Genetics Problems Key 2015
In human eye color, one can assume for practical purposes that brown is dominant over blue.
1. If a brown-eyed couple have a blue-eyed child, what are the chances that a second child will also be blue-eyed?
25%
2. If the couple have four blue-eyed children in a row, how could you explain the phenomenon? It was just luck.
3. If a brown-eyed man with two brown-eyed parents marries a brown-eyed woman who also has two brown-eyed
parents, is there any chance that the couple could have a blue-eyed child? Yes, both parents could be heterozygous.
4. One parent is homozygous for a certain trait and the other parent is heterozygous. Make a Punnett square to
determine what fraction of their offspring is expected to be heterozygous. You decide which traits and assign them
with letters like we did in the previous problem.
B
b
B
BB
Bb
B
BB
Bb
50% of their offspring will be heterozygous.
5. Draw a Punnett square to show the possible outcomes for a cross of two heterozygotes for a trait.
F
f
F
FF
Ff
f
Ff
ff
Alcaptonuria
Homer and Marge have three children. Two are healthy (Bart and Lisa). The third (Maggie) has an illness that
involves brown urine. The front of Maggie's diapers show a brown coloration characteristic of alcaptonuria, a trait
inherited as an autosomal recessive disorder. The parents do not have the disease.
6. Draw a pedigree of this family using the standard symbols. Indicate which members of the pedigree must be
carriers of alcaptonuria using a half-filled symbol.
The parents are each carriers.
7. If Homer and Marge have another child, what is the probability it will have alcaptonuria?
25% Since each parent must be heterozygous for the trait, there is a 25% chance of the next child getting the
trait.
8. The pedigree below shows how the disease familial
hypercholesterolemia (FHC) is passed from generation to
generation. Describe the type of inheritance it shows. Is it
autosomal or sex-linked? Is it recessive or dominant? Does
skip generations? Other observations?
Both males and females can get it. It does not appear to be
sex-linked. It seems to be dominant because there are no
it
carriers. The reason affected people do not always pass it on indicates that they must be heterozygous for the trait.
It does not skip generations.
9. In rabbits, if B = dominant black allele and b = recessive white allele, which of these genotypes could a white rabbit
have? Bb, BB, bb
A white rabbit could be only: bb
10. In horses, trotter (T) is dominant over pacer (t). A trotter is mated to a pacer and the offspring is a pacer. Give
the genotype of all horses.
Trotter parent: Tt
Pacer parent: tt
Offspring pacer: tt
11. In humans, freckles is dominant over no freckles. A man with freckles is married to a woman with freckles, but
the children have no freckles. What chance did each child have for freckles?
Man: Ff
Woman: Ff
Children: ff
Each child had a 75% chance of getting freckles.
12. In horses, B = black coat, b = brown coat; T = trotter, t = pacer. A black pacer mated to a brown trotter produces
a black trotter offspring. Give all possible genotypes for this offspring.
Black pacer: Bbtt or BBtt
Brown trotter: bbTt or bbTT
Offspring: BbTt
13. John has unattached earlobes like his father, but his mother has attached earlobes. What is John’s genotype?
Unattached=dominant; attached=recessive John’s father is heterozygous; his mother is homozygous recessive; John
is heterozygous
14. In humans, the allele for short fingers is dominant over that for long fingers. If a person with short fingers who
had one parent with long fingers marries a person with long fingers, what are the chances for each child to have short
fingers? F=short fingers; f=long fingers
f
f
F
Ff
Ff
f
ff
ff
50 % will have long fingers
15. Use a Punnett square and the laws of probability to show the results of a cross between two individuals who are
heterozygous for one trait. What are the chances of an offspring having the dominant phenotype? 75% The
recessive phenotype? 25%
16. A woman with blood type B gives birth to a child with blood type B. Could a man with blood type A be the father
of this child? Why or why not? Yes, he could have IA, i
dad
mom
IB
i
Child with
A
A B
I
I ,I
IA, i
B blood
i
IB, i
i,i
17. Investigators note that albino tigers usually have crossed eyes. What two inheritance pattern can account for a
syndrome such as this? pleiotropy
18. A woman is color blind. What are the chances that her sons will be color blind? All of her sons will be color blind.
If she is married to a man with normal vision, what are the chances that her daughters will be color blind?
None of her daughters will be color blind.
What are the chances that her daughters will be carriers?
All of her daughters will be carriers.
19. A man of blood type A and a woman of blood type B produce a child of type O.
a. What are the genotypes of the man, woman, and child?
b. If the couple were to have more children, what possible blood types could be produced?
a.
b.
Genotypes of man, woman, child 
Other blood types possible 
IA
i
IB
i
IA,IB
IB, i
IA, i
i,i
Man
Woman
IA, i
IB, i
Possible blood types: AB, A, B, O
Child
ii
All of them!
Match the phrases below (20-23) to the items in the key.
Key:
a. multiple alleles
Three or more alleles for a given trait (blood type).
b. incomplete dominance
c. Polygenes
One trait is controlled by several genes.
d. epistatic gene
Affect the phenotypic expression of alleles that are at a different
location (Bombay type of blood).
e. pleiotropic gene
Pleiotropic genes affect several different phenotypic characteristics.
20. A cross between oblong and round squash produced oval squash. b
21. Although most people have an IQ of about 100, IQ generally ranges from about 50-150. c
22. Investigators noted that whenever a particular species of plant had narrow instead of broad leaves,
was also short and yellow, instead of tall and green. e
23. In rabbits, the allele C is dominant to the allele cch, which is dominant to the allele ch, which is
allele c. Each individual has only two of these alleles. a
it
dominant to the
24. In humans, color blindness is a sex-linked, recessive trait, and widow’s peak (autosomal) is dominant over
continuous hairline. A man with normal vision and a widow’s peak married a woman who is color blind and has a
continuous hairline. The woman has a girl who has a widow’s peak and is color blind. What are the genotypes of all
persons involved? Is this man the girl’s father? How do you know?
B=normal vision; b=color blind
H=widow’s peak; h=continuous hairline
Man (BBHH or BbHh) X woman (bbhh)  bbHh
Since the man has normal vision and the trait is carried on the 23 rd chromosome, his daughter would definitely get his
dominant allele for vision. She could not possibly be color blind, although she would be a carrier. The girl could not be
the man’s daughter. Her father would also have to be color blind. All the discussion of the hairline is irrelevant. Isn’t
this fun!?
Match the conditions in the key with the description below:
Key:
a. Down syndrome
b. Turner syndrome
c. Klinefelter syndrome
d. Double Y male
__c___ 25. Male with underdeveloped testes and some development of the breasts.
__a___ 26. 47, XX, +21 (Trisomy 21)
__b___ 27. 45, X, -23 (XO female)
__d___ 28. 47, xyy, +23
29. What is the only known sex chromosomes monosomy in humans? Turner Syndrome
Name and describe two sex chromosome trisomies.
30. Klinefelter Syndrome (XXY, XXXY, XXXXY)
31. Triple X female: (XXX)
YY Male: (XYY)
Complete the table below
Disorder
32.
cystic
fibrosis
33.
Tay-Sachs
34.
PKU
Symptoms
A protein does not work that is supposed to remove chlorine from
cells, mucous builds up on cell, organs can fail, lungs are coated with
mucous, shortened life expectancy, recessive.
A protein does not work that is supposed to remove fats from the
brain, brain cells are crushed, short life expectancy, recessive.
A protein does not work that is supposed to break down the amino
acid phenylalanine, can cause mental retardation, recessive.
Chances of two
carriers having an
affected child
The chances are
25%.
The chances are
25%.
The chances are
25%.
More of those tricky double crosses! The symbols below apply to problems 33-37.
R = round seed
r = wrinkled seed
Y = yellow
y = green
35. Complete the Punnett square for a dihybrid cross of two flowering plants: TTRr X TtRr.
T= tall
TR
R = red
TR
TTRR
t = short
Tr
TTRr
r = white flower
tR
TtRR
Tr
TTRr
TTrr
TtRr
Ttrr
TR
TTRR
TTRr
TtRR
TtRr
Tr
TTRr
TTrr
TtRr
Ttrr
tr
TtRr
36. List the genotypes and phenotypes resulting from an RrYy plant cross-pollinated by an Rryy plant.
Make a Punnett square. Then list the genotypes and phenotypes.
Ry
ry
RY
RRYy
RrYy
Ry
RRyy
Rryy
rY
RrYy
rrYy
ry
Rryy
rryy
Genotypes:
RRYy-1; Rryy-6; RrYy-6; rryy-2
Phenotypes:
Round, yellow: 8; Round, green: 6; wrinkled, green: 2
37. List the genotypes and phenotypes resulting from an rrYy plant cross-pollinated by a RrYy.
RY
Ry
rY
ry
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Genotypes:
RrYY-2 ; RrYy-4; Rryy-2; rrYy-4; rryy-2; rrYY-2
Phenotypes:
Round, yellow: 6; Round, green: 2; wrinkled, green: 2; wrinkled, yellow: 6
You should be able to do this one in your head!
38. What genotypes and phenotypes would result from a cross of RRYY and rryy?
Genotypes: 100% RrYy
Phenotypes: 100% round and yellow
39. If a plant with an RrTt genotype is crossed with a plant that is rrtt, (R=red; r=gray; T=tall; t=short)
a. all the offspring will be tall with red fruit.
b. 75% will be tall with red fruit.
c. 50% will be tall with red fruit.
d. 25% will be tall with red fruit. (correct)
40. In guinea pigs, smooth coat (S) is dominant over rough coat (s) and black coat (B) is dominant over white coat (b).
In the cross SsBb X SsBb, how many of the offspring will have a smooth black coat?
a. 9 only
b. about 9 of every 16 (CORRECT)
c. only 1 out of 16
d. 6 of every 16
In fruit flies, long wings (L) is dominant over short wings (l), and gray body (G) is dominant over black body (g). In
each instance, what are the most likely genotypes of the previous generation, if a student gets the following
phenotypic results?
lg
Phenotypic Result
Parent 1
Parent 2
LG
LlGg
41. 1:1:1:1
LlGg
llgg
Lg
Llgg
42. 9:3:3:1
LlGg
LlGg
lG
llGg
lg
llgg
43. In humans, short fingers and widow’s peak are dominant over long fingers and continuous hairline. A heterozygote
is married to a heterozygote. What is the chance that any child will have the same phenotype as the parents?
Since each parent is heterozygous for two traits, the Punnett square show the classic cross that Mendel discovered.
The phenotypes are 9:3:3:1. This means that on average, of every 16 offspring, there will be 9 that have dominant
phenotype for both traits, 3 will be dominant for only one trait, 3 will be dominant for the other trait, and one will be
homozygous recessive for both traits.
9 of 16 will look like the parents.
44. In rabbits, black color is due to a dominant allele B and brown color to its recessive allele b. Short hair is due to
the dominant allele S and long hair is due to its recessive alleles. In a cross between a homozygous black, long-haired
rabbit and a brown, homozygous short-haired one, what would the first generation of offspring look like?
BBss X bbSS  F1 = BbSs
45. A woman whose husband worked at the Chernobyl nuclear reactor gives birth to a
hemophilic son.
A. She should blame the reactor accident because the radiation might have caused the
hemophilia.
B. She should not blame the accident because she carried the hemophilia allele.
C. Her husband should sue for divorce because it can't be his child.
D. Further genetic tests should be done to determine who is at fault.
E. Hemophilia is environmentally induced, not genetic.
46. Barring new mutation, and assuming complete penetrance and single-locus inheritance,
what are possible modes of inheritance of the trait shown in figure below?
A. autosomal recessive
C. autosomal dominant
B. X-linked recessive
D. X-linked dominant
47. To discover whether an animal showing the dominant trait is homozygous or heterozygous, it must be crossed with
an animal that is:
A. homozygous dominant
B. homozygous recessive
C. heterozygous
48. In hamsters, long tails (L) are dominant to short tails (l). A student wishes to perform a test cross to determine
whether a female long-tailed hamster is homozygous or heterozygous for long tail length. She mates the hamster with
a male long-tailed hamster & studies the offspring, which are 100% long-tailed. She concluded that the female
hamster's genotype is "LL". What mistake(s) did the student make?
A. She should have mated the female hamster with a male that was known to be hybrid.
B. She should have mated the female hamster with a short-tailed male hamster.
C. She should have mated the female hamster with another long-tailed female.
D. She has to mate members of the litter before she can make a conclusion.
49. A DNA molecule produced by combining DNA from different sources is known as ___ recombinant DNA
________.
50. On an electrophoresis gel, band B is closer to the positive end of the gel than is band A. Which of the following
statements is true?
a. Band B is more negatively charged than band A.
b. Band B moved faster than band A.
c. Band A is smaller than band B.
d. Band B consists of larger DNA fragments than does band A.
51. Suppose a restriction enzyme recognizes the six-base sequence
AAGCTT
TTCGAA
in a double strand of DNA. Between which two nucleotides on each strand would the enzyme have to cut to
produce a fragment with sticky ends that are four bases long?
a. GC
b. CT
c. AA
d. AG
52. Which
a.
b.
c.
d.
of the following steps is NOT essential in producing recombinant DNA?
Cut out a piece of DNA from a DNA molecule.
Splice a piece of DNA into DNA from another organism.
Use a restriction enzyme to form sticky ends in DNA.
Read the DNA sequence of the piece of DNA to be cut and spliced.
53. The genetic engineer uses restriction enzymes to
A. restrict the coding of amino acid sequences along a given strand of MRNA.
B. inhibit the action of endonucleases as nucleotide sequences are being organized.
C. function in the isolation of plasmids in bacteria.
D. serve as a vehicle for mobile genes.
E. serve as a molecular knife to cleave (cut) a specific DNA segment from a DNA strand.
54. A plasmid is
A. a ring of bacteria in a bacterial cell.
B. a fragmentary particle in a eukaryotic cell.
C. a vehicle that carries a human gene into a mouse cell.
D. all of the above
E. none of the above
55. DNA ligase is an enzyme that
A. strips away antagonistic nucleotide sequences from a DNA strand.
B. makes an inserted gene stick to the plasmid vehicle.
C. causes the loss of a specific nucleotide from a DNA segment.
D. does all of the above.
E. does none of the above.
56. Transformation refers to
A. the production of complementary strands of mRNA from DNA.
B. the movement of tRNA from one attachment site to another on a ribosome.
C. the conversion of the genetic information carried by a tRNA molecule into the amino acid sequence of a
polypeptide.
D. the transporting of a human gene into a bacterial cell
E. movement of a ribosome along mRNA.
57. A paleontologist has recovered a little bit of tissue from the 400-year-old preserved skin of an extinct dodo
(bird). The researcher would like to compare DNA from the sample with DNA from living birds. Which of the
following would be most useful for increasing the amount of DNA available for testing?
A. gel electrophoresis
B. polymerase chain reaction (PCR)
C. transformation
D. translation
E. create sticky ends with DNA ligase
58. In nature, restriction enzymes
A. transfer DNA between organisms.
B. protect bacteria from viruses
C. enable viruses to invade nuclei.
D. restrict the ability of bacteria to form clones.
E. promote cell division in restricted bacteria.
59. What is the name of DNA that is made from two different sources?
A. mutant DNA
C. mitochondrial DNA
B. plasmid DNA
D. recombinant DNA
60. A member of a population of genetically identical cells produced from a single cell is a
A. clone
B. plasmid
C. mutant
D. sequence
61. Organisms that contain genes from other organisms are called
A. transgenic
B. mutagenic
C. donor organisms
62. One function of gel electrophoresis is to
a. separate DNA fragments.
b. cut DNA.
c. recombine DNA.
d. extract DNA.
D. cloned organisms