Genetics Review Problems (with Answers)
Name ___________________________
1. Match the following terms and their definitions by writing the correct term next to its definition.
genetics
recessive
heredity
homozygous
cross
heterozygous
hybrid
phenotype
allele
genotype
dominant
gene
__recessi ve______ two copies of allele needed for phenotype to be observed
__homozygous____ having two of the same allele
__heterozygous___ having two different alleles
__genotype__ ____ genetic makeup of an organism
__allele__________ variant of a gene
__dominant ______ only one copy of allele needed for phenotype to be observed
__ hybrid________ offspring of two different varieties
__ cross _________ cross-fertilization/hybridization between two different varieties
__ genetics ______ science of heredity
__phenotype_____ physical traits of an organism
__heredity_______ transmission of qualities from parent to offspring
__gene_________ unit of hereditary information contained in an organism's DNA
2a. Define Mendel's two principles:
segregation: pairs of alleles separate during gamete (egg and sperm) formation and are paired again
during fertilization
independent assortment: each pair of alleles assorts independently in gamete formation.
2b. Complete the Punnett Square and explain how Mendel's principles of segregation and independent
assortment are demonstrated in the cross below.
E/e = unattached/attached earlobes
W/w = widow's peak/straight-hairline
Possible genotypes
of gametes (eggs)
from woman
EW
Ew
eW
ew
EeWw x EeWw
Possible genotypes of gametes (sperm) from man
EW
Ew
eW
ew
EEWW
EEWw
EeWW
EeWw
EEWw
EEww
EeWw
Eeww
EeWW
EeWw
eeWW
eeWw
EeWw
Eeww
eeWw
eeww
Seg - There is only one allele of each gene (E/e and W/w) in the gametes. There are two alleles for each
gene in the fertilized egg/offsrping. Ind. Assort. - In the gametes, the alleles for each gene assort
separately, so there are four possible combinations of the alleles, rather than just two (EW, ew or Ew,
eW).
3. Complete the following genetics problems, answering all questions completely. Show your Punnett
squares or calculations for full credit on the test!!
a. A man heterozygous for hitchhiker's thumb has two children with a woman homozygous for the trait
(hh). What is the probability that BOTH of their children have hitchhiker's thumb?
First, set up a Punnett Square showing the cross. Then use the rule of multiplication to calculate the
probability for both children to have the trait. Recall that if you are interested in finding out the probability
of something happening AND something else happening, multiply the two probabilities
h
h
H
Hh
Hh
There is a 50% chance that each child will have hh (hitchhiker's thumb). The
probability that both will have it is 0.5 x 0.5 = 0.25 or 25 percent. Expressed
as fractions: 1/2 x 1/2 = 1/4
h
hh
hh
b. Two pink (RW) flowers meet, fall in love, and have offspring. If this trait is incompletely dominant, what
percentage of offspring would you expect to have red flowers?
Recall that incomplete dominance means that the heterozygote has a phenotype intermediate of the two
homozygous phenotypes.
R
W
R
RR
RW
W
RW
WW
Approximately 25 % of offspring would be expected to have red flowers.
c. A man with type O blood has children with a woman with type B blood. Her father had type A blood.
What are the possible phenotypes of the children?
IB
i
i
IB i
ii
i
IB i
ii
The fact that the woman's father was type A is a hint that you need to
consider whether the woman is homozygous or heterozygous for type B
blood. If the father had an IB allele, he would have been type AB, not type A.
So he had to be heterozygous. And the woman, his daughter must be, too.
The children of a type O man and this woman could be either type O or type B
(heterozygous).
d. The gene for red eyes in fruit flies is X-linked. The allele for white eyes (Xr) is recessive to the allele
for red eyes (XR). Suppose a homozygous red-eyed fly had offspring with a white-eyed fly. What
percentage of offspring would you expect to be red-eyed?
Xr
Xr
XR
XR Xr
XR Xr
XR
XR Xr
XR Xr
Red is dominant to white, so 100 % of
offspring would be red-eyed.
e. Suppose you were interested in breeding gray, long-haired rabbits. You had one true-breeding gray,
long-haired rabbit (GGLL) and one heterozygote gray, long-haired rabbit. What percentage of offspring
would you expect to be gray and long-haired?
GL
GL
GL
GL
GL
GGLL
Gl
GGLl
gL
GgLL
gl
GgLl
When in doubt, draw a 4x4 Punnett Square to solve a dihybrid
cross. Remember that each gamete can contain only one allele
for each gene. So, there will be one G/g and one L/l in each
gamete in this example. Once you set up a 4x4 square, you may
realize that you don't need all rows or columns. For example,
here, every row will have the same result, so you can ignore the
bottom three rows. It turns out that 100 % of offspring would be
gray and long-haired.
f. Both Rita and Mark have a widow's peak, but their son Matt has a straight hairline.
What are their genotypes? Because Matt has a straight hairline, but his parents' have widow's peak,
straight hairline must be recessive and each parent must carry a recessive allele, masked behind a
dominant allele. Matt's genotype is ww and his parents' genotypes are both Ww.
Rita and Mark also have freckles, but Matt does not. What is the probability that their next child will have
freckles? What is the probability that their next child will have a straight hairline? What is the probability
that their child will have both freckles and a straight hairline?
Again, both of his parents are heterozygous freckles and Matt is homozygous for the recessive trait, not
having freckles. Two 2x2 Punnett squares can be used to predict the possible genotypes and
phenotypes of Rita and Mark's next child.
W
w
W
WW
Ww
w
Ww
ww
Their next child has a 25 % chance of having straight hairline.
F
f
F
FF
Ff
f
Ff
ff
Their next child has a 75 % chance of having freckles.
Use the rule of multiplication to determine the probability of having straight hairline AND freckles:
0.25 x 0.75 = 0.1875 or about 19 % (1/4 x 3/4 = 3/16). Or you could set up a 4x4 Punnett square.
g. Assume that three genes control eye color, each found on a
different chromosome, and that each of these genes has only
two alleles—one dominant and one recessive—represented by
the following letters: gene 1 B and b; gene 2 G and g; gene 3,
Y and y. The chart of the right shows how eye color is
determined by the number of dominant alleles in his or her
genotype. Determine the phenotype for children of the following cross.
Write the phenotype in the square with the genotype.
# dominant
alleles
6-5
4-3
2-1
0
eye
color
black
brown
green
blue
Type of inheritance? ___polygenic___
If a trait is polygenic, count the number of dominant alleles to determine phenotype.
bgy
Bgy
BGY
BbGgYy
brown
BBGgYy
brown
BGy
BbGgyy
green
BBGgyy
brown
BgY
BbggYy
green
BBggYy
brown
Bgy
Bbggyy
green
BBggyy
green
bGY
bbGgYy
green
BbGgYy
brown
bGy
bbGgyy
green
BbGgyy
green
bgY
bbggYy
green
BbggYy
green
bgy
bbggyy
blue
Bbggyy
green
4. If fur color in mice is a polygenic trait, how can a mouse be lighter or darker than either of its parents?
Suppose fur color is controlled by three genes. If the parents are heterozygous for each gene, each
parent would have 3 dominant and 3 recessive alleles and an intermediate phenotype. It is possible that
these mice can pass down 8 different combinations of dominant and recessive alleles, resulting in a range
of phenotypes among offspring. For example, if both pass down 3 dominant alleles, they could have a
very dark offspring. If both pass down down 3 recessive alleles, they could have a very light offspring.
See section 9.16 for a Punnett Square showing this possibility.
5. Explain why geneticists consider there to be no true biological race among humans. What type of
genetic variation is likely to be shared between people identified as being of a particular race? What type
of genetic variation is likely to be as different between people of the same race as between people of
different races?
People of the same race likely share genetic variation in genes that control appearance, for example hair
color, hair type, skin color, eye shape. Traits other than ones associated with appearance are just as
likely to vary between members of the same race as between people of different races. This has been
demonstrated by numerous studies.
6. Using the term antigen and antibody, explain why transfusion with an incompatible blood type can kill
the recipient.
For example, people who have type A blood have red blood cells with the type A antigen on the surface.
People who have type B blood have red blood cells with the type B antigen on the surface. If A blood is
transfused into a person with type B blood, that person would make anti-A antibodies that would attach to
the type A blood cells, causing them to clump and clot in blood vessels. This clotting can kill a person. If
type B blood were put into someone with type A blood, the type A person would make anti-B antibodies
and have a similar reaction.
Type AB people do not make antibodies to either A and B and type O people have red blood cells that
lack antigens. Type AB is the universal recipient and type O is the universal donor.
7. Why is sickle-cell anemia considered to be an example of pleiotropy?
The gene that causes sickle-cell anemia in homozygous recessives, in heterozygotes, confers resistance
against malaria while not causing sickle-cell disease. Thus, one gene affects two seperate phenotypes.
This phenomenon is observed because the parasite (the protist Plasmodium) that causes malaria cannot
survive well in sickled red blood cells.
8. Why do sons only inherit X-linked genes from their mother and why X-linked recessive conditions are
more often found in men.
Sons inherit a Y from their fathers and an X from their mothers, thus can only inherit X-linked genes from
mom. Because men only have one X, they are more likely to show X-linked recessive conditions than
women, whose second X may carry a dominant allele that would mask a recessive allele.
9. Assume five major genes (A, B, C, D, and S) interact to determine coat color of mice. The genetic
determination of coat color in other mammals is similar to that of mice. Coat color is affected by the color,
amount, and location of pigment in the mouse's hair.
A – affects whether pigment is present in the whole hair or not
B – determines the color of pigment produced
C – permits expression of any pigment at all; cc animals are albino
D – controls the amount of pigment present in any hair
S – controls distribution of pigment throughout body (presence/absence of spots)
Which of the five coat color genes would you say is epistatic to (blocks) the other genes? Explain.
C – because it determines whether there is any pigment at all. If there is no pigment, it does not matter
where the pigment would be or what color it would be.