Genetics Problems – Review
1. When black squirrels are mated with white squirrels, all of the offspring are gray.
A) Which allele is dominant? Explain. THIS IS A TRICK QUESTION! There are 3 phenotypes (black,
gray, and white). This is your clue that this is a case of INCOMPLETE DOMINANCE. The gray
results in the heterozygous squirrels.
B) Predict the fraction of each phenotype in the offspring when two gray squirrels are mated.
CB=black
CW=white
gray X gray
C CW X CBCW
B
CBCB is a black squirrel
CBCW is a gray squirrel
CWCW is a white squirrel
CB
CB
CW
¼ CBCB (Black)
½ CBCW (Gray)
¼ CWCW (White)
CBCB CBCW
CW CBCW CWC
W
C) Is it possible to have pure breeding gray squirrels? Explain.
No. The only way for a squirrel to be gray is to be heterozygous. To be pure breeding, a squirrel must
be homozygous.
2. In peas, green pods are dominant over yellow pods. Design an experimental plan to determine if a plant
with green pods is homozygous or heterozygous with a single cross. Assume you have a plentiful supply of
both green and yellow pod plants.
To determine if a plant with green pods (the dominant phenotype) is homozygous or heterozygous, the
easiest experiment is to perform a test cross. To do this you would mate the plant in question with a plant
that produces yellow pods (the recessive phenotype). The Punnett squares below summarize the expected
outcomes for the two possibilities:
G = green
g = yellow
If the test plant is homozygous, the cross is:
GG X gg
If the test plant is heterozygous, the cross is:
Gg X gg
g
g
G
Gg
Gg
G
G
Gg
Gg
g
The offspring will all have green pods.
g
g
Gg
Gg
gg
gg
½ of the offspring will have green pods.
½ of the offspring will have yellow pods.
3. One kind of eye color in Drosophila is due to a sex-linked gene with red eyes dominant over white eyes.
Wing length is not sex-linked. Normal wings are dominant over vestigial wings. A white-eyed, normalwinged male is mated to a red-eyed, vestigial-winged female. Among their offspring were white-eyed
normal-winged males, red-eyed normal-winged males, white-eyed vestigial-winged males, and red-eyed
vestigial-winged males.
A) Draw a pedigree chart to summarize these results.
white eyes
normal wings
red eyes
vestigial wings
white eyes
normal wings
red eyes
vestigial wings
red eyes
white eyes
normal wings vestigial wings
B) Use your pedigree chart to determine the genotypes of the parents and all of the offspring mentioned above.
XR = red eyes
Xr = white eyes
N = normal wings
n = vestigial
wings
XrY
Nn
white eyes
normal wings
red eyes
vestigial wings
XRXr
nn
XRY
red eyes
vestigial wings nn
XrY white eyes
Nn normal wings
red eyes
white eyes
normal wings vestigial wings
XRY
XrY
Nn
nn
4. Cystic Fibrosis is the most common fatal genetic disease in the United States. It is caused by a recessive
allele on chromosome number 7. If two people who are heterozygous for the disease marry, what are their
chances of having a child with the disease?
N = normal allele
n = Cystic fibrosis allele
heterozygous X heterozygous
Nn X Nn
n
n
NN Nn
N
N
n
Nn
There is a ¼ chance of having a child with the disease.
nn
¾ normal.
¼ cystic fibrosis
5. A geneticist has 2 pure breeding lines of plants. One produces plants that grow to 60 cm in height. The
second produces plants that grow to 40 cm in height. When these two lines are crossed the resulting F1
generation are uniformly 50 cm in height. Predict the resulting phenotypes and their percentages in the F2
generation.
Since there are 3 possible phenotypes and one of the phenotypes is midway between the other 2, you should
immediately think of incomplete dominance. Define your alleles by the 2 extreme phenotypes:
HT = 60 cm tall
HS = 40 cm tall
The F2 generation is obtained by mating 2 of the F1 plants, which must be
heterozygous:
heterozygous X heterozygous
HTHS X HTHS
HT
0
T
H
¼ 60 cm tall
½ 50 cm tall
¼ 40 cm tall
HS
n
HTHT HTHS
HS HTHS HSHS
6. Assume pointed ears are dominant over rounded ears in bunnies. Neither red or yellow ears are dominant
with the heterozygous individual having orange ears. The traits are not linked. A female bunny with
rounded orange ears is mated to a red-eared male bunny that is heterozygous for ear shape. Predict the
phenotypes of their offspring.
P = pointed ears
p = rounded ears
CR = red ears
CY = yellow ears
rounded orange ears X heterozygous shaped red ears
ppCRCY
X
PpCRCR
(Meiosis)
pCR
pCY
pCR
pCY
PCR
PCR
pCR
pCR
(cross out duplicates)
In this case, each parent can make only 2 different kinds of gametes, so you only need a 4 box Punnett Square:
pCR
PCR
pCR
PpCRCR
ppCRCR
Pointed, red
pCY
PpCRCY
Rounded, red
ppCRCY
Pointed, orange Rounded, orange
¼ Pointed red ears
¼ Pointed orange ears
¼ Rounded red ears
¼ Rounded orange ears
7. Two short-tailed cats are bred together. They produce 3 kittens with long tails, 5 kittens with short tails, and
2 kittens with no tail. Based on these results, explain how tail length is inherited in cats and predict the
genotype of each of the parents and offspring.
Once again, there are 3 possible phenotypes and one of the phenotypes is midway between the other 2. You
should immediately think of incomplete dominance. Define your alleles by the 2 extreme phenotypes:
Short tails would result from being heterozygous. Therefore, the short-tailed parents
must be TLTN:
short tail X short tail
TLTN X TLTN
So we would predict:
TN
¼ TLTL (long tailed offspring)
TL
½ TLTN (short tailed offspring)
n
L N
L TLTL
TT
T
¼ TNTN (no tail offspring)
TL = long tail
TN = no tail
TN TLTN
TNTN
This is a very close match to the 30% long tailed, 50% short tailed, and 20% no tail seen in the offspring!
8. A woman sues a man for support of her child. She has type A blood, her child has type O blood, and the
man has type B.
A) Is it possible for the man to be the child’s father? Explain.
Type A X Type B
IAi X IBi
IB
i
IA
IAIB
nA
I i
i
IBi
ii
There is a 1 in 4 chance of having a
Type O child.
B) The court orders further tests. The Rh factor is a separate blood factor with positive being dominant
over negative. Both the man and the woman are Rh negative. The baby is Rh positive. How would this
information affect the court’s decision?
R = Rh positive
r= Rh negative
This couple can only have Rh negative children.
Since the baby is Rh positive, the man is not the father.
Rh negative X Rh negative
rr X rr
r
r
r rr
rr
r rr
rr
n
100% of their children are
Rh negative.
9. In humans, muscular dystrophy is a condition in which the muscles waste away during early life, resulting
in death in the early teens. It is caused by a recessive, sex-linked gene. A couple has 5 children, 3 boys and
2 girls. The oldest child, a boy, begins to show symptoms of the disease. If you were their family
physician, what would you tell them about the chances that their other children would have the disease?
XN = Normal allele
XN Xn
Xn = Muscular dystrophy allele
XNY
?
?
?
?
Xn Y
The father must have a normal phenotype since he survived beyond his early teens. The son who has the
disease must have received an Xn from his mother, since his father gave him the Y chromosome. Therefore,
the mother must be heterozygous. We can now do a Punnett Square to predict their possible offspring:
XN Xn
XN
Y
XN
XNXN
XNY
Xn
X Xn
N
XNY
x
XnY
Normal males
Males with muscular
dystrophy
½ Normal Females
¼ Normal Males
¼ Males with muscular
dystrophy
Normal females
You could reassure the parents that none of their daughters will have the disease (although there is a 50%
that they will be carriers like their mother). There is a 50% chance that each of their sons will have the
disease.
10. A normal man of blood type AB marries a normal woman of blood type O whose father was colorblind.
Predict the possible phenotypes of their children and the frequencies of each with respect to blood type and
colorblindness.
XnY Colorblind
A
father
I = A allele
B
I = B allele
Normal woman,
Normal man,
i = O allele
Blood Type O
Blood Type AB
XN = Normal allele
XNY IA IB
XNXnii
Xn = Colorblind allele
Since her father was colorblind,
he must be a carrier for the
colorblind allele. The only way
to have blood type O is to be
homozygous for the i allele.
The man is normal so his X
chromosome must carry the
normal allele. The only way to
have blood type AB is to be IA IB.
So the cross we are looking at is:
XNXnii X
XNY IA IB
(Meiosis)
(cross out duplicates)
XNIA
XNIB
YIA
YIB
XNi
XNi
Xni
Xni
You will need an 8 square Punnet “rectangle”:
XNi
Xni
XNIA
XNXNIAi
Normal girl,
Type A
XNXnIAi
Normal girl,
Type A
The expected offspring would be:
1/4 Normal girls, Type A
1/4 Normal girls, Type B
1/8 Normal boys, Type A
1/8 Normal boys, Type B
1/8 Colorblind boys, Type A
1/8 Colorblind boys, Type B
XNIB
XNXNIBi
Normal girl,
Type B
XNXnIBi
Normal girl,
Type B
YIA
XNYIAi
Normal boy,
Type A
XnYIAi
Colorblind boy,
Type A
YIB
XNYIBi
Normal boy,
Type B
XnYIBi
Colorblind boy,
Type B