Chapter 5 Answers
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- Mutagenize a wild type (auxotrophic) strain and screen for mutations that fail to grow on minimal media, but grow well on minimal media supplemented with proline.
- Take mutants #1-#10 and characterize them, based on:
- genetic mapping of the mutants (different locations indicate different genes);
- different response to proline precursors (a different response suggests different genes);
- complementation tests among the mutations (if they complement then they are mutations in different genes).
- If the mutations are in different genes, the F1 progeny would be wild type (able to grow on minimal medium without proline).
- If the mutations are in the same gene, the F1 progeny would NOT be wild type (unable to grow on minimal medium without proline).
- There are many correct answers for this question. Here is one:
Mutant Strain 1 2 3 4 5 Mutant in Loci 1 – No value No value No value No value Mutant in locus (1,2) 2 – – No value No value No value Mutant in locus (1,2) 3 + + – No value No value Mutant in locus (3) 4 + + + – No value Mutant in locus (4) 5 + + – – – Mutant in locus (3 and 4) - The auxotrophic strain is mutant in one gene. This gene has both a HindIII and XhoI site within its sequence, but not an EcoRI Thus, the EcoRI library could contain a restriction fragment with an entire, intact gene, while the two other enzymes would break the gene into two fragments that would not be cloned together as a functional gene.
- E. coli cells do not normally import proteins from their environments, thus none of the Enzyme A proteins would enter the cells to affect a rescue. If the product of Enzyme A was added, then it could rescue the strain, but only if the product could be taken up by the cells.
