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- 2. Shown below is the DNA sequence of a eukaryotic gene that encodes a short peptide. The sequence of the final processed mRNA synthesized from this gene is given below. Genomic DNA sequence: 5'-AGCTCATGTGCGAGTCCTGACGCTGACTAGG-3' 3'-TCGAGTACACGCTCAGGACTGCGACTGATCC-5' Processed mRNA sequence: 5'-G*UCAUGUGCGAACGCUGACUAGGAAAAAAAA....-3' In the genomic DNA sequence shown above, draw a box around each of the two exons in the gene or write the two exon sequences. а. b. In the processed mRNA above, some nucleotides are present that are not coded for in the genomic DNA sequence. Name the two processes that have occurred to add these nucleotides to the mRNA.5. A DNA sequence of "ACG" will code for the amino acid - (LS1- 1)* Second MRNA base G UUU Phe UUC UCU UAU UGU . Tyr Cys UAC Ser UAA Stop UGA Stop A UGC UUA UCA Leu UUG A E UCG UAG Stop UGG Trp G CGU 7 His CAC CUU CCU CAU AGU CUC Leu CUA CCC Pro CCA CGC Cye (C) Ang CAA CGA Gin CUG CCG CAG CGG G. A C La AUU AAU Ash AAC The AAA AGU Ser AGC ACU AUC Te ACC UG ALIA ACA AGA Arg AG AUG M ACG AAG GCU GC Val GCA GAU Asp GAC Ala GAA Slu GAG GUU GGJ GUC Gly GUA GUG GCG GGG Cys Thr Tyr Pro OIt will not code for an amino acid First mRNA base (5' end of codon) Third MRNA base (3' end of codon)1. What would be the amino acid sequence encoded by the mRNA5' C C A U G A C G U C G G A U C A A U G A G C 3' 2. If the nucleotide bolded and underlined in red in part 1 changes from C to a G, what type of mutation would that be? 3. What would happen to the amino acid sequence if the C bolded in red in part 1 is changed to a G? 4. What would happen to the amino acid sequence if the C bolded in red is deleted?
- 3. DNA: TACGGGCCTATACGCTACTACT CA TG GATCGG MRNA: UC Codon: Anitcodon: Amino Acids: 4. DNA: G T ACGCGTATACCGACATTC MRNA: Codon: Anitcodon: Amino Acids: A8m Transcribe the following DNA strand into mRNA and translate that strand into a polypeptide chain, identifying the codons, anticodons, and amino acid sequence. DNA: CGAT ACAATGGACCCGGTATGCGATATCC DATAA6. A portion of a gene is shown below. 5'-ATGATTCGCCTCGGGGCTCCCCAGTCGCTGGTGCTGCTGACGCTGCTCGTCG-3' 3'-TACTAAGCGGAGCCCCGAGGGGTCAGCGACCACGACGACTGCGACGAGCAGC-5' The sequence of the mRNA transcribed from this gene has the following sequence: 5'-AUGAUUCGCCUCGGGGCUCCCCAGUCGCUGGUGCUGCUGACGCUGCUCGUCG-3′ a. Identify the coding and noncoding strands of the DNA. b. Explain why only the coding strands of DNA are commonly published in databanks.DNA Sequence: TAC TCC GGC TCT CCC AGT TGA ACT Mutated Sequence: TAC TCG GCT CTC CCA GTT GAA CT Original Amino acid: Mutated Amino Acid: What mutation have occurred in the sequence? How does it affect the expression of amino acids? 2. DNA Sequence: TAC TCC GGC TCT CCC AGT TGA ACT Mutated Sequence: TAC TCT GGC TCT CCA AGT TGA ACT Original Amino acid: Mutated Amino Acid: What mutation has occurred in the sequence? How does it affect the expression of amino acids? 3. DNA Sequence: TAC TCC GGC TCT CCC AGT TGA ACT Mutated Sequence: TAC TCC GGC TCT CCC ACT TGA ACT Original Amino acid: Mutated Amino Acid: What type of mutation has occurred in the sequence? How does it affect the expression of amino acids? 4. DNA Sequence: TAC TCC GGC TCT CCC AGT TGA ACT Mutated Sequence: TAC TCC GGC TCG CCC ACT TGA ACT Original Amino acid: Mutated Amino Acid: What of mutation mutation has occurred in the sequence? How…
- If the first G changes to A what kind of mutation will happen? Show the change in amino acid sequence. This is base substitutions involve the replacement of one nucleotide with another. And it changes one amino acid coding, producing a missense mutation TAC CTA GCA CAC ATGTAGGTGGGCAAAGTT TAC CTA ACACACATGTAGGTGGGCAAAGTT6. Similar to the class notes (Intro to Genetics), a segment of DNA (shown below) contains a promoter segment (the first 9 base pairs), a ribosome binding segment (the next 6 base pairs), and a segment that codes for protein synthesis which is started by the rest of the base pairs. ACTCCATTGAACCATTTCTATGATCCGCTAACG-... TGAGGTAACTTGGTAAAGATACTAGGCGATTGC-... A. When the DNA is induced to be copied to mRNA, the top strand is coding, meaning that the mRNA makes an identical copy of the lower strand (replacing T with U) The mRNA copy starts with the ribosome binding sequence. What is the sequence of the mRNA that will go to the ribosomes? B. What are the first 6 amino acids of the protein that are coded for by the mRNA? C. What would the amino acid sequence be if... i. a transition mutation occurred on the final G in the mRNA? ii. all of the G & C bases in the protein synthesis portion had transition mutations? iii. a point deletion mutation occurred in the ATA sequence (in the lower strand…Sickle cell hemoglobin DNA CA CG TAGACTGAGGACA C Sickle cell hemoglobin MRNA Sickle cell hemoglobin AA sequence ValoHis.lku thro proo Gily 4. What type of mutation is this? Please explain why.
- 6. How many amino acids will the mRNA sequence "AUG GAC CUG UCG UGA" produce? (LS1-1) * Second MRNA base C. UUU Phe UUC UCU UAU UGU Cps UGC U Tyr UAC UCC Ser UCA UUA Leu UUG UAA Stop UGA Stop A Gu (G) 19) UCG UAG Stop UGG Trp G. Tye AGU A CUU CCU CAU His CAC CGU CUC Leu CUA CC CGC Cys (C) Pro CAA Gin CAG Arg CGA CCA CUG. CCG CGG Arg R AU ACU A C AAU Asn AAC Thr Le AGU Ser AGC AUC e ACC C AUA ACA Lys IK AGA Lys AGG AAA AUG M ACG AAG The GUU GCU GAU GGU Asa GAC GCC Val GCA GGC Ala Gly GUA GAA GGA Glu GSG GUG GCG GAG First mRNA base (5' end of codon) Third mRNA base (3' end of codon)If we have the following mutations, find the type of the mutation (silent or missense or nonsense?) 17C=U 36G=A 49G=U 115A=C 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’V S X It F3 3 9999999 E D G Which letter indicates the amino end of the growing polypeptide? с F4 $ 4 A Q Search R F mRNA F5 er ge % UAC 5 V T F6 GGG AUGCCCACG G 6 B Y De F7 H ►/11 v lo & 7 7 F8 U 4 N UAG * 8 865 F9 8 1 5 G Мо Alt F10 ( 9 9 K 2 O < " ) 6 0 F11 1 P* L 3 • V A 30 F12 ; { I Ctrl = 332 PM 4/7/2023 ? + 1 } 1 11 A 1 G Backspace Home Delete Enter Shift PgUp PgDn