Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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- Transcribe an mRNA sequence from this TEMPLATE STRAND of DNA 3' CGTACGTGTATCCCATCC 5' 5' GCAUGCACAUAGGGUAGG 3' 5' GCAUGCACAUAGGGUAGG 3' 3' CGUACGUGUAUCCCAUCC 5' 5' GCATGCACATAGGGTAGG 3' 3' CGTACGUGTATCCCATCC 5'arrow_forwardIf the mRNA sequence is 5'-UUU-CCC-AAA-3'.The DNA coding strand should be 5'-TTT-CCC-AAA-3' ?The DNA template strand should be 3'-AAA-GGG-TTT-5' ?So in the 5 to 3 direction, the DNA template strand should be 5'TTT-GGG-AAA-3' ?arrow_forwardThe following DNA sequence is found on a chromosome in rice plants: 5’ ACCTTGCTCACATGTGGCGTACCTTTCCGTCTATCTGAACAC 3’ What is the amino acid sequence of the longest protein that could be made from this stretch of DNA, assuming that this strand of DNA is the non-template strand? (Remember that the start of translation requires the sequence AUG in the mRNA and “STOP” is not an amino acid.)arrow_forward
- If one DNA strand has the nucleotide sequence below, what is the nucleotide sequence on its complementary strand? 5’ACCGATTACGATTACG3’ If the template DNA strand has the nucleotide sequence below, what is the nucleotide sequence on mRNA? 5’ACCGATTACGATTACG3’ In the chart below, indicate the unique structural or functional characteristics of DNA and RNA, as well as their similarities. Be very specific. Unique DNA characteristics Similar characteristics Unique RNA characteristicsarrow_forwardThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'.TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'.AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCCTACGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. d Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 -41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?arrow_forwardAs you should recall, DNA, when not being actively transcribed, has a double helical structure. This portion of the DNA has had the two strands separated in preparation of transcribing for a needed protein. The following is one of the two complimentary strands of DNA: 3' - AACCAGTGGTATGGTGCGATGATCGATTCGAGGCTAAAATACGGATTCGTACGTAGGCACT - 5' Q: Based on written convention, i.e. the 3'-5' orientation, is this the coding strand or the template strand? ______________________________ Q: Assuming this strand extends from base #1 to #61 (going left to right), interpret the correctly transcribed mRNA and translated polypeptide for bases 24 - 47: mRNA: ___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___- polypeptide chain: ________--________--________--________--________--________--________--________arrow_forward
- 5’ GGACCTATCAAAATCCTTATGCGCTAGGATAGCTAACGCATCCAC3’ This is the template strand. The +1 transcription start site is bold. Transcribe template DNA to mRNA. Make sure you write mRNA in the 5’ to 3’ direction. This is tricky – don’t assume the polymerase knows right from left. It can only synthesize new DNA in 5’>3’ direction.arrow_forwardName each of the processes pictured: -RNA 5' TCAC CÁCTCAT 3' TACCACCTA 3' UUCAC CACUCAU U 5' DŇA RNA polymerase Primary RNA transcript Exon 1 Intron Exon 2 Intron Exon 3 Spliced RNA 000 Exon 1 Exon 2 Exon 3 AAAAAAA polypeptide chain Met Phe Arg P E UAC AAA GCU AUGUUUCGA Ribosome There are possible nucleotide "triplets" orarrow_forwardpcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:arrow_forward
- This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.arrow_forwardWhich primer would bind to this coding strand as a reverse primer? 5'- ATGGCCAAAT GATTCCCACG ATTTGGCCAT TGAGATCCGG - 3' O3'- ATGGCCAAAT - 5' O 5'- CCGGATCTCA - 3' O 5'- ATGCGCAGAT - 3' O3'- CCGGATCTCA - 5' Narrow_forward15)arrow_forward
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