IV. CENTRAL DOGMA OF MOLECULARBIOLOGYSuppose the followingbase sequence was found in a 30-basepolymer: 3' CAG TTA AGC CTC GGT TACCAG GAT ACG GGA 51. What would be the first 10 bases atthe 5' end of the complementarystrand?Second MRNA baseYour answerUUUUCUUAUUGUPheTyrUGCCysUUCUACSerUccUUAUCAUAA Stop UGA StopLeu2. What would be the correspondingMRNA sequence of the givenUUGucGUAG Stop UGG TrpCGUHisCUUCCuCAUCucpolymer?ccCACProCAALeuCUACuaArgCCACGAGinccGCAGCGYour answerAAUAGUAsnAGCAUUACUSerAUC le ACCAACThrACAAGALysAGGAUAAAAArgAUGACGAAG3. What would be the resulting aminoacid sequence? (Please use thegenetic code chart above.) Pleaseseparate amino acids with "dash"(e.g.phe-leu-ile).GUUGCUGAUGGUAspGGCGACAlaGAAGUCGCCGlyValGUAGGAGluGGGGCAGUGGCGGAGYour answerFirst MRNA base (5 end of codon)Third MRNA base (3' end of codon)

Central Dogma: Central dogma is a process by which information from DNA is converted into…

Answered: IV. CENTRAL DOGMA OF MOLECULAR Suppose…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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pcc300ATAAADATATAOOTTAA 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:
1. (a) What will be the newly synthesized DNA from the template given? DNA Template 3 - CGGATGCCCGTATAC -5 3 - GCCTACGGGCATATG -5 5 - GCCTACGGGCATAAG -3 5 - GCCTACGGGCATATG -3 3 - CGGATGCCCGTATAC -5 (b) Which is the DNA template given if the mRNA is 5 - CGGAUGCCCGUAUACGUA -3 ? 3 - GCCTACGGGCATATGGTA -5 5 - GCCTACGGGCATAAGGAT -3 5 - GCCTACGGGCATATGCAT -3 3 - CGGATGCCCGTATACCTA -5
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
www D le C 3⁰ A B Indicate True (T) or False (F) for the following statements. Only use the letter (T/F) in the space provided 1. The name of this process is best known as Rho dependent termination 2. The enzyme C called DNA polymerase incorporates ribonucleotides into B called the mRNA False 3. The DNA region A contains inverted palindrome sequences which results in formation of a stem-loops structure 4. During this process, the structure D called terminating hairpin forms and increases the enzyme affinity which terminates transcription
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5' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3' 3' CACGATCGCCCTTACTCGACCCTATGATCATCCCGA 5' Template Strand: 9. Using the template strand, transcribe the DNA above, Be sure you write your sequence 5 - 5 a indicate the 5' and 3' ends of any nucleic acid molecule(s). 10. Use the codon chart below to translate your mRNA into an amino acid sequence. Begin at the first codon. Third First position (5' end) Second position position (3'end) UGU Cys UAU Tyr Cc UGC Cys UGA Stop UGG Trp UCU Ser -Y UAC Tyr UAA Stop UAG Stop UUU Phe - F UUC Phe UUA Leu UUG Leu FL UCC Ser -- UCA Ser UCG Ser CGU Arg CGC Arg ER CGA Arg CGG Arg CCU Pro CAU His CUU Leu CUC Leu -- CAC His CAA Gln CAG Gln CCC Pro -P A - CUA Leu CUG Leu CCA Pro CCG Pro AAU Asn AAC Asn AGU Ser AGC Ser AGA Arg ACU Thr AUU lle AUC lle AUA lle AUG Met M ACC Thr -T ACA Thr ACG Thr A. AAA Lys K AAG Lys -R AGG Arg A. GAU Asp -D GAC Asp GGU Gly GGC Gly GCU Ala GUU Val GUC Val GCC Ala A -G GGA Gly GGG Gly A -V GUA Val GUG Val GCA Ala GCG Ala GAA Glu -E…
30. Below is a pre-mRNA sequence that contains 2 exons and an intronic region. Based on your knowledge of the spicing reaction, predict the ligated exon sequence. [ | (hint: there is a consensus sequence for this "cut n paste' rxn.] 5'-ACGACAGGAUGAAGGUAAAUCGGGUAGGGGCGGCUGACUCUCUUUUUCCCUCAGGUCGUAA-3'
5’ AUG UUA CGU AAU GCU GUC GAA UCU AUU UGC UUU ACA UAA 3' Write the sequence of the DNA template (antisense) strand from which the mRNA was synthesized. Write the sequence of the DNA coding (sense or informational) strand complementary to the template strand. Write the sequence of tRNA anticodon that corresponds to the given mRNA molecule. Write the amino acid sequence of the peptide synthesized from the given mRNA nucleotide sequence.
mRNA sequence of A gene Write the amino acid sequence of the gene A. 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCCUACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACGACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGCUGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGAGGUAGAAGCCGCUGGGGCUUGGGGCU-3’
17) Synthesis of the mRNA starts at the boxed A/T base pair indicated by the box and proceeds left to right on the sequence below. Transcribe and translate this bacterial gene. 5'-GGACCGCGGGGCAGGATTGCTCCGGGCTGTTTCATGACTIGICAGGTGGGATGACTTGGATGGAAAAGTAGAAGGTCATG-3 3'-CCTGGCGCCCCGTCCTAACGAGGCCCGACAAAGTACTGAACAGTCCACCCTACTGAACCTACCTTTTCATCTTCCAGTAC-5′ 1 -+--at - --+-- 80
You continue to study the expression of the hexose kinase gene and capture the following electron micrograph of the gene being expressed. MRNA 1 20 ORI 40 60 TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGIAATATĞGGGATGCACTATC 5' 3' AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCA'NTATAÇCCCTACGTGATAG CACTATC promoter RNA polymerase ribosome
Given the mRNA sequence below, 5'-UUCCCAAAGUAU-3'Answer the following below:1. What is the sequence of amino acids using the table of genetic code? (three-letter abbreviation)2. Given the situation that a mutation occurs by substitution of the 12th nucleotide with cytidine 5'-monophosphate, what is the resulting amino acid sequence?3. What type of mutation occured? Specific what type under frameshift and point mutations

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