Name:__________________________________ Section:_______ Date:___________
Intensive Biology
Chapter 9
Basic Genetics Worksheet
Terms to know:
P generation, F1 generation, F2 generation, alleles, traits, characteristics, homologous
pairs, homozygous, heterozygous, phenotype, genotype, principle of segregation,
principle of independent assortment, testcross, dominant, recessive, carriers, incomplete
dominance, codominance, multiple alleles, linked genes, sex-linked gene, pleiotropy,
polygenic inheritance
Variations of Punnett Square:
 Monohybird
 Dihybrid
 Testcross
 Incomplete dominance
 Codominance
 Multiple alleles
 Sex linked genes
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Name:__________________________________ Section:_______ Date:___________
Practice Problems
Problem #2 - In pea plants, yellow peas are dominant over green peas.
Use a Punnett square to predict the phenotypic and genotypic outcome (offspring) of a
cross between a plant heterozygous/hybrid for yellow (Yy) peas and a plant
homozygous/purebred for green (yy) peas.
Problem #3 - In pea plants, yellow peas are dominant over green peas.
Use a Punnett square to predict the phenotypic and genotypic outcome (offspring) of a
cross between two plants heterozygous for yellow peas.
Problem #4 - In pea plants, round peas are dominant over wrinkled peas.
Use a Punnett square to predict the phenotypic and genotypic outcome (offspring) of a
cross between a plant homozygous for round peas (RR) and a plant homozygous for
wrinkled peas (rr).
Problem #5 - In pea plants, round peas are dominant over wrinkled peas.
Use a Punnett square to predict the phenotypic and genotypic outcome (offspring) of a
cross between two plants heterozygous for round peas.
1. Within trible populations coat texture (stiff bristles verses soft short fine hair) and
toe type (clawed verses no claws) are autosomal genes. As a geneticist, you are
doing the following cross: male parental phenotype is soft, short fine hair and no
toe claws; female phenotype is stiff bristles and clawed toes. In looking at the
offspring of these matings, you find that all F1 are phenotypicNatasha stiff
bristled and have clawed toes. Were the parents heterozygous or homozygous?
Which are dominant alleles? If you now mate males to females within the F1
generation, what would you expect the phenotype ratio to be?
2. In Venutian newts, skin color (royal blue verses hunter green) and ruffle patterns
(fringed red verses spiked yellow) are geneticNatasha inherited. A royal blue,
fringed red ruffled female whose father was hunter green with a spiked yellow
ruffle is mated to a royal blue, fringed red ruffled male.
 In looking at all of the offspring from their matings, one finds that roughly
56% are royal blue with fringed red ruffles; roughly 6% are hunter green
with spiked yellow ruffles; roughly 18% are royal blue with spiked yellow
ruffles; roughly 18% are hunter green with fringed red ruffles. What were
the genotypes of the parents?
 If a royal blue with spiked yellow ruffles was mated with hunter green
with fringed red ruffles from this generation, what are the possible
phenotypes (and the ratio) and the possible genotypes of the offspring?
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Name:__________________________________ Section:_______ Date:___________
EXTRA PRACTICE PROBLEMS
All individuals featured on this worksheet are fictional characters. Any likeness or similarity that they
may have to persons living or dead is pure coincidence.
Basic Monohybrid Crosses
1. Max’s second toe is much longer than his thumb toe – at least an inch! This is a
recessive trait. If he marries Mia, who is a heterozygous for having her second toe
shorter than her thumb toe, answer the following questions. Use the letter T to
represent this allele.
a. Which trait is dominant – long second toe or short second toe?
b. What are Max and Mia’s genotypes?
c. If Max and Mia have children, what are the possibilities of their children’s
phenotype?
d. What are the possibilities of their children’s genotype?
2. Jonathan is heterozygous for folded eyelids. This is a dominant trait. He marries
Charlotte who does not have folded eyelids. Use the letter “E” to represent this allele
and answer the following questions:
a. What are Jonathan and Charlotte’s genotypes?
b. If they have children, what are the possibilities of their genotypes?
c. What are the possibilities of their phenotype?
3. Because of Ben’s lovely long fingers, he can play the piano with style. But if you
look at his parents, Ben’s father has long fingers and his mother has short fingers.
Knowing that long fingers is the dominant gene, what must Ben’s genotype be? Use
the letter “L” to represent the allele for long finger genes.
4. Mara is so double jointed that her thumb nail can bend over backwards to touch her
watch. Her father is double jointed also, but her mother is not. If being double
jointed is a recessive trait, what is the genotype of Mara and her parents? Use the
letter “D” to represent the allele for this trait.
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Name:__________________________________ Section:_______ Date:___________
Dihybrid Crossing
5. In rabbits, black hair depends on a dominant allele, B, and brown on a recessive allele,
b. Short hair is due to a dominant allele, H, and long hair to a recessive allele, h. If a
homozygous black, homozygous short-haired male is mated with a brown, long-haired
female, what will their offspring look like? What will be the genotypes of the offspring?
If two of these F1 rabbits mated, what phenotypes would you expect among their
offspring, in what proportion?
6. The alleles for the following two traits are expressed as either dominant or recessive.
For eye color, brown “B” is the dominant allele, and blue “b” is the recessive allele. For
the shape of the nose, pointy “N” is the dominant allele, and flat “n” is the recessive
allele.
Emma is heterozygous brown eyed and has a flat nose. She marries Noah who is
heterozygous brown eyed and a heterozygous pointy nosed person. What are the possible
genotypes and phenotypes for their children?
Test Cross
7. Alex’s German shepherd is an impressive black beauty. The black color for German
shepherd’s is a dominant trait. One day, Alex, struck with utter curiosity, is
determined to figure out the genotype of her pet. Being one of 12 brilliant prodigies
of Ms. Yun’s biology class, she decides to conduct a scientific experiment at home.
Her dog marries Tyler’s white German shepherd and their firstborn puppy turns out to
be the color white. “Ah ha!,” Alex shouts, “My dog’s genotype is…” (Use the letter
“b” to represent this trait)
8. Roma has perfect pitch. If you tell her to sing the note “A”, she can find the note
perfectly without any instrumental accompaniment. Having perfect pitch is a
recessive trait. She marries Tyler who does not have perfect pitch, and they eventually
have three children. Tyler Jr, their third son is fortunate to take after his mother’s
musical abilities and he has perfect pitch too. What are the genotypes for Roma, Tyler
, and Tyler Jr.? Use the letter “P” to represent the allele for this gene.
9. Polydactyly (extra fingers and toes) is due to a dominant gene. Isaac’s father is
polydactyl, the Isaac’s mother has the normal phenotype and they have Isaac who has
five fingers per hand. What is the genotype of Isaac’s father? Of Isaac’s mother?
What is the probability that Isaac’s sibling have the normal number of digits?
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Name:__________________________________ Section:_______ Date:___________
Incomplete Dominance
10. Brachydactyly is a disorder that causes people to have abnormally short fingers and
toes because of a missing bone. A person with homozygous alleles for normal fingers
will have normal fingers, a person with heterozygous alleles will develop the disorder
brachydactyly, and a person with homozygous alleles for the abnormal gene will be
severely crippled, missing some fingers and toes.
If Max is homozygous for the normal gene, and Mara is homozygous for the
brachydactyly gene: (Use the letter “B” to represent the brachydactyly allele)
a. What are the genotypes and phenotypes for Max and Mara?
b. What are the possible genotypes of their children?
c. What are the possible phenotypes of their children?
Jonathan and Mara have the same genotype. Jonathan marries Mia, who is heterozygous
for this gene.
d. What is Mia’s genotype?
e. What are the possible genotypes for their children?
f. What are the possible phenotypes for their children?
11. Hair type alleles express incomplete dominance. Emma is homozygous for pin
straight hair (HH) and Ben is heterozygous for wavy hair. The opposite for
homozygous pin straight hair is homozygous super curly hair.
a. What are the genotypes for Ben and Emma?
b. What are the possible genotypes of their children?
c. What are the possible phenotypes of their children?
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Name:__________________________________ Section:_______ Date:___________
Multiple Alleles
12. Charlotte loves gardening and has a beautiful garden filled with tulips of different
colors. If she cross pollinates a homozygous yellow flower with a heterozygous pink
flower, answer the following questions. Remember, alleles for flower colors express
incomplete dominance. Genotypes: yy = yellow; rw = pink; ry = orange; wy = cream.
a. What are the possible alleles found in the parents’ gametes?
b. What are the possible genotypes of the first filial generation?
c. What are the possible phenotypes of the first filial generation?
13. Two dominant alleles can be found for blood type = IA or IB. The recessive allele is i.
A person can be homozygous for IA IA (Type A) or IB IB (Type B) or ii (Type O);
heterozygous for IAi (Type A) or IBi (Type B) or IA IB (Type AB).
Alex is type A blood and Noah is type B blood. Their son has type O blood.
a. What are Alex’s and Noah’s genotype?
b. Is it possible for their next child to have type AB blood?
c. Is it possible for any of their children to have type A blood?
14. Roma has type O blood and Isaac has type AB blood.
a. Can they have a child with Isaac’s type blood?
b. Can they have a child with Roma’s blood?
c. What are the phenotypes of their children’s blood?
15. Blood typing has often been used as evidence in paternity cases, when the blood type
of the mother and child may indicate that a man alleged to be the father could not
possibly have fathered the child. For the following mother and child combinations,
indicate which blood groups of potential fathers would be exonerated.
Blood Group of Mother
Blood Group of Child
AB
O
A
O
B
A
B
AB
O
A
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Man Exonerated if he
belongs to blood group(s)
Name:__________________________________ Section:_______ Date:___________
Sex Linked traits practice problems
1.
A human female “carrier” who is heterozygous for the recessive, sex-linked trait red color blindness,
marries a normal male. What proportion of their female progeny will show the trait?
A. All
B. ½
C. ¼
D. 0
E. ¾
2.
Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a women’s
grandparents could not be the source of any of the genes on either of her X-chromosomes?
A. Mother’s Father.
B. Father’s Mother.
C. Mother’s Mother.
D. Father’s Father.
E. Mother’s Mother and Mother’s Father.
3.
Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a man’s
grandparents could not be the source of any of the genes on his Y-chromosome?
A. Father’s Mother.
B. Mother’s Father.
C. Father’s Father.
D. Mother’s Mother, Mother’s Father, and Father’s Mother.
E. Mother’s Mother.
4.
A human female “carrier” who is heterozygous for the recessive, sex-linked trait causing red-green
color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male
progeny will have red-green color blindness (or alternatively, will be hemophiliac)?
A. 100%
B. 75%
C. 50%
D. 25%
E. 0%
5.
Hemophilia in humans is due to an X-chromosome mutation. What will be the results of mating
between a normal (non-carrier) female and a hemophilac male?
a. half of daughters are normal and half of sons are hemophilic.
b. all sons are normal and all daughters are carriers.
c. half of sons are normal and half are hemophilic; all daughters are carriers.
d. all daughters are normal and all sons are carriers.
‘
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Name:__________________________________ Section:_______ Date:___________
Pedigree Problems
1. Attached earlobes are a recessive trait. The black circles or squares indicate individuals in this family
tree that have attached earlobes. Perform a pedigree analysis and indicate the genotype of every individual
in this family. If a particular case is ambiguous, list all possible genotypes for that individual.
If the couple on the top right of this pedigree chart were to have another child, what are the odds that that
child would be a boy with attached earlobes?
2. Refer to the pedigree below in which females are indicated by circles and males are indicated by squares.
This pedigree shows the pattern of inheritance of blood types through several generations.
The most likely genotype of the mate of the F1 AO female is __________.
BO
AB
OO
BB
AA
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