
sickle-cell
normal
MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
sickle-cell
normal
TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
sickle-cell
normal
PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
What does this mutation do to
hemoglobin?
Abnormal hemoglobin crystalizes when
oxygen is low, causing red blood cells to
become sickle shaped.
http://www.umass.edu/microbio/chi
me/hemoglob/2frmcont.htm
Normal and Sickle Cell
Hemoglobin
Heme
Valine
Mutation
Formation of
Hemoglobin
Crystals
Glutamic Acid is polar
Valine is non-polar
Sticks to other non-polar
region on hemoglobin
Sickle Cell Anemia
• Autosomal
Recessive.
1/10
1/10
1
2
• How do we identify
carriers?
3
• Analyze DNA to
determine genotype.
1/400
4
5
6
How is DNA Analyzed?
• Look at DNA sequences to determine mutation.
• Look for changes made by mutation in sequences
recognized by restriction enzymes.
• Purify DNA and analyze for presence or absence
of restriction site.
• Restriction Fragment Length Polymorhpism
(RFLP) Analysis
Align DNA sequences
What type of
mutation is
this?
Substitution
Restriction Enzymes
http://www.worthpublishers.com/lehninger3d/index.html
• Enzymes that recognize specific sequences
of nucleotides in DNA (words).
CAT vs. ACT
• Cut DNA at these sequences.
DdeI
Cuts at CTGAG
Won’t cut CTGTG
Which sequence will be cut by DdeI?
Cuts at CTGAG
Normal
Sickle
CCTGAGGAG
CUTS
CCTGTGGAG
DOESN’T CUT
How can we tell which sample was
cut with the restriction enzyme?
• Separate DNA fragments by size
• Gel Electrophoresis
• DNA is negatively charged
(page 374) Gel Electrophoresis of Macromolecules
(page 374) Gel Electrophoresis of Macromolecules (photo)
http://www.bio.umass.edu/biochem/mydna/modules/charge.html
Figure 20.7 Using restriction fragment patterns to distinguish DNA from different
alleles
25. The normal hemoglobin gene is cut
by DdeI and the sickle cell gene is not.
If we digest both DNAs with DdeI,
which will have the larger sized DNA
fragments?
a. Normal hemoglobin gene
b. Sickle cell hemoglobin gene
26. If we ran both digested DNA
samples on a gel, which would run
further?
a. Normal hemoglobin gene
b. Sickle cell hemoglobin gene
27. If a person is a carrier for sickle
cell anemia how many DNA fragments
would I see on the gel if DdeI only cut
once in the normal hemoglobin gene?
a. 0
b. 1
c. 2
d. 3
e. 4
SS
-
+
Ss
ss
Sickle Cell Anemia Test:
How can I detect carriers?
Sickle Cell Genetic Test
• DNA sequences for normal hemoglobin and
sickle cell hemoglobin are run through a
program that looks for DNA sequences
recognized by restriction enzymes.
• Use results to predict sizes of fragments
seen in SS, Ss and ss individuals.
Results
Normal Hemoglobin
DdeI 7 Fragments
37
50
68
84
89
139
159
Sickle Cell Hemoglobin
DdeI 6 Fragments
37
50
84
89
139
227
The 68 bp and 159 bp fragments have combined to form a
227 bp fragment
Restriction Fragment Length
Polymorhpism (RFLP) Analysis
Normal
Sickle
Cell
Sickle Cell Anemia RFLP
Determine the genotype of each family member.
1
2 3 4 5 6
227
159
1
2
139
89/84
68
3
50
4
5
37
Ss Ss ss Ss Ss SS
Which child would develop Sickle Cell Anemia?
6
Practice Questions:
Cystic Fibrosis
One of the most common autosomal recessive
disorders in Caucasians, with 1 in 25 being
carriers. In cystic fibrosis, 70% of all mutations
associated with the disease result in the loss of
three nucleotides TTC from the CFTR gene.
(TTC would encode Phenylalanine)
28. What impact will this mutation
have on transcription?
a. None, mRNA will still be formed
b. None, protein will still be formed
c. mRNA will not be formed
d. protein will not be formed
e. a premature stop codon might stop
transcription early
29. How will the CFTR protein
produced in most CF patients differ
from that in non-carriers?
a. It will not be made
b.It will have no phenylalanine
c. It will have one less phenylalanine
d.It will have extra phenylalanines
e. It will be the same
C is the normal CFTR gene and
c is the mutated CFTR gene.
30. An individual with cystic fibrosis would
have which genotype?
a.CC
b.Cc
c. cc
d.C
e. c
31. If two parents had the genotype Cc,
what percent of their children would
develop cystic fibrosis?
a. 0%
b.25%
c. 50%
d.75%
e. 100%