Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 16, Problem 7P
You have sequenced a
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Assume 2x108 reads of 75 bps long are obtained from a next-generation sequencing experiment to sequence a human genome. Suppose the length of the human genome is 3x109 bps. What is the depth (i.e., coverage) of the sequencing?
Draw the structure of a dideoxynucleotide that would be used for DNA sequencing, and explain why they result in chain termination. Write out the sequence of the first 20 nucleotides for the gene shown in the sequencing gel below (remember to start at the bottom of the gel and work upward, from smallest to largest fragments).
You have sequenced the genome of the bacterium Salmonella typhimurium and find
a protein that is 100 percent identical to a protein in the bacterium Escherichia coli.
When you compare nucleotide sequences of the S. typhimurium and E. coli genes,
you find that their nucleotide sequences are only 87 percent identical. How would
you interpret the observations? Please make sure to select ALL correct answer
options.
Because genetic code is redundant, changes in the DNA nucleotide sequence
can occur without change to its encoded protein.
Due to the flexibility in the third positions of most codons, the DNA sequence
can accumulate changes without affecting protein structure.
Natural selection will eliminate many deleterious amino acid changes. This will
reduce the rate of change in the amino acid sequence and lead to sequence
conservation of the proteins.
Protein sequences are expected to evolve and diverge more slowly than the
genes that encode them.
Chapter 16 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 16 - You have discovered a new species of Archaea from...Ch. 16 - 16.2 Repetitive DNA poses problems for genome...Ch. 16 - 16.3 When the whole-genome shotgun sequence of the...Ch. 16 - How do cDNA sequences facilitate gene annotation?...Ch. 16 - 16.5 How do comparisons between genomes of related...Ch. 16 - 16.6 You are designing algorithms for the...Ch. 16 - 16.7 You have sequenced a region of the Bacillus...Ch. 16 - You have just obtained 100-kb of genomic sequence...Ch. 16 - 16.9 The human genome contains a large number of...Ch. 16 - Based on the tree of life in Figure 16.12, would...
Ch. 16 - 16.11 When comparing genes from two sequenced...Ch. 16 - 16.12 What is a reference genome? How can it be...Ch. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - 16.16 Consider the phylogenetic tree below with...Ch. 16 - You have isolated a gene that is important for the...Ch. 16 - 16.18 When the human genome is examined, the...Ch. 16 - Symbiodinium minutum is a dinoflagellate with a...Ch. 16 - Substantial fractions of the genomes of many...Ch. 16 - 16.21 A modification of the system, called the ...Ch. 16 - 16.22 A substantial fraction of almost every...Ch. 16 - 16.23 In the globin gene family shown in Figure ,...Ch. 16 - You are studying similarities and differences in...Ch. 16 - In conducting the study described in Problem 24,...Ch. 16 - Prob. 26PCh. 16 - Prob. 27PCh. 16 - Prob. 28PCh. 16 - Prob. 29P
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- If you had the RNA sequence below: 5'UUUGGAG 3' and you were going to make a piece of DNA that would be a complement to it, what would the DNA sequence be? 5' 3' What 12-nucleotide primer would you use in the PCR technique when you want to amplify a gene whose end is as follows: 3' CGGCTCGACAAGGTG5' ? 5' 3'arrow_forwardNucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?arrow_forwardIn the practical you have been analysing a human genomic library. You know from your calculations that only a small proportion of the human genome is represented, even when the entire class results are considered. Therefore, the chance of finding a particular single-copy gene in your library is very small. Outline a strategy for constructing a genomic DNA library more representative of the entire human genome. You will need to consider alternative vectors and the efficiency of transformation of the bacterial cells.arrow_forward
- You obtain the DNA sequence of a mutant of a 2-kb gene in which you are interested and it shows base differences at three positions, all in different codons. One is a silent change, but the other two are missense changes (they encode new amino acids). How would you demonstrate that these changes are real mutations and not sequencing errors? (Assume that sequencing is about 99.9 percent accurate.)arrow_forwardYou were going to sequence a rice DNA fragment whose sequence was only know at one end, as shown below. 5’ AAACGATCGAGTCGCATCCAAAATCGATACCC—unknown region 3’ TTTGCTAGCTCTGCGTAGGTTTTAGCTATGGG—unknown region After several tries, you obtained a beautiful sequencing image as shown here: The worked out well partially because you had designed a primer for sequencing the unknown region according to the following guideline: Tm is 55 – 60°C. Ensures primer had a appropriate melting temperature for PCR ans sequencing. The GC content of the primer is the same as the genome/template (rice = 60%, human/Drosophila = 45-50%). A same nucleotide cannot be more than 2 in a row, e.g. CCC, GGGGG, AAA. The secondary structure of the primer must be none or weak. No primer dimers (The primer anneals to itself). 3’ end is the most important: it should not end in A, preferably ends in GG, GC, CG or CC This website can help you design the primer: http://www.oligoevaluator.com/OligoCalcServlet…arrow_forwardAbout 60% of the base pairs in the human genome are AT. If the human genome has 3.2 billion base pairs of DNA, about how many times will the following restriction sites be present? a. BamHI (recognition sequence is 5′–GGATCC–3′) b. EcoRI (recognition sequence is 5′–GAATTC–3′) c. HaeIII (recognition sequence is 5′–GGCC–3′)arrow_forward
- In order to target a specific region of genomic DNA with CRISPR, researchers must include a guide RNA containing a 20-basepair long spacer sequence that matches the DNA sequence at the target site. (i) How many possible guide RNA spacer sequences are there? (ii) One of the possible risks of genetic engineering methods is “off-target” editing, where a modification of the genome occurs in a part of the genome other than the target site. Imagine you design a 20-basepair guide RNA spacer sequence to target a specific portion of the Zebrafish genome, which is 1.7 billion nucleotides long. Assuming all nucleotides are equally common, estimate the probability that your spacer sequence occurs in at least one other position in the Zebrafish genome.arrow_forwardBelow 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_forwardDescribe the outcome of a chain-terminator sequencing procedure in which (a) too little ddNTP is added or (b) too much ddNTP is added.arrow_forward
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