1. When neither allele is dominant, so that a
heterzygote has a phenotype that is a blending of
each of the homozygous phenotypes (such as one
red color allele and one white color allele producing
pink flowers) it is called
25% 25% 25% 25%
A. Multiple alleles
B. Polygenic traits
C. Incomplete dominance
nc
e
al
Au
to
so
m
te
m
pl
e
In
co
in
h
do
m
ni
c
yg
e
Po
l
er
it a
in
an
tra
le
al
pl
e
ul
ti
M
ce
its
le
s
D. Autosomal inheritance
Incomplete dominance
2. The curly hair allele (HC) and the straight hair (HS)
alleles show incomplete dominance . If one parent
has curly hair and one parent has straight hair, what
is the probability that they have a child with curly
hair?
A. 100%
20%
20%
20%
20%
20%
B. 75%
C. 50%
D. 25%
0%
25
%
50
%
75
%
10
0%
E. 0%
If one parent has curly hair and the other has straight
hair, each offspring will inherit one of each allele from
parents. All offspring will have wavy hair.
C
H
C
H
S
H
C
S
HH
C
S
HH
S
H
C
S
HH
C
S
HH
3. If both parents have wavy hair, what is
the phenotype (and genotype) ratio for
their offspring?
25% 25% 25% 25%
A. 3:1 (curly to straight)
B. All wavy haired offspring
C. 1:2:1 (curly: wavy: straight)
9:
3:
3:
1
st
r
aig
g
(cu
rly
:w
re
d
1:
2:
1
yh
ai
av
lw
Al
av
y:
of
fsp
rin
ht
)
ig
st
ra
to
ur
ly
(c
3:
1
ht
)
D. 9:3:3:1
C
H
S
H
C
H
C
C
HH
Curly
C
S
HH
Wavy
S
H
C
S
HH
Wavy
S
S
HH
Straight
1 Curly : 2 Wavy: 1 Straight hair
4. Almost all sex-linked traits, such as
hemophilia, red-green colorblindness and
Duchene’s muscular dystrophy are caused
by
25% 25% 25% 25%
A. A gene on the Y chromosome
B. A gene on the X chromosome
C. A gene on chromosome 21
iff
er
en
c
21
e
al
d
Ho
rm
ch
ro
on
ne
ge
A
on
m
ro
m
ch
X
th
e
on
ge
ne
A
os
om
o.
..
o.
..
om
ch
r
Y
th
e
on
ne
ge
A
es
D. Hormonal differences
Females have two X chromosomes, so they have two copies of any genes on
the X chromosome. Females must inherit a recessive allele from both parents
to have the recessive phenotype.
Males only have one X chromosome. The Y chromosome does not have the
same genes. So males will express a recessive trait if there only X
chromosome has the recessive allele.
5. Red-green colorblindness is caused by a
recessive sex-linked trait (on the X
chromosome).
What is the genotype of a colorblind man?
20% 20% 20% 20% 20%
A. XBXB
B. XBY
C. XbY
D. XbXb
no
t
de
te
r
m
in
e
Xb
Xb
Ca
n
Xb
Y
Y
XB
XB
XB
E. Can not determine
Males only have one X chromosome.
A coloblind male has the recessive colorblindness
allele on his only X chromosome. (Xb)
6. If a man is colorblind (a recessive sexlinked trait),
A. All of his sons will be
A
an
d
C
B
th
Bo
Bo
th
A
ill
rs
w
ht
e
da
ug
fh
is
an
d
b.
..
..
ill
l. .
.
Al
lo
fh
is
da
ug
ht
er
sw
co
be
ill
sw
Al
lo
so
n
E.
fh
is
D.
lo
C.
colorblind
All of his daughters will be
colorblind
All of his daughters will be at
least carriers
Both A and B
Both A and C
Al
B.
20% 20% 20% 20% 20%
A Father passes his only
X chrosomes to all
daughters. So if he is
colorblind, all of his
daughters will be at
least carriers.
B
X
A father passes his Y
chromosome to all sons.
So a father being
colorblind doesn’t affect
as son’s probability of
being colorblind.
b
X
Y
B
X X
b
B
X Y
7. If a mother is colorblind and a father is not
colorblind, what is the probability of a daughter
being colorblind? (Find the probability just
among possible daughter outcomes)
20%
20%
20%
20%
20%
A. 100%
B. 75%
C. 50%
D. 25%
0%
25
%
50
%
75
%
10
0%
E. 0%
B
X
b
X
b
X
Y
B
b
b
X X XY
B
b
b
X X XY
Daughters
all have
normal
color
vision
The father’s only X chromosome always gets passed down to
any daughters he has.
Since this father does not have colorblindness, his X
chromosome must have the normal color vision allele.
All of his daughters will receive this dominant allele, and will
NOT have red-green colorblindness.
8. If a mother is colorblind and a father is
not colorblind, what is the probability of
son being colorblind? (Find probability just
among possible son outcomes)
A. 100%
20%
20%
20%
20%
20%
B. 75%
C. 50%
D. 25%
0%
25
%
50
%
75
%
10
0%
E. 0%
B
X
b
X
b
X
Y
B
b
b
X X XY
B
b
b
X X XY
All the
sons will
be
colorblind.
Colorblind mother is XbXb. Father with normal color vision is XBY.
With these parents, all the sons will inherit their only X from their
colorblind mother and will be colorblind.
All daughters will inherit a dominant color vision gene on the X
chromosome from their father, so they will have normal color vision
but will be carriers due to the X they receive from their mother.
9. The pedigree below represents a family history for
sickle cell anemia which is a recessive trait (ss). Shaded
in individual have this recessive genetic condition.
Choose the response that includes all of the possible
genotypes of individual I-2.
A. SS only
B. SS or Ss
C. Ss only
D. ss only
or
ss
y
nl
20%
Ss
y
20%
ss
o
20%
on
l
or
Ss
20%
SS
SS
on
l
y
20%
Ss
E. Ss or ss
10. The pedigree below represents a family history for
sickle cell anemia which is a recessive trait (ss). Shaded
in individual have this recessive genetic condition.
Choose the response that includes all of the possible
genotypes of individual III-2
A. SS only
B. SS or Ss
C. Ss only
D. ss only
or
ss
20%
Ss
y
nl
y
on
l
20%
ss
o
20%
Ss
y
on
l
SS
SS
E. Ss or ss
20%
or
Ss
20%
11. The pedigree below represents a family history for
sickle cell anemia which is a recessive trait (ss). Shaded in
individual have this recessive genetic condition. Choose
the response that includes all of the possible genotypes of
individual II-4.
or
ss
20%
Ss
nl
y
20%
ss
o
y
20%
on
l
or
Ss
20%
SS
on
l
y
20%
Ss
SS only
SS or Ss
Ss only
ss only
Ss or ss
SS
A.
B.
C.
D.
E.
Explanations
9. You can determine the
genotype of individual I-2
because individuals with the
recessive phenotype must
have two recessive alleles.
(ss)
10. In many cases, you
can not determine the
genotype of an individual
with the dominant
phenotype. Individual III-2
could be SS or Ss. There
is no way of determining
which genotype she has
from the pedigree.
11. You can determine that individual II-4
must be heterozygous (Ss) because they
have a child with the recessive
phenotype. This child must have inherited
a recessive allele from both parents.
(Also they have a parent with the
recessive phenotype- they must have
inherited the recessive allele from this
parent.)
12. The 4 main types of blood A, B, AB
and O are due to
C
B
an
d
C
d
an
A
an
c
he
r it
in
d
nk
e
xli
Se
e
le
le
s
al
le
tip
M
ul
ge
n
ic
in
he
r
ita
nc
e
Polygenic inheritance
Multiple alleles
Sex-linked inheritance
A and C
B and C
Po
ly
A.
B.
C.
D.
E.
20% 20% 20% 20% 20%
13. For blood types, the A allele (IA) and the B
allele (IB) are codominant. The O allele (i) is
recessive.
If a mother has blood type B and a father has
blood type A, which blood types are possible for
their children?
25%
25%
25%
O
B
or
or
B,
AB
A
A,
AB
,A
on
ly
,o
rB
AB only
AB, A, or B
A or B
A, B, AB or O
AB
A.
B.
C.
D.
25%
The A blood type parent could have the
genotype IA i.
The B blood type parent could have the
genotype IB i.
So all 4 blood types are possible.
IB
i
IA
IAIB
(AB)
Iai
(A)
i
Ibi
(B)
ii
(O)
14. A trait that is polygenic is
50%
50%
W
id
o
w
’s
pe
a
k
He
ig
ht
A. Height
B. Widow’s peak
There is a wide variation in
height because more than
one gene controls this trait.
15. An autosomal trait is one that
A. is equally likely in males
and females
B. is more likely in males
C. is more likely in females
D. is caused by a dominant
allele
E. is caused by a recessive
allele
is
.
ea
re
by
a
ca
us
ed
by
a
us
ed
l..
.. .
an
ta
do
m
el
yi
lik
is
ca
is
ce
ss
iv
n
in
el
y
m
or
e
lik
m
or
e
in
fe
m
al
m
...
es
m
al
n
is
el
yi
lik
ly
ua
l
eq
is
al
es
es
20% 20% 20% 20% 20%
The autosomal chromosomes are the chromosomes
that are found in homologous pairs in both males
and females. (chromosomes 1-22 for humans).
Since both genders have pairs, these traits are
equally likely in males and females.
16. Nondisjunction is
A. An extra round of DNA
replication in mitosis
B. An extra round of DNA
replication in meiosis
C. Improper separation of
homologous pairs in
meiosis
D. Fertilization of an egg
with multiple sperm
i..
eg
n
iza
tio
n
of
an
at
io
t il
Fe
r
pr
Im
gw
of
...
...
re
op
e
rs
ep
ar
of
DN
A
d
ou
n
ar
ex
tr
An
An
ex
tr
ar
ou
n
d
of
DN
A
re
...
25% 25% 25% 25%
17. Having an extra copy of a
chromosome is called
on
ov
al
M
20%
y
20%
en
t
en
t
va
l
Tr
iso
m
nt
le
iva
20%
M
on
os
om
20%
y
20%
Bi
Trivalent
Trisomy
Bivalent
Monovalent
Monosomy
Tr
A.
B.
C.
D.
E.
Down’s Syndrome is Trisomy 21
1. C
2. E
3. C
4. B
5. C
6. C
7. E
8. A
9. D
10. B
11. C
12. B
13. D
14.A
15.A
16.C
17.B