Genetics Practice Problems: KEY
Mendelian Genetics
1. Pattipan squash are either white or yellow. You start growing pattipans and find out that if you
want to get white pattipans then at least one of the parents must be white. Which color is
dominant? White alleles are dominant to yellow alleles. If yellow were dominant, then you
should be able to get white squash from a cross of two yellow heterozygotes.
2. True-breeding tall red-flowered plants are crossed with dwarf white flowered plants. F1
generation consists of all tall pink-flowered plants. Assuming that height and flower color are
each determined by a single gene locus, predict the results of an f1 cross of the TtRr plants. List
the phenotypes and predicted ratios for the F2 generation Since folwer color shows incomplete
dominance, there will be six phenotypic classes in the F2 instead of the normal four classes
(ie note 9:3:3:1) You could find the answer using Punnett squares or by probabilities
Tall red
T_RR = ¾ x ¼
3/16
Tall pink
T_Rr = ¾ x ½
3/8 or 6/16
Tall white
T_rr = ¾ x ¼
3/16
Dwarf red
ttRR = ¼ x ¼
1/16
Dwarf pink
ttRr = ¼ x ½
1/8 or 2/16
Dwarf white
ttrr = ¼ x ¼
1/16
3. 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 combination, indicate which blood groups of
potential fathers would be exonerated in court.
Blood group Blood group of
Exonerated
of mother
child
blood group
(s)
AB
A
No groups
O
B
A or O
A
AB
A or O
O
O
AB
B
A
B or O
4. In rabbits, CC = normal; Cc = deformed legs, cc = lethal
BB = black coat color; Bb = brown coat color, bb = white coat color
What are the phenotypic ratios of offspring from a deformed-leg, brown rabbit and a
deformed-leg, white rabbit? Parental cross CcBb x Ccbb
Note that cc offspring will die and no BB are possible. Therefore only 4 phenotypic
classes are possible.
Lethal (cc__) = ¼
Normal brown (CCBb) = ¼ x ½ = 1/8
Normal white (CCbb) ¼ x ½ = 1/8
1/8: 1/8: ¼: ¼ frequency
Deformed brown (CcBb) = ½ x ½ = ¼
Deformed white (Ccbb) = ½ x ½ = ¼
1:1:2:2 ratio
Or
1/6 normal brown: 1/6 normal white: 2/6 deformed brown: 2/6 deformed white
5. Fur color in rabbits is determined by a single gene locus for which there are four alleles. Four
phenotypes are possible: black, Chinchilla (gray color caused by white hairs with black tips),
Himalayan (white with black patches on extremities of rabbit), and white. The black allele (C) is
dominant over all other alleles, Chinchilla (Cch) is dominant over Himalayan (Ch), and white © is
recessive to all others.
a. A black rabbit is crossed with a Himalayan, and the F1 consists of a ratio of 2 black: 2
Chinchilla. What are the possible genotypes of the parents?
Parents are: black (C__) himalayan (Ch__), right away you can say that the black is (CCch)
because half of the offspring are Chinchilla. Because Cch is dominant over both Ch and c the
Himalayan parent could be either Ch Ch or Ch c. A test cross would be necessary to determine this
genotype
b. A black rabbit is crossed with a Chinchilla rabbit. The F1 ratio was 2 black: 1 Chinchilla :
1 Himalayan. From these results determine which genotypes the parents can not be.
Although you can not directly determine the parental genotypes, you can eliminate some. The
black parent could not be CC or C Cch , because both the C and Cch alleles are dominant to Ch, and
Himilayan offspring were produced. The Chinchilla parent could not be CchCch for the same
reason. One or both parents had to have Ch as their second allele; one, but not both, might have c
as their second allele.
6. For the following crosses, determine the probability of obtaining the indicated genotype in an
offspring
Cross
Offspring
Probability
AAbb x AaBb
AAbb
½ Aa x ½ bb = ¼
AaBB x AaBb
aaBB
¼ aa x ½ BB = 1/8
AABbcc x aabbCC
AaBbCc
1 Aa x ½ Bb x 1 Cc =1/2
AaBbCc x AaBbcc
aabbcc
¼ aa x1/4 bb x ½ cc = 1/32
7. A man who has type O blood has a child with a woman who has type A blood. The woman’s
mother has AB blood, and her father, type O. What is the probability that the child has each of the
following blood types? a. O 1/2
b. A 1/2
c. B 0
d. AB 0
Woman must be IAi man is ii
8. Polydactyly (extra fingers and toes) is due to a dominant gene. A father is polydactyl, the mother
has the normal phenotype, and they have had one normal child. What is the genotype of the
father? Of the mother? What is the probability that a second child will have the normal number
of digits? Father’s geneotype must be Pp since polydactyly is dominant and he has had one
normal child. Mother’s genotype is pp. The chance of the next child having normal digits is
½ or 50% because the mother can only donate a p allele and there is a 50% chance that the
father will donate a p allele.
9. Achondroplasia is a common form of hereditary dwarfism that causes very short limbs, stubby
hands, and an enlarged forehead. Above are four pedigrees depicting families with this specific
type of dwarfism. What is the most likely mode of inheritance? Cite a reason for your answer.
Autosomal dominant – it affects both sexes, and occurs every generation
10. Draw a pedigree to depict the following family:
One couple has a son and a daughter with normal skin pigmentation. Another couple has
one son and two daughters with normal skin pigmentation. The daughter from the first
couple has three children with the son of the second couple. Their son and one daughter
have albinism; their other daughter has normal skin pigmentation.
11. Caleb has a double row of eyelashes, which he inherited from his mother as a dominant trait. His
maternal grandfather is the only other relative to express the trait. Veronica, a woman with normal
eyelashes, falls madly in love with Caleb, and they marry. Their first child, Polly has normal
eyelashes. Now Veronica is pregnant again and hopes they will have a child who has double
eyelashes. What chance does a child of Veronica and Caleb have of inheriting double eyelashes
1/2 chance of that the child will have double eyelashes Draw a pedigree of this family.