Jakob_Quiz2b Key

Jakob_Quiz2b Key - MCB
104
–
Quiz
2
...

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Unformatted text preview: MCB
104
–
Quiz
2
 Section
106
 
 
 
 Name:
____________________________
 
 
 1A
(2
points).
What
is
the
phenotype
of
a
wildtype
hermaphrodite
C.
elegans
worm
in
 which
you
have
used
a
laser
to
destroy
the
anchor
cell
early
in
development?
Why?
 
 The
worm
will
be
vulvaless.
The
EGF
signal
from
the
AC
is
required
to
drive
P6.p
into
 the
vulval
fate.
In
the
absence
of
EGF
all
Pn.p
cells
adopt
the
tertiary
fate.
 

 1B
(2
points).
The
anchor
cell
normally
lies
above
the
P6.p
cell.
How
would
relocating
the
 anchor
cell
next
to
the
P7.p
cell
in
a
wild‐type
hermaphrodite
affect
Pn.p
cell
fates
and
what
 phenotype
would
the
worm
have?
 
 The
cell
closest
to
the
AC
adopts
the
primary
fate
and
adjacent
cells
adopt
the
 secondary
fate.
In
this
worm
the
P7.p
will
be
primary
and
P6.p
and
P8.p
will
be
 secondary.
The
worm
will
still
develop
a
vulva,
although
it
may
not
connect
to
the
 gonad
properly.
 
 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
 A‐
&
C‐
 Yes
 A‐
&
D‐
 No
 D‐
&
B‐
 Yes
 D‐
&
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
 A
and
E
are
in
the
same
 complementation
group
 A‐
&
E‐
 No
 A‐
&
D‐
&
E‐
 (triple
mutant
fusion)
 No
 A,D,
&
E
are
all
in
the
same
 complementation
group
 
 MCB
104
–
Quiz
2
 Section
106
 
 
 
 Name:
____________________________
 
 3A
(6
points).
Predict
the
sex
and
phenotypes
of
the
F1
progeny
resulting
from
the
C.
 elegans
cross
below.
The
dumpy
(dpy)
mutation
is
autosomal
recessive
and
the
 uncoordinated
(unc)
mutation
is
X‐linked
recessive.

 
 
 Hermaphrodites
 
 
 
 
 Males
 dpy/+
;
unc/unc

 
 


X
 








 






dpy/+;
+/0


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

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

 
 
 yellow
x
orange

3
yellow
:
1
orange
in
F2
 
 orange
x
blue

9
red
:
4
blue
:
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.
Yellow
and
orange
are
alleles
of
the
same
gene.
Blue
represents
the
second
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.

 
 The
yellow
mutation
could
either
be
dominant
or
recessive.
Either
way
blue
is
epistatic.

 If
recessive:
9
red
:
4
blue
:
3
yellow
 If
dominant:
9
yellow
:
4
blue
:
3
red
 
 
 ...
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This note was uploaded on 04/05/2010 for the course MCB 104 taught by Professor Urnov during the Spring '09 term at Berkeley.

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