Genetics Problem Set

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1 AP Biology Name: Genetics 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 b. AaBB c. Aabb d. AaBb 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 b. AaBBcc c. AaBBCc d. AaBbCc 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 F 1 generation? 4. A plant from the F 1 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?

2 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 (I A ) and B (I B ) codominant over O (i). Genotypes I A I A and I A i are type A blood, genotypes I B I B and I B i are type B blood, genotype I A I B 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?

3 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

4 Incomplete Dominance Problem 14. Red-flowering snapdragons are homozygous for allele R 1. White-flowering snapdragons are homozygous for allele R 2. Heterozygous plants (R 1 R 2 ) bear pink flowers. What phenotypes should appear among F 1 offspring of the crosses listed below? What are the expected probabilities for each phenotype? Red Pink White a. R 1 R 1 X R 1 R 2 b. R 1 R 1 X R 2 R 2 c. R 1 R 2 X R 1 R 2 d. R 1 R 2 X R 2 R 2 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?

5 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): 778 black, vestigial: 785 black, normal: 158 gray, vestigial: 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: axial (A) terminal (a) Stem length: tall (T) dwarf (t) Seed shape: round (R) 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

6 Solving Genetics Problems Punnett Squares When examining the results of a cross between two organisms, it is often useful to represent the possible gene combinations in the offspring by using a Punnett square. Below is a list of steps to use every time you make a one-factor Punnett square. Step 1: Choose a letter to represent the genes in the cross. Use a letter whose capital form does not look too similar to its lowercase form. This will make it easier for you to read your finished Punnett square. Except for that requirement, it is not important which letter you select. Step 2: Write the genotypes of the parents. This step is often written as an abbreviation of the cross being studied. The X between the parents genotypes is read is crossed with. Step 3: Determine the possible gametes that the parents can produce. Remember that alleles are segregated during the formation of gametes. Each gamete has ½ the number of alleles found in the parents. Step 4: Enter the possible gametes at the top and side of the Punnett square. Step 5: Complete the Punnett square by writing the alleles from the gametes in the appropriate boxes. This step represents the process of fertilization. The allele from the gamete above the box and the allele from the gamete to the side of the box are combined inside each of the four boxes. If there is a combination of capital letter and lowercase letter in a box, write the capital letter first. The letters inside the boxes represent the probable genotypes of the offspring resulting from the cross. Step 6: Determine the phenotypes of the offspring. Remember that phenotype refers to the physical appearance of an organism. The principle of dominance makes it possible to determine the phenotype that corresponds to each genotype inside the Punnett square.

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