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One map unit on a human chromosome corresponds roughly to


A) 10 base pairs.
B) 1 million base pairs.
C) 1 base pair.
D) 100,000 base pairs.
E) 1,000 base pairs.

F) A) and E)
G) A) and C)

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What process observed by Curt Stern resulted in twin spot formation in Drosophila?


A) Chromosome loss
B) Mitotic recombination
C) Chromosome nondisjunction
D) Meiotic recombination
E) Incomplete dominance

F) A) and C)
G) A) and B)

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Neurospora crassa, a mycelial-form fungus, is useful for genetic and biochemical studies because it


A) produces meiotic products contained in an ascus.
B) is haploid.
C) has a relatively short life cycle.
D) All of the above
E) None of the above

F) C) and D)
G) B) and E)

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A polymorphic DNA marker is one in which more than one type of sequence is found in the population at that site in the genome.

A) True
B) False

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What are heterokaryons?

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They are fungal cells in which...

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Mitotic rather than meiotic recombination analysis has been used to construct genetic maps in the fungus Aspergillus.

A) True
B) False

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If two genes are linked, then the frequency of parental ditype tetrads resulting from a two-point testcross will be


A) equal to the frequency of nonparental ditype tetrads.
B) equal to the frequency of the tetratype tetrads.
C) greater than the frequency of the nonparental ditype tetrads.
D) less than the frequency of tetratype tetrads.
E) none of the above.

F) A) and B)
G) B) and D)

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If 156 tetrads are analyzed following a series of two-point testcrosses in Neurospora, and 102 of them are PD, 40 are T, and the remaining 14 are NPD, what is the frequency of recombination between the two genes?

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The recombination frequency is...

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In the life cycles of yeast and Chlamydomonas, the haploid vegetative cells never undergo fusion to form a diploid cell.

A) True
B) False

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How can you calculate the recombination frequency between a and b in the previous question?

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We can assume that the grandmother is homozygous wild-type for the genes, since the recessive alleles are rare. Her daughter will then be a heterozygote (a+ b+/a b). When she marries a normal a+ b+/Y man, their male children will express all four possible phenotypic classes (since they are hemizygous): the female parental a+ b+ and ab and the recombinants a+ b and a b+. If a large number of such pedigrees are available, the recombination frequency can be calculated.

In the yeast Saccharomyces cerevisiae, when an MATa mating type fuses with a MATΞ± mating type, the resulting cell is diploid.

A) True
B) False

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In order to use tetrad analysis in an organism to map the distance between a gene and the centromere on a chromosome,


A) the organism may produce either ordered or unordered tetrads.
B) the organism must produce unordered tetrads.
C) the parental centromeres must be distinguishable.
D) the chromosome must have a cytological marker.
E) the organism must produce ordered tetrads.

F) C) and D)
G) A) and E)

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During the analysis of an ordered tetrad, 22 percent of the asci showed second-division segregation for a particular genetic locus. What is the distance between the gene and the centromere?

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Eleven map...

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DNA markers allow genetic mapping in humans on the basis of molecular phenotypes, rather than outwardly visible phenotypes.

A) True
B) False

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At what stage may mitotic recombination occur?

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When homologs align ...

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Explain how the two formulas for determining the distance between linked genes in diploid organisms and in tetrad analysis relate to each other.

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In diploid organisms, we calculate the recombination frequency by determining the percentage of recombinant progeny out of the total number of progeny. In tetrad analysis, we determine the proportion of recombinant spores in tetrads produced from crosses out of the total number of spores. Therefore, in diploid organisms, the formula is: (number of recombinants/total number of progeny) Γ— 100. In tetrad analysis, it is: (1/2T + NPD/total tetrads) Γ— 100.

Following a series of two-point testcrosses in Neurospora, 200 tetrads were analyzed. Of these, 100 were PD and 100 were NPD. These two genes are


A) 50 map units apart on the same chromosome.
B) 100 map units apart on the same chromosome.
C) epistatic.
D) on different chromosomes.
E) adjacent to each other on the same chromosome.

F) A) and B)
G) A) and C)

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X-linked genes in humans are easier to study than autosomal genes because


A) recombination frequencies are lower.
B) hemizygosity in males allows the effects of genotypic pairings to be observed.
C) Suitable pedigrees are easily available.
D) they do not have as high a recombination rate.
E) there are fewer genes to deal with.

F) All of the above
G) B) and E)

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How does the arrangement of ascospores in the tetrads of yeast differ from that in Neurospora crassa?

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The ascospores in a tetrad of yeast are unordered (random), whereas those of Neurospora are linearly ordered.

Describe the possible tetrad types resulting from a cross qr Γ— q+r+ in which the genes are located on separate chromosomes.

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Depending on the metaphase alignment of ...

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