IN-CLASS PRACTICE PROBLEMS - KEY (Mendelian Inheritance and Extensions of Mendel s First Law)

Transcription

1 BIO 184 Page 1 IN-CLASS PRACTICE PROBLEMS - KEY (Mendelian Inheritance and Extensions of Mendel s First Law) PART 1: MULTIPLE CHOICE QUESTIONS WITH ONLY ONE BEST ANSWER. For each question, select the best answer. There is only one best answer for each question. 1. An individual with the genotype Aa,Bb,cc,dd,Ee can make different types of gametes. a. one b. two c. three d. six e. eight Count up the number of heterozygous loci. Then raise 2 to this number. 2 3 = 8 2. At the end of a normal meiosis I in humans, the two daughter cells a. contain 23 chromosomes each b. are diploid c. will undergo DNA replication before entering meiosis II d. are gametes At the end of meiosis I, the cells are haploid. They contain one copy of each human autosome (22) and one sex chromosome (X or Y) in the duplicated state. There is no DNA replication between MI and MII. Gametes are not formed until meiosis II has been completed. 3. Most field mice have tails that are the same length as their bodies. However, occasionally a mutant mouse is observed that has a very short, stubby tail. When a field mouse that is pure-breeding for a normal tail is crossed with a mouse that is pure-breeding for a stubby tail, all of the F1 offspring have tails that are intermediate in length. When the F1s are bred to one another, the expected phenotypic ratio among the F2s will be: a. 3 wild : 1 stubby b. 3 intermediate: 1 stubby c. 2 intermediate : 1 stubby : 1 wild d. 9 wild: 4 intermediate: 3stubby The inheritance pattern suggests that the trait is under the control of a single gene with two alleles that are incompletely dominant to one another. T L = long, T S = short. T L T L = long tails; T S T S = short; T L T S = intermediate length tails. The initial cross is T L T L x T S T S to yield F1s, all of which have intermediate length tails: T L T S. According to Mendel's First Law, 1/2 of the F1s will be T L T S and have intermediate length tails, 1/4 will be T L T L and have long tails, and 1/4 will be T S T S and have short tails. This is a 2:1:1 ratio. For questions 4-5, refer to the paragraph below: Occasionally, Drosophila flies are born with curly wings. A genetics professor takes several of these unusual flies and crosses them to one another with the following result: 522 curly wings, 266 normal wings. 4. The mutation that causes curly wings is probably

2 BIO 184 Page 2 a. recessive and lethal in the homozygous state b. recessive and semi-lethal in the homozygous state c dominant and lethal in the homozygous state d. dominant and semi-lethal in the homozygous state 522/266 = 2.04 (very close to a 2:1 ratio). This suggests that the allele for curly wings is dominant to the wild-type allele and is lethal in the homozygous state (all CC flies die; Cc flies have curly wings; cc flies are wild-type.) 5. According to the answer you chose in question 9, all curly-winged flies must be a. homozygous for the mutant allele b. heterozygous c. homozygous for the normal allele d. dihybrid for the two genes that control the trait Because all CC flies die, living curly-winged flies must be heterozygous. 6. Cindy has blood type AB. Assuming that no one in the family has the Bombay Phenotype, which of the following blood types could her biological parents have? a. Father type A, mother type O b. Father type AB, mother type O c. Father type A, mother type B d. Father and mother both type O e. Father and mother both type A The I A and I B alleles are codominant. Therefore, if a parent has either of these alleles, it will be expressed in his/her blood type. One of Cindy's parents gave her A and the other one gave her B. The only possibility in the choices given is "c." 7. There are four different blood groups: A, B, AB, and O. How many different alleles govern this system? a. 1 b. 2 c. 3 d. 4 Three alleles: I A, I B, and i. For Questions 8-9, refer to the paragraph below: In a species of honey bee, S = striped body, s = stripeless; V = wild wings, v = vestigial wings. 8. If a dihybrid queen produces 100 male offspring, about how many would have stripeless bodies and vestigial wings? a. all of them b. 3/4 c. 1/2 d. 1/4 e. none

3 BIO 184 Page 3 In bees, sex is determined by the number of chromosome sets. Female bees are diploid and male bees are haploid. A dihybrid queen has the genotype SsVv. According to Mendel's Second Law, she will produce 4 gametes in equal numbers: SV, sv, Av, and sv. These gametes will then grow into male adults by mitosis. Therefore, 1/4 of the male offspring will be sv (stripeless bodies and vestigial wings). 9. All male bees with striped bodies and wild wings have the genotype a. S,V b. Ss,Vv c. SS,VV d. s,v Males are haploid. (See answer to 8, above.) 10. The unusual coat color of tortoise shell cats results from a. X-inactivation b. a temperature sensitive mutation c. incomplete penetrance d. overdominance Cats have two genes that govern coat color. The B gene is autosomal and has two alleles: B = black; b = white. The O gene is X-linked and has two alleles: X O = orange; X o = null (no function). B acts first in the pathway. If at least one copy of B is present, the O allele will turn the black pigment orange, while o will not alter the color. Cats with the genotype "bb" are always white, regardless of what allele they have at the O locus. Male cats can be B_X O Y (orange), B_X o Y (black), or bbx -- Y (white). Female cats can be B_X O X O (orange), B_X o X o (black), B_X O X o (tortoise shell) or bbx -- X -- (white). Due to X-inactivation, B_X O X o express patches of orange and black on their fur (tortoise shell). Males with a normal karyotype have only one X chromosome so they cannot be tortoise shell. (Rare X O X o Y males are an exception.) 11. Graham s father, paternal uncle and aunt, and paternal grandmother were all born with extra digits. This disorder is called polydactyly and is autosomal dominant. Graham was born with the normal number of fingers and toes, so he felt certain that he had not inherited the polydactyly mutation. Graham s wife has no family history of polydactyly, so the couple is surprised when their first daughter is born with two extra toes. The polydactyly mutation is probably a. overdominant b. incompletely penetrant c. lethal in the homozygous state d. located on the X chromosome This inheritance pattern suggests that Graham "carries" a dominant allele for polydactly. However, the allele was non-penetrant in him. When not everyone who has a mutant allele expresses the phenotype, the mutant allele is incompletely penetrant (when referring to the entire pedigree.) 12. In birds, sex is determined a. by the temperature at which the fertilized eggs are incubated b. by whether they are diploid or haploid c. by a chromosome-based system in which the males are homogametic d. by a chromosome-based system in which the males are heterogametic

4 BIO 184 Page 4 Sex is determined by chromosomes in birds, but the male birds have two copies of the same sex chromosome while females have two different chromosomes. (This is the opposite situation from mammals.) 13. In unicorns, horn color is under the control of two genes, A and B. When two AaBb unicorns are crossed, the result is 9 gold horn: 7 silver horn. This ratio indicates that a. all gold-horned unicorns are aabb b. all gold-horned unicorns are either aabb or aabb c. all gold-horned unicorns have at least one dominant allele at both loci d. all gold-horned unicorns are dihybrids A 9:7 ratio indicates epistasis, where "aa" or "bb" yields a silver horn. Only unicorns with the genotype A_B_ have gold horns. (Recall that, according to Mendel's Second Law, 9 of the offspring will be A_B_, 3 will be A_bb, 3 will be aab_, and 1 will be aabb (when considering phenotypic ratios in the offspring of a dihybrid cross). 14. The mutation that causes male pattern baldness a. is located on an autosome b. can be inherited from either parent c. is dominant in males and recessive in females d. is sex-influenced e. All of the above When solving these types of problems, I suggest you use B* to track the bald allele and B to track the non-bald allele. 15. Sickle cell anemia is an autosomal recessive disease that is common in many parts of Africa. The high frequency of the disease in Africa is explained by a. overdominance b. the high rate of malaria in the region c. heterozygote advantage d. the fact that the sickle cell allele protects its carriers against malaria e. All of the above We went over this several times in class. PART 2: MULTIPLE CHOICE QUESTIONS THAT MAY HAVE MORE THAN ONE CORRECT ANSWER. For each question, select all correct answers. 16. Which of the following is/are dihybrid(s)? a. AA,SS b. Aa,SS c. AA,Ss d. Aa,Ss e. aa,ss Dihybrid = double heterozygote. 17. Mendel s predicted 3:1 ratio from a monohybrid cross assumes: a. Gametes are haploid. b. Dihybrids produce 4 different gametes in equal numbers.

5 BIO 184 Page 5 c. All offspring are equally viable. d. Gametes mate randomly. e. Males are the heterogametic sex. "b" is incorrect because Mendel's First Law does not govern the behavior of the alleles for more than one gene. "e" is incorrect because Mendel never observed X-linkage. Pea plants do not have different sexes. Each pea flower has both male and female parts. 18. When a pure-breeding orchid with bright pink flowers is crossed with a pure-breeding orchid with yellow flowers, all of the F1 plants have yellow flowers. However, when the F1s are crossed to one another, the plants in the F2 generation produce either bright pink, yellow, or orange flowers. Which of the following is/are probably true? a. Two genes control the trait. b. The allele for bright pink flowers is incompletely dominant to the allele for yellow flowers. c. The phenotypic ratio in the F2 generation is 9 bright pink: 3 yellow: 4 orange. d. The least common phenotype in the F2 generation is orange. e. The phenotypic ratio in the F2 generation is 1 bright pink: 2 orange: 1 yellow This is another case of epistasis, but in this instance a third phenotype is observed. This type of epistasis would result in a 9:3:4 ratio, where the new phenotype (in this case orange) accounts for "3". This cannot be incomplete dominance because the F1 generation is yellow. 19. A pea plant with the genotype (Aa, Dd, Ff, Rr) is crossed with a pea plant with the genotype (aa, DD, Ff, Rr). Assuming independent assortment, which of the following will be true of the offspring? a. One in eight will have the genotype (Aa, DD, FF, rr). b. One in eight will have the phenotype (A_ D_ F_ R_). c. All will have at least one dominant allele. d. Some will be pure-breeding at all four loci. e. Some will be heterozygous at all four loci. To solve this type of problem, examine one locus at a time and then use the product rule. For example, in a cross between Aa and aa, 1/2 of the offspring will be Aa. In a cross between Dd and DD, 1/2 of the offspring will be DD. In a cross between Ff and Ff, 1/4 of the offspring will be FF. In a cross between Rr and Rr, 1/4 of the offspring will be rr. (1/2)(1/2)(1/4)(1/4) = 1/ A tomato plant that breeds true for deep green leaves is crossed with a tomato plant that breeds true for yellow-tinted leaves. All of the F1s have deep green leaves, but when the F1s are crossed to one another, the yellow-tinted leaves reappear, with about 1 plant with yellow-tinted leaves for every 15 with deep green leaves. Which of the following is possible? a. Leaf color is under the control of a single gene, where the allele for yellow-tinted leaves is recessive and semi-lethal. b. Leaf color is under the control of two genes that have redundant functions. c. Leaf color is under the control of a single gene, where the allele for deep green leaves is codominant with the allele for yellow-tinted leaves. d. Leaf color is under the control of two genes acting along a pathway. The default is yellow-tinted leaves and the dominant alleles for two different genes are both needed for the plant to have deep green leaves. e. Leaf color in tomato plants is controlled by a mechanism identical to the one that controls coat color in Labrador Retrievers. With a 15:1 ratio, there are two possibilities. This is the only instance (for the mechanisms described in lecture) where the ratio gives an ambiguous result.

6 BIO 184 Page The human Y chromosome a. is much larger than the human X chromosome b. carries the SRY gene c. carries holandric genes d. never crosses over with the X chromosome e. is usually present in 2 copies per sperm cell We discussed this in detail in class.