AP Biology
Genetic Problems
1. One gene has alleles A and a . Another has the alleles B and b. For each genotype listed, what
type (s) of gametes will be produced?
Genotypes
a. AABB
b. AaBB
c. Aabb
d. AaBb
2. Still referring to Problem 1, what will be the genotypes of the offspring from the following matings?
(Indicate the frequencies of each genotype among them.)
a. AABB x aaBB
b. AaBB x AABb
c. AaBb x aabb
d. AaBb x AaBb
3. In one experiment, Mendel crossed a pea plant that bred true for green pods with one that bred true
for yellow pods. All the F1 plants had green pods. Which form of the trait (green or yellow pods) is
recessive? Explain how you arrived at your conclusion.
4. Return to Problem 1, and assume you now study a third gene having alleles C and c. For each genotype
listed, what type(s) of gametes will be produced?
Genotypes
a. AABBCC
b. AaBBcc
c. AaBBCc
d. AaBbCc
5. Mendel crossed a true-breeding tall, purple-flowered pea plant with a true-breeding dwarf, whiteflowered plant. All F1 plants were tall and had purple flowers. If an F1 plant self-fertilizes, then what is
the probability that a randomly selected F2 offspring will be heterozygous for the genes specifying height
and flower color?
6. At a certain gene location on a human chromosome, a dominant allele controls tongue rolling, an ability to
curl up the sides of the tongue. People who are homozygous for a recessive allele at the locus cannot roll
the tongue. At a different gene locus, a dominant allele controls whether the earlobes will be attached or
detached. The two pairs of genes assort independently. Suppose a tongue-rolling, detached-earlobe woman
marries a man who has attached earlobes and cannot roll his tongue. Their first child has the father’s
phenotype. Given this outcome;
a. What are genotypes of the mother, father, and child?
b. What is the probability that a second child of theirs will have detached earlobes and won’t be a
tongue roller?
Test Crosses
7. A tall (T) pea plant is crossed with a short (t) pea plant. All 55 of the offspring are tall. What is the
genotype of the tall plant?
8. In watermelon, the allele for short fruit (S) is dominant to the allele for long fruit (s) and the allele for
green fruit (G) is dominant to the allele for mottled fruit color (g). Each of four short green plants was
crossed with a long mottled plant. The results of these crosses are shown below.
plant
short green
short mottled
long green
long mottled
1
28
30
33
31
2
71
-60
-3
109
---4
49
56
--What are the genotypes of plants?
1 _______________
2 _______________
3 _______________
4 _______________
9. Red - flowering snapdragons are homozygous for the allele R1. White - flowering snapdragons are
homozygous for a different allele (R2). Heterozygous plants (R1R2) bear pink flowers. What phenotypes
should appear among F1 offspring of the crosses listed, and what are the expected proportions for each
phenotype?
a. R1R1 X R1R2
b. R1R1 X R2R2
c. R1R2 X R1R2
d. R1R2 X R2R2
10. DNA fingerprinting is a method of identifying individuals based on locating unique base sequences in
their DNA. Before researchers refined the method, attorneys often relied on the ABO blood - typing
system to settle disputes over paternity. Suppose, as a geneticist, you were called on to testify during a
paternity case in which the mother has type A blood, the child has type O blood and the alleged father
has type B blood. How would you respond to the following statements?
a. Attorney of the alleged father: “the mother‘s blood is type A, so the child‘s type O blood must have
come from the father. Because my client has type B blood, he simply could not be the father.”
b. Mother’s attorney: “Because the further tests prove this man is heterozygous, he must be the father.”
11. In chickens, two pairs of genes affect the comb type. When both genes are recessive, a chicken will
have a single comb. A dominant allele of one gene, P, gives rise to a pea comb. A dominant allele of the
other (R) gives rise to a rose comb. An epistatic interaction occurs when a chicken has at least one of both
dominants, P__ R __, which give rise to a walnut comb. Predict the F1 ratio resulting from a cross between
two walnut - combed chickens that are heterozygous for both genes (PpRr).
12. A mutated allele gives rise to an abnormal form of hemoglobin (HbS instead of HbA ). Homozygotes
(HbS HbS) develop sickle-cell anemia. But heterozygote ( HbS HbA ) show few outward symptoms.
Suppose a female’s mother is unaffected by this genetic disorder, but her father is homozygous for the
HbS allele. She marries a male who is heterozygous for the allele, and they plan to have children. For each
pregnancy, state the probability that this couple will have a child who is:
a. homozygous for the HbS allele
b. homozygous for the HbA allele
c. heterozygous for the HbS HbA
13. A recessive allele a is responsible for albinism, an inability to produce or deposit melanin in tissues.
Humans and some other organisms can have this phenotype. In each of the following cases, what are the
possible genotypes of the father, of the mother, and of their children?
a. Both parents have normal phenotypes; some of their children are albino and others are unaffected.
b. Both parents are albino and have only albino children
c. The women is unaffected, the man is albino, and they have one albino child and three unaffected
children. (What gives rise to this 3:1 ratio? )
[SOLUTIONS]
1. a. AB
b AB & aB
c. Ab & ab
d. AB, aB, Ab, & ab
2. a. All offapring will be AaBB
b. 25% AABB; 25% AaBB; 25% AABb; 25% AbBb
c. 25% AaBb; 25%Aabb; 25% aaBb; 25% aabb
d. 1/16 AABB 1/8 AaBB
1/16 aaBB
1/8 AABb
1/4 AaBb
1/8 aaBb
1/16 AAbb
1/8 Aabb
1/16 aabb
3. Yellow is recessive. ... the 1st generation plants have green phenotype & must be heterozygous,
green must be dominant over the recessive yellow.
4. a. ABC
c. ABC, aBC, ABc, & aBc
b. ABc and aBc
d. ABC,aBC, AbC, abC, ABc, aBc, Abc, and abc.
5. F1 plants must all be heterozygous for both genes. When these plants self pollinate, 1/4 of F2 of plants will be
heterozygous for both genes.
6. a. Mother heterozygous for both genes; father homozygous recessive for both genes. The 1st child is homozygous
recessive for both genes
b. Probability of 2nd child being a tongue roller with detached earlobes is ¼
7. homozygous tall
8. 1. SsGg
2. SsGG
9. a. 1/2 red flowers & 1/2 pink
3. SSGG
4. SSGg
b. all pink
c. 1/4 red, 1/2 pink, 1/4 white d. 1/2 pink, 1/2 white
10. a. Mother IAi Male with blood type B could have fathered the child if he, too is heterozygous I Bi
11. 9/16 walnut (pea-rose); 3/16 rose; 3/16 pea; 1/16 single
12. Both parents heterozygous (HbS
HbS)
a. 1/4
b. 1/4
c. 1/2
13. a. Both parents (Aa). They may produce (aa) and unaffected (AA or Aa) children
b. Both parents and all children must be (aa)
c. Albino male (aa). Female must be (Aa) due to albino child. If she were AA, then no albino children. The one albino
child is (aa); three unaffected children must be (Aa). 50% chance any child will be albino. 3:1 ratio not surprising
given the small number of progeny.