Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 11, Problem 20PDQ
The human genome contains approximately 106 copies of an Alu sequence, one of the best-studied classes of short interspersed elements (SINEs), per haploid genome. Individual Alus share a 282–
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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?
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
The human genome contains thousands of sequences known as small open reading frames, some of which encode proteins of about 30 amino acids. What is the minimum number of nucleotides required to encode such a protein?
Chapter 11 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 11 - CASE STUDY | Art inspires learning A genetics...Ch. 11 - Prob. 2CSCh. 11 - Prob. 3CSCh. 11 -
HOW DO WE KNOW?
1. In this chapter, we focused on...Ch. 11 - Review the Chapter Concepts list on p. 199. These...Ch. 11 - Prob. 3PDQCh. 11 - Describe how giant polytene chromosomes are...Ch. 11 - What genetic process is occurring in a puff of a...Ch. 11 - Prob. 6PDQCh. 11 - Why might we predict that the organization of...
Ch. 11 -
8. Describe the sequence of research findings...Ch. 11 - Prob. 9PDQCh. 11 - Prob. 10PDQCh. 11 - Provide a comprehensive definition of...Ch. 11 - Prob. 12PDQCh. 11 - Define satellite DNA. Describe where it is found...Ch. 11 - Prob. 14PDQCh. 11 -
15. Mammals contain a diploid genome consisting...Ch. 11 - Prob. 16PDQCh. 11 - Prob. 17PDQCh. 11 - Prob. 18PDQCh. 11 - Prob. 19PDQCh. 11 - The human genome contains approximately 106 copies...Ch. 11 - Prob. 21PDQ
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- The genomes of most multicellular eukaryotes encode~25,000 genes, yet their proteomes contain over 200,000proteins. Propose two processes that, taken together, account for this discrepancyarrow_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_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_forward
- 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.arrow_forwardYou would like to add a nuclear localization sequence (NLS) of Lys-Lys-Lys-Arg-Lys to a protein that is usually found in the cytoplasm of a yeast cell. To accomplish this, you introduce the nucleotide sequence encoding the NLS into the gene that encodes the cytoplasmic protein of interest. a. What is the size of the nucleotide insert that will encode the NLS? Briefly explain. 5' 3' b. Below is a diagram of the gene encoding the cytoplasmic protein of interest in the yeast genome. If your goal is to put the NLS at the carboxyl (C) terminus of the protein, at which location (A-E) should the NLS be inserted? Briefly explain. A TATAA ATATT promoter +1 B ATG TAC D TAA ATT stop codon E 3' 5'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_forward
- Given 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 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_forwardA 210-bp sequence within the CFTR gene on human chromosome 7 is shown below. The three bold underlined nucleotides are deleted in a common cystic fibrosis (CF) mutation, removing a phenylalanine amino acid from the CFTR protein. 1 AGAGGGTAAA ATTAAGCACA GTGGAAGAAT TTCATTCTGT TCTCAGTTTT 51 CCTGGATTAT GCCTGGCACC ATTAAAGAAA ATATCATCTT TGGTGTTTCC 101 TATGATGAAT ATAGATACAG AAGCGTCATC AAAGCATGCC AACTAGAAGA 151 GGTAAGAAAC TATGTGAAAA CTTTTTGATT ATGCATATGA ACCCTTCACA 201 CTACCCAAAT PCR primers have been designed to amplify fragments within this sequence: Forward: GGATTATGCCTGGCACCATT Reverse: AGTGTGAAGGGTTCATATGC DNA from a CF patient is tested with a PCR assay using a pair of these primers, and the PCR product is found to be 3 bp shorter than that expected from the sequence shown above. What length PCR products (in bp) would you expect in the mother of the CF patient? A. 95 and 92 B. 149 C. 133 and 130 D. 149 and 146 E. 146arrow_forward
- In relation to central dogma of molecular biology answer the following questions: The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference?arrow_forwardAs part of a project investigating potential new drug targets in the fight against malaria, you are seeking to clone the gene for a protein from the malaria parasite Plasmodium falciparum. You wish to express this protein in BL21 (DE3) cells, a standard laboratory strain of Escherichia coli. After purification of your protein, you run an SDS-PAGE gel and notice that the major band has lower molecular weight than expected, so you fear you are getting a truncated version. 1. What technique could you use to confirm that you are obtaining a shortened version of your intended protein? explainarrow_forwardAs part of a project investigating potential new drug targets in the fight against malaria, you are seeking to clone the gene for a protein from the malaria parasite Plasmodium falciparum. You wish to express this protein in BL21 (DE3) cells, a standard laboratory strain of Escherichia coli. After purification of your protein, you run an SDS-PAGE gel and notice that the major band has lower molecular weight than expected, so you fear you are getting a truncated version. (a) Give TWO possible causes of your protein becoming truncated. explainarrow_forward
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Genome Annotation, Sequence Conventions and Reading Frames; Author: Loren Launen;https://www.youtube.com/watch?v=MWvYgGyqVys;License: Standard Youtube License