The following eukaryotic DNA is transcribed into RNA and the mRNA transcript is translated into protein. 1 GGC GAC GACA TATAAAGCGA 41 TATCCATGGA 81 AATAAAACGC 121 GTCGACTGC 5' UGGAC 3' 5' TGGAC 3' 3' UGGAC 3' 11 O 3' TGGAC 3' 51 CGCGCCATGG 21 91 CGACTGTAGA 61 Useful sequences: TATA box: TATAAA, Kozak sequence: A/GccAUGG, Poly-adenylation signal: AAUAAA What are the first bases in the mRNA transcript? TCACATAAGC GGTGC GAT GC 111 GTATCAGTCA TTCAGCGTAG TCTGATGC CA 31 CT GAT GAGCC 71 101

The following eukaryotic DNA is transcribed into RNA and the mRNA transcript is translated into protein. 1 GGC GAC GACA TATAAAGCGA 41 TATCCATGGA 81 AATAAAACGC 121 GTCGACTGC 5' UGGAC 3' 5' TGGAC 3' 3' UGGAC 3' 11 O 3' TGGAC 3' 51 CGCGCCATGG 21 91 CGACTGTAGA 61 Useful sequences: TATA box: TATAAA, Kozak sequence: A/GccAUGG, Poly-adenylation signal: AAUAAA What are the first bases in the mRNA transcript? TCACATAAGC GGTGC GAT GC 111 GTATCAGTCA TTCAGCGTAG TCTGATGC CA 31 CT GAT GAGCC 71 101

Name: Thx! The following eukaryotic DNA is transcribed into RNA and the mRNA
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Description: Thx! The following eukaryotic DNA is transcribed into RNA and the mRNA transcript is translated intoprotein.1GGC GAC GACA TATAAAGCGA41TATCCATGGA81AATAAAACGC121GTCGAC TGC5' UGGAC 3'5' TGGAC 3'3' UGGAC 3'11O 3' TGGAC 3'51CGCGCCATGG2191CGACTGTAGA61Usefu

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 17QP: Given the following mRNA, write the double-stranded DNA segment that served as the template....

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Which of the following represents the sequence of an RNA transcript for which the coding strand (also known as non-template strand) of DNA has the sequence: GTACTGGCTAGCTGCTAGAA? Note all sequences are written 5'-3'. OA. AAGAUCGUCGAUCGGUCAUG OB. AAGATCGTCGATCGG TCATG OC. GTACTGGC TAGCTGC TAGAA OD. GUACUGGCUAGCUGCUAGAA
For the messenger RNA sequence below, find the beginning of the amino acid coding sequence and translate the sequence using the genetic code provided below. 5' - AAUUAUGGGCAAUAUGCCGGGCcGGUUAAGCG - 3' Second Letter A UGU cys u UGC Phe UCU UU U UUC UUA UAU Tyr Ser UAC UAA UAG Leu UCA Stop UGA Stop UUG UCG Stop UGG Trp CUU CU CAU His CGU c cuc Leu ccc ССА CCG Pro CÁC CAA CAG CGC CGA CGG Arg CUA CUG Gin 1st 3rd letter Ser u letter AUU ACU AAU AAC AAA AAG Asn A AUC AUA AUG lle ACC ACA AGU AGC AGA AGG Thr Lys Arg Met ACG GUU G GUC GUA GUG GCU GAU GAC GAA Asp GGU Val GCC Ala GGC Gly GCA GGA Glu GCG GAG GGG G
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 letter
The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt-5’ Answer the following questions: Q1. Which is the coding strand? Which is the template strand? [10%] Top-bottom. Bottom-Top. Both can be used as either coding or template for this gene.
The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt What is the length in AA’s of the Sh protein? Assume fMet is NOT CLEAVED [10%] 39 AA 13 AA 12 AA 21 AA
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’
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'
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 77 90 5' AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 110 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5' 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. (i) Identify both the hexameric sequences of the promoter region in the coding strand above.
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)
The DNA sequence of a short gene from a sea slug, and the mature RNA synthesized from this gene, are shown below. DNA sequence: 5’ - AGCATCTCATGTGCGAGTCCTGACGCTGACTAGC – 3’ 3’ - TCGTAGAGTACACGCTCAGGACTGCGACTGATCG – 5’ mature mRNA: 5’ – cap-AUCUCAUGUGCGAACGCUGACUAGAAAAAAAAAA- 3’ Draw boxes around any structures or bases that were added to the RNA during processing.
The DNA sequence of a short gene from a sea slug, and the mature RNA synthesized from this gene, are shown below. DNA sequence: 5’ - AGCATCTCATGTGCGAGTCCTGACGCTGACTAGC – 3’ 3’ - TCGTAGAGTACACGCTCAGGACTGCGACTGATCG – 5’ mature mRNA: 5’ – cap-AUCUCAUGUGCGAACGCUGACUAGAAAAAAAAAA- 3’ How many amino acids are in the peptide encoded by the gene? _____________