MCB 421 HOMEWORK #2 ANSWERS
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1. The Salmonella typhimurium strain TR248 is auxotrophic for both histidine and cysteine due to a mutation in the hisC gene and a mutation in the cysA gene. His+ revertants are found at a frequency of 1 per 107 cells. Cys+ revertants are also found with a frequency of 1 per 107 cells. a.) How would you select for His+ revertants only or Cys+ revertants only? (What kind of medium would you plate the cells on?) ANSWER: Plate greater than 107 cells on a minimal plate without His (for His+ revertants) or without Cys (for Cys+). Any cells growing on the plate are revertants. b.) At what frequency would you expect to find revertants that are both His+ and Cys+? How could you directly select for such double revertants? ANSWER: If the reversions are the result of independent events, then the frequency will be the product of the two independent frequencies, i.e. 10 -7 x 10-7 equals one in 1014 cells. c.) His+ Cys+ revertants are actually found at a frequency of 1 per 108 cells. Propose an explanation for this result. ANSWER: The higher than expected frequency tells us that reversion at both loci must occur through a single event. One explanation for this is that the two mutants were both nonsense mutants with the same introduced stop codon, and you have isolated a mutation of a tRNA gene to become a suppressor tRNA for that codon. Thus, one mutation (in a tRNA gene) can revert both phenotypes. Everybody gets credit for this question since we did not get to tRNA suppressors in lecture. d.) Describe a simple genetic experiment to test your hypothesis. ANSWER: Try growing defective phage on your double revertant strain. You should use phage that have lethal defects due to nonsense mutations. If the phage are able to plaque your strain, you have not only demonstrated the presence of a nonsense suppressor but also specified which type.
MCB 421 HOMEWORK #2 ANSWERS
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2. In the following table, briefly diagram or indicate the common properties of each type of mutation. ANSWER:
Mutation effect on DNA Missense base substitution Nonsense base substitution resulting in a stop codon Frameshift insertion or deletion of 1 or 2 base pairs Deletion loss of multiple base pairs Insertion addition multiple pairs of base
effect on Protein
substituted amino acid
truncated polypeptide; inactive protein
altered amino acid sequence downstream of mutation; usually truncated polypeptide; inactive protein null
usually absent; usually inactive protein
may insert extra amino acid (if in frame) or cause premature truncation; usually inactive protein usually null; if insertion is in frame and doesn’t cause premature termination and is in a permissive site, sometimes protein remains functional
effect Phenotype
on
may result in loss of function or conditional phenotype
usually null
null
MCB 421 HOMEWORK #2 ANSWERS
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3. An elegant genetic experiment to determine the number of bases required to code for each amino acid took advantage of a large collection of frameshift mutations in the rII gene of phage T4 [Crick, F., L. Barnett, S. Brenner, and R. Watts-Tobin. 1961. Nature 192: 1227-1232]. The wild-type DNA and amino acid sequence corresponding to the first portion of the rII gene are shown below.
bp 1 30
ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC Met tyr asn ile lys cys leu thr lys asn
31
60 GAA CAA GCT GAA ATT GTT AAA CTG TAT TCA glu gln ala glu ile val lys leu tyr ser AGT GGT AAT TAC ACC CAA CAG GAA TTG GCT ser gly asn tyr thr gln gln glu leu ala
61
Four frameshift mutations isolated were FCO (an insertion of an A at base 50), FC1 (a deletion of the A at base 32), FC40 (an insertion of an A at base 60), and FC88 (a deletion of the C at base 75). a.) Write out the amino acid sequence for the double mutant FCO FC1. ANSWER: AAA AAC GAc aag ctg aaa ttg tta aAA CTG Lys Asn Asp Lys Leu Lys Leu Leu Lys Leu b.) Write out the amino acid sequence for the double mutant FC1 FC40. ANSWER: AAA AAC GAc aag ctg aaa ttg tta aac tgt att caa AGT Lys Asn Asp Lys Leu Lys Leu Leu Asn Cys Ile Gln Ser c.) Write out the amino acid sequence for the double mutant FCO FC88. ANSWER: GTT Aaa act gta ttc aag tgg taa tta cac CAA Val Lys Thr Val Phe Lys Trp STOP d.) Write out the amino acid sequence for the double mutant FC40 FC88. ANSWER: TCA aag tgg taa tta cac CAA Ser Lys Trp STOP
MCB 421 HOMEWORK #2 ANSWERS
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e.) The double mutants FCO FC1 and FC1 FC40 both produce functional rII protein, but FCO FC88 is inactive. What does this indicate about the structure and function of the wild-type rII protein? ANSWER: The N-terminus and the C-terminus of the wild-type rII protein are essential for function. The region in the middle of the protein, between base 30 and 75, does not appear to be important, as frameshift mutations do not alter protein function. In the double mutant FCO FC88, there is a premature stop codon that results in a truncated (non-functional) protein. For both double mutants FCO FC1 and FC1 FC40, the correct reading frame is restored at base 50 and base 61, respectively, resulting in a mutant rII protein with a functional C-terminus. f.) Would the double mutant FCO, FC40 produce a functional rII protein? ANSWER: GTT Aaa act gta ttc aaa gtg gta att aca ccc aac agg Val Lys Thr Val Phe Lys Val Val Ile Thr Pro Asn Arg aat tgg ct Asn Trp The protein would not be active because of the frameshift caused by the two base insertions.
4). In 1977, F. Sanger and his colleagues published the first DNA sequence of an entire genome. They determined the sequence of the phage PhiX 174 (Nature 265:687-695). One of the astonishing findings of the analysis was that the DNA sequence encoding the D gene also contained the entire coding sequence of the E gene. How could this be explained, based on your knowledge of the genetic code and protein synthesis? The coding sequences overlap so the proteins must be translated in different reading frames. From the data we can’t tell if translation is from the 2 (of the 3) different reading frames of the same mRNA or from opposite directions on the complementary mRNA transcripts. What would you predict would happen if you isolated a substitution mutation in the D gene that knocked out activity of the D protein?
MCB 421 HOMEWORK #2 ANSWERS
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Most likely the mutation would also cause a codon change in the E gene reading frame too. This would likely lead to a mutant E protein. From the data presented we can’t tell if the D mutant would have a phenotype.
Homework assignment #2 is due at the beginning of class on Wednesday, September 10th. Please note the “rules” for homework in the course syllabus. You must show your work for credit. If you have any questions please come see us during office hours or discussion section. The following questions will not be graded but will give you additional practice solving problems like those you may encounter on exams. The answers to these additional problems can be found under “sample problems” in the course website. The best way to benefit from the additional problems is to try to solve them on your own, then to consult the answers on the web to check your answers. If you are unsure of any terminology, please see the Microbial Genetics glossary on the course web site. More sample problems and solutions are available on the course web site.
1. The antibiotic rifampicin inhibits RNA polymerase. Propose a genetic experiment designed to learn which subunit of the RNA polymerase is the binding site (target) for rifampicin. 2. When grown on MacConkey-Lactose plates, lac+ cells from red colonies and lac cells form white colonies. How could you screen for strains with a mutator phenotype using MacConkey-Lactose plates? [Hint: you would want to start with a lac mutant. Explain why.] 3. How could you isolate temperature sensitive lethal mutants? [Describe the approach you would use and how you would identify the temperature sensitive lethal mutants.] 4. Mutations in genes that repair DNA damage often result in a “mutator” phenotype. Such mutator strains accumulate mutations at a much higher frequency than that observed in wild-type cells. a. Some investigators do not like to use mutator strains to isolate mutations in genes within the bacterial chromosome. Instead they use mutator strains to isolate mutations in phage chromosomes or in plasmids. Suggest a reason for the preference. If you did use a mutator strain, what would be a high priority for the next step in your project? (You may not know how to do what should be done, but what would you propose to do?) b. You grow phage lambda lytically on a wild-type strain and on a mutD strain. You then check the phage particles from each infection for phage mutants that affect the morphology of the phage plaques on normal bacteria. Which host strain would give you the highest yield of mutants? Why? c. Often, investigators grow multiple independent cultures inoculated from inoculated from independent colonies of mutator strains for mutagenesis. Why?