A) The phenotype presented is X-linked dominant because II-3 has the same phenotype as I-1.
B) The phenotype presented is X-linked recessive because III-3 and III-4 do not have the same phenotype as II-5.
C) The phenotype cannot be Y-linked because II-2 doesn't have the same phenotype as I-1.
D) The phenotype could be autosomal recessive as III-3 and III-4 don't have the same phenotype as II-5.
E) The phenotype could be autosomal dominant if II-5 and II-6 are heterozygotes, and III-3 and III-4 are homozygous recessive.
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Multiple Choice
A) 2
B) 4
C) 5
D) 6
E) 7
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Multiple Choice
A) Both the mitochondria and the chloroplast generate ATP.
B) A single eukaryotic cell may contain thousands of copies of the mitochondrial genome.
C) According to the endosymbiotic theory, chloroplasts are thought to have evolved from cyanobacteria.
D) The mutation rate of mitochondrial DNA is higher than the mutation rate of nuclear DNA.
E) Oxidative phosphorylation capacity is constant throughout a person's lifetime.
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Multiple Choice
A) zero
B) one
C) two
D) three
E) four
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Multiple Choice
A) Negative supercoiling makes it more difficult for strands to separate while positive supercoiling would allow the strands to separate more readily.
B) At high temperatures, the condition is more conducive for the strands to denature.
C) The high temperature would increase the formation of the hydrogen bonds between bases.
D) Positive supercoiling would allow the DNA to maintain its double-stranded structure at higher temperature.
E) Positive supercoiling would allow the DNA to readily separate for transcription and replication.
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Multiple Choice
A) 0
B) 5
C) 10
D) 15
E) 100
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Multiple Choice
A) The defect in the cellular growth comes from an inability to generate enough ATP.
B) The growth defect is known to come from having excessive copies of mitochondria, resulting in toxicity from excess ATP.
C) The mutations on the mtDNA can result in deficiency of the enzymes involved in aerobic respiration.
D) They have no means to make any ATP because of mtDNA defects that affect the normal mitochondrial functions.
E) The petite mutants only have to rely on anaerobic processes such as fermentation and glycolysis.
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Multiple Choice
A) homoplasmy.
B) heteroplasmy.
C) hemiplasmy.
D) pseudoplasmy.
E) paraplasmy.
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Not Answered
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Multiple Choice
A) transcription
B) replication
C) metabolism
D) destabilization
E) translation
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Multiple Choice
A) It would not be stable due to the lack of a eukaryotic-specific origin of replication; hence, it could not replicate properly in a eukaryotic cell.
B) It would be generally stable because the chemical nature of DNA is the same regardless of the cell type.
C) Due to the lack of centromeres on prokaryotic chromosomes, the chromosomes will not segregate normally during cell division.
D) The prokaryotic chromosome can be induced to be stabilized by cleavage of circular form to mimic linear eukaryotic chromosome.
E) The bacterial chromosome would be lost and eventually degraded.
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Multiple Choice
A) alanine
B) arginine
C) leucine
D) valine
E) serine
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Multiple Choice
A) 0
B) 5
C) 10
D) 15
E) 100
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Multiple Choice
A) 1
B) 2
C) 3
D) 4
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Multiple Choice
A) It remains in a highly condensed state throughout the cell cycle.
B) It makes up most chromosomal material and is where most transcription occurs.
C) It exists at the centromeres and telomeres.
D) It occurs along one entire X chromosome in female mammals when this X becomes inactivated.
E) It is characterized by the absence of crossing over and replication late in the S phase.
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