1.a. In F factor conjugation, only the F factor is transferred to the F- strain; in HFr conjugation, the chromosomal DNA is transferred first, the F factor last.
b. Conversion is the transfer of an F factor to a F- strain, which converts the strain to F+. Recombination is the transfer of bacterial genes from an Hfr or F’ strain to an F- strain.
c. F (fertility) factors are cytoplasmic episomes needed for conjugation. F’ factors are F factors that contain some genes from the bacterial chromosome.
d. Prototrophs are strains of organisms that do not require a given compound for growth; they can survive on minimal media. Auxotrophs require media supplemented with specific substances not required by wild-type organisms.
e. An F+ cell that has a free fertility (F) factor, while an Hfr cell has the fertility factor integrated into its bacterial genome.
2.
1) Interrupted mating study: Mate Hfr and F- strains together and look at the times at which markers are transferred (Griffiths 212).
2) Use Hfr and F- for reciprocal crosses and look at the relative frequency of the triple mutant (or wild type).
3) Use generalized transduction and determine the frequency of cotransduction (Griffiths 225-226).
3. a. A = E >> B = C = D
Conversion occurs most frequently from F+ strains where the F factor is not integrated into the bacterial chromosome (A and E); the independent F factor plasmid is easily replicated and transferred during conjugation.
Hfr strains (F+ strains with an integrated F factor, such as B, C, and D) will have a much lower frequency of gene conversion. During conjugation, the F factor is the last part of the bacterial chromosome to be transferred. Because the mating process is often interrupted early, gene conversion does not frequently occur.
b. E >> B > C > D > A
E will have the highest frequency of lac transfer because it has an F’ factor that carries the lac gene. B will have the next highest frequency because it is an Hfr with an F factor located immediately before its lac gene. Because the F factor enters last during Hfr conjugation, lac will be the first gene entered into the F- strain. C will transfer lac at a lower frequency, and D at an even lower frequency because of the lac gene's position relative to start of transfer. A will have the lowest frequency of lac transfer, because Hfr cells must first be generated to transfer lac.
c. D or B
Hfr strains B and D are most likely to produce the F’ plasmid seen in F because of their proximity of lac to the fertility factor. For a description of how an F’ strain is produced, see pages 218-219 and Figure 7-14 in Griffiths.
d. i. To select for lac+ conjugates, grow on lactose as the sole carbon source.
ii. No individuals of the lac+met-pro- phenotype are observed because their generation requires four crossovers to occur.
iii. To determine map order, determine the map distance between genes via their frequency of recombination. The parental genotypes are lac+ mec- pro- and lac-met+ pro-; all the exconjugates selected for are lac+. The number of lac mec recombinants is 53, making the map distance between lac and mec 19.4 m.u. (53/273). The number of lac pro recombinants is 43, so the map distance between lac and pro is 15.8 m.u. (43/273). Finally, there are 10 met pro recombinants, so the map distance between met and pro is 2.7 m.u. Since we know that lac enters last, the gene order is met – 2.7 – pro - 15.8 – lac.
iv. Separated by a map distance of 2.7 m.u., pro and met are closest genes together on the map.
4. a. Because the most crossovers give the rarest bacteria (a-b-c+d+), this genotype must be made by having crossovers between every marker. Thus, the order is either bcad or acbd. We can distinguish between these two possibilities by comparing a-b+c+ clones (270) versus a+b-c+ clones (20). The smaller number for the latter indicates that the order is acbd.
b. Although one could get an accurate map distance by looking at double crossovers events, we can estimate the distances simply by looking at bacteria with single crossovers. These are
Crossover between d and b: a-b-c- 50
Crossover between b and c: a-b+c- 160
Crossover between c and a: a-b+c+ 270
So d and b are closest together (actually 10%).
c. By the same reasoning, c and a are furthest apart (actually 30%).
d. Interrupted matings can be used to map the genes. These are described in the text (Griffiths p. 212). You expect to get a+ cells first, then a+c+, then a+c+b+, then a+c+b+d+.
5. a. In the table provided, a “+” means that recombinants were formed from a mating of the two strains. If such a mating could occur, it means that the deletions of the two strains did not overlap. If mating of the two strains did not produce recombinants (“-“), then the deletions must overlap in some way. For instance, the deletions in strains 1 and 2 must not overlap, as recombinants were produced. However, the deletion in 1 must overlap somewhere with the deletions in strains 3-7, as their crosses were unable to produce recombinants. By observing which strains can produce recombinants with each other and which strains cannot, you can create a map of these deletions.
Below is a map showing the location of deletions for each strain; "-----" signifies a deletion
1 --------------------------- --------------------- 2
3 ---------- --------------------------------------- 4
5 --------------------- ---------------------------- 6
7 ----------------------------------------------------
(from part b we find additionally that #7 doesn't extend as much to the left as #1, #3, or #5)
b. Recombinants can only be recovered when the mutants are mated to strains that do not have a deletion in the same region as the point mutation. Below is a map showing where the mutations must be located based upon the data showing which strains can yield recombinants:
e b c d a
1 --------------------------- --------------------- 2
3 ---------- --------------------------------------- 4
5 --------------------- ---------------------------- 6
7 ------------------------------------------------
Therefore, the order of these point mutations is e-b-c-d-a.
6. a. I. To select for pur+ transductants, use pro and his media.
II. To select for pro+ transductants, use pur and his media.
III. To select for his+ transductants, use pro and pur media.
b. Because pur+pro+his- bacteria are not found, it is not possible to obtain pur+ and pro+ without his+. Therefore, his must map between pur and pro.
c. In Experiment I, many more (87%) of the bacteria are only pur+ compared with the number that are pur+ his+ (10%) or pur+ his+ pro+ (3%). Therefore, the distance between pur and his must be greater than that between his and pro. Examination of the other experiments gives the same result.
d. The most informative experiment is Experiment III because it results from the selection of the middle gene.
e. Of the results in Experiment III, the pur+ pro+ and the pur- pro- results say how many times the phage did or did not pick up the flanking genes. The remaining results indicate that the phage is 4 times more likely to pick up pro+ then pur+ (compare 60% to 15%), so pro is 4 times closer to his than pur.
7. a. Recombination - relatively rare wild-type phage are produced by recombination between two phage with mutations in the same cistron.
Complementation - relatively frequent production of wild-type phage activity because of mutations in two different cistrons
b. Lysis - a type of phage growth where cells are disrupted to release phage.
Lysogeny - a type of phage growth where the phage integrates into the host chromosome.
c. Transduction is DNA entry into the cell mediated by phage. If any sequence can be packaged into the phage particle (e.g., with P22), then the transduction is general; if only certain DNAs can be package (e.g., with lambda) then transduction is specific.
8. To determine if mutations are in the same cistron, perform a cis-trans test. If the two mutations do not produce wild-type phage in trans, then they fail to complement and are in the same cistron. If two mutations within the same cistron produce wild-type phage in cis conformation, then the two mutations must both be recessive.