Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 12, Problem 4CONQ
Summary Introduction
To review:
The consensus sequence of the given sequences.
Introduction:
The branch of science that deals with biology, computer science, statistics, mathematics, and statistics is called bioinformatics. Bioinformatics tools are useful in finding out the consensus sequence by looking at their alignment. The process of conversion of a particular DNA (deoxyribonucleic acid) strand to mRNA (messenger ribonucleic acid) with the help of RNApolymerase enzyme is called transcription.
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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
This is part of the Escherichia coli DNA sequence that contains an inverted repeat.
(Note: top strand is the coding strand).
5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3'
3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5'
(i)
Draw the structure of hairpin loop that will be formed during the end of transcription.
(ii) Describe the function of the hairpin loop during transcription.
The sequence of the coding strand of a bacterial gene is given below. The positions of the first nine bases are numbered for your convenience. A missense mutation was introduced at position seven where the C was changed to a T resulting a mutant gene.
123456789
5'- ATGGCCCGACCGCAACTTTTCCGAGCTCTGGTGTCTGCGCAGTGACC-3
a. Write the template DNA (complementary strand) sequence for the wild type gene above
b. Write the DNA sequence of the mutant gene (Both DNA strands)
c. Write the sequence of mRNA produced from the mutant gene
d. Write the sequence of the mutant protein using the codon usage table provided in the end of this document.
Chapter 12 Solutions
Genetics: Analysis and Principles
Ch. 12.1 - 1. Which of the following base sequences is used...Ch. 12.1 - Prob. 2COMQCh. 12.2 - With regard to a promoter, a transcriptional start...Ch. 12.2 - Prob. 2COMQCh. 12.2 - 3. Sigma factor is needed during which stage(s) of...Ch. 12.2 - A uracil-rich sequence occurs at the end of the...Ch. 12.3 - Which RNA polymerase in eukaryotes is responsible...Ch. 12.3 - Prob. 2COMQCh. 12.3 - Prob. 3COMQCh. 12.3 - Prob. 4COMQ
Ch. 12.4 - Which of the following are examples of RNA...Ch. 12.4 - A ribozyme is a. a complex between RNA and a...Ch. 12.4 - Prob. 3COMQCh. 12.4 - Prob. 4COMQCh. 12.5 - 1. Which of the following is not a key difference...Ch. 12 - Prob. 1CONQCh. 12 - Prob. 2CONQCh. 12 - Prob. 3CONQCh. 12 - Prob. 4CONQCh. 12 - 5. Mutations in bacterial promoters may increase...Ch. 12 - Prob. 6CONQCh. 12 - 7. In Chapter 9, we considered the dimensions of...Ch. 12 - 8. A mutation within a gene sequence changes the...Ch. 12 - Prob. 9CONQCh. 12 - At the molecular level, describe how factor...Ch. 12 - Prob. 11CONQCh. 12 - What is the complementarity rule that governs the...Ch. 12 - 13. Describe the movement of the open complex...Ch. 12 - 14. Describe what happens to the chemical bonding...Ch. 12 - Prob. 15CONQCh. 12 - Prob. 16CONQCh. 12 - Prob. 17CONQCh. 12 - Mutations that occur at the end of a gene may...Ch. 12 - If the following RNA polymerases were missing from...Ch. 12 - 20. What sequence elements are found within the...Ch. 12 - 21. For each of the following transcription...Ch. 12 - 22. Describe the allosteric and torpedo models for...Ch. 12 - Which eukaryotic transcription factor(s) shown in...Ch. 12 - 24. The initiation phase of eukaryotic...Ch. 12 - A eukaryotic protein-encoding gene contains two...Ch. 12 - 26. Describe the processing events that occur...Ch. 12 - Prob. 27CONQCh. 12 - Prob. 28CONQCh. 12 - Prob. 29CONQCh. 12 - Prob. 30CONQCh. 12 - 31. In eukaryotes, what types of modifications...Ch. 12 - Prob. 32CONQCh. 12 - Prob. 33CONQCh. 12 - 34. Figure 12.21 shows the products of alternative...Ch. 12 - 35. The processing of ribosomal RNA in eukaryotes...Ch. 12 - Prob. 36CONQCh. 12 - Prob. 37CONQCh. 12 - After the intron (which is in a lariat...Ch. 12 - Prob. 1EQCh. 12 - 2. Chapter 21 describes a technique known as...Ch. 12 - Prob. 3EQCh. 12 - As described in Chapter 21 and in experimental...Ch. 12 - Prob. 5EQCh. 12 - Prob. 6EQCh. 12 - 1. Based on your knowledge of introns and pre-mRNA...Ch. 12 - Discuss the types of RNA transcripts and the...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- 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?arrow_forwardThe following DNA sequences found on the sense strand belong to the same eukaryotic gene: Sequence 1: 5'-GATTCAATAAAGCTCAGATCGCTCACGTCGCGACTC-3' Sequence 2: 5'-TCCGAGGTCACTAGATACTCGTCGATCGTATAAATG-3' a) Which sequence is likely to be found upstream from the coding sequence? Justify your answer. b) Which sequence is likely to be found downstream from the coding sequence? Justify your answer. c) Which sequence will not be transcribed into an mRNA transcript? Justify your answer.arrow_forwardTGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTTarrow_forward
- Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…arrow_forwardThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.arrow_forwardBelow is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortest ORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longest ORF (from N to C-terminal end)?arrow_forward
- 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.arrow_forwardThe 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 AAarrow_forwardwhat is the anticodon sequence that would build this protein? AUGUUUGUACAUUUGUGUGGGAGUCACCUGGUUGAGGCGUUGUAUUUGGUUUGUGGCGAGCGCUUUUACCAGUUAGAGAAUUACUGAarrow_forward
- For this double stranded DNA: 3’-AGGTCCAGTCTTCGGCGGATGATCGGGCCAACCCCCGTAACTGGTCACCGGAATGCC-5’ 5’-TCCAGGTCAGAAGCCGCCTACTAGCCCGGTTGGGGGCATTGACCAGTGGCCTTACGG -3’ , while using the bottom strand is the template for transcription. If the START CODON is AUG and the STOP CODON IS UGA, how many amino acids are in the encoded protein and what is the primary structure?arrow_forwardFor this double stranded DNA: 3’-AGGTCCAGTCTTCGGCGGATGATCGGGCCAACCCCCGTAACTGGTCACCGGAATGCC-5’ 5’-TCCAGGTCAGAAGCCGCCTACTAGCCCGGTTGGGGGCATTGACCAGTGGCCTTACGG -3’ , while using the bottom strand is the template for transcription. If the START CODON is AUG and the STOP CODON IS UGA, what would it look like drawn out?arrow_forwardWhich of the following DNA strands, the top or bottom, would serve as a template for RNA transcription if the DNA molecule were to unwind in the indicated direction? 5′ ACGGACTGTACCGCTGAAGTCATGGACGCTCGA 3′ 3′ TGCCTGACATGGCGACTTCAGTACCTGCGAGCT 5′ ⎯⎯⎯⎯→ Direction of DNA unwinding What would be the resulting RNA sequence (written 5′→3′ )?arrow_forward
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