Genetics of Bacteria Practice Test

Isolation of Bacterial Mutants Questions

1. Approximately 10⁸ E. coli cells of a triple auxotrophic strain (arg^–lys^–ser^–) are plated on complete medium to form a bacterial lawn. Replica plates are prepared containing minimal medium supplemented by the amino acids arginine, lysine, and serine. What is the genotype of the colonies that appear on the replica plates?



2. Two triple auxotrophic bacterial strains are allowed to conjugate in broth, diluted, and plated onto complete agar (master plate). Replica plates containing various supplements are then made from the master. From the position of each clone and the type of media on which it is found, determine its genotype.



Replica plates: Each dish contains minimal medium plus the supplements shown at the bottom.



3. A bacterial strain unable to synthesize methionine (met^–) is transduced by a phage from a bacterial strain unable to synthesize isoleucine (ile^–). The broth culture is diluted and plated on minimal medium supplemented with isoleucine. An equivalent amount of diluted broth culture is plated on minimal medium. Eighteen clones appeared on the minimal plates and 360 on the isoleucine plates. Calculate the standardized recombination ratio.
4. DNA damage (mutation) is an essential initiation event for a cell to transform into a cancer ous state, but it is not the only event causing cancer (see Chapter 11). Therefore, DNA-damaging agents (mutagens) are only potential carcinogens (agents causing cancer). Most chemical carcinogens are not biologically active in their original form; they must first be metabolized to carcinogenic metabolites. The Ames test (named after the inventor, Bruce Ames) is one of the standard tests for a quantitative estimate of the mutagenic potency of a chemical. This test employs an auxotrophic strain of Salmonella typhimurium that cannot make the amino acid histidine (his^–). To increase the sensitivity of the tester strain (1) it carries a mutation that makes the cell envelope more permeable to allow penetration of the test chemicals, (2) its capacity for excision repair is eliminated so that most of the primary lesions remain unhealed, and (3) a genetic element that makes DNA replication more error prone is introduced via a plasmid. Rat liver extract is added to a minimal medium culture plate coated with a thin layer of these bacteria. A disk of filter paper is impregnanted with the chemical to be tested; the paper is placed in the center of the plate. After 2 days of incubation, the number of colonies are counted.   (a) What events are being scored by the colony counts?   (b) Why was mammalian liver extract added to the test?   (c) Diagram the expected distribution of colonies on a plate containing a known carcinogen. Explain why this distribution develops.   (d) Suppose that the test chemical (e.g., nitrosoguanidine) is mixed with the bacteria prior to plating at two dosages (low and high). A control is run simultaneously with these two doses. Diagram the expected distribution of colonies on these three plates.

Bacterial DNA Replication and Cell Division Questions

5. Several lines of evidence suggest that the circular chromosome of E. coli has two replication forks. The length of one whole unreplicated chromosome is 1300 μm (about 500 times longer than the E. coli cell). There are ten base pairs per one complete turn of the DNA double helix, equivalent to 34Å or 3.4 × 10 ^–3μm.   (a) How many nucleotide base pairs are in the E. coli DNA complement or genome?   (b) If the E. coli genome is replicated in 40 min at 37°C by two replicating forks, how many revolutions per minute (rpm) must the parental double helix make to allow separation of its complementary nucleotide strands during replication?

Regulation of Bacterial Gene Activity Questions

6. In addition to the lacI⁺ allele, producing repressor for the lactose system in E. coli and the constitutive lacI^– allele, a third allele I^s has been found, the product of which is unable to combine with the inducer (lactose). Hence, the repressor ("superrepressor") made by Is remains unbound and free to influence the operator locus.   (a) Order the three alleles of the lacI locus in descending order of dominance according to their ability to influence the lactose operator.   (b) Order the four alleles of the promoter locus (p) in descending order of dominance according to their ability to bind RNA polymerase.   (c) Using + for production and 0 for nonproduction of the enzymes permease (P) and β- galactosidase (β-gal), complete the following table. (Hint: See Example 10.7.)



7. 
8. In the lactose operon of E. coli, lacY⁺ makes permease, an enzyme essential for the rapid transportation of galactosides from the medium to the interior of the cell. Its allele lacY_– makes no permease. The galactoside lactose must enter the cell in order to induce the lacZ⁺ gene to produce the enzyme b-galactosidase. The allele lacZ^–makes a related but enzymatically inactive protein called lacCZ. Predict the production or nonproduction of each of these products with a normal operator O⁺ by placing + or 0, respectively, in the table below



9. In genotype (1) of the table below, the lacI^– allele allows constitutive enzyme production by the lacY⁺ and lacZ⁺ genes in an operon with a normal operator gene, O⁺. The action of lacI^– might be explained by one of two hypothesis: (1) lacI^– produces an internal inducer, thus eliminating the need for lactose in the medium to induce enzyme synthesis; lacI⁺ produces no internal inducer, or (2) lacI⁺ produces a repressor substance that, in the absence of lactose inducer, blocks enzyme formation, but in the presence of lactose inducer the repressor becomes inactivated to allow enzyme synthesis; lacI^– produces no repressor.   (a) Assuming dominance of the lacI ^– allele under the first hypothesis in an E. coli partial diploid of the constitution lacI⁺/lacl^–, would internal inducer be produced?   (b) Under the conditions of part (a), would enzymes be produced constitutively or inductively in a wild-type lac operon?   (c) Assuming dominance of the lacI⁺ allele under the second hypothesis in a partial diploid of the constitution lacI ⁺/lacI^–, would repressor be produced?   (d) Under the conditions of part (c), would enzymes be produced constitutively or inductively in a wild-type lac operon?   (e) From the pattern of reactions exhibited by genotypes (2) and (3) in the table below, determine which of the two hypotheses is consistent with the data.   (f) Is the repressor substance diffusible, or can it only act on loci in cis position with the lacI locus? How can this be determined from the information in the table?



10. List two kinds of single mutations that can change the function of an operator.
11. A bacterial mutation renders a cell incapable of fermenting many sugars (e.g., lactose, sorbitol, xylose) simultaneously. The operons of genes specifying the respective catabolic enzymes are wild type (unmutated). Offer an explanation for this phenomenon.
12. Shown below is a hypothetical biosynthetic pathway subject to feedback inhibition; letters represent metabolites; numbers represent enzymes. Identify the enzymes that are most likely to be subject to feedback inhibition and their inhibitor(s). Note: The inhibitor may consist of more than one metabolite.

Mapping The Bacterial Chromosome Questions

13. A cross is made between the streptomycin-resistant (str^r) F^– strain of genotype gal^– thr^– azi^r lac^–Ton^r mal^– xyl–leu^– and the prototrophic Hfr strain that has opposite characters. After 60 min of contact, samples are transferred to plates with minimal medium plus streptomycin. The original mixture is in the ratio 2 × 10⁷ Hfr to 4 × 10⁸ F ^–. The percentages of each Hfr gene transferred are: 72% Tons, 0% mal⁺, 27% gal⁺, 91% azi^s, 0% xyl ⁺, 48% lac⁺.     (a) How many F^– cells exist in the original mixture for every Hfr cell?   (b) What is the counterselective agent that prevents Hfr individuals from obscuring the detection of recombinants?   (c) In what order are these genes probably being transferred by the Hfr strain?
14. Four Hfr strains of E. coli are known to transfer their genetic material during conjugation in different sequences. Given the time of entry of the markers into the F^– recipient, construct a genetic map that includes all of these markers and label the time distance between adjacent gene pairs.



15. Two mutants at the tryptophan locus, trp^– _A and trp^–_B, are known to be close to a cysteine locus (cys). A bacterial strain of genotype cys⁺ trp^–_A is transduced by phage from a bacterial strain that is cys^– trp–_B. The reciprocal cross is also made wherein the strain cys^–trp^–_B is transduced by phage from a strain that is cys⁺ trp^–_A. In both cases, the numbers of prototrophic recombinants are equivalent. Determine the order of the tryptophan mutants relative to the cysteine marker.
16. Five point mutations (a–e) were tested for wild-type recombinants with each of the five deletions shown in the topological map below. The results are listed in the table below (+ = recombination, 0 = no recombination). Determine the order of the point mutations.

Answers

Vocabulary

1. auxotrophic
2. clone
3. theta (θ)
4. lawn
5. conjugation
6. transduction
7. merozygote
8. plasmid
9. DNA gyrase
10. helicase

Multiple-Choice

1. c
2. d
3. c
4. a
5. d
6. c
7. d
8. a
9. e
10. c

Isolation of Bacterial Mutants

1. Plate 1 contains a mutation to ser⁺; plate 3 contains a mutation to arg⁺.
2.  



3. 20% recombination
4. (a) Back mutations (reverse mutations) from his^– to his⁺.   (b) It supplies the mammalian metabolic functions that are usually required to convert a chemical into its carcinogenic metabolites.  (c) After 2 days, most of the his- bacteria have died for lack of histidine. Back-mutation rates are expected to be proportional to concentration of the chemical that forms a radially diminishing concentration gradient around the paper disk. Close to the disk there is a zone in which no cells grow because of toxic levels of the chemical. Beyond this zone there may be so many his⁺ revertants that the cells almost form a continuous lawn. At the periphery are a few larger clones (because they are isolated) representing spontaneous his⁺ mutants that have not been exposed to the chemical.

Bacterial DNA Replication and Cell Division

Regulation of Bacterial Gene Activity

6. (a) I^s, I⁺, I^–  (b) p^s, p⁺, p^{i cr}, p^–



7. (a),   (d) = inductive;   (b),   (c),   (e) = constitutive
8. 
9. (a)Yes   (b) Constitutively   (c)Yes   (d) Inductively   (e) Note that in genotypes (2) and (3) the lacI^– allele fails to produce enzymes in the absence of the external inducer (lactose); it fails to exhibit dominance. Therefore, the first hypothesis is incorrect. Under the second hypothesis, lacI⁺ is dominant and produces a repressor that (in the absence of lactose inducer) blocks enzyme synthesis as seen in genotypes (2) and (3). (f) Genotype (3) has lacY⁺ and lacZ⁺ on one DNA molecule and lacI⁺ on a different DNA molecule. Yet in the absence of lactose inducer, the repressor made by lacI⁺ still prevents the production of enzymes by lacY⁺ and lacZ⁺. Therefore, the repressor must be able to act at a distance (it behaves as a diffusible substance) on genes that are either on the same DNA molecule (cis position) or on a different DNA molecule (trans position).
10. (1)A change that prevents repressor binding. (2) Modifications that increase repressor binding so that operons cannot be derepressed even when inducer is bound.
11. The mutation could be in the gene for adenyl cyclase or in the gene for catabolite activator protein (CAP).
12. I inhibits 7; J inhibits 9; G alone or (I, J) together inhibits 5; E inhibits 3; enzyme 1 could be inhibited by (I, J, E), (G, E), (I, J, C), or (C, G).

Mapping the Baterial Chromosome

13. (a) 20   (b) Streptomycin; Hfr is streptomycin-sensitive (str^s).   (c) Origin-(thr⁺ Leu⁺)-azi^s-Ton^s-lac⁺-gal⁺-str^s-(mal⁺ xyl⁺). The genes for synthesizing the amino acids threonine and leucine must have entered first, otherwise none of the other recombinants could survive on unsupplemented medium. Note: The order of markers within parentheses has not been determined.
14. 
15. cys-trp_B-trpA
16. a-c-d-e-b