Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT b) Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. When you get to the stop codon – you may write an asterisk (i.e. a “*”) to denote the stop codon. c) Does the sense strand DNA sequence have 5’ and 3’ UTR sequences? If so – write them in the space below 5’ UTR: 3’ UTR:

Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT b) Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. When you get to the stop codon – you may write an asterisk (i.e. a “*”) to denote the stop codon. c) Does the sense strand DNA sequence have 5’ and 3’ UTR sequences? If so – write them in the space below 5’ UTR: 3’ UTR:

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 3QP: If the genetic code used 4 bases at a time, how many amino acids could be encoded?

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Similar questions

The template strand of a double helical segment of DNA consists of the following sequence: 5’-GTAGCCTTAAGCGATCACCGTCCGTATTACTAGTGGCCAGACTCTTTTCACTCTCATGTATAGTTG-3’ Following translation process, what is the amino acid sequence that will be coded for? (show your answer using THREE-letter amino acid code starting from N-terminus to C-terminus)
Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many codons are in the longest ORF? What is the frame of the shortest ORF? How many AA are in the shortest ORF?
What is the consensus sequence of the following six DNA sequences?GGCATTGACTGCCATTGTCACGCATAGTCAGGAAATGGGAGGCTTTGTCAGGCATAGTCA
For this double stranded DNA: 3’-AGGTCCAGTCTTCGGCGGATGATCGGGCCAACCCCCGTAACTGGTCACCGGAATGCC-5’ 5’-TCCAGGTCAGAAGCCGCCTACTAGCCCGGTTGGGGGCATTGACCAGTGGCCTTACGG -3’ , while using the bottom strand is the template for transcription. If the START CODON is AUG and the STOP CODON IS UGA, how many amino acids are in the encoded protein and what is the primary structure?
Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…
Consider the following DNA template: 5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’ 3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’ If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation. Met-Ala-Leu-Thr-Gln-Glu-Gly Met-Gly-Ser-Leu-Asn-Ser-Gln Met-Thr-Asn-Ser-Leu-Ala-Gln Met-Gln-Ala-Leu-Thr-Asn-Ser Met-Glu-Ala-His-Trp-Ser-Tyr
the one above: Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTT 2. Using this mutated DNA strand, express it as a polypeptide by using the correct reading frame. When you get to the stop codon – you may write an “*” to denote the stop codon. 3. How many amino acids were changed in the mutated polypeptide?
Given the template DNA sequence: 3’ - TAC - CAG - GTT - ACC - ATC - 5’ A.) What will be the mRNA requence corresponding to the template DNA sequence? B.) What is the amino acid sequence in letter A? ( e.g. Arg, Phe, etc.) C.) If the coding sequence of the dsDNA will "serve" as the template for transcription, what is the corresponding mRNA sequence? D.) With the mRNA transcript in letter C, what will be the amino acid sequence? ( e.g. Arg, Phe, etc.)
For this double stranded DNA: 3’-AGGTCCAGTCTTCGGCGGATGATCGGGCCAACCCCCGTAACTGGTCACCGGAATGCC-5’ 5’-TCCAGGTCAGAAGCCGCCTACTAGCCCGGTTGGGGGCATTGACCAGTGGCCTTACGG -3’ , while using the bottom strand is the template for transcription. If the START CODON is AUG and the STOP CODON IS UGA, what would it look like drawn out?
Based on these sequences. Remove codons 24 to 66, inclusive. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
Below is a sequence of DNA.5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many "reading frames" can be identified for this sequence? How many "open reading frames" can be identified for this sequence? What is the frame of the longest ORF? How many codons are in the longest ORF? What is the frame of the shortest ORF? How many AA are in the shortest ORF? Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortestORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longestORF (from N to C-terminal end)?
Place sequences a, b and c into the correct order based on start and stop codons. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence B TCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTC AATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTC GCGCCGAAAAAGATATGG