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
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.
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?
daughter; for first son.
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?
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?
What pattern of inheritance would lead a geneticist to
suspect that an inherited disorder of cell metabolism is due to a defective
disorder would always be inherited from the .
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
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.
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.
- - -
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.
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.
and A, %; between A and S, %.
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
and S, %, sequence of genes is - -
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?
blue-oval and white-round (parentals) and each of blue-round and white-oval
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.
Banana plants, which are triploid, are seedless and
therefore sterile. Propose a possible
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