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 11, Problem 16P
A strain of
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A scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below.
Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000.
Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000.
The scientist determines the nucleotide sequence of the coding strand of the GS1 gene from strain A. It is identical to the GS1 sequence from the wild type gene except for a single change occurring approximately 1⁄3 of the way into the GS1 open reading frame. A small region of the GS1 sequence (including the site where the mutation occurs) from the wild type and mutant strains is shown below.
Wild type TGTCCTCGGCCACAAGTTCTCTATC
Strain A TGTCCTCGGCCACTAGTTCTCTATC
How has this…
A scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below.
Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000.
Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000.
What type(s) of mutation may have occurred in the GS1 gene in strain A?
how will mutation (with - ) will affect E.coli grown in lactose medium.
what is the implication of the following genotypes?
i+ p+ o+ z- y+
Chapter 11 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 11 - 11.1 Identify two general ways chemical mutagens...Ch. 11 - 11.2 Nitrous acid and (BU) alter DNA by different...Ch. 11 - 11.3 What is the difference between a transition...Ch. 11 - What is the difference between a synonymous...Ch. 11 - 11.5 UV irradiation causes damage to bacterial...Ch. 11 - Ultraviolet (UV) radiation is mutagenic.
What...Ch. 11 - Researchers interested in studying mutation and...Ch. 11 - The effect of base - pair substitution mutations...Ch. 11 - Describe the purpose of the Ames test. How are...Ch. 11 - 11.10 In numerous population studies of...
Ch. 11 - 11.11 Two different mutations are identified in a...Ch. 11 - What is the phenotype effect of inserting a Ds...Ch. 11 - 11.13 Answer the following questions concerning...Ch. 11 - Several types of mutation are identified and...Ch. 11 - 11.15 A sample of the bacterium is exposed to...Ch. 11 - 11.16 A strain of is identified as having a null...Ch. 11 - Describe the difference between DNA transposons...Ch. 11 - 11.18 How are flanking direct repeat sequences...Ch. 11 - 11.19 Using the adeninethymine base pair in this...Ch. 11 - The partial amino acid sequence of a wild-type...Ch. 11 - Prob. 21PCh. 11 - 11.22 Many human genes are known to have homologs...Ch. 11 - The fluctuation test performed by Luria and...Ch. 11 - In this chapter, three features of genes or of DNA...Ch. 11 - Briefly compare the production of DNA double -...Ch. 11 - During mismatch repair, why is it necessary to...Ch. 11 - 11.27 Following the spill of a mixture of...Ch. 11 - 11.28 In an Ames test using Salmonella bacteria a...Ch. 11 - A wild - type culture of haploid yeast is exposed...Ch. 11 - A fragment of a wild - type polypeptide is...Ch. 11 - Prob. 31PCh. 11 - Alkaptonuria is a human autosomal recessive...Ch. 11 - 11.33 In an experiment employing the methods of...Ch. 11 - Using your knowledge of DNA repair pathways choose...Ch. 11 - 11.35 Ataxia telangiectasia is a human inherited...Ch. 11 - A geneticist searching for mutations uses the...Ch. 11 - 11.37 In a mousebreeding experiment a new mutation...Ch. 11 - 11.38 Considering the Dumbo mutation in a Problem,...Ch. 11 - 11.39 Thinking back to the discussion of...Ch. 11 - 11.40 Common baker’s yeast () is normally grown at...Ch. 11 - 11.41 The two gels illustrated below contain...
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- A scientist is researching GS1, an enzyme with a relative molecular mass (Mr) of 78,000 present in a bacterium. The scientist has isolated two mutant strains of the bacterium as described below. Strain A: In this strain the GS1 protein is completely non-functional. Analysis of strain A shows that it produces a shortened GS1 protein with an Mr of only 38,000. Strain B: This produces functional GS1, but the Kcat is somewhat reduced. Analysis shows it produces a lengthened form of GS1, with an Mr of about 86,000. Sequencing of the GS1 gene from strain B shows that it is identical to the wild type gene except for a single alteration (the replacement of one nucleotide by another). How might this account for the features of the GS1 protein produced by strain B?arrow_forwardGiven the following phenotypes, explain how the mutation (identified by a (-) superscript) will affect the E. coli grown in lactose medium. l+p+o+z-y+ i-p+o+z+y+ l+p+o-z+y+ l+p-o+z+y+arrow_forward1. a)What would happen if the aminoacyl tRNA synthetase responsible for charging alanine tRNAs also charged methionine tRNAs with alanine? b)What would happen if an individual was homozygous for mutant alleles of the gene encoding the aminoacyl tRNA synthetase responsible for charging leucine tRNAs?arrow_forward
- A glycine residue is in position 210 of the tryptophan synthetase enzyme of wild-type E. coli. If the codon specifying glycine is GGA, how many single-base substitutions will result in an amino acid substitution at position 210? What are they? How many will result if the wild-type codon is GGU?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 studying the tryptophan synthetase gene that Yanofsky also examined to determine the relationship between the nucleotide sequence and the amino acid sequence of the gene. Yanofsky found a large number of mutations that affected the tryptophan synthetase gene. A) If you took this mutant E. Coli line (that has an Arginine at this location) and exposed it to a mutagen that could potentially change bases, what are the second mutations you would most likely discover that would restore the activity of the tryptophan synthetase gene and where would it be located? B) Most of the mutations that Yanofsky recovered were missense mutations. However, Yanofsky also recovered a nonsense mutation that changed amino acid number 15 into a stop codon. This codon normally encodes Lysine. Does the recovery of this mutation support the hypothesis that this Lysine residue is critical in the function of the tryptophan synthetase protein?arrow_forward
- Cystic fibrosis (CF) is an inherited disorder caused by different types of mutations, many of which prevent ions from moving across cell membranes. Normally there are channel proteins that allow passage of the ions, but in patients with one kind of CF these proteins seem odd. Closer examination shows that these proteins display the correct amino acid sequence. However, they fail to do their job. A) Given that the primary structure of the protein is correct, what can you infer about the DNA sequence for the gene coding this protein on this patient, is there a mutation? Explain. B) Why is the primary structure insufficient to guarantee the proper function of the protein?arrow_forwardNeurospora crassa can synthesize the amino acid arginine from a precursor as follows: precursor ornithine citrulline arginine George Beadle and Edward Tatum identified various mutants (arg mutants) unable to synthesize arginine from this precursor. One such class (Class I) could synthesize arginine if either ornithine or citrulline was supplied. A second mutant class (Class II) could synthesize arginine if citrulline was added, but not if ornithine was added. A third class (Class III) could not synthesize arginine from either ornithine or citrulline. Which class or classes of arg mutants will grow on complete media?arrow_forwardConsider a stretch of DNA (a hypothetical gene) that has the sequence 5’ ATG-CTA-TCA-TGG-TTC-TAA 3’ A) Transcribe and translate this gene using the genetic code table. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations. B) Now, our hypothetical gene has undergone a mutation. The mutant sequence is....3’ TAC-GAT-AGT-ACC-AAT-ATT 5’5’ ATG-CTA-TCA-TGG-TTA-TAA 3’ Transcribe and translate the mutant sequence. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations. C) Indicate the type of mutation (nonsense, missense, silent, or frame shift) present. D) How severe of a consequence will this mutation likely be in terms of protein function (none, mild, moderate or severe)? Why?arrow_forward
- We have two specific strains of E. coli that have shown horizontal gene transfer (HGT) when mixed. To experimentally determine the method of HGT that is happening, the following conditions are set up in different tubes of culture media: A) Donor and recipient strain mixed together (control - no treatment). B) Donor and recipient strains mixed together, DNase added (can digest DNA in solution, not within cells).C) Special tube containing a membrane filter (with pores that allow DNA and viruses to pass through, but not bacterial cells) that separates two compartments. Donor strain is added on one side, the recipient strain on the other (they are separated by the filter).D) Donor and recipient strains mixed together, with chemical that inactivates viruses (chemical affects bacteriophages in solution so they are unable to attach to cells). The results: Tubes A, B, and D: HGT was observed. Tube C: HGT was NOT observed. Based on this, which type of HGT was occurring? Conjugation,…arrow_forwardXeroderma 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_forwardWhat are the similarities and differences in the synthesis of DNA in the leading and lagging strands in E. coli?arrow_forward
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