CHAPTER 9
Experimental Questions
E1.
In the experiment of Figure 9.1, a met– bio– thr+ leu+ thi+ cell could become
met+ bio+ thr– leu– thi– by a (rare) double mutation that converts the met– bio– genetic
material into met+ bio+. Likewise, a met+ bio+ thr– leu– thi– cell could become met+
bio+ thr+ leu+ thi+ by three mutations that convert the thr– leu– thi– genetic material
into thr+ leu+ thi+. From the results of Figure 9.1, how do you know that the occurrence
of 100 met+ bio+ thr+ leu+ thi+ colonies is not due to these types of rare multiple
mutations?
Answer: Because no colonies were observed when 108 cells of the met– bio– thr+ leu+
thi+ or met+ bio+ thr– leu– thi– strains were plated alone.
E2.
In the experiment of Figure 9.1, Lederberg and Tatum could not discern whether
met+ bio+ genetic material was transferred to the met– bio– thr+ leu+ thi+ strain or if
thr+ leu+ thi+ genetic material was transferred to the met+ bio+ thr– leu– thi– strain.
Let’s suppose that one strain is streptomycin resistant (say, met+ bio+ thr– leu– thi–)
while the other strain is sensitive to streptomycin. Describe an experiment that could
determine whether the met+ bio+ genetic material was transferred to the met– bio– thr+
leu+ thi+ strain or if the thr+ leu+ thi+ genetic material was transferred to the met+ bio+
thr– leu– thi– strain.
Answer: Mix the two strains together and then put some of them on plates containing
streptomycin and some of them on plates without streptomycin. If mated colonies are
present on both types of plates, then the thr+, leu+, and thi+ genes were transferred to
the met+ bio+ thr– leu– thi– strain. If colonies are found only on the plates that lack
streptomycin, then the met+ and bio+ genes are being transferred to the met– bio– thr+
leu+ thi+ strain. This answer assumes a one-way transfer of genes from a donor to a
recipient strain.
E3.
Explain how a U-tube apparatus can distinguish between genetic transfer
involving conjugation and genetic transfer involving transduction. Do you think a U-tube
could be used to distinguish between transduction and transformation?
Answer: The U-tube can distinguish between conjugation and transduction because of the
pore size. Because the pores are too small for the passage of bacteria, this prevents direct
contact between the two bacterial strains. However, viruses can pass through the pores
and infect cells on either side of the U-tube.
It might be possible to use a U-tube to distinguish between transduction and
transformation if a filter was used that had a pore size that was too small for viruses but
large enough to allow the passage of DNA fragments. However, this might be difficult,
because the difference in sizes between DNA fragments and phages are relatively small
compared to the differences in sizes between bacteria and phages.
E4.
What is an interrupted mating experiment? What type of experimental
information can be obtained from this type of study? Why is it necessary to interrupt
mating?
Answer: An interrupted mating experiment is a procedure in which two bacterial strains
are allowed to mate, and then the mating is interrupted at various time points. The
interruption occurs by agitation of the solution in which the bacteria are found. This type
of study is used to map the locations of genes. It is necessary to interrupt mating so that
you can vary the time and obtain information about the order of transfer; which gene
transferred first, second, and so on.
E5.
In a conjugation experiment, what is meant by the term time of entry? How is it
determined experimentally?
Answer: The time of entry is the time it takes for a gene to be initially transferred from
one bacterium to another. To determine this time, measurements are made at various
lengths of time and then these data are extrapolated back to the x-axis.
E6.
In your laboratory, you have an F – strain of E. coli that is resistant to
streptomycin and is unable to metabolize lactose, but it can metabolize glucose.
Therefore, this strain can grow on plates that contain glucose and streptomycin, but it
cannot grow on plates containing lactose. A researcher has sent you two E. coli strains in
two separate tubes. One strain, let’s call it strain A, has an F factor that carries the genes
that are required for lactose metabolism. On its chromosome, it also has the genes that are
required for glucose metabolism. However, it does not have the genes that confer
streptomycin resistance. This strain can grow on plates containing lactose or glucose, but
it cannot grow if streptomycin is added to the plates. The second strain, let’s call it strain
B, is an F – strain. On its chromosome, it has the genes that are required for lactose and
glucose metabolism. Strain B is also sensitive to streptomycin. Unfortunately, when
strains A and B were sent to you, the labels had fallen off the tubes. Describe how you
could determine which tubes contain strain A and strain B.
Answer: Mate unknown strains A and B to the F – strain in your lab that is resistant to
streptomycin and cannot use lactose. This is done in two separate tubes (i.e., strain A plus
your F – strain in one tube, and strain B plus your F – strain in the other tube). Plate the
mated cells on growth media containing lactose plates plus streptomycin. If you get
growth of colonies, the unknown strain had to be strain A, the F+ strain that had lactose
utilization genes on its F factor.
E7.
As mentioned in solved Problem S2, origins of transfer can be located in many
different locations, and their direction of transfer can be clockwise or counterclockwise.
Let’s suppose a researcher mated six different Hfr strains that were thr+ leu+ tons str r
azi s lac+ gal + pro+ met+ to an F–strain that was thr– leu– tonr str s azi r lac– gal – pro–
met–, and obtained the following results:
Strain
1
Order of Gene Transfer
tons azi s leu+ thr+ met+ strr gal+ lac+ pro+
2
leu+ azi s tons pro+ lac+ gal+ strr met+ thr+
3
lac+ gal+ strr met+ thr+ leu+ azi s tons pro+
4
leu+ thr+ met+ strr gal+ lac+ pro+ tons azi s
5
tons pro+ lac+ gal+ strr met+ thr+ leu+ azi s
6
met+ strr gal+ lac+ pro+ tons azi s leu+ thr+
Draw a circular map of the E. coli chromosome and describe the locations and
orientations of the origins of transfer in these six Hfr strains.
Answer:
E8.
An Hfr strain that is hisE+ and pheA+ was mated to a strain that is hisE– and
pheA–. The mating was interrupted and the percentage of recombinants for each gene was
determined by streaking on plates that lacked either histidine or phenylalanine. The
following results were obtained:
[Insert Text Art]
A.
Determine the map distance (in minutes) between these two genes.
B.
In a previous experiment, it was found that hisE is 4 minutes away from
the gene pabB. PheA was shown to be 17 minutes from this gene. Draw a genetic map
describing the locations of all three genes.
Answer:
A.
If we extrapolate these lines back to the x-axis, the hisE intersects at about
3 minutes and the pheA intersects at about 24 minutes. These are the values for the times
of entry. Therefore, the distance between these two genes is 21 minutes (i.e., 24 minus 3).
B.________________________________________________

hisE
4

pabB
17

pheA
E9.
Acridine orange is a chemical that inhibits the replication of F factor DNA but
does not affect the replication of chromosomal DNA, even if the chromosomal DNA
contains an Hfr. Let’s suppose that you have an E. coli strain that is unable to metabolize
lactose and has an F factor that carries a streptomycin-resistant gene. You also have an
F– strain of E. coli that is sensitive to streptomycin and has the genes that allow the
bacterium to metabolize lactose. This second strain can grow on lactose plates. How
would you generate an Hfr strain that is resistant to streptomycin and can metabolize
lactose? (Hint: F factors occasionally integrate into the chromosome to become Hfr
strains, and occasionally Hfr strains excise their DNA from the chromosome to become
F+ strains that carry an F factor.)
Answer: First, mate the streptomycin-resistant strain to the strain that has the genes that
allow the cell to metabolize lactose. You can select for mated cells on growth media
containing lactose and streptomycin. These cells will initially have an F factor with the
streptomycin-resistant gene. Expose these cells to acridine orange in a media that also
contains streptomycin. This will select for the survival of rare cells in which the F factor
has become integrated into the chromosome to become Hfr cells.
E10. In a P1 transduction experiment, the P1 lysate contains phages that carry pieces of
the host chromosomal DNA, but the lysate also contains broken pieces of chromosomal
DNA (see Figure 9.10). If a P1 lysate is used to transfer chromosomal DNA to another
bacterium, how could you show experimentally that the recombinant bacterium has been
transduced (i.e., taken up a P1 phage with a piece of chromosomal DNA inside) versus
transformed (i.e., taken up a piece of chromosomal DNA that is not within a P1 phage
coat)?
Answer: One possibility is that you could treat the P1 lysate with DNase I, an enzyme
that digests DNA. (Note: If DNA were digested with DNase I, the function of any genes
within the DNA would be destroyed.) If the DNA were within a P1 phage, it would be
protected from DNase I digestion. This would allow you to distinguish between
transformation (which would be inhibited by DNase I) versus transduction (which would
not be inhibited by DNase I). Another possibility is that you could try to fractionate the
P1 lysate. Naked DNA would be smaller than a P1 phage carrying DNA. You could try to
filter the lysate to remove naked DNA, or you could subject the lysate to centrifugation
and remove the lighter fractions that contain naked DNA.
E11. Can you devise an experimental strategy to get P1 phage to transduce the entire
lambda genome from one strain of bacterium to another strain? (Note: The general
features of phage lambda’s life cycle are described in Chapter 10.) Phage lambda has a
genome size of 48,502 nucleotides (about 1% of the size of the E. coli chromosome) and
can follow the lytic or lysogenic life cycle. Growth of E. coli on minimal growth media
favors the lysogenic life cycle, whereas growth on rich media and/or UV light promotes
the lytic cycle.
Answer: You could infect E. coli with lambda and then grow under conditions that
promote the lysogenic cycle (i.e., on minimal media). You could then infect the lysogenic
E. coli strain with P1. On occasion, the P1 phage packages a piece of the bacterial
chromosome and therefore it could package the lambda DNA that has been integrated
into the bacterial chromosome. (Note: The size of the lambda genome is small enough to
fit inside of P1.) The P1 phages isolated from the lysogenic E. coli strain could then be
used to infect a nonlysogenic strain.
E12. Let’s suppose a new strain of P1 has been identified that packages larger pieces of
the E. coli chromosome. This new P1 strain packages pieces of the E. coli chromosome
that are 5 minutes long. If two genes are 0.7 minutes apart along the E. coli chromosome,
what would be the cotransduction frequency using a normal strain of P1 and using this
new strain of P1 that packages large pieces? What would be the experimental advantage
of using this new P1 strain?
Answer:
Cotransduction frequency = (1 – d/L)3
For the normal strain:
Cotransduction frequency = (1 – 0.7/2)3 = 0.275, or 27.5%
For the new strain:
Cotransduction frequency = (1 – 0.7/5)3 = 0.64, or 64%
The experimental advantage is that you could map genes that are farther than 2
minutes apart. You could map genes that are up to 5 minutes apart.
E13. If two bacterial genes are 0.6 minutes apart on the bacterial chromosome, what
frequency of cotransductants would you expect to observe in a P1 transduction
experiment?
Answer:
Cotransduction frequency = (1 – d/L)3
= (1 – 0.6 minutes/2 minutes)3
= 0.34, or 34%
E14. In an experiment involving P1 transduction, the cotransduction frequency was
0.53. How far apart are the two genes?
Answer:
Cotransduction frequency = (1 – d/L)3
0.53  (1  d / 2 minutes)3
(1  d / 2 minutes)  3 0.53
(1  d / 2 minutes)  0.81
d  0.38 minutes
E15. In a cotransduction experiment, the transfer of one gene is selected for and the
presence of the second gene is then determined. If 0 out of 1,000 P1 transductants that
carry the first gene also carry the second gene, what would you conclude about the
minimum distance between the two genes?
Answer: You would conclude that the two genes are farther apart than the length of 2%
of the bacterial chromosome.
E16. In a cotransformation experiment (see solved Problem S4), DNA was isolated
from a donor strain that was proA+ and strC+ and sensitive to tetracycline. (The proA
and strC genes confer the ability to synthesize proline and confer streptomycin resistance,
respectively.) A recipient strain is proA– and strC – and resistant to tetracycline. After
transformation, the bacteria were first streaked on plates containing proline,
streptomycin, and tetracycline. Colonies were then restreaked on plates containing
streptomycin and tetracycline. (Note: Both types of plates had carbon and nitrogen
sources for growth.) The following results were obtained:
70 colonies grew on plates containing proline, streptomycin, and tetracycline,
while only 2 of these 70 colonies grew when restreaked on plates containing
streptomycin and tetracycline but lacking proline.
A.
If we assume that the average size of the DNA fragments is 2 minutes,
how far apart are these two genes?
B.
What would you expect the cotransformation frequency to be if the
average size of the DNA fragments was 4 minutes and the two genes are 1.4 minutes
apart?
Answer:
A.
We first need to calculate the cotransformation frequency, which equals
2/70, or 0.029.
Cotransformation frequency = (1-d / L)3
0.029  (1  d / 2 minutes)3
d  1.4 minutes
B.
Cotransformation frequency = (1  d / L)3
 (1  1.4 / 4)3
 0.27
As you may have expected, the cotransformation frequency is much higher when
the transformation involves larger pieces of DNA.