Molecular Biology Transcription and Translation Chapter 17 Genotype

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 m. RNA 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. 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 (r. RNA) 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 m. RNA? 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 (r. RNA) and protein. By reading the order of codons the ribosome knows which amino acids to insert into the growing protein.

Transfer RNA (t. RNA) • The amino acids are transferred to the ribosome by transfer RNA (t. RNA) • These t. RNA molecules can bind to the m. RNA 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

3 D Structure of a Ribosome (spaghetti and meatballs) http: //www. bio. cmu. edu/Courses/Biochem. Mols/ribosome/70 S. 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, TAG

Figure 17. 18 Polyribosomes Multiple ribosomes can translate a m. RNA 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

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 Thr. Phe. Ala. Ser. Asp. Thr His. Leu. Thr. Gln Ile. Cys. Phe. Stp. His. Asn. Ala. Glu. Ser. Val Met. Gln. Lys. Gln. Cys. Leu Cys. Lys. Ser. Arg. Val. Cys 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 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 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 m. RNA sequence, translate it into amino acids. 5’ CACCAUGGUGCACCUGACUCCUGAGGAGAAG 3’ N-Met-Val-His. Leu-Thr-Pro-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. 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 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-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. x 3 DNA sequencers

Hemoglobin sequences >normal B-hemoglobin 626 base pairs ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAGAAGTCTGC GGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGG ACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCA AGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGCTGCA CTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGC AAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACT AAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGA AGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTTCATTGC >sickle-cell B-hemoglobin 626 base pairs ACATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGTGGAGAAGTCTGC GGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCTTGG ACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCAGTTATGGGCAACCCTAAGGTGAAGGCTCATGGCA AGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGCTGCA CTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCACTTTGGC AAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAGTATCACT AAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGATATTATGA AGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTTCATTGC 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 CCTAATAAAAAACATTTTCATTGC

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 Sequences Where is the mutation? normal B-hemoglobin MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFF ESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFA TLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQ KVVAGVANALAHKYH sickle-cell B-hemoglobin MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFF ESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFA TLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQ KVVAGVANALAHKYH Study by Hand? Alignment Program

Program aligns amino acid sequences sickle-cell normal MVHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLS sickle-cell normal TPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVD sickle-cell 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/chi me/hemoglob/2 frmcont. 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. • How do we identify carriers? • Analyze DNA to determine genotype. 3 1/400 1/10 1 2 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/lehninger 3 d/index. html • Enzymes that recognize specific sequences of nucleotides in DNA (words). CAT vs. ACT • Cut DNA at these sequences. Dde. I Cuts at CTGAG Won’t cut CTGTG

Which sequence will be cut by Dde. I? Cuts at CTGAG Normal CCTGAGGAG CUTS Sickle 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 Dde. I and the sickle cell gene is not. If we digest both DNAs with Dde. I, 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 Dde. I 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 Dde. I 7 Fragments 37 50 68 84 89 139 159 Sickle Cell Hemoglobin Dde. I 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 1 159 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, m. RNA will still be formed b. None, protein will still be formed c. m. RNA 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%