Name: __________________________________ Date: ___________________ Class: _______
KEY FOR Genetics Review Problems: Monohybrid and Dihybrid
Complete Dominance, Incomplete Dominance, Codominance, Sexlinked, Pedigrees
These problems should be completed on a separate sheet of paper. Skip the ones you don’t
know, but leave a blank space for them so you can do them later!
1. A type of blindness known as aniridia occurs in humans and is due to an allele, A, which is
dominant to the allele for normal vision, a. Migraine headaches are due to a dominant allele,
M, which is dominant to the allele for the absence of migraine headaches, m. Each gene is
found on a different pair of Autosomes. A man with normal vision and without migraines
marries a woman with normal vision and migraines. The man’s father had aniridia but the
woman’s did not. The woman’s father had normal headaches, but never a migraine.
Migraines: MM, Mm
No Migraines: mm
Aniridia: aa
Normal vision: AA, Aa
Man: normal vision, no migraines
AA or Aa
mm
(father had aniridia, so the man’s genotype must be Aamm)
Woman: Normal vision, migraines
AA or Aa
MM or Mm
(father had no migraines, so the woman’s genotype must be AAMm or AaMm )
a. Was the man’s father homozygous or heterozygous for aniridia? homozygous
b. Did the woman’s mother suffer from migraines? We can’t tell from the information
given. YES
c. What is the probability of this couple producing a child who is normal for both traits?
3/8
Father: Aamm x Mother: AaMm
1/15
2. In humans, wavy hair (CS) results from the co-dominant situation of curly hair (C) and
straight hair (S). What are the possible results (genotypes and phenotypes and their
percentages of the offspring) if a curly-haired man and a wavy-haired woman have children?
Genotype
CC
CS
%
50%
50%
Phenotype
Curly Hair
Wavy Hair
50%
50%
%
3. Hemophilia is a sex-linked disorder. If you have
the dominant gene (XH), you will have normal
blood. If you have ONLY the recessive gene (Xh),
your blood will not clot. Sarah’s blood clots
normally. What are all of the possible genotypes
of her parents?
2/15
4. In chimpanzees, straight fingers (F) are dominant to bent fingers (f). Complete a Punnett
square to show the genotypes and phenotypes expected for the following cross: A
homozygous straight female and a homozygous bent male?
Genotype
Ff
Phenotype
Straight Fingers
%
100%
%
100%
5. What kind of surface molecule (antigen) is on type O blood?
There are no antigens on type O blood.
3/15
6. In dogs, erect ears (E) is dominant over droopy ears (e). What are the results (genotypes
and phenotypes and their percentages of the offspring) if a homozygous erect dog has a
litter of puppies with a heterozygous dog?
Genotype
EE
Ee
Phenotype
Erect ears
%
50%
50%
%
100%
7. Tony has hemophilia, a sex-linked recessive disorder in which someone’s blood does not
clot properly. One day, he becomes angry at his parents, and blames his father, Chris, for
giving him the disease. His mother, Anna, defends Chris, saying that he can’t blame him for
that.
a. Who gave Tony hemophilia: Both parents, just Anna, or just Chris?
b. Explain your answer by drawing a Punnett square and using it to support your
conclusion.
Tony inherited his Y allele from his father.
He inherited hemophilia (carried on the X
chromosome) from his mother.
8. In humans, straight toes (S) is dominant over curled toes (s). What would be the result
(genotypes and phenotypes and their percentages of the offspring) of a cross between a
recessive male and a heterozygous female?
Genotype
Ss
ss
Phenotype
Straight toes
Curled toes
%
50%
50%
%
50%
50%
9. What kind of antibodies does type A blood have?
4/15
Type B antibodies
10. What is a sex-linked trait? A trait that is carried on one of the sex chromosomes
(usually on the X)
11. Muscular dystrophy in humans is caused by a recessive, sex-linked gene (Xm). Suppose
that a woman who is a carrier has a child with a normal man. What are the chances that the
child is a…
a. Normal Female (phenotypically normal female can be a carrier)
50%
b. Normal Male 25%
c. Carrier Female25%
d. Carrier Male 0%
e. Affected Female 0%
f. Affected Male 25%
12. In humans, the ability to roll your tongue (T) is dominant over not being able to roll your
tongue (t). What would be the result (genotypes and phenotypes and their percentages of
the offspring) of a cross between a heterozygous
male and a homozygous recessive female.
Genotype
Tt
tt
%
50%
50%
Phenotype
Can roll tongue
50%
Can not roll tongue
%
50%
5/15
13. Mary, who has type A blood has children with Juan, who has type B blood. They have four
children: Anita (genotype IAi), Marisa (genotype ii), Stephen (genotype IA IA), and Marc
(genotype IA IB). The youngest child has type O blood. The second-oldest has blood type A.
The oldest child is from Mary’s previous marriage to Adam, who had blood type A.
Mary: Type A (IA IA or IAi)
Juan: Type B (IB IB or IBi)
Children: Anita (IAi), Marisa (ii), Stephen (IA IA), and Marc (IA IB)
Youngest: Type O
Second-oldest: Type A (either Anita or Stephen)
Oldest: Mary and Adam’s child (has type A) (either Anita or Stephen)
Since Mary and Juan (not Adam) had Marisa (ii), Their blood types must be: (Mary: I A i )
and (Juan: IB i)
a. What is Mary’s genotype? IA i
b. What is Juan’s genotype? IB i
c. Arrange Mary’s children from youngest to oldest below. Marisa, Marc, Anita,
Stephen
WORK:
Youngest: Type O – Marisa
Oldest: Mary & Adam’s child (type A) is Stephen b/c Mary and Juan can not make
blood type (IA IA),
Second-oldest: Type A – Must b Anita because Stephen is Mary and Juan’s child
This leaves the Second-youngest to be Marc
ANSWER:
Arranged from youngest to oldest: Marisa, Marc, Anita, Stephen
14. In cattle, when a cow with Red (R) coat color and cow with White (W) coat color are mated
together, the offspring have a blended coat color that looks pink. It is called “roan.” What
type of inheritance does this show? Choose from: complete dominance, incomplete
dominance, codominance, sex-linked.
15. What is the female sex chromosome called? The male sex chromosome?
X
Y
16. A person with blood type A can successfully donate blood to people with which blood
type(s)? A, AB
17. In Andalusian chickens, black (B) and white (W) chickens mate to make grey chickens in a
pattern of inheritance called incomplete dominance. Show a Punnett square for a cross
between a grey chicken and a white chicken. What
are the expected genotypes and phenotypes and
their percentages of the offspring) of the cross?
Genotype
%
BW
50%
WW
50%
6/15
Phenotype
Gray
White
%
50%
50%
18. What does the “P” in “P generation” stand for? Parental
19. In humans, wavy hair (CS) results from the co-dominant situation of curly hair (C) and
straight hair (S). What are the possible results (genotypes and phenotypes and their
percentages of the offspring) if two curly-haired
people have children?
Genotype
CC
Phenotype
Curly Hair
%
100%
%
100%
20. A person with blood type AB can NOT successfully donate blood to people with which blood
type(s)?
A, B, O
21. In cats, one type of fur color is sex-linked. Gene XW produces a yellow coat, and XB
produces a black coat. The heterozygote, XBXW has a mixture of black and yellow hairs and
is called a tortoise shell.
a. There are two types of inheritance going on in the example above. What are they?
Choose from: Complete dominance, Incomplete dominance, Co-dominance, Sexlinked.
b. Show a cross between a tortoise shell cat and a yellow cat. What are the results of
the cross (genotypes and phenotypes with percentages)
Genotype
XB XW
XW XW
XB Y
XW Y
Phenotype
Tortoise Shell Female
Yellow Female
Black Male
Yellow Male
%
25%
25%
25%
25%
%
25%
25%
25%
25%
7/15
c. A black cat and a yellow cat have kittens. 50% of the cats produced are tortoise shell
and the other 50% are yellow male cats. What are the genotypes of the parents? XW
XW and XB Y
22. What is the difference between incomplete dominance and codominance?
-In incomplete dominance, the heterozygote is a blend of both homozygous traits.
-In codominance, the heterozygote shows BOTH homozygous traits. NO blending
occurs. Both traits are equally present in the heterozygote
23. A person who has sickle cell anemia has the genotype SS and a person who does not have
the sickle cell trait has the genotype RR. This disease is inherited through incomplete
dominance. What is the percent chance that a man who is a carrier for sickle cell can have
children with sickle cell disease with a woman who is normal?
50% RS
50% RR
0% chance these two individuals will have a child with
sickle cell
8/15
24. Which blood types are:
a. Dominant? A, B, AB
b. Recessive? O
c. Co-dominant to each other? A and B
25. What is a “carrier”? An individual who has one copy of the allele for a recessive
disorder and does not exhibit symptoms
26. A person who can successfully donate blood to ANY person is considered the “universal
donor.” Which blood type is the universal donor? Which blood type is the universal
recipient? DONOR: “O” RECIPIENT: “AB”
27. Cara, a woman who does not show any symptoms of sickle cell disease has three children
with Mark, who also does not show any symptoms of sickle cell disease. One of their
children has sickle cell disease. What are the genotypes of Cara and her husband, Mark?
Genotype for Sickle Cell is SS. Genotype for Normal blood is RR.
Cara and her husband Mark are both RS.
28. What is the difference between complete dominance and incomplete dominance?
-In complete dominance, the heterozygote has the dominant phenotype.
-In codominance, the heterozygote shows BOTH homozygous traits. NO blending
occurs. Both traits are equally present in the heterozygote
29. Can a type O father and a type AB mother have a type AB baby? Why or why not?
Complete a Punnett square to show your results.
No, A type O father and a type AB mother can not have a
Type AB baby because the father can only give an O allele.
He does not have an A or a B allele to give. So, the child can
get an A or a B allele from Mom, but always an O from dad.
They can only have children with type A or type B blood.
30. Color blindness in humans is due to a
recessive, sex-linked gene (Xb). A man who
is normal marries a woman who is colorblind. What are the chances of having a ….
a. Colorblind female? (0%)
b. Colorblind male? (50%)
c. Normal female? (phenotypically
normal female can be a carrier) (50%)
9/15
d. Normal male? (0%)
e. Carrier female? (50%)
f. Carrier male? (0%)
31. Explain the principle of segregation using Alyssa as an example. Alyssa has freckles, and
her genotype is Ff. When Alyssa’s body makes eggs, it will use the principle of segregation.
Explain this according to how it will happen to Alyssa.
a. Alyssa also has brown eyes, genotype Bb. Explain the principle of independent
assortment as it pertains to her body making eggs.
When Alyssa makes eggs, her genes for freckles F and f will sort into her eggs
independently of how her genes for eye color will sort into her eggs (the way one
sorts doesn’t depend on the other). Some of her eggs will end up with the genes
FB, others Fb, some fB and some fb.
32. When is a test-cross used? To determine the genotype of a phenotypically dominant
individual.
33. What does the F in “F1” stand for? Filal (it’s Latin for “son”)
34. How do male and female sex chromosomes look different? The male chromosome (Y) is
smaller than the female chromosome (X).
35. In the jimsonweed plant, the allele for smooth seed pods, s, is recessive to the allele for
spiny pods, S. At another independently assorting locus, the allele for white flowers, w is
recessive to the allele for purple flowers, W.
a. Two plants of unknown genotype are crossed and the following progeny result:
smooth and white: 27. spiny and white: 85. spiny and purple: 256. Smooth and
purple: 93. Identify possible genotypes for the parent plants.
Smooth ss
Spiny: SS, Ss
White: ww
Purple: WW, Ww
Smooth and white: (ssww) 27
Spiny and white: (SSww or Ssww) 85
Spiny and purple: (SSWW, SSWw, SsWw, SsWw) 256
Smooth and purple: (ssWW or ssWw) 93
Two ways to figure out what the genotype of the plants are:
 Think about ratios: What is the one ratio you needed to remember?
(9:3:3:1). Notice that the plants represent this ratio 256 (9) : 93(3) :
85 (3) : 27 (1)
 What do the genotypes of the plants HAVE to be in order to have smooth
and white plants as well as dominant plants?
10/15
b. In another cross, two plants of unknown genotype are crossed and produce the
following progeny: smooth and white: 75, spiny and white:82. Spiny and purple: 69;
smooth and purple:77. Identify possible genotypes for the parent plants.
Smooth and white: (ssww) 75
Spiny and white: (SSww or Ssww) 82
Spiny and purple: (SSWW, SSWw, SsWw, SsWw) 69
Smooth and purple: (ssWW or ssWw) 77
o Think about ratios. If you DO NOT see the 9:3:3:1, then you know that the
gentypes are NOT heterozygous. It appears that each plant is represented in about
equal number.
o You will most likely need to do a Punnett square here. This one is very tricky, so if
you got it, good for you! You know that BOTH parents can not be heterozygous
(otherwise you would see a 9:3:3:1 ratio). Look at the ratio again. Notice it is
1:1:1:1. This means that you will have equal numbers of each plant, and four types
of plants. Draw your Punnett square to match what you need (only four types of
plants in equal number) – so you only need four boxes.
o One of the plants must be heterozygous, and one must be recessive (trial and
error).
The genotypes of the parents are: SsWw and ssww
11/15
36. Use the pedigree for Huntington’s disease below to answer the following questions. You
may answer the Pedigree questions on this paper.
a. Is the individual #1 a male or female? Female
b. Determine whether Huntington’s is dominant or recessive to normal (non-disease)
phenotype.
12/15
c. Does individual #6 have Huntington’s or do they have a normal phenotype?
d. What is the genotype of the individual #8? It must be Hh to have hh kids.
e. If individual #11 was crossed with an individual who has Huntington’s, what is the
percent chance that they will have a child with a normal phenotype? (Do a Punnett
square)
13/15
14/15
37. Use the pedigree below to answer the following questions. You may answer the Pedigree
questions on this paper.
a. Is the individual III-4 a male or female?
b. Determine whether the disease above is dominant or recessive to normal (nondisease) phenotype.
c. Does individual IV-2 have the disease or do they have a normal phenotype?
d. What is the genotype of the individual II-3?
e. If individual III-1 was crossed with individual III-6, what is the percent chance that
they will have a child with a normal phenotype? (Do a Punnett square(s))
15/15