ch 14 practice problems
85) Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.
1/9
19) Which of the plants will be true-breeding? A) 1 and 4 B) 2 and 3 C) 1-4 D) 1 only E) none
A Topic: Concept 14.1 Skill: Application
14) A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism? A) HT B) Hh C) HhTt D) T E) tt
A Topic: Concept 14.1 Skill: Comprehension
22) What are Punnett squares used for? A) predicting the result of genetic crosses between organisms of known genotypes B) determining the DNA sequence of a given gene C) identifying the gene locus where allelic variations are possible D) testing for the presence of the recessive allele E) more than one of the above
A Topic: Concept 14.1 Skill: Knowledge
6) The F1 offspring of Mendel's classic pea cross always looked like one of the two parental varieties because A) one allele was completely dominant over another. B) each allele affected phenotypic expression. C) the traits blended together during fertilization. D) no genes interacted to produce the parental phenotype. E) different genes interacted to produce the parental phenotype.
A Topic: Concept 14.1 Skill: Knowledge
40) A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce A) all sharp-spined progeny. B) 50% sharp-spined, 50% dull-spined progeny. C) 25% sharp-spined, 50% dull-spined, 25% spineless progeny D) all spineless progeny E) It is impossible to determine the phenotypes of the progeny.
A Topic: Concept 14.3 Skill: Application
34) If the intermediate F1 progeny were allowed to self-pollinate, and 25% of the F2 progeny were tall, 50% were intermediate in size, and 25% were short, this would suggest A) incomplete dominance. B) polygenic inheritance. C) complete dominance. D) pleiotropy. E) multifactorial inheritance.
A Topic: Concept 14.3 Skill: Application
55) Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by A) environmental factors such as soil pH. B) the allele for blue hydrangea being completely dominant. C) the alleles being codominant. D) the fact that a mutation has occurred. E) acknowledging that multiple alleles are involved.
A Topic: Concept 14.3 Skill: Comprehension
52) the phenotype of the heterozygote differs from the phenotypes of both homozygotes A. incomplete dominance B. multiple alleles C. pleiotropy D. epistasis
A Topic: Concept 14.3 Skill: Knowledge
64) When a disease is said to have a multifactorial basis, it means that A) many factors, both genetic and environmental, contribute to the disease. B) it is caused by a gene with a large number of alleles. C) it affects a large number of people. D) it has many different symptoms. E) it tends to skip a generation.
A Topic: Concept 14.4 Skill: Knowledge
68) This is caused by a dominant single gene defect and generally does not appear until the individual is 35-45 years of age. A. Huntington's disease B. Tay-Sachs disease C. phenylketonuria D. cystic fibrosis E. sickle-cell disease
A Topic: Concept 14.4 Skill: Knowledge
75) Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted ratio of their offspring would be: A) 1 black eyes, green skin : 1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin B) 3 black eyes, green skin : 3 black eyes, white skin : 9 orange eyes, green skin : 1 orange eyes, white skin C) 1 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 9 orange eyes, white skin D) 9 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 1 orange eyes, white skin E) There is insufficient information to determine Sam's genotype.
A Topic: Web/CD Activity: Dihybrid Crosses
74) Andalusian chickens with the genotype CBCB are black, those with the genotype CBCW are gray. What is the relationship between the CB and the CW alleles? A) CB is dominant to CW. B) CB is recessive to CW. C) CW is dominant to CB. D) The relationship is one of incomplete dominance. E) CB and CW are codominant.
A Topic: Web/CD Activity: Incomplete Dominance
84) What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.) A) AABBCC × aabbcc → AaBbCc B) AABbCc × AaBbCc → AAbbCC C) AaBbCc × AaBbCc → AaBbCc D) aaBbCC × AABbcc → AaBbCc
A) 1 B) 1/32 C) 1/8 D) 1/2
83) The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? A) aabbccdd B) AaBbCcDd C) AABBCCDD D) AaBBccDd E) AaBBCCdd
A) 1/256 B) 1/16 C) 1/256 D) 1/64 E) 1/128
78) Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows: If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.) A) homozygous for the three dominant traits B) homozygous for the three recessive traits C) heterozygous for all three characters D) homozygous for axial and tall, heterozygous for seed shape
A) 1/64 B) 1/64 C) 1/8 D) 1/32
82) Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? A) All three children are of normal phenotype. B) One or more of the three children have the disease. C) All three children have the disease. D) At least one child is phenotypically normal. (Note: Remember that the probabilities of all possible outcomes always add up to 1.)
A) 3/4 × 3/4 × 3/4 = 27/64 B) 1 - 27/64 = 37/64 C) 1/4 × 1/4 × 1/4 = 1/64 D) 1 - 1/64 = 63/64
80) 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). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring: A) 318 one-pod, normal leaf : 98 one-pod, wrinkled leaf B) 323 three-pod, normal leaf : 106 three-pod, wrinkled leaf C) 401 one-pod, normal leaf D) 150 one-pod, normal leaf : 147 one-pod, wrinkled leaf : 51 three-pod, normal leaf : 48 three-pod, wrinkled leaf E) 223 one-pod, normal leaf : 72 one-pod, wrinkled leaf : 76 three-pod, normal leaf; 27 three-pod, wrinkled leaf
A) PPLl × PPLl, PpLl, or ppLl B) ppLl × ppLl C) PPLL × any of the 9 possible genotypes or PPll × ppLL D) PpLl × Ppll E) PpLl × PpLl
79) A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino is crossed with a second black one, 7 blacks and 5 albinos are obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.
Albino (b) is a recessive trait; black (B) is dominant. First cross: parents BB × bb; gametes B and b; offspring all Bb (black coat). Second cross: parents bb × Bb; gametes 1/2 B and 1/2 b (heterozygous parent) and b; offspring 1/2 Bb and 1/2 bb.
10) How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE? A) 4 B) 8 C) 16 D) 32 E) 64
B Topic: Concept 14.1 Skill: Application
15) It was important that Mendel examined not just the F1 generation in his breeding experiments, but the F2 generation as well, because A) he obtained very few F1 progeny, making statistical analysis difficult. B) parental traits that were not observed in the F1 reappeared in the F2, suggesting that the traits did not truly disappear in the F1. C) analysis of the F1 progeny would have allowed him to discover the law of segregation, but not the law of independent assortment. D) the dominant phenotypes were visible in the F2 generation, but not in the F1. E) all of the above
B Topic: Concept 14.1 Skill: Comprehension
20) P = purple, pp = white. The offspring of a cross between two heterozygous purple-flowering plants (Pp × Pp) results in A) all purple-flowered plants. B) purple-flowered plants and white-flowered plants. C) two types of white-flowered plants: PP and Pp. D) all white-flowered plants. E) all pink-flowered plants.
B Topic: Concept 14.1 Skill: Comprehension
7) What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants? A) There is considerable genetic variation in garden peas. B) Traits are inherited in discrete units, and are not the results of "blending." C) Recessive genes occur more frequently in the F1 than do dominant ones. D) Genes are composed of DNA. E) An organism that is homozygous for many recessive traits is at a disadvantage.
B Topic: Concept 14.1 Skill: Comprehension
13) A 9:3:3:1 phenotypic ratio is characteristic of which of the following? A) a monohybrid cross B) a dihybrid cross C) a trihybrid cross D) linked genes E) both A and D
B Topic: Concept 14.1 Skill: Knowledge
9) What is genetic cross between an individual showing a dominant phenotype (but of unknown genotype) and a homozygous recessive individual called? A) a self-cross B) a testcross C) a hybrid cross D) an F1 cross E) a dihybrid cross
B Topic: Concept 14.1 Skill: Knowledge
28) Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers? (Assume independent assortment.) A) 65 B) 190 C) 250 D) 565 E) 750
B Topic: Concept 14.2 Skill: Application
33) If the intermediate F1 progeny were allowed to self-pollinate, and the F2 progeny were also intermediate in size, but following a normal distribution, this would suggest A) incomplete dominance. B) polygenic inheritance. C) complete dominance. D) a strong environmental influence. E) codominance.
B Topic: Concept 14.3 Skill: Application
38) In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white? A) red × white B) roan × roan C) white × roan D) red × roan E) The answer cannot be determined from the information provided.
B Topic: Concept 14.3 Skill: Application
39) The relationship between genes S and N is an example of A) incomplete dominance. B) epistasis. C) complete dominance. D) pleiotropy. E) codominance.
B Topic: Concept 14.3 Skill: Comprehension
51) the ABO blood group system A. incomplete dominance B. multiple alleles C. pleiotropy D. epistasis
B Topic: Concept 14.3 Skill: Knowledge
67) Individuals with this disorder are unable to metabolize certain lipids, affecting proper brain development. Affected individuals die in early childhood. A. Huntington's disease B. Tay-Sachs disease C. phenylketonuria D. cystic fibrosis E. sickle-cell disease
B Topic: Concept 14.3 Skill: Knowledge
56) What is the probability that their first child will be an albino? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1
B Topic: Concept 14.4 Skill: Application
57) If their first two children have normal pigmentation, what is the probability that their third child will be an albino? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1
B Topic: Concept 14.4 Skill: Application
73) What is the expected phenotypic ratio of a cross between two orange-eyed MendAliens? A) 3 black-eyed:1 orange-eyed B) 0 black-eyed:1 orange-eyed C) 1 black-eyed:3 orange-eyed D) 1 black-eyed:0 orange-eyed E) 1 black-eyed:1 orange-eyed
B Topic: Web/CD Activity: Monohybrid Crosses
16) When crossing a homozygous recessive with a heterozygote, what is the chance of getting an offspring with the homozygous recessive phenotype? A) 0% B) 25% C) 50% D) 75% E) 100%
C Topic: Concept 14.1 Skill: Application
5) A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates A) the blending model of genetics. B) true-breeding. C) dominance. D) a dihybrid cross. E) the mistakes made by Mendel.
C Topic: Concept 14.1 Skill: Comprehension
21) Mendel accounted for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that A) new mutations were frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1. B) the mechanism controlling the appearance of traits was different between the F1 and the F2 plants. C) traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1. D) the traits were lost in the F1 due to blending of the parental traits. E) members of the F1 generation had only one allele for each character, but members of the F2 had two alleles for each character.
C Topic: Concept 14.1 Skill: Comprehension
2) A plant with purple flowers is allowed to self-pollinate. Generation after generation, it produces purple flowers. This is an example of A) hybridization. B) incomplete dominance. C) true-breeding. D) the law of segregation. E) polygenetics.
C Topic: Concept 14.1 Skill: Knowledge
4) What is the difference between a monohybrid cross and a dihybrid cross? A) A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents. B) A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny. C) A monohybrid cross involves organisms that are heterozygous for a single character, whereas a dihybrid cross involves organisms that are heterozygous for two characters. D) A monohybrid cross is performed only once, whereas a dihybrid cross is performed twice. E) A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.
C Topic: Concept 14.1 Skill: Knowledge
30) Given the parents AABBCc × AabbCc, assume simple dominance and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent? A) 1/4 B) 1/8 C) 3/4 D) 3/8 E) 1
C Topic: Concept 14.2 Skill: Application
26) In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short? A) 1/2 B) 1/4 C) 0 D) 1 E) 1/6
C Topic: Concept 14.2 Skill: Comprehension
44) Three babies were mixed up in a hospital. After consideration of the data below, which of the following represent the correct baby and parent combinations? (image not found) A) I-3, II-1, III-2 B) I-1, II-3, III-2 C) I-2, II-3, III-1 D) I-2, II-1, III-3 E) I-3, II-2, III-1
C Topic: Concept 14.3 Skill: Application
43) Two blue budgies were crossed. Over the years, they produced 22 offspring, 5 of which were white. What are the most likely genotypes for the two blue budgies? A) yyBB and yyBB B) yyBB and yyBb C) yyBb and yyBb D) yyBB and yybb E) yyBb and yybb
C Topic: Concept 14.3 Skill: Application
47) Which of the following is a possible phenotype for the father? A) A B) O C) B D) AB E) impossible to determine
C Topic: Concept 14.3 Skill: Application
37) Skin color in a fish is inherited via a single gene with four different alleles. How many different types of gametes would be possible in this system? A) 1 B) 2 C) 4 D) 8 E) 16
C Topic: Concept 14.3 Skill: Comprehension
31) A 1:2:1 phenotypic ratio in the F2 generation of a monohybrid cross is a sign of A) complete dominance. B) multiple alleles. C) incomplete dominance. D) polygenic inheritance. E) pleiotropy.
C Topic: Concept 14.3 Skill: Knowledge
50) the ability of a single gene to have multiple phenotypic effects A. incomplete dominance B. multiple alleles C. pleiotropy D. epistasis
C Topic: Concept 14.3 Skill: Knowledge
53) cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections A. incomplete dominance B. multiple alleles C. pleiotropy D. epistasis
C Topic: Concept 14.3 Skill: Knowledge
49) Which of the following is a possible phenotype of the father? A) A negative B) O negative C) B positive D) A positive E) O positive
C Topic: Concept 14.3, Concept 14.4 Skill: Application
61) What is the likelihood that the progeny of D-3 and D-4 will have wooly hair? A) 0% B) 25% C) 50% D) 75% E) 100%
C Topic: Concept 14.4 Skill: Application
60) What is the genotype of individual B-5? A) WW B) Ww C) ww D) WW or ww E) ww or Ww
C Topic: Concept 14.4 Skill: Application
58) Huntington's disease is caused by a dominant allele. If one of your parents has the disease, what is the probability that you, too, will have the disease? A) 1 B) 3/4 C) 1/2 D) 1/4 E) 0
C Topic: Concept 14.4 Skill: Knowledge
59) A woman has six sons. The chance that her next child will be a daughter is A) 1. B) 0. C) 1/2. D) 1/6. E) 5/6.
C Topic: Concept 14.4 Skill: Knowledge
69) Effects of this recessive disorder can be completely overcome by regulating the diet of the affected individual. A. Huntington's disease B. Tay-Sachs disease C. phenylketonuria D. cystic fibrosis E. sickle-cell disease
C Topic: Concept 14.4 Skill: Knowledge
18) Which of the boxes correspond to plants with a heterozygous genotype? A) 1 B) 1 and 2 C) 1, 2, and 3 D) 2 and 3 E) 2, 3, and 4
D Topic: Concept 14.1 Skill: Application
24) The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment means that A) none of the traits obeyed the law of segregation. B) the diploid number of chromosomes in the pea plants was 7. C) all of the genes controlling the traits were located on the same chromosome. D) all of the genes controlling the traits behaved as if they were on different chromosomes. E) the formation of gametes in plants occurs by mitosis only.
D Topic: Concept 14.1 Skill: Comprehension
1) Pea plants were particularly well suited for use in Mendel's breeding experiments for all of the following reasons except that A) peas show easily observed variations in a number of characters, such as pea shape and flower color. B) it is possible to completely control matings between different pea plants. C) it is possible to obtain large numbers of progeny from any given cross. D) peas have an unusually long generation time. E) many of the observable characters that vary in pea plants are controlled by single genes.
D Topic: Concept 14.1 Skill: Comprehension
3) Which of the following statements about Mendel's breeding experiments is correct? A) None of the parental (P) plants were true-breeding. B) All of the F2 progeny showed a phenotype that was intermediate between the two parental (P) phenotypes. C) Half of the F1 progeny had the same phenotype as one of the parental (P) plants, and the other half had the same phenotype as the other parent. D) All of the F1 progeny resembled one of the parental (P) plants, but only some of the F2 progeny did. E) none of the above
D Topic: Concept 14.1 Skill: Comprehension
11) Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. This suggests A) that the parents were true-breeding for contrasting traits. B) incomplete dominance. C) that a blending of traits has occurred. D) that the parents were both heterozygous. E) that each offspring has the same alleles.
D Topic: Concept 14.1 Skill: Knowledge
25) Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt × BBtt will have black fur and long tails? A) 1/16 B) 3/16 C) 3/8 D) 1/2 E) 9/16
D Topic: Concept 14.2 Skill: Application
27) A couple has three children, all of whom have brown eyes and blond hair. Both parents are homozygous for brown eyes (BB), but one is a blond (rr) and the other is a redhead (Rr). What is the probability that their next child will be a brown-eyed redhead? A) 1/16 B) 1/8 C) 1/4 D) 1/2 E) 1
D Topic: Concept 14.2 Skill: Application
36) Tallness (T) is dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous condition results in pink (Rr) flower color. A dwarf, red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are the genotype and phenotype of the F1 individuals? A) ttRr-dwarf and pink B) ttrr-dwarf and white C) TtRr-tall and red D) TtRr-tall and pink E) TTRR-tall and red
D Topic: Concept 14.3 Skill: Application
42) A blue budgie is crossed with a white budgie. Which of the following results is not possible? A) green offspring B) yellow offspring C) blue offspring D) A and B E) A, B, and C
D Topic: Concept 14.3 Skill: Application
45) Which of the following is a possible genotype for the son? A) IBIB B) IBIA C) ii D) IBi E) IAIA
D Topic: Concept 14.3 Skill: Application
46) Which of the following is a possible genotype for the mother? A) IAIA B) IBIB C) ii D) IAi E) IAIB
D Topic: Concept 14.3 Skill: Application
32) This could be an example of A) incomplete dominance. B) polygenic inheritance. C) complete dominance. D) A and B E) B and C
D Topic: Concept 14.3 Skill: Comprehension
48) Which of the following is the probable genotype for the mother? A) IAIARR B) IAIARr C) IAirr D) IAiRr E) IAiRR
D Topic: Concept 14.3, Concept 14.4 Skill: Application
70) This results from a defect in membrane proteins that normally function in chloride ion transport. A. Huntington's disease B. Tay-Sachs disease C. phenylketonuria D. cystic fibrosis E. sickle-cell disease
D Topic: Concept 14.4 Skill: Knowledge
71) Which of the following techniques involves the preparation of a karyotype? A) amniocentesis B) chorionic villus sampling C) fetoscopy D) A and B only E) A, B, and C
D Topic: Concept 14.4 Skill: Knowledge
76) All the offspring of a cross between a red-flowered plant and a white-flowered plant have pink flowers. This means that the allele for red flowers is ________ to the allele for white flowers. A) dominant B) codominant C) pleiotropic D) incompletely dominant E) recessive
D Topic: Web/CD Activity: Incomplete Dominance
72) All the offspring of a cross between a black-eyed MendAlien and an orange-eyed MendAlien have black eyes. This means that the allele for black eyes is ________ the allele for orange eyes. A) codominant to B) recessive to C) more aggressive than D) dominant to E) better than
D Topic: Web/CD Activity: Monohybrid Crosses
17) Which of the boxes marked 1-4 correspond to plants with dark leaves? A) 1 only B) 1 and 2 C) 2 and 3 D) 4 only E) 1, 2, and 3
E Topic: Concept 14.1 Skill: Application
12) Two characters that appear in a 9:3:3:1 ratio in the F2 generation should have which of the following properties? A) Each of the characters is controlled by a single gene. B) The genes controlling the characters obey the law of independent assortment. C) Each of the genes controlling the characters has two alleles. D) Only A and C are correct. E) A, B, and C are correct.
E Topic: Concept 14.1 Skill: Comprehension
23) Which of the following is false, regarding the law of segregation? A) It states that each of two alleles for a given trait segregate into different gametes. B) It can be explained by the segregation of homologous chromosomes during meiosis. C) It can account for the 3:1 ratio seen in the F2 generation of Mendel's crosses. D) It can be used to predict the likelihood of transmission of certain genetic diseases within families. E) It is a method that can be used to determine the number of chromosomes in a plant.
E Topic: Concept 14.1 Skill: Knowledge
8) Which of the following is (are) true for alleles? A) They can be identical or different for any given gene in a somatic cell. B) They can be dominant or recessive. C) They can represent alternative forms of a gene. D) Only A and B are correct. E) A, B, and C are correct.
E Topic: Concept 14.1 Skill: Knowledge
29) In a cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC? A) 1/4 B) 1/8 C) 1/16 D) 1/32 E) 1/64
E Topic: Concept 14.2 Skill: Application
35) In snapdragons, heterozygotes have pink flowers, whereas homozygotes have red or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring will have pink flowers? A) 0% B) 25% C) 50% D) 75% E) 100%
E Topic: Concept 14.3 Skill: Application
41) If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios? A) 3 sharp-spined : 1 spineless B) 1 sharp-spined : 2 dull-spined : 1 spineless C) 1 sharp spined : 1 dull-spined : 1 spineless D) 1 sharp-spined : 1 dull-spined E) 9 sharp-spined : 3 dull-spined : 4 spineless
E Topic: Concept 14.3 Skill: Application
54) Which of the following is an example of polygenic inheritance? A) pink flowers in snapdragons B) the ABO blood groups in humans C) Huntington's disease in humans D) white and purple flower color in peas E) skin pigmentation in humans
E Topic: Concept 14.3 Skill: Knowledge
62) What is the probability that individual C-1 is Ww? A) 3/4 B) 1/4 C) 2/4 D) 2/3 E) 1
E Topic: Concept 14.4 Skill: Application
63) People with sickle-cell trait A) are heterozygous for the sickle-cell allele. B) are usually healthy. C) have increased resistance to malaria. D) produce normal and abnormal hemoglobin. E) all of the above
E Topic: Concept 14.4 Skill: Knowledge
65) Which of the following terms is least related to the others? A) pedigree B) karyotype C) amniocentesis D) chorionic villus sampling E) epistasis
E Topic: Concept 14.4 Skill: Knowledge
66) Substitution of the "wrong" amino acid in the hemoglobin protein results in this disorder. A. Huntington's disease B. Tay-Sachs disease C. phenylketonuria D. cystic fibrosis E. sickle-cell disease
E Topic: Concept 14.4 Skill: Knowledge
81) A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these individuals? What other genotypes, and in what frequencies, would you expect in offspring from this marriage?
Man IAi; woman IBi; child ii. Other genotypes for children are 1/4 IAIB ,1/4 IAi, 1/4 IBi
86) In 1981, a stray black cat with unusual rounded, curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the curl cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you obtain true-breeding curl cats? How could you be sure they are true-breeding?
Matings of the original mutant cat with true-breeding noncurl cats will produce both curl and noncurl F1 offspring if the curl allele is dominant, but only noncurl offspring if the curl allele is recessive. You would obtain some true-breeding offspring homozygous for the curl allele from matings between the F1 cats resulting from the original curl × noncurl crosses whether the curl trait is dominant or recessive. You know that cats are true-breeding when curl × curl matings produce only curl offspring. As it turns out, the allele that causes curled ears is dominant.
77) In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: CRCR (red) × CWCW (white) →CRCW (pink). If flower position (axial or terminal) is inherited as it is in peas (see Table 14.1 in your text), what will be the ratios of genotypes and phenotypes of the F1 generation resulting from the following cross: axial-red (true-breeding) × terminal-white? What will be the ratios in the F2 generation?
Parental cross is AACRCR × aaCWCW. Genotype of F1 is AaCRCW, phenotype is all axial-pink. Genotypes of F2 are 4 AaCRCW : 2 AaCRCR : 2 AACRCW : 2 aaCRCW : 2 AaCWCW : 1AACRCR : 1 aaCRCR : 1 AACWCW : 1 aaCWCW. Phenotypes of F2 are 6 axial-pink : 3 axial-red : 3 axial-white : 2 terminal-pink : 1 terminal-white : 1 terminal-red.
