Concept explainers
In an Ames test using
a. What is the reason for the different experimental results described?
b. Is the test compound still considered to be a potential mutagen? Explain why or why not.
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Genetic Analysis: An Integrated Approach (3rd Edition)
- After performing analysis of whole genome sequencing from a candidate that grew on SC-H in the initial screen, you are unable to identify a causative mutation. Which of the following options is a likely explanation for why this mutant passed the initial screen? Select all that apply. A. The mutation is a truncation mutation and Illumina sequencing only provides information for missense mutations. B. One of the missense mutations you analyzed is actually causative, but the factor has not been previously linked to transcription. C. The causative mutation lies in a sequence coverage gap. D. The mutation is in a hypothetical protein and uncategorized genes cannot be sequenced by Illumina technology.arrow_forwardYou can carry out matings between an Hfr and F strain by mixing the two cell types in a small patch on a plate and then replica plating to selective medium. This methodology was used to screen hundreds of different cells for a recombination-deficient recA - mutant. Why is this an assay for RecA function? Would you be screening for a recA mutation in the F or Hfr strain using this protocol? Explain.arrow_forwardIf you use the pUC18 vector to clone in the MCS region, predict the following: a) Do bacteria that are blue in color have a cloned insert? b)Do bacteria that are white in color have a cloned insert? c) If you were to grow these cells on Chloramphenicol (an antibiotic), would the bacteria with the pUC plasmid grow? Why or Why not?arrow_forward
- Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. (a) The dnaE gene encodes the a subunit of DNA polymerase III. What effect is expected from a mutation in this gene? How could the mutant strain be maintained? (b) The dnaQ gene encodes the e subunit of DNA polymerase. What effect is expected from a mutation in this gene?arrow_forwardA small DNA molecule was cleaved with several different restriction nucleases, and the size of each fragment was determined by gel electrophoresis.The following data were obtained. (a) Is the original molecule linear or circular?(b) Draw a map of restriction sites (showing distances between sites) that isconsistent with the data given.(c) How many additional maps are compatible with the data?(d) What would have to be done to locate the cleavage sites unambiguouslywith respect to each other?arrow_forwarda)What two restriction sites are you going to use to clone your PCR product into the pL4440 plasmid? What are their DNA sequence? b) State the primer sequence you will use to amplify the F27C1.7 gene ready to be cloned into the pL4440 plasmid? c) How would you go about cloning this amplified DNA into pL4440? Using your knowledge of cloning list 5 important aspects of the method.arrow_forward
- Xeroderma pigmentosum is a genetic disease caused by an error in the nucleotide excision repair process that fixes damage to DNA by ultraviolet light. Studies have shown that it can result from mutations in any one of seven genes. What can you infer from this finding? A) There are seven genes that produce the same protein B) These seven genes are the most easily damaged by ultraviolet light. C) There are seven enzymes involved in the nucleotide excision repair process. D) These mutations have resulted from translocation of gene segments.arrow_forwardLet’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.arrow_forwardYou are using the restriction enzyme HAEIII to digest different samples of the taster gene isolated from cheek cells of different people and amplified by PCR. When viewing the bands on the electrophoresis gel, one would expect that a taster (homozygote) would have---------band(s), whereas a carrier (heterozygote) would show--------band(s), and a non-taster would show------band(s).arrow_forward
- In the Avery, McLeod, McCarty Experiment where supernatant from heat killed, virulent S Strain pneumonia solutions were added to non-virulent R Strain pneumonia cell cultures and allowed to grow in liquid media (i.e., broth). In tubes where DNAase was added to the supernatant prior to cell culture, what was the observed effect when plating and growing the S. pneumonia cells to solid media? a All RNA was degraded and Transformation of the R Strain to S Strain occurred. b All Protein was degraded and Transformation of the R Strain to S Strain occurred. c All DNA was degraded and Transformation of the R Strain to S Strain did not occur. d All RNA was degraded and no Transformation occurred indicating RNA is the molecule of Transformation inheritance e None of the above are truearrow_forwardKnowing that you are using HindIII and EcoRI to cut your plasmids, and that those two enzymes cut within the MCS, use the map of pUC19 provided below to compute: What will be the sizes of the 2 restriction fragments if NO insert is present in pUC19? What will be the sizes of the 2 restriction fragments if the approximately 317 bp RT-PCR product (insert from WT satC dimer) was ligated successfully into the SmaI site? What will be the sizes of the 2 restriction fragments if TWO approximately 317 bp RT-PCR products (2 ligated inserts from WT satC dimer) were ligated into the SmaI site?arrow_forwardYou are engineering a new vector that contains a screenable marker that can be used for blue/white screening of successful clones. For each site (1, 2, and 3) on the cloning vector below, describe why it would or would not be a good place for you to put the polylinker to facilitate blue/white screening. You can assume that the polylinker itself will not interfere with coding sequence in that region. In other words, the polylinker length will be a multiple of 3 nucleotides, will not contain a stop codon, and any amino acids translated will not affect the activity of the protein in that region. The arrows indicate the direction of transcription for the gene.arrow_forward
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