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Genetics A Conceptual Approach 5th Edition Pierce Test Bank

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Genetics A Conceptual Approach 5th Edition Pierce Test Bank

ISBN-13: 978-1464109461

ISBN-10: 146410946X

 

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Genetics A Conceptual Approach 5th Edition Pierce Test Bank

ISBN-13: 978-1464109461

ISBN-10: 146410946X

 

 

 

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Chapter 11: Chromosome Structure and Organelle DNA

 

Multiple-Choice Questions

 

  1. Which of the following is not true of negatively supercoiled DNA?

 

  1. Eases the separation of nucleotide strands during replication and transcription.
  2. Allows DNA to be packed into small spaces.
  3. Has less than 10 bp per turn of its helix.
  4. Is more negatively charged due to additional phosphates per turn of the helix.
  5. Is found in most cells.

 

Answer: c

Section: 11.1

Comprehension Question

 

  1. How many complete rotations would most likely correspond to a positively supercoiled DNA molecule that is 100 bp in length?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: d

Section: 11.1

Comprehension Question

 

  1. How many complete rotations would most likely correspond to a negatively supercoiled DNA molecule that is 100 bp in length?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. How many complete rotations would most likely correspond to a relaxed DNA molecule that is 100 bp in length?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: c

Section: 11.1

Comprehension Question

 

  1. How many base pairs per turn of the helix would most likely correspond to a relaxed DNA molecule?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: c

Section: 11.1

Comprehension Question

 

  1. How many base pairs per turn of the helix would most likely correspond to a positively supercoiled DNA molecule?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. How many base pairs per turn of the helix would most likely correspond to a negatively supercoiled DNA molecule?

 

  1. 0
  2. 5
  3. 10
  4. 15
  5. 100

 

Answer: d

Section: 11.1

Comprehension Question

 

  1. Which of the following is not true of bacterial DNA?

 

  1. Most bacterial genomes consist of a single circular DNA molecule.
  2. Bacterial DNA is not attached to any proteins that help to compact it.
  3. Bacterial DNA is confined to a region in the cell called the nucleoid.
  4. Many bacteria contain additional DNA in the form of small circular molecules called plasmids.
  5. About 3 to 4 million base pairs of DNA are found in a typical bacterial genome.

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. You are studying a small eukaryotic gene of about 2000 bp in length. Estimate how many copies of histone H4 you would find along this region of the chromosome.

 

  1. 10
  2. 20
  3. 40
  4. 80
  5. 100

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. You are studying a small eukaryotic gene of about 2000 bp in length. Estimate how many copies of histone H1 you would find along this region of the chromosome.

 

  1. 10
  2. 20
  3. 40
  4. 80
  5. 100

 

Answer: a

Section: 11.1

Comprehension Question

 

  1. The human Y chromosome is about 50 million base pairs long. About how many nucleosomes would you expect to find associated with this chromosome?

 

  1. 2,500
  2. 50,000
  3. 250,000
  4. 1,000,000
  5. 50,000,000

 

Answer: a

Section: 11.1

Comprehension Question

 

  1. Which of the following amino acids has a positive charge that helps to hold the DNA in contact with the histones?

 

  1. Alanine
  2. Arginine
  3. Leucine
  4. Valine
  5. Serine

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. Which of the following is not true of heterochromatin?

 

  1. Remains in a highly condensed state throughout the cell cycle
  2. Makes up most chromosomal material and is where most transcription occurs
  3. Exists at the centromeres and telomeres
  4. Occurs along one entire X chromosome in female mammals when this X becomes inactivated
  5. Is characterized by the absence of crossing over and replication late in the S phase

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. How does histone acetylation affect chromatin?

 

  1. It loosens the chromatin and allows increased transcription.
  2. It allows DNA to become resistant to damage.
  3. It helps the histones have a greater affinity for DNA.
  4. It inhibits DNA replication by making it more difficult to separate the DNA strands.
  5. It causes the chromatin to become more condensed in preparation for metaphase.

 

Answer: a

Section: 11.1

Comprehension Question

 

  1. Which of the following is an example of an epigenetic change in eukaryotes?

 

  1. A loss of an AT base pair from a gene
  2. The addition of methyl groups to cytosines in the promoter region of a gene
  3. The substitution of an AT base pair by a GC base pair in a gene as a result of a mistake during DNA replication
  4. A deletion that simultaneously removes two genes from the genome
  5. None of the above represent epigenetic changes.

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. The agouti locus helps determine coat color in mice, and this phenotype can vary from light to dark between genetically identical individuals. You have discovered a drug that reduces the variation in the agouti  What is a likely explanation for this drug’s mechanism of action.

 

  1. Inhibits DNA polymerases
  2. Inhibits DNA methyl transferases
  3. Activates shelterin proteins
  4. Activates mitochondrial transcription
  5. Causes DNA damage

 

Answer: b

Section: 11.1

Comprehension Question

 

  1. Which of the following has repetitive DNA and heterochromatin?

 

  1. Telomere
  2. Centromere
  3. Mitochondria
  4. Chloroplast
  5. a and b

 

Answer: e

Section: 11.2

Comprehension Question

 

  1. Where would you expect to find the variant histone CenH3?

 

  1. telomere
  2. euchromatin
  3. centromere
  4. mitochondria
  5. chloroplast

 

Answer: c

Section: 11.2

Comprehension Question

 

  1. Telomeres exist to help with the _________ of the ends of eukaryotic chromosomes.

 

  1. transcription
  2. replication
  3. metabolism
  4. destabilization
  5. translation

 

Answer: b

Section: 11.2

Comprehension Question

 

  1. A normal chromosome in a higher eukaryotic species would be expected to contain all of the following except:

 

  1. One centromere
  2. One telomere
  3. Two copies of histone 2A per nucleosome
  4. Satellite DNA
  5. Tandem repeat sequences

 

Answer: b

Section: 11.2

Comprehension Question

 

  1. An Alu sequence is an example of which type of DNA sequence in eukaryotes?

 

  1. Moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: c

Section: 11.3

Comprehension Question

 

  1. A telomere is an example of which type of DNA sequence in eukaryotes?’

 

  1. moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: b

Section: 11.3

Comprehension Question

 

  1. A gene-encoding sequence is an example of which type of DNA sequence in eukaryotes?

 

  1. Moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: e

Section: 11.3

Comprehension Question

 

  1. A tRNA gene is an example of which type of DNA sequence in eukaryotes?

 

  1. Moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: a

Section: 11.3

Comprehension Question

 

  1. A ribosomal RNA gene is an example of which type of DNA sequence in eukaryotes?

 

  1. Moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: a

Section: 11.3

Comprehension Question

 

  1. A centromere is an example of which type of DNA sequence in eukaryotes?

 

  1. Moderately repetitive DNA
  2. Highly repetitive DNA
  3. Short interspersed elements
  4. Long interspersed elements
  5. Unique-sequence DNA

 

Answer: b

Section: 11.3

Comprehension Question

 

  1. Copies of a gene that arose by gene duplication are part of a gene _______.

 

  1. Complex
  2. Family
  3. Tandemoplex
  4. Structure
  5. Chromosome

 

Answer: b

Section: 11.3

Comprehension Question

 

  1. Which of the following statements is true?

 

  1. Most proteins in the human mitochondrion are encoded by nuclear genes.
  2. One piece of evidence supporting the endosymbiotic theory is the extreme similarity between mitochondrial DNAs from different organisms.
  3. Heteroplasmy refers to the presence of different alleles in a single organelle.
  4. Plants contain chloroplasts, not mitochondria.
  5. cpDNA evolves faster than nuclear DNA.

 

Answer: a

Section: 11.4

Comprehension Question

 

  1. Which of the following statements is not true?

 

  1. Both the mitochondria and the chloroplast generate ATP.
  2. A single eukaryotic cell may contain thousands of copies of the mitochondrial genome.
  3. According to the endosymbiotic theory, chloroplasts are thought to have evolved from cyanobacteria.
  4. The mutation rate of mitochondrial DNA is higher than the mutation rate of nuclear DNA.
  5. Oxidative phosphorylation capacity is constant throughout a person’s lifetime.

 

Answer: e

Section: 11.4

Comprehension Question

 

  1. The __________ membrane of the chloroplast bears the enzymes and pigments required for photophosphorylation.

 

  1. Outer
  2. Middle
  3. Thylakoid
  4. Plasma
  5. Double

 

Answer: c

Section: 11.4

Comprehension Question

 

  1. How many membranes separate the mitochondrial matrix from the cytoplasm?

 

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4

 

Answer: c

Section: 11.4

Comprehension Question

 

  1. How many different types of histones are found in the nucleosome that packages mitochondrial DNA?

 

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4

 

Answer: a

Section: 11.4

Comprehension Question

 

  1. The __________ theory states that the ancestors of mitochondria and chloroplasts were free-living bacteria.

 

  1. Phylogenetic
  2. Endosymbiotic
  3. Cell
  4. Cytoplasmic inheritance
  5. Old world

 

Answer: b

Section: 11.4

Comprehension Question

 

  1. The presence of more than one type of DNA in the organelles of a single cell is called __________.

 

  1. Homoplasmy
  2. Heteroplasmy
  3. Hemiplasmy
  4. Pseudoplasmy
  5. Paraplasmy

 

Answer: b

Section: 11.4

Comprehension Question

 

  1. Pea plants produce both pollen and eggs. A pea plant inherits a mutation for cytoplasmic male sterility. How will this affect the plant and/or its progeny?

 

  1. The plant will be able to reproduce only by self-fertilization.
  2. The plant will be able to reproduce only by cross-fertilization.
  3. The plant will be unable to produce progeny.
  4. The plant will produce progeny, but the progeny will not be able to reproduce.

 

Answer: b

Section: 11.4

Comprehension Question

 

  1. Paternal transmission of mitochondria is common in which group?

 

  1. humans
  2. mice
  3. most gymnosperms
  4. most flowering plants
  5. insects

 

Answer: c

Section: 11.4

Comprehension Question

 

  1. Which statement about mitochondrial genomes is not true?

 

  1. In most animals, the mitochondrial genome consists of a single circular DNA molecule.
  2. Plant mitochondrial genomes often include multiple circular DNA molecules.
  3. Each mitochondrion typically contains many copies of the mitochondrial genome.
  4. All copies of the mitochondrial genome within a cell are identical.

 

Answer: d

Section: 11.4

Comprehension Question

 

  1. What kind of gene would not be found in a chloroplast genome?

 

  1. a tRNA gene
  2. a gene for a subunit of the photosynthesis enzyme RuBisCO
  3. a gene for a ribosomal protein
  4. a gene for ribosomal RNA
  5. a gene for a histone protein

 

Answer: e

Section: 11.4

Comprehension Question

 

  1. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite yeast.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The peak farthest to the right in Figure 1 is:

 

  1. mitochondrial DNA.
  2. genomic DNA.
  3. chloroplast DNA.

 

Answer: d

Section: 11.4

Comprehension Question

 

  1. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The large peak to the left in all the figures is:

 

  1. mitochondrial DNA.
  2. genomic DNA.
  3. chloroplast DNA.

 

Answer: c

Section 11.4

Comprehension Question

 

  1. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Which figure has the DNA profile for the petite yeast?

 

  1. 1
  2. 2
  3. 3
  4. 4

 

Answer: b

Section: 11.4

Comprehension Question

 

  1. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Which figure has the DNA profile for the plant cell?

 

  1. 1
  2. 2
  3. 3
  4. 4

 

Answer: a

Section: 11.4

Comprehension Question

 

  1. DNA can be isolated from cells, and the different types of DNA from the cell can be separated by density gradient centrifugation. The following figures show the density gradient centrifugation profiles for DNA isolated from four different cell types: plant, animal, wild-type yeast with a plasmid, and petite

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Which figure has the DNA profile that would most closely match DNA from a human cell?

 

  1. 1
  2. 2
  3. 3
  4. 4

 

Answer: c

Section 11.4

Comprehension Question

 

 

  1. Assuming there are no heteroplasmic individuals but that people from different families have different DNA, how many different mitochondrial DNAs are there in the pedigree

 

 

 

 

 

 

 

  1. 2
  2. 4
  3. 5
  4. 6
  5. 7

 

Answer: c

Section: 11.4

Application Question

 

  1. Which pair in the pedigree shares the same mitochondrial DNA?

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. I-2 and III-1
  2. II-2 and III-2
  3. III-1 and III-2
  4. II-2 and III-1
  5. I-3 and II-6

 

Answer: b

Section 11.4

Application Question

 

Short-Answer Questions

 

  1. When chromatin from any eukaryote is digested with micrococcal nuclease (an endonuclease) and fractionated using electrophoresis, DNA fragments of approximately 200 base pairs in length are observed. Explain this result.

 

Answer: Chromatin is composed of repeating nucleosomes that consist of about 200 base pairs of DNA wrapped around a histone octamer complex.  DNA wrapped around the octamer core is protected from endonuclease degradation.  However, because the linker DNA between adjacent nucleosomes is exposed, it can be cleaved by endonucleases.  Thus, chromatin digested by endonucleases generates 200 base-pair-long fragments of DNA.

Section: 11.1

Application Question

 

  1. Describe the structure and packing of a bacterial chromosome.

 

Answer: Bacterial cells typically contain a single, circular DNA molecule associated with various proteins to make a compact structure called a nucleoid.  Bacteria do not contain histones.   Chromosome packing is thought to organize the genetic material into discrete looped domains containing genes that can be selectively activated by relaxing the degree of supercoiling within given domains.

Section: 11.1

Application Question

 

  1. If a bacterial chromosome were inserted into a eukaryotic cell, would it be stable? Would it segregate like eukaryotic chromosomes do during mitosis and meiosis?  Why or why not?

 

Answer: No, it would not be stable because: (1) it would not contain a eukaryotic-specific origin of replication, and therefore it could not replicate in a eukaryotic cell, and (2) bacterial chromosomes don’t have centromeres,   so they can’t segregate in eukaryotic cells (no kinetochores, no spindle fibers, etc.).   The bacterial chromosome would be lost and eventually degraded.

Section: 11.1

Application Question

 

  1. While working on Drosophila you find a new mutant strain with an abnormal histone H3 gene. This novel histone mutant is predicted to cause the nucleosomes to bind an extra 15 bp of DNA compared to wild type. If you digest isolated chromatin with a nuclease to release the core nucleosomes, what size DNA fragments would you expect from wild type and mutant flies?

 

Answer: The core nucleosome from wild type flies wraps approximately 145 bp of DNA; therefore, if the mutant histone is predicted to wrap an extra 15 bp, the mutant nucleosomes should wrap 160 bp.

Section: 11.1

Application Question

 

  1. How does the organization of the eukaryotic chromosome differ from the organization of a bacterial chromosome? Include in your answer:

 

  1. Classes of DNA sequences.
  2. Special features of the chromosome.
  3. Organization of genes within the chromosome.
  4. Proteins that interact with chromosomal DN.A

 

Answer:

  1. Bacterial chromosomes generally consist of mainly unique DNA sequences with very little moderately repetitive or highly repetitive DNA. In contrast, eukaryotic chromosomes typical contain unique, moderately repetitive DNA and highly repetitive DNA. The unique component contains many of the protein-encoding genes as well as other poorly defined unique sequences. The moderately repetitive sequences consist of some gene families, telomeric DNA, and many of the transposable elements. The highly repetitive DNA, which is unlikely to be transcribed, can be found within the centromeric regions in some species and elsewhere. Repetitive DNA sequences may be tightly coiled and may stain darkly (heterochromatin).

 

  1. The typical bacterial genome is assumed to consist of one circular chromosome while exceptions are known. Eukaryotic genomes consist of multiple linear chromosomes with a characteristic number for each species. Eukaryotic chromosomes consist of two telomeres and usually, but not always, a single centromere. Telomeres are essential for protecting the ends of linear chromosomes and allowing a way of completing chromosome replication without resulting in the shortening of the linear chromosome (see telomerase in Chapter 12). Since bacterial chromosomes are usually circular, telomeres are not needed. The centromere is the portion of the chromosome where microtubules attach during mitosis and meiosis. It often consists of highly repetitive DNA although a specific sequence is not essential to its function. The bacterial chromosome doesn’t possess a centromere although it does have a mechanism to insure the accurate segregation of daughter chromosomes to daughter cells during cell division. The details of this mechanism are still not completely understood.

 

  1. The bacterial chromosome consists of mainly genes with some transposable elements. The DNA sequences between these structures are usually quite short so that the gene density (number of genes per a particular number of nucleotides of DNA) is high. In contrast, eukaryotic chromosomes are often “gene sparse” which few genes per the same number of nucleotides and many sequences not involved in forming genes. Many eukaryotic genes are part of gene families whereas bacterial genes are usually unique with the main exception being the rRNA genes.

 

  1. Both types of chromosomes interact with a variety of proteins. With eukaryotes, the nucleosome with its set of histone proteins is a basic structural unit of the chromosome, involved in both the packaging of the DNA and the regulation of gene activity. Additional proteins also interact with eukaryotic DNA to provide additional gene regulation and higher levels of chromosome organization. Bacterial chromosomes also interact with certain proteins, although not histones, and such proteins are likely involved in gene regulation and chromosome structure and organization, but the details are not yet known.

Section: 11.1

Application Question

 

  1. There are some genomes that have been reported to be positively coiled instead of negatively supercoiled, which is the status of most genomes that we have studied. The genomes that are positively supercoiled seem to belong to viruses and cells that exist at very high temperatures. Why might positive supercoiling be an advantage at high temperatures?

 

Answer: Negative supercoiling makes it easier for the two strands of DNA to separate

(denature), whereas positive supercoiling makes it more difficult for strands to separate. At high temperatures the heat energy is already producing a tendency for the strands to separate by breaking the hydrogen bonds. While separation is needed for replication and transcription to occur, it is necessary for DNA to retain its double-stranded structure at other times and positive supercoiling likely facilitates this.

Section: 11.1

Application Question

 

  1. Explain why DNA-DNA hybridization might be useful in helping to assess evolutionary relationships.

 

Answer: Two single-stranded DNA molecules will anneal if complementary; the strength of the association will be greater the more the two strands are complementary. We assume that the greater the similarity of DNA sequence, the more closely related the two species.

Section: 11.3

Application Question

 

  1. What is “satellite” DNA? Explain why satellite DNA anneals rapidly after it is denatured.

 

Answer: Satellite DNA, also called highly repetitive DNA, is one of three major classes of DNA sequences found in eukaryotic genomes.  This class of DNA is present in hundreds of thousands to millions of copies.   Highly repetitive fractions of DNA exhibit rapid reassociation kinetics because they are characterized by having numerous (up to millions) copies of identical sequence repeats; thus, once melted, complementary partner strands are readily available and can rapidly reanneal.

Section: 11.3

Application Question

 

  1. List and describe the three major classes of DNA sequences in the eukaryotic genome.

 

Answer:

  • Single-copy sequences (1-10 copies per genome)
  • Moderately repetitive sequences (10-100,000 copies per genome; repeats average 150-300 base pairs)
  • Highly repetitive sequences (100,000 millions of copies per genome; repeats usually, but not always, 10 base pairs or fewer)

Section: 11.3

Application Question

 

  1. While doing field work you discover two new closely related species of oysters. You measure the DNA content per cell of each species and find that the second species has significantly more DNA than the first species.  DNA hybridization analysis of both species yields the following results. Based on the graphs shown, suggest an explanation for the difference in C-value between the species.

 

 

 

 

 

 

 

 

 

 

 

 

Answer: Both species appear to have the same amount of highly repetitive and unique DNA, but the second species appears to have significantly more moderately repetitive DNA.  One possible explanation for the presence of more moderately repetitive DNA in the second species is accumulation of replicative transposons in that species, but not in the first.

Section: 11.3

Application Question

 

  1. “Mitochondrial Eve” is the name given to the idea that all humans alive today can trace their mitochondrial ancestry to a single African female alive around 150,000 years ago. What feature of mitochondrial inheritance makes this a reasonable conjecture?

 

Answer: Mitochondrial Eve is the most recent common matrilineal ancestor of all humans as measured through the mitochondria. mtDNA is maternally inherited, but more importantly, there is no recombination between parental genomes as is the case with nuclear genes. Therefore, it is possible to easily follow the inheritance of the mitochondria from generation to generation.

Section: 11.4

Application Question

 

  1. Two haploid strains of petite yeast mutants are obtained independently. Each is crossed to a wild-type strain, and the resulting diploid is sporulated (goes through meiosis to produce haploid spores).  Use the following results to explain the difference between the two strains, and why the crosses give different results.

 

Strain petite A ´ wild type                             Strain petite B ´ wild type

 

 

 

non-petite diploid                                           non-petite diploid

 

 

 

50% non-petite haploid spores                        81% non-petite haploid spores

50% petite haploid spores                               19% petite haploid spores

 

Answer: Strain A has a recessive nuclear mutation that causes the petite phenotype.  Crossing with a wild type gives a heterozygous diploid.  When sporulated, half the haploid spores have the normal allele and half have the recessive allele that causes the petite phenotype.

 

Strain B has a mitochondrial mutation.  Crossing with a wild type gives a heteroplasmic diploid.  When sporulated, spores randomly receive either normal mitochondria, mitochondria with the mutant petite allele, or a mixture.  Those with normal mitochondria or a mixture will be non-petite, but the exact proportion of wild-type, non-petites is random.

Section: 11.4

Application Question

 

  1. A new Drosophila phenotype is investigated with a series of crosses. P (parental) organisms are true-breeding. The following is the first cross:

 

P      wild-type female ´ mutant male

 

 

 

F1                          all wild type

 

 

 

F2                                  ?

 

Predict the F2 results if the allele that causes the mutant phenotype is X-linked recessive.  Then, predict the results if the allele that causes the mutant phenotype is mitochondrial.

 

Answer: If X-linked recessive, 3/4 of the F2 will be wild type.  Two-thirds of the wild type will be female and 1/3 will be male.  One-quarter of the F2 will be mutant, and all will be male.

If mitochondrial, all F2, male and female, will be wild type.

Section: 11.4

Application Question

 

  1. A new Drosophila phenotype is investigated with a series of crosses. P (parental) organisms are true-breeding. The following is the first cross:

 

P      wild-type female ´ mutant male

 

 

 

F1                          all wild type

 

 

 

F2                                  ?

 

 

A reciprocal cross to the one just shown is performed. Predict the results in the F1 and F2 generations, if the mutant phenotype is X-linked recessive. Then, predict the results if the allele that causes the phenotype is mitochondrial.

Answer: If X-linked recessive, 1/2 the F1 will be wild-type females and 1/2 will be mutant males.  In the F2, 1/4 will be wild-type males; 1/4 will be wild-type females; 1/4 will be mutant males; and 1/4 will be mutant females.

If mitochondrial, all F1 and all F2 will be mutant.

Section: 11.4

Application Question

 

  1. Using the following pedigree, explain whether each of the following inheritance patterns is possible for the phenotype being followed, citing specific individuals in your answer: X-linked dominant, X-linked recessive, Y-linked, mitochondrial, autosomal recessive, autosomal dominant.

 

 

 

 

I                                         1                       2

 

 

 

 

 

 

II    1                      2               3                  4 5                      6

 

 

 

 

 

 

 

 

III            1                                                           2                 3                      4

 

 

Answer:

  • The phenotype cannot be X-linked dominant because II-4 doesn’t have the same phenotype as I-1.
  • The phenotype cannot be X-linked recessive. III-3 and 4 don’t have the same phenotype as II-5.
  • The phenotype cannot be Y-linked. II-2 doesn’t have the same phenotype as I-1; III-3 and III-4 don’t have the same phenotype as the II-6.
  • The phenotype cannot be mitochondrial. II-3 and II-5 don’t have the same phenotype as I-2. III-3 and III-4 don’t have the same phenotype as II-5.
  • The phenotype cannot be autosomal recessive. III-3 and III-4 don’t have the same phenotype as II-5.
  • The phenotype could be autosomal dominant if II-5 and II-6 are heterozygotes, and III-3 and III-4 are homozygous recessive.

Section: 11.4

Application Question

  1. Draw a pedigree using the following information, filling in symbols for the mitochondrial disease: In generation I, the mother is affected with a mitochondrial disease but the father is not. They have two children.  The oldest, a male, has a son with an unaffected woman.  The youngest, a female, has a daughter with an unaffected man.

 

Answer:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Section: 11.4

Application Question

 

  1. Explain why the symptoms for human mitochondrial diseases usually appear in adulthood, and become worse with age.

 

Answer: Mitochondrial DNA mutations affect oxidative phosphorylation and the production of ATP.  This process declines with ageRemember, someone with weakened mitochondrial function would show symptoms when his or her capacity for oxidative phosphorylation declined past a critical threshold.  Symptoms would become worse as the ability to perform oxidative phosphorylation declined.

Section: 11.4

Application Question

 

  1. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leavesRemember,me of the progeny have green and yellow leaves and some have all green leaves. When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. Explain the phenotype of the plant’s leaves.

 

Answer: The plant is heteroplasmic for a mutation in the chloroplast genome that affects chloroplast development and/or photosynthesis.  The yellow spots are cells that, by replicative segregation, have received only mutant chloroplast genomes.  The green areas are cells that are either homoplasmic for normal chloroplasts or heteroplasmic.

Section: 11.4

Application Question

 

  1. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leavesRemember,me of the progeny have green and yellow leaves and some have all green leaves. When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves. Explain the results of the crosses with the plant.

 

Answer: The chloroplasts are inherited maternally in this plant.  When the plant is the egg donor, by replicative segregationRemember,me eggs will contain only normal chloroplasts and the progeny will have all green leaves.  Some eggs will be heteroplasmic and have the same phenotype as the maternal parent.  Presumably, eggs that are homoplasmic for mutant chloroplasts will not produce viable plants. When the plant is the pollen donor, the plant with nonmutant chloroplast DNA will contribute the chloroplasts, and all progeny will have all-green leaves.

Section: 11.4

Application Question

 

  1. A plant has green leaves with multiple yellow spots. When used as an egg donor in a cross with a normal plant that has all green leavesRemember,me of the progeny have green and yellow leaves and some have all green leaves. When used as the pollen donor in a cross with a normal plant, all the progeny have all green leaves.

 

One of the progeny with green and yellow leaves has one branch with all green leaves and one branch with all yellow leaves. A flower on the all-green branch is fertilized with pollen from a flower on the all-yellow branch. Predict the results of this cross and the reciprocal cross.

 

Answer: When the all-green branch supplies the egg, the progeny will all have normal chloroplasts and have all green leaves.  When the all-yellow branch supplies the egg, the progeny will all have mutant chloroplasts, and will not survive.

Section: 11.4

Application Question

 

  1. At the end of the nineteenth century, American bison were bred with domestic cattle in an attempt to rescue their declining populations. The resulting hybrids were bred with true bison. After many generations, animals that look like bison may still contain some ancestral cattle DNA. Ward et al. (Animal Conservation 2: 51-57, 1999) tested current bison herds for a mitochondrial DNA marker specific to cattle.  Of the North American bison they tested, 5.2% had the cattle-specific mitochondrial DNA marker. Does this under- or overestimate the number of bison-cattle hybrids? Explain your answer.

 

Answer: It underestimates the number of bison with cattle ancestry because it detects only bison with maternal cattle ancestry.

Section: 11.4

Application Question

 

  1. At the end of the nineteenth century, American bison were bred with domestic cattle in an attempt to rescue their declining populations. The resulting hybrids were bred with true bison. After many generations, animals that look like bison may still contain some ancestral cattle DNA. Conservation geneticists have considered isolating bison that show no cattle ancestry by the mitochondrial DNA test, and breeding them to maintain a group of “pure” bison. The number of bison with no cattle genes is estimated at 15,000 animals (The New York Times, 23 April 2002, “Genetically, bison don’t measure up to frontier ancestors”). List one reason this might not be a good idea.

 

Answer: Bison were already experiencing a population “bottleneck” at the end of the nineteenth century, when the number of breeding adults was low. Their genetic diversity was reduced at that timeRemember, selecting only pure animals returns the population to bottleneck numbers. Selecting bison by the mitochondrial DNA test only might mean including bison that have cattle genes inherited from paternal ancestors.

Section: 11.4

Application Question

 

  1. Jack and Jill’s son Jake has a severe case of myclonic epilepsy and ragged-red fiber (MERFF) syndrome. His case includes frequent and disabling myclonic seizures (involuntary twitching of the muscles) along with hearing loss, exercise intolerance, and poor night vision. Like most cases of MERFF, his is associated with a mitochondrial mutation that he inherited from his mother Jill. His mother didn’t know that she harbors the MERFF mutation among her mtDNA molecules, but she has experienced occasional mild muscle twitching throughout her life and she does not see very well at night.

 

  1. What is the most likely explanation for the difference in the severity of MERFF between Jake and his mother?
  2. Jack and Jill would like to have another child. They want to know the probability that their next child will also suffer from MERRF syndrome. How would you counsel them?

 

Answer:

  1. The most likely explanation is heteroplasmy for mtDNA molecules in the cells of Jake and his mother. When a female is heteroplasmic for mtDNA, the proportion of mutant mtDNA that gets passed to an offspring is random.  Therefore, it is likely that Jake has a higher proportion of mutant mtDNA molecules in his cells compared to his mother, producing a more severe syndrome.
  2. Counseling in this situation is highly problematic. This chapter cited the case of a 20-year-old man who carried the MERRF mutation in 85% of his mtDNAs and was normal and his cousin who carried it in 96% of his mitochondria and was severely affected.  There appears to be a sharp threshold in the percentage of abnormal mtDNAs between having the syndrome and not having it.  Later in the chapter the author describes a bottleneck during which the mtDNAs within the cells are reduced to just a few copies.  During this bottleneck, the percentage of abnormal mtDNAs can increase or decrease randomly and significantly, making it impossible to predict with confidence the percentage of abnormal mtDNAs that will be passed to the next generation.  Therefore, it is impossible to predict with confidence the probability that the next child will be affected.

Section: 11.4

Application Question

 

 

 

 

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