Molecular Biology

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Molecular Biology

• Transcription and Translation
• (Chapter 17)

2
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.

3
Central Dogma

• DNA (deoxyribonucleic acid)
• Transcription
• RNA (ribonucleic acid)
• Translation
• Protein (amino acid)

4
Figure 17.2  Overview the roles of transcription
and translation in the flow of genetic information
5
Central Dogma

• DNA (deoxyribonucleic acid)
• Transcription
• RNA (ribonucleic acid)
• Translation
• Protein (amino acid)

6
2 things required for RNA synthesis
(transcription)

• 1. Single stranded DNA
• 2. Enzyme and nucleotides
• Use U instead of T
• ribose instead of deoxyribose

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Enzyme is RNA polymerase

• Binds to specific DNA sequence called promoter.
• Only transcribes DNA into RNA in one direction
  on gene.

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Figure 17.6  The stages of transcription
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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.
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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.
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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
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Movie 17-06
13
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
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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

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10. Transcription uses which enzyme?

• a. DNA polymerase
• b. RNA polymerase
• c. Ribosome
• d. Ligase
• e. It doesn't need an enzyme

16
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

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12. RNA polymerase binds to...

• a. RNA
• b. introns
• c. exons
• d. a promoter
• e. a ribosome

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13. In splicing _____ are removed from RNA.

• a. RNA
• b. introns
• c. exons
• d. a promoter
• e. a ribosome

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14. The portion of a gene that encodes protein is
found on ____.

• a. RNA
• b. introns
• c. exons
• d. a promoter
• e. a ribosome

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Central Dogma

• DNA (nucleic acid)
• Transcription
• RNA (nucleic acid)
• Translation
• Protein (amino acid)

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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.

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Figure 17.3  The triplet code
Messenger RNA
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Figure 17.4  The dictionary of the genetic code
Each amino acid is encoded by a three letter
combination of nucleotides called codons.
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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
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Figure 17.2  Overview the roles of transcription
and translation in the flow of genetic information
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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.
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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.

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Figure 17.12  The structure of transfer RNA
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Figure 17.14  The anatomy of a ribosome
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3D Structure of a Ribosome(spaghetti and
meatballs)

• http//www.bio.cmu.edu/Courses/BiochemMols/ribosom
  e/70S.htm
• http//www.umass.edu/molvis/pipe/ribosome/tour/ind
  ex.htm

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Figure 17.15  Initiation of translation
Translation begins at an ATG codon. ATG
Methionine
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Figure 17.16  The elongation cycle of translation
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Figure 17.17  Termination of translation
There are three stop codons that terminate
translation. TGA, TAA, TAG
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Figure 17.18  Polyribosomes
Multiple ribosomes can translate a mRNA
simultaneously
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Figure 17.23  A summary of transcription and
translation in a eukaryotic cell
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Movie 17-10
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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

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16. Translation uses which enzyme?

• a. DNA polymerase
• b. RNA polymerase
• c. Ribosome
• d. Ligase
• e. It doesn't need an enzyme

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17. Which are found in a ribosome?

• a. DNA
• b. RNA
• c. Protein
• d. A and C
• e. B and C

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18. A codon contains how many bases?

• a. 1
• b. 2
• c. 3
• d. 4
• e. 5

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Molecular Biology

• Understanding Genetic Diseases
• (Chapter 17)

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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

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If we look at a section of DNA, we dont know
which strand will be transcribed into RNA.
5' 3'
ACATTTGCTTCTGACACAAC
tgtaaacgaagactgtgttg 3'
5' 5' 3 3'
5' ACAUUUGCUUCUGACACAAC uguaaacgaagacuguguug

?
?
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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' 5' 3 3'
5' ACAUUUGCUUCUGACACAAC uguaaacgaagacuguguug
1 ThrPheAlaSerAspThr 4
AsnAlaGluSerValVal 2 HisLeuLeuLeuThrGln 5
MetGlnLysGlnCysLeu 3 IleCysPheStpHisAsn 6
CysLysSerArgValCys
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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 AC
• 1 Thr Phe Ala Ser Asp Thr
• A CAU UUG CUU CUG ACA CAA C
• 2 His Leu Leu Leu Thr Gln
• AC AUU UGC UUC UGA CAC AAC
• 3 Ile Cys Phe Stp His Asn

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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

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Given the b-hemoglobin mRNA sequence, translate
it into amino acids. 5 CACCAUGGUGCACCUGACUCCUGAG
GAGAAG 3
N-Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-C
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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

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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
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Figure 17.22  Categories and consequences of
point mutations
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19. How many potential reading frames are present
on a double stranded DNA?

• a. 1
• b. 2
• c. 3
• d. 6
• e. 9

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20. How many nucleotides in a codon?

• a. 1
• b. 2
• c. 3
• d. 6
• e. 9

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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

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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. What
type of mutation would be represented in this
RNA? AUG CCU AAC UGA UUA
Met
Pro
Asn
Stop
FRAMESHIFT
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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. What
type of mutation would be represented in this
RNA? AUG CCU AAU CAU UAA
Met
Pro
Asn
Stop
His
SUBSTITUTION
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Translate the normal and sickle cell b-hemoglobin
genesNormal5 CACCAUGGUGCACCUGACUCCUGAGGAGAAG
35 CACCAUGGUGCACCUGACUCCUGTGGAGAAG 3Sickle
CellN-Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-CN-Me
t-Val-His-Leu-Thr-Pro-Val-Glu-Lys-C
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Figure 17.21  The molecular basis of sickle-cell
disease
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22. Sickle cell anemia is caused by?

• a. A substitution
• b. A deletion
• c. A translocation
• d. Non-disjunction
• e. A frameshift

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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

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Figure 20.x3 DNA sequencers
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Hemoglobin sequences

• gtnormal B-hemoglobin 626 base pairs
• ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACC
  ATGGTGCACCTGACTCCTGAGGAGAAGTCTGCGGTTACTGCCCTGTGGGG
  CAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGG
  TGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCC
  ACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAA
  AGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGG
  GCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGAT
  CCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCA
  TCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAG
  TGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGC
  TTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAA
  CTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCC
  TAATAAAAAACATTTATTTTCATTGC
• gtsickle-cell B-hemoglobin 626 base pairs
• ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACC
  ATGGTGCACCTGACTCCTGTGGAGAAGTCTGCGGTTACTGCCCTGTGGGG
  CAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGG
  TGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCC
  ACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAA
  AGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGG
  GCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGAT
  CCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCA
  TCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAG
  TGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGC
  TTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAA
  CTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCC
  TAATAAAAAACATTTATTTTCATTGC
• Study by hand?

No Way, Use Computers - Bioinformatics
62
Biology Workbench

• Free bioinformatics program
• Anyone can generate an account
• http//workbench.sdsc.edu/

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Hemoglobin DNA was 626 base pairs long.

• How large a protein could this DNA encode?

626/3 208 amino acids
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Open Reading Frames (ORF)Have a start codon
AUGand a stop codon UAA, UGA, UAG
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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.
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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
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Normal b-hemoglobin gene sequence showing start
and stop codons

• ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACC
  ATGGTGCACCTGACTCCTGAGGAGAAGTCTGCGGTTACTGCCCTGTGGGG
  CAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGG
  TGGTCTACCCTTGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCC
  ACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCAAGAA
  AGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGG
  GCACCTTTGCCACACTGAGTGAGCTGCACTGTGACAAGCTGCACGTGGAT
  CCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCA
  TCACTTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAG
  TGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACTAAGCTCGC
  TTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAA
  CTACTAAACTGGGGGATATTATGAAGGGCCTTGAGCATCTGGATTCTGCC
  TAATAAAAAACATTTATTTTCATTGC

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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

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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

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Translated Hemoglobin SequencesWhere is the
mutation?

• normal B-hemoglobin
• MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
  TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
  PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
• sickle-cell B-hemoglobin
• MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS
  TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD
  PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
• Study by Hand?

Alignment Program
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Program aligns amino acid sequences

• sickle-cell MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYP
  WTQRFFESFGDLS
• normal MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYP
  WTQRFFESFGDLS
• sickle-cell TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFA
  TLSELHCDKLHVD
• normal TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFA
  TLSELHCDKLHVD
• sickle-cell PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAG
  VANALAHKYH
• normal PENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAG
  VANALAHKYH

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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/2frmc
  ont.htm

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Normal and Sickle Cell Hemoglobin
Valine Mutation
Heme
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Formation of Hemoglobin CrystalsGlutamic Acid
is polarValine is non-polarSticks to other
non-polar region on hemoglobin
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Sickle Cell Anemia

• Autosomal Recessive.
• How do we identify carriers?
• Analyze DNA to determine genotype.

1/10
1/10
1/400
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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

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Align DNA sequences
What type of mutation is this?
Substitution
78
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
• Wont cut CTGTG

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Which sequence will be cut by DdeI?

• Cuts at CTGAG
• Normal CCTGAGGAG
• Sickle CCTGTGGAG

CUTS
DOESNT CUT
80
How can we tell which sample was cut with the
restriction enzyme?

• Separate DNA fragments by size
• Gel Electrophoresis
• DNA is negatively charged

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(page 374) Gel Electrophoresis of Macromolecules
82
(page 374) Gel Electrophoresis of Macromolecules
(photo)
http//www.bio.umass.edu/biochem/mydna/modules/cha
rge.html
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Figure 20.7  Using restriction fragment patterns
to distinguish DNA from different alleles
84
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

85
26. If we ran both digested DNA samples on a
gel, which would run further?

• a. Normal hemoglobin gene
• b. Sickle cell hemoglobin gene

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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

87
Sickle Cell Anemia TestHow can I detect
carriers?
88
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.

89
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

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Restriction Fragment Length Polymorhpism (RFLP)
Analysis
Normal Sickle Cell
91
Sickle Cell Anemia RFLPDetermine the genotype of
each family member.
227 159 139 89/84 68 50 37
SS
Ss
Ss
ss
Ss
Ss
Which child would develop Sickle Cell Anemia?
92
Practice QuestionsCystic 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)

93
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

94
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

95
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

96
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