Concept explainers
a)
Interpretation:
The mRNA strand from the given DNA template strand has to be predicted.
Concept Introduction:
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
Illustrated relationships are:
DNA informational strand: 5’ ATG CCA GTA GGC CAC TTG TCA 3’
DNA Template strand: 3’ TAC GGT CAT CCG GTG AAC AGT 5’
mRNA: 5’ AUG CCA GUA GGC CAC UUG UCA 3’
b)
Interpretation:
The mRNA strand from the given DNA template strand has to be predicted.
Concept Introduction:
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
Illustrated relationships are:
DNA informational strand: 5’ ATG CCA GTA GGC CAC TTG TCA 3’
DNA Template strand: 3’ TAC GGT CAT CCG GTG AAC AGT 5’
mRNA: 5’ AUG CCA GUA GGC CAC UUG UCA 3’
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FUND.OF GEN CHEM CHAP 1-13 W/ACCESS
- Nonearrow_forwardUsing the codon table, identify a 5’-3’ sequence of nucleotides in the dna template strand for mRna coding for the polypeptide sequence NH2-PHe-Pro-lys-COOH.arrow_forwardConsider 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?arrow_forward
- The BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINYarrow_forwardc) Based on your answer to part b above, determine the polypeptide sequence produced by the ribosome from the mRNA which you transcribedarrow_forwardGiven the following DNA sequence of the template (i.e. noncoding) strand for a given gene: 5'ATTGGCTGTTAGAGCGGCCGTCTAAACATCGTTGGA3' Part A) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends) Part B) Use the genetic code to write the peptide sequence translated in a cell from the mRNA synthesized in part A. Please use the 3 letter abbreviation for each amino acid.arrow_forward
- Given the following DNA sequence from the template strand of a given gene: 5'CTTGCGTCACCTAAGACCTGTCATCG3' a) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends) b) Write the peptide sequence translated from the mRNA produced in part a.arrow_forwardWhat 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'arrow_forwardConsider a template strand of DNA with the following sequence: 3 '–CAA TGT ATT TTT GCT–5 '. (a) What is the informational strand of DNA that corresponds to this template? (b) What mRNA is prepared from this template? (c) What polypeptide is prepared from the mRNA?arrow_forward
- A) Based on the mRNA sequence below, provide the corresponding DNA template (5'-3') and protein sequences (N-C terminus) using the single letter abbreviations for each 5' GCA UAU CCU UGU GAU 3' B) Identify the two unique amino acids in the protein sequence above, provide their full names and brief explanation why you chose them C) Draw the two amino acids from 3. connected with a peptide bond to each other (with free amino and carboxy termini) at physiological pH|arrow_forwardFor the following DNA sequence: 3’–CGATACGGCTATGCCGGCATT–5’, what is the sequence of the corresponding segment of mRNA formed using the DNA segment above as the template?arrow_forwardThe DNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U A G U UUU1 UUC UUA LOU Leu CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala CAU His CAC CAA CAG Gin AAU Asn AAC AAA 1 Lys AAG LYS G {}a UAA Stop UGA Stop A UAG Stop UGG Trp GAU 1 GAC Asp GAA GIU Glu GAGJ UGU UGC CGU CGC CGA CGG AGU AGC AGA AGG Cys GGU GGC GGA GGG Arg Ser Arg DOA DOA DOA DUTO Third letter Glyarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning