Jakob_PS6

Jakob_PS6 - Problem
Set
6
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Unformatted text preview: Problem
Set
6
 MCB
104
–
Section
105/106
 
 Things
to
know
for
Quiz
II
 
 1. C.
elegans
genetics
(hermaphrodites
vs.
males;
cross‐progeny
vs.
self‐progeny)
 2. Cells
and
signaling
pathways
underlying
vulva
development
 3. Characterizing
a
mutation
(see
below)
 a. Dominant
vs.
recessive
 b. Type
of
allele
(loss
of
function,
dominant
negative,
etc.)
 c. Allele
series
 d. Penetrance
vs.
expressivity
 4. Biological
pathway
analysis
using
genetics
 a. Auxotroph
experiments
(arg‐)
 b. Complementation
to
identify
mutations
in
same
gene
 c. Epistasis
to
determine
the
order
 i. Epistasis
in
yeast/neurospora
 ii. Epistasis
in
mendelian
genetics
 iii. Epistasis
in
signaling
pathways
(NOT
INCLUDED
IN
QUIZ)
 
 Characterizing
a
mutation
 
 m
=
mutant
allele
of
interest
 +
=
wild‐type
allele
 null
or
Df
=
complete
loss
of
function
 lf
=
recessive
loss
of
function
 
 First
analyze
the
heterozygote
to
determine
if
mutation
is
dominant
or
recessive,
then
 compare
to
other
mutants
to
further
characterize
the
type
of
mutation.
 
 1. +/m
=
mutant
phenotype

dominant
 a. +/m
=
dead
→
dominant
lethal
 b. +/m
=
lf/lf
→
dominant
negative
 c. +/m
=
+/null
→
haploinsufficient
 d. else
→
gain
of
function
 
 2. +/m
=
wild‐type
phenotype

recessive
 a. m/m
=
dead
→
recessive
lethal
 b. m/m
=
Df/Df
→
null
mutation
 c. else
→
weak
loss
of
function
allele
 
 3. +/m
=
intermediate
phenotype

incompletely
dominant
 
 
 
 Mendelian
Genetics
&
Gene
Interactions
 
 Analyzing
the
ratio
of
phenotypes
in
the
F2
generation
in
a
cross
between
two
mutants
can
 give
you
information
about
the
underlying
genes
and
alleles.
The
results
can
be
interpreted
 as
follows:
 
 3:1
phenotypic
ratio

One
gene
involved;
two
alleles
of
the
same
gene.
The
allele
giving
 rise
to
the
3x
phenotype
is
dominant.
(e.g.
alleles
of
albino
gene
in
mice
determine
fur
 color)
 
 9:3:3:1
phenotypic
ratio

Two
genes
involved;
no
gene
interaction.
(e.g.
pea
shape
vs.
pea
 color)
 
 9:7
phenotypic
ratio

Two
genes
involved;
no
epistasis.
Genes
are
in
same
pathway
and
 both
give
rise
to
the
same
phenotype
when
mutated.
(e.g.
gene
1
is
required
for
growth
and
 gene
2
is
required
to
turn
on
gene
1)
 
 9:3:4

Two
genes
involved;
recessive
epistasis.
Gene
that
gives
rise
to
the
4x
phenotype
 when
homozygous
mutant
is
epistatic
when
both
genes
are
homozygous
mutant.
(e.g.
two
 genes
required
to
synthesize
intermediates
in
a
pigmentation
pathway)
 
 12:3:1

Two
genes
involved;
dominant
epistasis.
Gene
that
gives
rise
to
the
12x
 phenotype
is
epistatic
as
long
as
at
least
one
dominant
allele
is
present.
(e.g.
Gene1
shuts
 off
pigment
expression
as
long
as
one
functional
allele
is
present,
gene2
synthesizes
the
 pigment.)
 
 Practice
Problems
 
 1.
Predict
if
each
mutation
below
will
most
likely
lead
to
a
loss
of
function,
dominant
 negative,
or
gain
of
function
allele.
 
 Frameshift
mutation
in
Ras
 
 
 Nonsense
mutation
in
the
EGFR
that
eliminates
the
cytosolic
tail
of
the
receptor
 
 
 Mutation
in
the
promoter
region
of
Ras
that
creates
a
novel
transcription
factor
 binding
site
 
 
 Mutation
in
the
splice
acceptor
site
between
the
middle
two
exons
of
Ras
 
 
 Point
mutation
in
Ras
that
destroys
the
binding
site
of
Ras‐GAP
 2.
You
are
studying
recessive
mutations
in
two
unlinked
C.
elegans
genes.
One
mutation
 gives
rise
to
a
dumpy
phenotype
(Dpy)
and
the
other
causes
the
worms
to
be
 uncoordinated
(Unc).
For
all
the
questions
below
assume
that
worms
are
homozygous
for
 all
alleles.
 
 You
mate
an
Unc
hermaphrodite
with
a
wild‐type
male.
You
get
a
mix
of
wild‐type
and
Unc
 hermaphrodites,
but
all
the
males
are
Unc.
What
does
this
tell
you
about
the
location
of
the
 unc
gene?
 
 
 
 Next
you
mate
a
hermaphrodite
that
is
both
Unc
and
Dpy
with
a
wild‐type
male.
Predict
the
 phenotype(s)
of
the
resulting
offspring.
Which
hermaphrodites
are
self
progeny
and
which
 are
cross
progeny?
 
 
 
 

 Finally,
you
mate
a
Dpy
hermaphrodite
with
a
Dpy
and
Unc
male.
Predict
the
phenotype(s)
 of
the
resulting
offspring.
Can
you
distinguish
the
self
progeny
from
the
cross
progeny?
 
 
 
 
 3.
You
are
studying
the
assembly
of
the
car
protein
in
neurospora
(haploid),
which
is
 required
for
growth.
You
isolate
four
car
auxotrophs
(car1,
car2,
car3,
car4)
that
fail
to
 grow
on
minimal
media
but
grow
on
minimal
media
+
car.

 
 How
could
you
show
that
these
mutants
are
all
mutated
in
different
genes?
 
 
 +
=
growth;
‐
=
no
growth
 
 Car1­
 Car2­
 Car3­
 Car4­
 Min
media
 ‐
 ‐
 ‐
 ‐
 






+
wheels
 ‐
 +
 +
 +
 






+
seats
 ‐
 +
 ‐
 +
 






+
engine
 ‐
 ‐
 ‐
 ‐
 






+
gas
 ‐
 +
 ‐
 ‐
 






+
car
 +
 +
 +
 +
 

 Using
the
table
above,
draw
the
biosynthetic
pathway
of
cars
in
neurospora.
Which
gene
is
 responsible
for
each
step
of
the
pathway?
 
 
 Next
you
generate
the
heterokaryons
listed
below
(fusion
of
2
or
3
haploid
neurospora
to
 generate
one
organism
with
a
shared
cytoplasm).
Predict
if
each
heterokaryon
will
or
will
 not
grow
on
minimal
media.
For
each
mutant
that
you
predict
would
fail
to
grow,
name
one
 supplement
(other
than
car)
that
could
restore
growth.
 
 Heterokaryon
 Growth?
 Supplement
 Car1‐
&
Car3‐
 
 
 Car1‐,Car2‐
&
Car2‐
 
 
 Car4‐,
Car3‐
&
Car4‐,
Car3‐
 
 
 Car1‐,
Car2‐
&
Car3‐,
Car4‐
&
Car3‐
 
 
 
 
 4.
You
are
studying
a
plant
that
has
red
flowers
when
all
genes
are
wild‐type.
You
isolate
 three
mutants
that
give
rise
to
white,
green,
and
blue
flowers
respectively.
You
perform
the
 following
crosses:
 
 Cross
 F1
Phenotypes
 F2
Phenotypes
 white
x
green
 all
red
 9
red
:
3
green
:
4
white
 green
x
blue
 all
red
 9
red
:
3
blue
:
4
green
 
 What
ratio
of
phenotypes
would
you
predict
from
a
white
x
blue
cross?
What
is
the
order
of
 these
genes
in
the
synthetic
pathway
for
red?
 
 
 
 5.
Name
the
component
of
the
RTK
signaling
pathway
encoded
by
each
of
the
following
C.
 elegans
genes.
 
 lin‐1
__________________________

 let‐23
_________________________
 
 
 let‐60
________________________
 
 lin‐3
__________________________
 
 Which
gene(s)
is
required
in
the
anchor
cell
for
vulval
development?
Which
gene(s)
is
 required
in
the
Pn.p
cells
for
vulval
development?
 
 
 
 What
type
of
mutation
in
let‐60
leads
to
a
multi‐vulva
phenotype?
What
type
of
mutation
in
 let‐23
leads
to
a
vulvaless
(bag
of
worms)
phenotype?

 
 
 
 What
would
be
the
phenotype
of
a
lin‐1
loss
of
function,
lin‐3
gain
of
function
double
 mutant?
 ...
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