Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 8, Problem 33P
The sequence of a complete eukaryotic gene encoding the small protein Met Tyr Arg Gly Ala is shown here. All of the written sequences on the template strand are transcribed into RNA.
5′ CCCCTATGCCCCCCTGGGGGAGGATCAAAACACTTACCTGTACATGGC 3′
3′ GGGGATACGGGGGGACCCCCTCCTAGTTTTGTGAATGGACATGTACCG 5′
| a. | Which strand is the template strand? In which direction (right-to-left or left-to-right) does RNA polymerase move along the template as it transcribes this gene? |
| b. | What is the sequence of the |
| c. | A single base mutation in the gene results in synthesis of the peptide Met Tyr Thr. What is the sequence of nucleotides making up the mRNA produced by this mutant gene? |
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Chapter 8 Solutions
Genetics: From Genes to Genomes
Ch. 8 - For each of the terms in the left column, choose...Ch. 8 - Match the hypothesis from the left column to the...Ch. 8 - How would the artificial mRNA 5GUGUGUGU . . . 3 be...Ch. 8 - An example of a portion of the T4 rIIB gene in...Ch. 8 - Consider Crick and Brenners experiments in Fig....Ch. 8 - The HbSsickle-cell allele of the human -globin...Ch. 8 - The following diagram describes the mRNA sequence...Ch. 8 - The amino acid sequence of part of a protein has...Ch. 8 - The results shown in Fig. 8.5 may have struck you...Ch. 8 - Identify all the amino acid-specifying codons in...
Ch. 8 - Before the technology existed to synthesize RNA...Ch. 8 - A particular protein has the amino acid sequence...Ch. 8 - How many possible open reading frames frames...Ch. 8 - Prob. 14PCh. 8 - Charles Yanofsky isolated many different trpA-...Ch. 8 - The sequence of a segment of mRNA, beginning with...Ch. 8 - You identify a proflavin-generated allele of a...Ch. 8 - Using recombinant DNA techniques which will be...Ch. 8 - Describe the steps in transcription that require...Ch. 8 - Chapters 6 and 7 explained that mistakes made by...Ch. 8 - The coding sequence for gene F is read from left...Ch. 8 - If you mixed the mRNA of a human gene with the...Ch. 8 - Prob. 23PCh. 8 - The Drosophila gene Dscam1 encodes proteins on the...Ch. 8 - Describe the steps in translation that require...Ch. 8 - Locate as accurately as possible the listed items...Ch. 8 - Concerning the figure for Problem 26: a. Which...Ch. 8 - a. Can a tRNA exist that has the anticodon...Ch. 8 - For parts a and b of Problem 28, consider the DNA...Ch. 8 - Remembering that the wobble base of the tRNA is...Ch. 8 - Prob. 31PCh. 8 - The yeast gene encoding a protein found in the...Ch. 8 - The sequence of a complete eukaryotic gene...Ch. 8 - Arrange the following list of eukaryotic gene...Ch. 8 - Prob. 35PCh. 8 - The human gene for 2 lens crystallin has the...Ch. 8 - In prokaryotes, a search for genes in a DNA...Ch. 8 - a. The genetic code table shown in Fig. 8.2...Ch. 8 - a. Very few if any eukaryotic genes contain tracts...Ch. 8 - Explain how differences in the initiation of...Ch. 8 - Do you think each of the following types of...Ch. 8 - Null mutations are valuable genetic resources...Ch. 8 - The following is a list of mutations that have...Ch. 8 - Considering further the mutations described in...Ch. 8 - Adermatoglyphia described previously in Problem 18...Ch. 8 - Prob. 46PCh. 8 - You learned in Problem 21 in Chapter 7 that the...Ch. 8 - When 1 million cells of a culture of haploid yeast...Ch. 8 - Why is a nonsense suppressor tRNATyr, even though...Ch. 8 - A mutant B. adonis bacterium has a nonsense...Ch. 8 - You are studying mutations in a bacterial gene...Ch. 8 - Another class of suppressor mutations, not...Ch. 8 - Yet another class of suppressor mutations not...Ch. 8 - At least one nonsense suppressing tRNA is known...Ch. 8 - An investigator was interested in studying UAG...Ch. 8 - Prob. 56PCh. 8 - In certain bacterial species, pyrrolysine Pyl,...Ch. 8 - Canavanine is an amino acid similar to arginine...
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- 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 double stranded DNA sequence shown contains the promoter for the transcription of a bacterial gene. GGCACCTGCGATGCATGAATATATCGATCGGGAATCGCTATGTCAAGCCATGGCTAGATTA CCGTGGACGCTACGTACTTATATAGCTAGCCCTTAGCGATACAGTTCGGTACCGATCTAAT Draw a box around each of the promoter elements and identify each. Identify which strand will be used as the template strand by putting a vertical line between the -1/+1 start site nucleotides and underlining in the direction of transcription on the template strand as the example below indicates. ATCGG\GAATCGC TAGCCCTTAGCG Give the sequence of the RNA createdarrow_forwardFor 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 Garrow_forward
- 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 Garrow_forwardIn 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 CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGarrow_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_forward
- b) Shown below is a short gene of an unknown bacteria genome (Figure 2). (Note: Transcription starts at Transcription Start Site (TSS).) 5'TATTATAACGCATGACGAGCCATGCATTATCGGTATATGCACTGACCCGGAAAGGCTCCTTTTGGAGCCTTTTTT-3' 3'-ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTTCCGAGGAAAACCTCGGAAAAA-5' Promoter (i) (ii) TSS (iii) Terminator Figure 2 Which DNA strand (the top or the bottom) is used by polymerase as a DNA template? List the mechanistic steps that can trigger the initiation of transcription by the Sigma Factor. What are the amino acids translated from the resulting mRNA? Indicate the amino (NH₂*) and carboxyl (COO) termini of the polypeptide chain.arrow_forwardA segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)arrow_forwardConsider 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-Tyrarrow_forward
- 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.arrow_forwardhe Sequence below comes from the alpha-2 globin of the human hemoglobin gene cluster found in chromosome 16. The globin region of the hemoglobin protein itself consists of 2 alpha chains and 2 beta chains. 1 actcttctgg tccccacaga ctcagagaga acccaccatg gtgctgtctc ctgccgacaa 61 gaccaacgtc aaggccgcct ggggtaaggt cggcgcgcac gctggcgagt atggtgcgga 121 ggccctggag aggatgttcc tgtccttccc caccaccaag acctacttcc cgcacttcga 181 cctgagccac ggctctgccc aggttaaggg ccacggcaag aaggtggccg acgcgctgac 241 caacgccgtg gcgcacgtgg acgacatgcc caacgcgctg tccgccctga gcgacctgca 301 cgcgcacaag cttcgggtgg acccggtcaa cttcaagctc ctaagccact gcctgctggt 361 gaccctggcc gcccacctcc ccgccgagtt cacccctgcg gtgcacgcct ccctggacaa 421 gttcctggct tctgtgagca ccgtgctgac ctccaaatac cgttaagctg gagcctcggt 481 agccgttcct cctgcccgct gggcctccca acgggccctc ctcccctcct tgcaccggcc 541 cttcctggtc…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
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