AP Biology – Genetics Practice Answer Key
Note: anywhere that allele letter is not specified, you may choose your own. Which side of a square
holds maternal vs. paternal gametes is also up to you.
MONOHYBRID AND POLYHYBRID CROSSES
1.
A
a
A
AA
Aa
a
Aa
aa
The trees must both be heterozygous. Some
offspring are ff; therefore, both parents must
have at least one f allele. Both parents are
fuzzy, therefore, they also have at least one F
allele.
Chance normal skin: 75%
Chance albino: 25%
If normal, 66% chance carrier
6.
2.
H
h
h
Hh
hh
h
Hh
hh
50% Hh, 50% hh
50% horns, 50% no horns
R
r
R
RR
Rr
r
Rr
rr
White is dominant. If red is dominant, as shown
above, you could cross two red trees and yet
get white offspring (25% chance). However, if
white is dominant, a white tree must have at
least one dominant allele, which means a
parent would’ve had that allele, making the
parent white.
3.
B
b
B
BB
Bb
b
Bb
bb
25% BB, 50% Bb, 25% bb
75% brown, 25% blue
4.
W
W
W
WW
WW
w
Ww
Ww
0% short whiskers
5.
F
f
F
FF
Ff
f
Ff
ff
7. Cow A x bull
h
h
h
hh (D)
Cow B x bull
H
h
h
hh (E)
Cow C x bull
h
h
H
Hh (F)
A: hh, B: Hh, C: hh, D: hh, E: hh, F: Hh, bull: Hh
8.
Sickle-cell (in this scenario) must be recessive.
If it were dominant, their parents would have to
have it in order to have produced sickle-celled
children.
So the siblings’ genotypes must be ss. Karen &
Steve’s parents must therefore be Ss.
S
s
S
SS
Ss
s
Ss
ss
Karen and Steve both have a 66% chance of
being carriers. Only if they are both carriers
could they have a child with sickle-cell anemia.
If Karen has a 66% chance of being a carrier,
and so does Steve, then the odds under the
Rule of Multiplication that they are both
carriers are .66 x .66 = .44.
S
s
S
SS
Ss
s
Ss
ss
Odds of child with sickle-cell: 25%.
The odds, then, of a child with sickle-cell =
.25*.44 = .11 or 11% probability.
9.
TG
Tg
tG
tg
TG
TTGG
TTGg
TtGG
TtGg
Tg
TTGg
TTgg
TtGg
Ttgg
tG
TtGG
TtGg
ttGG
ttGg
tg
TtGg
Ttgg
ttGg
ttgg
Tall/Green = 9/16 or 56.25%
Tall/Yellow = 3/16 or 18.75%
Short/Green = 3/16 or 18.75%
Short/Yellow = 1/16 or 6.25%
10.
T
t
T
TT
Tt
t
Tt
tt
G
g
G
GG
Gg
g
Gg
gg
Height
Tall/Green
.75
Tall/Yellow
.75
Short/Green .25
Short/Yellow .25
Color
.75
.25
.75
.25
P
.5625
.1875
.1875
.0625
11a. Hh
b. H or h
c. hh
d. h or h
e. 50%
f. 50%
12a. Ab, Ab, ab, ab
b. AB, AB, aB, aB
c. AB, Ab, aB, ab
d. AB, AB, Ab, Ab
e. Ab, Ab, Ab, Ab
f. ab, ab, ab, ab
13a. 50%
b. 50%
c. 25%
d. 0%
e. 0%
14. Perform a test cross. Mate him with a
female with a white coat. Her genotype is
guaranteed to be bb. If any of their offspring
are white, he must be heterozygous, because a
bb baby must have received a b allele from him
as well as from Mom. If all the babies are blackcoated, he’s probably BB, but it isn’t a
guarantee.
15.
Parental
Genotype
Aa x Aa
Aa x aa
AaBb x AaBB
AaBb x AABb
AaBb x AaBb
Offspring
Genotype
Aa
Aa
AABB
aabb
AaBb
Probability
.5
.5
.25 x .5 = .125
0
.5 x .5 = .25
16. AA x Aa, Aa x Aa, Aa x aa, AA x aa
17. p of being tall (at least one T) = .5, p of
being colored (at least one C) = .75
The parents must be Cc x Cc with respect to
color, it’s the only monohybrid cross that can
produce a .75 probability at all. This eliminates
options c and e.
With respect to tallness, the parents must be Tt
x tt. At least one parent must have a T. If
they’re TT, all offspring will be tall. If they’re Tt,
crossing them with another heterozygote yields
.75 tall, not .5. Crossing them with a tt, though,
yields .5. This eliminates options b, d, and e.
The option that fits is option a.
18a. 25%.
b. 66%.
c. 75%.
19. We seek progeny that will be dominant for
A, B, and D, but recessive for C and E.
P dominant phenotype for A: .75
P dominant phenotype for B: .5
P recessive phenotype for C: .5
P dominant phenotype for D: .75
P recessive phenotype for E: 1.0
.75*.5*.5*.75*1 = .14 or 14% chance.
20a. .25*.25*.25*.25 = .0039 or .39%
b. .5*.5*.5*.5 = .0625 or 6.25%
c. .25*.25*.25*.25 = .0039 or .39%
d. .5*.25*.25*.5 = .015 or 1.5%
e. .5*.25*.25*.25 = .0078 or .78%
21. (3/8 = .375, 1/8 = .125)
A good way to start is with the
recessive/recessive puppies. A .125 frequency
could only be obtained via .5*.25 operation.
.25 only occurs in heterozygote vs heterozygote
crosses, so the parents are both heterozygous
for one of the two genes. For the other gene,
one parent must be heterozygous and the other
must be homozygous recessive in order to have
a .5 chance of homozygous recessive puppies.
Our options are therefore either BbSs x Bbss or
BbSs x bbSs.
The former produces a homozygous dominant
chance of (.75*.5) = .375, which fits the data.
But so does the latter, so that isn’t helpful.
However, we know that there is a .375 chance
of black spotted, and a .125 chance of chestnut
solid. If the parents are BbSs x Bbss, the chance
of black spotted is .75*.5 = .375, and the chance
of chestnut solid is .25*.5 = .125, which fits the
data.
If the parents are BbSs x bbSs, the chance of
black spotted is .5 * .25 = .125, which is does
not match the data.
The parents’ genotypes are BbSs x Bbss.
22. If all F1 have red terminal flowers, red and
terminal must be dominant. F1 plants are all
heterozygous for both genes, so the odds of a
red F2 flower = .75 and the odds of an axial F2
flower = .25. The odds of a red axial flower are
therefore .75*.25 = .1875 or 18.75%.
23. Wherever there is a ?, it means either allele
could be present there.
a. One/normal = .764, One/wrinkled = .236 At
least one parent must be homozygous
dominant for number in order to guarantee all
offspring have one pod. This leaves a 75%
dominant/25% recessive split for shape.
Parents’ genotypes are PPLl x ??Ll.
b. Three/normal = .753, Three/wrinkled = .247
Both parents must be recessive to guarantee
recessive offspring for number. This leaves a
75%/25% split for shape. Parents’ genotypes
are ppLl x ppLl.
c. One/normal = 1.0 At least one parent must
be homozygous dominant for number, and at
least one parent must be homozygous
dominant for shape. Parents’ genotypes could
be PPLL x ???? or PP?? x ??LL.
d. One/normal = .379, One/wrinkled = .371,
Three/normal = .129, Three/wrinkled = .121
See problem #21 where these same proportions
appear. The rationale for parental genotypes
here is identical to that problem. Parents’
genotypes must be PpLl x Ppll.
e. One/normal = .56, One/wrinkled = .181,
Three/normal = .191, Three/wrinkled = .068.
Notice that recessive/recessive is close to .0625,
which is .25*.25. .25 recessive can only be
obtained by crossing heterozygotes. Therfore,
parents’ genotypes are PpLl x PpLl.
24. This suggests that both genes follow a
simple Mendelian pattern of inheritance. The
F1 offspring are heterozygous, AaBb, and are
smelly non-singers. Therefore, being smelly is
dominant while fragrant is recessive, and being
a non-singer is dominant while singing is
recessive. The smelly singer parents must have
been AAbb and the fragrant non-singer parent
was aaBB.
25. The man and woman are both either MM
or Mm for methanethiol, and also BB or Bb for
excreting betamin. Their son is mmbb, on the
other hand. The parents must therefore be
heterozygous for both (MmBb x MmBb). In
subsequent progeny, the odds are 56.25% no
methanethiol/betamin, 18.75% no
methanethiol/no betamin, 18.75%
methanethiol/ betamin, 6.25%
methanethiol/no betamin
26. To determine whether curling is dominant
or recessive, the curl cat should be mated with
a true-breeding non-curl cat. If curling is
recessive, no offspring will show curling. If
curling is dominant, either half or all of the
offspring will show curling.
An organism can be shown to be true-breeding
if, when mated with other organisms of its
same phenotype, it always produces offspring
with that phenotype. For true-breeding
recessive, this means you’re mating aa xaa,
which will always produce aa offspring. For
true-breeding dominant, this means that a truebreeding AA individual will always have
offspring with at least one A, no matter
whether mated to AA or Aa. If the individual
had not been true-breeding (Aa), then mating it
with other dominant individuals would
sometimes produce all dominant offspring (if
the partner is AA), but would sometimes
produce a few recessive offspring (Aa). This
method has the advantage of being possible to
carry out even if you don’t know whether the
phenotype is dominant or recessive; mating like
with like will still demonstrate whether an
individual is true-breeding.
INCOMPLETE DOMINANCE, CODOMINANCE,
MULTIPLE ALLELES, LETHALITY
27. Purple = RW, white = WW
P
W
W
PW
WW
W
PW
WW
1:1 Purple:White, 1:1 PW:WW
28a.
R
r
R
RR
Rr
r
Rr
rr
.25 RR (red), .5 Rr (pink), .25 rr (white)
A
a
A
AA
Aa
a
Aa
aa
.25 AA, .5 Aa, .25 aa or .75 axial, .25 terminal
Y
W
Y
YY
YW
W
YW
WW
1:2:1 Yellow:Cream:White, 1:2:1 YY:YW:WW
b.
Y
Y
Y
YY
YY
W
YW
YW
1:1 Yellow:Cream, 1:1 YY:YW
29. (Typo, should read codominance in coat
color, roan has some red hairs and some white
hairs)
R
W
R
RR
RW
W
RW
WW
1:2:1 Red:Roan:White, 1:2:1 RR:RW:WW
30. (Typo, should be in incomplete dominance
section) Color exhibits incomplete dominance
between the black and white alleles. A gray
rooster and black hen would have a 50% chance
of gray and 50% chance of black offspring.
31. (Typo, should be in incomplete dominance
section) If P generation is AARR x aarr, the F1
generation will be entirely AaRr, which is 100%
axial-pink. In the F2 generation…
Axial-Red: .75 * .25 = .1875 or 3/16
Axial-Pink: .75 * .5 = .375 or 3/8
Axial-White: .75 * .25 = .1875 or 3/16
Terminal-Red: .25 * .25 = .0625 or 1/16
Terminal-Pink: .25 * .5 = .125 or 1/8
Terminal-White: .25 * .25 = .0625 or 1/16
The ratio of axial-red:axial-pink:axialwhite:terminal-red:terminal-pink:terminalwhite is therefore 3:6:3:1:2:1
32. Four phenotypes are possible: H-color, Icolor, HI-color, and i-color.
33a. The parents are C? x ChCh or Chc. If parent
one is CC, all offspring in F1 would be black, so
that’s not correct. Possible parents are
therefore:
CCh x ChCh =1:1 black:chinchilla, fits the data
Cc x ChCh = 1:1 black:chinchilla, also fits
CCh x Chc = 1:1 black:chinchilla, also fits
Cc x Chc = 2:1:1 black:chinchilla:white, no fit
So the parents can be narrowed down to three
pairs, but not definitively determined.
b. This time, the parents must definitely be Cc
x Chc for reasons outlined in part a.
34.
IB
IO
IA
IAIB
IAIO
IB
IBIB
IBIO
1:1:1:1 IAIB: IAIO: IBIB: IBIO
1:1:2 Type AB:Type A: Type B
35. Their children cannot have type O blood;
their father doesn’t have an IO allele to
bequeath them.
Silvia & Juan
IO
IO
IA
IAIO
IAIO
36. All four: A, AB, B, and O.
IB
IBIO
IBIO
37. No. He could be genotype IBIO and she
could be genotype IAIO, giving them a ¼ chance
of type O offspring.
David & Nicole
IB
?
38. He is right. He would have to have an IO
allele for their child to be genotype IOIO. Either
she or the hospital has some explaining to do…
IA
IAIB
IA?
IB
IBIB
IB?
39. A, AB, and B.
41.
Blood type of
Mother
Blood Type of
Child
AB
A
O
A
O
B
B
AB
O
A
40. Baby 1 cannot belong to Duke/Betty. Baby
2 cannot belong to Kelly/Mohammed,
Duke/Betty, or Silvia/Juan. Baby 3 cannot
belong to Silvia/Juan or David/Nicole. Baby 4
cannot belong to Kelly/Mohammed or
Duke/Betty.
Parents
Possible Babies
Kelly & Mohammed
1, 3
Duke & Betty
3
Silvia & Juan
1,4
David & Nicole
1,2,4
So Baby 1: Kelly & Mohammed
Baby 2: David & Nicole
Baby 3: Duke & Betty
Baby 4: Silvia & Juan
Kelly & Mohammed
IO
IO
A
I
?
A O
II
O
?I
Duke & Betty
IO
O
I
O
I
OO
I I
OO
I I
42. Aa x Aa
Offspring could be achondroplastic or not, Aa or
aa. 66% chance Aa, 33% chance aa.
43. CcBb x Ccbb
C
c
C
CC
Cc
c
Cc
cc
A O
II
1/3 normal, 2/3 deformed
O
?I
B
b
b
Bb
bb
b
Bb
bb
O
I
OO
I I
Exonerated
Man Blood
Types
None (BB
only)
A
A, O
AB
B, O
1/4 black, 1/2 brown, 1/4 white
OO
I I
CCBB = 1/3 * 1/4 = 1/12
CCBb = 1/3 * 1/2 = 1/6
CCbb = 1/3 * 1/4 = 1/12
CcBB = 2/3 * 1/4 = 1/6
CcBb = 2/3 * 1/2 = 1/3
Ccbb = 2/3 * 1/4 = 1/6
1:2:1:2:4:2
Normal/Black:Normal/Brown:Normal/White:De
formed/Black:Deformed/Brown:Deformed/Whi
te
44. Hairlessness must be the heterozygous
condition, normal hair must be the homozygous
dominant condition, and homozygous recessive
is lethal. The crosses are:
B
b
B
BB
Bb
B
BB
Bb
½ normal hair, ½ hairless, matching the data.
B
b
B
BB
Bb
b
Bb
bb
Because the bb zygote dies, 2/3 of offspring of
the second cross are hairless, 1/3 are normal
hair.
SEX-LINKAGE, POLYGENIC TRAITS, EPISTASIS
45.
XH
Y
XH
XH XH
XHY
Xh
XH Xh
XhY
48a. The mother must be a carrier, XH Xh. The
son receives his X chromosome from her, and
his X chromosome bears the hemophilia allele,
therefore she has a recessive allele. She also
has a dominant allele because she is not a
hemophiliac.
b.
Xh
Y
Normal daughter (XH XH, XH Xh), normal son
(XHY), hemophiliac son (XhY).
XH
XH Xh
XHY
Xh
Xh Xh
XhY
46.
Polydactyly is autosomal dominant,
colorblindness is X-linked recessive.
D
d
Their odds are 50% for either a hemophiliac
son, a hemophiliac daughter, or a hemophiliac
child in general.
d
Dd
dd
d
Dd
dd
Xh
Y
XH
XH Xh
XHY
Xh
Xh Xh
XhY
Percent colorblind + five-fingered: .5 * .5 = .25
or 25%.
Percent normal vision + five-fingered: .5 * .5 =
25%.
47. When not otherwise specified, assume that
sex-linked means X-linked. Regardless, we
know that the calico cat is the female, because
males cannot be heterozygous for sex-linked
traits. Furthermore, a Punnett Square is actually
unnecessary, but here it is anyways:
Xh
Y
XH
XH Xh
XHY
Xh
Xh Xh
XhY
As was just stated “males cannot be
heterozygous for sex-linked traits,” therefore
the odds of a calico male kitten are 0%.
49.
Xr
Y
XR
XR Xr
XR Y
X?
X? Xr
X ?Y
They will certainly have some red-eyed females
(XR Xr) and some red-eyed males (XRY).
However, without knowing the mother’s full
genotype, we don’t know whether they could
also have white-eyed females and males (Xr Xr
and XrY).
50a. No, he cannot. His son can have a
receding hairline, but the allele didn’t come
from Dad.
b. His mother’s father would be the best
indication, as she has his X-chromosome. If he
had a receding hairline, his mother is at least a
carrier. Her grandfathers can also be helpful, as
can her brothers.
51. He must also be colorblind; she received a
colorblind X-chromosome from him, and he
only has the one.
52. Nothing at all.
53a. Two genes. (Notice, that the extreme
classes are about 1/16th of the total – this is the
proportion of aabb in a dihybrid cross)
b. Tricky tricky! Each gene must add or
subtract a certain length to the piglet’s tail, with
a 15 cm baseline if length is being added per
allele or a 35 cm baseline if length is being
subtracted. It doesn’t actually matter in this
case whether A and B add or whether A and B
subtract; for the sake of clarity, A and B will add
while a and b code for no addition, i.e. baseline
tail length.
So, the 15 cm pig has genotype aabb. The 35
cm pig has genotype AABB. 35 is 4 x 5 cm
greater than 15, so it seems that each A allele
adds 5 cm, as does each B allele.
This gives us the following parents: aabb (15
cm) and either AABb or AaBB (30 cm, doesn’t
matter which, AABb will be used for squares).
A
A
a
Aa
Aa
a
Aa
Aa
B
b
b
Bb
bb
b
Bb
bb
Offspring could have the genotypes AaBb or
Aabb. (Had the other possible parent #2 been
used, genotypes would be AABb or AaBb.)
Possible phenotypes are 20 cm or 25 cm tails.
54a. To start, we know E??? x E???. Of nine
offspring, seven are golden, which is closest to
75%. The parents are most likely heterozygous
for E. Then, of the ee puppies, half are brown
and half are black. The parents are most likely
Bb x bb, although they could be Bb x Bb
because a sample size of two is difficult to draw
a good conclusion from. So the answer is most
likely EeBb x Eebb.
b. To start, we know eeB? X E???. Not all
offspring are golden, so the male must be Ee.
Of non-golden offspring, about ¾ are dominant
black and about ¼ are recessive brown, so both
parents are most likely heterozygous (eeBb x
EeBb). But the male could be homozygous
recessive instead (Eebb).
55. The colored rooster could be CCpp or Ccpp.
The white hen could be any genotype except
CCpp or Ccpp. 400 chicks all being born colored
powerfully suggests that she must be pp. The
only way should be pp and white is to be ccpp.
We’ve now narrowed it down to a or c as the
answer.
If a were true and the rooster were Ccpp, half
the offspring would be white (ccpp). So the
answer is C, the rooster is CCpp and the hen is
ccpp, guaranteeing all offspring are Ccpp –
colored.
PEDIGREES
Pedigee #1
a) Yes
b) No
c) Yes
d) No
e) No
Pedigree #2
a) No
b) Yes
c) No
d) Yes
e) No
Pedigree #3
a) Yes
b) No
c) No
d) No
e) No
Pedigree #4
a) Yes
b) Yes
c) Yes
d) Yes
e) No
Aa x Aa
XA Y x XA Xa
Aa x Aa
XA Y x XA Xa
Aa x Aa
aa x Aa
Aa x aa
Xa Y x XA Xa
XA Y x XA Xa
Pedigree A: Deaf Mutism
Deaf mutism must be autosomal recessive. See
future pages for completed pedigree.
1. Because they’re both homozygous for the
deaf-mute allele. It’s impossible (mutation
aside) for them to have children who hear
normally.
2. They’re both carriers of the deaf-mute allele
(Dd).
3. Either parents 9 and 12 are DD, ensuring all
their children received a D allele, or it was
simple luck that their children all received their
D allele rather than their d allele. Each child
had a 50/50 chance.
4. II-3, II-6, III-1, III-4, III-5, III-6, III-7, IIII-9, III10, III-12, IV-4, IV-5, IV-6
5. The pattern is that they didn’t have either a
parent or a child with deaf mutism.
Pedigree B: Hemophilia
Hemophilia is X-linked recessive. See future
pages for completed pedigree.
6. If the mother is a carrier (XH Xh) and the
father is healthy (XH Y), they can have an
afflicted son (XhY).
7. Carriers (XH Xh), although they could have
also been afflicted (Xh Xh) but didn’t happen to
be in this case. This is because, in order for a
female to be affected, she must receive Xh from
both parents. This means her mother must be
either a carrier of the allele, or homozygous for
the allele and therefore hemophiliac.
8. They all have an affected father. This is
because they must receive Xh from both
parents. If her father has Xh then he’s
hemophiliac, since he doesn’t have a second
copy of the gene that could mask it.
9. II-3, II-7, III-4, III-11, III-12, III-15, III-18, IV-1,
V-2, IV-5, IV-7. The common denominators are
that all of them are women, none of them have
an affected mother or father, and none of them
have an affected child. This means that they’re
all known to possess one dominant allele, and
because they’re women they must have a
second allele, but the second cannot be
determined, just as in the case of an autosomal
recessive trait.
Pedigree C: Brachydactyly
10. Brachydactyly is most likely autosomal
dominant. All brachydactalic individuals had at
least one parent who also had brachytactyly,
and there’s no clear trend in terms of sex.