I. Ron:

Seminar 1]1-4-74

A. Discussion about the Adenine marker

l.

2.

log of rhe
Molecular 25 M;

Weight

6.

7.

Always check competent cells before using. -I£ cells have been frozen
thaw them & use immediately for transformation.

Four markers have already been mapped:

Tryp-Tyr biological activity appears at 2,3- em from the origin

Histidine " " " t  =§3.38 em non w
Methionine " 7 " "3.8 cm noon Wo vith 1
Uracil " " i" * 4,5 om "oof '

_ marker at

Now the marker Adenine has been mapped; it seems to survive ata towers, _3 22.
rate. One experiment gave 0.5% survival & another 4-5% survival of

the marker after Ry digestion.

A careful mapping for the adenine marker was done in the heavy region
& #t was not found. The entire gel was cut up into 5 mm slices. The
adenine marker was located several em beyond the other four markers,

6.5«7 em From the origin in the lighter region of the material, when methionine
is at 3 - 3.5 cm.

It was suggested that instead of expressing the distance traveled
by the markers from the origin, description should be made in terms
of M.W. (by use of a phage DNA of known M.W. as a marker & comparing
its distance from the origin with that of other markers). To do this
there must be a linear relationship between the M.wW. & the distance
traveled by the marker. The heavier the marker, the less distance it
would travel.

experimental data from other sources
| \
|

<— bend ( a non-linear relationship between velocity &
loz molecular weight)

 

5M,
1 Mi
distance traveled
——_->
The bend occurs somewhere between 25 and 5 Mill yew.

The range of Ry digested DNA is SearB5sMibtton-Mews 1 - 15 million.
This may indicate thati Re tabiegt pasasas ten, be in the linear range.
Run the electrophoresis for different lengths of times & compare the
distances traveled over the different periods of time by two, light
bands (e.g., SV 40 as one),which are probably in the linear range, with
the heavy band. See if there is a single linear relationship for
the movement of the three bands.
There is a retardation effect on large molecules (a function of weight),
but this can be controlled by changing the % agarose. By diluting the
agarose (using a lower conc.), the larger molecules may move more freely.

Use the electron microscope for statistically good information on size.
Seminar 11-4-74, p.2

8. Transformation Efficiency

Cc.

1.

3.

With the His, marker, there is a difference in transformation
efficiency according to dilution. There is no Linear relationship

in the number of transformed colonies between 10 and 1071. plates.

The 10 olates have more transformed colonies than should occur.

All the other markers show linear relationships between the number of
transformed colonies & the dilution used. Perhaps in the case of

His, there is an oversaturation with DNA; by using more cells, more
DNA can be soaked up & more transformants obtained.

Nontransformed cells may be helping transformed cells in the case of
the His, marker. Test for this in a reconstruction experiment.
Combine cells + DNA for a half hour transformation. Then add DNdAase.
Make dilutions of the transformed cells. Add untransformed cells to
the diluted transformed cells to see if there is an increase in the
transforming activity = helper effect. Trance amounts of His may be
added to the medbum ~ just enough to give the cells a head start.

Reversion is not a problem with histidine.

Transforming activity of markers that survive Ry treatment decreases.

1.

The reason why this occurs is not known.
a. This phenomenon may be connected to size change.
b. It may be due to loss of specific sequences on DNA that are
needed for recombination.
c. New kinds of terminals may be made.
d. With this R, treated DNA there may be a different kind of
transformat fon - Ectopic Insertion.

Use DNA sheared by passage through a needle. Treat with R,.

Shear first - test for biological activity. Then Ry treat & test for
biological activity. Compare the two. Is there any correlation
between the survival of the marker with the starting size of DNA?,

If yes, the phenomenon may be primarily because of decrease in size.
If one shears to a size about the same as Ry pieces, then checks
transformation for Tryp-Tyr, it will be low but measurable. R
treatment of this DNA should not decrease the Tryp-Tyr frequency of
transformation, because the size will be relatively unchanged.

Shear to 20 million M.W. Shear to 10 million M.W. Then in each case
treat with R.. The averaze M.W. change of the Tryp-Tyro piece

will be greater in the 20 million case.

What does exonuclease do to activity after R, treatment?
Exonuclease changes the erds by breaking DNA down from the ends.

If Ry- generated ends are zausing the low transformation efficiency,
then exonuclease I brief treatment may restore efficiency.

There are Rec™ mutants in 3.s.. Contact Don Glazer at Berkely for
these specific strains which lack exonucleases.

Effect of shearing?

a. Starting DNA material is 80 million M.wW.
Sheared material is 40 million M.W.
Ry fragment that contains Tryp marker is about 20 million M.W.
Ry fragment that contains Aro -Tyro marker is less than 13 million.
End

a.

Seminar 11-4-74, p.3

R, treatment only:
Survival,compared to unsheared parentyis 37% for Tryp marker
26% for His marker

Shearing first, then R dizestion:

Survival, compared to the sheared parent ( whose activity was
probably 1/2 that of the unsheared parent), , is 57% for
Tryp and 71% for His.

R, digestion first & then shearing: Transformatéon activity
should be reduced, if the shearing is done with a very fine needle,

Shearing does not operate randomly but tends to break the DNA in
half each time.

Shear to 19 million (half of what is expected from Ry treatment).
Then test the decrease of transforming activity of Tryp-Tyro,
before & after subsequent R digestion.

Or shear to 20 million & do size fractionation, discarding the
smaller pieces, and then test the decrease of Cransforming activity
of Tryp-Tyro, before & after subsequent Ry digestion.

Shearing events are not independent of one another in a large
molecule. The location of the shear cuts relative to the RF
cuts should be random.

Independent of the Location of the marker, the size of the DNA
introduces another efficiency factor. If the size is gotten dow
far enough by shearing, then R_ treatment (on the assumption that
Ry acts only on certain specific sites) should have relatively
less effect on transformation.

Compare the activity of fragments of 10 million obtained by
shearing & by Ry treatment. These fragments can be fractionated
by sucrose gradient or electrophoresis, depending on the distance
traveled by the individual fragments. The survival of Single
markers is approximately the same, excdpt for certain markers
such as Lysine, Glycine, and Aroz. There is less than 107
Survival. A change in weight (size) affects biological activity.

ffects:

Methiontne marker survives 3 times as much as Tryp. This may
be due to end effects.

Use of Ry + certain markers are always at end or always far
away from end.

Use of shearing - the markers'* position is random.
Markers far from an end survive better.

vifferent markers behave differently. The time of integration may
be different for different markers,

Geometry - location of marker on the chromosome.

a.

Though there is experimental evidence that biological activity
decrease is due to decrease irithe size of DNA, size is not all,
we believe.
If.

Seminar 11-4-74, p.4

b. If one treats DNA with Rr, electrophoreses it on gel, & assays
the gel for single markers, it is suggested that geometry
differences could account for the difference in survival.

i. Survival of Tyr alone is twice as high as Tryp-Tyr together.
ii. Survival of Tryp alone is 7-8 times as high as Tryp-Tyr together.
lii. The lowest survival is for Tryp-Tyr.

iv. His, has the highest survival (20 X)3; it is the most
centrally located marker.

c. This suggests a very specific geometry: Biological activity is
related to the marker's position on the chromosome.

d. Sheared DNA shows no such systematic effect.

e. Rr treatment gives a definite location of a marker.

Dusko: Densitometry

l.

2
3.

To record the band pattern obtained from gel electrophoresis.
The scan is very reproducible.
SV40 and T7 markers were introduced for positions, to help calculate
M.W.s , and to evaluate the noise level; each gives the band width for
a homogeneous DNA - a single narrow peak each.
To lose the very high noise signals, gel must be removed from quartz tubes.
The noise is noise of film & not of instrument. Take the average of many
runs & noise signals will cancel out.
How much DNA is present between bands is another problem.
Do a standard curve using known DNA concentrations as control.
In the region between T7 - SV40, about 20 resolvable bands are seen.
The relative intensity of the bands is not seen too well:
a. because the base line is too high.
b. maybe there are changes in the dye contrast; perhaps dye is washed
out at one end of gel,
¢. maybe there are changes in the light concentration; UV intensity
may not be uniform from one end to the other.
What is shown is that the bands do exist and that they are reproducible.
Troughs are places with fewer species of DNA; peaks are sites with a
greater number of species of DNA.
Histidine is in a trough - in the biggest interband.
Factors in band width:
a. diffusion of the DNA - there is a relationship between M.W. &
diffusion coefficient.
b. optical setup which causes scattering
c. artifacts
d. band width as a function of M.W.- <2iffusion is dependent on M.W. &
artifacts are not.
Seminar 11-4-74, p.5

III. Hela:

A. Beneckea Doubling Time
In minimum medium + 2% NaCl + 0.25% glucose: 48 minutes
In complete medium of TB + 2% NaCl 30-35 minutes

B. Crossing Experiments:
Parents: B.s. $B 363 Lyst+ Tryp- Tyr- Aroj- leu- cys- His,- Strep®

B.s. SD8 Lys- Tryp+ Tyr+ Aroj+ leut+ cys- Hise Strep>

1. Plates with minimal medium + required amino acids

Ss -
a. SD8 alone (10, 1o7l, 1072, 10 3 concentrations)
without cysteine ~ no colonies
with cysteine- no colonies

b. 863 alone, with or without cysteine - 0 colonies
c. 853 + SD8 without cysteine- 0 colonies
d. 863 + SD8 + cysteine

10 combined concentration of both kinds of cells - heavy overgrowth ofedt:

197! tt w ry tl 2) ty LS still te "w t ew
but some individual
2 colonies.
1o7* " " " " " " . individual colonies
1073 ve et Hh t af i Ut 143 " - (not as many:
e. Minimum medium + leu + his: Parents - crossing -
" " + leuthistcys: " - " +

When the two kinds of cells were grown together in mautrient broth, when plated,
a positive result was obtained - colonies grew on plates with cysteine.
Possibly this represents:
i, a recombination of Tryp+ Tyr+ Arog+ of SD8 with Tryp-Tyr-Aroz- of 863
or Lyst+ of 863 with Lys- of SD8.
ii. a mutual transformation.

2. Suggestions:

a. The bacteria were grown together in broth with shaking for 48 hours
becaus2 B.s. lyses without such aeration. But shaking is detrimental
to conjugatioh. Therefore, grow each parent separately with shaking
& then do the conjugation for a limited period of time in fresh medium
without or with gently shaking.

b. Repeat the experiment with a shorter period of interaction between the
cells.

ce. See which markers may be segregated (look at many cells using replica
plating). Purify the colonies and check out their phenotypes.

d. The parents may be releasing DNA while growing (in broth or/& agar)
& the DNA may be performing transformation. Therefore, add DNAase.
Seminar 11-4-74, p.6

3. Testing of Single Markers:

The two kinds of bacteria were grown in nutrient broth with shaking

for 48 hours.

They were then plated together for 12 days on plates
with amino acids minus Tryp, Tyro, Leuj, Lys.

The four bacterial

colonies that grew from this first crossing experiment of 8653 ® $D3

 

 

were tested for single markers (the colonies were unpurified). tte
Minimal nutrient
tHis,Tyr,
Tryp,Lys, 434F
all essential all essential all essential all essential nee ic +
Colonv# AAs put minus AAs but minus AAs ‘but minus AAs but minus bae Arsds 400%/ml
ny YS Phe Tryp His Cys Strep.
L - + + + _- -
2 - + + + - 7
3 + + + + -
4 + + + + - +
+ indicates growth of bacteria

Ww re

ve

indicates bacteria doesn't grow
and 2 are like parent $D8

and 4 indicate recombination (7)

If all essential amino acids minus cysteine are used, alot of reversion occurs

(cell growth).

Consequently, in testing for the cysteine marker,

minimal medium + only the amino acids required by 863 and SD8 arevused, minus

cysteine.

2c