Given many mutations in a gene, how could you identify deletion mutations?

ANSWER: Deletion mutations typically do not revert, so failure to revert is a hint that a mutation may be a deletion. Failure to produce wild-type recombinants with several other mutations that can recombine with each other is stronger evidence that a mutation is a deletion.

Given a collection of mutations that affect a gene, including several deletion mutations, how would you construct a deletion map?

ANSWER: Recombination tests with pairwise combinations of each of the mutations -- failure to produce wild-type recombinants indicates that two mutations affect the same base pair. This could be due to two point mutations that affect the base pair, a point mutation and a deletion that affect an overlapping DNA sequence, or two deletion mutations that affect an overlapping DNA sequence. However, if two mutations that recombine with each other both fail to recombine with a third mutation, the third mutation is likely to be a deletion that removes the DNA corresponding to the mutant base pairs in each of the other two mutations.

A deletion map of the nadB gene is shown below (data from Cookson et al.1987. J. Bacteriol. 169: 4285).

The black boxes represent the region removed by each deletion and the numbers above represent nadB alleles that cannot repair the indicated deletion. Based upon this deletion map, answer the following questions and explain your answer.

1. Could nad-110 repair nadB164?

   ANSWER: No. Deletion removes interval containing wild-type equivalent of allele 164.

2. Could nad-114 repair nadB95?

   ANSWER: Yes. Deletion does not remove wild-type equivalent of allele 95.

3. Could nad-110 repair nad-114?

   ANSWER: No. Region corresponding to �110 would still be missing.

4. Could nadB172 repair nadB95?

   ANSWER: Yes. Each point mutant contains the wild-type interval for the other allele.

5. Could nadB172 repair nadB62?

   ANSWER: Maybe. Even though both alleles map within the same deltion interval, you do not know whether the mutations in alleles nadB172 and nadB62 affect the same base pair or not. If the mutations affect the same base pair, recombinational repair would not be possible. However, if the mutations affect adjacent base pairs or base pairs that are further apart, recombinational repair is possible -- the frequency of recombinational repair would depend upon how far apart the two mutations are, so if the two mutations are very close this would be a rare event.

6. Why are there so many mutations in some deletion intervals and not in others?

   ANSWER: There are several reasons why the number of mutations that map within a deletion will not be uniform.
   

   • The larger a deletion interval the more targets there are per round of mutagenesis.
   • The nature of the mutagen used will affect the spectrum of mutations recovered, each mutagen will may "hot spots" for mutagenesis which may not be evenly distributed throughout the gene.
   • Not all residues of the gene product are essential for proper structure and/or function. The codons corresponding to the critical residues will be the only mutable sites recoverable as missense mutants. Such critical residues are often unevenly distributed within the gene.

7. You isolate a new mutant that had a phenotype expected for a nadB mutation but could repair all of the known nadB deletion mutations. Suggest two potential explanations for this result?

   ANSWER:
   

   • The mutant may be affected in a gene other than nadB but which confers a NadB^- phenotype when mutated. Other genes in a common pathway or trans-acting regulatory factors for nadB might have this phenotype. In this case the nadB gene in the donor would be wild-type and, hence, would be able to repair all deletion intervals of nadB in the recipient by recombinational repair.
   • The new mutant allele may be located within the nadB gene but maps outside of the end points of the deletion intervals, for example in the same interval as allele 55.

A deletion map of the nadB gene is shown below (data from Cookson et al. 1987. J. Bacteriol.169: 4285). The black boxes represent the region removed by each deletion.

The map position of several nadB mutations was determined by deletion mapping. The results are shown in the following table. Given these results, indicate the map position of the mutations.

ANSWERS:

• nad-171 maps between the endpoints of Del(nad-111) and Del(nad-107).
• nad-172 maps between the endpoints of Del(nad-126) and Del(nad-105).
• nad-173 maps between the endpoints of del-nad-111 and Del(nad-107) (same as nad-171).
• nad-174 maps between the endpoints of del-nad-107 and Del(nad-108).
• nad-118 cannot repair any of the deletions, so it is probably itself a deletion mutation that removes some or all of the DNA that is removed by every deletion.

Proteins encoded by the put operon allow cells to use proline as a sole carbon or nitrogen source. A deletion map of part of the put operon is shown below. (The open bar shows the region deleted in the indicated mutant.)

1. A new Put^- mutant was isolated that can revert to Put⁺ but cannot repair any of these deletions. Based upon these results, what can you infer about the properties of the mutation. Propose a genetic recombination experiment to test your idea. Specify the donor(s) and recipient(s) and how you would select for recombinants.

   ANSWER:

   • The new mutant can revert so it is probably NOT a deletion (i.e. it is probably a point mutant).
   • The new mutant cannot repair any of the deletions so it probably lies within the region spanned by every deletion (that is, the interval including mutations put-838, put-1147, etc).
   • There are a variety of ways you could test your idea. For example, you could test for recombination with other point mutations in the same deletion interval, selecting for repair of the Put^- phenotype -- unless every single mutation affected the exact same base pair, then the new mutation should recombine with some of these point mutations but maybe not all (you could use the new mutant either as donor or recipient in these experiments). Three-factor crosses against a recipient with two point mutations in adjacent intervals would also test this idea (if you had a way to construct the double Put^- mutant -- e.g. a double mutant carrying both put-1055 and put-1019).

2. A second new Put^- mutant was isolated that cannot revert to Put⁺ and cannot repair any of these deletions. Based upon these results, what can you infer about the properties of the mutation. Propose a genetic recombination experiment to test your idea. Specify the donor(s) and recipient(s) and how you would select for recombinants.

   ANSWER:
   

   • The new mutant cannot revert so it probably is a deletion, not a point mutation. (Note that other types of polar mutations including insertion, nonsense, and frameshift mutations CAN revert! Furthermore, not all deletion mutations are polar.)
   • The new mutant cannot repair any of the deletions so it probably includes the region spanned by every deletion (that is, the interval including mutations put-838, put-1147, etc) -- it may remove additional DNA as well, but it probably removes at least some DNA from this deletion interval.
   • One good test would be to look for recombination with other point mutations located in each of the different deletion intervals, selecting for repair of the Put^- phenotype -- if the mutation is a relatively large deletion, it will probably be unable to repair multiple point mutations (again you could use the new mutant either as donor or recipient in these experiments but it is usually a good idea to use the deletion mutant as a recipient because deletions do not undergo true reversion).

3. In the deletion map shown above, some deletion intervals contain many point mutations but some deletion intervals only contain a single point mutation. List three reasons why point mutations may be much rarer in some deletion intervals.

   ANSWER:

   • Some deletion intervals may have mutation hot spots.
     
   • Some deletion intervals may span regions of the gene product that are not very important for proper structure and function.
     
   • Although they are drawn as being the same size (because based on genetic tests it is not usually easy to know whether one interval is larger than another) remember that the amount of DNA deleted may be different from one interval to another -- thus, the "target size" for mutations may be larger for some deletion intervals than others.

Salmonella typhimurium can use ethanolamine as a sole carbon or nitrogen source if vitamin B12 is available. A large number of ethanolamine utilization (eut) mutants have been isolated (Roof and Roth. 1988. J. Bacteriol. 170: 3855-3863). Some of the eut mutants are unable to use ethanolamine as a carbon or nitrogen source (N^-C^-) and some of the eut mutants are unable to use ethanolamine as a carbon source but can still use ethanolamine as a nitrogen source (N⁺C^-). The position of the (N⁺C^-) mutations was determined by mapping against the eut deletions shown in the following figure. The results are shown in the following table (+ indicates that the mutation can repair the deletion and - indicates that the mutation cannot repair the deletion).

1. Indicate the position of the (N⁺C^-) mutations on the deletion map.

2. Some (N^-C^-) point mutations map in the same deletion intervals as the (N⁺C^-) mutations, but most of the (N^-C^-) point mutations map between deletions eut-729 and eut-237. Suggest an explanation for these results.

A new putP mutation was mapped against a set of putP deletion mutations. The region removed by each deletion mutations is indicated by an open box below the putP map. The results indicating whether or not recombinants were obtained are shown to the right of each deletion.

Based on these results, where does the new putP mutation map with respect to the deletion intervals?

ANSWER: The mutation maps in deletion interval #8 because it cannot repair any deletion that removes this interval, but it can repair all the other deletions.

The nadC gene encodes an enzyme involved in NAD biosynthesis (Hughes et al. 1991. Genetics 127: 657-670). The nadC gene maps between the aroP and guaC genes on the S. typhimurium chromosome. Over 100 nadC point mutations were obtained by hydroxylamine mutagenesis. To facilitate construction of a fine structure genetic map of these point mutations, a set of deletions were isolated by selecting for mutants that simulteneously become either aroP nadC or nadC guaC. These deletions were then crossed with nadC point mutations. The results are shown in the table below.

Given these results, draw the best possible deletion map of the nadC mutations.

ANSWER:

A deletion map of the maltose transport operon from Salmonella typhimurium is shown below [adapted from Schneider, E., L. Bishop, E. Schneider, V. Alfandary, and G. F. Ames. 1989. J. Bacteriol. 171: 5860-5865]. The region removed by each deletion is shown by a line below the genes and the corresponding allele number is indicated above each line.

The map position of several mal mutations was determined by mapping against these deletions. The results are shown in the following table. Given these results, indicate the map position for each of the mutations on the deletion map. Indicate precisely which deletion interval the mutations map within. If it is not possible to determine whether a mutation maps in a single interval, be sure to comment on this in your answer and suggest a reason that might cause such a result. [Hint: use a ruler to draw the endpoints of the deletion intervals on the map, then you can identify each deletion interval with a number or letter.]

ANSWER:

Please send comments, suggestions, or questions to smaloy@sciences.sdsu.edu
Last modified October 18, 2004