The 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?

The 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?

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter19: Genomes And Proteomes

Section: Chapter Questions

Problem 1ITD: Below is a sequence of 540 bases from a genome. What information would you use to find the...

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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?
Given 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?
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
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…
Eukaryotic 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…
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?
If 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?
You 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'
Based on the alignment of these protein sequences, which [pairs] of the genes appear to be most similar to each other? And give reasons why?
Calculate the number of clones required to obtain with a probability of 0.99 a specifi c 5-kb fragment from C. elegans (Table).
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.