Punnett Square Practice: Mice
Name_____________
In mice, the ability to run normally is a dominant trait. Mice with this trait are called “running mice” (R). The recessive trait causes
mice to run in circles only. Mice with this trait are called “waltzing mice” (r).
Hair color is also inherited in mice. Black hair (B) is dominant over brown hair (b).
1 Trait Crosses
1.
2.
3.
Cross a heterozygous running mouse with a homozygous running mouse.
a.
What are the parent genotypes? __________ x ____________
b.
Draw the punnett square.
c.
What is the genotype ratio?
d.
What is the phenotype ratio?
Cross a homozygous brown mouse with a homozygous brown mouse.
a.
What are the parent genotypes? __________ x ____________
b.
Draw the punnett square.
c.
What is the genotype ratio?
d.
What is the phenotype ratio?
Cross a homozygous waltzing mouse with a heterozygous running mouse.
a.
What are the parent genotypes? __________ x ____________
b.
Draw the punnett square.
c.
What is the genotype ratio?
d.
What is the phenotype ratio?
2 Trait Crosses
4.
5.
6.
Cross a homozygous waltzing, heterozygous black mouse with a homozygous waltzing, heterozygous black mouse.
a.
What are the parent genotypes? __________________ x ___________________ (then do FOIL……..)
b.
Draw the punnett square (4x4 square!!).
c.
What is the phenotype ratio?
Cross a homozygous running, heterozygous black mouse with a waltzing, brown mouse.
a.
What are the parent genotypes? __________________ x ___________________ (then do FOIL……..)
b.
Draw the punnett square (4x4 square).
c.
What is the phenotype ratio?
Cross a heterozygous running, homozygous brown mouse with a heterozygous running, homozygous black mouse.
a.
What are the parent genotypes? __________________ x ___________________ (then do FOIL……..)
b.
Draw the punnett square (4x4 square).
c.
What is the phenotype ratio?
1-Trait Genetics Practice Problems
Name:
In humans, albinism is controlled by a single recessive allele ( c ), and normal skin pigmentation is
controlled by the dominant allele ( C ).
1.
If a homozygous normal male marries a woman with albinism, what are their chances they will have a baby
with albinism? Show the Punnett Square for your answer.
2. If a heterozygous normal female marries a man with albinism, what are their chances they will have a baby
with albinism? Show the Punnett Square for your answer.
3.
If both the mom and dad are heterozygous normal (do NOT have albinism), what are their chances they will
have a baby with albinism? Show the Punnett Square for your answer.
4. A man and woman plan to marry and wish to know the probability of their having any children with albinism.
What could you tell them about their chances for having a baby with albinism if (use punnett square if possible):
a. Both are normally pigmented, but each has one parent with albinism?
b. The man has albinism, the woman is normal, but her father has albinism?
c. The woman has albinism and the man’s family includes no one with albinism?
Phenylketonuria (PKU) is a heritable condition in humans involving inability to break down the amino acid
phenylalanine because of lack of a certain enzyme. If not diagnosed and treated very shortly after birth, PKU’s
develop severe mental disabilities and usually do not reproduce. Almost all PKU children, therefore, are born
to parents who are not PKU’s.
5. Is the gene responsible for PKU dominant or recessive? Explain how you know.
6. The normal (non-PKU) brother of a PKU seeks the advice of a genetic counselor before a contemplated marriage.
a. Neither of his parents have PKU. What are the genotypes of his parents? (use any letter you choose)
b. What is the chance that he is a carrier for PKU (heterozygous)? Explain how you know. Use a Punnett
Square if possible.
In fruit flies, wing length is controlled by one gene. Body color is controlled by another gene. See the picture
below.
Long wing = L
Gray body = G
Vestigial (short) wing = l
Black body = g
Scenario 1: Cross a heterozygous gray-body and a homozygous black body.
Genotype ratio: ______ : _____ : ______
Phenotype ratio: _____ : _____
GG
Gg
gg
Gray
black
Scenario 2: Cross a heterozygous gray-body and a heterozygous gray-body.
Genotype ratio: ______ : _____ : ______
Phenotype ratio: _____ : _____
GG
Gg
gg
Gray
black
Scenario 3: Cross a homozygous long-wing and a homozygous vestigial-winged
Genotype ratio: ______ : _____ : ______
Phenotype ratio: _____ : _____
LL
Ll
ll
Long
vestigial
Scenario 4: Cross a homozygous long-wing and a homozygous long-wing.
Genotype ratio: ______ : _____ : ______
Phenotype ratio: _____ : _____
LL
Ll
ll
Long
vestigial
Scenario 5: Cross a heterozygous long-wing and a heterozygous long wing.
Genotype ratio: ______ : _____ : ______
LL
Ll
ll
Phenotype ratio: _____ : _____
Long
vestigial
Scenario 6: Make a cross between 2 flies that are expected to produce 50% homozygous gray-bodied offspring
and 50% heterozygous gray-bodied offspring.
Parent genotypes: _______ and _________
Parent phenotypes: __________ and __________
Scenario 7: Make a cross between 2 flies that are expected to produce 50% heterozygous long-winged
offspring and 50% vestigial-winged offspring.
Parent genotypes: _______ and _________
Parent phenotypes: __________ and ___________
Scenario 8: Make a cross between 2 flies that are expected to produce all heterozygous gray-bodied offspring.
Parent genotypes: _______ and _________
Parent phenotypes: __________ and ___________
Scenario 9: Make a cross between 2 flies that are expected to have all vestigial-winged offspring.
Parent genotypes: _______ and _________
Parent phenotypes: __________ and ___________
2-Trait Genetics Practice Problems
1.
Name:
A homozygous short hair, homozygous stripeless cat and a homozygous long hair, homozygous tabby cat have
kittens.
a. Draw the Punnett Square (4x4 square).
b. What is the phenotype ratio?____________________________
c. What percentage of the offspring will have short hair and be stripeless? _____________
d. What percentage of the offspring will have long hair and be tabby? _______________
2. A heterozygous tabby, heterozygous normal cat and a homozygous stripeless, homozygous colorpoint cat have
kittens.
a.
Draw the Punnett Square (4x4 square).
b. What is the phenotype ratio?____________________________
c. What percentage of the offspring will be tabby and colorpoint? _____________
d. What percentage of the offspring will have stripeless and normal? _______________
3. In turkeys, a dominant gene, B, produces bronze color. Its recessive allele, b, results in a red color. Another
dominant gene, F, results in normal feathers; its recessive allele, f, produces feathers without webbing.
Two bronze turkeys with normal feathers were mated. Their offspring consisted of 9 bronze with normal
feathers, 1 red with webless feathers, 3 red with normal feathers, and 3 bronze with webless feathers.
What were the genotypes of the parents (bronze/normal)? ________________________________________
Explain how you know. ________________________________________________________________ _____
4.
A solid colored, short-haired female rabbit (ssHH) is mated to a spotted, long-haired male rabbit (SShh).
Draw the Punnett Square.
How many out of 16 will be expected to be spotted with long hair? ________________________
How many out of 16 will be expected to be spotted with short hair? ________________________
5. In Coleus plants, the following traits are found:
D = deeply scalloped leaves
d = shallow scalloped leaves
W = white at base of leaf
w = no white at base of leaf
Cross a heterozygous deep scallop / homozygous no white plant with a
heterozygous deep scallop / heterozygous white plant.
Show the Punnett Square.
What is the phenotype ratio?
You want to get a homozygous deep scallop, homozygous white plant. Can you get that from this cross? _____
What genotypes of parents could you cross to get this type of plant? _______________ and _______________