Bsci110b Chapter 15 Review Questions

14 Questions | Total Attempts: 87

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BSCI110B Chapter 15 Review Questions


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  • 1. 
    A man with hemophilia (a recessive, sex-linked condition) has a daughter of normal phenotype.  She marries a man who is normal for the trait.  What is the probability that a daughter of this mating will be a hemophiliac?  That a son will be a hemophiliac?  If the couple has four sons, what is the probability that all four will be born with hemophilia? ;  ; 
  • 2. 
    Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles.  It is seen almost exclusively in boys born to apparently normal parents and usually results in death in the early teens.  Is this disorder caused by a dominant or recessive allele?  Is its inheritance sex-linked or autosomal?  How do you know?  Explain why this disorder is almost never seen in girls.             , if the disorder were , it would affect at least one parent of a child born with the disorder.  The disorder’s inheritance is sex-linked because it is seen only in .  For a girl to have the disorder she would have to inherit recessive alleles fro parent.  This would be very rare, since males with the recessive allele on their X chromosome die in their early teens.
  • 3. 
    Red-green color blindness is caused by a sex-linked recessive allele.  A color-blind man marries a woman with normal vision whose father was color-blind.  What is the probability that they will have a color-blind daughter?  What is the probability that their first son will be color-blind?            for each daughter;  for first son.
  • 4. 
    A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings.  The offspring have the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162.  What is the recombination frequency between these genes for body color and wing size?             %
  • 5. 
    In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes.  The offspring are as follows: wild type, 721; black-purple, 751; gray-purple, 49; black-red, 45.  What is the recombination frequency between these genes for body color and eye color?             %
  • 6. 
    What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?             The disorder would always be inherited from the .
  • 7. 
    Women born with an extra X chromosome (XXX) are healthy and phenotypically indistinguishable from normal XX women.  What is a likely explanation fro this finding?  How could you test this explanation?             The inactivation of the  X chromosome in XXX women would leave them wit  genetically active X chromosome, as in women with the normal number of chromosomes.  Microscopy should reveal two  in XXX women.
  • 8. 
    Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units: A-C, 28 map units; A-D, 25 map units; B-C, 20 map units; B-D, 33 map units.        - - -
  • 9. 
    Assume that genes A and B are linked and are 50 map units apart.  An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci.  What percentage of the offspring will show phenotypes resulting from crossovers?  If you did not know that genes A and B were linked, how would you interpret the results of this cross?             % of the offspring would show phenotypes that resulted from crossovers.  These results would be the same as those from a cross where A and B were not linked.  Further crosses involving other genes on the same chromosome would reveal the linkage and map distances.
  • 10. 
    A space probe discovers a planet inhabited by creatures that reproduce with the same hereditary patterns seen in humans.  Three phenotypic characters are height (T=tall, t=dwarf), head appendages (A=antennae, a=no antennae), and nose morphology (S=upturned snout, s=downturned snout).  Since the creatures are not “intelligent,” Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses.  For tall heterozygotes with antennae, the offspring are: tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5.  For heterozygotes with antennae and an upturned snout, the offspring are: antennae-upturned snout, 47; antennae-downturned snout, 2; no antennae-downturned snout, 48; no antennae=upturned snout, 3.  Calculate the recombination frequencies for both experiments.             Between T and A, %; between A and S, %.
  • 11. 
    Using the information from problem 11, scientists do a further testcross using a heterozygote for height and nose morphology.  The offspring are: tall-upturned snout, 40; dwarf-upturned snout, 9; dwarf-downturned snout, 42; tall-downturned snout, 9.  Calculate the recombination frequency from these data; then use your answer from problem 11 to determine the correct sequence of the three linked genes             Between T and S, %, sequence of genes is  - -
  • 12. 
    Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart.  You cross a homozygous blue-oval plant with a homozygous white-round plant.  The resulting F1 progeny are crossed with homozygous white-oval plants, and 1,000 F2 progeny are obtained.  How many F2 plants of each of the four phenotypes do you expect?              each of blue-oval and white-round (parentals) and  each of blue-round and white-oval (recombinants)
  • 13. 
    You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome.  Gene a has recombination frequencies of 14% with the vestigial-wing locus and 26% with the brown-eye locus.  Where is a located on the chromosome?             About  the distance from the vestigial-wing locus to the brown-eye locus.
  • 14. 
    Banana plants, which are triploid, are seedless and therefore sterile.  Propose a possible explanation.             Because bananas are triploid,  pairs cannot line up during meiosis.  Therefore, it is not possible to generate gametes that can fuse to produce a zygote with the triploid number of chromosomes.
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