Coding DNA 5’- GTG ACT CGT TGT GCC ATT GCA GCT AAA CAC TTC GAG CCC TGT- 3’ mRNA 5’- GUG ACU CGU UGU GCC AUU GCA GCU AAA CAC UUC GAG CCC UGU- 3’ What is the polypeptide sequence?
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Coding DNA 5’- GTG ACT CGT TGT GCC ATT GCA GCT AAA CAC TTC GAG CCC TGT- 3’
mRNA 5’- GUG ACU CGU UGU GCC AUU GCA GCU AAA CAC UUC GAG CCC UGU- 3’
What is the polypeptide sequence?
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- CTA GCC CTC CGT TAC TAG TTA CCT ACT TAT TCA ATT TTG TAA ACG CTC ATC CGA ACC CGC TTT TAA TTG CCC ACT TAG TCG ATT ACC CGT TTA TGT TAA TTA CCT ATC 1. Build the mRNA molecule matching the RNA nucleotides to the DNA nucleotides properly letter by letter. (assume that the mRNA is bacterial there are not intros to cut out) 2. Figure out the tRNA triplet (codons) that would fit the mRNA triplets. (letter by letter) 3. Look up for each tRNA codon and find the corresponding symbol and amino acid abbreviations. The symbols should spell out a meaningful English message.CTA GCC CTC CGT TAC TAG TTA CCT ACT TAT TCA ATT TTG TAA ACG CTC ATC CGA ACC CGC TTT TAA TTG CCC ACT TAG TCG ATT ACC CGT TTA TGT TAA TTA CCT ATC 1. Build the mRNA molecule matching the RNA nucleotides to the DNA nucleotides properly letter by letter. (assume that the mRNA is bacterial there are not intros to cut out)The following sequence represents the dsDNA code for a short peptide 5' -CTT TCC CAT CAC CGC ATG CAT CCT CCC TCC TTT CTT TAA TAT TGG-3' 3'-GAA AGG GTA GTG GCG TAC GTA GGA GGG ACC AAA GAA ATT ATA ACC-5' Transcribe the DNA strand given above to write the sequence of the mRNA strand in the 5’ to 3’ direction. (1) Use the table and write the sequence of the resulting peptide. (1) Is it possible for a codon to code for another amino acid? (1) What will be the effect if a mutation changes the codon UAU to UAA? (1) What is a reading frame? (1) If you are given a nucleotide sequence, how would you find Open Reading Frames? (1) DISCUSS the reason why there are leading and lagging strands in replication?
- AGUCAGUCAG The codon chart is shown below, what amino acid sequence does the MRNA sequence: 3' UUUCCUCAA 5' code for? RNA Codon Chart 31 Alanine G U A Tyrosine Stop Valine A G Cysteine U 3' U G Stop G Tryptophan 3' G 5' Arginine G U A Leucine Serine A Lysine A U Proline G Asparagine leucine-threonine-glutamic acid asparagine-serine-phenylalanine serine-histidine-glutamine phenylalanine-proline-histidine phenylalanine-proline-glutamine Serine Glutamic acid Aspartic acid Histidine Glutamine Arginine Isoleucine 2 Threonine Methionine QGiven Sequence: 3’-TACGGTCCGGATTCGGTAGCTAGCATC-5’ Provide: Complementary Strand: Transcript Amino Acid Sequence 2.Given Sequence: 5’-GGGCATATGCCGTTTACCGGTTTGACTAAATAACCA-3’ Provide: Complementary Strand: Transcript Amino Acid Sequence 3.Given Sequence: 3’-AAC CAA TAC GTG AGG ATA CCA AGT AAC ACT CCC-5’ Provide: Complementary Strand: Direct Transcript: Transcript for Translation: Amino Acid Sequence:3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ Make vertical lines between codons to make this assignment easier to do. Which strand is the template strand? The first strand is the template strand as it going 3-5 Copy the template strand in mRNA. Label the 5’ and 3’ ends. 5' AUG UUA CCC GCU GCG CGA AGC AAA GUC UAA 3” Write out the amino acid (you can use the short form) that this protein would be made of (keep in mind proteins are usually a minimum of 50 proteins. Met-Leu-Pro-Ala-Ala-Arg-Ser-Lys-Val What do you notice about the mRNA strand compared to the non template strand of DNA? 2. What amino acid does the second codon code for on the DNA template strand? ______Assume the 6th amino acid in the strand is changed from a T to a C. What amino acid does it code for now? _______ What type of mutation is this? 3. Assume the 6th amino acid is changed from T to G on the DNA template strand. What type of mutation is this? What effect would…
- Consider the following mature mRNA from a human cell: 5' UAAUGUCGCAAUAACC 3¹ What is the sequence of amino acids in the translated protein? Second letter A First letter С U d A G บบบ UUC P U UUA UUG Leu CUU CUC CUA CUG GUU GUC GUA GUG -Phe Met-Ser-Gln Met-Arg-Lys-Ser Leu Stop-Cys-Arg-Asn-Asn C AUU AUC lle AUA ACA AUG Met ACG Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC GCU GCC GCA GCGJ Ser Pro Thr Ala UAU UAC Tyr UGU UGC. UAA Stop UGA Stop UAG Stop UGG Trp CAC CAGGin CAU His Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. AAU Asn AAA Lys AAG. There is no start codon, resulting in no translated protein. GAU ASP GAC Glu GAA GAG G CGU CGC CGA CGG Cys AGU AGC AGA AGG Arg GGU GGC GGA GGGJ Arg ser Gly MCAG U UCAG с А SCAG U SCAG U Third letterWhat are the correct codons of the MRNA from the given DNA strand that needs to be transcribed? * Sense strand 5' AAT GCC AGT GGT3' Antisense strand MRNA 000 000 TRNA (anticodon) Figure 1. Amino acid (peptide) 3' TTA CGG TCA CCA 5' 5' TTA CGG TCA CCA 3' 3' UUA CGG UCA CCA 5' 3' AAU GCC AGU GGU 5'AKS 5c1: Using codon wheel below, which of the models correctly represents the usage of the base pairing rule, the correct sequence of events, and creation of proteins at the ribosomes? * UGPO Alanine GU Tyrosíne Stop GU AC Valine Cysteine C GA Stop Tryptophan G A Arginine Leucine Serine Lysine UG Proline Asparagine ACU GAC U Glycine o Glutamic Phenyl- Leucine Serine acid Aspartic acid alanine Histidine Glutamine Arginine a aujonajo Threonine ethionine
- Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5' TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5' 46 5 AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5 77 90 110 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3' ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. Illustrate how termination of transcription occurs in the gene above. (Hint: position from 156 to 180)What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'thank you