PCB5065 - Fall 2013 – Exam 4 – Chase
Name___Key______________________
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. 2.5 pt each for any two:
1 – all examples of transmission are mother to offspring
2 – offspring are segregating for the trait
3 – also accepted non-Mendelian progeny ratios of affected and unaffected progeny, although in
a small pedigree, it' hard to make a case for non-Mendelian ratios. (Did anyone try a chisquare? I'll try one when I get a few minutes!)
4 – Increased frequency of affected individuals in generation III compared to II
1b [10 pt] Briefly explain how each of the features you listed is consistent with a mitochondrial
mutation.
1 – human mitochondrial inheritance is strictly maternal [3 pt]
2 – in the case of mitochondrial inheritance, segregating offspring result from somatic
segregation of mutant and normal mitochondrial genotypes in the germline of hetroplasmic
moms [7 pt]
or
2 – in the case of mitochondrial inheritance, increasing numbers (and severity) of affected
individuals can result from somatic segregation of mutant and normal mitochondrial genotypes
in the germline of heteroplasmic moms [7 pt]
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PCB5065 - Fall 2013 – Exam 4 – Chase
Name___Key______________________
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 pt] Explain how we identify chlamydomonas mutants that do not have functional
chloroplasts.
These mutants require acetate to grow
or
These mutants are light sensitive so grow much better in dark than in light
2b [3 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?
Simply cross mutant with wild type and look at the segregation of mutant and wild type in the 4
progeny spores
2:2 = nuclear
4:0 or 0:4 = plastid (depending on whether the mutant was from the mating type + or mating
type - parent)
I accepted introducing wild-type plastids and looking for somatic segregation of mutant and
wild types, which would imply a plastid mutation; in practice this is very difficult due to the
chlamydomonas life cycle, but it works in theory.
I also accepted sequencing the plastid genome – but crossing would be cheaper, faster and
easier! (The grant reviewers would be quick to point this out to you!!!)
2c [2 pt] Explain how we identify saccharomyces mutants that do not have functional
mitochondria.
These mutants grow on glucose but not on glycerol
or
These mutants give a small colony as compared to wild type when grown on glucose /glycerol
combined medium
2d [3 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.
Simply cross mutant with wild type and look at the segregation of mutant and wild type in the 4
progeny spores: 2:2 = nuclear; 4:0 or 0:4 = mitochondrial
I also accepted introducing wild type mitochondria and looking for somatic segregation for
mutant and wild-type, which would imply a mitochondrial mutation (note that in yeast,
crossing is the best way to introduce wild-type mitochondria!)
I also accepted crossing with rho0 and looking for complementation, which would imply a
nuclear mutation, since rho0 carries no mitochondrial genome.
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PCB5065 - Fall 2013 – Exam 4 – Chase
Name___Key______________________
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.
Cross 2; based on the genotypes of the parents we expect a 1:1 segregation of CLPG/+ (mutant)
: +/+ (wild-type) progeny, but instead we see all wild types, based on phenotype. (Note you
don't have any genotype information for the progeny.)
3b [5 pt] Based only on the observations above, could CLPG be a maternal effect mutation?
Explain why or why not.
No, because in a maternal effect, all the progeny have the same phenotype, determined by the
mother's genotype. The expectation for cross 1 would be all normal and for cross 2 all mutant.
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.
Yes, if the bias was against the CLPG allele through the female; recall that bias is for one allele
over the other and usually occurs through one sex or the other
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.
Yes, if the paternal allele is the expressed allele at the CLPG locus; in cross 1 we see expression
of the paternally contributed CLPG or the paternally contributed + allele; in cross 2, the paternal
parent is homozygous so we see expression of the paternally contributed + allele; the
maternally contributed allele is not expressed so the effects of the maternally contributed CLPG
mutant allele are not observed.
3e [5 pt] Based only on the observations above, could paramutation at the Callipyge locus
account for these observations? Explain why or why not.
No because silencing by paramutation is not sex-dependent, so we don't see differences in
reciprocal crosses
HOWEVER, I neglected to specify that all of the + alleles were identical in both crosses and that
the CLPG alleles were identical in both crosses, so if you said yes, because the + allele could be
silencing the CLPG allele in cross 2, I accepted that as a correct answer.
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