EBK CONCEPTS OF GENETICS
12th Edition
ISBN: 9780134818979
Author: Killian
Publisher: YUZU
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 12, Problem 23ESP
Summary Introduction
To determine: The number of base pairs presents between each Alu sequence in the human genome.
Introduction: Alu is a SINE (short interspersed nuclear element) and accounts for 10% of the haploid genome. SINEs are non-autonomous and non-coding transposable elements. They belong to the class of retrotransposons. These sequences are defective predominantly and can be added to the new location of the genome through insertion mutation.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
The human genome contains approximately 106 copies of an Alusequence, one of the best-studied classes of short interspersedelements (SINEs), per haploid genome. Individual Alus share a282-nucleotide consensus sequence followed by a 3'-adeninerichtail region. Given that there are approximately 3 * 109bp per human haploid genome, about how many base pairs arespaced between each Alu sequence?
Given the partial transposons DNA sequence 5’-ACCGTATTCGGT-3’ upstream from the central region, assuming both terminal inverted repeats and flanking direct repeats have 6 base pairs, hypothetically write the transposon structure downstream from the central region.
The DNA-binding domain of each CREB protein subunit recognizes the sequence 5′–TGACGTCA–3′. Due to random chance, how often would you expect this sequence to occur in the human genome, which contains approximately 3 billion base pairs? Actually, only a few doze genes are activated by the CREB protein. Does the value of a few dozen agree with the number of random occurrences expected in the human genome? If the number of random occurrences of the sequence in the human genome is much higher than a few dozen, provide at least one explanation why the CREB protein is not activating more than a few dozen gene Actually, only a few doze genes are activated by the CREB protein. Does the value of a few dozen agree with the number of random occurrences expected in the human genome? If the number of random occurrences of the sequence in the human genome is much higher than a few dozen, provide at least one explanation why the CREB protein is not activating more than a few dozen gene
Chapter 12 Solutions
EBK CONCEPTS OF GENETICS
Ch. 12 - In bacteriophages and bacteria, the DNA is almost...Ch. 12 - After salivary gland cells from Drosophila are...Ch. 12 - If a human nucleus is 10 m in diameter, and it...Ch. 12 - Roberts syndrome is a rare inherited disorder...Ch. 12 - Prob. 2CSCh. 12 - Roberts syndrome is a rare inherited disorder...Ch. 12 - HOW DO WE KNOW? In this chapter, we focused on how...Ch. 12 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 12 - Contrast the size of the single chromosome in...Ch. 12 - Describe the structure of giant polytene...
Ch. 12 - What genetic process is occurring in a puff of a...Ch. 12 - During what genetic process are lampbrush...Ch. 12 - Why might we predict that the organization of...Ch. 12 - Describe the sequence of research findings that...Ch. 12 - Describe the molecular composition and arrangement...Ch. 12 - Describe the transitions that occur as nucleosomes...Ch. 12 - Provide a comprehensive definition of...Ch. 12 - Mammals contain a diploid genome consisting of at...Ch. 12 - Assume that a viral DNA molecule is a 50-m-long...Ch. 12 - How many base pairs are in a molecule of phage T2...Ch. 12 - Examples of histone modifications are acetylation...Ch. 12 - Contrast the structure of SINE and LINE DNA...Ch. 12 - Variable number tandem repeats (VNTRs) are...Ch. 12 - It has been shown that infectious agents such as...Ch. 12 - Cancer can be defined as an abnormal proliferation...Ch. 12 - In a study of Drosophila, two normally active...Ch. 12 - Prob. 21ESPCh. 12 - An article entitled Nucleosome Positioning at the...Ch. 12 - Prob. 23ESPCh. 12 - Following is a diagram of the general structure of...Ch. 12 - Microsatellites are currently exploited as markers...Ch. 12 - At the end of the short arm of human chromosome 16...
Knowledge Booster
Learn more about
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
- How many binary sequences of length n contain at most five 1 digits? The genetic code specifies an amino acid through a sequence of three nucleotides. Each nucleotide can be of one of the four types T, A, C and G, beingrepetitions allowed. How many amino acids can be encoded in this way?And if there are n types. Comparearrow_forwardIf the bandicoot genome is 3.62 x 109 base pairs, and the "highly repetitive DNA" fraction is composed entirely of copies of sequence 5'TGCGTGTGTGC3' and its complement, how many copies of this sequence are present in the bandicoot genome?arrow_forwardin the human gene for the beta chain of hemoglobin, the first 30 nucleotides in the amino acid coding region is represented by the sequence 3'TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? If the DNA duplex for the beta chain of hemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.arrow_forward
- The human RefSeq of the entire first exon of a geneinvolved in Brugada syndrome (a cardiac disordercharacterized by an abnormal electrocardiogram andan increased risk of sudden heart failure) is:5′ CAACGCTTAGGATGTGCGGAGCCT 3′The genomic DNA of four people (1–4), three ofwhom have the disorder, was subjected to singlemolecule sequencing. The following sequences represent all those obtained from each person. Nucleotidesdifferent from the RefSeq are underlined. Individual 1:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGCGGAGACT 3′Individual 2:5′ CAACGCTTAGGATGTGAGGAGCCT 3′Individual 3:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGGCGGAGCCT 3′Individual 4:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGTGGAGCCT 3′a. The first exon of the RefSeq copy of this gene includes the start codon. Write as much of the aminoacid sequence of the encoded protein as possible,indicating the N-to-C polarity.b. Are any of these individuals homozygotes? If so,which person and what allele?c. Is…arrow_forwardEukaryotic genomes are replete with repetitive sequences that make genome assembly from sequencereads difficult. For example, sequences such asCTCTCTCTCT . . . (tandem repeats of the dinucleotide sequence CT) are found at many chromosomallocations, with variable numbers (n) of the CT repeating unit at each location. Scientists can assemblegenomes despite these difficulties by using the pairedend sequencing strategy diagrammed in Fig. 9.9. Inother words, they can make libraries with genomicinserts of defined size, and then sequence both endsof individual clones. Following are 12 DNA sequence reads from sixcloned fragments analyzed in a genome project. 1Aand 1B represent the two end reads from clone 1, 2Aand 2B the two end reads from clone 2, etc. Clones1–4 were obtained from a library in which the genomic inserts are about 2 kb long, while the inserts inclones 5 and 6 are about 4 kb long. All of these sequences have their 5′ ends at the left and their 3′ endsat the right. To simplify…arrow_forwardGiven the fact that 1 fg of DNA = 9.78 * 105base pairs (on average), you can convert the amount of DNA per cell to the lengthof DNA in numbers of base pairs. (a) Calculate the number of basepairs of DNA in the haploid yeast genome. Express your answer inmillions of base pairs (Mb), a standard unit for expressing genomesize. Show your work. (b) How many base pairs per minute weresynthesized during the S phase of these yeast cells?arrow_forward
- What is the length in AA’s of the LilP protein? Assume fMet is NOT CLEAVED. Enter just the number, nothing else! Write out the sequence of the polypeptide in AA: use the three letter notation, e.g. Met-Ser-Pro- A lilP mutant called lilPXS is isolated that produces a truncated polypeptide of only 6 AA in length. Describe a single basepair DNA change that would lead to this truncated version of the protein. Multiple options are possible (100 words max.)arrow_forwardIn the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence: 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? 4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region. 4C. In NOT more than 200 words, explain how eukaryotic RNA synthesized by RNA polymerase II is modified before leaving the nucleus?arrow_forwardIn the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence: 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? 4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.arrow_forward
- In the human genome for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotide in the amino acid coding region is represented by the sequence 3’-TACCACHTGGACTGAGGACTCCTCTTCAGA-5' What is the sequence for the partner strand?arrow_forwardGiven the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs ( what letter changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - AGGCTCGCCTGT - 5' Mutated DNA Sequence: 3' - AGTCTCGCCTGT - 5'arrow_forwardThe consensus sequence of the protospacer adjacent motif (PAM) sequence in bacterial strain A is 5'-NGG-3, whereas in bacterial strain B the consensus PAM sequence is 5'-NGRRN-3: Calculate the frequency at which each of these PAM sequences will occur in DNA using N = any nucleotide and R = purine base. Bacterial strain A: PAM sequence occurs once in every 16 nucleotides Bacterial strain B: PAM sequence occurs once in every 32 nucleotides Bacterial strain A: PAM sequence occurs once in every 32 nucleotides Bacterial strain B: PAM sequence occurs once in every 16 nucleotides Bacterial strain A: PAM sequence occurs once in every 32 nucleotides Bacterial strain B: PAM sequence occurs once in every 32 nucleotides Bacterial strain A: PAM sequence occurs once in every 16 nucleotides Bacterial strain B: PAM sequence occurs once in every 16 nucleotidesarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
Genome Annotation, Sequence Conventions and Reading Frames; Author: Loren Launen;https://www.youtube.com/watch?v=MWvYgGyqVys;License: Standard Youtube License