Genetic Analysis 2nd Edition by Sanders – Test Bank

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Genetic Analysis 2nd Edition by Sanders – Test Bank

 

Sample  Questions

 

Genetics: An Integrated Approach (Sanders)

Chapter 2   Transmission Genetics

2.1   Multiple-Choice Questions

1)    Mendel performed many types of crosses, including those in which the same genotypes are crossed but the sexes of the parents are switched. These are known as __________.

  1. A) replicate crosses
  2. B) reciprocal crosses
  3. C) test crosses
  4. D) monohybrid crosses
  5. E) dihybrid crosses

Answer:  B

Section:  2.1

Skill:  Knowledge/Comprehension

2)    Crosses in which F1 plants heterozygous for a given allele are crossed to generate a 3:1 phenotypic ratio in the F2 generation are known as __________.

  1. A) replicate crosses
  2. B) reciprocal crosses
  3. C) test crosses
  4. D) monohybrid crosses
  5. E) dihybrid crosses

Answer:  D

Section:  2.2

Skill:  Knowledge/Comprehension

3)    In peas, the round allele is dominant over the wrinkled allele. A plant with round peas was crossed to a plant with wrinkled peas and all of the resulting plants had round peas. What is the genotype of the parents in this cross?

  1. A) RR × rr
  2. B) RR × Rr
  3. C) Rr × rr
  4. D) Rr × Rr
  5. E) rr × rr

Answer:  A

Section:  2.2

Skill:  Application/Analysis

4)    The blending theory would predict that the phenotype seen in the F1 generation from a cross between a pure breeding plant with dark purple flowers and a pure breeding plant with white flowers would be __________.

  1. A) dark purple
  2. B) light purple
  3. C) white
  4. D) a 3:1 ratio of purple to white flowers
  5. E) a 1:1 ratio of purple to white flowers

Answer:  B

Section:  2.1

Skill:  Application/Analysis

5)    In peas, the yellow allele is dominant over the green allele. A plant with yellow peas was crossed to a plant with green peas. The resulting plants were 50% yellow and 50% green. What is the genotype of the parents in this cross?

  1. A) YY × yy
  2. B) YY × Yy
  3. C) Yy × yy
  4. D) Yy × Yy
  5. E) yy × yy

Answer:  C

Section:  2.2

Skill:  Application/Analysis

6)    Assuming independent assortment, what phenotypic ratio would you expect to see if an individual with the genotype RrGg is self-crossed?

  1. A) 1:3
  2. B) 9:3:3:1
  3. C) 1:2:1
  4. D) 1:3:2:1
  5. E) 3:1

Answer:  B

Section:  2.3

Skill:  Application/Analysis

7)    What genotypic ratio would you expect to see among the progeny of a monohybrid cross?

  1. A) 1:3
  2. B) 9:3:3:1
  3. C) 1:2:1
  4. D) 1:3:2:1
  5. E) 3:1

Answer:  C

Section:  2.2

Skill:  Application/Analysis

8)    A couple has four children. What is the probability that they have four boys?

  1. A) 1/2
  2. B) 1/4
  3. C) 1/8
  4. D) 1/16
  5. E) 1/32

 

Answer:  D

Section:  2.4

Skill:  Application/Analysis

9)    Humans have a gene, T, that is involved in muscle formation of the tongue. Individuals homozygous for one allele can roll their tongues, while individuals homozygous for the other allele cannot. If both parents can roll their tongues, but their child cannot, what can be said about the mode of inheritance?

  1. A) Tongue rolling is dominant.
  2. B) Tongue rolling is recessive.
  3. C) The parents were both homozygous, but the child was heterozygous.
  4. D) Tongue rolling is dominant, and both parents were heterozygous (Tt).
  5. E) Tongue rolling is recessive, and both parents were heterozygous (Tt).

Answer:  D

Section:  2.6

Skill:  Synthesis/Evaluation

10)  In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would show the dominant phenotype for all three traits?

  1. A) 3/64
  2. B) 9/64
  3. C) 27/64
  4. D) 32/64
  5. E) 64/64

Answer:  C

Section:  2.3

Skill:  Application/Analysis

11)  In peas, axial (A) flower position is dominant to terminal (a), and tall (L) is dominant to short (l).
If a plant that is heterozygous for both traits is allowed to self-fertilize, how many of the offspring would also be heterozygous for both traits?

  1. A) 9/16
  2. B) 1/4
  3. C) 3/16
  4. D) 1/8
  5. E) 1/16

Answer:  B

Section:  2.3

Skill:  Application/Analysis

12)  The law of segregation would predict that the F2 progeny of F1 heterozygous plants will exhibit a __________.

  1. A) 3:1 phenotypic ratio
  2. B) 9:3:3:1 phenotypic ratio
  3. C) 1:2:1 genotypic ratio

 

  1. D) 9:3:3:1 phenotypic ratio and 1:2:1 genotypic ratio
  2. E) 3:1 phenotypic ratio and 1:2:1 genotypic ratio

Answer:  E

Section:  2.2

Skill:  Knowledge/Comprehension

13)  The law of independent assortment would predict that the F2 progeny of F1 heterozygous plants will exhibit a __________.

  1. A) 3:1 phenotypic ratio
  2. B) 9:3:3:1 phenotypic ratio
  3. C) 1:2:1 genotypic ratio
  4. D) 9:3:3:1 phenotypic ratio and 1:2:1 genotypic ratio
  5. E) 3:1 phenotypic ratio and 1:2:1 genotypic ratio

Answer:  B

Section:  2.3

Skill:  Knowledge/Comprehension

14)  What phenotypic ratio would you expect as a result of a test cross between a dihybrid organism and one that is homozygous recessive for alleles at two independent loci?

  1. A) 3:1
  2. B) 1:2:1
  3. C) 1:1:1:1
  4. D) 9:3:3:1
  5. E) 9:4:2:1

Answer:  C

Section:  2.3

Skill:  Application/Analysis

15)  How many different types of gametes can be produced by a short plant with yellow, round peas with a heterozygous genotype (YyRrSs)?

  1. A) 3
  2. B) 6
  3. C) 8
  4. D) 10
  5. E) 12

Answer:  C

Section:  2.3

Skill:  Application/Analysis

16)  By convention, when an observed experimental outcome has a probability of occurrence of less than 5% (<0.05), the experimental results are considered to be __________.

  1. A) within normal expected range
  2. B) statistically significant and different from the expected outcome
  3. C) not significant
  4. D) less than one standard deviation from the mean
  5. E) equal to the mean

Answer:  B

Section:  2.5

Skill:  Knowledge/Comprehension

17)  The statistical interpretation of a chi-square value is determined by identifying the __________.

  1. A) mean
  2. B) degrees of freedom
  3. C) average
  4. D) P value
  5. E) joint probability

Answer:  D

Section:  2.5

Skill:  Knowledge/Comprehension

18)  The P value is a quantitative expression of the probability that the results of another experiment of the same size and structure will deviate from expected results as much as or more than by chance. The greater the difference between observed and expected results of an experiment, __________.

  1. A) the lower the χ2 value and the lower the P value
  2. B) the greater the χ2 value and the greater the P value
  3. C) the greater the χ2 value and the lower the P value
  4. D) the lower the χ2 value and the greater the P value
  5. E) the greater the χ2 value; but the P value is unaffected

Answer:  C

Section:  2.5

Skill:  Knowledge/Comprehension

19)  The statistical interpretation of a χ2 value is determined by identifying the P value for each experiment, and the P value is dependent on the number of degrees of freedom (df) in the experiment being examined. For a coin flip experiment, the df value is equal to 1. You perform an experiment to determine how many times out of 100 die rolls that you roll a “1” in a single roll. What would the df value be equal to?

  1. A) 1
  2. B) 2
  3. C) 3
  4. D) 4
  5. E) 5

Answer:  E

Section:  2.5

Skill:  Application/Analysis

20)  The genes responsible for some of the traits that Mendel observed have been recently identified and have helped in determining how molecular variation produces morphologic variation in pea plants. Allelic variation in the Sbe1 gene, which produces starch-branching enzyme 1, is responsible for which trait in peas?

  1. A) round and wrinkled pea shape
  2. B) yellow and green pea color
  3. C) purple and white flowers
  4. D) tall and short plant height
  5. E) axial and terminal flower position

Answer:  A

Section:  2.6

Skill:  Knowledge/Comprehension

21)  In 1997, a gene called Le was discovered by two research groups led by David Martin and Diane Lester. Allelic variation in the Le gene, which controls elongation of the plant stem between branches, is responsible for which trait in peas?

  1. A) inflated and constricted pod shape
  2. B) yellow and green pod color
  3. C) purple and white flowers
  4. D) tall and short plant height
  5. E) axial and terminal flower position

Answer:  D

Section:  2.6

Skill:  Knowledge/Comprehension

22)  The gene L determines hair length in rabbits. The gene B determines hair color. A rabbit with long, black hair is crossed to a rabbit with short, white hair. All the offspring have long, black hair. What are the genotypes of the parents?

  1. A) LLBB × llbb
  2. B) LlBb × LlBb
  3. C) LlBb × llbb
  4. D) Llbb × llBb
  5. E) Impossible to determine from the information given

Answer:  A

Section:  2.3

Skill:  Application/Analysis

23)  In rabbits, long hair and black fur are produced by the dominant alleles L and B, which assort independently. The genotype ll produces short hair and the genotype bb produces white fur. A cross between a male with short, black fur and a female with long, white fur produces four offspring with short, black fur, four offspring with long, white fur, four offspring with short, white fur, and four offspring with long, black fur. What are the genotypes of the parents?

  1. A) llBB × LLbb
  2. B) LlBb × LlBb
  3. C) llBb × Llbb
  4. D) LLBB × llbb
  5. E) Impossible to determine from the information given.

Answer:  C

Section:  2.3

Skill:  Application/Analysis

 

24)  You count 1000 F2 seeds from a monohybrid cross. How many do you expect to display the dominant phenotype?

  1. A) 1000
  2. B) 750
  3. C) 500
  4. D) 250
  5. E) 0

Answer:  B

Section:  2.5

Skill:  Application/Analysis

25)  In the accompanying figure, the chance that individual III-2 is a heterozygous carrier is __________.

  1. A) 0%
  2. B) 25%
  3. C) 50%
  4. D) 75%
  5. E) 100%

Answer:  E

Section:  2.6

Skill:  Application /Analysis

26)  In the accompanying figure, the chance that individual IV-7 is a heterozygous carrier is __________.

  1. A) 1/4
  2. B) 1/3
  3. C) 1/2
  4. D) 2/3
  5. E) 3/4

Answer:  D

Section:  2.6

Skill:  Application/Analysis

2.2   Short-Answer Questions

1)    Mendel performed numerous controlled genetic crosses to obtain strains that consistently produced a single phenotype without variation. What are these strains that consistently produce the same phenotype called?

Answer:  pure-breeding or true-breeding strains

Section:  2.1

Skill:  Knowledge/Comprehension

2)    In a test cross, a pure-breeding plant is crossed with a plant suspected to be heterozygous (Aa). What is the genotype of the pure-breeding plant?

Answer:  aa

Section:  2.1

Skill:  Application/Analysis

3)    Why did Mendel cut off the nascent anthers during the process of artificial cross-fertilization?

Answer:  to prevent self-fertilization or to prevent uncontrolled crosses

Section:  2.2

Skill:  Knowledge/Comprehension

4)    What simple type of cross that investigates the inheritance of only one trait could be used to illustrate Mendel’s law of segregation?

Answer:  monohybrid cross

Section:  2.2

Skill:  Knowledge/Comprehension

5)    If an affected individual is born to parents who are unaffected, what is the likely mode of inheritance?

Answer:  autosomal recessive

Section:  2.6

Skill:  Knowledge/Comprehension

6)    What type of cross would be used to illustrate Mendel’s law of independent assortment?

Answer:  dihybrid cross (or test cross)

Section:  2.3

Skill:  Knowledge/Comprehension

 

7)    A cross between a short pea plant and a tall pea plant results in a 1:1 genotypic and phenotypic ratio in the offspring. What are the genotypes of the parent plants?

Answer:  Ss × ss (heterozygous × homozygous recessive)

Section:  2.2

Skill:  Application/Analysis

8)    What is the probability of rolling one six-sided die and obtaining a 1 or a 2?

Answer:  1/6 + 1/6 = 2/6 = 1/3

Section:  2.4

Skill:  Application/Analysis

9)    What is the probability of rolling one six-sided die and obtaining any number but 6?

Answer:  1 – 1/6 = 5/6

Section:  2.4

Skill:  Application/Analysis

10)  What is the probability of rolling two six-sided dice and obtaining two 4’s?

Answer:  1/6 × 1/6 = 1/36

Section:  2.4

Skill:  Application/Analysis

11)  What is the probability of rolling two six-sided dice and obtaining at least one 3?

Answer:  Probability of die 1 being a 3 and die 2 not: 1/6 × 5/6 = 5/36

Probability of die 2 being a 3 and die 1 not: 1/6 × 5/6 = 5/36

Probability of die 1 and 2 being a 3: 1/6 × 1/6 = 1/36

Probability of any of these possibilities = addition rule: 5/36 + 5/36 + 1/6 = 11/36

Section:  2.4

Skill:  Application/Analysis

12)  What is the probability of rolling two six-sided dice and obtaining an odd number on at least one die?

Answer:  9/36 + 9/36 + 9/36 = 27/36 = 3/4

Probability of rolling odd number the first die only = 3/6 (odd) × 3/6 (even) = 9/36

Probability of rolling odd number the second die only = 3/6 (even) × 3/6 (odd) = 9/36

Probability of rolling odd number both dice = 3/6 (odd) × 3/6 (odd) = 9/36

Probability of any one of these three possible scenarios = addition rule

Section:  2.4

Skill:  Application/Analysis

13)  When calculating the probability of a given genotype in a trihybrid cross, you can generate a Punnett square. Which of the rules of probability can be used to calculate the joint probability of simultaneous inheritance of multiple alleles?

Answer:  the product rule

Section:  2.4

Skill:  Synthesis/Evaluation

14)  In a cross between individuals who are both heterozygous (carriers) for a recessive disease such as albinism, you would like to determine the risk of one or more children to inherit the recessive phenotype. Which of the rules of probability can be used to calculate the probability of a particular combination of events that each have two alternative outcomes?

Answer:  binomial probability

Section:  2.4

Skill:  Synthesis/Evaluation

15)  You have self-fertilized a plant with round seeds that is heterozygous, and you want to determine what proportion of the offspring will be not only dominant, but also true-breeding. Which of the rules of probability can be used to calculate the probability of obtaining a particular outcome when specific information about that outcome modifies the probability calculation?

Answer:  conditional probability

Section:  2.4

Skill:  Synthesis/Evaluation

16)  In a dihybrid cross, you want to calculate the probability that an F2 progeny of the cross will inherit both dominant phenotypes. Which of the rules of probability can be used to calculate the probability of obtaining that combination of alleles?

Answer:  the product rule

Section:  2.4

Skill:  Synthesis/Evaluation

17)  The statistical value obtained from a chi-square analysis refers to the probability that the deviations between the observed numbers and the expected numbers are caused by what?

Answer:  random chance

Section:  2.5

Skill:  Knowledge/Comprehension

18)  A normal distribution curve contains all the possible experimental outcomes in graph form. The tall central segment of the curve represents the outcomes with the highest probability of occurrence. The average outcome, represented by the center of the data distribution, is known as what?

Answer:  the mean (µ)

Section:  2.5

Skill:  Knowledge/Comprehension

19)  Geneticists must be able to compare the outcomes they obtain in their experiments to the outcomes that might be expected to occur. Which test would they use to confirm that the difference between observed and expected outcomes can be attributed to chance?

Answer:  chi-square test

Section:  2.5

Skill:  Knowledge/Comprehension

 

2.3   Fill-in-the-Blank Questions

1)    One key to Mendel’s success was choosing to observe ________ traits, which exhibit one of two possible phenotypes.

Answer:  dichotomous

Section:  2.1

Skill:  Knowledge/Comprehension

2)    The hereditary particles that are passed from one generation are called alleles in modern terminology. This term had not been invented in Mendel’s time; instead, he determined that two “________” (alleles) were present for each trait in a plant and together determined the phenotype
of the trait.

Answer:  elementen

Section:  2.2

Skill:  Knowledge/Comprehension

3)    A ratio of 9:3:3:1 is expected among the F2 progeny of a dihybrid cross as a result of ________ of alleles at two loci.

Answer:  independent assortment

Section:  2.3

Skill:  Knowledge/Comprehension

4)    Binomial expansion is a complex genetic calculation requiring repetition and precision in the use of the product rule and the sum rule. A shortcut called ________ eliminates the need for these repetitive calculations and can be used for any number of expansions between 0 and the nth power to yield the size of each possible class and the total number of classes possible.

Answer:  Pascal’s triangle

Section:  2.4

Skill:  Knowledge/Comprehension

5)    The P value is dependent on the number of ________, which is equal to the number of independent variables in an experiment.

Answer:  degrees of freedom (df)

Section:  2.5

Skill:  Knowledge/Comprehension

2.4   Essay Questions

1)    How did the study of physics with Professors Doppler and Ettinghausen influence Mendel’s understanding of genetics?

Answer:  Doppler, an experimental physicist famous for the Doppler effect, espoused a “particulate” view of physics and taught Mendel how to separate individual characteristics from one another in experiments. Professor Ettinghausen taught Mendel the mathematics of combinatorial analysis. Mendel would apply each of these lessons to his later research. Mendel’s superior insight came principally from his familiarity with quantitative thinking and his understanding of the particulate nature of matter learned through the study of physics with Doppler. Central to Mendel’s experimental success was counting the number of progeny with specific phenotypes. This logical and now routine component of data-gathering was the key to Mendel’s ability to formulate the hypotheses that explained his results. Under Doppler and Ettinghausen, Mendel learned to study individual properties of matter separately and to think in quantitative terms about combinations of outcomes.

Section:  2.1

Skill:  Synthesis/Evaluation

2)    Describe the traits that make Pisum sativum an ideal organism for genetic studies. Why did Mendel ultimately decide not to include exterior seed coat color (gray vs. white) as one of the traits he analyzed?

Answer:  There are many varieties of peas with distinct, heritable features in the form of dichotomous phenotypes that can be easily observed and quantified. In addition, mating of plants can be closely controlled. Since each pea plant has both sperm-producing (stamens) and egg-producing (carpels) organs, they can be self-crossed to generate true-breeding plants. After creating these true-breeding plants, Mendel could test for dominant or recessive inheritance patterns by cross-pollination (fertilization between different plants). Mendel initially selected an eighth trait producing either gray or white exterior seed coats. Early in his analysis, he saw that plants with purple flowers always had gray seed coats and that those with white flowers always had white seed coats. He speculated that flower color and seed-coat color were determined by the same genetic mechanism, and he was correct. (The pigment anthocyanin is produced by plants that have purple flower color and gray seed coats, but a mutation eliminates anthocyanin production in plants with white flowers and white seed coats.)

Section:  2.1

Skill:  Synthesis/Evaluation

3)    Describe the blending theory of heredity and how Mendel’s results help to reject this theory.

Answer:  The blending theory viewed the traits of progeny as a mixture of the characteristics possessed by the two parental forms. Under this theory, progeny were believed to display characteristics that were approximately intermediate between those of the parents. Mendel reasoned that if the blending theory were true, he would see evidence of it in each trait. If no blending were seen in individual traits, the blending theory would be disproved. F1 experimental results reject the blending theory of heredity because all F1 progeny have the same phenotype (i.e., the dominant phenotype) that is indistinguishable from the phenotype of one of the pure-breeding parents. This specifically contradicts the blending theory prediction that the F1 would display a mixture of the parental phenotypes. The persistence of the dominant phenotype and the reemergence of the recessive phenotype in the F2 also contradict the blending theory.

Section:  2.1

Skill:  Synthesis/Evaluation

4)    What are Mendel’s first and second laws, and what do they state?

Answer:

First Law: Law of Segregation—The two alleles for each trait will separate from one another during gamete formation, and each allele will have an equal probability (1/2) of inclusion in a gamete. Random union of gametes at fertilization will unite one gamete from each parent to produce progeny in ratios that are determined by chance.

Second Law: Law of Independent Assortment—During gamete formation, the segregation of alleles at one locus is independent of the segregation of alleles at another locus.

Sections:  2.2 and 2.3

Skill:  Knowledge/Comprehension

5)    In Guinea pigs, short hair (S) is dominant over long hair (s), rough coat (R) is dominant over smooth coat (r), and black hair (B) is dominant over white hair (b). List all the different possible gametes that can be produced by each of the individuals below.

  1. SSRRbb
  2. ssRrBB
  3. SsRrbb
  4. SsRrBb

Answer:

  1. SSRRbb: SRb
  2. ssRrBB: sRB, srB
  3. SsRrbb: SRb, Srb, sRb, srb
  4. SsRrBb: SRB, SRb, SrB, Srb, sRB, sRb, srB, srb

Section:  2.3

Skill:  Application/Analysis

Genetics: An Integrated Approach (Sanders)

Chapter 4   Gene Interaction

4.1   Multiple-Choice Questions

1)    Which mode of inheritance produces heterozygotes with phenotypes that differ from either homozygote but typically more closely resembles one homozygous phenotype than the other?

  1. A) complete dominance
  2. B) incomplete dominance
  3. C) codominance
  4. D) epistasis
  5. E) incomplete penetrance

Answer:  B

Section:  4.1

Skill:  Knowledge/Comprehension

2)    Which mode of inheritance results in the phenotype of a heterozygote being indistinguishable from that of an organism homozygous for the dominant allele?

  1. A) complete dominance
  2. B) incomplete dominance
  3. C) codominance
  4. D) epistasis
  5. E) incomplete penetrance

Answer:  A

Section:  4.1

Skill:  Knowledge/Comprehension

3)    Which mode of inheritance results in both alleles being detected equally in the heterozygous phenotype?

  1. A) complete dominance
  2. B) incomplete dominance
  3. C) codominance
  4. D) epistasis
  5. E) incomplete penetrance

Answer:  C

Section:  4.1

Skill:  Knowledge/Comprehension

4)    A mutation results in an enzyme that is partially active compared to the wild-type allele. This type of “leaky” mutation is classified as __________.

  1. A) amorphic
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) neomorphic
  5. E) dominant negative

Answer:  B

Section:  4.1

Skill:  Knowledge/Comprehension

5)    A mutation resulting in an inactive gene product is classified as __________.

  1. A) amorphic
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) neomorphic
  5. E) dominant negative

Answer:  A

Section:  4.1

Skill:  Knowledge/Comprehension

6)    Two proteins interact to form a multimeric complex. When one of the proteins is mutated, there is a substantial loss of functional activity in the multimeric protein. This type of mutation is classified as __________.

  1. A) amorphic
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) neomorphic
  5. E) dominant negative

Answer:  E

Section:  4.1

Skill:  Knowledge/Comprehension

7)    Many oncogenes result from mutations that cause a protein to be expressed in cells where it is normally not expressed or is expressed at inappropriate times during development. This type of mutation can be described as __________.

  1. A) amorphic
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) neomorphic
  5. E) dominant negative

Answer:  C

Section:  4.1

Skill:  Application/Analysis

8)    A mutation results in a gene product with a novel function that is not normally found in wild-type organisms. This type of mutation is known as __________.

  1. A) amorphic
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) neomorphic
  5. E) dominant negative

 

Answer:  D

Section:  4.6

Skill:  Knowledge/Comprehension

9)    Most combinations of different ABO alleles result in complete dominance of one allele. Which combination results in codominance?

  1. A) IAi
  2. B) IAIB
  3. C) IBi
  4. D) ii
  5. E) either IAi or IBi

Answer:  B

Section:  4.1

Skill:  Application/Analysis

10)  What is the genotype for individuals with blood type O?

  1. A) IAi
  2. B) IAIB
  3. C) IBi
  4. D) ii
  5. E) either IAi or IBi

Answer:  D

Section:  4.1

Skill:  Knowledge/Comprehension

11)  What type of allele is often detected as a distortion in segregation ratios, where one class of expected progeny is missing?

  1. A) dominant negative allele
  2. B) temperature-sensitive allele
  3. C) lethal allele
  4. D) partial dominance
  5. E) incomplete penetrance

Answer:  C

Section:  4.1

Skill:  Knowledge/Comprehension

12)  You discover a new allele of a gene important for tail formation in mice. WT mice have long tails, but mice heterozygous for the allele have short tails. When you cross two heterozygous mice together, you obtain a 2:1 ratio of short-tailed mice to long-tailed mice. None of the short-tailed progeny are homozygous. What type of allele results in short tails?

  1. A) dominant negative allele
  2. B) temperature-sensitive allele
  3. C) lethal allele
  4. D) partially dominant allele
  5. E) incompletely penetrant allele

Answer:  C

Section:  4.1

Skill:  Application/Analysis

13)  The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What type of allele is the mutant allele?

  1. A) null
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) dominant negative
  5. E) neomorphic

Answer:  C

Section:  4.1

Skill:  Application/Analysis

14)  The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What is the dominance relationship between the WT and mutant allele for the phenotype of amount of enzyme per cell?

  1. A) The WT allele is dominant.
  2. B) The mutant allele is dominant.
  3. C) The WT and mutant alleles are codominant.
  4. D) The WT and mutant alleles show incomplete dominance.
  5. E) It is impossible to determine from the information given.

Answer:  D

Section:  4.1

Skill:  Application/Analysis

15)  The amount of enzyme activity in a cell that is homozygous for a mutant allele is 0 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount
of enzyme activity in a heterozygote is 0 units. What type of allele is the mutant allele?

  1. A) null
  2. B) hypomorphic
  3. C) hypermorphic
  4. D) dominant negative
  5. E) neomorphic

Answer:  D

Section:  4.1

Skill:  Application/Analysis

16)  Most people with the dominant mutant polydactyly allele have extra digits, but at least 25% have the normal number of digits. What is the genetic explanation for this observation?

  1. A) dominant negative allele
  2. B) temperature-sensitive allele
  3. C) lethal allele

 

  1. D) partial dominance
  2. E) incomplete penetrance

Answer:  E

Section:  4.2

Skill:  Knowledge/Comprehension

17)  Gene interactions in which an allele of one gene modifies or prevents expression of alleles of another gene is known as __________.

  1. A) complete dominance
  2. B) incomplete dominance
  3. C) codominance
  4. D) epistasis
  5. E) incomplete penetrance

Answer:  D

Section:  4.3

Skill:  Knowledge/Comprehension

18)  Bateson and Punnett crossed two white-flowered lines and saw all purple flowers in the F1 generation. They concluded this was an example of complementary gene interactions because
a cross of the F1 plants yielded what ratio in the F2 generation?

  1. A) 8 purple to 8 white
  2. B) 16 purple to 0 white
  3. C) 0 white to 16 purple
  4. D) 7 purple to 9 white
  5. E) 9 purple to 7 white

Answer:  E

Section:  4.3

Skill:  Application/Analysis

19)  If a dominant allele of one gene completely suppresses the phenotypic expression of alleles of another gene, this is an example of __________.

  1. A) recessive interaction
  2. B) dominant interaction
  3. C) recessive epistasis
  4. D) dominant epistasis
  5. E) dominant suppression

Answer:  E

Section:  4.3

Skill:  Knowledge/Comprehension

20)  The 9:6:1 ratio seen in the dihybrid cross of summer squash indicates what genetic relationship between the two genes controlling fruit shape?

  1. A) recessive interaction
  2. B) dominant interaction
  3. C) recessive epistasis
  4. D) dominant epistasis
  5. E) dominant suppression

Answer:  B

Section:  4.3

Skill:  Knowledge/Comprehension

21)  In the biosynthetic pathway for conversion from homoserine to methionine, you identify a Neurospora crassa double mutant Met1/Met2. This mutant will grow only if which supplement(s) are added to the minimal media?

  1. A) homoserine only
  2. B) cysteine only
  3. C) homocysteine only
  4. D) cystathionine and homocysteine
  5. E) methionine only

Answer:  E

Section:  4.3

Skill:  Synthesis/Evaluation

22)  Which step is catalyzed by the enzyme responsible for the Met 3 mutant?

  1. A) homoserine → cysteine
  2. B) cysteine → cystathionine
  3. C) cystathionine → homocysteine
  4. D) homocysteine → methionine
  5. E) methionine → homoserine

Answer:  B

Section:  4.3

Skill:  Synthesis/Evaluation

 

23)  Independent assortment predicts a 9:3:3:1 ratio with four different phenotypes in the F2 progeny.
If the alleles are epistatic, what would you predict?

  1. A) more than four phenotypes
  2. B) fewer than four phenotypes
  3. C) heterozygotes with a novel phenotype between the dominant and recessive homozygotes
  4. D) 1/3 of the progeny with the dominant phenotypes and 2/3 recessive
  5. E) no change in the 9:3:3:1 ratio

Answer:  B

Section:  4.3

Skill:  Application/Analysis

24)  Two pure-breeding mutant plants produce white flowers. When they are crossed, all of the progeny have wild-type purple flowers. What does this genetic complementation tell you?

  1. A) The genes are part of distinct biosynthetic pathways.
  2. B) The two lines exhibit different mutations in the same gene.
  3. C) More than one gene is involved in determining the phenotype.
  4. D) The allele is pleiotropic.
  5. E) The allele exhibits incomplete dominance.

Answer:  C

Section:  4.3

Skill:  Application/Analysis

25)  Deafness is caused by recessive mutations in any one of at least five genes. Two deaf individuals have nine children, all of whom have normal hearing. Which of the following can you conclude?

  1. A) The parents have mutations in the same gene.
  2. B) The parents have mutations in different genes.
  3. C) An epistatic interaction has occurred.
  4. D) The mutations are codominant to the normal allele.
  5. E) The mutations are incompletely dominant to the normal allele.

Answer:  B

Section:  4.4

Skill:  Application/Analysis

26)  In yeast, there are three gene products required to synthesize the amino acid lysine. You cross two haploid lysine auxotrophs to form a diploid. The diploid fails to grow on plates lacking lysine. Which of the following can you conclude?

  1. A) The haploid strains have mutations in the same gene.
  2. B) The haploid strains have mutations in different genes.
  3. C) The haploid strains form a complementation group.
  4. D) The haploid strains do not complement each other.
  5. E) Both (A) and (D) are true.

Answer:  E

Section:  4.4

Skill:  Synthesis/Evaluation

 

4.2   Short-Answer Questions

1)    A metabolic reaction requires 40 units of enzymatic activity to proceed. If a dominant allele D can generate 40 units of enzyme and a mutant allele d” generates 20 units of enzyme, what can be said of the dominant wild-type allele?

Answer:  D is haplosufficient.

Section:  4.1

Skill:  Application/Analysis

2)    A metabolic reaction requires 10 units of enzymatic activity to proceed. If a dominant allele D can generate 8 units of enzyme and a recessive allele d can generate 2 units of enzyme, what can be said of the dominant wild-type allele?

Answer:  D is haploinsufficient.

Section:  4.1

Skill:  Application/Analysis

3)    Are loss-of-function mutations more likely to be dominant or recessive?

Answer:  recessive

Section:  4.1

Skill:  Knowledge/Comprehension

4)    What are the three categories of loss-of-function mutations?

Answer:  amorphic (null), hypomorphic (leaky), dominant negative

Section:  4.1

Skill:  Knowledge/Comprehension

5)    What are the two categories of gain-of-function mutations?

Answer:  neomorphic and hypermorphic

Section:  4.1

Skill:  Knowledge/Comprehension

6)    You cross a pure-breeding white flower with a pure-breeding red flower, and the offspring are all pink. This is an example of what type of inheritance?

Answer:  incomplete dominance

Section:  4.1

Skill:  Application/Analysis

7)    You cross a pure-breeding white flower with a pure-breeding red flower, and the offspring are white with red spots. This is an example of what type of inheritance?

Answer:  codominance

Section:  4.1

Skill:  Application/Analysis

 

8)    Alleles of the Sbe1 gene, which controls pea shape, can be detected by DNA analysis. In heterozygous plants, two distinct bands of DNA are visible. What is the relationship between the alleles of the heterozygote at the molecular level?

Answer:  codominant

Section:  4.1

Skill:  Knowledge/Comprehension

9)    The four different human blood types are caused by how many different alleles? What are the alleles?

Answer:  three: IA, IB, and i

Section:  4.1

Skill:  Knowledge/Comprehension

10)  Individuals with blood type A have which blood antigen(s)?

Answer:  A antigen

Section:  4.1

Skill:  Knowledge/Comprehension

11)  Individuals with blood type AB have which blood antibody(ies)?

Answer:  none; neither anti-A nor anti-B

Section:  4.1

Skill:  Knowledge/Comprehension

12)  An individual who is a “universal acceptor,” meaning he or she will not have an adverse reaction to a transfusion with any type of blood, has which blood type?

Answer:  AB

Section:  4.1

Skill:  Application/Analysis

13)  Which antigen, expressed on the surface of all red blood cells, is modified by A- or B-transferases?

Answer:  H antigen

Section:  4.1

Skill:  Knowledge/Comprehension

14)  The allele responsible for the Siamese coat-color pattern produces an unstable tyrosinase enzyme. This type of gene product is an example of what type of allele?

Answer:  temperature-sensitive

Section:  4.1

Skill:  Knowledge/Comprehension

15)  King George III of England and other members of the royal family were afflicted with a series of strange, seemingly unrelated symptoms including abdominal pain, rapid pulse, convulsions, and insanity. It has been determined that he likely suffered from porphyria, caused by a mutation in a single allele. What is the genetic term describing the alteration of multiple, distinct traits of an organism by a mutation in a single gene?

 

Answer:  pleiotropy

Section:  4.2

Skill:  Application/Analysis

16)  Which type of pathway is responsible for melanin production: a biosynthetic pathway, signal transduction pathway, or developmental pathway?

Answer:  biosynthetic pathway

Section:  4.3

Skill:  Application/Analysis

17)  In familial hypercholesterolemia, individuals who are homozygous for the mutation lack LDL cholesterol receptors in the liver, causing high levels of serum cholesterol. This is a defect in which type of gene pathway: a biosynthetic, signal transduction, or developmental pathway?

Answer:  signal transduction pathway

Section:  4.3

Skill:  Application/Analysis

18)  To better understand which genes are involved in developmental pathways, geneticists use experimental analyses of mutant phenotypes. What is this analytic approach called?

Answer:  genetic dissection

Section:  4.3

Skill:  Knowledge/Comprehension

19)  Numerous genes contribute to the wild-type red eye color in Drosophila. Two genes, brown and vermilion, produce eye pigments. What does the white gene do?

Answer:  transports pigments to eye cells

Section:  4.3

Skill:  Knowledge/Comprehension

20)  Mendel studied tall and short pure-breeding lines of pea plants. Inherited genetic variation would dictate that one line would produce tall plants and the other short plants, but the genes and phenotypes are influenced by what external factor?

Answer:  environmental influence

Section:  4.3

Skill:  Knowledge/Comprehension

4.3   Fill-in-the-Blank Questions

1)    ________ mutations decrease or eliminate gene activity.

Answer:  Loss-of-function

Section:  4.1

Skill:  Knowledge/Comprehension

2)    ________ mutations cause overexpression or result in new functions.

Answer:  Gain-of-function

Section:  4.1

Skill:  Knowledge/Comprehension

3)    IA and IB alleles exhibit ________ (complete or incomplete) dominance over the i allele.

Answer:  complete

Section:  4.1

Skill:  Knowledge/Comprehension

4)    If an organism with a particular genotype fails to produce the corresponding phenotype, the organism is said to be ________ for the trait.

Answer:  nonpenetrant

Section:  4.1

Skill:  Knowledge/Comprehension

5)    In Labrador retrievers, coat color is controlled by gene interaction in which homozygosity for a recessive allele can mask the phenotypic expression of a second gene. This genetic interaction is known as ________ and has a characteristic ________ ratio.

Answer:  recessive epistasis; 9:3:4

Section:  4.3

Skill:  Knowledge/Comprehension

4.4   Essay Questions

1)    Proper cross-matching of blood type is essential for safe blood transfusion. What is the general rule for safe blood transfusion, and what are the consequences if a patient receives the incorrect blood type by accident? If a patient’s blood type is unknown, what is the safest course of action?

Answer:  The general rule for safe blood transfusion is that the recipient blood must not contain an antibody that reacts with an antigen in the donated blood. When such a reaction occurs, blood clots produced by clumping blood cells form at the site of transfusion. The clots block circulation, deprive tissues of oxygen, and can potentially cause life-threatening complications. Thus, if a patient’s blood type is not known, they are given type O-negative blood because it contains no antigens for the patient’s own blood to react with.

Section:  4.1

Skill:  Synthesis/Evaluation

2)    You have crossed two Mexican hairless dogs, and the offspring are 1/3 hairy and 2/3 hairless. Given this phenotypic ratio, draw the Punnett square for this cross. What are the genotypes of the P1 and F1 dogs in this cross? List the predicted genotype as well as the phenotype for each of the offspring. Which genotypes are hairless, and which are hairy? Can you design a genetic cross that would yield a true-breeding hairless line (where all offspring are hairless)?

Answer:  The 1:2 ratio in the offspring provides strong evidence that this trait involves a lethal allele. Thus, the Punnett square for the cross would be

You cannot design a true-breeding cross, because the hh genotype is lethal. Thus, a maximum of 2/3 of all of the offspring from any cross are predicted to be hairless.

Section:  4.1

Skill:  Synthesis/Evaluation

3)    Describe the difference between incomplete penetrance and variable expressivity. It is often difficult to pinpoint the cause of incomplete penetrance or variable expressivity. What possible interactions may be responsible?

Answer:  In incomplete penetrance, a genotype is not expressed by every organism in which it is present. In variable expressivity, the organisms that share a genotype express the corresponding phenotype to different degrees. Both are explained by genetic or nongenetic interactions that modify or prevent the consistent expression of a genotype.

Three kinds of interactions may be responsible: (1) other genes interactin ways that modify the expression of the mutant allele, (2) environmental or developmental (i.e., nongenetic) factors interact with the mutant allele to modify its expression, and (3) some combination of other genes and environmental factors interact to modify expression of the mutation. In inbred laboratory strains of model genetic organisms, variation in genetic factors can be eliminated experimentally to allow separation of gene–gene and gene–environment variability, something that cannot be done in organisms such as humans.

Section:  4.2

Skill:  Synthesis/Evaluation

4)    Explain how the ch (Himalayan) allele and tyrosinase control the Siamese coat-color pattern. What is the underlying reason that certain parts of a Siamese cat (e.g., tail and ears) are darker than the cat’s trunk?

Answer:  The tyrosinase enzyme produced by the hypomorphic ch (Himalayan) allele is unstable and is inactivated at a temperature very near the normal body temperature of most mammals. Thus, ch is a temperature-sensitive allele. The parts of cats that are farthest away from the core of the body (the paws, ears, tail, and tip of the nose) at most times tend to be slightly cooler than the trunk. At these cooler extremities, the temperature-sensitive tyrosinase produced by the ch allele remains active, producing pigment in the hairs there. However, in the warmer central portion of the body, the slightly higher temperature is enough to cause the tyrosinase produced by the ch allele to denature, or unravel. This inactivates the enzyme and leads to an absence of pigment in the central portion of the body. Animals that are chch or chc have the Himalayan phenotype. The final allele in the series, c, is a null allele that does not produce functional tyrosinase.

Section:  4.3

Skill:  Synthesis/Evaluation

5)    Phenylketonuria (PKU) is caused by the absence of the enzyme phenylalanine hydroxylase, which catalyzes the first step of the pathway that breaks down the amino acid phenylalanine, a common component of dietary protein. Explain how environmental intervention is commonly practiced to prevent the development of this human autosomal recessive condition.

Answer:  PKU damages neurons because the body is unable to break down dietary phenylalanine, ultimately causing irreversible mental retardation and neuron death. There are two main ways in which the environment has been changed to affect the development of PKU. First, newborns are routinely screened for PKU so they can be placed on a phenylalanine-free diet and will avoid PKU complications. Second, individuals living with PKU are warned of foods containing phenylalanine by strict labeling of foods (e.g., foods and beverages containing the artificial sweetener aspartame). Thus, while they will always have the mutated allele, they are able to avoid developing the disease through strict control of diet.

Section:  4.3

Skill:  Synthesis/Evaluation

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