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 -hemoglobin gene sequence, draw the mRNA that would be synthesized
during transcription.
5’ CACCATGGTGCACCTGACTCCTGAGGAGAAG 3’3’ GTGGTACCACGTGGACTGAGGACTCCTCTTC 5’
5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’
Coding Strand
Non-Coding Strand
9. In transcription _____ is used as a template to form ____.
a. DNA, RNA
b. DNA, protein
c. RNA, DNA
d. RNA, protein
e. 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
5' 3' ACATTTGCTTCTGACACAAC tgtaaacgaagactgtgttg 3' 5'
5' 3’ 3' 5'ACAUUUGCUUCUGACACAAC uguaaacgaagacuguguug
If we look at a section of DNA, we don’t know which strand will be transcribed into RNA.
DNA
RNA
? ?
5' 3' ACATTTGCTTCTGACACAAC tgtaaacgaagactgtgttg 3' 5'
5' 3’ 3' 5'ACAUUUGCUUCUGACACAAC uguaaacgaagacuguguug
1 ThrPheAlaSerAspThr 4 AsnAlaGluSerValVal 2 HisLeuLeuLeuThrGln 5 MetGlnLysGlnCysLeu 3 IleCysPheStpHisAsn 6 CysLysSerArgValCys
Each strand of RNA could be translated in three different reading frames. Thus there are 6 possible reading frames.
DNA
Protein
RNA
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.
ACA UUU GCU UCU GAC ACA AC1 Thr Phe Ala Ser Asp Thr A CAU UUG CUU CUG ACA CAA C2 His Leu Leu Leu Thr Gln
AC AUU UGC UUC UGA CAC AAC3 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.
ug uaa acg aag acu gug uug
4 Asn Ala Glu Ser Val Val
u gua aac gaa gac ugu guu g
5 Met Gln Lys Gln Cys Leu
ugu aaa cga aga cug ugu ug
6 Cys Lys Ser Arg Val Cys
Given the -hemoglobin mRNA sequence, translate it into amino acids. 5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’
N-Met-Val-His-Leu-Thr-Pro-Glu-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
DeletionT 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. 2
c. 3
d. 6
e. 9
20. How many nucleotides in a codon?
a. 1
b. 2
c. 3
d. 6
e. 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
Met Pro Asn Stop
We saw before that the RNA
AUG CCU AAU GAU UAA
was translated to the amino acids.
What type of mutation would be represented in this RNA?
AUG CCU AAC UGA UUA
Met Pro Asn Gly Stop
FRAMESHIFT
Figure 17.4 The dictionary of the genetic code
Met Pro Asn Stop
We saw before that the RNA
AUG CCU AAU GAU UAA
was translated to the amino acids.
What type of mutation would be represented in this RNA?
AUG CCU AAU CAU UAA
Met Pro Asn Gly Stop
SUBSTITUTION
His
Translate the normal and sickle cell -hemoglobin genes
Normal5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’5’ CACCAUGGUGCACCUGACUCCUGTGGAGAAG 3’Sickle Cell
N-Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-CN-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 -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 pairsACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAGAAGTCTGCGGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGC
>sickle-cell B-hemoglobin 626 base pairsACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGTGGAGAAGTCTGCGGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGC
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 AUGand 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?
• The longest ORF was 147 amino acids
• We predicted the gene could encode a protein of 208 amino acids. Why do we see a difference?
3/64 about every 20 amino acids
Not likely to occur by chance
Normal -hemoglobin gene sequence showing start and stop codons
ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAGAAGTCTGCGGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGC
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. 200
c. 150
d. 100
e. 50
Translated Hemoglobin SequencesWhere is the mutation?
normal B-hemoglobinMVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
sickle-cell B-hemoglobinMVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
Study by Hand?Alignment Program
Program aligns amino acid sequences
sickle-cell MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
normal MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
sickle-cell TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
normal TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
sickle-cell PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
normal 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/chime/hemoglob/2frmcont.htm
Normal and Sickle Cell Hemoglobin
Heme
ValineMutation
Formation of Hemoglobin
Crystals
Glutamic Acid is polar
Valine is non-polar
Sticks to other non-polar region on hemoglobin
Sickle Cell Anemia
1 2
3 4 5 6
1/10 1/10
1/400
• Autosomal Recessive.
• How do we identify carriers?
• Analyze DNA to determine genotype.
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 Enzymeshttp://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
NormalCCTGAGGAG
Sickle CCTGTGGAG
CUTS
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.
ResultsNormal Hemoglobin
DdeI 7 Fragments
37 50 68 84 89 139 159
Sickle Cell HemoglobinDdeI 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
1 2 3 4 5 6
1 2
3 4 5 6
Sickle Cell Anemia RFLPDetermine the genotype of each family member.
Ss Ss ss Ss Ss SS
Which child would develop Sickle Cell Anemia?
227
159
139
89/84
68
50
37
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 formede. 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%
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