Essentials of Genetics - Masteringgenetic
9th Edition
ISBN: 9780134143699
Author: KLUG
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 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?
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 11 Solutions
Essentials of Genetics - Masteringgenetic
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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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
- 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_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?arrow_forwardThe human genome contains approximately 106 copies of an Alusequence, one of the best-studied classes of short interspersed elements(SINEs), per haploid genome. Individual Alu units share a282-nucleotide consensus sequence followed by a 3 @adenine@richtail region [Schmid (1998)]. Given that there are approximately3 * 10^9 base pairs per human haploid genome, about how manybase pairs are spaced between each Alu sequence?arrow_forward
- 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_forwardAlthough DNA transposons are abundant in the genomes of multicellular eukaryotes, class 1 elements usually make up the largest fraction of very large genomes such as those from humans (~2500 Mb), maize (~2500 Mb), and barley (~5000 Mb). Given what you know about class 1 and class 2 elements, what is it about their distinct mechanisms of transposition that would account for this consistent difference in abundance?arrow_forwardIn not more than 200 words, explain how the human genome of 3.4 Gb would be in 2.3 meters long when stretched linearly if this size is fit into a cell of minuscule proportions.arrow_forward
- Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.arrow_forwardIn addition to the standard base-paired helical structures, DNA can form X-shaped hairpin structures called cruciforms in which most bases are involved in Watson–Crick pairs. Such structures tend to occur at sequences with inverted repeats. Draw the cruciform structure formed by the DNA sequence TCAAGTCCACGGTGGACTTGC.arrow_forwardThe E. coli genome contains 1009 Chi sequences. Do these sequences occur at random, and, if not, how much more or less frequently than random do they occur?arrow_forward
- Approximately what portion of the human genome is composed of repeat sequences?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_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_forward
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