Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 9, Problem 18P
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The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post-transcriptional processing which is essentially way too different from the original code in the DNA template.
Given: GUC-CAC-UUA-ACC-CCU-GAG-GAG-AAA-UCG-GCC (Protein with known amino acid sequence)
Requirement: Original DNA code. Itemize the steps you would take to get to know the original DNA code of the protein in focus.
Given the following DNA sequence of the template strand for a given gene:
5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3'
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 in part A. Please use the 3 letter abbreviation for each amino acid.
Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?
The following RNA sequence represents a small messenger which can be translated in a prokaryotic cell:
5'-ACGAAUGCACAGUAAAACUGGCUAGCGUAGGCUGA-3
Assume that the messenger RNA is translated in the cell, using the correct machinery and signals
required for accurate protein synthesis. Using this RNA sequence and the Genetic Code Dictionary (see
your textbook for the dictionary), solve the following problems
A. Write the sequence of a protein that would be translated from this mRNA, using the appropriate stop and start
signals, and indicating the correct termini of the protein product.
B. Suppose that the underlined A in the sequence is changed to a U. Write the expected protein product of this
mRNA.
Chapter 9 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 9 - 9.1 Some proteins are composed of two or more...Ch. 9 - In the experiments that deciphered the genetic...Ch. 9 - 9.3 Several lines of experimental evidence pointed...Ch. 9 - Outline the events that occur during initiation of...Ch. 9 - 9.5 A portion of a DNA template strand has the...Ch. 9 - Describe three features of tRNA molecules that...Ch. 9 - Prob. 7PCh. 9 - For each of the anticodon sequences given in the...Ch. 9 - What is the role of codons UAA, UGA and UAG in...Ch. 9 - Compare and contrast the composition and structure...
Ch. 9 - Consider translation of the following mRNA...Ch. 9 - Prob. 12PCh. 9 - Third-base wobble allows some tRNAs to recognize...Ch. 9 - The genetic code contains 61 codons to specify the...Ch. 9 - 9.15 The three major forms of (,, and ) interact...Ch. 9 - The accompanying figure contains sufficient...Ch. 9 - 9.17 The line below represents a mature eukaryotic...Ch. 9 - 9.18. After completing Problem, carefully draw a...Ch. 9 - 9.19 Define and describe the differences in the...Ch. 9 - 9.20. Describe the roles and relationships...Ch. 9 - 9.21 In an experiment to decipher the genetic...Ch. 9 - Identify and describe the steps that lead to the...Ch. 9 - Prob. 23PCh. 9 - Har Gobind Khorana and his colleagues performed...Ch. 9 - 9.25 An experiment by Khorana and his colleagues...Ch. 9 - Prob. 26PCh. 9 - 9.27 The mature transcribed from the human gene is...Ch. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - 9.31 A portion of the coding strand of for a gene...Ch. 9 - A eukaryotic mRNA has the following sequence. The...Ch. 9 - Diagram a eukaryotic gene containing three exons...Ch. 9 - Prob. 34PCh. 9 - 9.35 Table lists and gene sequences for or ...Ch. 9 - Prob. 36PCh. 9 - In terms of the polycistronic composition of mRNAs...Ch. 9 - Prob. 38PCh. 9 - 9.39 Answer the following questions about the...Ch. 9 - 9.40 for each of the following anticodon...Ch. 9 - Prob. 41PCh. 9 - Prob. 42P
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- Consider the following DNA sequence, which codes for a short polypeptide: 5'-ATGGGCTTAGCGTAGGTTAGT-3' Determine the mRNA transcript of this sequence. You have to write these sequences from the 5' end to the 3' end and indicate those ends as shown in the original sequence in order to get the full mark. How many amino acids will make up this polypeptide? Determine the first four anticodons that will be used in order to translate this sequence.arrow_forwardRefer to the double stranded DNA molecule with the sequence below to answer the following questions: 5’ATATGGGTCTCGATAGGGCTGTTTTCTCCGGC 3’ 3’TATACCCAGAGCTATCCCGACAAAAGAGGCCG 5’ Which strand functions as the transcription template, the top one or the bottom one? Explain your reasoning. What is the mRNA transcript and polypeptide from this strand? In the space below, copy the DNA strand that is transcribed, and write the mRNA transcript and polypeptide chain below it. Align the mRNA and polypeptide so that it is clear which DNA bases they came from. DNA strand: mRNA: amino acid sequence:arrow_forwardListed below are five amino acids. Use the genetic code to determine the exact codon for each amino acid. A point mutation at the genetic level in each codon results in the change indicated. For each mutation, indicate whether it is due to a transition or a transversion, and then indicate the effect of each mutation at the protein (amino acid level) (i.e. silent, nonsense, missense). In addition, Please note, each of the three lines above an amino acid represents a single RNA base. For example, when you look at the codon chart AUG would stand for Met (methionine) Lys 1 Glu Ile 3 Stop Ile 4.arrow_forward
- The following four mutations have been discovered in a gene that has more than 60 exons and encodes a very large protein of 2532 amino acids. Indicate which mutation would likely cause a detectable change in the size of the mRNA and/or the size of the protein product. Consider a detectable change to be >10% of the wild-type size. A table of the genetic code is shown below. First letter 0 00 U O A บบบ UUC UUA UUG U CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu >Leu AUU AUC lle AUA AUG Met >Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Second letter C Ser Pro Thr Ala CAU CAC CAA CAG UAU UGU Tyr UAC UGC UAA Stop UGA UAG Stop UGG AAU AAC AAA AAG A GAU GAC GAA GAG His Gin Asn Lys Asp G Glu CGU CGC CGA CGGJ AGU AGC AGA AGG GGU GGC GGA GGG O AAG576UAG (changes codon 576 from AAG to UAG) Cys Stop Trp O GUG326AUG (changes codon 326 from GUG to AUG) Arg Ser Arg Gly DUAG DUA G DCAG DO AG deletion of codon 779 insertion of 1000 base pairs into the sixth intron (this particular…arrow_forwardThe following is a DNA sequence of gene Z. The underlined sequence represents the promoter for gene Z and the underlined and italicized sequence encodes the gene Z ribosome binding (RBS) site. Transcription begins at and includes the T/A base pair at position 60 (bold) a. What are the nucleotides of the mRNA from gene Z?b. What are the amino acids encoded by gene Z? (A codon chart is found on the finalpage)arrow_forwardGiven the following sequence of a coding DNA strand: AGTTGCGCATGCCAGAGAGGTTCGAGTGCACATAACTTGAG The template DNA strand will have the following sequence written in the conventional manner with no orientation indicators: The corresponding tRNA sequence written in the conventional manner with no orientation indicators will be: After translation was complete, the resulting polypeptide underwent post-translational modification wherein the N-terminal Met residue was removed from the polypeptide. Using one-letter abbreviations, the sequence of the resulting polypeptide isarrow_forward
- Template strand of DNA is: 3’ TACATAACCGGGCCCATATCGGCCATTTGC5’. 2a). Following transcription, what is the total number of codons in the mRNA transcript? 2 b). Where is the start codon located in this mRNA transcript? 2c). Following translation of this mRNA transcript, how many amino acids will the proteincontain and identify the amino acids sequence of this gene from a genetic code table*.*Note= using a genetic code tablearrow_forwardConsider a stretch of DNA (a hypothetical gene) that has the sequence 5’ ATG-CTA-TCA-TGG-TTC-TAA 3’ A) Transcribe and translate this gene using the genetic code table. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations. B) Now, our hypothetical gene has undergone a mutation. The mutant sequence is....3’ TAC-GAT-AGT-ACC-AAT-ATT 5’5’ ATG-CTA-TCA-TGG-TTA-TAA 3’ Transcribe and translate the mutant sequence. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations. C) Indicate the type of mutation (nonsense, missense, silent, or frame shift) present. D) How severe of a consequence will this mutation likely be in terms of protein function (none, mild, moderate or severe)? Why?arrow_forwardThe genetic code was solved partly by the use of in vitro systems to translate synthetic RNAs into peptides. In these systems, ribosomes, amino acids, and buffers that support translation are added and there is no control of where translation begins. AAA = Lys; AUA = Ile; AAU = Asn; UAA = stop. What peptides would NOT be produced in an in vitro system if the following oligonucleotide were added: AAAAAAAAAUAAAAAAAA Select one: a) Lys-Lys-Lys-Lys-Lys-Lys-Lys-Lys b) Lys-Lys-Ile-Lys-Lys c) Lys-Lys-Asn-Lys-Lysarrow_forward
- The following is a portion of an mRNA sequence: 3’ –AUCGUCAUGCAGA-5’ a)During transcription, was the adenine at the left-hand side of the sequence the first or the last nucleotide used to build the portion of the mRNA shown? Explain how you know. b)Write out the sequence and polarity of the DNA duplex that encodes this mRNA segment. Label the template and coding DNA strands. c)Identify the direction in which the promoter region for this gene will be located.arrow_forwardThe following is a DNA sequence of gene Z. The underlined sequence represents the promoter for gene Z and the underlined and italicized sequence encodes the gene Z ribosome binding (RBS) site. Transcription begins at and includes the T/A base pair at position 60 (bold). a. What are the nucleotides of the mRNA from gene Z?b. What are the amino acids encoded by gene Z? (A codon chart is found on the final page)arrow_forwardThe length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the MRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference? Explain in detail. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). A Ix BIUS Paragraph Arial 14pxarrow_forward
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