Jakob_Quiz2a Key - MCB
104
–
Quiz
2
...

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Unformatted text preview: MCB
104
–
Quiz
2
 Section
105
 
 
 
 
 Name:
____________________________
 
 1A
(2
points).
You
are
studying
a
hermaphrodite
C.
elegans
worm
with
a
gain
of
function
 mutation
in
let‐60(Ras).
What
would
be
the
phenotype
of
this
worm
if
you
used
a
laser
to
 destroy
the
anchor
cell
early
in
development?
Why?
 
 Multivulva.
In
a
let­60
gain
of
function
worm
the
pathway
promoting
vulval
fate
is
 constitutively
turned
on,
independent
of
the
EGF/EGFR
signal.
Therefore
it
doesn’t
 make
a
difference
if
the
AC
is
present
or
not.
 
 1B
(2
points).
The
anchor
cell
normally
lies
above
the
P6.p
cell.
How
would
relocating
the
 anchor
cell
next
to
the
P4.p
cell
in
a
wild‐type
hermaphrodite
affect
Pn.p
cell
fates
and
what
 phenotype
would
the
worm
have?
 
 The
P4.p
cell
would
adopt
a
primary
fate
and
P3.p
and
P5.p
would
become
secondary.
 These
cells
would
develop
into
a
vulva,
although
it
probably
would
not
connect
to
the
 gonad
and
therefore
me
non­functional.
 
 2
(6
pts).
You
have
isolated
five
neurospora
mutants
(A‐E)
that
fail
to
grow
in
the
absence
 of
vitamin
A.
You
make
the
following
double‐mutants
by
fusing
two
haploid
neurospora
 into
a
heterokaryon,
and
test
for
growth
on
minimal
media
(lacking
vitamin
A).
The
“&”
 indicates
fusion.
 Double
Mutant
 Grows
on
Minimal
Media
 B‐
&
D‐
 Yes
 B‐
&
C‐
 No
 C‐
&
A‐
 Yes
 C‐
&
E‐
 No
 
 Fill
out
the
table
below
with
your
predictions
for
the
mutant
combinations
listed.
In
10
 words
or
less,
explain
your
answers.

 
 Mutant
Combination
 Grows
on
 Why
(10
words
or
less)
 Minimal
Media?
 A‐
&
B‐
 Yes
 A
and
B
are
in
different
 complementation
groups.
 B
and
E
are
in
the
same
 complementation
group.
 B‐
&
E‐
 No
 B‐
&
C‐
&
E‐
 (triple
mutant
fusion)
 No
 B,
C,
and
E
are
all
in
the
same
 complementation
group.
 
 MCB
104
–
Quiz
2
 Section
105
 
 
 
 
 Name:
____________________________
 
 3A
(6
points).
Predict
the
sex
and
phenotypes
of
the
F1
progeny
resulting
from
the
C.
 elegans
cross
below.
The
uncoordinated
(unc)
mutation
is
autosomal
recessive
and
the
 dumpy
(dpy)
mutation
is
X‐linked
recessive.

 
 
 Hermaphrodites
 
 
 
 
 Males
 unc/+
;
dpy/dpy

 
 


X
 








 






unc/+;
+/0


 
 
 Remember
that
the
self
cross
here
is
the
same
as
unc/+;dpy/dpy
X
unc/+;
dpy/dpy,
so
 you
get
three
different
genotypes
and
two
different
phenotypes
resulting
from
the
self
 cross.

 F1
hermaphrodites:
Dpy
(self),
DpyUnc
(self),
Unc
(cross),
wt
(cross)
 
 F1
males:
DpyUnc
and
Dpy
 
 3B
(3
points).
Can
you
distinguish
the
self
progeny
from
the
cross
progeny?
If
so,
what
are
 the
phenotypes
of
each?
 
 Yes.
All
the
hermaphrodites
resulting
from
the
self
will
be
dumpy
while
all
the
 hermaphrodites
from
the
cross
will
be
wild­type
for
dumpy.

 
 4.
You
are
studying
a
strain
of
red
roses.
Using
mutagenesis
you
isolate
three
pure‐ breeding
mutants
that
have
yellow,
orange,
or
blue
flowers.
To
better
characterize
your
 mutants
you
perform
the
crosses
below,
cross
the
F1
progeny
to
each
other,
and
analyze
 the
F2
progeny.

 
 
 blue
x
orange

3
blue
:
1
orange
in
F2
 
 orange
x
yellow

9
red
:
4
yellow
:
3
orange
in
F2
 
 4A
(2
points).
How
many
distinct
genes
have
you
identified
in
your
screen?
Assume
all
of
 the
genes
are
unlinked.
 
 Two
genes.
Blue
and
orange
are
two
alleles
of
the
same
gene.
Yellow
is
in
a
different
 gene.
 
 4B
(4
points).
Predict
the
phenotypes
and
ratios
in
the
F2
generation
of
blue
x
yellow
cross.
 Both
parents
in
this
cross
are
pure‐breeding.

 
 Orange
and
yellow
are
definitely
recessive
mutations
and
yellow
is
epistatic
to
orange.
 The
blue
mutation
could
be
dominant
or
recessive,
but
either
way
yellow
should
be
 epistatic
since
blue
is
a
mutation
in
the
same
gene
as
orange.
 
 If
blue
recessive:
9
red
:
4
yellow
:
3
blue
 If
blue
dominant:
9
blue
:
4
yellow
:
3
red
 
 ...
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