12. Lab 7 Chromosomal Mutations Answer Key

12. Lab 7 Chromosomal Mutations Answer Key -...

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Unformatted text preview: Lab
7

Chromosomal
Mutations

































Name:_____________________________________
 
 SSID:_________________________
 1. (3
marks
total)
 Why
do
most
polyploidy
species
tend
to
have
even
numbers?
(E.g.,
 Why
are
triploid
 species
 almost
always
sterile?)
b)
 How
would
you
account
for
the
appearance
of
individuals
with
an
 odd
number
of
chromosomes
(that
is,
21,
39,
etc.)
in
series
of
polyploidy
species
with
2n
=
20,
40,
60,
 80?
 
 a) (2 marks total) Polyploids
with
an
even
number
of
chromosomes
are
more
likely
to
have
a
regular
 meiosis
than
those
with
an
odd
number,
and
so
are
more
likely
to
be
fertile.
If
chromosome
number
is
 an
 odd
 number
 –
 meiosis
 will
 produce
 mostly
 unbalanced
 gametes.
 (e.g., 
 3
 sets
 of
 chromosomes
 must
segregate
into
2
daughter
cells;
most
likely
one
daughter
will
end
up
with
2
chr
&
the
other
will
 have
 only
 one
 chromosome
 from
 any
 set
 of
 homologs.
 (1 mark).
 This
 is
 because
 each
 of
 the
 chromosomes
 will
 have
 a
 homologue
 with
 which
 to
 pair
 and
 align
 at
 the
 metaphase
 plate
 during
 meiosis
I. Many
combinations
of
incorrect
number
of
chromosomes
will
occur.(1 mark)
 b) (1 mark) These
 specimens
 represent
 the
 occasional
 aneuploid.
 For
 instance,
 the
 one
 with
 21
 chromosomes
would
have
one
extra
chromosome,
and
the
one
with
39
chromosomes
would
lack
one.
 Nondisjunction
would
be
the
cause
in
both
cases. 
 2. (3
marks
total)The
following
chromosomes
are
related
to
each
other
by
a
series
of
inversions
and
are
 listed
 in
 the
 order
 in
 which
 they
 arose.
 
 For
 each
 of
 these
 changes,
 indicate
 on
 the
 preceding
 chromosome
(the
one
above)
the
breakpoints
that
would
give
rise
to
each
inversion.
 
 
 
 I:
A

B

C

D

E

F

*

G

H

I

J

 II:
A

B

F

E

D

C

*

G

H

I

J
 III:
A

B

F

E

H

G

*

C

D

I

J
 IV:
A

B

F

E

H

G

*

I

D

C

J
 These
inversions
are
overlapping:

 I:
A

B
↓
C

D

E

F
↓
*

G

H

I

J

 II:
A

B

F

E
↓
D

C

*

G

H
↓
I

J
 III:
A

B

F

E

H

G

*
↓
C

D

I
↓
J
 IV:
A

B

F

E

H

G

*

I

D

C

J
 
 
 
 
 
 3. (3
marks
total)
Distinguish
between
pseudodominance,
pseudogenes,
and
pseudolinkage.
 Lab
7

Chromosomal
Mutations

































Name:_____________________________________
 (3 marks total; 1 for each term) Pseudodominance
 is
 when
 a
 normally
 recessive
 allele
 appears
 dominant
due
to
deletion
of
the
masking
dominant
allele.
A
pseudogene
is
an
inactive
gene
derived
 from
 an
 ancestral
 active
 gene.
 This
 usually
 occurs
 through
 gene
 duplication
 followed
 by
 mutation.

 Pseudolinkage
 is
 apparent
 linkage
 of
 genes
 known
 to
 be
 on
 nonhomologous
 chromosomes.
 This
 results
from
a
reciprocal
translocation.


 4. (1
 mark)
 Gene
 mapping
 experiments
 on
 a
 species
 of
 plant
 show
 that
 all
 genes
 belong
 to a
 single
 linkage
 group
 even
 though
 the
 plant
 is
 known
 to
 have
 four
 pairs
 of
 chromosomes.
 What
 could
 account
for
this?
 
 (1
mark)
A
series
of
reciprocal
translocations
involving
all
four
pairs
of
chromosomes
would
account
 5. (1
mark)
Consider
the
following
wild‐type
&
mutant
sequences:
 
 
 WT
 
 …CTTGCAAGCGAATC…
 
 
 Mutant
 …CTTGCTAGCGAATC…
 The
substitution
shown
seems
to
have
created
a
stop
codon.
What
further
information
do
you
need
 to
be
confident
that
it
has
done
so?
 
 (1 mark) You
need
to
know
the
reading
frame
of
the
possible
message.
 
 6. (3
mark
total)
 Hypothetically
speaking,
a
particular
protein
named
AWAKE
is
1
000
amino
acid
&
is
 known
to
be
involved
in
producing
the
phenotype
‘awake
in
8:30
am
classes’.

Some
individuals
were
 genotyped
&
had
a
deletion
of
2
nucleotides
near
the
5’
end
of
the
AWAKE
gene.

Other
individuals
 have
a
deletion
of
99
bases
in
the
same
region.
a)
(1
mark)
What
is
the
length
of
the
AWAKE
gene?
b)
 (2
marks)
 Explain
which
mutation
 you
think
would
be
more
likely
to
be
the
most
deleterious
to
the
 AWAKE
phenotype
 
 a)(1
mark)
 The
coding
region
of
this
gene
is
3
000
base
pairs
in
length.

b)
(2
marks)
 The
2
bp
deletion
 would
likely
be
more
deleterious
because
it
will
change
the
reading
frame
of
the
gene
while
the
99
bp
 deletion
 is
 a
 multiple
 of
 3
 and
 the
 protein
 will
 essentially
 be
 missing
 a
 33
 amino
 acid
 region.
 The
 effects
of
this
latter
mutation
would
depend
on
the
involvement
of
this
section
of
the
protein
in
the
 protein’s
function
–
it
could
be
fairly
unimportant
to
crucial
for
function.

 7. (6
marks
total)
 Certain
mice
called
 waltzers
have
a
recessive
mutation
that
causes
them
to
execute
 bizarre
 steps.
 W.H.
 Gates
 crossed
 waltzers
 with
 homozygous
 WT
 mice
 and
 found,
 among
 several
 hundred
 normal
 progeny,
 a
 single
 waltzing
 female
 mouse.
 When
 mated
 with
 a
 waltzing
 male,
 she
 produced
 all
 waltzing
 offspring.
 
 When
 mated
 with
 a
 homozygous
 normal
 male,
 she
 produced
 all
 normal
 progeny.
 
 Some
 males
 &
 females
 of
 this
 group
 of
 normal
 progeny
 were
 intercrossed,
 and
 there
were
no
waltzing
offspring
among
their
offspring.

T.S.
Painter
examined
the
chromosomes
of
 waltzing
mice
that
were
derived
from
some
of
Gates’s
crosses
&
that
showed
a
breeding
behaviour
 Lab
7

Chromosomal
Mutations

































Name:_____________________________________
 similar
 to
 that
 of
 the
 original,
 unusual
 waltzing
 female.
 He
 found
 that
 these
 mice
 had
 40
 chromosomes,
just
as
in
normal
mice
or
the
usual
waltzing
mice.

In
the
unusual
waltzers,
however,
 one
member
of
a
chromosome
pair
was
abnormally
short.

 Diagram
the
crosses
(3
marks),
and
then
 interpret
 these
 observations
 as
 completely
 as
 possible,
 both
 genetically
 &
 cytogenetically
 (noting
 what
results
you
would
have
gotten)
(3
marks).
 
 I
mark
for
idea
that
it
is
likely
that
the
female
has
a
deletion,
uncovering
expr
of
recessive
phenotype.
 1
mark
for
justification
that
normal
progeny
from
mating
with
normal
male
would
have
been
 heterozygous
&
thus,
would
have
produced
some
waltzers
in
their
offspring.
1
mark
for
explaining
 results
if
was
deletion
(observations).

 
 
 Lab
7

Chromosomal
Mutations

































Name:_____________________________________
 8. (6
 marks)
 Six
 bands
 in
 a
 salivary‐gland
 chromosome
 of
 Drosophila
 are
 shown
 in
 the
 following
 illustration,
along
with
the
extent
of
5
deletions
(Del
1

Del
5):

 Recessive
alleles
a,
b,
c,
d,
e,
and
f,
are
known
to
be
in
the
region,
but
their
order
is
unknown.
When
the
 deletions
are
combined
with
each
allele,
the
following
results
are
obtained:
 
 
 
 
 In
this
table,
a
minus
means
the
deletion
is
missing
the
corresponding
WT
allele
(i.e.,
the
deletion
 uncovers
the
recessive),
and
a
plus
sign
indicates
that
the
corresponding
WT
allele
is
still
present.

Use
 these
data
to
infer
which
band
on
the
salivary
chromosome
contains
each
gene.
(Determine
the
order
of
 the
genes.)

 
 Give
1
mark
for
each
gene
that
is
correctly
matched
with
the
chromosomal
band.
 
 
 
 
 
 
 
 
 
 Lab
7

Chromosomal
Mutations

































Name:_____________________________________
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ...
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This note was uploaded on 01/10/2010 for the course BIOLOGY biol2040 taught by Professor Tamarakelly during the Fall '09 term at York University.

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