PROBLEM SET
Chapter 7-- Linkage and Mapping in Prokaryotes

P. 163, #2. What genotypic notation indicates alleles that make a bacterium.
a. resistant to penicillin? penr
pen is the abbreviation for penicillin and the r indicates resistance.
b. sensitive to azide? azis
azi is the abbreviation for azide and the s indicates sensitive.
c. require histidine for growth? his-
his is the abbreviation for histidine and the - indicates that the cell can not synthesize histidine and thus requires histidine for growth.
d. unable to grow on galactose? gal-
gal is the abbreviation for galactose and the - indicates that the cell can not utilize galactose as an energy source.
e. able to grow on glucose? glu+
glu is the abbreviation for glucose and the + indicates that the cell can utilize glucose as an energy source.
f. susceptible to page T1 infection? ton+
ton is the abbreviation for T1 phage infection and the + indicates that the cell is susceptible to T1 phage infection.

P. 163, #3. What are the differences between a heterotroph and an auxotroph?

A heterotroph requires an organic form of carbon to grow. In general, this means that the organism can not carry out photosynthesis and thus can not use the energy from sunlight directly. On the other hand an auxotroph has some specific nutritional requirement, such as a specific amino acid, to grow.
What are the differences between a minimal and a complete medium?

Minimal medium only contains the nutritional requirements of the wildtype strain of an organism (prototroph could grow on minimal medium), while complete medium has all the nutritional requirements that are needed by a certain strain of an organism (for example the auxotroph his- would require a complete medium, that is minimal medium plus histidine).
What are the differences between an enriched and a selective medium?

An enriched medium is the same as a complete medium (see above), while a selective medium is designed to selected for the growth of a particular strain of organism. Selective medium would be minimal medium with only one substance added.

P. 163, #5. What is a plasmid?

An independent, self-replicating particle of DNA. For example, the F factor that allows a bacterium to carry out conjugation is a plasmid. The particle of DNA is autonomous from the bacterial chromosome and it replicates on its own separate from the bacterial chromosome.
How does one integrate into a host's chromosome?

The integration of the F factor requires that the two circles of DNA (the plasmid and the bacterial chromosome) touch and then a single crossover cause both strands to break and then the repair process causes the plasmid to be attached at either end into the bacterial chromosome. The F factor is now an integral part of the bacterial chromosome.
How does it leave?

The F factor leaves by a reverse of how it goes in. It simply forms a loop and another crossover event breaks the ends and then the repair process attaches one end of the plasmid to its other end, which makes a circle. The repair process also connects the free ends of the bacterial chromosome so that it is also a circle. Occasionally, the process makes mistakes and a piece of the bacterial chromosome goes with the plasmid.

P. 163, #10. The DNA from a prototrophic strain of E. coli is isolated and used to transform an auxotrophic strain deficient in the synthesis of purines (purB-), pyrimidines (pyrC-), and the amino acid tryptophan (trp-). Tryptophan was used as the marker to determine whether transformation had occurred (the selected marker). What are the gene order and the map distance between the loci, given the following data?

trp+ pyrC+ purB+      86
trp+ pyrC+ purB-       4
trp+ pyrC- purB+      67
trp+ pyrC- purB-      14

This transformation has three loci, though one (tryptophan) is always trp+, because it was used as the marker. The transformants had to be able to grow on media without tryptophan added to it to be recognized as a transformant. If you look at the frequency of each class you notice that one is in very low frequency (trp+ pyrC+ purB-) with only 4 found. This class is due to a double crossover, one between each of the three loci. This leaves the locus that is in the middle untransformed. As purB is the locus that was untransformed, it is the locus in the middle.

The map distance is given as a function of the relative rate of recombination, just as it is in fruit flies or corn. The recombination rate between purB and pyrC is given by dividing the number of recombinants between them by the total number of recombinants. Ignore the class trp+ pyrC- purB- as this class did not transform at either of the pyrC or purB loci.

                            single transformants
recombination rate (m.u.) = ----------------------------- x 100
                            total number of transformants
                         67 + 4
recombination rate = ------------ x 100 = 45 m.u.
                       67 + 4 + 86

As the purB locus is the one in the middle, then it must be closer to the trp locus that the pyrC locus is close to the trp locus. You can also see this if you calculate the cotransfer index (r). Notice that in the transformants, trp+ pyrC+ purB+ and trp+ pyrC- purB+, the loci trp and purB are contransformed while in the other two transformants only one locus was transformed.

                        number with trp+ and purB+
cotransfer index (r) =  ---------------------------------------
                        total number of tranformants
                     67 + 86
cotransfer index =  --------- = 0.89
                      171

Similarly, the cotransfer index for trp and pyrC can be calculated.

                        number with trp+ and pyrC+
cotransfer index (r) =  ---------------------------------------
                        total number of tranformants
                     86 + 4
cotransfer index =  --------- = 0.53
                      171

As the contransfer index for trp and purB (0.89) is higher than for trp and pyrC (0.53), we can conclude that purB is closer to trp than pyrC is close to trp.

P. 163, #12. Three Hfr strains of E. coli (P4X, KL98, and Ra-2) are mated individually with an auxotrophic F- strain, using interrupted mating techniques. Using these data construct a map of the E. coli chromosome, including distance in minutes.
Approximate Time of Entry ---------------------------------- Donor Loci Hfr P4X Hfr KL98 Hfr Ra-2 ----------------------------------------------------- gal+ 11 67 70 thr+ 94 50 87 xyl+ 73 29 8 lac+ 2 58 79 his+ 38 94 43 ilv+ 77 33 4 argG+ 62 18 19 -----------------------------------------------------
How many different petri plates and selective media are needed?

The first step is to put the time of transfer for each locus in order for each one of the strains. For Hfr P4X, the order is:
lac(2)--gal(11)--his(38)--argG(62)--xyl(73)--ilv(77)--thr(94)
For Hfr KL98, the order is:
agrG(18)--xyl(29)--ilv(33)--thr(50)--lac(58)--gal(67)--his(94)
For Hfr Ra-2, the order is:
ilv(4)--xyl(8)--argG(19)--his(43)--gal(70)--lac(79)--thr(87)
Notice that the direction of transfer for Ra-2 is backwards from that of the other two strains. Notice also that the time of transfer between the loci is constant from one Hfr strain to the next. For example, between lac(2) and gal(11) is 9 minutes for Hfr P4X and the time of transfer between gal(70) and lac(79) for Hfr Ra-2 is also 9 minutes, though the direction of transfer is opposite.

The best way to show the map is with a diagram of the bacterial chromosome. We will arbitrarily assign the lac locus the value of 0 minutes, with a total of 100 minutes for complete transfer of the bacterial chromosome. The 100 minutes is calculated from the 92 minutes between lac(2) and thr(94) for Hfr P4X plus 8 minutes between lac(79) and thr(87) for Hfr Ra-2.

P. 165, #28. In a transformation experiment, an a+ b+ c+ strain is used as the donor, and a a- b- c- strain as the recipient. One hundred a+ transformants are selected and then replica-plated to determine whether b+ and c+ are present. The genotypes of the transformants appear following. What can you conclude about the relative position of the genes?
a+ b- c- 21 a+ b- c+ 69 a+ b+ c- 3 a+ b+ c+ 7
This problem is just like problem 10 above. If you look at the frequency of each class you notice that one is in very low frequency (a+ b+ c-) with only 3 found. This class is due to a double crossover, one between each of the three loci. This leaves the locus that is in the middle untransformed. As c is the locus that was untransformed, it is the locus in the middle.

The map distance is given as a function of the relative rate of recombination, just as it is in fruit flies or corn. The recombination rate between b and c is given by dividing the number of recombinants between them by the total number of recombinants. Ignore the class a+ b- c- as this class did not transform at either of the b or c loci. However, you notice that the number of single transformants (69 + 3) is much greater than the number of double transformants (7). As the number of single transformants is greater than the number of double transformants, the distance between the b locus and the c locus is two far to map.

You can also look at the relative distance among the loci by calculating the cotransfer index. Notice that in the transformants, a+ b+ c+ and a+ b- c+, the loci a and c are contransformed while in the other two transformants only one locus was transformed.

                        number with a+ and b+
cotransfer index (r) =  ---------------------------------------
                        total number of tranformants
                     69 + 7
cotransfer index =  --------- = 0.76
                      100

Similarly, the cotransfer index for a and b can be calculated.

                        number with a+ and b+
cotransfer index (r) =  ---------------------------------------
                        total number of tranformants
                      7 + 3
cotransfer index =  --------- = 0.10
                      100

As the contransfer index for a and c (0.76) is higher than for a and b (0.10), we can conclude that c is closer to a than b is close to a.

P. 165, #30. A mating between his+, leu+, thr+, pro+, strs cells (Hfr) and his-, leu-, thr-, pro-, strr cells (F-) is allowed to continue for twenty-five minutes. The mating is stopped and the genotypes of the recombinants determined. What is the first gene to enter and what is the probable gene order?

Genotype           number of colonies
  his+       0
  leu+      12
  thr+      27
  pro+       6

The conjugant in highest frequency represents the locus that was transfered first, the conjugant in second highest frequency represents the locus transferred second. Thus if we simply rank the conjugants from most frequent to least frequent, the loci will be in that order on the bacterial chromosome from first to last. Note that no colonies that were his+ were isolated, which means that it would be transferred last of these four and that it is further along the bacterial chromosome that 25 minutes. The order of the loci is thr -- leu -- pro -- his.

P. 166, #34. A bacterial strain that is lys+ his+ val+ is used as a donor, and lys- his- val- is the recipient. Initial transformants are isolated on minimal medium + histidine + valine.
a. What genotype will grow on this medium?

Any genotype that can synthesize lysine, as this is the nutrient that is needed. Thus, lys+ his- val-, lys+ his+ val+, lys+ his- val+, and lys+ his+ val- will all grow on this medium.
b. These colonies are replicated to minimal medium + histidine, and 75% of the original colonies grow. What genotypes will grow on this medium?

Any of the original colonies that can also synthesize valine as that nutrient along with lysine is missing. Thus, lys+ val+ his- and lys+ val+ his+ will grow on this medium. c. The original colonies are also replicated to minimal medium + valine, and 6% of the colonies grow. What genotypes will grow on this medium?

Any of the original colonies that can also synthesize histidine as that nutrient along with lysine is missing. Thus, lys+ val- his+ and lys+ val+ his+ will grow on this medium.
d. Finally, the original colonies are replicated to minimal medium. No colonies grow. From this information, what genotypes will grow on minimal medium + histidine and on minimal medium + valine.

Based on these results, none of the original colonies were transformants at all three loci, that is lys+ val+ his+ was not seen. Thus the genotype from this transformation that will grow on minimal medium + histidine is lys+ val+ his-, and the genotype that will grow on minimal medium + valine is lys+ val- his+.
e. Based on this information, which gene is closer to lys?

First, we need to summarize the above information. The various transformants that were isolated and their relative percentages are given in the table below.

lys+ val+ his+         0
lys+ val+ his-        75
lys+ val- his+         6

From the table, we can see that lys and val are co-transformed with a much greater frequency (75%) than is lys with his. The order could be lys - val - his or val - lys - his. his must be on the edge because it is further away from lys.
f. The original transformation is repeated, but the original plating is on minimal medium + lysine + histidine. Fifty colonies appear. These colonies are replicated to determine their genotypes and the following results are recorded.
val+ his+ lys+ 0 val+ his- lys+ 37 val+ his+ lys- 3 val+ his- lys- 10
Based on all the results, what is the most likely gene order?

The single transformant in very low frequency (3) is val+ his+ lys- is due to a double-crossover, which would indicate that lys is in the middle.

Last updated on 22 August 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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PROBLEM SET Chapter 7-- Linkage and Mapping in Prokaryotes

PROBLEM SET
Chapter 7-- Linkage and Mapping in Prokaryotes