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Biology 10th Edition Raven Johnson Mason Losos Singer Test Bank

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Biology 10th Edition Raven Johnson Mason Losos Singer Test Bank

ISBN-13: 978-0073383071

ISBN-10: 0073383074

 

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Biology 10th Edition Raven Johnson Mason Losos Singer Test Bank

ISBN-13: 978-0073383071

ISBN-10: 0073383074

 

 

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Chapter 13

Chromosomes, Mapping, and the Meiosis–Inheritance Connection

 

 

Multiple Choice Questions

  1. A genetic _____ indicates the distances between gene loci measured in terms of the frequency of recombination.
    A. map
    B.  profile
    C.  pedigree
    D.  clone
    E.  karyotype

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. Of the 23 pairs of human chromosomes, 22 pairs are homologous and are found in both males and females. These are called ________.
    A. bivalents
    B.  autosomes
    C.  recombinant chromosomes
    D.  somatic chromosomes

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. Traits that are controlled by genes located on the X chromosome are said to be ________________.
    A. autosomal
    B.  gametal
    C.  sex-linked
    D.  pleiotropic

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. Allele pairs are most likely to assort independently of one another when
    A. they control unrelated traits.
    B.  they control related traits.
    C.  they are on the same chromosome.
    D.  they are sex linked.
    E.  they are on different chromosomes.

 

Blooms Level: 2. Understand
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance

 

  1. The number of allele pairs that assort independently in an organism is generally much higher than the number of chromosome pairs. This phenomenon is due to
    A.independent assortment.
    B. segregation.
    C. crossing over.
    D. sex-linkage.
    E. chromosome inactivation.

 

Blooms Level: 2. Understand
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. The theory of chromosomal inheritance was first proposed by
    A. Mendel.
    B.  Morgan.
    C.  Knight.
    D.  Sutton.
    E.  Stern.

 

Blooms Level: 1. Remember
LO: 13.01.02 Explain the evidence for genes being on chromosomes.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

 

  1. In Drosophila, the sex of an individual is determined by
    A. 1 pair of alleles.
    B.  the number of X chromosomes.
    C.  the number of Y chromosomes.
    D.  1 pair of autosomes.
    E.  2 pairs of alleles.

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. In Morgan’s experiments, the white eye allele in Drosophila was shown to be
    A. located on the X chromosome.
    B.  located on the Y chromosome.
    C.  dominant.
    D.  located on an autosome.
    E.  codominant.

 

Blooms Level: 1. Remember
LO: 13.01.01 Describe sex-linked inheritance in fruit flies.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

 

  1. The geneticist who discovered the white eye mutation in Drosophila and helped establish that genes are carried on chromosomes was
    A. Mendel.
    B.  Sutton.
    C.  Sturtevant.
    D.  Janssens.
    E.  Morgan.

 

Blooms Level: 1. Remember
LO: 13.01.02 Explain the evidence for genes being on chromosomes.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

  1. Genetic exchange between 2 homologous chromosomes is called
    A. synapsis.
    B.  pleiotropy.
    C.  crossing over.
    D.  allelic exchange.
    E.  independent assortment.

 

Blooms Level: 1. Remember
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance

 

  1. Occasionally, chromosomes fail to separate during meiosis, leading to daughter cells that have an abnormal number of chromosomes. This phenomenon is called
    A. epistasis.
    B.  nondisjunction.
    C.  crossing over.
    D.  pleiotropy.
    E.  chromosome inactivation.

 

Blooms Level: 1. Remember
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. Humans who have lost one copy of an autosome are called
    A. haploid.
    B.  trisomic.
    C.  bisomic.
    D.  monosomic.
    E.  monoploid.

 

Blooms Level: 1. Remember
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. In humans, individuals with trisomy of the ______ chromosome are most likely to survive until adulthood.
    A. X.
    B.  13th.
    C.  15th.
    D.  18th.
    E.  21st.

 

Blooms Level: 1. Remember
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. If a human female has 2 Barr bodies per cell, it is almost certain that
    A. her father had 1 Barr body per cell.
    B.  her mother also had 2 Barr bodies per cell.
    C.  she developed from a fertilized egg with 3 X chromosomes.
    D.  she is genetically a male with female characteristics.
    E.  she is genetically a normal fertile female.

 

Blooms Level: 2. Understand
LO: 13.02.02 Explain the genetic consequences of dosage compensation in mammals.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. A human female with only one X chromosome is said to have a condition called
    A. X chromosome inactivation.
    B.  Angelman syndrome.
    C.  Turner syndrome.
    D.  Klinefelter syndrome.
    E.  Down syndrome.

 

Blooms Level: 1. Remember
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. The most common fatal genetic disorder of Caucasians is
    A. cholera.
    B.  cystic fibrosis.
    C.  hemophilia.
    D.  sickle cell anemia.
    E.  muscular dystrophy.

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. In sickle cell anemia, the defective hemoglobin differs from the normal hemoglobin by
    A. the color of the pigment.
    B.  the size of the molecule.
    C.  a single amino acid substitution.
    D.  the total number of amino acids.
    E.  the type of blood cell it is found in.

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. Hemophilia is caused by a
    A. recessive allele on the X chromosome.
    B.  dominant allele on the X chromosome.
    C.  codominant allele on the X chromosome.
    D.  recessive allele on an autosome.
    E.  dominant allele on an autosome.

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. _______________ is a human hereditary disease that is caused by a dominant allele but does not show up in affected individuals until they are in middle age.
    A. Cystic fibrosis
    B.  Sickle cell anemia
    C.  Tay-Sachs disease
    D.  Huntington’s disease
    E.  Hemophilia

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. Amniocentesis is a procedure that is normally used
    A. to reduce the risk of genetic disease.
    B.  for gene therapy.
    C.  to change the sex of the fetus.
    D.  for diagnosis of genetic disorders.
    E.  for nourishing the fetus.

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. Huntington’s disease is caused by a single dominant allele. It is a lethal disease, yet it persists in the human population. Which of the following statements best describes why?
    A. Huntington’s disease is sex-linked and every human has at least one X chromosome; thus, the chances are extremely high for this allele to be maintained in the human population.
    B.  Huntington’s disease can present symptoms so mild that they appear to lack dominant expression of the allele in some individuals; in those cases, the allele is passed on to the offspring.
    C.  While lethal to a parent, Huntington’s disease will not be lethal to the offspring since it can skip a generation.
    D.  Huntington’s disease presents symptoms in mid-life, after most people have already had offspring.
    E.  Even though Huntington’s disease is lethal, it improves chances for reproduction before the person dies.

 

Blooms Level: 5. Evaluate
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. In humans, the male has an X and a Y sex chromosome. The human female has two X chromosomes. In birds, the female has a Z and a W sex chromosome while the male has two Z chromosomes. Which of the following statements is accurate about which parent determines the gender of the offspring?
    A. In humans and birds, the male determines the gender of all the offspring.
    B.  In humans and birds, the female determines the gender of all the offspring.
    C.  In humans, the male determines the gender of the offspring, and in birds the female determines the gender.
    D.  In humans, the female determines the gender of the offspring, and in birds the male determines the gender.
    E.  Determination of the gender of any human or bird offspring is related to the environmental conditions at the time of conception.

 

Blooms Level: 3. Apply
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. Sickle cell anemia is caused by a defect in the
    A. oxygen-carrying pigment hemoglobin.
    B.  ability of the blood to clot.
    C.  ability of red blood cells to fight infection.
    D.  chloride ion transport protein.

 

Blooms Level: 1. Remember
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. How many Barr bodies does a normal human female contain in each diploid cell?
    A. 0
    B.  1
    C.  2
    D.  3

 

Blooms Level: 1. Remember
LO: 13.02.02 Explain the genetic consequences of dosage compensation in mammals.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. A test cross can be used to do all of the following except
    A. determine whether an individual that displays a dominant phenotype is homozygous for the trait.
    B.  determine whether an individual that displays a dominant phenotype is heterozygous for the trait.
    C.  gather genotype information from phenotype information.
    D.  identify the chromosome on which a gene is located.

 

Blooms Level: 2. Understand
LO: 13.01.02 Explain the evidence for genes being on chromosomes.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

  1. Which of the following animals is a genetic male?
    A. bird ZW
    B.  grasshopper XO
    C.  honeybee diploid
    D.  Drosophila XXY

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. In humans, if non-disjunction led to an individual with a genotype of XO, that person would
    A. be female because each cell lacks a Y chromosome.
    B.  be male because each cell has only one X chromosome.
    C.  display both male and female characteristics.
    D.  not survive.

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. In humans, if non-disjunction led to an individual with a genotype of XXY, that person would
    A. be female because each cell has two X chromosomes.
    B.  be male because each cell has one Y chromosome.
    C.  display both male and female characteristics.
    D.  not survive.

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. In some species, sex determination is influenced by environmental temperature during development.  If you wanted to determine the temperature at which one would obtain a 1:1 sex ratio in a particular species of turtle, which of the following experiments would best address this question?
    A. Grow the turtles in two different incubators at temperatures of 22ºC and 30ºC.
    B.  Grow the turtles in three different incubators at temperatures of 26ºC, 28ºC, and 30ºC.
    C.  Grow the turtles in four different incubators at temperatures of 24ºC, 26ºC, 28ºC, and 30ºC.
    D.  Grow the turtles in five different incubators at temperatures of 22ºC, 24ºC, 26ºC, 28ºC, and 30ºC.

 

Blooms Level: 3. Apply
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. Any genetic differences between individuals in a population are called
    A. markers.
    B.  alleles.
    C.  polymorphisms.
    D.  SNPs.

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

 

  1. The classic experiments performed by Creighton and McClintock in Maize
    A. provided the initial evidence for genetic recombination.
    B.  provided evidence that genes located on the same chromosome do not assort independently.
    C.  allowed for the establishment of the first genetic map.
    D.  provided evidence for the physical exchange of genetic material between homologues.

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. In humans, if an XY individual had a deletion of the SYR gene, that person would
    A. develop as a female.
    B.  have both male and female characteristics.
    C.  have ambiguous genitalia.
    D.  develop as a male.

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. Which statement about calico cats is false?
    A. Calico cats can be male or female.
    B.  The different colored fur is due to the inactivation of one X chromosome.
    C.  The variation in coat color is an example of an epistatic interaction.
    D.  Calico cats are genetic mosaics.

 

Blooms Level: 1. Remember
LO: 13.02.02 Explain the genetic consequences of dosage compensation in mammals.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. If an XY human had a genetic disorder that causes insensitivity to androgens, that person’s genotype and phenotype would be
    A. XX, female.
    B.  XX, male.
    C.  XY, female.
    D.  XY, male.

 

Blooms Level: 1. Remember
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

  1. Which offspring inherit all their mitochondrial DNA from their mother and none from their father?
    A. daughters
    B.  sons
    C.  both sons and daughters
    D.  neither sons nor daughters

 

Blooms Level: 1. Remember
LO: 13.03.01 Describe the inheritance pattern for genes contained in a chloroplast or mitochondrion DNA.
Section: 13.03 Exceptions to the Chromosomal Theory of Inheritance
Topic: Inheritance

  1. Nondisjunction of a single pair of autosomes can lead to all of the following except
    A. aneuploidy.
    B.  monosomy.
    C.  trisomy.
    D.  euploidy.

 

Blooms Level: 2. Understand
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. If you needed to determine the order of genes on a chromosome, you should perform
    A. a test cross.
    B.  a two-point cross.
    C.  a three-point cross.
    D.  a SNP test.

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. A 39-year-old woman is in her sixth week of pregnancy. Due to her advanced age, she is at higher risk for having a baby with Down’s syndrome than younger pregnant women. She would like to find out as early as possible whether or not her baby has Down’s syndrome. Her doctor should suggest
    A. amniocentesis.
    B.  genetic counseling.
    C.  chorionic villi sampling.
    D.  a pedigree analysis.

 

Blooms Level: 2. Understand
LO: 13.05.02 Describe the consequences of nondisjunction in humans.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. In Drosophila, dosage compensation is controlled by the male-specific lethal (MSL) complex consisting of MSL proteins and roX RNAs. Based on what you know about dosage compensation, the role of the MSL complex in males would be to
    A. double the level of expression of genes on the X chromosome.
    B.  increase the level of expression of genes on the X chromosome by 50%.
    C.  decrease the level of expression of genes on the X chromosome by 50%.
    D.  decrease the level of expression of genes on the X chromosome by 100%.
    E.  double the level of expression of genes on the Y chromosome.

 

Blooms Level: 3. Apply
LO: 13.02.02 Explain the genetic consequences of dosage compensation in mammals.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

In Drosophila, the allele red eyes (bw+) is dominant to the allele for brown eyes (bw).  At another gene locus on the same chromosome, the allele for thin wing veins (hv+) is dominant to the allele for heavy wing veins (hv). Flies homozygous for bw and hv+ are crossed to flies homozygous for bw+ and hv to obtain doubly heterozygous F1 progeny.

  1. Given that these 2 gene loci are very closely linked, the genotypic ratio in the F2 generation should be closest to
    A. 1:2:1
    B.  1:1:1:1
    C.  9:3:3:1
    D.  3:1

 

Blooms Level: 4. Analyze
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning

 

  1. Given that these 2 gene loci are very closely linked, the phenotypic ratio in the F2 generation should be closest to
    A. 1 brown, heavy: 2 red, thin: 1 red, thin
    B.  1 brown, thin: 2 red, thin: 1 red, heavy
    C.  3 red, thin: 1 brown heavy
    D.  1 brown: 1 red: 1 heavy: 1 thin

 

Blooms Level: 4. Analyze
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning

  1. What would be the results of a test cross with the F1 flies?
    A. 1 brown, thin: 1 red, heavy
    B.  1 brown, heavy: 1 red, thin
    C.  1 brown, thin: 2 red, thin: 1 red, heavy
    D.  1 brown, heavy: 2 red, thin: 1 red, thin

 

Blooms Level: 4. Analyze
LO: 13.04.01 Describe how genes on the same chromosome will segregate.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning

 

  1. What is the relationship between recombination frequency and the actual physical distance on a chromosome?
    A. As physical distance increases, the recombination frequency increases in a linear fashion.
    B.  As physical distance increases, the recombination frequency decreases in a linear fashion.
    C.  As physical distance increases, the recombination frequency first increases in a linear fashion, but gradually levels off to a frequency of 0.5.
    D.  As physical distance increases, the recombination frequency first increases, but then decreases.

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. In a two-point cross to map genes A and B, you obtained 98 recombinant types and 902 parental types among the offspring. How far apart are these genes?
    A. 9.8 cM
    B.  0.98 cM
    C.  90.2 cM
    D.  9.02 cM
    E.  .098 cM

 

Blooms Level: 3. Apply
LO: 13.04.03 Calculate map distances from the frequency of recombinants in testcrosses.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning

 

  1. Morgan’s student Sturtevant demonstrated that the recombination frequencies between a series of linked genes is additive. Examine the following recombination data from Sturtevant, and determine the proper order of the genes on the Drosophila X chromosome. Assume y is in the 0.0 position.

    A.  y m v w
    B.  y w v m
    C.  y m w v
    D.  y w m v

 

Blooms Level: 5. Evaluate
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

You are a forensic technician working on a DNA sample obtained from a crime scene. Your job is to compare the unknown DNA sample with known DNA samples collected from five different suspects. Preliminary analysis using only a few DNA markers revealed that the unknown sample could possibly match two of the suspects. After consulting the case file, you find out that these two suspects are brothers (but not twins). You realize that you will have to do a more detailed analysis on the samples so that you can distinguish between the brothers and determine which brothers’ DNA matches the unknown sample.

 

  1. Which of the following will help you distinguish between the two final suspects?
    A. single nucleotide polymorphisms (SNPs)
    B.  human genetic map
    C.  linkage data
    D.  anonymous markers

 

Blooms Level: 4. Analyze
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

  1. Why can’t you use mitochondrial DNA to distinguish between these two suspects?
    A. The sequence of mitochondrial DNA has not yet been determined.
    B.  The brothers share the same mitochondrial DNA.
    C.  There are no molecular techniques available that allow one to analyze mitochondrial DNA.
    D.  Because mitochondrial DNA is inherited in a paternal pattern.

 

Blooms Level: 3. Apply
LO: 13.03.01 Describe the inheritance pattern for genes contained in a chloroplast or mitochondrion DNA.
Section: 13.03 Exceptions to the Chromosomal Theory of Inheritance
Topic: Inheritance

 

  1. In some human populations, the proportion of individuals who are heterozygous for the sickle cell allele is much higher than would be expected by chance alone. Why?
    A. Individuals with two normal alleles have an advantage over heterozygous individuals.
    B.  Individuals with two harmful alleles have an advantage over heterozygous individuals.
    C.  Individuals with two harmful alleles have an advantage over individuals with two normal alleles.
    D.  Heterozygous individuals have an advantage over individuals with two normal alleles.

 

Blooms Level: 2. Understand
LO: 13.05.01 Explain how mutations can cause disease.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

  1. A deletion of a particular stretch of chromosome 15 can cause either Prader-Willi syndrome or Angelman syndrome, depending on
    A. the parental origin of the normal and deleted chromosome.
    B.  whether or not the region is methylated properly.
    C.  whether a translocation event has occurred.
    D.  whether a nondisjunction event has occurred.

 

Blooms Level: 1. Remember
LO: 13.05.03 Recognize how genomic imprinting can lead to non- Mendelian inheritance.
Section: 13.05 Selected Human Genetic Disorders
Topic: Inheritance

 

  1. How did the development of anonymous markers aid in the production of a human genetic map?
    A. Anonymous markers are genetic markers that do not cause a detectable phenotype, but can be detected by molecular techniques. The markers correspond to specific and unique chromosomal regions, thereby allowing for the identification and ordering of particular segments of DNA. Such information was essential to the generation of a human genetic map.
    B.  Anonymous markers are genetic markers that cause a detectable phenotype and can’t be detected by molecular techniques. The markers correspond to specific and unique chromosomal regions, thereby allowing for the identification and ordering of particular segments of DNA. Such information was essential to the generation of a human genetic map.
    C.  Anonymous markers are genetic markers that do not cause a detectable phenotype, but can be detected by molecular techniques. The markers correspond to specific and unique genetic regions, thereby allowing for the identification and ordering of particular segments of the chromosome. Such information was essential to the generation of a human genetic map.

 

Blooms Level: 1. Remember
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance

 

  1. Why isn’t mitochondrial DNA a unique identifier?
    A. Mitochondrial DNA is inherited through the paternal lineage. All offspring inherit their father’s mitochondria, and therefore the same mitochondrial DNA. As a result, all family members that share a paternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person’s relationship within a paternal line, but cannot be used to identify a specific individual.
    B.  Mitochondrial DNA is inherited through the maternal lineage. All offspring inherit their mother’s mitochondria, and therefore the same mitochondrial DNA. As a result, all family members that share a maternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person’s relationship within a maternal line, but cannot be used to identify a specific individual.
    C.  Mitochondrial DNA is inherited through the maternal lineage. All female offspring inherit their mother’s mitochondria, and therefore the same mitochondrial DNA. As a result, all female family members that share a maternal lineage would have the same mitochondrial DNA. Mitochondrial DNA can therefore be used to confirm or eliminate a person’s relationship within a maternal line, but cannot be used to identify a specific individual.

 

Blooms Level: 1. Remember
LO: 13.03.01 Describe the inheritance pattern for genes contained in a chloroplast or mitochondrion DNA.
Section: 13.03 Exceptions to the Chromosomal Theory of Inheritance
Topic: Inheritance

 

Numeric Response Questions

  1. In humans, hemophilia is caused by a recessive allele on the X chromosome. Suppose a man with hemophilia marries a normal woman whose mother had hemophilia.  What is the probability that their second child will have hemophilia?  ( Enter the probability as a percent.  Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5 )
    50

 

Blooms Level: 3. Apply
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance

 

  1. In humans, hemophilia is caused by a recessive allele on the X chromosome. Suppose a man with hemophilia marries a normal woman whose mother had hemophilia. If an ultrasound test shows that their first child is a girl, what is the probability that she has hemophilia?  (Enter the probability as a percent. Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5 )
    50

 

Blooms Level: 3. Apply
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance
Type: Quantitative Reasoning

  1. In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus on the same chromosome, there is a dominant gene for red eyes and its recessive allele for purple eyes. A male that was heterozygous at both gene loci was mated with a female that was homozygous for both recessive alleles and the following results were observed among the offspring:

    Normal wings and red eyes – 420
    Vestigial wings and red eyes – 80
    Normal wings and purple eyes – 70
    Vestigial wings and purple eyes – 430

    According to these data, what is the distance, in centimorgans, between these 2 gene loci? (Enter the number only without the units.  For example, 100 cM would be entered as 100)

    15

 

Blooms Level: 3. Apply
LO: 13.04.03 Calculate map distances from the frequency of recombinants in testcrosses.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning

 

  1. In fruit flies () there is a dominant allele for red eyes and a recessive allele for white eyes. These alleles are located on the X chromosome.  If a heterozygous red-eyed female is mated with a white-eyed male, what percentage of the offspring are expected to be white-eyed females?  ( Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5 )

    25

 

Blooms Level: 3. Apply
LO: 13.01.01 Describe sex-linked inheritance in fruit flies.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance
Type: Quantitative Reasoning

  1. At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision.  A heterozygous brown-eyed woman who is a carrier of colorblindness marries a blue-eyed man who is not colorblind.  What is the probability that their first child will be a blue-eyed female who has normal color vision?  (Enter the probability as a percent. Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5)
    25

 

Blooms Level: 4. Analyze
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance
Type: Quantitative Reasoning

 

  1. At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision.  A heterozygous brown-eyed woman who is a carrier of colorblindness marries a blue-eyed man who is not colorblind.  The woman becomes pregnant and an ultrasound test shows that the child is a girl.  What is the probability that she will be colorblind?  (Enter the probability as a percent. Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5)
    0

 

Blooms Level: 4. Analyze
LO: 13.02.01 Describe the relationship between sex chromosomes and sex determination.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance
Type: Quantitative Reasoning

 

Multiple Choice Questions

  1. Suppose you are carrying out a series of crosses with an insect where the mechanism of sex determination is unknown.  You discover a mutant female with short bristles and decide to cross it with a wild type male that has normal bristles.  Half of the F1 progeny have short bristles but all of these short-bristled F1 progeny are males.  Based on these results, a valid hypothesis would be
    A. Males are ZW, females are ZZ, and short bristles are caused by a dominant allele on the Z chromosome
    B.  Males are ZZ, females are ZW, and short bristles are caused by a recessive allele on the Z chromosome
    C.  Males are ZZ, females are ZW, and short bristles are caused by a dominant allele on the W chromosome
    D.  Males are ZZ, females are ZW, and short bristles are caused by a dominant allele on the Z chromosome

 

Blooms Level: 5. Evaluate
LO: 13.01.02 Explain the evidence for genes being on chromosomes.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

 

  1. In 1910, Morgan did a series of experiments with the fruit fly Drosophila, an organism where females are XX and males are XY.  When a mutant male fly with white eyes was crossed with a wild type female with red eyes, none of the F1 progeny had white eyes but 18% of the F2 progeny had white eyes.  Unexpectedly, all of these white-eyed F2 flies  were males.  Based on these results, Morgan concluded that white eyes is caused by a recessive X-linked allele.  Suppose Morgan has found that half of the F1 progeny had white eyes but all of these white-eyed F1 flies were females.  In this case, a valid hypothesis would be
    A. White eyes is caused by a recessive Y-linked allele
    B.  White eyes is caused by a dominant Y-linked allele
    C.  White eyes is caused by a dominant X-linked allele
    D.  White eyes is caused by a dominant autosomal allele

 

Blooms Level: 5. Evaluate
LO: 13.01.02 Explain the evidence for genes being on chromosomes.
Section: 13.01 Sex Linkage and the Chromosomal Theory of Inheritance
Topic: Inheritance

 

Numeric Response Questions

  1. In humans, red-green colorblindness is caused by a recessive allele on the X chromosome.  Therefore, if a female who is a carrier for the allele that causes colorblindness marries a male who is colorblind, the probability that their first child will be a girl who is colorblind is 25%.  This assumes normal dosage compensation where the X that is inactivated in females varies randomly from cell to cell.  But suppose the normal mechanism of dosage compensation is altered so that the X that is inactivated in any given female is random but the same X is inactivated in every cell of her body.  In this case, what is the probability that there first child would be a girl with hemophilia?  ( Express the probability as a percent.  Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5 )
    37.5

 

Blooms Level: 4. Analyze
LO: 13.02.02 Explain the genetic consequences of dosage compensation in mammals.
Section: 13.02 Sex Chromosomes and Sex Determination
Topic: Inheritance
Type: Quantitative Reasoning

 

 

Multiple Choice Questions

  1. Genetic maps are based on recombination frequencies.  Because both odd and even numbers of crossovers can occur between any 2 gene loci, as the physical distance between two loci increases, the maximum recombination frequency levels off at 50%.  However, suppose you discovered a species where only an even number of crossovers can occur between any two gene loci.  In this case, as the physical distance between two loci increases, you would expect the maximum recombination frequency to
    A. remain at zero
    B.  increase with no limit
    C.  level off at 25%
    D.  level off at 75%
    E.  level off at 100%

 

Blooms Level: 4. Analyze
LO: 13.04.02 Explain the relationship between frequency of recombinant progeny and map distance.
Section: 13.04 Genetic Mapping
Topic: Inheritance
Type: Quantitative Reasoning