Jakob_PS7 Key

Jakob_PS7 Key - Problem
Set
7
 


Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem
Set
7
 
 1.
Bicoid
is
one
of
several
maternal
effect
genes
required
for
the
successful
development
of
 Drosophila
embryos.

You
have
isolated
a
recessive
mutation
in
the
bicoid
gene
(bc‐).
For
 each
cross
below,
list
the
genotypes
of
all
F1
progeny
that
you
predict
will
survive
to
 adulthood.
 
 bc‐/+
female
X
bc‐/+
male
 
 all
progeny
(bc­/+;
bc­/bc­;
+/+)
survive
because
mother
is
het
 
 bc‐/+
female
X
bc‐/bc‐
male
 
 
 all
progeny
(bc­/+;
bc­/bc­)
survive
because
mother
is
het
 
 bc‐/bc‐
female
X
bc‐/+
male
 
 
 all
progeny
(bc­/bc­;
bc­/+)
die
because
mother
is
homozygous
mutant
 
 2.
You
are
studying
a
recessive
loss
of
function
mutation
in
a
fly
gene
called
extensin.The
 mRNA
for
extensin
is
deposited
in
the
egg
by
the
mother
and
is
required
for
normal
 proportioning
of
the
adult
fly.
The
endogenous
gene
for
extensin
is
turned
on
later
in
 development
and
is
required
for
flies
to
grow
to
their
normal
size.
Predict
the
genotype
and
 the
phenotype
of
the
progeny
from
the
crosses
listed
below.
Choose
from
the
following
 phenotypes:
(1)
disproportionate;
(2)
small;
(3)
small
&
disproportionate;
(4)
wild‐type.
 
 If
the
mother
is
mutant
the
offspring
will
be
disproportionate.
If
the
offspring
are
mutant
they
 will
be
small.
In
other
words,
the
proportion
is
determined
by
maternal
effect
while
size
is
 determined
just
like
any
autosomal
recessive
mutation.
 
 Ex‐/+
female
X
Ex‐/+
male
 
 
 
 Wild­type
(Ex­/+;
+)
and
small
(Ex­/Ex­)
 
 Ex‐/Ex‐
female
X
Ex‐/+
male
 
 
 
 Disproportionate
(Ex­/+)
and
Small
&
disproportionate
(Ex­/Ex­)
 
 +/+
female
X
Ex‐/Ex‐
male
 
 
 
 All
wild­type
(Ex­/+)
 
 MCB
104
–
Section
105/106
 3.
One
of
the
characteristics
of
mitochondrial
diseases
is
that
their
phenotype
can
change
 significantly
over
time.
What
unique
feature
of
organelle
genetics
likely
explains
this
 phenomenon?
 
 “Heteroplasmy”
–
a
mix
of
wild­type
and
mutant
mitochondria
in
each
cell.
In
most
cases,
in
 order
to
express
the
mutant
phenotype
a
cell
must
reach
a
threshold
ratio
of
mutant
 mitochondria
to

wild­type
mitochondria.
Every
time
a
cell
divides
this
ratio
can
shift
based
on
 random
segregation
of
the
mitochondria.
Therefore,
as
the
cells
in
a
given
tissue
turn
over
 they
may
go
from
wild­type
to
mutant
or
vice
versa.
 
 4.
What
determines
sex
in
humans?
What
about
in
flies?
 
 In
humans
the
presence
of
a
Y
chromosome
determines
maleness.
In
flies
the
number
of
X
 chromosomes
determines
sex.
The
Y
chromosome
has
no
influence
in
flies.

 
 5.
Why
do
XXX
mammals
survive
but
XXX
flies
die?
 
 Because
dosage
compensation
in
mammals
turns
OFF
extra
copies
of
the
X
chromosome,
no
 matter
how
many
extra
copies
there
are.
In
flies
there
is
no
mechanism
for
turning
OFF
 expression
of
X­linked
genes
so
they
will
be
overexpressed
in
a
XXX
fly
and
lead
to
death.


 
 6.
For
each
mutation
listed,
give
an
example
of
a
possible
extragenic
suppressor
mutation.
 For
each
suppressor,
indicate
if
it
would
be
an
informational
or
interaction
suppressor,
and
 ifit
is
gene
specific
or
non‐specific.
There
may
be
more
than
one
possible
answer.

 
 
 A
mutation
in
the
poly‐adenylation
signal
of
the
Ras
mRNA.
 
 A
mutation
in
the
polyadenylation
machinery
that
allows
it
to
recognize
the
mutant
 poly­adenylation
signal.
Informational
suppressor.
Gene
non­specific.
 
 
 A
mutation
in
EGF
that
disrupts
the
EGF‐receptor
binding
site
 
 A
mutation
in
EGFR
that
compensates
for
the
mutation
in
EGF
and
restores
binding.
 Interaction
suppressor.
Gene
specific.

 
 
 A
point
mutation
in
the
ribosome
binding
site
of
the
EGF
mRNA
that
disrupts
the
 recruitment
of
ribosomes
to
the
mRNA
 
 A
mutation
in
the
ribosomes
that
allows
for
recognition
of
the
mutated
ribosome
 binding
site.
Informational
suppressor.
Gene
non­specific.
 
 7a.
Imagine
that
you
have
generated
a
transgenic
mouse
that
has
a
copy
of
the
gene
for
 green
fluorescent
protein
(GFP)
integrated
into
the
X
chromosome
(XGFP).
You
cross
a
 female
mouse
homozygous
for
the
GFP
transgene
with
a
wild‐type
male.
What
will
the
 pattern
of
GFP
expression
be
in
male
offspring?
 
 All
cells
will
be
GFP+.
The
males
get
their
X
chromosome
from
the
transgenic
mother
and

 because
there
is
no
dosage
compensation
in
male
mammals
the
single
copy
of
X
is
fully
 expressed.

 
 7b.
What
will
the
pattern
of
GFP
expression
be
in
female
offspring?
 
 50/50
mix
of
GFP+
and
GFP­
cells.
The
female
offspring
have
one
X
chromomsome
from
mom
 that
includes
the
GFP
gene
and
one
X
chromosome
from
dad
that
dose
not
have
GFP.
The
 phenotype
of
each
cell
depends
on
which
X
chromosome
was
randomly
inactivated.
 
 7c.
Next
you
delete
the
Xist
locus
in
the
transgenic
mouse
and
repeat
the
cross
described
 above.
How,
if
at
all,
will
this
change
the
expression
of
GFP
in
the
F1
offspring?
 
 No
change
in
male
offspring.
Female
offspring
will
now
express
GFP
in
all
cells.
Because
only
 the
X
chromosome
from
dad
has
a
functional
Xist
locus
it
will
always
be
inactivated
while
the
 maternal
X
(with
the
GFP)
will
always
remain
active.
 
 7d.
As
the
XGFPXGFP
female
mice
age,
some
of
them
develop
tumors.
You
isolate
the
cells
 from
the
tumor
of
one
mouse
and
find
that
100%
of
the
cells
are
green.
When
you
isolate
 cells
from
a
tumor
in
another
mouse
you
find
that
none
of
the
cells
are
green.
How
can
this
 be?

 
 Tumors
are
usually
clonal,
arising
from
a
single
progenitor
cell.
Since
the
X­inactivation
state
 is
passed
stably
from
parent
cell
to
daughter
cell
all
of
the
cells
in
the
tumor
will
inactivate
 the
same
X
chromosome
as
the
original
progenitor.
Any
given
tumor
has
a
50/50
change
of
 being
GFP+
or
GFP­,
but
either
way
all
cells
in
the
tumor
are
the
same.
 
 
 
 
 8.
You
have
discovered
a
point
mutation
in
the
protease
caspase‐1
that
converts
a
cysteine
 (TGT)
codon
into
a
serine
(TCT)
codon
at
position
189.
Because
this
cysteine
normally
 forms
the
core
of
the
catalytic
site
in
caspase‐1,
the
point
mutant
has
lost
all
protease
 activity.
In
a
screen
for
suppressor
mutations
you
find
that
a
mutation
in
a
tRNA
can
 restore
caspase‐1
to
its
wild‐type
function.

 
 In
which
tRNA
would
you
expect
to
find
this
suppressor
mutation?
 
 The
suppressor
mutation
would
have
to
be
in
a
cysteine
tRNA.
Becasuse
this
cysteine
is
 so
important
to
the
function
of
caspase­1,
replacing
the
mutant
serine
with
any
amino
 acid
other
than
cysteine
will
not
rescue
the
function.

 
 Draw
the
base‐pairing
that
occurs
between
the
mRNA
of
caspase‐1
and
tRNAs
at
 position
189
in
a
wild‐type
background
vs.
a
suppressed
background.

 
 
 
 
 
 Wild­type
 
 
 Suppressed
 
 
 9.

You
perform
the
following
fly
cross:
w+/w‐

female
X

w+/Y
male.

 One
of
the
F1
females
is
white‐eyed.

Which
parent
underwent
a
meiotic
error?

In
which
 stage
of
meiosis
did
the
error
occur?
 
 Since
w
is
a
recessive
mutation,
the
white­eyed
female
must
be
w­/w­/Y.
Only
the
mother
has
 the
w­
allele,
so
the
error
must
have
been
in
the
mother.
If
the
error
occurred
during
meiosis
I,
 when
homologous
chromosomes
segregate,
the
offspring
would
be
w+/w­/Y
and
therefore
 still
red­eyed.
The
error
occurred
in
meiosis
II
when
sister
chromatids
(w­/w­)
failed
to
 segregate.



 tRNA
 
 ACA
 
 
 
 AGA
 
 mRNA
 
 UGU
 
 
 
 UCU
 
 *Remember
that
T
in
DNA
is
replaced
by
U
in
RNA
 
 ...
View Full Document

Ask a homework question - tutors are online