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 1EQ
Summary Introduction
To review:
The position of the intron in the genomic DNA (deoxyribonucleic acid), and the identification of the normal consensus sequence for splicing. Also, the identification of the splice site sequence.
Introduction:
The cDNA (complementary DNA) is the DNA which is made from the single-stranded RNA (ribonucleic acid). The cDNA is used for the cloning of the eukaryotic genes in the given prokaryotes. Genomic DNA is the chromosomal DNA which helps to form the hereditary material which is passed from one generation to the next. The introns are the sequences which are removed by the splicing of the premature RNA.
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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
The 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 created
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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- Below is the CODING sequence of a gene. An A is inserted at position 131. Insert the variant and write the mature mRNA sequence. Then write the amino acid sequence of the resulting transcript. AGAGCTACCAGCTTTAGCGGGATATAAACACACTTATCCTACTATCCCTA 51 TCTTCCCCCAAAGAGGGATCATGCCGTCGACGGCCAAACCGATCGAGCGG 101 AGGTTAAGGGCCATTGGAAAAGGACACTCTGTACGACTAAGAAAATAGC 151 AGCGACTGAGCTGAACGCTACGGGGGGGGGGGTATATACCCCCCCCCCCC 201 CCAGCTACGAGCGACTTTTTATTTTTTTTTTTTATTTTTTTTCTTTTTTA 251 TTTTTTTTGTTTTTTTTTTTCGCGACTACAGAGACTTCTAGCGATCGACG 301 AGCTAGCTAGCGCGGCATCGGACGCGATCGCGATCCTAGACTCTACGATarrow_forwardThe DNA sequence of a short gene from a sea slug, and the mature RNA synthesized from this gene, are shown below. DNA sequence: 5’ - AGCATCTCATGTGCGAGTCCTGACGCTGACTAGC – 3’ 3’ - TCGTAGAGTACACGCTCAGGACTGCGACTGATCG – 5’ mature mRNA: 5’ – cap-AUCUCAUGUGCGAACGCUGACUAGAAAAAAAAAA- 3’ How many amino acids are in the peptide encoded by the gene? _____________arrow_forwardThe DNA sequence of a short gene from a sea slug, and the mature RNA synthesized from this gene, are shown below. DNA sequence: 5’ - AGCATCTCATGTGCGAGTCCTGACGCTGACTAGC – 3’ 3’ - TCGTAGAGTACACGCTCAGGACTGCGACTGATCG – 5’ mature mRNA: 5’ – cap-AUCUCAUGUGCGAACGCUGACUAGAAAAAAAAAA- 3’ In the DNA sequence, draw boxes around two exons in the gene. The splicing machinery does NOT pay attention to codons. The splicing of this DNA demonstrates that. How?arrow_forward
- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGarrow_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_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
- 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_forwardWhich of the following represents the sequence of an RNA transcript for which the coding strand (also known as non-template strand) of DNA has the sequence: GTACTGGCTAGCTGCTAGAA? Note all sequences are written 5'-3'. OA. AAGAUCGUCGAUCGGUCAUG OB. AAGATCGTCGATCGG TCATG OC. GTACTGGC TAGCTGC TAGAA OD. GUACUGGCUAGCUGCUAGAAarrow_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' (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.arrow_forward
- 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 Garrow_forwardb) Shown below is a short gene of an unknown bacteria genome (Figure 2). (Note: Transcription starts at Transcription Start Site (TSS).) TSS 5'-TATTATAACGCATGAGGAGCCATGCATTATCGGTATATGCACTGACCCGGAAAGGCTCCTTTTGGAGCCTTTTTT-3' 3-ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTTCCGAGGAAAACCTCGGAAAAA-5 Promoter ********* Terminator Figure 2 Based on the DNA sequence of terminator, draw the structure of the hairpin loop that will be formed during the end of transcription.arrow_forwardFor the following sequence design the forward and reverse primer... explain and justify your answer. Gene of Interest: a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg…arrow_forward
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