Download Name: AP Biology - Unit 6: Patterns of Inheritance Mendelian

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AP Biology - Unit 6: Patterns of Inheritance
Mendelian Problem Set
Independent Assortment Problems
1. One gene has alleles A and a. Another has alleles B and b. For each genotype listed, what type(s) of
gametes will be produced? (Assume independent assortment occurs before gametes form.)
Genotypes
Possible Gametes
a. AABB
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b. AaBB
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c. Aabb
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d. AaBb
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2. Assume that you now study a third gene having alleles C and c. For each genotype, what type(s) of
gametes will be produced?
Genotypes
Possible Gametes
a. AABBCC
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b. AaBBcc
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c. AaBBCc
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d. AaBbCc
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Monohybrid Cross Problems
For the following problems, SHOW ALL OF YOUR WORK. For each cross, be sure to give the genotype(s) of
the parents and the resulting genotype(s) and phenotype(s) of the offspring. For consistency, all
probabilities should be expressed as percentages (e.g. 50% homozygous dominant, 50% heterozygous).
3. The gene for tall (T) is dominant over dwarf (t) in the garden pea plant used by Mendel. A pea plant
that comes from a line of true-breeding tall plants is crossed with a dwarf pea plant. What are the
phenotype(s) and genotype(s) of the F1 generation?
4. A plant from the F1 generation of problem 3 is crossed with a plant from a true-breeding dwarf
lineage. What are the genotype(s) and phenotype(s) of the offspring?
5. In humans, dimples are a trait that exhibits simple dominance. The two alleles for this trait are
dimples (D) and no dimples (d). A dimpled man whose mother has no dimples marries a woman with
no dimples. What is the probability that their first child will have dimples?
6. Sickle cell anemia (SCA) is a human genetic disorder known to be caused by a recessive allele. A
couple plans to marry and wants to know the probability that they will have an affected child. In the
following scenarios, use your knowledge of Mendelian inheritance to give them advice.
a. The man and woman do not have SCA. However, each has one parent with the disorder.
b. The man is affected by the disorder. The woman has no family history of SCA.
Multiple Alleles & Codominace Problems – ABO Blood Types
7. The alleles for blood type in humans are inherited as A (IA) and B (IB) codominant over O (i).
Genotypes IAIA and IAi are type A blood, genotypes IBIB and IBi are type B blood, genotype IAIB is type
AB blood, and genotype ii is type O blood. A man with type B blood marries a woman with type A
blood. They have six children, all with type AB blood. What are the most likely genotypes of the
father, mother, and children?
8. A father with blood type B and a mother with blood type A have a son with blood type A. The son
marries a woman with blood type A whose mother was type O. What are the genotypes of the son
and his wife? What is the probability that their first child will be blood type A?
9. A man with blood type A marries a woman with blood type A. They have eight children with blood
type A and one child with blood type O. What are the genotypes of the father, mother, the eight
type A children, and the one type O child?
10. A father with blood type AB and a mother with blood type B have a daughter with blood type A. She
marries a man with blood type B whose father was blood type A. What are the genotypes of the
daughter and her husband? What is the probability that their first child will have type AB blood?
11. DNA fingerprinting is a method used to identify individuals by locating unique base sequences in their
DNA molecules. Before researchers refined the method, attorneys often relied on ABO blood-typing
to settle disputes over paternity. Suppose that you, as a geneticist, are asked 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 for the alleged father: “The mother’s blood is type A, so the child’s type O blood
must have come from the father. My client has type B blood; he could not be the father.”
b. Mother’s attorney: “Further tests show that the man is heterozygous. Therefore, he must be
the father.”
Dihybrid Cross Problems
12. John can roll his tongue and is heterozygous for this dominant trait (Rr). John’s second toe is shorter
than his first. His wife Rita cannot roll her tongue and has a longer second toe. Rita is heterozygous
for this dominant trait (Tt). Use a Punnett square to predict the possible genotype(s) and
phenotype(s) of their first child.
13. In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p) and normal
leaf (L) is dominant to wrinkled leaf (l). These traits are influenced by genes on different
chromosomes and therefore inherited independently. Determine the genotypes for the two parent
plants that produced the following offspring:
a. 308 one-pod, normal / 98 one-pod, wrinkled
b. 323 three-pod, normal / 106 three-pod, wrinkled
c. 401 one-pod, normal
d. 150 one-pod, normal / 147 one-pod, wrinkled / 51 three-pod, normal / 48 three-pod, wrinkled
e. 223 one-pod, normal / 72 one-pod, wrinkled / 76 three-pod, normal / 27 three-pod, wrinkled
Incomplete Dominance Problem
14. Red-flowering snapdragons are homozygous for allele R1. White-flowering snapdragons are
homozygous for allele R2. Heterozygous plants (R1R2) bear pink flowers. What phenotypes should
appear among F1 offspring of the crosses listed below? What are the expected probabilities for each
phenotype?
Red
Pink
White
a. R1R1 X R1R2
b. R1R1 X R2R2
c. R1R2 X R1R2
d. R1R2 X R2R2
Epistasis Problem
15. In Labrador Retrievers, coat color is determined by the interaction between two genes. The alleles
for one gene influence the color of the pigment in the hair (B is black and b is chocolate brown). The
alleles for a second gene influence whether the pigment is deposited in the hair (E is full deposition
and e is reduced deposition). Yellow Labrador Retrievers are homozygous recessive for the second
gene. A female chocolate Lab (bbEe) is crossed with a male black Lab (BbEe). What are the possible
genotypes and phenotypes of the puppies? (Use a Punnett square to show the results.)
Sex-Linked Inheritance Problems
16. A man with hemophilia (a recessive, X-linked condition) has a daughter who is not affected by
hemophilia. She marries an unaffected man. What is the probability that a daughter of this mating
will be hemophiliac? What is the probability that a son will be a hemophiliac? If the couple has 4
sons, what is the probability that all four will be born with hemophilia?
17. Red-green color blindness is a recessive, X-linked condition. A color-blind man marries a woman who
is a carrier for color-blindness. What is the probability that they will have color-blind sons? What is
the probability of color-blind daughters?
Other Problems
18. A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly
with vestigial wings. The offspring have the following phenotypic distribution:
gray, normal (wild type):
black, vestigial:
black, normal:
gray, vestigial:
778
785
158
162
Based on the above data, what is the recombination frequency between these genes?
19. Determine the sequence of genes along a chromosome based on the following recombination
frequencies:
A-B - 8 map units
A-C - 28 map units
A-D - 25 map units
B-C - 20 map units
B-D - 33 map units
20. Flower position, stem length, and seed shape were three characteristics that Mendel studied. Each is
controlled by an independently assorting gene and has dominant and recessive expression as follows:
Flower position:
Stem length:
Seed shape:
axial (A)
tall (T)
round (R)
terminal (a)
dwarf (t)
wrinkled (r)
If a plant that is heterozygous for all three characteristics is allowed to self-fertilize, what
percentage of the offspring would you expect to be as follows?
(Note: use the rules of probability instead of a Punnett square.)
a. homozygous for the 3 dominant traits
b. homozygous for the 3 recessive traits
c. heterozygous for all 3 traits
d. homozygous for axial and tall, heterozygous for seed shape