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- Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?Given the following stretch of mRNA, what would be the sequence of the corresponding non-template DNA? 5' - UUG-CAA-UCG-CAG-UGC-CGC-AUA-GAU - 3' Group of answer choices 3' - AAC-GTT-AGC-GTC-ACG-GCG-TAT-CTA - 5' 5' - TTG-CAA-TCG-CAG-TGC-CGC-ATA-GAT - 3' 5' - AAC-GTT-AGC-GTC-ACG-GCG-TAT-CTA - 3' 3' - AAC-GUU-AGC-GUC-ACG-GCG-UAU-CUA - 5' 3' - TTG-CAA-TCG-CAG-TGC-CGC-ATA-GAT - 5'The sequence of part of an mRNA transcript is What is the sequence of the DNA coding strand? 5'- 5' - AUGGGGAACAGCAAGAGUGGGGCCCUGUCCAAGGAG - 3' 5'- ATGGGGAACAGCAAGAGTGGGGCCCTGTCCAAGGAG What is the sequence of the DNA template strand? TACCCCTTGTCGTTCTCACCCCGGGACAGGTTCCTC Incorrect -3' -3'
- Below is a double stranded DNA that contains an ORF. Identify the ORF and in the spaces below, first provide the mRNA sequence that would result (in the 5' to 3' direction). Then give the protein sequence this mRNA would encode (in the N to C direction using the 3-letter abreviations and a single space in between). 5'-CTGCGC TAGGCGCAGTAGGGCATCCCACGACA-3' 3'-GACGCGATCCGCGTCATCCCG TAGGG TGC TGT-5' Codon Table: บบบ UUC JUA UUG CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val mRNA sequence: UCU UCC UCA UCG protein sequence: CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala UAU UAC UAA UAG CAU CAC CAA CAG Tyr Stop GAU GAC His Gin AAU Asn AAC AAAG} LUS Asp GAA GAG} Glu UGU Cys UGC UGA Stop UGG Trp CGU CGC CGA CGG AGU AGC Arg Ser AGG} Arg GGU GGC GGA GGG (DO NOT include the 5' and 3' numbers or the N and C. Write only the nucleotides and the abreviations for the amino acids. Include no spaces in the mRNA and a single space in between the amino acid…Here is the nucleotide sequence of an imaginary strand of DNA: 5’ AATTGGCCATGC 3’. If this strand of DNA was transcribed, the resulting messenger mRNA molecule would be: Met-Val-Tyr-Lys 3’ TACGTACG 5’ 3’ TTAACCGGTACG 5’ 3’ UUAACCGGUACG 5’Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'
- Here is the sequence of a portion of a bacterial gene. The template strand is on the bottom: 5’-ATGCTGCGTGCATGGGATATAGGTAGCACACGTCC-3’ 3’-TACGACGCACGTACCC TATATCC ATCGTGTGCAGG-5’ Assuming that transcription starts with the first C in the template strand, and continues to the end, what would be the sequence of the mRNA derived from this fragment? Q) Would there be an effect on translation of changing the fourthT in the template strand to a C? If so, what effect?Give only typing answer with explanation and conclusion which of these choices represents one possible corresponding mRNA sequence that can be transcribed by RNA polymerase, and later translated by ribosomes from the following DNA template? 5'- CTGTATCCTAGCACCCAAATCGCAT - 3'; A. 5'- CTA GCA CCC AAA TCG CAT TAG - 3', B. 5' - AUG CGA UUU GGG UGC UAG - 3', C. 5' - AUG CGA UUU GGG UGC - 3', D. 5- ATG CGA TTT GGG TCG TAG - 3'Answer the following whether it is TRUE or FALSE: 1. For each DNA segment 3'-ACCTGCCTACCCG-5' the sequence of the mRNA molecule synthesized is 5'-TGGACGGATGGGC-3' 2. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-ATGGCTCCATACATG-3'. 3. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-AUGGCUCCAUACAUG-3'. 4. The template strand is the strand of DNA used for RNA synthesis. 5. Transcription forms a messenger RNA molecule with a sequence that is identical to the DNA template from which it is prepared.
- Use a codon chart determine the amino acid sequence. Remember to read through the strand and ONLY start after the promoter and STOP when it tells you to stop. Follow example below: Example: DNA AGA TATA TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC MRNA O protein AUG GAG GCC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG start-glu-ala-thre-hist - asp-glu-threo-stop met DNA CCT ATA TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC mRNA DGGA UAU) AUG uGul Gcc nccl cAul GCol protein ly Tur MeT cys AlA ser HIJ Ala 2 3 4 DNA AGA ACT ATA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA protein DNA TAT ATAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTC GCC ATC mRNA protein D DNA TAA ACT ATA TAC CTA GCT TAG ATC TAA TTA CCC ATC mRNA protein Auu UGA UAU AGU GAUCGA AUC MAG Auu AAU leu Stop. TRY-Met-Asp- ARG-Isle-Stop-Ile. Asn DNA CTA TTT ATA TAC TAG AGC GAA TAG AAA CTT ATC ATC mRNA protein D DNA CAT ATA TAC CTT AGT TAT CCA TTG ACT CGA ATT GTG CGC TTG…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'Here is the sequence of a portion of a bacterial gene. The template strand is on the bottom:5’-ATGCTGCGTGCATGGGATATAGGTAGCACACGTCC-3’3’-TACGACGCACGTACCC TATATCC ATCGTGTGCAGG-5’(a) Assuming that transcription starts with the first C in the template strand, and continues to the end, what would be the sequence of the mRNA derived from this fragment? (b) Find the initiation and stop codons in this mRNA. (c) Would there be an effect on translation of changing the fourth T in the template strand to a C? If so, what effect? (d) Would there be an effect on translation of changing the fourth A in the template strand to a C? If so, what effect?