Genetics Problems – Monohybrid Crosses
1. In peas, yellow pod color (Y) is dominant to green pod color (y). Give the genotypic and
phenotypic ratio of the following crosses:
a. Heterozygous yellow x homozygous green
b. Heterozygous yellow x heterozygous yellow
c. Homozygous yellow x homozygous green
2. In people, curly hair (C) is dominant to straight hair (c). What are the phenotypic and genotypic
rations of:
a. Heterozygous curly x homozygous straight hair
b. Homozygous curly x homozygous curly
3. If in people, well spaced teeth (B) is dominant to bucked teeth (b), what would be the genotypes
and phenotypes of children resulting from a marriage between a man and woman heterozygous
for well-spaced teeth.
4. In tomatoes, red fruit “R” is dominant to yellow fruit “r”. A tomato plant homozygous for red
fruit is crossed with a plant homozygous yellow fruit.
a. What will be the expected genotypes and phenotypes of the F1 generation?
b. Cross two members of the F1 generation. What are the genotypic and phenotypic ratios
for the F2 generation?
c. Which of the F2 generation are homozygous? Heterozygous?
5. An extra finger in man occurs rarely. The condition is carried on a dominant gene. Suppose one
parent has the normal number of fingers and the other parent has an extra finger but is
heterozygous for the trait. What is the probability that their first child will be normal? Their
second child? Their third child?
In humans, normal pigmented skin is due to the dominant gene “C”. Albinism is due to being
homozygous recessive “c”.
6. Two normally pigmented adults marry and have an albino child. What is the probability that
their next child will have normal pigmentation?
7. A normally pigmented man marries an albino woman. Their first child is an albino.
a. What are the genotypes of the mom, dad, and child?
b. If the couple were to have more children, what would their phenotype(s) be?
8. An albino man marries a normally pigmented woman. They proceed to have nine children, all of
whom are normally pigmented.
a. What are the genotypes of the mom, dad, and children?
9. A normally pigmented man (whose father was an albino) marries an albino woman (whose
parents were both normally pigmented). They have three children, two whose phenotype is
normal pigment and one whose phenotype is albino.
a. List the genotypes of the dad, his parents, mom, her parents, and the three children.
10. A spotted rabbit is crossed with a solid-colored rabbit and produces all spotted offspring. When
these F1 rabbits were crossed among themselves. The resulting F2 generation produced 30
spotted rabbits and 10 solid-colored rabbits.
a. Which of these characteristics is dominant?
b. How many of the spotted rabbits in the F2 generation would you expect to be
homozygous?
c. How many of the solid rabbits in the F2 generation would be homozygous?
d. What could you do to prove whether the F2 spotted rabbits are homozygous or
heterozygous?
11. In sheep, a white coat of fur is dominant to a black coat of fur. White fur is worth more than
black fur. If a black sheep is born, how could the farmer eliminate the genes for the black coat of
fur from his flock?
12. Two black female mice are crossed with a brown male mouse. Female number 1 was mated
with the male brown mouse and they produced 9 black mice and 7 brown mice. Female number
2 was mated with the male brown mouse and they produced 57 black mice. Which color is
dominant? What are the genotypes of the two females and male mice?
Genetics Problems – Incomplete Dominance and Codominance
With incomplete dominance, a cross between organisms with two different phenotypes produces
offspring with a third phenotype that is a blending of the parental traits. It's like mixing paints, red +
white will make pink. Red doesn't totally block (dominate) the white, instead there is incomplete
dominance, and we end up with something in-between.
Apparent phenotype blending can occur when there is incomplete dominance resulting in an
intermediate expression of a trait in heterozygous individuals. For example, in primroses, four-o'clocks,
and snapdragons, red or white flowers are homozygous while pink are heterozygous. The heterozygous
pink flowers result because the single red allele is unable to code for the production of enough red
pigment to make the petals dark red due to the presence of the white allele.
The meaning of the prefix "co-" is "together". Cooperate - work together. Coexist - exist
together. Cohabitate - habitate together. Codominance is similar to incomplete dominance in that they
both show a third phenotype --- not the usual dominant or recessive trait, but a third, different
phenotype. However, in codominance, the recessive and dominant traits appear together in the
phenotype of hybrid organisms.
Example (codominance): In shorthorn cattle, when a red bull (RR) is crossed with a white cow (rr),
their heterozygous offspring (Rr) are neither red nor white, but roan. Roan cattle have intermingled red
and white hairs. If a roan bull (Rr) is crossed with a roan cow (Rr) the calves have a one in four chance
of being white (rr), two in four chances of being roan (Rr), and a one in four chance of being red (RR).
1. In 4 o’clock plants, the genotype RR results in red flowers, the genotype rr results in white
flowers, and the genotype Rr results in pink flowers. If a white flowered 4 o’clock plant (rr) is
crossed with a red flowered 4 o’clock plant (RR),
a. What will be the phenotypic ratio of the F1 generation?
2. If a four o’clock plant having red flowers (RR) is crossed with a white flowered (rr) four o’clock,
the resulting plants are all pink flowered (Rr).
a. What phenotypes would result from a red crossed with a red?
b. What phenotypes would result from a red crossed with pink?
c. What phenotypes would result from a pink crossed with white?
d. What phenotypes would result from a white crossed with white?
3. In poppies Red flowers (RR) are incompletely dominant to white flowers (rr). If they are bred,
they will produce pink flowered poppies (Rr).
a. If two pink flowered poppies were crossed, what phenotypes and genotypes would result?
b. If a horticulturist wanted to get a garden of nothing but pink poppies, what are some ways
they might go about this?
4. Make a punnett square to show the cross of a roan cow (Rr) with a red bull (RR).
a. List the genotypic and phenotypic ratios.
b. A farmer wants to establish a pure strain of roan cattle that breeds true. Why is this
genetically impossible?
5.
Black plumage (AA) and white plumage (aa) are codominant in Andalusian fowls to produce
blue (Aa) plumage. What genotypic and phenotypic rations will result from the following
crosses?
a. Blue x black
b. Blue x blue
c. Blue x white
Dihybrid Inheritance
1.
In tomatoes, red fruit (R) is dominant over yellow fruit (r) and tall vines (T) are dominant
over short vines (t). Cross a homozygous yellow fruit, tall vine tomato plant with a
homozygous red fruit, dwarf vine tomato plant.
a. What will be the genotypic and phenotypic ratios of the F1 generation?
b. Cross an F1 generation plant with a homozygous yellow fruit, tall vine plant. List the
genotypic and phenotypic ratios of the offspring.
c. Cross a heterozygous red fruit, tall vine plant with a homozygous red fruit, dwarf vine
plant. List the genotypic and phenotypic ratios of the offspring.
2.
In black rabbits, black hair is due to the presence of a dominant gene “B” and brown is a
result of homozygous recessive alleles “bb”. Short hair is due to the dominant gene “S” and
long hair results from the homozygous recessive alleles “ss”. Cross a homozygous black
short haired male and a homozygous brown long haired female,
a. What will be the genotypic and phenotypic ratios of the F1 generation?
b. Cross two members of the F1 generation. What are the resulting genotypic and
phenotypic ratios of the F2 generation?
c. Cross a homozygous black long-haired male and a homozygous brown short-haired
female. What will be the resulting genotypic and phenotypic ratios of the F1 generation?
d. In letter b, how many of the black, short haired rabbits would be homozygous for both
pairs of genes?
3.
Suppose you had a black short-haired male rabbit and a brown long-haired female rabbit and
wished to develop a homozygous strain of black long-haired rabbits from them. Outline the
breeding procedure necessary to establish such a strain.
4.
In guinea pigs, black coat is dominant to white coat and short hair is dominant to long hair.
A man was given a black, short-haired guinea pig. In an effort to find out the animal’s
genotype, the man performed a test cross with a white, long-haired animal several times. The
results of the crosses were as follows: 18 black, short haired and 21 black, long-haired. Use
B for black, bb for white, S for short, and ss for long hair. What is the genotype of the man’s
black guinea pig? Of the test cross animal? Of the offspring?
5.
In cattle, the hornless allele (H) is dominant to horned (h) and the black allele (B) is
dominant to red (b). Give all possible genotypes, phenotypes and the ratios expected for the
following crosses.
a. Horned red cow x homozygous hornless homozygous black bull
b. Horned red cow x heterozygous hornless heterozygous black bull
c. Heterozygous hornless homozygous red cow x homozygous horned heterozygous black
bull
6.
In moose, the antlered allele (H) is dominant to the non-antlered allele (h). Additionally,
bearded (B) is dominant to non-bearded (b). What parents would a non-antlered, nonbearded moose have if both parents were antlered and bearded? List their genotypes.
7.
Mr. And Mrs. Jones had just had a child and went to visit Mr. and Mrs. Smith who also just
had a child. Each couple, in great egocentric pride, wanted to show off their child to the
other couple. The suffering babies were passed back and forth from parent to parent for
many hours until suddenly, in the shuffle, everyone forgot whose baby was whose. Mr.
Jones was a genetics teacher and so he worked out a solution. He knew that for eye color,
brown eyes (B) are dominant to blue eyes (b). Mr. Jones had blue eyes, as did his wife. Mr.
Smith had brown eyes and his wife had blue eyes. Upon close inspection, Mr. Jones found
the baby A had blue eyes and Baby B had brown eyes. Mr. Jones, in his infinite wisdom,
knew from the data which child was his. Which baby belonged to the Joneses and which
baby belonged to the Smiths?
8.
Now that Mr. Jones had finally calmed the two women and assured them that they had their
own children, along came the Browns, who also had just had a baby. The ecstatic wives
passed their not too amused babies around again, and pretty soon the wives were all weeping
and wailing over whose baby was whose. The babies did not like this, and so they also
started to wail. Mr. Jones rapidly became bald. Genetics again proved invaluable, and just
by strange coincidence, Mr. Jones happened to have on him some PTC to test the three
babies for the “taster gene.” Taking advantage of the three widely gaping mouths, Mr. Jones
performed some taster tests and found out who was a taster and who was not. Using some
other gathered data, Mr. Jones got the following results.
BROWN
taster
taster
Baby I. non-taster
curler
non-curler
curler
albino
normal pigment
normal pigment
SMITH
tastser
non-curler
normal pigment
taster
non-curler
normal pigment
Baby II.taster
curler
normal pigment
JONES
non-taster
non-taster
Baby III.taster
curler
curler
non-curler
normal pigment
normal pigment
albino
Taster (T) is dominant to non-taster(t), Curler (C) is dominant to non-curler (c), and normal
pigment (A) is dominant to albino (a).
9.
Tall pea plants (T) are dominant to short plants (t). Smooth seeds (S) are dominant to
wrinkled seeds (s). Cross a homozygous tall smooth plant with a short wrinkled plant. What
are the genotypic and phenotypic ratios of the F1 generation?
a. Cross two offspring from the F1 generation. Give the phenotypic and genotypic ratios of
the F2 generation.
10.
In horses, black fur (B) is dominant to the recessive chestnut coloring (b). The trotting gate
will present itself due to a dominant gene (T), and the pacing gait is due to its recessive allele
(t).
a. If a homozygous black pacer is mated to a homozygous chestnut trotter, what will be the
phenotype of the F1 generation?
b. If two F1 individuals were mated, what would be the genotypic and phenotypic ratios of
the F2 generation?
c. If a F1 male is mated to a female homozygous black pacer, what is the predicted
genotypic and phenotypic ratio of their offspring?
Blood Type Problems – Multiple Alleles
Blood Type uses multiple alleles. IA is dominant over i. IB is dominant over i. However, IA and IB are
codominant to produce type AB blood. Both IAIA and IAi present themselves as type A blood. IBIB and
IBi are both type B blood. If an individual is homozygous ii, then their blood type is O.
1.
If the genotypes of the parents are IAi and IAIA, what are the resulting genotypic and
phenotypic ratios?
2.
A man’s maternal grandfather has type AB blood. All of his other grandparents are type O.
What are the chances of the man being type A? Type B? Type AB? Type O?
3.
A woman with type A blood marries a man with type O blood. The wife’s father is type O.
What is the probability that their expected child will be type O?
4.
John has type O blood. His father has type B blood and his mother has type A. What are the
genotypes of John’s parents?
5.
If one parent is homozygous for type A blood, the other parent has type O blood. What are
the resulting genotypic and phenotypic ratios?
6.
A child has type O blood. What are the possible genotypes of his parents?
7.
One parent is heterozygous for type A blood and the other is heterozygous for type B blood.
What are the possible blood types of their children? (The genotypic and phenotypic ratios?)
8.
Suppose a man with type A blood marries a woman with type AB blood. What blood types
would you expect to find among their children? In this problem, we are not told whether the
man is homozygous or heterozygous for the IA allele. What would tell us which genotypes he
has?
9.
Suppose two newborn babies were accidentally mixed up in the hospital. From the following
blood types, determine which baby belongs to which parents. Then determine the genotype
of each of the six people in the problem.
Baby 1
Baby 2
Type O
Type A
Mrs. Brown
Mr. Brown
Type B
Type AB
Mrs. Smith
Mr. Smith
Type B
Type B
Genetics Problems – Sex-Linked Traits
The genes for these traits are on the X chromosome; because boys only receive one X chromosome they
are more likely to inherit disorders passed to them from their mother who would be a carrier.
Hemophilia and Colorblindness are sex-linked traits; the punnet square below shows how a woman who
is a carrier passes the trait to her son, but not her daughters.
Human somatic cells, including zygotes, have _________ pairs of chromosomes (total). The first 22
pairs of these are called _______________. The last pair are called ___________ chromosomes.
Human females (symbol = _______) are written as: ____________. Human males (symbol = _______)
are written as: __________.
Gametes produced by females carry only the ______ sex chromosome. Gametes produced by males
carry either the _______ or _______ sex chromosome. This is in addition to __________ (#) other
autosomes.
Question 1:
Which gamete determines the sex of the baby? Why?
Question 2:
Use correct genetic problem-solving (cross, Punnett square, explanation) to determine the probability
that a couple will have a female child.
Sex-Linked Disorders
Because a female has 2 X chromosomes, she has 2 sex-linked genes (one on each X chromosome). But a
male can only have 1 sex-linked gene because the Y chromosome does not carry a copy of the genes
located on the X chromosome. Therefore, males often suffer from more sex-linked disorders. Also, sexlinked disorders are usually caused by _______________ (not dominant) genes.
Example: Hemophilia (recessive sex-linked disorder)
Normal female = _____________
Carrier female = ______________
Normal male = ______________
Male with hemophilia = _____________
Because the male only has sex-linked genes on the ________ chromosome, there is nothing on the
______ chromosome to mask the disorder on the _______ chromosome. So if a male has even 1 copy of
the disordered gene, he suffers from the disorder. A female can carry 1 disordered gene on one _______
chromosome but still have a normal gene to mask it on the other ________ chromosome. A "carrier
female" usually doesn't suffer from the disorder, she just carries it and can pass it on .... usually to her
SONS as you'll soon see!
Question 3:
Cross a carrier female (for hemophilia) with a normal male. Of all their offspring, what is the probability
they will produce a hemophiliac son? (XH = normal blood, Xh = hemophilia)
Question 4:
Cross a hemophiliac female with a normal male. Of all their offspring, what is the probability they will
produce a hemophiliac daughter? (XH = normal blood, Xh = hemophilia)
Question 5:
Cross a carrier female (for color blindness) with a normal male. Of all their offspring, what is the
probability they will produce a carrier daughter? (XB = normal vision, Xb = color blind)
How Disorders Can Skip a Generation
You've all heard how some disorders can skip a generation and then show up. Here's how it works!
Question 6:
Cross a normal female (grandma) with a hemophiliac male (grandpa). Of all their offspring, what is the
probability they will produce a hemophiliac son? A hemophiliac daughter? A carrier daughter? Now
cross one of the daughters (mom) with a normal male (dad). What sex is the child that gets the disorder?
See how the disorder skipped a generation? (XH = normal blood, Xh = hemophilia)
Dominant Sex-Linked Disorders
So what happens if a dominant gene instead of a recessive gene causes a sex-linked disorder? The
presence of a single dominant gene on the X chromosome causes the disorder.
Question 7:
Mom suffers from a disorder of the eye called retinoblastoma caused by a dominant sex-linked gene
(XD). Dad does not have the trait.. Half of their daughters and half of their sons do not inherit the
disease. Figure out the genotypes of the people involved and complete the Punnett square.
Practice Problems:
1. Women have sex chromosomes of XX, and men have sex chromosomes of XY.
a. What percentage of their children are expected to be male? Female?
b. Which of a man's grandparents could not be the source of any of the genes on his Ychromosome?
c. Which of a woman's grandparents could not be the source of any of the genes on either of
her X-chromosomes?
Colorblindness is a sex-linked recessive trait where XN gives normal vision and is dominant to the Xn
colorblindness allele.
2. A boy, whose parents and grandparents had normal vision, is color-blind. What are the genotypes
for his mother and his maternal grandparents? Use XN for the dominant normal condition and
Xn for the recessive, color-blind phenotype.
3. If a normal woman with homozygous normal vision marries a colorblind man, draw a Punnett
square showing the genotype(s) of children that could be expected from this marriage. Write
the phenotypic and genotypic ratios.
4. A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green
color blindness, marries a normal male. What proportion of their male children will have redgreen color blindness?
5. A woman with red-green color-blindness has a mother with normal vision. Knowing that colorblindness is a sex-linked recessive gene, can you determine what her father's phenotype is?
a. If so, what is it?
b. The woman marries a man with normal vision. What is the probability they will have
sons who are red-green color-blind?
c. What is the probability they will have daughters who are red-green color-blind?
6. A man and his wife both have normal color vision, but a daughter has red-green color blindness, a
sex-linked recessive trait. The man sues his wife for divorce on grounds of infidelity. Can
genetics provide evidence supporting his case?
Hemophilia is a sex-linked trait where XH gives normal blood clotting and is dominant to the hemophilia
allele Xh.
7. Give the genotype of a woman with normal blood clotting whose father had hemophilia and a
normal man whose father had hemophilia.
a. What is the probability that a mating between these two individuals will produce a child,
regardless of sex, that has hemophilia?
b. If this couple has a daughter, what is the probability that the daughter will be a carrier of
the hemophilia trait?
c. What is the probability a daughter would have hemophilia?
d. If this couple has a son, what is the probability he will have hemophilia?
8.
In a cross between a white-eyed female fruit fly (XwXw) and red-eyed male (XWY), What
percent of the female offspring will have white eyes? (White eyes are X-linked, recessive)
9.
The bison herd on Konza Prairie has begun to show a genetic defect. Some of the males have a
condition known as "rabbit hock" in which the knee of the back leg is malformed slightly. We
do not yet know the genes controlling this trait but for the sake of our question, we shall
assume it is a sex-linked gene and that it is recessive. Now, suppose that the herd bull (the
dominant one which does most of the breeding) that is normal (XN) mates with a cow that is a
carrier for rabbit hock. What are his chances of producing a normal son?
a. If he mates with this cow every year, what percentage of their daughters are expected to
have normal knees?
b. What percentage of their daughters will be carriers of rabbit hock?
10.
A rancher owns a bull with many desirable characteristics. Unfortunately, he also has a sexlinked trait that in the recessive form leads to no pigment formation in the iris of the eye. This
makes the bull very sensitive to sunlight and could lead to blindness. The rancher wishes to
breed him to a cow that will minimize the chances of any offspring showing this trait. She
would especially like to produce another bull with most of his sire's desirable qualities but
without the non-pigmented eye. Two cows with the dominant normal colored eyes (XN) are
available that have been genetically typed for this particular trait. Cow 1 has a genotype of
XNXN and cow 2 is XNXn. Which of these two cows should the rancher choose as a mate to her
bull if she wishes to minimize the occurrence of the non-pigmented eye in his offspring?
a. What percentage of the male offspring from the preferred cross will have non-pigmented
eyes?
b. Will crossing the bull with this cow eliminate the trait from the herd?
11.
Clouded leopards are a medium sized, endangered species of cat, living in the very wet cloud
forests of Central America. Assume that the normal spots (XN, pictured here) are a dominant,
sex-linked trait and that dark spots are the recessive counterparts. Suppose as a Conservation
Biologist, you are involved in a clouded leopard-breeding program. One year you cross a male
with dark spots and a female with normal spots. She has four cubs and, conveniently, two are
male and two female. One each of the male and female cubs has normal spots and one each
have dark spots. What is the genotype of the mother?
a. Suppose a few years later, you cross the female cub that has normal spots with a male
that also has normal spots. How many of each genotype will be found in the cubs
(assume she has 4 cubs)?
b. Will any of the cubs from this latest cross have dark spots?
c. If so, how many and of what sex will they be?
12.
Coat color in cats is a codominant trait and is also located on the X chromosome. Cats can be
black, yellow or tortoise (calico). A calico cat has black, yellow, and white splotches. In order
to be a calico, the cat must have an allele for the black color and an allele for the yellow color.
Use a punnet square to show why there are no male calico cats.
13.
Black and yellow fur in cats are both X-linked. Possession of the allele B produces black, and
Y produces yellow. The combination BY produces tortoise shell or tricolor (black, yellow,
and white markings).
a. Cross a black male with a tortoise shell (calico) female and describe the phenotypes
expected.
b. Cross a black male with a yellow female and give the genotypic and phenotypic ratios
expected.
14.
Use XN for normal vision and Xn for the colorblind allele. Sam has normal vision. His oldest
sister is colorblind and his youngest sister has normal vision. What are the mother’s and
father’s genotypes?
15.
A man’s paternal grandfather has normal vision; his maternal grandfather was colorblind. His
mother is colorblind and his father has normal vision.
a. What are the genotypes for the parents and grandparents mentioned?
b. What type of vision does the man have?
c. What would be the genotypes of his sisters?
d. If this man were to marry a woman who has the same genotype as his sisters, what type
of vision would you expect their children to have?
Challenge Problems
16.
Albinism is not X-linked while hemophilia is X-linked. An albino, non-hemophiliac man
marries a woman who is a normally pigmented, non-hemophiliac. Her father was a
hemophiliac and her mother was an albino (what a mess). What would be the expected
genotypic and phenotypic ratios of their children?
17.
If a blue eyed, color blind woman marries a man with normal vision who is homozygous for
brown eye color, what traits would be the expected in their children?
a. If one of their sons marries a heterozygous brown eyed, heterozygous normal visioned
woman, what traits would you expect to see in their children?
b. What if the woman was homozygous for brown eyes and normal vision?
c. What if the woman was heterozygous for brown eyes and homozygous for normal
vision?
d. What if the woman was homozygous for brown eyes and heterozygous for normal
vision?