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%