Answers Part 3 a

Answers Part 3 a - Question 1. A c onju g atio n e xp e...

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Unformatted text preview: Question 1. A c onju g atio n e xp e rime nt is p e rformed i n Escherichia c oli using t he par e n ts Hfr arg+ lys+ met+ h is+ strS and F- a rg- lys- m et- h is- strR. Th e m atin g was interrupted at several times an d plated on several diffe rent me dia with the following r esults. T h e t able below shows the n umb e r o f colo nies o btained o n differe nt me dia as i n dicated . Time (min .) M i n i mal me dium plus streptomy cin plus Histidine Ar ginine Ar ginine L ysine H istidine H istidine Methio nine Methio nine Lysine 0 0 0 0 0 0 20 0 15 1 70 60 2 00 600 3 50 5 50 0 10 20 30 40 Ar ginine Lysine Methio nine 0 50 4 00 1 0 00 1 0 00 What
is
the
order
of
the
four
auxotrophic
genes
with
respect
to
the
starting
 point
of
transfer
(0)?
 
 A.
0
arg /met 
his 
lys
 
 B.
0
arg /met 
lys 
his
 
 C.
0
his 
 
arg /met 
lys
 Answer:
 1
 Question 2. 
 A
cross
is
made
in
E.
coli
between
an
Hfr
strain
that
is
leu+
his+
 gly+
Strs
and
an
F‐
strain
that
is
leu‐
his‐
gly‐
StrR.

Interrupted
 mating
studies
show
that
his+
enters
the
recipient
last.

 
 
In
an
interrupted
mating
between
the
same
two
strains,
what
medium

 would
you
select
the
conjugates
on
before
screening
for
recombinants?
 A .  B.  C.  D.  Answer:
 Medium
containing
his
and
lacking
streptomycin
 Medium
containing
his
and
containing
streptomycin
 Medium
lacking
his
and
containing
streptomycin
 Medium
lacking
his
and
lacking
streptomycin
 2
 Question 3. You
select
on
medium
lacking
his
and
containing
streptomycin
and

 then
test
by
replica
plating
for
the
presence
of
the
leu+
and
gly+
alleles.
 The
following
numbers
of
individuals
are
found
for
each
genotype:
 
 gly+
leu+
his+ 
= 
259
 
 gly‐

leu+
his+ 
= 




1
 
 gly+
leu‐
his+ 
= 


11
 
gly‐

leu‐
his+ 
= 


29
 
 
 
 















300
total
 What
is
the
gene
order?
 A.
gly
leu
his
 B.
leu
gly
his
 C.
gly
his
leu
 Answer:
 3
 Question 4. You
select
on
medium
lacking
his
and
containing
streptomycin
and

 then
test
by
replica
plating
for
the
presence
of
the
leu+
and
gly+
alleles.
 The
following
numbers
of
individuals
are
found
for
each
genotype:
 
 gly+
leu+
his+ 
= 
259
 
 gly‐

leu+
his+ 
= 




1
 
 gly+
leu‐
his+ 
= 


11
 
gly‐

leu‐
his+ 
= 


29
 leu
gly
his
 
 
 
 
















300
total
 You
have
determined
the
order
as
leu
gly
his.
 Now,
what
are
the
map
distances
between
these
three
genes?
 A.  leu
 

4mu






gly



10mu



his

 B.


leu
 
10mu 
gly




4mu



his
 C.


leu 

10mu 
gly
 

10mu


his
 Answer:
 4
 Question 5. In
Griffiths’
transformation
experiment,
suppose
that
the
genes
needed
to
make
the
 harmless
R
strain
become
a
virulent
S
strain
are
called
enzB+,
tox1+
and
attC+.
 
 In
your
transformation
experiments,
you
find
that
transforming
R
to
S
occurs
very
rarely.

 Looking
more
closely
at
the
individual
genes,
you
find
that
many
transformants
are
 enzB+
attC+,
but
rarely
are
there
enzB+
tox1+
or
attC+
tox1+
transformants.
 
 What
does
this
suggest
about
relative
placement
of
the
enzB,
tox1
and
attC
genes?
 a.
tox1
is
far
from
enzB
but
not
attC
 b.
enzB
is
close
to
attC
 c.
attC
is
inbetween
enzB
and
tox1
 d.
enzB
is
close
to
attC
and
tox1
is
far
from
enzB
and
attC
 
 Answer
is
d.

For
enzB
and
attC
to
be
transformed
together
often,
they
must
be
relatively
 
close
together.
tox1
must
distant
from
both,
as
tox1
is
rarely
transformed
along
with

 either
of
the
other
two
genes.
 5
 ...
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