Using the table above and the mRNA transcript AUG-CUC-UAC-AAG-UAG, choose the false statement: XXX The protein encoded for by the transcript above would appear as met-leu-tyr-glu. The segment of DNA from which the mRNA segment above was transcribed would have had the following sequence: TAC-GAG-ATG-TTC-ATC. a mutation that changes the codon UAC to UAU wouldn't have any effect on the primary structure of the protein it codes for. A TRNA with the anticodon UUU will carry the amino acid lysine (Lys).

Using the table above and the mRNA transcript AUG-CUC-UAC-AAG-UAG, choose the false statement: XXX The protein encoded for by the transcript above would appear as met-leu-tyr-glu. The segment of DNA from which the mRNA segment above was transcribed would have had the following sequence: TAC-GAG-ATG-TTC-ATC. a mutation that changes the codon UAC to UAU wouldn't have any effect on the primary structure of the protein it codes for. A TRNA with the anticodon UUU will carry the amino acid lysine (Lys).

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,...

See similar textbooks

Similar questions

Using the genetic code table provided below, identify the open reading frame in this mRNA sequence, and write out the encoded 9 amino acid long peptide sequence: 5'- CGACAUGCCUAAAAUCAUGCCAUGGAGGGGGUAACCUUUU C A G U UUU Phe UCU Ser UUC Phe UCC Ser UAC UCA Ser UAA UCG Ser UAG UUA Leu Leu G C CUU Leu CUC Leu CCC CUA Leu CUG Leu AUU lle AUC lle AUA lle AUG Met ACG ACU Thr ACC Thr ACA Thr Thr A UAU Tyr UGU Cys Tyr UGC Cys CCU Pro CAU His CGU Arg Pro CAC His Pro CAA Gln CGC Arg CGA Arg CCA CCG Pro CAG Gln CGG Arg GUU Val GCU Ala GAU GUC Val GCC Ala GAC GUA Val GCA Ala GAA GUG Val GCG Ala GAG Stop UGA Stop UGG AAU Asn AAC AAA AAG AGU Asn AGC G Lys Lys Asp Asp Glu Glu Stop A Trp Ser Ser AGA Arg AGG Arg GGU Gly GGC Gly UCAG GGA Gly GGG Gly с U C A G U C A G U C A G
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
The amino acids, in one-letter symbols and no spaces, coded by the following mRNA sequence is 5’ AAUGGAACGUCGGUACUGCCAUCGCAUUAGUACCAUGGCAAGCUGAAGC 3’
Suppose the codon sequence GUGCAAUUCGAGGCC has a single base pair mutation to GUGCAAUUCAAGGCC. If the old protein sequence was Val-Gln-Phe-Glu-Ala, what will be the new sequence encoded by the mutant 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-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
what is the anticodon sequence that would build this protein? AUGUUUGUACAUUUGUGUGGGAGUCACCUGGUUGAGGCGUUGUAUUUGGUUUGUGGCGAGCGCUUUUACCAGUUAGAGAAUUACUGA
BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple
If we have the following mutations, find the type of the mutation (silent or missense or nonsense?) 17C=U 36G=A 49G=U 115A=C 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’
Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?
In the copies of each sequence below, divide the sequences into codons (triplets) by putting a slash between each group of three bases. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
In the human genome for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotide in the amino acid coding region is represented by the sequence 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'