Download GENETICS PRACTICE PROBLEMS

GENETICS PRACTICE PROBLEMS - KEY
SIMPLE DOMINANT / RECESSIVE MONOHYBRID CROSSES
1. Tallness is dominant over dwarfism in pea plants. A tall plant of unknown genotype is
test-crossed. Of the offspring, 869 are dwarf and 912 are tall. What is the genotype
of the unknown parent?
a. Must be heterozygous because otherwise, none of the offspring could
have the recessive phenotype.
2. The genes for dark eyes (black and brown) usually dominate over genes for blue or
gray eyes. A man with black eyes marries a woman with light gray eyes. They have
two children, a boy with black eyes and a girl with blue eyes. What are the
genotypes of the man, his wife, the boy and the girl?
a. Man = Bb, must be heterozygous because he gave a b gene to his
daughter
b. Woman = bb, because she has the recessive phenotype
c. Boy = Bb, because he must have gotten B from dad in order to be
dark-eyed and could only get a b from mom
d. Girl = bb because she has the recessive phenotype
3. A brown-eyed man with a blue-eyed mother marries a brown-eyed woman with a
blue-eyed father. What is the probability that their first child will be brown-eyed?
That the second child will also be brown-eyed?
a. first child has a 75% chance of being brown eyed (cross between two
heterozygous parents)
b. second child also has a 75% chance, so the chance of both children
having brown eyes is ¾ x ¾ = 9/16
4. A man and a woman have 7 children. Of the children, 6 have brown eyes and 1 has
blue eyes. What are the genotypes of the parents?
a. Bb
5. In goats, a recessive gene causes the goats to "faint" when they are startled. A
farmer breeds two goats (that have never fainted) and their first offspring faints two
days after its birth. What must the parent's genotypes have been?
a. Ff or they couldn’t have produced a ff offspring
6. In the fruit fly Drosophila, the allele for normal body (B) is dominant to the allele for
hairy body (b). When two normal-bodied fruit flies were mated, they produced 81
hairy-bodied flies and 319 normal-bodied flies. The genotypes of the parents are
most likely
a. Bb x Bb
7. In humans, tongue rolling is a dominant trait, while those with the recessive
condition cannot roll their tongues. Bob can roll his tongue, but his mother could not.
He is married to Sally, who cannot roll her tongue. What is the probability that their
first-born child will not be able to roll his/her tongue?
a. Bob = Rr, Sally = rr; 50% chance offspring will be able to roll tongue.
CODOMINANCE and INCOMPLETE DOMINANCE
8. A man with type A blood marries a woman with type A blood. Their first child has
blood type O. What are the genotypes of the mother, father and baby?
a. Man could be AA or AO to have blood type A, same for the woman. If
their child is OO, both parents must be AO
9. A man with type AB blood marries a woman with type O blood, but whose father had
type A blood. What genotype would you expect their first child to have?
a. Man = AB, woman = OO. Their first child would have a 50% chance of
being AO (type A) or BO (type B).
10.A man with type B blood marries a woman with type A blood. They have six type AB
children. What are the genotypes of the father, mother and children?
a. Man could be BB or BO, woman could be AA or AO. Since all their
children are AB, probably the man is BB and his wife is AA.
11.A man with type A blood marries a woman with type A blood. They have eight type A
children and one type O child. What are the genotypes of everyone in the family?
a. Man could be AA or AO, same with the woman. The one OO child has
the genotype OO, so both parents must be AO. The type A children
could be AA (probably ¼ of the children) or AO (probably ½ of the
children).
12.In snapdragon plants that display incomplete dominance, the allele CR produces red
flowers and CW produces white flowers. If a homozygous red-flowered snapdragon is
crossed with a homozygous white-flowered snapdragon, the percent ratio of the
offspring will be
a. red-flowered snapdragon = CRCR, white-flowered snapdragon = CWCW,
all offspring will be CRCW, which will be pink.
13.Madeup grasshoppers are black with stripes. A red-striped individual and a yellowstriped individual are crossed. 100% of the offspring have orange stripes.
a. What type of inheritance is this?
i. incomplete dominance
b. In a second cross of two grasshoppers, 35 of the offspring have red stripes
and 45 of the offspring have orange stripes. What were the genotypes of the
two parents?
i. Red stripes = RR, Orange stripes = RY, so one parent must be
RR (red) and the second must be RY (orange).
14.In Portland, there is a creature know as a blazer. It comes in three colors: red, black,
and speckled. Upon close examination, speckled blazers show a mixture of red and
black hairs.
a. What kind of inheritance is seen in blazers?
i. codominance
b. What would be the genotypes and phenotypes of the offspring if a red blazer
were crossed with a black one?
i. Red = RR, Black = BB, 100% of offspring will be speckled.
c. What would be the genotypes and phenotypes of the offspring if a red blazer
was crossed with a speckled one?
i. Red = RR, Speckled = RB, 50% of offspring will be red (RR) and
50% will be speckled (RB)
15.In a certain cactus, prickly spines can be two pronged or one pronged. If a true
breeding one-pronged cactus is crossed with a true breeding two-pronged cactus,
plants in the F1 generation each have a mixture of spines, some are two-pronged,
some are one-pronged.
a. Is this an example of codominance or incomplete dominance?
i. codominance because both phenotypes are seen together.
b. Show the F2 generation (a cross between the two F1's). What are the
phenotypes of the offspring and in what proportion?
i. One prong parent = OO, two prong parent = TT. All the offspring
will be mixed-prong = OT.
SEX LINKED INHERITANCE
16.In humans, the gene from normal blood clotting (H) is dominant to the gene for
hemophilia. This is a sex-linked trait found on the X chromosome. A woman with
normal blood clotting has four children. They are a normal son, a hemophiliac son,
and two normal daughters. The father has normal blood clotting. What is the
probably genotype for each member of the family?
a. woman = XHXh
b. man = XHY
c. normal son = XHY
d. hemophiliac son = XhY
e. normal daughters = could be XHXh or XHXH
17.Colorblindness in humans is a sex-linked trait. A boy, whose parents and
grandparents have normal vision, is color-blind. What are the genotypes for the boy,
his parents and his grandparents?
a. boy = XbY
b. mom = XBXb
c. dad = XBY
d. maternal grandparents = XBXb , XBY
e. paternal grandparents = XBXb or XBXB, XBY
18.The bison herd on Konza Prairie has begun to show a genetic defect. Some of the
males have a deformity known as “rabbit hock” in which the knee of the back leg is
malformed slightly. The deformity has not been observed in female bison and is
sometimes seen in bison with two normal parents. How is this trait most likely
inherited and WHY?
a. it’s sex-linked recessive. It’s sex-linked because it’s seen more
commonly in males than in females and it must be recessive if it’s
seen in offspring of two normal parents.
19.Suppose that the herd bull (the dominant one who does most of the breeding) mates
with a cow that is a carrier for rabbit hock.
a. What are the chances of the baby being a normal male bison?
i. male = XRY, female = XRXr . There’s a 25% chance that the baby
will be a normal male. There’s a 50% chance that the baby will
be male (vs female) and of the males, half will be normal.
b. If the herd bull continues to mate with this cow, what percentage of their
daughters will have normal knees?
i. All their daughters will have normal knees because they have to
get a normal X chromosome from the male.
c. What percentage of their daughters will be carriers of rabbit hock?
i. 50% of their daughters will be carriers because they have a
50% chance of getting the rabbit hock gene from their mother.
20.Coat color in cats is a codominant trait and is located on the X chromosome. Cats
can be black, yellow or calico. A calico cat has black and yellow splotches. In order to
be calico, the cat must have an allele for the black color and an allele for the yellow
color. Use a Punnett square to show why there are no male calico cats.
a. male cats can’t have two alleles for anything on the X chromosome,
they can only have one.
21.A female calico cat is crossed with a male black cat. What are the phenotypes of the
offspring and in what proportion.
a. female = XBXY, male = XBY
b. ¼ female black - XBXB
c. ¼ female calico - XBXY
d. ¼ male black - XBY
e. ¼ male yellow – XYY
DIHYBRID CROSSES
22.A man with a dimple and brown eyes (whose father had blue eyes but no dimples)
marries a woman with a dimple and brown eyes (whose father had blue eyes but no
dimple). What is the probability that their first child will be blue-eyed and without a
dimple?
a. man = DdBb, woman = DdBb, child = ddbb
b. ¼ chance of getting dd x ¼ chance of getting bb = 1/16 chance of
getting ddbb
23.In some flowers, a true-breeding red-flowered strain gives all pink flowers when
crossed with a white-flowered strain. Axial flower position (middle of stem) is
dominant to terminal flower position (end of stem).
a. What will be the ratio of genotypes and phenotypes from the following cross:
axial-red (true-breeding) and terminal-white?
i. axial red = AARR, terminal white = aaWW
ii. 100% of offspring will be AaRW = axial pink
b. What will be the ratios in the F2 generation (crossing two F1 plants)?
i. 3/16 axial-red (A_RR)
ii. 6/16 axial-pink (A_RW)
iii. 3/16 axial-white (A_WW)
iv. 1/16 terminal-red (aaRR)
v. 2/16 terminal-pink (aaRW)
vi. 1/16 terminal-white (aaWW)
24.In sesame plants, the one-pod condition is dominant to the three-pod condition, and
normal leaf is dominant to wrinkled leaf. These traits are inherited independently.
Determine the genotypes for the two parents for all the possible matings producing
the following offspring:
a. 318 one-pod normal, 98 one-pod wrinkled
i. This is close to a 3:1 ratio and we’re only looking at one trait
(normal vs wrinkled). The parents are both OONn (homozygous
one-pod, heterozygous wrinkled leaf).
b. 323 three-pod normal, 106 three-pod wrinkled
i. Again, a 3:1 ratio and we’re only looking at one trait. The
parents are both ooNn (homozygous three-pod, heterozygous
wrinkled leaf).
c. 401 one-pod normal
i. The most likely answer is that one parent is OONN (homozygous
one-pod, homozygous normal leaf). With this parent, the other
parent could be anything.
ii. Other possible answers include: OOnn x ooNN, OONn x ooNN,
OOnn x OoNN (and there may be other combinations that will
also work)
d. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 threepod wrinkled
i. One parent was OoNn (heterozygous one-pod, heterozygous
normal-leaf), the second parent was Oonn (heterozygous onepod, homozygous wrinkled-leaf)
e. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod
wrinkled
i. both parents were OoNn (heterozygous one-pod, heterozygous
normal-leaf). This is roughly a 9:3:3:1 ratio.
PROBABILITY PROBLEMS – Answer using math, not a Punnett Square
25.Assuming that the probability of having a female child is 50% and having a male
child is 50%, what is the probability of a couple’s first-born child being female and
the second-born child being male?
a. probability of 1st child being female = ½
b. probability of 2nd child being male = ½
c. probability of both events occurring together = ½ x ½ = 1/4
26.An organism has three independently assorting traits AaBbCc. What fraction of its
gametes will contain the recessive genes abc?
a. ½ a x ½ b x ½ c = 1/8 chance of abc
27.An organism with the genotype AaBbCCdd is crossed with one that is AAbbccDd.
What are the odds of having an offspring with the genotype AabbCcDd. In this
problem, do not use punnett squares. Solve the problem using math and statistics.
b. for A gene: ½ of getting a from first parent, 1 of getting A from second
parent
c. for B gene: ½ of getting b from first parent, 1 of getting b from second
parent
d. for C gene: 1 of getting C from first parent, 1 of getting c from second
parent
e. for D gene: 1 of getting d from first parent, 1/2 of getting D from
second parent
f. TOTAL: ½ x ½ x ½ = 1/8
28.Both parents in a trihybrid cross are AaBbCc.
g. What is the probability of their producing an offspring that is recessive for all
three traits?
i. probability of getting recessive genes: ½ x ½ (for a) x ½ x ½
(for b) x ½ x ½ (for c) = 1/64
h. What is the probability of producing an offspring with the genotype A-bbC(where “-“ means the gene could be either dominant or recessive)?
i. ½ x 1 (for a) x ½ x ½ (for b) x ½ x 1 (for c) = 1/16