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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
This is a closed book, closed notes exam. Please confine your answers to the space provided.
1. In the human pedigree shown above, shaded individuals are affected by vision loss in young
adulthood. Squares represent males and circles females. Roman numerals indicate generations
and Arabic numerals indicate individuals.
1a [5 pt] List two features of the inheritance pattern for vision loss that are consistent with
vision loss due to a mitochondrial mutation.
1b [10 pt] Briefly explain how each of the features you listed is consistent with a mitochondrial
mutation.
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
2. Lower organisms that can survive without functional mitochondria or without functional
chloroplasts have been used extensively in genetic studies of organelle function.
2a [2.5 pt] Explain how we identify chlamydomonas mutants that do not have functional
chloroplasts.
2b [2.5 pt] Once we have a chlamydomonas mutant that does not have functional chloroplasts,
explain how we can determine whether the mutation is located in the chloroplast genome or
the nuclear genome?
2c [2.5 pt] Explain how we identify saccharomyces mutants that do not have functional
mitochondria.
2d [2.5 pt] Once we have a saccharomyces mutant that does not not have functional
mitochondria, explain how we can determine whether the mutation is located in the
mitochondrial genome or in the nuclear genome.
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
3. The dominant CLPG mutant at the Callipyge locus produces a desirable skeletal muscle
overgrowth phenotype in sheep. This mutation exhibits unusual genetics as follows:
Cross 1: When CLPG/+ males exhibiting the overgrowth phenotype were mated to +/+ females,
half of the progeny exhibited the overgrowth phenotype. The remaining half exhibited normal
muscle phenotype.
Cross 2: When CLPG/+ females exhibiting the overgrowth phenotype were mated to normal
(+/+) males, all of the progeny had normal muscle phenotype.
3a [5 pt] Which of these observations does not conform to Mendelian expectations? Explain
your answer.
3b [5 pt] Based only on the observations above, could CLPG be a maternal effect mutation?
Explain why or why not.
3c [5 pt] Based only on the observations above, could a transmission bias at the Callipyge locus
account for these observations? Explain why or why not.
3d [5 pt] Based only on the observations above, could genomic imprinting at the Callipyge locus
account for these observations? Explain why or why not.
3e [5 pt] Based only on the observations above, could paramutation at the Callipyge locus
account for these observations? Explain why or why not.
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
Exam - developmental genetics (50 pts)
1. (15 pts) For the following crosses,
what are the expected percentages of larvae
having the phenotypes shown on the
right (lowercase is mutant)?
Female fly:
Male fly:
BCD/bcd
BCD/bcd
:
bcd/bcd
BCD/BCD
:
bcd/bcd
BCD/bcd
:
BCD/bcd
bcd/bcd
:
BCD/BCD
bcd/bcd
:
Kn/kn
Kn/kn
:
kn/kn
Kn/Kn
:
kn/kn
Kn/kn
:
Kn/kn
kn/kn
:
Kn/Kn
kn/kn
:
Oops, this was a trick question! Some of the crosses listed above are not feasible by conventional
genetics. Which ones (circle them)? Why?
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
2. (15 pts) We discussed two essential features of a genetic module, i.e. a genetic system that
can function as a switch. Describe them. Include an explanation of the underlying molecular
basis for each of the two features using the Bicoid/Hunchback system as an example.
Given a suitable reporter gene for the Hunchback promoter that can be modified and put into
transgenic flies, how might you alter the system experimentally to separately measure the effects
of these two components?
What fundamental role does the Bicoid/Hunchback module play in patterning of the fly embryo
along the anterior/posterior axis? (Hint: a picture might help.)
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
3. (5 pts) What homeologous structures are transformed by the following homeotic genes of
Drosophila and plants?
AntP (fly):
Ubx (fly):
Agamous (Arabidopsis, C class gene):
True or false: Agamous, AntP and Ubx proteins are all homeoxbox type DNA binding proteins.
True or false: Typically, mutations in fly homeotic genes cause segments where the mutant gene
is normally expressed to acquire features normally found in the segment immediately anterior to
the effected segment.
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PCB5065 - Fall 2013 – Exam 4 – Chase/McCarty
Name_________________________
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