Chapter 44

BACTERIAL GENETICS: SEXUAL REPRODUCTION

PRE-LECTURE ASSIGNMENT
1. Quickly review notes for the previous lec-
ture.
2. Suggested readings:

a. General genetics textbooks
Altenburg: Chap. 22, pp. 377-385, 392-
393.
Sinnott, Dunn, and Dobzhansky: Chap.
23, pp. 315-318.
Snyder and David: Chap. 26, pp. 407-
409.
Winchester: Chap. 23, pp. 326-327.

b. Additional references
Jacob, F., and Wollman, E. L. 1958.
Genetic and physical determination of
chromosomal segments in Escherichia
coli. Sympos. Soc. exper. Biol., 12:
75-92,
Lederberg, J. 1947. Gene recombina-
tion and linked segregations in Escheri-
chia coli. Genetics, 32: 505-525. Re-
printed in "Papers in microbial genetics",
selected by J. Lederberg. 1951. Madi-
son: University of Wisconsin Press.
Lederberg, J. 1956. Conjugal pairing
in Escherichia coli. J. Bact., 71: 497-
498.
Lederberg, J. 1959. Bacterial repro-
duction. Harvey Lect., 53: 69-82.
Wollman, E. L., and Jacob, F. 1956.
Sexuality in bacteria. Scient. Amer.,
195: 109-118.

LECTURE NOTES
A. Genetic recombination in bacteria
1. Genetic analysis based only upon mutations
in asexually reproducing lines is severely
limited (see also Chap. 6).
2. The infrequency of sexual processes shows
their detection under the microscope is or-
dinarily improbable.

B.

Lecturer—J. LEDERBERG

3. The search for sexuality is greatly expe-

dited by the selective isolation of specific

genotypes, using mutants for different nu-

tritional defects.

a, The wild-type Escherichia coli strain
K-12 is nutritionally sufficient, i.e., is
a prototroph (M* T*) (see central part of
Fig. 44-1).

b. Two nutritionally-dependent mutants, or
auxotrophs, were obtained -- one requir-
ing the amino acid methionine (M7 T*),
and the other requiring the amino acid
threonine (Mt T7).

c. Neither auxotroph can produce colonies
when plated separately on a basal, mini-
mal culture medium containing neither
amino acid.

d. If the two mutant strains are mixed and
genetic recombination produces Mt tt
prototrophs, these will be the only cells
forming colonies when plated on agar
containing the minimal medium.

e. Such evidence for genetic recombination
was first obtained by Lederberg and
Tatum (1946).

. The test for prototrophs derived from dif-

ferent auxotrophs is very efficient for de-
tecting genetic recombination.

The fertilization process

1.

Early experiments used mating type F*,
and gave a very low frequency of recom-
bination.

. Later, other strains were found giving very

high frequencies of recombination, hence
called Hfr strains, which were useful in
learning the details of fertilization.

. As demonstrated, ac’ (Hfr) cell is seen un-

der the microscope to form a conjugation
bridge with a 9 (F ) cell once these make a
random contact (see also left side of Fig.
44-1).

251
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Figure 44-1

. When Wollman and Jacob used a Waring

blender to separate the cells at different
times after mating started, they found

a. bacteria so separated were viable.

b, there was a progressive linear transfer

of genetic markers from ¢' to 9.

1) This transfer required 100 minutes for
completion. (The total DNA length
transferred is about 22u, or 5 to 10
times the bacteria's length. )

2) The DNA to be transferred must have
unwound,

3) One end of the DNA string was prefer-
entially transferred first.

ce. Our most detailed map of bacterial genes
was made this way.

. The rate of transfer is 1, 000 nucleotide

units, or n'its, per second (6 x 108 nits in
100 minutes).

. The genetic unit (cistron) specifying a typi-

cal protein is about 2, 000 n'‘its.

a. Levinthal and Garen studied a variety of
mutants defective for alkaline phosphat-
ase, using procedures of this sort.

b. They found it required about 2, 000 ntits
to code this enzyme, which contains a-
bout 400 amino acids.

c. Such evidence supports a 4:1 (n'it: amino
acid) coding ratio (see Chap. 41).

. The injected o DNA synapses with that in the

2.

. Recombination between the o’ and 2 DNA oc-

curs, producing recombinant strands with
markers from both parents. The mechan-

ism for this is not known, and could result

from

a. breakage and cross-unions between par-
ental strands.

b. a copy-choice mechanism, in which the
daughter DNA alternates in using ma-
ternal and paternal DNA as a template.

C. Markers for genetic recombination

1. Usually 6 to 12 markers are studied in a
single cross.

2. The top center plate in Fig. 43-1 shows, us-
ing two markers, how recombinants typi-
cally are detected (see also Chap. 43 F2a).
a. A loop of streptomycin was brushed ver-

tically on the agar, then four clones were
streaked horizontally across this region.

b. Suppose one parental type (top colony in
Fig. 43-1 top center plate) was lactose-
negative (light colored) and streptomy-
cin-sensitive (interrupted streak).

c. Suppose the other parental type (third
colony down on this plate) was lactose-
positive (dark colored) streptomycin-
resistant (uninterrupted streak).

d. The other two colonies would be recom-
binants -- lactose-positive streptomy-
cin-sensitive (bottom one), and lactose-
negative streptomycin-resistant (colony
next to top).

3. Traits of mutants may involve their nutri-
tion, bacteriophages, specific antigens,
motility, ete.

D. Basis for sexual differentiation
1. Male sexuality is an infectious phenomenon
a. One Ft (¢) cell can rapidly convert all F7
(?) cells in a culture to Ft.

b. The new Ft cells transmit this trait to
their progeny.

c. The infective factor must multiply at
least twice as fast as the typical cell.

d. This factor, called F, is extra-chromo-
somal and not isolable as a cell-free
virus particle (see right side of Fig. 44-
1, where the chromosome is diagrammed
as a single line).

2. Properties of the F particle

a. It is transferred from ¢ to ? in a transi-
ent mating.

b. Such matings are more unstable but more
frequent than matings which involve chro-~
mosome transfer.

c. The dye, acridine orange, inhibits re-
production of F but has no apparent ef-
fect on chromosomal genes. Treatment
with this dye results in converting Ft to
F cells.

Mating types

1. Ft and F- types have been described al-
ready.

2. The F particle

a, must modify the o cell wall so as to re-
cognize and react with a ? cell it contacts,

b. must form a bridge between oc’ and g,

c. probably confers motility to the Ft ¢o
chromosome by attaching to it at least
temporarily (Fig. 44-2, small circle at-
tached to chromosome in F* cell).

3. Hfr mating type carries F on the chromo-
some end transmitted last in fertilization

(Fig. 44-2).

 

Figure 44-2

a. Hfr strains do not transmit F particles
contagiously.

b. All Hfr strains are derived from Ft
celis (which may have come from F-
cells),

c. Many Hfr strains are unstable and re-
vert to F* type, simultaneously losing
their high fertility and gaining the pro-
perty of infectious F particles,

d. The o character of Hfr is not inhibited by
acridine orange.

Episomal cycle refers to the facultative parti-
cipation of a factor as an extra-chromosomal
or as a chromosomal element.

1. Fis an episome

a. In F* cells, Fis normally extra-chro-
mosomal. It is suggested that the low
fertility of F+ is due to a transient con-
nection of F with the chromosome.

b. In Hfr cells F has a chromosomal locus.

2, The episome lambda (see also Chaps. 45

and 46)

a. E. coli strain K-12 normally carries a
symbiotic bacteriophage, lambda, to
which it is relatively insensitive.

b. A sensitive mutant, which had lost
lambda, was found. This strain is used
to test for lambda.

c. In the normal strain, lambda is temper-
ate since it does not cause conspicuous
lysis.

d. Since the normal insensitive strain can
potentially produce infective lambda, it
is said to be lysogenic.

e. Crosses between lysogenic and sensitive
bacteria showed that lambda, as a pro-
phage, has a definite locus on the chro-
mosome, Lp, closely linked to Gal, a
locus responsible for galactose fermen-
tation.

f. This close linkage was observed in the
haploid segregants from diploid excep-
tional clones of genotype Gal* Lp*/
Gal Lp’. (That is, ability to use galac-
tose, Galt, was closely linked to capa-
city to produce bacteriophage, Lp*, as
was Gal” to Lp§).

g. Occasionally the Lp* chromosomal fac-
tor enters a cycle of vegetative multi-
plication in the cytoplasm, after which
it matures as intact virus.

This occurs quite frequently when
Lp* cells are treated with ultraviolet
light (Lwoff).

h. Once Lp* forms mature virus the cell
lyses and free virus is liberated.

i. The free lambda can enter other cells
either to multiply as a parasitic virus,
or to reenter the chromosome to again
produce the lysogenic state.

2538
. Other particles in E. coli also act as epi-

somes.

. Itis possible that cytoplasmic factors in

other organisms are due to genic factors
which can enter the cytoplasm.

POST-LECTURE ASSIGNMENT

254

1. Read the notes immediately after the lec-

ture or as soon thereafter as possible,
making additions to them as desired.

. Review the reading assignment.
- Be able to discuss or define orally or in

writing the items underlined in the lecture
notes,

- Complete any additional assignment.
QUESTIONS FOR DISCUSSION

44, 1. What kinds of genetic information obtained
from sexually reproducing organisms cannot
be obtained from organisms reproducing only
asexually ?

44, 2. Why is it futile to search cytologically for
mating couples in an ordinary culture of E.
coli strain K-12 in which F* fertilizes F-
once per million bacteria?

44, 3. How do you suppose it was possible to
prove that the results from mixing auxotrophs
were not the consequence of mutation but ra-
ther of genetic recombination?

44, 4. Suppose anHfrclone has the normal genes
R O TS V while an F” clone is mutant for
these markers.

The clones are mixed at10 a.m. At the
times specified below the "happy couples"
were separated, and analysis showed the nor-
mal genes indicated there had been trans-

ported into the F7 cells.

10:02 a.m. -- none
10:05 a.m. -- T

10:15 a.m. -- OT
10:25 a.m. --SOTV
10:35 aam. --VSOT

Make a genetic map for the marker genes
which is as complete as the data allow.

44. 5. If the DNA in a bacterial nucleus which is
transferred is 22. long, approximately how
much of this is transferred 10 minutes after
fertilization begins ?

Approximately how long does it take for an
average gene to be transferred in a bacterial
fertilization ?

44. 6. Several independently-arising, auxotro-
phic, mutants for the same trait were ob-
tained and crossed to each other in pairs.

What conclusions could you reach from the
results following, from which mutational
events have been excluded, on the supposition
that bacteria can be mated and can be sepa-
rated instantaneously ?

a. Mutant A x B never gave prototrophs,
even when fertilization was permitted
to be completed.

44,

44,

44,

44,

44,

44,

44,

44,

44,

44.

44,

b. Mutant A x C never gave prototrophs
when the pairs were separated before
1,000 seconds, but gave increasing
percentages of prototrophs up to 1, 002
seconds, after which this maximum
frequency remained unchanged.

c. Mutant C x D gave no prototrophs be-
fore 1,001 seconds, but reached its
maximum frequency beginning 1, 002
seconds after mating.

7. Draw a suitably labeled diagram showing
a daughter DNA strand and the maternal and
paternal DNA strands from which it was pro-
duced by a copy-choice mechanism.

8. List as many differences as you can be-
tween Ft and F~ cells. ~

9. List as many differences as you can be-
tween Ft and Hfr cells.

10. Specify the kinds of particulate genetic
matter which may be transferred from one
bacterium to another.

11. In bacteria, is sex type determined geneti-
cally? chromosomally? Explain.

12. Describe how one might obtain experimen-
tal evidence that F is transferred extra-chro-
mosomally from Ft to F~ cells.

13. What would you predict about the chemical
composition of the F particle? Justify your
answer.

14. Describe how you would proceed to obtain
a Gal* Lp* strain from a Gal~ Lp* strain.
How would you test to show the desired
genotype was obtained ?

15. What are the similarities and differences
between F and lambda?

16. What experimental evidence can you give
in support of the schematic diagram at the

left of Tig. 44-1?

17. Which cytoplasmic factors discussed in
Chapters 34 and 35 might be episomes ?

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