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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Question
Chapter 11, Problem 11P
Summary Introduction
Introduction:
Primers may be defined as the oligonuceotide strands of
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Check out a sample textbook solutionStudents have asked these similar questions
Design 6 bp primers to amplify the region of this sequence that is highlighted in yellow.
attatatttt atattatata ctctgggctc agagcagccc
40
41
atattatata tatatatttt aaaatattat aaatttattt
80
81
cagtcacgcg tcctgatgac attatatttt ataatttttt 120
121 ttttattttt attatatttt aaaatattat aaatttattt 160
161 aaaatattat tatatattta aaatttattt attataaaat 200
201 aaaatattat ttttattttt gagatcagga cggctgcatg 240
Forward
primer
Reverse
primer
For the following sequence design the forward and reverse primer... explain and justify your answer.
Full sequence would be:
1 tctagagtca 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…
For 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…
Chapter 11 Solutions
Genetics: From Genes to Genomes
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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Similar questions
- (i) For the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAATTGT TA T CCGC T CA CA AT T C CACA CA A CATA CGAGC CGGAAG CA TA A 110 120 130 140 150 160 (ii) An allele of a gene has the following change in it's sequence ATG GTG CÁC CTG ACT CCT GTG GAG AAG TCT compared to the wild type ATG GTG CAC CTG ACT CT GAG GAG AAG TCT With reference to the sequence; there is a codon, resulting in a change from is a mutation in the to which mutation.arrow_forwardYou are studying a protein that contains the peptide sequence RDGSWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below. 5'-CGTGACGGCTCGTGGAAGCTAGTCATC-3' 3'-GCACTGCCGAGCACCTTCGATCAGTAG-5' This sequence does not contain any BamHI restriction enzyme sites. The target sequence for the BamHI restriction nuclease is GGATCC. Your goal is to create a BamHI site on this plasmid by manipulating the DNA sequence, without changing the coding sequence of the protein. How would you do this, ie what would the new sequence be?arrow_forwardThe chain terminator method was used to sequence the following DNA fragment: ACTGGGCATAAGCGGGAACTTTGCAGAACTGGCTGGCCTCAGAGCAGGGA. 1. Predict a band pattern in a gel after sequencing this DNA fragment using a radioactively labeled primer [32P]-5’- TCTGAGGCCAGCCAGTTCTGCAAAGTTC. 2. Due to an experimental mistake, dATP was not added in all four reaction mixtures. How does the band pattern change?arrow_forward
- Each of the following pairs of primers has a problem with it. Tell why the primers would not work well. (a) Forward primer 5'GCCTCCGGAGACCCATTGG 3' Reverse primer 5'TTCTAAGAAACTGTTAAGG 3' (b) Forward primer 5'GGGGCCCCTCACTCGGGGCCCC 3'Reverse primer 5'TCGGCGGCCGTGGCCGAGGCAG 3' (c) Forward primer 5'TCGAATTGCCAATGAAGGTCCG 3'Reverse primer 5'CGGACCTTCATTGGCAATTCGA 3'arrow_forwardDesign primers that will amplify the following region of DNA (assume this is one strand from a double stranded region of DNA). The primers should be 15 bases in length. Indicate the 5' and 3' ends of the primers. 5' GGATCGATCAAGAACAATGACAGGATCGAGGAATTCAGCCTACGCAGCCCGTAGCTGGAGGGA 3'arrow_forwardThe following DNA sequence was determined by Sanger sequencing, using a 20 nt long sequencing primer that ended ...AGTACAACAA-3'. 5'-agtacaacaa ctctcggtc tacggtacgc ctgcgggcgc gtagccaatc tagcacttcg-3' 3'-tcatgttgtt gagagccag atgccatgcg gacgcccgcg catcggttag atcgtgaagc-5′ A. If the technician forgot to add ddNTPs to the reaction, what would the sequencing chromatogram look like? Blank Many peaks, but only one at each position Overlapping peaks at every position All peaks are black There is only one peak, at 60 nt B.When the reaction is done correctly, ddCTP is labeld with a yellow fluorescent tag. When the Sanger sequencing reaction is complete, what will be the lengths, in nucleotides, of the three shortest products that have the yellow tag? C. Could you perform Illumina sequencing using ddNTPs? Why or why not? Explain.arrow_forward
- 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.arrow_forwardBelow is a DNA sequence of the coding strand for a small gene. This gene has no introns. +1 5'- TATAAGATGCGTAGGATGCAGCTGTTTCAGCAGCCACGGTCTCGGCCCAGATAGCAGATAATAAACACGC GTA-3 a. Is this gene for an eukaryote or a prokaryote? Give one reason (. b. How many amino acids are expected to be coded by this gene? c. There are five underlined nucleotide sequences, interpret the purpose of three of them ONLY?arrow_forwardBM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadplearrow_forward
- The DNA sequence of one strand of a gene from threeindependently isolated mutants is given here (5′ endsare at left). Using this information, what is the sequence of the wild-type gene in this region?mutant 1 ACCGTAATCGACTGGTAAACTTTGCGCGmutant 2 ACCGTAGTCGACCGGTAAACTTTGCGCGmutant 3 ACCGTAGTCGACTGGTTAACTTTGCGCGarrow_forwarda) Complete the table below. Assume that reading is from left to right and that the columns represent transcriptional and translational alignments. Label the 5’ and 3’ ends of DNA and RNA and carboxy and amino acid ends of protein. (You may fill in this chart by hand writing- no typing necessary here.) 2. b) Is the top or bottom DNA strand the template strand?arrow_forwardThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3’ 3'...AAGCTCGAGAGCAGCAGCTСТАТGCGCTАСТАТААTGACCATTATАССССТАСGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?arrow_forward
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