ExamView - Genetics · When roan colored cattle are mated, 25% of the offspring are red, 50% are...
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Name: ________________________ Class: ___________________ Date: __________ ID: A 1 Genetics Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____. a. homozygous recessive c. homozygous dominant b. heterozygous d. haploid ____ 2. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F 1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F 2 chicks? a. 100% rose comb b. 75% rose comb and 25% single comb c. 100% single comb d. 50% rose comb and 50% single comb Figure 10-7 ____ 3. What fraction of this cross will be recessive for both traits? a. 1/2 c. 1/8 b. 1/4 d. 1/16 ____ 4. Suppose an animal is heterozygous AaBb. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits? a. 2 c. 6 b. 4 d. 8 ____ 5. A true-breeding tall pea plant is crossed with a true-breeding short pea plant, and all the offspring are tall. What is the most likely genotype of the offspring? a. tt c. TT b. Tt d. TT or tt ____ 6. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white? a. 0 c. 50 b. 25 d. 75
Transcript of ExamView - Genetics · When roan colored cattle are mated, 25% of the offspring are red, 50% are...
Name: ________________________ Class: ___________________ Date: __________ ID: A
1
Genetics
Multiple ChoiceIdentify the choice that best completes the statement or answers the question.
____ 1. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____.a. homozygous recessive c. homozygous dominantb. heterozygous d. haploid
____ 2. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F2 chicks?a. 100% rose combb. 75% rose comb and 25% single combc. 100% single combd. 50% rose comb and 50% single comb
Figure 10-7
____ 3. What fraction of this cross will be recessive for both traits?a. 1/2 c. 1/8b. 1/4 d. 1/16
____ 4. Suppose an animal is heterozygous AaBb. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits?a. 2 c. 6b. 4 d. 8
____ 5. A true-breeding tall pea plant is crossed with a true-breeding short pea plant, and all the offspring are tall. What is the most likely genotype of the offspring?a. tt c. TTb. Tt d. TT or tt
____ 6. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white?a. 0 c. 50b. 25 d. 75
Name: ________________________ ID: A
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____ 7. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022 yellow seeds.Which of these is the most likely estimate of the number of green seeds he collected from the F2 generation?a. 0 c. 6000b. 2000 d. 18000
____ 8. A heterozygous organism is best described as which of these?a. dominant c. hybridb. genotype d. true-breeding
____ 9. Of the following species used in agriculture, which is most likely a polyploid?a. cow c. henb. goat d. wheat
____ 10. Gregor Mendel used pea plants to studya. flowering.b. gamete formation.c. the inheritance of traits.d. cross-pollination.
____ 11. Offspring that result from crosses between true-breeding parents with different traitsa. are true-breeding.b. make up the F2 generation.c. make up the parental generation.d. are called hybrids.
____ 12. The factors that determine traits are calleda. proteins.b. traits.c. genes.d. characters.
____ 13. Gregor Mendel concluded that traits area. not inherited by offspring.b. inherited through the passing of factors from parents to offspring.c. determined by dominant factors only.d. determined by recessive factors only.
____ 14. When Gregor Mendel crossed a tall plant with a short plant, the F1 plants inheriteda. an allele for tallness from each parent.b. an allele for tallness from the tall parent and an allele for shortness from the short parent.c. an allele for shortness from each parent.d. an allele from only the tall parent.
____ 15. The principle of dominance states thata. all alleles are dominant.b. all alleles are recessive.c. some alleles are dominant and others are recessive.d. alleles are neither dominant nor recessive.
Name: ________________________ ID: A
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____ 16. A tall plant is crossed with a short plant. If the tall F1 pea plants are allowed to self-pollinate,a. the offspring will be of medium height.b. all of the offspring will be tall.c. all of the offspring will be short.d. some of the offspring will be tall, and some will be short.
____ 17. The principles of probability can be used toa. predict the traits of the offspring produced by genetic crosses.b. determine the actual outcomes of genetic crosses.c. predict the traits of the parents used in genetic crosses.d. decide which organisms are best to use in genetic crosses.
____ 18. Organisms that have two identical alleles for a particular trait are said to bea. hybrid.b. homozygous.c. heterozygous.d. dominant.
Tt
T t
TT
T TT Tt
T TT Tt
T = tall
t = short
Figure 11-1
____ 19. In the Punnett square shown in Figure 11-1, which of the following is true about the offspring resulting from the cross?a. About half are expected to be short.b. All are expected to be short.c. About half are expected to be tall.d. All are expected to be tall.
Name: ________________________ ID: A
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____ 20. A Punnett square shows all of the following EXCEPTa. all possible results of a genetic cross.b. the genotypes of the offspring.c. the alleles in the gametes of each parent.d. the actual results of a genetic cross.
____ 21. What principle states that during gamete formation genes for different traits separate without influencing each other’s inheritance?a. principle of dominanceb. principle of independent assortmentc. principle of probabilitiesd. principle of segregation
____ 22. If a pea plant that is heterozygous for round, yellow peas (RrYy) is crossed with a pea plant that is homozygous for round peas but heterozygous for yellow peas (RRYy), how many different phenotypes are their offspring expected to show? Draw a dihybrid punnett square if needed.a. 2b. 4c. 8d. 16
____ 23. Situations in which one allele for a gene is not completely dominant over another allele for that gene are calleda. multiple alleles.b. incomplete dominance.c. polygenic inheritance.d. multiple genes.
____ 24. A cross of a red cow (RR) with a white bull (WW) produces all roan colored offspring (RW). (Roan color is a mixture of red and white hair) This type of inheritance is known asa. incomplete dominance.b. polygenic inheritance.c. codominance.d. multiple alleles.
____ 25. Variation in human skin color is a result ofa. incomplete dominance.b. codominance.c. polygenic traits.d. multiple alleles.
____ 26. Gregor Mendel’s principles of genetics apply toa. plants only.b. animals only.c. pea plants only.d. all organisms.
____ 27. Linked genesa. are never separated.b. assort independently.c. are on the same chromosome.d. are always recessive.
Name: ________________________ ID: A
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Figure 11-1
____ 28. What type of inheritance pattern does the trait represented by the shaded symbols in Figure 11-1 illustrate?a. incomplete dominance c. codominanceb. multiple alleles d. sex-linked
____ 29. For the trait being followed in the pedigree, individuals II-1 and II-4 in Figure 11-1 can be classified as ____.a. homozygous dominant c. homozygous recessiveb. mutants d. carriers
____ 30. What is the relationship between individual I-1 and individual III-2 in Figure 11-1?a. grandfather-granddaughter c. great aunt-nephewb. grandmother-grandson d. mother-son
____ 31. If a female fruit fly heterozygous for red eyes (XRXr) crossed with a white-eyed male (XrY), what percent of their offspring would have white eyes?a. 0% c. 50%b. 25% d. 75%
____ 32. When roan colored cattle are mated, 25% of the offspring are red, 50% are roan colored, and 25% are white. Upon examination, it can be seen that the coat of a roan colored cow consists of both red and white hairs. This trait is one controlled by ____.a. multiple alleles c. sex-linked genesb. codominant alleles d. polygenic inheritance
____ 33. A phenotype that results from a dominant allele must have at least _____ dominant allele(s) present in the parent(s).a. one c. threeb. two d. four
Name: ________________________ ID: A
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____ 34. Examine the graph in Figure 11-3, which illustrates the frequency in types of skin pigmentation in humans. Another human trait that would show a similar inheritance pattern and frequency of distribution is ____.
Figure 11-3a. height c. number of fingers and toesb. blood type d. incidence of cystic fibrosis
____ 35. A man heterozygous for blood type A marries a woman heterozygous for blood type B. The chance that their first child will have type O blood is ____.a. 0% c. 50%b. 25% d. 75%
____ 36. What phenotype is depicted in Figure 11-5?
Figure 11-5
a. O c. Ab. AB d. B
____ 37. Nondisjunction is related to a number of serious human disorders. How does nondisjunction cause these disorders? a. alters the number of gametes producedb. alters the number of zygotes producedc. alters the chromosome structured. alters the chromosome number
Name: ________________________ ID: A
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____ 38. What occurs during the process of meiosis in humans that can lead to a child with the condition of Down Syndrome?a. inhertinece of a complete set of chromosomesb. production of gametes which are diploidc. inheritence of an extra chromosome 21d. production of gametes with one duplicate sex chromosome
____ 39. A pea plant homozygous for the trait of smooth seeds is crossed with a pea plant that is homozygous for the trait of wrinkled seeds. The first generation produces seeds that are all smooth. What percent of the second-generation plants will have smooth seeds when the F1 generation is self-fertilized? a. 100% c. 50%b. 75% d. 25%
____ 40. A man with a certain syndrome marries a woman who is normal for that trait. They have 6 children, three girls and three boys. All of the girls have the same syndrome as the father whereas none of the boys is affected. Which type of heredity is not possible here? a. Y-linked c. X-linked dominantb. X-linked recessive d. autosomal recessive
ID: A
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GeneticsAnswer Section
MULTIPLE CHOICE
1. ANS: AThe most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it would likely produce white offspring when mated with a brown mouse.
FeedbackA Correct.B If the white mouse is heterozygous, then this would mean white is dominant and its
brown mate is homozygous recessive. Such a cross would yield about half white and brown offspring.
C If the white mouse were homozygous dominant, then its offspring would have to be white.
D Sperm and eggs are haploid. Individual mice are not.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
2. ANS: BThis is the classic situation in which the F1 generation (all heterozygous) are crossed to produce offspring in a 3 to 1 ratio of dominant to recessive.
FeedbackA Remember that the F1 individuals are heterogygous.B Well done.C Remember that the F1 individuals are heterogygous.D About 3/4 of the offspring will have rose comb.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
3. ANS: DThis is a classic F1 cross for two independent traits. The ratio is 9:3:3:1. Of the sixteen possible combinations, only one or 1/16 has a genotype with completely recessive alleles.
FeedbackA Check page 282.B See page 282.C Only one in 16 are expected to have both recessive traits.D Well done.
PTS: 1 DIF: Bloom's Level C REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
ID: A
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4. ANS: BThere are four combinations: AB, Ab, aB, and ab.
FeedbackA Remember the traits are not linked.B Correct.C Check page 275.D Find all possible combinations of A with B and b and a with B and b.
PTS: 1 DIF: Bloom's Level D REF: 275NAT: LS_2b STA: SB2.c TOP: 10-5
5. ANS: BSince the parents are true breeding, they are most likely homozygous (TT and tt). This means the offspring are most likely heterozygous, Tt.
FeedbackA See page 280.B Correct.C But one parent was short.D If all offspring are tall, then they cannot be TT and tt.
PTS: 1 DIF: Bloom's Level D REF: 280NAT: LS_2b STA: SB2.c TOP: 10-6
6. ANS: CThe original parents must be heterozygous (Bb) since they produced both black and white offspring. The white offspring must be homozygous recessive (bb). If the white offspring is crossed with its parent then half the offspring will be white.
FeedbackA The black parent is heterozygous.B The parents of this cross are Bb and bb.C Correct.D Review pages 277–281.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
ID: A
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7. ANS: BIn the F2 generation, we expect a 3 to 1 ratio of yellow to green seeds. Thus, if there are approximately 6000 yellow seeds, then we expect about 2000 green seeds.
FeedbackA Check page 278.B Correct.C Try again.D There should be 3 times more yellow seeds than green seeds.
PTS: 1 DIF: Bloom's Level D REF: 278NAT: LS_2b STA: SB2.c TOP: 10-4
8. ANS: CA heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa) individuals.Thus the heterozgous individual is a hybrid.
FeedbackA This description refers to a trait or allele.B This description refers to the genetic makeup of an individual.C Correct.D This refers to a homozygous individual.
PTS: 1 DIF: Bloom's Level B REF: 279NAT: LS_2b STA: SB2.c TOP: 10-5
9. ANS: DPlants are often polyploids, while animals are not.
FeedbackA Animals are not usually polyploids.B Polyploid animals are not usually robust.C Check page 284.D Correct.
PTS: 1 DIF: Bloom's Level B REF: 284NAT: LS_2c TOP: 10-9
10. ANS: C PTS: 1 DIF: B OBJ: 11.1.1 11. ANS: D PTS: 1 DIF: A OBJ: 11.1.1 12. ANS: C PTS: 1 DIF: B OBJ: 11.1.2 13. ANS: B PTS: 1 DIF: A OBJ: 11.1.2 14. ANS: B PTS: 1 DIF: E OBJ: 11.1.2 15. ANS: C PTS: 1 DIF: B OBJ: 11.1.3 16. ANS: D PTS: 1 DIF: B OBJ: 11.1.4 17. ANS: A PTS: 1 DIF: A OBJ: 11.2.1 18. ANS: B PTS: 1 DIF: B OBJ: 11.2.2 19. ANS: D PTS: 1 DIF: E OBJ: 11.2.2 20. ANS: D PTS: 1 DIF: A OBJ: 11.2.2
ID: A
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21. ANS: B PTS: 1 DIF: B OBJ: 11.3.1 22. ANS: A PTS: 1 DIF: E OBJ: 11.3.1 23. ANS: B PTS: 1 DIF: B OBJ: 11.3.2 24. ANS: C PTS: 1 DIF: A OBJ: 11.3.2 25. ANS: C PTS: 1 DIF: E OBJ: 11.3.2 26. ANS: D PTS: 1 DIF: B OBJ: 11.3.3 27. ANS: C PTS: 1 DIF: A OBJ: 11.5.1 28. ANS: D
FeedbackA You are on the right track.B Please consider again all the factors.C Please refer to pages 307-308.D You are correct.
PTS: 1 DIF: Bloom's Level D REF: 299–300 | 307–308NAT: LS_2b STA: SB2.c TOP: 11-5
29. ANS: D
FeedbackA Please consider again the reference to the individual, not the allele.B Please consider the factors again.C Please refer to pages 299-300.D You are correct.
PTS: 1 DIF: Bloom's Level B REF: 299–300NAT: LS_2b STA: SB2.c TOP: 11-3
30. ANS: B
FeedbackA You are on the right track.B You are right.C Please consider again the symbols in creating a pedigree.D Please refer to page 299.
PTS: 1 DIF: Bloom's Level B REF: 299STA: SB2.c TOP: 11-3
ID: A
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31. ANS: C
FeedbackA Please consider all the factors.B You are on the right track.C You are correct.D Please refer to pages 296-300.
PTS: 1 DIF: Bloom's Level C REF: 296–300NAT: LS_2b STA: SB2.c TOP: 11-1
32. ANS: B
FeedbackA You are on the right track.B You are right!C Please consider again the inheritance pattern.D You are heading in the right direction.
PTS: 1 DIF: Bloom's Level C REF: 299–300NAT: LS_2b STA: SB2.c TOP: 11-4
33. ANS: A
FeedbackA You are correct!.B Think about the effect of a dominant allele on the phenotype.C Try working a cross.D Review the concept of a dominant allele.
PTS: 1 DIF: Bloom's Level B REF: 299–300NAT: LS_2b TOP: 11-1
34. ANS: A
FeedbackA You are correct.B Please review the inheritance pattern of blood types.C Please consider again the frequency of people in your class who have the same number
of fingers and toes.D Please consider the rarity of this disease in the population.
PTS: 1 DIF: Bloom's Level D REF: 309TOP: 11-4
ID: A
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35. ANS: B
FeedbackA Be careful to distinguish between genotype and phenotype.B You are correct.C Try writing out the cross.D Review the information on human blood types.
PTS: 1 DIF: Bloom's Level D REF: 304NAT: LS_2b STA: SB2.c TOP: 11-4
36. ANS: D
FeedbackA Please refer to page 304.B That's a red blood cell in the diagram.C Please review page 204 again.D You are correct!
PTS: 1 DIF: Bloom's Level A REF: 304STA: SB2.c TOP: 11-4
37. ANS: DNondisjunction occurs during cell division in which the homologous chromosomes fail to separate properly in meiosis I or in which the sister chromatids fail to separate in meiosis II. This results in gametes with an inaccurate number of chromosomes.
FeedbackA Please review the process of meosis.B Please review the formation of gametes.C You are on the right track.D You are correct!
PTS: 1 DIF: Bloom's Level C REF: 312–314NAT: LS_2c STA: SB2.c TOP: 11-9
38. ANS: CNondisjunction of chromosome 21 occurs during meiosis I, resulting in gametes with two copies of that chromosome.
FeedbackA You are on the right track.B This would produce a polyploid zygote.C You are correct.D You are on the right track.
PTS: 1 DIF: Bloom's Level C REF: 312–314NAT: LS_2c STA: SB2.c TOP: 11-9
ID: A
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39. ANS: BSmooth is the dominant allele. The first generation would be heterozygous, Rr. In the second generation, 75% would have the genotype Rr or RR, or smooth seeds, whereas 25% would have the recessive genotype, rr, and be wrinkled.
FeedbackA Did you consider all the factors?B You are correct.C Please refer to page 301.D Please review dominant and recessive inheritance patterns.
PTS: 1 DIF: Bloom's Level D REF: 301NAT: LS_2b STA: SB2.c TOP: 11-1
40. ANS: AAll modes of inheritance are possible except Y linkage, which is not possible because female offspring are affected.
FeedbackA That's right!B Draw the pedigree and assign possible genotypes.C Draw the pedigree and assign possible genotypes.D Draw the pedigree and assign possible genotypes.
PTS: 1 DIF: Bloom's Level F REF: 305–308NAT: LS_2b STA: SB2.c TOP: 11-5
Name: ________________________ Class: ___________________ Date: __________ ID: B
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Genetics
Multiple ChoiceIdentify the choice that best completes the statement or answers the question.
____ 1. Variation in human skin color is a result ofa. incomplete dominance.b. multiple alleles.c. codominance.d. polygenic traits.
____ 2. Offspring that result from crosses between true-breeding parents with different traitsa. are true-breeding.b. are called hybrids.c. make up the parental generation.d. make up the F2 generation.
____ 3. Gregor Mendel concluded that traits area. determined by recessive factors only.b. not inherited by offspring.c. determined by dominant factors only.d. inherited through the passing of factors from parents to offspring.
Name: ________________________ ID: B
2
Tt
T t
TT
T TT Tt
T TT Tt
T = tall
t = short
Figure 11-1
____ 4. In the Punnett square shown in Figure 11-1, which of the following is true about the offspring resulting from the cross?a. About half are expected to be tall.b. All are expected to be short.c. About half are expected to be short.d. All are expected to be tall.
____ 5. The principles of probability can be used toa. decide which organisms are best to use in genetic crosses.b. predict the traits of the parents used in genetic crosses.c. predict the traits of the offspring produced by genetic crosses.d. determine the actual outcomes of genetic crosses.
____ 6. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022 yellow seeds.Which of these is the most likely estimate of the number of green seeds he collected from the F2 generation?a. 18000 c. 6000b. 0 d. 2000
Name: ________________________ ID: B
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Figure 11-1
____ 7. What is the relationship between individual I-1 and individual III-2 in Figure 11-1?a. great aunt-nephew c. grandfather-granddaughterb. grandmother-grandson d. mother-son
____ 8. What type of inheritance pattern does the trait represented by the shaded symbols in Figure 11-1 illustrate?a. incomplete dominance c. sex-linkedb. codominance d. multiple alleles
____ 9. For the trait being followed in the pedigree, individuals II-1 and II-4 in Figure 11-1 can be classified as ____.a. homozygous recessive c. mutantsb. homozygous dominant d. carriers
____ 10. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F2 chicks?a. 100% rose combb. 75% rose comb and 25% single combc. 100% single combd. 50% rose comb and 50% single comb
____ 11. A phenotype that results from a dominant allele must have at least _____ dominant allele(s) present in the parent(s).a. one c. fourb. two d. three
Name: ________________________ ID: B
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____ 12. What phenotype is depicted in Figure 11-5?
Figure 11-5
a. B c. Ob. AB d. A
____ 13. Examine the graph in Figure 11-3, which illustrates the frequency in types of skin pigmentation in humans. Another human trait that would show a similar inheritance pattern and frequency of distribution is ____.
Figure 11-3a. height c. blood typeb. incidence of cystic fibrosis d. number of fingers and toes
____ 14. A true-breeding tall pea plant is crossed with a true-breeding short pea plant, and all the offspring are tall. What is the most likely genotype of the offspring?a. TT or tt c. TTb. Tt d. tt
____ 15. A cross of a red cow (RR) with a white bull (WW) produces all roan colored offspring (RW). (Roan color is a mixture of red and white hair) This type of inheritance is known asa. incomplete dominance.b. multiple alleles.c. polygenic inheritance.d. codominance.
____ 16. A man heterozygous for blood type A marries a woman heterozygous for blood type B. The chance that their first child will have type O blood is ____.a. 75% c. 0%b. 25% d. 50%
Name: ________________________ ID: B
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____ 17. If a pea plant that is heterozygous for round, yellow peas (RrYy) is crossed with a pea plant that is homozygous for round peas but heterozygous for yellow peas (RRYy), how many different phenotypes are their offspring expected to show? Draw a dihybrid punnett square if needed.a. 16b. 4c. 2d. 8
____ 18. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white?a. 50 c. 0b. 75 d. 25
____ 19. A pea plant homozygous for the trait of smooth seeds is crossed with a pea plant that is homozygous for the trait of wrinkled seeds. The first generation produces seeds that are all smooth. What percent of the second-generation plants will have smooth seeds when the F1 generation is self-fertilized? a. 75% c. 25%b. 50% d. 100%
____ 20. What occurs during the process of meiosis in humans that can lead to a child with the condition of Down Syndrome?a. production of gametes which are diploidb. inheritence of an extra chromosome 21c. production of gametes with one duplicate sex chromosomed. inhertinece of a complete set of chromosomes
____ 21. Gregor Mendel’s principles of genetics apply toa. animals only.b. pea plants only.c. plants only.d. all organisms.
____ 22. A Punnett square shows all of the following EXCEPTa. all possible results of a genetic cross.b. the genotypes of the offspring.c. the actual results of a genetic cross.d. the alleles in the gametes of each parent.
____ 23. If a female fruit fly heterozygous for red eyes (XRXr) crossed with a white-eyed male (XrY), what percent of their offspring would have white eyes?a. 0% c. 50%b. 25% d. 75%
____ 24. The principle of dominance states thata. all alleles are recessive.b. some alleles are dominant and others are recessive.c. all alleles are dominant.d. alleles are neither dominant nor recessive.
Name: ________________________ ID: B
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____ 25. What principle states that during gamete formation genes for different traits separate without influencing each other’s inheritance?a. principle of segregationb. principle of independent assortmentc. principle of probabilitiesd. principle of dominance
____ 26. A tall plant is crossed with a short plant. If the tall F1 pea plants are allowed to self-pollinate,a. all of the offspring will be short.b. some of the offspring will be tall, and some will be short.c. all of the offspring will be tall.d. the offspring will be of medium height.
____ 27. Linked genesa. are never separated.b. assort independently.c. are on the same chromosome.d. are always recessive.
____ 28. When Gregor Mendel crossed a tall plant with a short plant, the F1 plants inheriteda. an allele for tallness from each parent.b. an allele for tallness from the tall parent and an allele for shortness from the short parent.c. an allele for shortness from each parent.d. an allele from only the tall parent.
____ 29. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____.a. haploid c. homozygous recessiveb. heterozygous d. homozygous dominant
Figure 10-7
____ 30. What fraction of this cross will be recessive for both traits?a. 1/2 c. 1/4b. 1/16 d. 1/8
____ 31. Nondisjunction is related to a number of serious human disorders. How does nondisjunction cause these disorders? a. alters the chromosome structureb. alters the number of zygotes producedc. alters the number of gametes producedd. alters the chromosome number
Name: ________________________ ID: B
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____ 32. When roan colored cattle are mated, 25% of the offspring are red, 50% are roan colored, and 25% are white. Upon examination, it can be seen that the coat of a roan colored cow consists of both red and white hairs. This trait is one controlled by ____.a. sex-linked genes c. multiple allelesb. polygenic inheritance d. codominant alleles
____ 33. The factors that determine traits are calleda. proteins.b. genes.c. traits.d. characters.
____ 34. Of the following species used in agriculture, which is most likely a polyploid?a. cow c. goatb. wheat d. hen
____ 35. Suppose an animal is heterozygous AaBb. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits?a. 2 c. 4b. 6 d. 8
____ 36. A heterozygous organism is best described as which of these?a. true-breeding c. genotypeb. hybrid d. dominant
____ 37. Organisms that have two identical alleles for a particular trait are said to bea. dominant.b. heterozygous.c. hybrid.d. homozygous.
____ 38. A man with a certain syndrome marries a woman who is normal for that trait. They have 6 children, three girls and three boys. All of the girls have the same syndrome as the father whereas none of the boys is affected. Which type of heredity is not possible here? a. X-linked dominant c. autosomal recessiveb. Y-linked d. X-linked recessive
____ 39. Situations in which one allele for a gene is not completely dominant over another allele for that gene are calleda. multiple genes.b. polygenic inheritance.c. incomplete dominance.d. multiple alleles.
____ 40. Gregor Mendel used pea plants to studya. flowering.b. gamete formation.c. the inheritance of traits.d. cross-pollination.
ID: B
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GeneticsAnswer Section
MULTIPLE CHOICE
1. ANS: D PTS: 1 DIF: E OBJ: 11.3.2 2. ANS: B PTS: 1 DIF: A OBJ: 11.1.1 3. ANS: D PTS: 1 DIF: A OBJ: 11.1.2 4. ANS: D PTS: 1 DIF: E OBJ: 11.2.2 5. ANS: C PTS: 1 DIF: A OBJ: 11.2.1 6. ANS: D
In the F2 generation, we expect a 3 to 1 ratio of yellow to green seeds. Thus, if there are approximately 6000 yellow seeds, then we expect about 2000 green seeds.
FeedbackA There should be 3 times more yellow seeds than green seeds.B Check page 278.C Try again.D Correct.
PTS: 1 DIF: Bloom's Level D REF: 278NAT: LS_2b STA: SB2.c TOP: 10-4
7. ANS: B
FeedbackA Please consider again the symbols in creating a pedigree.B You are right.C You are on the right track.D Please refer to page 299.
PTS: 1 DIF: Bloom's Level B REF: 299STA: SB2.c TOP: 11-3
8. ANS: C
FeedbackA You are on the right track.B Please refer to pages 307-308.C You are correct.D Please consider again all the factors.
PTS: 1 DIF: Bloom's Level D REF: 299–300 | 307–308NAT: LS_2b STA: SB2.c TOP: 11-5
ID: B
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9. ANS: D
FeedbackA Please refer to pages 299-300.B Please consider again the reference to the individual, not the allele.C Please consider the factors again.D You are correct.
PTS: 1 DIF: Bloom's Level B REF: 299–300NAT: LS_2b STA: SB2.c TOP: 11-3
10. ANS: BThis is the classic situation in which the F1 generation (all heterozygous) are crossed to produce offspring in a 3 to 1 ratio of dominant to recessive.
FeedbackA Remember that the F1 individuals are heterogygous.B Well done.C Remember that the F1 individuals are heterogygous.D About 3/4 of the offspring will have rose comb.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
11. ANS: A
FeedbackA You are correct!.B Think about the effect of a dominant allele on the phenotype.C Review the concept of a dominant allele.D Try working a cross.
PTS: 1 DIF: Bloom's Level B REF: 299–300NAT: LS_2b TOP: 11-1
12. ANS: A
FeedbackA You are correct!B That's a red blood cell in the diagram.C Please refer to page 304.D Please review page 204 again.
PTS: 1 DIF: Bloom's Level A REF: 304STA: SB2.c TOP: 11-4
ID: B
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13. ANS: A
FeedbackA You are correct.B Please consider the rarity of this disease in the population.C Please review the inheritance pattern of blood types.D Please consider again the frequency of people in your class who have the same number
of fingers and toes.
PTS: 1 DIF: Bloom's Level D REF: 309TOP: 11-4
14. ANS: BSince the parents are true breeding, they are most likely homozygous (TT and tt). This means the offspring are most likely heterozygous, Tt.
FeedbackA If all offspring are tall, then they cannot be TT and tt.B Correct.C But one parent was short.D See page 280.
PTS: 1 DIF: Bloom's Level D REF: 280NAT: LS_2b STA: SB2.c TOP: 10-6
15. ANS: D PTS: 1 DIF: A OBJ: 11.3.2 16. ANS: B
FeedbackA Review the information on human blood types.B You are correct.C Be careful to distinguish between genotype and phenotype.D Try writing out the cross.
PTS: 1 DIF: Bloom's Level D REF: 304NAT: LS_2b STA: SB2.c TOP: 11-4
17. ANS: C PTS: 1 DIF: E OBJ: 11.3.1
ID: B
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18. ANS: AThe original parents must be heterozygous (Bb) since they produced both black and white offspring. The white offspring must be homozygous recessive (bb). If the white offspring is crossed with its parent then half the offspring will be white.
FeedbackA Correct.B Review pages 277–281.C The black parent is heterozygous.D The parents of this cross are Bb and bb.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
19. ANS: ASmooth is the dominant allele. The first generation would be heterozygous, Rr. In the second generation, 75% would have the genotype Rr or RR, or smooth seeds, whereas 25% would have the recessive genotype, rr, and be wrinkled.
FeedbackA You are correct.B Please refer to page 301.C Please review dominant and recessive inheritance patterns.D Did you consider all the factors?
PTS: 1 DIF: Bloom's Level D REF: 301NAT: LS_2b STA: SB2.c TOP: 11-1
20. ANS: BNondisjunction of chromosome 21 occurs during meiosis I, resulting in gametes with two copies of that chromosome.
FeedbackA This would produce a polyploid zygote.B You are correct.C You are on the right track.D You are on the right track.
PTS: 1 DIF: Bloom's Level C REF: 312–314NAT: LS_2c STA: SB2.c TOP: 11-9
21. ANS: D PTS: 1 DIF: B OBJ: 11.3.3 22. ANS: C PTS: 1 DIF: A OBJ: 11.2.2
ID: B
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23. ANS: C
FeedbackA Please consider all the factors.B You are on the right track.C You are correct.D Please refer to pages 296-300.
PTS: 1 DIF: Bloom's Level C REF: 296–300NAT: LS_2b STA: SB2.c TOP: 11-1
24. ANS: B PTS: 1 DIF: B OBJ: 11.1.3 25. ANS: B PTS: 1 DIF: B OBJ: 11.3.1 26. ANS: B PTS: 1 DIF: B OBJ: 11.1.4 27. ANS: C PTS: 1 DIF: A OBJ: 11.5.1 28. ANS: B PTS: 1 DIF: E OBJ: 11.1.2 29. ANS: C
The most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it would likely produce white offspring when mated with a brown mouse.
FeedbackA Sperm and eggs are haploid. Individual mice are not.B If the white mouse is heterozygous, then this would mean white is dominant and its
brown mate is homozygous recessive. Such a cross would yield about half white and brown offspring.
C Correct.D If the white mouse were homozygous dominant, then its offspring would have to be
white.
PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
30. ANS: BThis is a classic F1 cross for two independent traits. The ratio is 9:3:3:1. Of the sixteen possible combinations, only one or 1/16 has a genotype with completely recessive alleles.
FeedbackA Check page 282.B Well done.C See page 282.D Only one in 16 are expected to have both recessive traits.
PTS: 1 DIF: Bloom's Level C REF: 277–281NAT: LS_2b STA: SB2.c TOP: 10-6
ID: B
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31. ANS: DNondisjunction occurs during cell division in which the homologous chromosomes fail to separate properly in meiosis I or in which the sister chromatids fail to separate in meiosis II. This results in gametes with an inaccurate number of chromosomes.
FeedbackA You are on the right track.B Please review the formation of gametes.C Please review the process of meosis.D You are correct!
PTS: 1 DIF: Bloom's Level C REF: 312–314NAT: LS_2c STA: SB2.c TOP: 11-9
32. ANS: D
FeedbackA Please consider again the inheritance pattern.B You are heading in the right direction.C You are on the right track.D You are right!
PTS: 1 DIF: Bloom's Level C REF: 299–300NAT: LS_2b STA: SB2.c TOP: 11-4
33. ANS: B PTS: 1 DIF: B OBJ: 11.1.2 34. ANS: B
Plants are often polyploids, while animals are not.
FeedbackA Animals are not usually polyploids.B Correct.C Polyploid animals are not usually robust.D Check page 284.
PTS: 1 DIF: Bloom's Level B REF: 284NAT: LS_2c TOP: 10-9
35. ANS: CThere are four combinations: AB, Ab, aB, and ab.
FeedbackA Remember the traits are not linked.B Check page 275.C Correct.D Find all possible combinations of A with B and b and a with B and b.
PTS: 1 DIF: Bloom's Level D REF: 275NAT: LS_2b STA: SB2.c TOP: 10-5
ID: B
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36. ANS: BA heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa) individuals.Thus the heterozgous individual is a hybrid.
FeedbackA This refers to a homozygous individual.B Correct.C This description refers to the genetic makeup of an individual.D This description refers to a trait or allele.
PTS: 1 DIF: Bloom's Level B REF: 279NAT: LS_2b STA: SB2.c TOP: 10-5
37. ANS: D PTS: 1 DIF: B OBJ: 11.2.2 38. ANS: B
All modes of inheritance are possible except Y linkage, which is not possible because female offspring are affected.
FeedbackA Draw the pedigree and assign possible genotypes.B That's right!C Draw the pedigree and assign possible genotypes.D Draw the pedigree and assign possible genotypes.
PTS: 1 DIF: Bloom's Level F REF: 305–308NAT: LS_2b STA: SB2.c TOP: 11-5
39. ANS: C PTS: 1 DIF: B OBJ: 11.3.2 40. ANS: C PTS: 1 DIF: B OBJ: 11.1.1
